[{"categories":["differential-geometry"],"content":" In this note we collect some references that discuss the notion of connection on vector bundle\nDifferential geometry by Loring Tu Geometry of Differential forms by Shigeyuki Morita Global Calculus by S Ramanan From Calculus to Cohomology by Madsen Natural Operations in differential geometry by Kolar, Michor, Slovak Foundations of Differential geometry by Kobayashi and Nomizu Differential geometry by Taubes Geometry of Physics by Theodore Frankel Modern differential geometry for Physicists by Chris Isham Differential Geometry by Loring Tu Let \\(E\\rightarrow M\\) be a \\(C^\\infty\\) vector bundle over \\(M\\). A connection on \\(E\\) is a map \\[\\nabla:\\mathfrak{X}(M)\\times \\Gamma(E)\\rightarrow \\Gamma(E)\\]\nsuch that for \\(X\\in \\mathfrak{X}(M)\\) and \\(s\\in \\Gamma(E)\\),\n\\(\\nabla_Xs\\) is \\(\\mathfrak{F}\\)-linear in \\(X\\) and \\(\\mathbb{R}\\)-linear in \\(s\\) (Leibniz rule) if \\(f\\) is a \\(C^\\infty\\) function on \\(M\\), then \\(\\nabla_X(fs)=X(f)s+f\\nabla_Xs\\) Since \\(X(f)=(df)(X)\\), the Leibniz rule may be written as \\(\\nabla_X(fs)=(df)(X)s+f\\nabla_Xs,\\) or, suppressing \\(X\\), \\[\\nabla(fs)=df\\cdot s+f\\nabla s\\].\nDifferential Geometry (Connections, Curvature and Characteristic classes) by Loring Tu Geometry of Differential forms by Shigeyuki Morita defines Connection on a vector bundle \\(E\\rightarrow M\\) as a map \\(\\nabla:\\mathfrak(M)\\times \\Gamma(E)\\rightarrow \\Gamma(E)\\) satisfying certain conditions\nGeometry of Differential forms by Shigeyuki Morita Global Calculus by Ramanan defines connection on a vector bundle \\(E\\rightarrow M\\) as a splitting of associated first order symbol sequence\nGlobal Calculus by Ramanan (Chapter 5) In the next page they give an equivalent description. This may look familiar than the \"first order symbol notion\"\nGlobal Calculus by Ramanan (Chapter 5) From Calculus to Cohomology by Madsen and Tornehave\ndefines connection on a vector bundle \\(\\xi \\rightarrow M\\) as an \\(\\mathbb{R}\\)-linear map \\(\\nabla:\\Omega^0(\\xi)\\rightarrow \\Omega^1(M)\\otimes_{\\Omega^0(M)}\\Omega^0(\\xi)\\) satisfying certain conditions From Calculus to Cohomology by Madsen Natural Operations in differential geometry by Kolar, Michor, Slovak\nfirst discuss notion of connection on a fibre bundle Natural Operations in Differential geometry by Kolar, Michor, Slovak Chapter 9 Then the discuss connection on principal bundle. After that they discuss connection on vector bundle, along with equivalent description using \"connector \\(K:TE\\rightarrow E\\)\"\nTwo interesting things that appear here are the following:\nIn previous discussion we said there is no obvious map \\(TE\\rightarrow E\\) (other than the usual projection, which is useless for us). In this case, connection is described using \"connector\". They already discussed connection on principal bundle. Frame bundle of a vector bundle is an example of a principal bundle. They are highlighting here that, connection on a vector bundle has to come from connection on the associated principal bundle. This was a very big relief for me when I was reading connections for first time. Foundations of differential geometry by Kobayashi and Nomizu\nunfortunately does not describe connection on vector bundle in an independent way. In first chapter they describe the notion of a principal \\(G\\)-bundle.\nThey also discuss the notion of associated vector bundle for a principal \\(G\\)-bundle \\(P(M,G)\\) and a representation \\(\\rho:G\\rightarrow {\\rm{GL}}(\\mathbb{R}^n)\\). Stating with a connection on principal bundle \\(P(M,G)\\), they associate \"parallel displacement\" on fibers of \\(E\\rightarrow M\\). Using this they define \"covariant derivative\". All this data fit nicely with our notion of connection on vector bundle. But, I expected they would discuss it independently.\nFoundations of Differential geometry by Kobayashi and Nomizu Foundations of differential geometry by Kobayashi and Nomizu Foundations of Differential geometry by Kobayashi and Nomizu Foundations of Differential geometry by Kobayashi and Nomizu Differential geometry by Taubes\nAs far as I understand, in this book, Taubes want to reserve the term covariant derivative for vector bundles and connections for principal bundles. Given a principal bundle \\(E\\rightarrow M\\), a covariant derivative is a map \\(\\nabla:C^\\infty(M;E)\\rightarrow C^\\infty(M;T^*M\\otimes E)\\)\nDifferential geometry by Taubes Some may think this version looks straightforward than that of Madesen calculus to Cohomology book definition. We are aware of differential forms taking values in vector bundle, such vector bundles are \\(E\\) and \\(E\\otimes T^*M\\) and a connection is just a map \\(\\nabla:C^\\infty(M;E)\\rightarrow C^\\infty(M;T^*M\\otimes E)\\). Geometry of Physics by Theodore Frankel Definition of connection on vector bundle is written in a simple way. Not all term is clear from the definition, but, it gives an idea, with less notation.\nGeometry of Physics by Theodore Frankel Modern differential geometry for Physicists by Chris Isham\nAs in the book of Kobayashi and Nomizu, they first define connection on principal bundles and use it to define connection on vector bundles.\nConnection on principal bundle gives \"parallel transpor\" for fibers of associated vector bundle (we assume we already have a representation \\(\\rho:G\\rightarrow {\\rm{GL}}(\\mathbb{R}^n)\\). This parallel transport on vector bundle is used to define connection on vector bundle. Modern Differential Geometry for Physicists by C J Isham Modern Differential Geometry for Physicists by C J Isham Modern Differential Geometry for Physicists by C J Isham ","permalink":"https://praphulla-koushik.github.io/2026/02/14/equivalent-definitions-of-connections-on-vector-bundle/","summary":"\u003c!-- wp:paragraph --\u003e\n\u003cp\u003eIn this note we collect some references that discuss the notion of connection on vector bundle\u003c/p\u003e\n\u003c!-- /wp:paragraph --\u003e\n\n\u003c!-- wp:list {\"ordered\":true} --\u003e\n\u003col class=\"wp-block-list\"\u003e\u003c!-- wp:list-item --\u003e\n\u003cli\u003eDifferential geometry by Loring Tu\u003c/li\u003e\n\u003c!-- /wp:list-item --\u003e\n\n\u003c!-- wp:list-item --\u003e\n\u003cli\u003eGeometry of Differential forms by Shigeyuki Morita\u003c/li\u003e\n\u003c!-- /wp:list-item --\u003e\n\n\u003c!-- wp:list-item --\u003e\n\u003cli\u003eGlobal Calculus by S Ramanan\u003c/li\u003e\n\u003c!-- /wp:list-item --\u003e\n\n\u003c!-- wp:list-item --\u003e\n\u003cli\u003eFrom Calculus to Cohomology by Madsen\u003c/li\u003e\n\u003c!-- /wp:list-item --\u003e\n\n\u003c!-- wp:list-item --\u003e\n\u003cli\u003eNatural Operations in differential geometry by Kolar, Michor, Slovak\u003c/li\u003e\n\u003c!-- /wp:list-item --\u003e\n\n\u003c!-- wp:list-item --\u003e\n\u003cli\u003eFoundations of Differential geometry by Kobayashi and Nomizu\u003c/li\u003e\n\u003c!-- /wp:list-item --\u003e\n\n\u003c!-- wp:list-item --\u003e\n\u003cli\u003eDifferential geometry by Taubes\u003c/li\u003e\n\u003c!-- /wp:list-item --\u003e\n\n\u003c!-- wp:list-item --\u003e\n\u003cli\u003eGeometry of Physics by Theodore Frankel \u003c/li\u003e\n\u003c!-- /wp:list-item --\u003e\n\n\u003c!-- wp:list-item --\u003e\n\u003cli\u003eModern differential geometry for Physicists by Chris Isham\u003c/li\u003e\n\u003c!-- /wp:list-item --\u003e\u003c/ol\u003e\n\u003c!-- /wp:list --\u003e\n\n\u003c!-- wp:heading {\"level\":1} --\u003e\n\u003ch1 class=\"wp-block-heading\"\u003e\u003cstrong\u003eDifferential Geometry \u003c/strong\u003e\u003c/h1\u003e\n\u003c!-- /wp:heading --\u003e\n\n\u003c!-- wp:paragraph --\u003e\n\u003cp\u003e\u003cstrong\u003eby Loring Tu \u003c/strong\u003e\u003c/p\u003e","tags":["connections","kobayashi-nomizu","principal-bundles"],"title":"Equivalent definitions of connections on vector bundle","type":"wp-import"},{"categories":["differential-geometry"],"content":" Let \\(\\pi:P\\rightarrow M\\) be a principal \\(G\\) bundle.\nLet \\(F\\) be a smooth manifold with an action of \\(G\\) from left (note that action of \\(G\\) on \\(P\\) is from right). Given this we want to associate a fiber bundle over \\(M\\). This action is same thing as giving a smooth map \\(G\\times F\\rightarrow F\\).\nWe look for a fiber bundle with fibre \\(G\\times F\\) and see if we can construct another fibre bundle with fibre \\(F\\) from the map \\(G\\times F\\rightarrow F\\).\nGiven a principal \\(G\\) bundle \\(P\\rightarrow M\\), a smooth manifold \\(F\\) with an action of \\(G\\), we want to associate a fiber bundle whose fiber is \\(F\\) and base is \\(M\\).\nAction of \\(G\\) on \\(P, F\\) induce action of \\(G\\) on \\(P\\times F\\) with \\((g,(p,f)\\mapsto (pf, gf)\\).\nThe map \\(P\\rightarrow M\\) induce map \\(P\\times F\\rightarrow M\\) with \\((p,f)\\mapsto \\pi(p)\\) for \\((p,f)\\in P\\times F\\).\nThis further induce map \\(\\pi_F:(P\\times F)/G\\rightarrow M\\) with \\(([p,f])\\mapsto \\pi(p)\\).\nIs this well defined?\nIf \\((p,f)\\sim (q,f')\\) then, is \\(\\pi(p)=\\pi(q)\\)?\n\\((p,f)\\sim (q,f')\\) when there is a \\(g\\in G\\) such that \\(q=pg, f'=fg\\). In that case of \\(q=pg\\), we know that \\(\\pi(p)=\\pi(q)\\) (property of principal bundle). So, the map is well defined.\nNow, what is the fiber of an element \\(m\\in M\\) via the map \\((P\\times F)/G\\rightarrow M\\)?\nAn element in fiber is \\([(p,f)]\\) such that \\(\\pi(p)=m\\).\nFix \\(p\\in \\pi^{-1}(m)\\).\nTake \\(f\\in F\\) and assign \\([(p,f)]\\). This gives a map \\(\\psi:F\\rightarrow \\pi_F^{-1}(m)\\).\nlet us check one-one nature.\nSuppose \\(f_1,f_2\\in F\\) be such that \\([(p,f_1)]=[(p,f_2)]\\) that means, there exists \\(g\\in G\\) such that \\(pg=p\\) and \\(f_2=f_1g\\). Free action of \\(G\\) on \\(P\\) says \\(g=1\\) which in turn says \\(f_2=f_1\\).\nLet us check onto nature.\nLet \\([(q,f)]\\in \\pi_F^{-1}(m)\\). We need to find \\(f'\\in F\\) such that, \\(\\psi(f')=[(q,f)]\\); that is, \\([(p,f')]=[(q,f)]\\). The condition \\(q\\in \\pi^{-1}(m)\\) says that, there exists \\(g\\in G\\) such that, \\(p=qg\\). Consider \\(f'=gf\\). For this, we have \\([(p,f')]=[(q,f)]\\), concluding that \\(\\psi\\) is surjective.\nThus, the map \\(\\psi:F\\rightarrow \\pi_F^{-1}(m)\\) is a bijection. It is more than a bijection. More details about this we will see some other time.\n","permalink":"https://praphulla-koushik.github.io/2026/02/10/vector-bundle-associated-to-a-principal-bundle/","summary":"\u003c!-- wp:paragraph --\u003e\n\u003cp\u003eLet \\(\\pi:P\\rightarrow M\\) be a principal \\(G\\) bundle.\u003c/p\u003e\n\u003c!-- /wp:paragraph --\u003e\n\n\u003c!-- wp:paragraph --\u003e\n\u003cp\u003eLet \\(F\\) be a smooth manifold with an action of \\(G\\) from left (note that action of \\(G\\) on \\(P\\) is from right). Given this we want to associate a fiber bundle over \\(M\\). This action is same thing as giving a smooth map \\(G\\times F\\rightarrow F\\).\u003c/p\u003e\n\u003c!-- /wp:paragraph --\u003e\n\n\u003c!-- wp:paragraph --\u003e\n\u003cp\u003eWe look for a fiber bundle with fibre \\(G\\times F\\) and see if we can construct another fibre bundle with fibre \\(F\\) from the map \\(G\\times F\\rightarrow F\\).\u003c/p\u003e","tags":["mathematics","physics","principal-bundles"],"title":"Vector bundle associated to a principal bundle","type":"wp-import"},{"categories":["differential-geometry"],"content":" We will understand the notion of a connection on a vector bundle in the following steps:\nGive the\u0026nbsp;definition of a connection Explain the\u0026nbsp;objects appearing in the definition\u0026nbsp;(sections and their algebraic structure) Study the\u0026nbsp;trivial bundle case, which motivates the axioms Examine the\u0026nbsp;tangent bundle case\u0026nbsp;and test familiar operations Explain why the\u0026nbsp;usual differential of a section\u0026nbsp;does not give what we want Let \\(E\\rightarrow M\\) be a vector bundle.\nA connection on \\(E\\rightarrow M\\) is defined to be a map \\[\\nabla:\\Gamma(M,TM)\\times \\Gamma(M,E)\\rightarrow \\Gamma(M,E)\\]\nsatisfying the following conditions\n\\(\\nabla\\) is \\(\\mathbb{R}\\)-bilinear when \\(\\Gamma(M,TM)\\) and \\(\\Gamma(M,E)\\) are seen as \\(\\mathbb{R}\\)-vector spaces; that is, \\(\\nabla(aX_1+bX_2,s)=a\\nabla(X_1,s)+b\\nabla(X_2,s)\\) \\(\\nabla(X,as_1+bs_2)=a\\nabla(X,s_1)+b\\nabla(X,s_2)\\) \\(\\nabla\\) is \\(C^\\infty(M)\\)-compatible when \\(\\Gamma(M,TM)\\) and \\(\\Gamma(M,E)\\) are seen as \\(C^\\infty(M)\\)-modules; that is, \\(\\nabla(fX,s)=f\\nabla(X,s)\\) \\(\\nabla(X,fs)=f\\nabla(X,s)+X(f)s\\) for all \\(a,b\\in \\mathbb{R}\\), all \\(f\\in C^\\infty(M)\\), all \\(X,X_1,X_2\\in \\Gamma(M,TM)\\), all \\(s,s_1,s_2\\in \\Gamma(M,E)\\).\nMany books define connection in a different way. This version of connection definition is taken from the book Differential Geometry -- Connections Curvature and Characteristic Classes by Loring W Tu\nFor more details about references and equivalent definitions of connections, you may see\n/2026/02/14/equivalent-definitions-of-connections-on-vector-bundle/\nBefore this definition becomes meaningful, let us recall what sections are and what algebraic structure their space carries.\nSections of vector bundle Recall that, a section of a vector bundle \\(\\pi: E\\rightarrow M\\) is a smooth map \\(s:M\\rightarrow E\\) satisfying the condition \\(\\pi\\circ s=1_M\\).\nWhen learning for the first time, there may be some confusion regarding order of composition.\nDoes it ask for \\(\\pi\\circ s=1_M\\) or \\(s\\circ \\pi=1_E\\)??\nSuppose it is \\(s\\circ \\pi=1\\), that would mean that \\(\\pi\\) is injective. Let \\(a_1,a_2\\in E\\) be such that \\(\\pi(a_1)=\\pi(a_2)\\). This implies \\(s(\\pi(a_1))=s(\\pi(a_2)\\). The assumption \\(s\\circ \\pi=1\\) along with the observation \\(s(\\pi(a_1))=s(\\pi(a_2))\\) implies \\(a_1=a_2\\). The condition of \\(\\pi\\) being injective is equilavelnt to the condition of fibers being singletons, which are far from being an (interesting) vector spaces.\nSo, for a section we are asking for \\(\\pi\\circ s=1_M\\) and not the other way.\nThat is all ok, but, what does it mean to ask for the condition \\(\\pi\\circ s=1_M\\)?\nWhen we see a section \\(s:M\\rightarrow E\\) as a collection \\(\\{s(m)\\}_{m\\in M}\\) the condition \\(\\pi\\circ s=1_M\\) means that \\(\\pi(s(m))=1\\); in other words \\(s(m)\\in \\pi^{-1}(m)=E_m\\) for all \\(m\\in M\\). So, a section \\(s:M\\rightarrow E\\) can be seen as a collection \\(\\{s(m)\\}_{m\\in M}\\) with the property that \\(s(m)\\in E_m\\) for all \\(m\\in M\\).\nAt this point it should be reminded that, not any random collection \\(\\{t(m)\\}_{m\\in M}\\) of elements \\(E\\) would assure that the associated map \\(t:M\\rightarrow E\\) is smooth.\nNow, we know what is a section of a vector bundle. As mentioned before, the set of sections of a vector bundle \\(E\\rightarrow M\\) is denoted by \\(\\Gamma(M,E)\\).\nIt is mentioned that connection \\(\\nabla\\) is \\(\\mathbb{R}\\)-bilinear.\nLet us unravel what is the \\(\\mathbb{R}\\)-vector space structure on \\(\\Gamma(M,E)\\). As \\(E_m\\) is a vector space for each \\(m\\in M\\) there is a distinguished element in it, the zero element \\(0_m\\in E_m\\). This gives a map \\(s:M\\rightarrow E\\) defined by \\(m\\mapsto 0_m\\) for \\(m\\in M\\).\nOnce you check this assignment is a smooth function, this gives an example of a section, which usually goes by the name of ''zero section of vector bundle \\(E\\rightarrow M\\)''. So, there is no confusion about zero element of \"vector space\".\nLet \\(s,t:M\\rightarrow E\\) be two sections of \\(\\pi:E\\rightarrow M\\). As mentioned before, we can see these sections as collections \\(\\{s(m)\\}_{m\\in M}\\) and \\(\\{t(m)\\}_{m\\in M}\\), with \\(s(m),t(m)\\in E_m\\) for all \\(m\\in M\\). Fix \\(m\\in M\\). The vector space structure on \\(E_m\\) allows us to add two elements of \\(E_m\\). In particular, the elements \\(s(m),t(m)\\in E_m\\) adds up to give \\(s(m)+t(m)\\in E_m\\). This collection \\(\\{s(m)+t(m)\\}_{m\\in M}\\) combines to give (no need to believe, you may prove) a smooth section \\(s+t:M\\rightarrow E\\). Let \\(a\\in \\mathbb{R}\\) and \\(s:M\\rightarrow E\\) be a section of \\(\\pi:E\\rightarrow M\\). As mentioned before, we can see this as a collection \\(\\{s(m)\\}_{m\\in M}\\) with \\(s(m)\\in E_m\\) for all \\(m\\in M\\). Fix \\(m\\in M\\). The vector space structure on \\(E_m\\) allows us to consider scalar multiplication of an element of \\(E_m\\) with a real number. In particular, the elements \\(a\\in \\mathbb{R}\\) and \\(s(m)\\in E_m\\) gives \\(as(m)\\in E_m\\). This collection \\(\\{as(m)\\}_{m\\in M}\\) combines to give (no need to believe, you may prove) a smooth section \\(as:M\\rightarrow E\\). With this, \\(\\Gamma(M,E)\\) is seen as an \\(\\mathbb{R}\\)-vector space. In particular (as we knew before), \\(\\Gamma(M,TM)\\) is also seen as an \\(\\mathbb{R}\\)-vector space.\nNote that, there is no reason to restrict multiplication of elements of \\(\\{s(m)\\}_{m\\in M}\\) by one fixed real number. As \\(s(m)\\) is sitting in different vector space for different \\(m\\), we may as well choose different scalar for different \\(m\\in M\\).\nGiven a collection \\(\\{f(m)\\}_{m\\in M}\\) of real numbers and a section \\(\\{s(m)\\}_{m\\in M}\\) we get a collection \\(\\{f(m)s(m)\\}_{m\\in M}\\). This gives a map \\(fs:M\\rightarrow E\\). But, there is no guarantee why this function \\(fs:M\\rightarrow E\\) is smooth. It would be an interesting exercise to see that if the collection \\(\\{f(m)\\}_{m\\in M}\\) is a smoothly varying family of real numbers, in other words, the associated map \\(f:M\\rightarrow E\\) is smooth, then, the map \\(fs:M\\rightarrow E\\) is a smooth map.\nThis gives an action of \\(C^\\infty(M)\\) on \\(\\Gamma(M,E)\\) with \\((f,s)\\mapsto fs\\) for \\(f\\in C^\\infty(M)\\) and \\(s\\in \\Gamma(M,E)\\).\nWith this, \\(\\Gamma(M,E)\\) becomes a \\(C^\\infty(M)\\)-module. In particular, \\(\\Gamma(M,TM)\\) also becomes a \\(C^\\infty(M)\\)-module.\nIt would be useful to recall/prove that,\nsections \\(\\Gamma(M,M\\times \\mathbb{R})\\) of trivial bundle \\(M\\times \\mathbb{R}\\rightarrow M\\) are precisely the smooth functions on \\(M\\). sections \\(\\Gamma(M,TM)\\) of tangent bundle \\(TM\\rightarrow M\\) are precisely the smooth vector field on \\(M\\). With these \\(\\mathbb{R}\\)-vector space structures on \\(\\Gamma(M,TM)\\) and \\(\\Gamma(M,E)\\) we are asking that \\(\\nabla\\) is \\(\\mathbb{R}\\)-blinear.\nWith these \\(C^\\infty(M)\\)-module structures on \\(\\Gamma(M,TM)\\) and \\(\\Gamma(M,E)\\) we are asking that \\(\\nabla\\) to be \\(C^\\infty(M)\\)-compatible.\nFor anything related to sections of vector bundles that we are going to see from now, we keep asking if there is \"compatibility\" with \\(C^\\infty(M)\\)-module structure.\nOk.\nBefore asking bigger questions, let us ask a simple question.\nWhat does connection mean in trivial vector bundles?? Let us look at the basic example of trivial bundle and see what does connection mean in that and what does it remind? Consider trivial bundle \\(M\\times \\mathbb{R}\\rightarrow M\\). A connection on this would be a map \\[\\Gamma(M,TM)\\times \\Gamma(M, M\\times \\mathbb{R})\\rightarrow \\Gamma(M,M\\times \\mathbb{R})\\]\nin other words\n\\[\\Gamma(M,TM)\\times C^\\infty(M)\\rightarrow C^\\infty(M)\\].\nThe requirement of such map should not to be confused with \\(C^\\infty(M)\\) action on \\(\\Gamma(M,TM)\\) which is a map \\(\\Gamma(M,TM)\\times C^\\infty(M)\\rightarrow \\Gamma(M,TM)\\).\nGiven a vector field \\(\\Gamma(M,TM)\\) and an element \\(f\\in C^\\infty(M)\\), is there a way to get an element in \\(C^\\infty(M)\\)?\nWe do know that vector fields evaluated on smooth functions gives smooth functions, right?\nGiven \\(X\\in \\Gamma(M,TM)\\) and \\(f\\in C^\\infty(M)\\) we have \\(Xf:M\\rightarrow \\mathbb{R}\\) defined as \\((Xf)(m)=f_{*,m}(X(m))\\). This gives a map \\[\\nabla: \\Gamma(M,TM)\\times C^\\infty(M)\\rightarrow C^\\infty(M)\\]\nwith \\((X,f)\\mapsto Xf\\). But, the notation \\(Xf\\) is not saying immediately what it means. What we are doing in this case is looking at differential of \\(f\\) and evaluating at \\(X\\). We do have a different notation for that, namely \\(df(X)\\). So, we have a map \\(d:\\Gamma(M,TM)\\times C^\\infty(M)\\rightarrow C^\\infty(M)\\)\nwith \\((X,f)\\mapsto (df)(X)\\) for \\(X\\in \\Gamma(M,TM)\\) and \\(f\\in C^\\infty(M)\\).\nLet us ask if there is any similarity with properties of \\((X,f)\\mapsto (df)(X)\\) with the properties mentioned for the connection on vector bundle.\nlinearity in first variable; that is, \\(df(aX_1+bX_2)=a(df)(X_1)+b(df)(X_2)\\) follows from linearity of the differential \\(f_{*,-}\\). linearity in second variable; that is, \\(d(af+bg)(X)=a (df)(X)+b(dg)(X)\\) follows from linearity of differentiation operation \\(f'+g'=(f+g)'\\). So, this is matching with \\(\\mathbb{R}\\)-bilinearity conditions of connection. Let us look at \\(C^\\infty(M)\\)-compatibility.\nGiven \\(g\\in C^\\infty(M)\\), \\(f\\in \\Gamma(M,M\\times \\mathbb{R})=C^\\infty(M)\\) and \\(X\\in \\Gamma(M,TM)\\) let us look at\n\\(df(gX)\\), \\(d(gf)(X)\\). We have, \\[df(gX)(m)=f_{*,m}((gX)(m))=f_{*,m}(g(m)X(m))=g(m)f_{*,m}(X(m))=g(df)(X)(m)\\]\nWe have \\[d(gf)(X)=X(gf)=X(g)f+gX(f)=X(g)f+g(df)(X)\\].\nThis is exactly what we are asking for connection to satisfy. Note that, we are not concluding the equation \\[d(gf)(X)=X(gf)=X(g)f+gX(f)=X(g)f+g(df)(X)\\].\nwith \\(f(dg)(X)+g(df)(X)\\) because, here we are seeing \\(g\\) as element of \\(C^\\infty(M)\\) and we want to put \\(dg\\) for only elements \\(g\\) of \\(\\Gamma(M,M\\times \\mathbb{R})\\).\nIf you dont understand what I said just now, give it sometime, you will get. So, what ever we are asking for differentiation of smooth maps (sections on trivial bundle) to satisfy, we are asking them as conditions for connection on vector bundle.\nWhat is a connection in simple terms? In this way, a connection on a vector bundle \\(E\\rightarrow M\\) can be seen\nas a prescription for differentiation of sections of vector bundle along sections of tangent bundle (vector fields) What does connection mean in case of tangent bundle? Consider tangent bundle \\(TM\\rightarrow M\\). Let \\(X:M\\rightarrow TM\\) be a section of the vector bundle \\(TM\\rightarrow M\\).\nWe said connection will prescribe \"derivative of \\(X\\).\"\nThe question now is, how to think about this notion of derivative of \\(X\\)? Let us recall derivative of section of trivial bundle \\(M\\times \\mathbb{R}\\rightarrow M\\).\nLet \\(f:M\\rightarrow \\mathbb{R}\\) be a section of trivial bundle \\(M\\times \\mathbb{R}\\rightarrow M\\).\nGiven a point \\(m\\in M\\), the function \\(f\\) evaluated at \\(m\\in M\\) is a real number. For such, the derivative of \\(f\\) at \\(m\\) evaluated at a tangent vector \\(v\\) is also a real number (under standard identification).\nSame we expect in the case of vector field \\(X:M\\rightarrow TM\\). Given a point \\(m\\in M\\), the function \\(X\\) evaluated at \\(m\\) is a tangent vector at \\(m\\). So, we expect similar behaviors from \"derivative of \\(X\\)\".\nWe ask \"derivative of \\(X\\)\" at point \\(m\\) evaluated at a tangent vector \\(v\\) to be also a tangent vector at the point \\(m\\).\nIn other words, given a vector field \\(X:M\\rightarrow TM\\), we are looking for a map \\[\\{v\\in T_mM\\}_{m\\in M}\\rightarrow \\{v'\\in T_mM\\}_{m\\in M}\\].\nBut, this collection \\(\\{v\\in T_mM\\}_{m\\in M}\\) reminds us of a vector field on \\(M\\), where this collection is asked to vary smoothly.\nAs the vector field \\(X\\) is smooth, it is reasonable to ask the associated map\n\\[\\{v\\in T_mM\\}_{m\\in M}\\rightarrow \\{v'\\in T_mM\\}_{m\\in M}\\]\ntakes a smoothly varying collection to a smoothly varying collection.\nGiven \\(X\\in \\Gamma(M,TM)\\) we are asking for a map\n\\[\\tilde{X}:\\Gamma(M,TM)\\rightarrow \\Gamma(M,TM)\\].\nThis is precisely what we asked in definition of connection.\nA connection on the tangent bundle \\(TM\\rightarrow M\\) would be a map\n\\[\\nabla:\\Gamma(M,TM)\\times \\Gamma(M,TM)\\rightarrow \\Gamma(M,TM)\\].\nWe have already come across such a map; the Lie bracket.\nBut, we do not know if Lie bracket satisfies requirements of connection.\nLet us check.\nLet \\(a,b\\in \\mathbb{R}\\) and \\(X,X_1,X_2,Y,Y_1,Y_2\\in \\Gamma(M,TM)\\).\nWe have \\[[aX_1+bX_2,Y](f)=(aX_1+bX_2)(Y(f))-Y ((aX_1+bX_2)(f))=--=a[X_1,Y](f)+b[X_2,Y](f)\\]\nAs Lie bracket is skew-symmetric, checking linearity in one coordinate is sufficient to conclude linearity in the other variable.\nSo, there is a hope that Lie bracket operation \\[\\nabla:\\Gamma(M,TM)\\times \\Gamma(M,TM)\\rightarrow \\Gamma(M,TM)\\]\nwith \\(\\nabla(X,Y)=[X,Y]\\) for \\(X,Y\\in \\Gamma(M,TM)\\) is a connection.\nLet us see for the \\(C^\\infty(M)\\) compatibility.\nLet \\(f\\in C^\\infty(M)\\) and \\(Y\\in \\Gamma(M,TM)\\). We have\n\\[[X,fY](g)=X(fY)(g)-fY(X)(g)=X(f Y(g))-fYX(g)=X(f) Y(g)+fX(Y(g))-fY(X(g))=X(f)Y(g)+f [XY-YX](g)=X(f)Y(g)+f[X,Y](g)\\].\nAs this is true for all \\(g\\in C^\\infty(M)\\), we have \\[[X,fY]=X(f)Y+f[X,Y]\\],\nThis observation makes us both happy and sad at the same time. We can be happy because this is one of the conditions for connections, namely,\n\\[\\nabla(X,fs)=f\\nabla(X,s)+X(f)s\\].\nBut, due to skew-symmetric nature of Lie bracket, this would mean that, \\[\\nabla (fX,s)=-\\nabla(s,fX)=-f\\nabla(s,X)-***=f\\nabla(X,s)+***\\]\nwhich is not what we ask in second \\(C^\\infty(M)\\)-compatibility. Thus, \\(\\nabla:\\Gamma(M,TM)\\times \\Gamma(M,TM)\\rightarrow \\Gamma(M,TM)\\) defined by \\((X,Y)\\mapsto [X,Y]\\) is not a connection.\nWhy cant we use same differential idea in tangent bundle? In case of trivial bundle \\(M\\times M\\times \\mathbb{R}\\), given a section \\(f:M\\rightarrow M\\times \\mathbb{R}\\) (seen as \\(f:M\\rightarrow \\mathbb{R}\\)) we just took differential. This gave an example of connection on bundle \\(M\\times \\mathbb{R}\\rightarrow M\\).\nWhy cant we do the same thing in case of tangent bundle \\(TM\\rightarrow M\\)?\nGiven a section \\(X:M\\rightarrow TM\\), ignoring that it is a section of tangent bundle, why can't we just consider it as a smooth function and look at the differential \\(dX\\)?\nWe have \\(dX:TM\\rightarrow T(TM)\\) with \\((dX)(Y)(m)=X_{*,m}(Y(m))\\). Forget about this being \\(\\mathbb{R}\\)-bilinear or \\(C^\\infty(M)\\) compatible.\nWe do not even have the right candidate. We wanted a map \\[\\Gamma(M,TM)\\times \\Gamma(M,TM)\\rightarrow \\Gamma(M,TM)\\],\nbut, here, with this \"usual differential'' idea, we are getting a map\n\\[\\Gamma(M,TM)\\times \\Gamma(M,TM)\\rightarrow \\Gamma(M,T(TM))\\].\nThus, usual derivative operation that we did in case of sections of trivial bundle, that we used to get a connection on trivial bundle, can not be used to get a connection on tangent bundle. So, it looks like a genuinely extra structure on tangent bundle (and more generally on any vector bundle).\nIt should be noted that, in case of section \\(f\\in \\Gamma(M,M\\times \\mathbb{R})\\), we asked for a map\n\\[\\tilde{f}:\\Gamma(M,TM)\\rightarrow \\Gamma(M,M\\times \\mathbb{R})\\].\nSo, the domain of \\(\\nabla(X)\\) in the above case should not be seen as section of the vector bundle in the consideration; instead, they should be seen as sections of tangent bundle.\nIn other words, what ever may be the vector bundle, the domain of differential of section should always be \\(\\Gamma(M,TM)\\). Why \\(ds\\) is not a connection on vector bundle? I think it is already clear why differential of \\(X:M\\rightarrow TM\\) is not the one we are looking for.\nFor more clarity, let us look at the case of sections of an arbitrary vector bundle \\(E\\rightarrow M\\).\nGiven a function \\(s:M\\rightarrow E\\), consider the usual differential \\(ds:TM\\rightarrow TE\\).\nThis map \\(ds\\) acts on \\(\\Gamma(M,TM)\\), similar to that of \\(\\nabla(s)(X)\\). Then, where is the issue?\nThe issue is, \\(ds(X)\\) is a map \\(M\\rightarrow TE\\), where as \\(\\nabla(s)(X)\\) is a map \\(M\\rightarrow E\\). Our intention is to assign a map \\(\\nabla(s)\\) for \\(s:M\\rightarrow E\\) which when evaluated on vector fields behaves like section \\(M\\rightarrow E\\) \"of same type as that of \\(s:M\\rightarrow E\\)\".\nSo, \\(ds\\) is not a correct candidate for this purpose.\nThat's ok, but, why not compose with the projection \\(TE\\rightarrow E\\) to land in \\(E\\)?\nLet us look at that as well.\nFix \\(m\\in M\\). Consider \\((\\pi_E\\circ (ds)(X))(m)\\). This gives, \\[(\\pi_E\\circ (ds)(X))(m)=\\pi_E(s_{*,m}(X(m))=s(m)\\].\nWhat ever may be \\(X(m)\\in T_mM\\) is, the image \\(s_{*,m}(X(m))\\) is always in \\(s(m)\\). So, \\(\\pi_E(s_{*,m}(X(m))=s(m)\\). So, when composed with \\(TE\\rightarrow E\\), we get \\(\\pi_E\\circ ds(X)=s\\); for all \\(X\\). When there is no relevance of \\(X\\), may be we are going in wrong diretcion or we are going to hit something boring. So, this also is a bad choice. We just can not do anything interesting to land in \\(E\\). There is no obvious choice.\nWhen it is not obvious and happens sometimes, it becomes important and gets a name. The name about to appear here is : \"connection on vector bundle\".\nDefinition of connection on vector bundle A connection on \\(E\\rightarrow M\\) is defined to be a map\n\\[\\nabla:\\Gamma(M,TM)\\times \\Gamma(M,E)\\rightarrow \\Gamma(M,E)\\]\nsatisfying the following conditions\n\\(\\nabla\\) is \\(\\mathbb{R}\\)-blinear when \\(\\Gamma(M,TM)\\) and \\(\\Gamma(M,E)\\) are seen as \\(\\mathbb{R}\\)-vector spaces; that is, \\(\\nabla(aX_1+bX_2,s)=a\\nabla(X_1,s)+b\\nabla(X_2,s)\\) \\(\\nabla(X,as_1+bs_2)=a\\nabla(X,s_1)+b\\nabla(X,s_2)\\) \\(\\nabla\\) is \\(C^\\infty(M)\\)-compatible when \\(\\Gamma(M,TM)\\) and \\(\\Gamma(M,E)\\) are seen as \\(C^\\infty(M)\\)-modules; that is, \\(\\nabla(fX,s)=f\\nabla(X,s)\\) \\(\\nabla(X,fs)=f\\nabla(X,s)+X(f)s\\) for all \\(a,b\\in \\mathbb{R}\\), all \\(f\\in C^\\infty(M)\\), all \\(X,X_1,X_2\\in \\Gamma(M,TM)\\), all \\(s,s_1,s_2\\in \\Gamma(M,E)\\).\nOf course there will be some differences in notation with other books. Some books may call this with other names. Some books says connection as something else.\nIn next session, we will try to address some of those questions and ask some more questions (and answer some of them).\nBy the way, when we say something is connection, is it connecting any thing? What does it connect?\nSee you later.\n","permalink":"https://praphulla-koushik.github.io/2026/02/06/connection-on-vector-bundle-introduction/","summary":"\u003c!-- wp:paragraph --\u003e\n\u003cp\u003eWe will understand the notion of a connection on a vector bundle in the following steps:\u003c/p\u003e\n\u003c!-- /wp:paragraph --\u003e\n\n\u003c!-- wp:list {\"ordered\":true} --\u003e\n\u003col class=\"wp-block-list\"\u003e\u003c!-- wp:list-item --\u003e\n\u003cli\u003eGive the\u0026nbsp;\u003cstrong\u003edefinition of a connection\u003c/strong\u003e\u003c/li\u003e\n\u003c!-- /wp:list-item --\u003e\n\n\u003c!-- wp:list-item --\u003e\n\u003cli\u003eExplain the\u0026nbsp;\u003cstrong\u003eobjects appearing in the definition\u003c/strong\u003e\u0026nbsp;(sections and their algebraic structure)\u003c/li\u003e\n\u003c!-- /wp:list-item --\u003e\n\n\u003c!-- wp:list-item --\u003e\n\u003cli\u003eStudy the\u0026nbsp;\u003cstrong\u003etrivial bundle case\u003c/strong\u003e, which motivates the axioms\u003c/li\u003e\n\u003c!-- /wp:list-item --\u003e\n\n\u003c!-- wp:list-item --\u003e\n\u003cli\u003eExamine the\u0026nbsp;\u003cstrong\u003etangent bundle case\u003c/strong\u003e\u0026nbsp;and test familiar operations\u003c/li\u003e\n\u003c!-- /wp:list-item --\u003e\n\n\u003c!-- wp:list-item --\u003e\n\u003cli\u003eExplain why the\u0026nbsp;\u003cstrong\u003eusual differential of a section\u003c/strong\u003e\u0026nbsp;does not give what we want\u003c/li\u003e\n\u003c!-- /wp:list-item --\u003e\u003c/ol\u003e\n\u003c!-- /wp:list --\u003e\n\n\u003c!-- wp:separator --\u003e\n\u003chr class=\"wp-block-separator has-alpha-channel-opacity\" /\u003e\n\u003c!-- /wp:separator --\u003e\n\n\u003c!-- wp:paragraph --\u003e\n\u003cp\u003e Let \\(E\\rightarrow M\\) be a vector bundle.\u003c/p\u003e","tags":["connections"],"title":"Connection on vector bundle (Introduction)","type":"wp-import"},{"categories":["Algebraic geometry"],"content":" Let \\(M\\) be a skew-symmetric matrix (with real entries). Let \\(\\lambda\\) be an eigenvalue of \\(M\\). This means, there exists vector \\(v\\) such that \\(Mv=\\lambda v\\). To relate with ``skew-symmetricness'' of \\(M\\), we apply transpose on both sides of previous equation, to get \\(v^TM^T=\\lambda v^T\\). As \\(M\\) is skew-symmetric, we see that \\(v^TM^T=\\lambda v^T\\) is equivalent to \\(-v^TM=\\lambda v^T\\). Now, multiply by \\(v\\) on both sides of the above equation to get \\(-v^TMv=\\lambda v^Tv\\). As \\(v\\) is eigenvector associated to eigenvalue \\(\\lambda\\), we get \\(-v^T\\lambda v=\\lambda v^Tv\\) which is samething as \\(-\\lambda v^Tv=\\lambda v^Tv\\).\nBut, we know \\(v^Tv\\neq 0\\). So, cancelling \\(v^Tv\\) of both sides tells that \\(-\\lambda=\\lambda\\) which means \\(\\lambda =0\\).\nThere is no assurance that \\(\\lambda\\) is a real number to begin with, but, it turns out \\(\\lambda\\) has to be real number. But, we have heard multiple times and repeated multiple times in high pitch that skew-symmetric matrices has eigenvalues as \\(0\\) or purely imaginary number.\nWhat went wrong here??\nEven if we start with \\(\\lambda\\) is a complex eigenvalue, the above observation can be used line by line.\nIs it? Are we sure that \\(v^Tv\\neq 0\\) irrespective of what kind of \\(v\\) is, as long as \\(v\\neq 0\\)? Let \\(v=(1,i)\\). Then, \\(v^Tv=1+i^2=0\\). So, we can not cancel the component \\(v^Tv\\) as per our interest. That is the issue in above discussion.\nWe can not say that \\(v^Tv\\) is nonzero if \\(v\\neq 0\\), but, we can always say \\(\\bar{v}^Tv\\neq 0\\) as far as \\(v\\neq 0\\). So, instead of applying \\(v^T\\), it may be smart idea to apply \\(\\bar{v}^T\\). This is done by considering conjugate and transport. As \\(M\\) is a real matrix, conjugate does not any effect. We have \\[Mv=\\lambda v\\]\n\\[\\bar{v}^T\\bar{M}^T=\\bar{\\lambda}\\bar{v}^T\\]\n\\[\\bar{v}^TM^T=\\bar{\\lambda}\\bar{v}^T\\]\n\\[\\bar{v}^T(-M)=\\bar{\\lambda}\\bar{v}^T\\]\n\\[-\\bar{v}^TM=\\bar{\\lambda}\\bar{v}^T\\]\n\\[-\\bar{v}^TMv=\\bar{\\lambda}\\bar{v}^Tv\\]\n\\[-\\bar{v}^T\\lambda v=\\bar{\\lambda}\\bar{v}^T\\]\n\\[-\\lambda \\bar{v}^Tv=\\bar{\\lambda}\\bar{v}^T\\]\nAs \\(v\\) is such that \\(\\bar{v}^Tv\\neq 0\\). So, we have \\(\\lambda =-\\bar{\\lambda}\\) that is, real part of \\(\\lambda\\) is zero. So, any eigenvalues of real skew symmetric matrix is zero or with a complex number with real part zero (purely imaginary complex number). ","permalink":"https://praphulla-koushik.github.io/2025/05/18/is-it-true-that-eigenvalues-of-skew-symmetric-matrices-are-always-zero/","summary":"\u003c!-- wp:paragraph --\u003e\n\u003cp\u003eLet \\(M\\) be a skew-symmetric matrix (with real entries). \u003c/p\u003e\n\u003c!-- /wp:paragraph --\u003e\n\n\u003c!-- wp:paragraph --\u003e\n\u003cp\u003eLet \\(\\lambda\\) be an eigenvalue of \\(M\\). This means, there exists vector \\(v\\) such that \\(Mv=\\lambda v\\). \u003c/p\u003e\n\u003c!-- /wp:paragraph --\u003e\n\n\u003c!-- wp:paragraph --\u003e\n\u003cp\u003eTo relate with ``skew-symmetricness'' of \\(M\\), we apply transpose on both sides of previous equation, to get \\(v^TM^T=\\lambda v^T\\). \u003c/p\u003e\n\u003c!-- /wp:paragraph --\u003e\n\n\u003c!-- wp:paragraph --\u003e\n\u003cp\u003eAs \\(M\\) is skew-symmetric, we see that \\(v^TM^T=\\lambda v^T\\) is equivalent to \\(-v^TM=\\lambda v^T\\). \u003cbr\u003e\u003cbr\u003eNow, multiply by \\(v\\) on both sides of the above equation to get \\(-v^TMv=\\lambda v^Tv\\). \u003c/p\u003e","tags":[],"title":"Is it true that eigenvalues of skew-symmetric matrices are always zero?","type":"wp-import"},{"categories":["homological-algebra","Algebraic geometry"],"content":" On a Saturday morning, I was thinking about sylow theorems. The question I asked myself is, do I know how to apply sylow theorems? Only application I was aware about, of sylow theorem, is to assure if a group of finite order is simple or not. As a first step, I thought to check for groups of order less than 100. Question is simple :\nFor which \\(n\u0026lt;100\\), a group of order \\(n\\) will be simple?\nA standard result that we (are supposed to) learn in first half a group theory course is that abelian simple groups are boring. Any abelian simple group is cyclic of prime order. So, I rephrased the question to as following question. For which \\(n\u0026lt;100\\), a non abelian group of order \\(n\\) will be simple?\nThe first thing to do is list down the numbers from 1 to 100\nSure! Here are the numbers from 1 to 100:\n1, 2, 3, 4, 5, 6, 7, 8, 9, 10,\n11, 12, 13, 14, 15, 16, 17, 18, 19, 20,\n21, 22, 23, 24, 25, 26, 27, 28, 29, 30,\n31, 32, 33, 34, 35, 36, 37, 38, 39, 40,\n41, 42, 43, 44, 45, 46, 47, 48, 49, 50,\n51, 52, 53, 54, 55, 56, 57, 58, 59, 60,\n61, 62, 63, 64, 65, 66, 67, 68, 69, 70,\n71, 72, 73, 74, 75, 76, 77, 78, 79, 80,\n81, 82, 83, 84, 85, 86, 87, 88, 89, 90,\n91, 92, 93, 94, 95, 96, 97, 98, 99, 100.\nIf we are looking for non abelian groups, we can safely remove groups of prime order. Now the question boils down to which numbers are prime numbers. We may not know many things about prime numbers but we know for sure even numbers are not prime numbers (except 2). So, we can safely focus our attention to the odd numbers\n1, 2, 3, 4, 5, 6, 7, 8, 9, 10,\n11, 12, 13, 14, 15, 16, 17, 18, 19, 20,\n21, 22, 23, 24, 25, 26, 27, 28, 29, 30,\n31, 32, 33, 34, 35, 36, 37, 38, 39, 40,\n41, 42, 43, 44, 45, 46, 47, 48, 49, 50,\n51, 52, 53, 54, 55, 56, 57, 58, 59, 60,\n61, 62, 63, 64, 65, 66, 67, 68, 69, 70,\n71, 72, 73, 74, 75, 76, 77, 78, 79, 80,\n81, 82, 83, 84, 85, 86, 87, 88, 89, 90,\n91, 92, 93, 94, 95, 96, 97, 98, 99, 100.\nThen, multiples of 3 can not be prime numbers (except 3). This is easy to do. Start from 3 and skip two numbers and strike the next number. We get\n1, 2, 3, 4, 5, 6, 7, 8, 9, 10,\n11, 12, 13, 14, 15, 16, 17, 18, 19, 20,\n21, 22, 23, 24, 25, 26, 27, 28, 29, 30,\n31, 32, 33, 34, 35, 36, 37, 38, 39, 40,\n41, 42, 43, 44, 45, 46, 47, 48, 49, 50,\n51, 52, 53, 54, 55, 56, 57, 58, 59, 60,\n61, 62, 63, 64, 65, 66, 67, 68, 69, 70,\n71, 72, 73, 74, 75, 76, 77, 78, 79, 80,\n81, 82, 83, 84, 85, 86, 87, 88, 89, 90,\n91, 92, 93, 94, 95, 96, 97, 98, 99, 100.\nThere is no need to look for multiple of 4 as they are already covered in multiples of 2. Next is to see multiples of 5 (of course ignoring 5). This is simple. Just strike out numbers that ends with 5 and 0. We get 1, 2, 3, 4, 5, 6, 7, 8, 9, 10,\n11, 12, 13, 14, 15, 16, 17, 18, 19, 20,\n21, 22, 23, 24, 25, 26, 27, 28, 29, 30,\n31, 32, 33, 34, 35, 36, 37, 38, 39, 40,\n41, 42, 43, 44, 45, 46, 47, 48, 49, 50,\n51, 52, 53, 54, 55, 56, 57, 58, 59, 60,\n61, 62, 63, 64, 65, 66, 67, 68, 69, 70,\n71, 72, 73, 74, 75, 76, 77, 78, 79, 80,\n81, 82, 83, 84, 85, 86, 87, 88, 89, 90,\n91, 92, 93, 94, 95, 96, 97, 98, 99, 100.\nThere is no need to look for multiple of 6 as they are already covered in multiples of 2 and 3. Next is to see multiples of 7. One way to do is tell 7 table loudly (seven vanjaar seven, seven tujaar forteen). We get 1, 2, 3, 4, 5, 6, 7, 8, 9, 10,\n11, 12, 13, 14, 15, 16, 17, 18, 19, 20,\n21, 22, 23, 24, 25, 26, 27, 28, 29, 30,\n31, 32, 33, 34, 35, 36, 37, 38, 39, 40,\n41, 42, 43, 44, 45, 46, 47, 48, 49, 50,\n51, 52, 53, 54, 55, 56, 57, 58, 59, 60,\n61, 62, 63, 64, 65, 66, 67, 68, 69, 70,\n71, 72, 73, 74, 75, 76, 77, 78, 79, 80,\n81, 82, 83, 84, 85, 86, 87, 88, 89, 90,\n91, 92, 93, 94, 95, 96, 97, 98, 99, 100.\nThere is no need to look for multiples of 8 as they are already covered in multiples of 2. There is no need to look for multiples of 9 as they are already covered in multiples of 3. All multiples of 10 are already strike out. If we ignore 1, the remaining numbers are supposed to be primes.\n2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.\nGroups of the above orders are of prime order, so, cyclic, thus abelian. So, we ignore those (along with 1 of course). We are left with 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49, 50, 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68, 69, 70, 72, 74, 75, 76, 77, 78, 80, 81, 82, 84, 85, 86, 87, 88, 90, 91, 92, 93, 94, 95, 96, 98, 99, 100\nAnother interesting result that assure abelian property of group based on number of elements is the following :\nany group of order \\(p^2\\) for a prime \\(p\\) is abelian. This is very simple to compute. Just remove \\(2^2,3^2,5^2,7^2\\) that is, \\(4,9,25,49\\). We are left with 6, 8, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 26, 27, 28, 30, 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 50, 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68, 69, 70, 72, 74, 75, 76, 77, 78, 80, 81, 82, 84, 85, 86, 87, 88, 90, 91, 92, 93, 94, 95, 96, 98, 99, 100\nAnother interesting result assures that, any \\(p\\)-group will have a non-trivial centre. In other words, centre of non abelian $p$-group is always non trivial (apart from being a normal subgroup). So, we can safely ignore all powers of primes.\nJust remove \\(2^2,2^3,2^4,2^5,2^6,3^2,3^3,3^4,5^2,7^2\\) that is, \\(4,8,16,32,64,9,27,81,25,49\\). We are left with 6, 10, 12, 14, 15, 18, 20, 21, 22, 24, 26, 28, 30, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 50, 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 65, 66, 68, 69, 70, 72, 74, 75, 76, 77, 78, 80, 82, 84, 85, 86, 87, 88, 90, 91, 92, 93, 94, 95, 96, 98, 99, 100\nNote that, all the numbers above has more than one prime factor. We do have an interesting result about groups of order which has exactly two prime factors. A group \\(G\\) with \\(|G|=pq\\) for distinct primes \\(p,q\\) can not be a simple group. Sylow theorem says number of sylow \\(p\\)-subgroups should be \\(1\\) or \\(q\\). Suppose number of sylow \\(p\\)-subgroups is \\(1\\), it is a normal subgroup so, \\(G\\) can not be a simple group. Suppose number of sylow \\(p\\)-subgroups is \\(q\\). As they can not have a non-trivial intersection, the number of non-identity elements in these sylow $p$-subgroups is \\(q(p-1)\\) (with \\(p-1\\) non identity elements from each of the \\(q\\) subgroups). So, for sylow \\(q\\)-subgroup we are left with only \\(pq-(pq-q)=q\\) elements. With \\(q\\) elements, we can construct exactly one sylow \\(q\\)-subgroup of \\(G\\). So, sylow \\(q\\)-subgroup is unique, hence normal. In any case \\(G\\) will have either a sylow \\(p\\)-subgroup or a sylow \\(q\\)-subgroup; hence can not be simple.\nWith this observation, we can safely remove numbers which are of the form \\(pq\\) for distinct primes \\(p,q\\). They are given by \\[2\\times 3, 2\\times 5, 2\\times 7, 2\\times 11, 2\\times 13, 2\\times 17, 2\\times 19, 2\\times 23, 2\\times 29, 2\\times 31, 2\\times 37, 2\\times 41, 2\\times 43, 2\\times 47\\]\n\\[3\\times 5, 3\\times 7, 3\\times 11, 3\\times 13, 3\\times 17, 3\\times 19, 3\\times 23, 3\\times 29, 3\\times 31\\]\n\\[5\\times 7, 5\\times 11, 5\\times 13, 5\\times 17, 5\\times 19\\]\n\\[7\\times 11, 7\\times 13\\]\nThe numbers coming in above procedure are\n6, 10, 14, 15, 21, 22, 26, 33, 34, 35, 38, 39, 46, 51, 55, 57, 58, 62, 65, 69, 74, 77, 82, 85, 86, 87, 91, 93, 94, 95.\nRemoving these numbers (some of them are already gone in previous filtering) from the previous sequence gives us, 12, 18, 20, 24, 28, 30, 36, 40, 42, 44, 45, 48, 50, 52, 54, 56, 60, 63, 66, 68, 70, 72, 75, 76, 78, 80, 84, 88, 90, 92, 96, 98, 99, 100\nAnother interesting result about groups of order which has exactly two prime factors says that, any group of order \\(p^2q\\) will have either a sylow $p$-subgroup or a sylow \\(q\\)-subgroup, hence not simple.\nThe proof is almost same as that of the other result we saw above. So, it is time to look for numbers of the form \\(p^2q\\) for distinct primes \\(p,q\\). They are\n\\[2^2\\times 3, 2^2\\times 5, 2^2\\times 7, 2^2\\times 11, 2^2\\times 13, 2^2\\times 17, 2^2\\times 19, 2^2\\times 23, 2^2\\times 29\\]\n\\[3^2\\times 2, 3^2 \\times 5, 3^2\\times 7, 3^2\\times 11\\]\n\\[5^2\\times 2, 5^2\\times 3\\]\n\\[7^2\\times 2\\]\nThe numbers coming in above procedure are 12, 18, 20, 28, 44, 45, 50, 52, 63, 68, 75, 76, 92, 98, 99.\nRemoving these numbers (some of them are already gone in previous filtering) from the previous sequence gives us, 24, 30, 36, 40, 42, 48, 54, 56, 60, 66, 70, 72, 78, 80, 84, 88, 90, 96, 100\nDo we know anything about groups of order of the form \\(p^2q^2\\) for distinct primes \\(p,q\\)? Some of the numbers in above collection has exactly three prime factors with each prime factor appearing exactly once. As we saw every group of order \\(pq\\) for distinct primes \\(p,q\\) is not simple, we can also see every group of order \\(pqr\\) for distinct primes \\(p,q,q\\) is also not simple (sylow theorem says one of sylow subgroups is normal).\nThe numbers of the form \\(pqr\\) for distinct primes \\(p,q,r\\) are\n\\[30=2\\times 3\\times 5\\]\n\\[42=2\\times 3\\times 7\\]\n\\[66=2\\times 3\\times 11\\]\n\\[70=2\\times 5\\times 7\\]\n\\[78=2\\times 3\\times 13\\]\nRemoving these numbers (some of them are already gone in previous filtering) from the previous sequence gives us, 24, 36, 40, 48, 54, 56, 60, 72, 80, 84, 88, 90, 96, 100\nSome of the above numbers are of the form \\(p^3q\\) for distinct primes \\(p,q\\); namely \\(24,40,54,56,88\\). Let us see if we can make some quick observations.\nFor \\(G\\) with \\(|G|=24=2^3\\times 3\\), the only interesting case is when \\(n_2=3\\). A trick gives us a group homomorphism \\(\\Phi:G\\rightarrow S_3\\) with a bound on kernel of \\(\\Phi\\), namely \\(8\\leq |Ker(\\Phi)|\u0026lt; 24\\). As kernel of a group homomorphism is a normal subgroup, we can see such \\(G\\) with \\(n_2=3\\) will have a normal subgroup and hence not simple. So, any group of order $24$ is not simple. One lesson to learn from this example is, it is possible for a group to have a normal subgroup, which is not equal to any of the sylow subgroups. For \\(G\\) with \\(|G|=40=2^3\\times 5\\) it is straight forward to see that only possible value for \\(n_5\\) (number of sylow \\(5\\)-subgroups) is \\(1\\) as we need \\(n_5=1+5k\\) to divide \\(8\\) and only possibility is when \\(n_5=1\\). So, any group of order \\(40\\) is not simple. For \\(G\\) with \\(|G|=54=2\\times 3^3\\) it is straight forward to see that only possible value for \\(n_3\\) (number of sylow \\(3\\)-subgroups) is \\(1\\) as we need \\(n_3=1+3k\\) to divide \\(2\\) and only possibility is when \\(n_3=1\\). So, any group of order \\(54\\) is not simple. For \\(G\\) with \\(|G|=56=2^3\\times 7\\), only choice for \\(n_7\\) is \\(1\\) or \\(8\\). Suppose \\(n_7=1\\), then it is normal and we are done. Suppose \\(n_7=8\\), due to the order of sylow \\(7\\) subgroup, no two sylow $7$-subgroups will have a non trivial intersection. So, sylow \\(7\\)-subgroups constitute \\(8\\times 6=48\\) non identity elements of \\(G\\) (with \\(6\\) non-identity elements from each of the \\(8\\) sylow $7$-subgroups). Out of \\(56\\), if \\(48\\) are reserved for sylow \\(7\\)-subgroups, we are left with only \\(8\\) elements and with those \\(8\\) elements we can only construct one sylow $2$-subgroup for \\(G\\); which means sylow \\(2\\)-subgroup of \\(G\\) is normal. So, any group of order \\(56\\) is not simple. For \\(G\\) with \\(|G|=88=2\\times 3^11\\) it is straight forward to see that only possible value for \\(n_{11}\\) (number of sylow \\(11\\)-subgroups) is \\(1\\) as we need \\(n_{11}=1+11k\\) to divide \\(8\\) and only possibility is when \\(n_{11}=1\\). So, any group of order \\(88\\) is not simple. Removing the above five numbers (some of them are already gone in previous filtering) from the previous sequence gives us,\n36, 48, 60, 72, 80, 84, 90, 96, 100\nOut of the above numbers, only three of them have three distinct prime factors, namely \\(60=2^2\\times 3\\times 5, 84=2^2\\times 3\\times 7, 90=2\\times 3^2\\times 5\\). For \\(G\\) with \\(|G|=84=2^2\\times 3\\times 7\\) it is straight forward to see that only possible value for \\(n_{7}\\) (number of sylow \\(7\\)-subgroups) is \\(1\\) as we need \\(n_{7}=1+7k\\) to divide \\(12\\) and only possibility is when \\(n_{7}=1\\). So, any group of order \\(84\\) is not simple. For \\(G\\) with \\(|G|=90=2\\times 3^2\\times 5\\) it is not straight forward. It would be fun to try on your own. As in the case of group of order \\(24\\), we need to think slightly differently. In that case we were able to associate a group homomorphism from \\(G\\) and the kernel turned out to be non-trivial normal subgroup of \\(G\\). In this case of \\(G\\) with \\(|G|=90\\), there are many ways to see existence of a normal subgroup, one of which is to involve the notion of \"normalizer of a subgroup\". With a small effort, we can see any group of order \\(90\\) has a non trivial normal subgroup, thus not a simple group.\nA special care needs to be taken when doing similar observation for groups of order \\(60\\), which we will push towards the end of this discussion.\nRemoving the above five numbers (some of them are already gone in previous filtering) from the previous sequence gives us,\n36, 48, 60, 72, 80, 96, 100\nFor \\(G\\) with \\(|G|=36\\), as in the case of \\(G\\) with \\(|G|=90\\), thinking in terms of normaliser of a subgroup would be useful. Try this on your own.\nFor \\(G\\) with \\(|G|=48=2^4\\times 3\\) only options for \\(n_3\\) are \\(1\\) and \\(16\\). If \\(n_3=16\\), then, we have total \\(2\\times 16=32\\) non identity elements (with \\(2\\) non identity elements coming from \\(16\\) sylow $3$-subgroups). As the order of \\(G\\) is \\(48\\), all sylow \\(2\\)-subgroups combined can have only \\(48-32=16\\) elements. But, each sylow \\(2\\)-subgroup will have exactly \\(16\\) elements. So, there can be exactly one sylow \\(2\\)-subgroup. As mentioned before, this means \\(G\\) is not simple.\nFor \\(G\\) with \\(|G|=72\\), it is not straight forward. As in the case of \\(G\\) with \\(|G|\\in \\{24,90\\}\\), we can associate a homomorphism from \\(G\\) whose kernel gives us a non trivial normal subgroup. Thus, any group of order \\(72\\) is not simple.\nFor \\(G\\) with \\(|G|=80=2^4\\times 5\\), only interesting case for \\(n_5\\) is \\(16\\). If this is the case, we will have contribution of \\(4\\times 16=64\\) non identity elements from sylow \\(5\\)-subgroups. As the order of \\(G\\) is \\(80\\), all sylow \\(2\\)-subgroups combined can have only \\(80-64=16\\) elements. But, each sylow \\(2\\)-subgroup will have exactly \\(16\\) elements. So, there can be exactly one sylow \\(2\\)-subgroup. As mentioned before, this means \\(G\\) is not simple.\nFor \\(G\\) with \\(|G|=96\\), as in the case of \\(G\\) with \\(|G|\\in \\{36,90\\}\\), thinking in terms of normaliser of a subgroup would be useful. Try this on your own.\nWe are left with only two options namely 60 and 100.\nWe said we are looking for groups of order less than 100. So, we should not really count 100.\nWe are left with only one possibility, namely 60. There are examples of groups of order 60 some of which are non-simple. But, there is exactly one simple group (upto isomorphism) of order \\(60\\), namely, the alternating group \\(A_5\\). So, we have the following result\nany non abelian simple group of order less than \\(100\\) is isomorphic to \\(A_5\\).\nOut of first 100 numbers, if only one number has the possibility of being order of a simple group, does it mean simple groups are interesting?? or uninteresting??\n","permalink":"https://praphulla-koushik.github.io/2024/11/18/non-abelian-simple-group-of-order-less-than-100/","summary":"\u003c!-- wp:paragraph --\u003e\n\u003cp\u003eOn a Saturday morning, I was thinking about sylow theorems. \u003c/p\u003e\n\u003c!-- /wp:paragraph --\u003e\n\n\u003c!-- wp:paragraph --\u003e\n\u003cp\u003eThe question I asked myself is, do I know how to apply sylow theorems? \u003c/p\u003e\n\u003c!-- /wp:paragraph --\u003e\n\n\u003c!-- wp:paragraph --\u003e\n\u003cp\u003eOnly application I was aware about, of sylow theorem, is to assure if a group of finite order is simple or not. \u003c/p\u003e\n\u003c!-- /wp:paragraph --\u003e\n\n\u003c!-- wp:paragraph --\u003e\n\u003cp\u003eAs a first step, I thought to check for groups of order less than 100. \u003c/p\u003e","tags":[],"title":"non-abelian simple group of order less than 100","type":"wp-import"},{"categories":["Analysis"],"content":" Let us consider a problem where you are asked to find infimum of the set\n\\[\\{\\int_0^{1}\\sqrt{1+f'(x)^2}dx\\}_{f\\in S}\\]\nwhere \\(S\\) is the set of all \\(f\\in C^1(\\mathbb{R})\\) with the property that \\(f(0)=10\\) and \\(f(1)=0\\).\nWhen we see integral and differential together, that should remind us the famous fundamental theorem of calculus, which says that \\[\\int_a^b g'(x)dx=g(b)-g(a)\\].\nIt is unfortunate (yet interesting) that the integrand \\(\\sqrt{1+f'(x)^2}\\) is not (immediately) of the form \\(g'(x)\\) for some \\(g\\). If it is just \\(f'(x)\\) instead of \\(\\sqrt{1+f'(x)^2}\\), we could have just applied the fundamental theorem of calculus directly. But, that is not the case here. Nevertheless, there is an obvious relation between \\(f'(x)\\) and \\(\\sqrt{1+(f'(x))^2}\\), namely,\n\\[f'(x)\\leq \\sqrt{1+(f'(x))^2}\\]\nwhich implies that\n\\[\\int_0^1f'(x)dx\\leq \\int_0^1\\sqrt{1+(f'(x))^2} dx,\\]\nwhich, when applied fundamental theorem of calculus, gives us\n\\[f(1)-f(0)\\leq \\int_0^1\\sqrt{1+(f'(x))^2} dx,\\]\nthat is,\n\\[-10\\leq \\int_0^1\\sqrt{1+(f'(x))^2} dx,\\]\nwhich is one of the most useless observations one can ever make.\nThe function \\(\\sqrt{1+(f'(x))^2}\\) is non-negative and so the corresponding integral \\(\\int_0^1\\sqrt{1+(f'(x))^2} dx\\). So, observing \\[\\int_0^1\\sqrt{1+(f'(x))^2} dx\\geq -10\\]\nis of no use for us; we already know that.\nInstead of \\(f(0)=10\\) and \\(f(1)=0\\), if we had \\(f(0)=0\\) and \\(f(1)=10\\), we would have realized that \\[f(1)-f(0)\\leq \\int_0^1\\sqrt{1+(f'(x))^2} dx,\\]\nwhich implies\n\\[10\\leq \\int_0^1\\sqrt{1+(f'(x))^2} dx,\\]\nwhich may be treated as a non-trivial observation in our situation.\nInstead of crying for what we do not have, we can focus on what other things can we do with what we have. It is only a matter of plus/minus. Instead of starting with the observation \\(f'(x)\\leq \\sqrt{1+(f'(x))^2}\\), we can start with the observation \\(-f'(x)\\leq \\sqrt{1+(f'(x))^2}\\) which implies that\n\\[-\\int_0^1f'(x)dx\\leq \\int_0^1\\sqrt{1+(f'(x))^2} dx,\\]\nwhich, when applied fundamental theorem of calculus, gives us\n\\[-(f(1)-f(0))\\leq \\int_0^1\\sqrt{1+(f'(x))^2} dx,\\]\nthat is,\n\\[10\\leq \\int_0^1\\sqrt{1+(f'(x))^2} dx,\\]\nwhich is a reasonably non-trivial observation we have made from the given data.\nAfter patting your back for this wonderful observation, we will proceed to the next step.\nAs \\(10\\leq \\int_0^1\\sqrt{1+(f'(x))^2} dx\\) for all \\(f\\in S\\), we can conclude that\n\\[10\\leq \\inf\\{\\int_0^{1}\\sqrt{1+f'(x)^2}dx\\}_{f\\in S}.\\]\nIf we able to do some magic and show\n\\[11\\leq \\int_0^1\\sqrt{1+(f'(x))^2} dx\\]\nfor all \\(f\\in S\\), we can conclude that \\[11\\leq \\inf\\{\\int_0^{1}\\sqrt{1+f'(x)^2}dx\\}_{f\\in S},\\]\nwhich is better than saying \\[10\\leq \\inf\\{\\int_0^{1}\\sqrt{1+f'(x)^2}dx\\}_{f\\in S}.\\]\nI am assuming you know why it is better? Do not prove me wrong. Give an attempt, you will figure out.\nKeeping that aside, we have not computed what is \\(\\int_0^1\\sqrt{1+(f'(x))^2} dx\\) for a fixed \\(f\\in S\\). Forget about computing integral, do we know atleast one concrete example of a function \\(f\\in C^1(\\mathbb{R})\\) with the property that \\(f(0)=10\\) and \\(f(1)=0\\)? It may be the case that \\(S\\) is an empty set and we are just looking for infimum of an empty set.\nForget about \\(C^1\\) function and all that. Can we find a function \\(f\\) with the property that \\(f(0)=10\\) and \\(f(1)=0\\); in other words, a function whose graph has the points \\((0,10)\\) and \\((1,0)\\).\nHigh school mathematics suggest us that there is a line passing through \\((0,10)\\) and \\((0,10)\\) given by \\(y=mx+c\\) with \\(c=10\\) and \\(m=-10\\); that is the line \\(y=10-10x\\). I need not highlight this function is \\(C^1\\).\nLet us compute \\(\\int_0^1\\sqrt{1+(f'(x))^2} dx\\) for \\(f(x)=10-10x\\). Note that, \\(f'(x)=-10\\). So, \\[\\sqrt{1+(f'(x))^2}=\\sqrt{101}.\\]\nWhich implies that,\n\\[\\int_0^1\\sqrt{1+(f'(x))^2} dx=\\int_0^{1}\\sqrt{101}dx=\\sqrt{101}\\int_0^1dx=\\sqrt{101}(1-0)=\\sqrt{101}.\\]\nIt is unfortunate to realize that, we do not have \\(11\\leq \\int_0^1\\sqrt{1+(f'(x))^2} dx\\) for all \\(f\\in S\\). So, \\[11\u0026gt; \\inf\\{\\int_0^{1}\\sqrt{1+f'(x)^2}dx\\}_{f\\in S}.\\]\nAs we have got one function \\(f\\) with the property that \\(\\int_0^{10}\\sqrt{1+f'(x)^2}dx=\\sqrt{101}\\), we see that,\n\\[\\inf\\{\\int_0^{1}\\sqrt{1+f'(x)^2}dx\\}_{f\\in S}\\leq \\sqrt{101}\\]\nAdding the previous observation \\[10\\leq \\inf\\{\\int_0^{1}\\sqrt{1+f'(x)^2}dx\\}_{f\\in S}\\]\nto the current observation, we see that,\n\\[10\\leq \\inf\\{\\int_0^{1}\\sqrt{1+f'(x)^2}dx\\}_{f\\in S}\\leq \\sqrt{101},\\]\nwhich can be seen as\n\\[\\sqrt{100}\\leq \\inf\\{\\int_0^{10}\\sqrt{1+f'(x)^2}dx\\}_{f\\in S}\\leq \\sqrt{101}\\]\nThe gap between \\(100\\) and \\(101\\) is \\(1\\) and the gap between \\(\\sqrt{100}\\) and \\(\\sqrt{101}\\) is much less than \\(1\\). So, our area of search is reduced to the area between \\(10=\\sqrt{100}\\) and \\(\\sqrt{101}\\).\nAs the question is asked to you and not to me, it is not fair for me to finish it for you. So, I will stop it here. You may want to enjoy finishing this step.\nApologies for spoiling your fun in attempting this problem. See you in next problem.\n","permalink":"https://praphulla-koushik.github.io/2024/09/21/computing-infimum-by-an-example/","summary":"\u003c!-- wp:paragraph --\u003e\n\u003cp\u003eLet us consider a problem where you are asked to find infimum of the set\u003c/p\u003e\n\u003c!-- /wp:paragraph --\u003e\n\n\u003c!-- wp:quote --\u003e\n\u003cblockquote class=\"wp-block-quote\"\u003e\u003c!-- wp:paragraph --\u003e\n\u003cp\u003e\\[\\{\\int_0^{1}\\sqrt{1+f'(x)^2}dx\\}_{f\\in S}\\]\u003c/p\u003e\n\u003c!-- /wp:paragraph --\u003e\u003c/blockquote\u003e\n\u003c!-- /wp:quote --\u003e\n\n\u003c!-- wp:paragraph --\u003e\n\u003cp\u003ewhere \\(S\\) is the set of all \\(f\\in C^1(\\mathbb{R})\\) with the property that \\(f(0)=10\\) and \\(f(1)=0\\).\u003c/p\u003e\n\u003c!-- /wp:paragraph --\u003e\n\n\u003c!-- wp:paragraph --\u003e\n\u003cp\u003eWhen we see integral and differential together, that should remind us the famous fundamental theorem of calculus, which says that \u003c/p\u003e","tags":[],"title":"computing infimum by an example","type":"wp-import"},{"categories":["category-theory","Analysis"],"content":" Let us check for limit/limsup/liminf of the sequence \\(\\frac{n}{10^{\\lceil \\log_{10}n \\rceil}}\\), where the notation \\(\\lceil x \\rceil\\) means the smallest integer greater than or equal to \\(x\\). For example, \\(\\lceil 0.1 \\rceil=1, \\lceil 0.9 \\rceil=1, \\lceil -1.2 \\rceil=-1, \\lceil -2.5 \\rceil=-2\\)\nTo compute limit (to have a hope of computing), we need to know it converge (which we can check by checking it is Cauchy sequence). To compute limsup/liminf, we do not need it to be Cauchy or we do not need to know if it is Cauchy. All that matters is if it is bounded. Is it bounded? Bounded below, for sure, as every element is non-negative. Is it bounded above?\nLet's check. If it is not bounded above, then, for 11924, there exists an \\(n\\in \\mathbb{N}\\) with the property that, \\[\\frac{n}{10^{\\lceil \\log_{10}n \\rceil}}\u0026gt;11924\\]\nwhich is equivalent to saying (or at least imply that) \\[\\log_{10}(n)-\\log_{10}(10^{\\lceil \\log_{10}n \\rceil})\u0026gt;\\log(11924)\\]\nwhich is equivalent to saying (or at least imply that) \\[\\log_{10}(n)-\\lceil \\log_{10}n \\rceil\u0026gt;\\log(11924)\\]\nwhich is just not true. The difference between \\(x\\) and \\(\\lceil x \\rceil\\) is less than \\(1\\). To be precise, \\(\\lceil x \\rceil-x\u0026lt;1\\). So, we can not have \\(\\log_{10}(n)-\\lceil \\log_{10}n \\rceil\u0026gt;\\log(11924)\\) as \\(\\log_{10}(11924)\u0026gt;0\\) where as \\(\\log_{10}(n)-\\lceil \\log_{10}n \\rceil\u0026lt;0\\).\nOr for that matter, we can not have \\(\\log_{10}(n)-\\lceil \\log_{10}n \\rceil\u0026gt;\\log_{10}(1924)\\) as \\(\\log(1924)\u0026gt;0\\) where as \\(\\log_{10}(n)-\\lceil \\log_{10}n \\rceil\u0026lt;0\\).\nWe can not even have \\(\\log_{10}(n)-\\lceil \\log_{10}n \\rceil\u0026gt;\\log_{10}(10)\\) as \\(\\log(1924)\u0026gt;0\\) where as \\(\\log_{10}(n)-\\lceil \\log_{10}n \\rceil\u0026lt;0\\).\nHow about \\(\\log_{10}(n)-\\lceil \\log_{10}n \\rceil\u0026gt;\\log_{10}(1)\\)?? That is also not true, as \\(\\log_{10}(1)\\geq 0\\) where as \\(\\log_{10}(n)-\\lceil \\log_{10}n \\rceil\u0026lt;0\\).\nSo, there is no \\(n\\in \\mathbb{N}\\) with the property that \\(\\log_{10}(n)-\\lceil \\log_{10}n \\rceil\u0026gt;\\log_{10}(1)\\) which means, for every \\(n\\in \\mathbb{N}\\), we have \\(\\log_{10}(n)-\\lceil \\log_{10}n \\rceil\u0026lt;\\log_{10}(1)\\)\nwhich is equivalent to saying (or at least imply that)\n\\[\\log_{10}(n)\u0026lt;\\log_{10}(1)+\\lceil \\log_{10}n \\rceil=\\log_{10}(1)+\\log_{10}(10^{\\lceil \\log_{10})n \\rceil}=\\log_{10}(10^{\\lceil \\log_{10})n \\rceil}\\]\nwhich is equivalent to saying (or at least imply that),\n\\[n\u0026lt;10^{\\lceil \\log_{10}n \\rceil}\\] for all \\(n\\in \\mathbb{N}\\),\nwhich is equivalent to saying (or at least imply that),\n\\[\\frac{n}{10^{\\lceil \\log_{10}n \\rceil}}\u0026lt;1\\] for all \\(n\\in \\mathbb{N}\\),\nThus the sequence \\((a_n)\\) is bounded above by \\(1\\).\nThis gives us idea about what to expect as \\(\\lim(a_n)\\) (if someone who we believe tells us that \\((a_n)\\) converges, but cant say anything more than that). As \\((a_n)\\) is bounded above by \\(1\\), the limit (if it exists) can be almost \\(1\\). If the limit does not exists, the \\(\\limsup (a_n)\\) is also almost \\(1\\).\nThere is a thing about dealing with log base 10. It behaves in a certain way when \\(n\\) is a power of \\(10\\). \\[\\log_{10}10=1, \\log_{10}(10^2)=2, \\log_{10}(10^3)=3,\\cdots\\],\nwhich assures us that, \\[\\lceil \\log_{10}10 \\rceil=1, \\lceil \\log_{10}(10^2) \\rceil=2, \\lceil \\log_{10}(10^3) \\rceil=3,\\cdots\\],\nwhich imply that \\[10^{\\lceil \\log_{10}10 \\rceil}=10^1, 10^{\\lceil \\log_{10}(10^2) \\rceil}=10^2, 10^{\\lceil \\log_{10}(10^3) \\rceil}=10^3,\\cdots\\],\nor, more generally, \\(10^{\\lceil \\log_{10}(10^k) \\rceil}=10^k\\) for all \\(k\\in \\mathbb{N}\\), which imply that,\n\\[\\frac{10^k}{10^{\\lceil \\log_{10}(10^k) \\rceil}}=\\frac{10^k}{10^k}=1\\] for all \\(k\\in \\mathbb{N}\\)\nGreat. It is not just that \\(1\\) is an upper bound for the sequence, but, it actually appears at multiple places. We actually have a subsequence of \\((a_n)\\) that is just the constant sequence \\(1\\)\nBoom!!! \\(1\\) is the limsup of \\((a_n)\\). If \\((a_n)\\) is bounded above by \\(L\\in \\mathbb{R}\\) and we are able to get a subsequence \\((a_{n_k})\\) that converges to \\(L\\), then, \\(\\limsup(a_n)=L\\). This does not say anything about the non-convergence of the sequence \\((a_n)\\). It may happen that the sequence converges to \\(1\\). It would not happen if we are able to find another subsequence that converges to anything else but \\(1\\). We know the behavior of the subseqeunce \\((a_{10^k})\\). Keeping it as the reference subsequence, let us check the subsequences to the immediate left and immediate right of \\((a_{10^k}\\). Consider the subsequence \\(a_{10^k-1}\\). We have \\[a_{10^k-1}=\\frac{10^k-1}{10^{\\lceil \\log_{10}(10^k-1) \\rceil}}\\]\nAs \\(10^k-1\u0026lt;10^k\\), we would have \\(\\log_{10}(10^k-1)\u0026lt;\\log_{10}(10^k)=k\\). As there is not much of a gap between \\(10^k-1\\) and \\(10^k\\), we can safely believe (you should prove) that, \\(\\lceil \\log_{10}(10^k-1) \\rceil=k\\). Thus, \\[a_{10^k-1}=\\frac{10^k-1}{10^{\\lceil \\log_{10}(10^k-1) \\rceil}}=\\frac{10^k-1}{10^k}=1-\\frac{1}{10^k}\\]\nIt is unfortunate that this subsequence also converges to \\(1\\). So, we can not conclude if \\((a_n)\\) converges or not. Let's continue with the subsequence that is immediate right to \\((a_{10^k})\\), the subsequence \\(a_{10^k+1}\\). We have \\[a_{10^k+1}=\\frac{10^k+1}{10^{\\lceil \\log_{10}(10^k+1) \\rceil}}\\].\nUnlike \\(10^k-1\\), the component \\(10^k+1\\) behaves in a slightly different way. As \\(10^k\u0026lt;10^k+1\\), we have \\(\\log_{10}(10^k)\u0026lt;\\log_{10}(10^{k}+1)\\). Which implies, \\(\\lceil \\log_{10}(10^k) \\rceil\u0026lt;\\lceil \\log_{10}(10^{k}+1) \\rceil\\); that is, \\(k\u0026lt;\\lceil \\log_{10}(10^{k}+1) \\rceil\\), which is same as saying \\(\\lceil \\log_{10}(10^{k}+1) \\rceil=k+1\\). So, \\[a_{10^k+1}=\\frac{10^k+1}{10^{\\lceil \\log_{10}(10^k+1) \\rceil}}=\\frac{10^k+1}{10^{k+1}}=\\frac{10^k+1}{10^{k+1}}=\\frac{1}{10}+\\frac{1}{10^k}\\].\nThis subsequence converges to \\(\\frac{1}{10}\\) as the sequence \\(\\frac{1}{10^k}\\) converges to \\(0\\). It gives comfort when we realize there is a subsequence that does not converge to \\(1\\). This assures the sequence \\((a_n)\\) is not convergent. We were able to find \\(\\limsup(a_n)\\) just from one subsequence. That is because an upper bound is matching with the limit of subsequence. We can not conclude \\(\\liminf\\) is \\(\\frac{1}{10}\\) as we may not be able to assure there is no subsequence that converges to an element less than \\(\\frac{1}{10}\\). We can assure if somehow we can prove that \\(\\frac{1}{10}\\) is a lower bound for the sequence. If \\((a_n)\\) is bounded below by \\(L\\in \\mathbb{R}\\) and we are able to get a subsequence \\((a_{n_k})\\) that converges to \\(L\\), then, \\(\\liminf(a_n)=L\\). It is easy to check if \\(\\frac{1}{10}\\) is a lower bound or nor. All we need to see is if, for all \\(n\\in \\mathbb{N}\\), we have \\[\\frac{1}{10}\\leq \\frac{n}{10^{\\lceil \\log_{10}n \\rceil}}\\]\nwhich is same as saying \\[10^{\\lceil \\log_{10}n \\rceil}\u0026lt;10 n\\],\nwhich is same as saying \\[\\log_{10}(10^{\\lceil \\log_{10}n \\rceil})\u0026lt;\\log_{10}(10 n)=\\log_{10}10+\\log_{10}n\\]\nwhich is same thing as saying \\(\\lceil \\log_{10}n \\rceil\u0026lt; 1+\\log_{10}n\\),\nwhich is as true as the statement \"sun rises in the east\". So, the sequence \\((a_n)\\) is bounded below by \\(\\frac{1}{10}\\) and we have a subsequence that converges to \\(\\frac{1}{10}\\). So, \\(\\liminf a_n=\\frac{1}{10}\\). ","permalink":"https://praphulla-koushik.github.io/2024/09/13/limit-limsup-liminf-of-a-sequence-by-an-example/","summary":"\u003c!-- wp:paragraph --\u003e\n\u003cp\u003eLet us check for limit/limsup/liminf of the sequence \\(\\frac{n}{10^{\\lceil \\log_{10}n \\rceil}}\\), where the notation \\(\\lceil x \\rceil\\) means the smallest integer greater than or equal to \\(x\\). \u003c/p\u003e\n\u003c!-- /wp:paragraph --\u003e\n\n\u003c!-- wp:paragraph --\u003e\n\u003cp\u003eFor example, \\(\\lceil 0.1 \\rceil=1, \\lceil 0.9 \\rceil=1, \\lceil -1.2 \\rceil=-1, \\lceil -2.5 \\rceil=-2\\)\u003c/p\u003e\n\u003c!-- /wp:paragraph --\u003e\n\n\u003c!-- wp:paragraph --\u003e\n\u003cp\u003eTo compute limit (to have a hope of computing), we need to know it converge (which we can check by checking it is Cauchy sequence).  \u003c/p\u003e","tags":[],"title":"limit/limsup/liminf of a sequence (by an example)","type":"wp-import"},{"categories":["differential-geometry"],"content":" Let \\(M\\) be a smooth manifold and \\(E\\rightarrow M\\) a vector bundle over \\(M\\). A connection on the vector bundle \\(E\\rightarrow M\\) is usually defined as a map \\[\\nabla : \\Gamma(M,TM)\\times \\Gamma(M,E)\\rightarrow \\Gamma(M,E)\\]\nsatisfying the following conditions:\n\\(\\nabla\\) behaves very well with the \\(\\mathbb{R}\\)-vector space structure on \\(\\Gamma(M,TM)\\) and \\(\\Gamma(M,E)\\); in the sense that, \\(\\nabla\\) is an \\(\\mathbb{R}\\)-bilinear map, \\(\\nabla\\) behaves reasonably well with the \\(C^\\infty(M)\\)-module structure on \\(\\Gamma(M,TM)\\) and \\(\\Gamma(M,E)\\); in the sense that, \\[\\nabla(fX,s)=f\\nabla(X,s)\\] for \\(X\\in \\Gamma(M,TM)\\) and \\(s\\in\\Gamma(M,E)\\)\n\\[\\nabla(X,fs)=f\\nabla (X,s)+X(f)s\\] for \\(X\\in \\Gamma(M,TM)\\) and \\(s\\in \\Gamma(M,E)\\)\nThe \\(C^\\infty(M)\\)-linearity of the map \\(\\nabla(-,s):\\Gamma(M,TM)\\rightarrow \\Gamma(M,E)\\) for each section \\(s\\in \\Gamma(M,E)\\) implies we have an \\(\\mathbb{R}\\)-linear map \\[\\Gamma(M,E)\\rightarrow \\hom_{C^\\infty(M)}(\\Gamma(M,TM),\\Gamma(M,E))\\]\nWe know that \\(\\hom (V,W)\\cong V^*\\otimes W\\) for vector spaces \\(V,W\\). More generally, \\(\\hom_R(V,W)=V^*\\otimes_RW\\) for \\(R\\)-modules \\(V,W\\) for a commutative ring \\(R\\).\nIn the case of \\(R=C^\\infty(M)\\), we have \\[\\hom_{C^\\infty(M)}(\\Gamma(M,TM),\\Gamma(M,E))\\cong \\Gamma(M,TM)^*\\otimes_{C^\\infty(M)}\\Gamma(M,E)\\],\nequivalently, \\[\\hom_{C^\\infty(M)}(\\Gamma(M,TM),\\Gamma(M,E))\\cong \\Gamma(M,T^*M)\\otimes_{C^\\infty(M)}\\Gamma(M,E)\\].\nWe also have a relation between \"tensor product of sections\" and section of tensor product of vector bundles; given by \\[\\Gamma(M,T^*M)\\otimes_{C^\\infty(M)}\\Gamma(M,E)\\cong \\Gamma(M,T^*M\\otimes E)\\]. For this reason, some people write a connection on a vector bundle \\(E\\rightarrow M\\) as an \\(\\mathbb{R}\\)-linear map \\(\\nabla:\\Gamma(M,E)\\rightarrow \\Gamma(M,T^*M\\otimes E)\\) satisfying the condition \\[\\nabla(fs)=(df)\\otimes s+f\\nabla(s)\\]\nfor all \\(f\\in C^\\infty(M)\\) and \\(s\\in \\Gamma(M,E)\\).\n","permalink":"https://praphulla-koushik.github.io/2024/07/09/alternative-description-of-connection-on-vector-bundle/","summary":"\u003c!-- wp:paragraph --\u003e\n\u003cp\u003eLet \\(M\\) be a smooth manifold and \\(E\\rightarrow M\\) a vector bundle over \\(M\\). \u003c/p\u003e\n\u003c!-- /wp:paragraph --\u003e\n\n\u003c!-- wp:group {\"layout\":{\"type\":\"constrained\"}} --\u003e\n\u003cdiv class=\"wp-block-group\"\u003e\u003c!-- wp:paragraph --\u003e\n\u003cp\u003eA connection on the vector bundle \\(E\\rightarrow M\\) is usually defined as a map \u003c/p\u003e\n\u003c!-- /wp:paragraph --\u003e\n\n\u003c!-- wp:quote --\u003e\n\u003cblockquote class=\"wp-block-quote\"\u003e\u003c!-- wp:paragraph --\u003e\n\u003cp\u003e\\[\\nabla : \\Gamma(M,TM)\\times \\Gamma(M,E)\\rightarrow \\Gamma(M,E)\\]\u003c/p\u003e\n\u003c!-- /wp:paragraph --\u003e\u003c/blockquote\u003e\n\u003c!-- /wp:quote --\u003e\n\n\u003c!-- wp:paragraph --\u003e\n\u003cp\u003esatisfying the following conditions:\u003c/p\u003e\n\u003c!-- /wp:paragraph --\u003e\n\n\u003c!-- wp:list --\u003e\n\u003cul class=\"wp-block-list\"\u003e\u003c!-- wp:list-item --\u003e\n\u003cli\u003e\\(\\nabla\\) behaves very well with the \\(\\mathbb{R}\\)-vector space structure on \\(\\Gamma(M,TM)\\) and \\(\\Gamma(M,E)\\); in the sense that, \\(\\nabla\\) is an \\(\\mathbb{R}\\)-bilinear map,\u003c/li\u003e\n\u003c!-- /wp:list-item --\u003e\n\n\u003c!-- wp:list-item --\u003e\n\u003cli\u003e\\(\\nabla\\) behaves reasonably well with the \\(C^\\infty(M)\\)-module structure on \\(\\Gamma(M,TM)\\) and \\(\\Gamma(M,E)\\); in the sense that, \u003c/li\u003e\n\u003c!-- /wp:list-item --\u003e\u003c/ul\u003e\n\u003c!-- /wp:list --\u003e\n\n\u003c!-- wp:quote --\u003e\n\u003cblockquote class=\"wp-block-quote\"\u003e\u003c!-- wp:paragraph --\u003e\n\u003cp\u003e\\[\\nabla(fX,s)=f\\nabla(X,s)\\]  for \\(X\\in \\Gamma(M,TM)\\) and \\(s\\in\\Gamma(M,E)\\)\u003c/p\u003e","tags":["connections","tensor-product"],"title":"(Alternative description of) Connection on vector bundle","type":"wp-import"},{"categories":["Linear algebra"],"content":" Let us look at the first class of multilinear maps; the bilinear maps. We want to study bilinear maps. The notion of \"study\" will have different meanings as we move forward (or backward) in the course. Let \\(V,W,T\\) be vector spaces and \\(\\varphi:V\\times W\\rightarrow T\\) be a bilinear map. The feeling that \"we are good at linear algebra\" suggests us to ask the question :\nCan we associate a linear map for the bilinear map \\(\\varphi:V\\times W\\rightarrow T\\), so that (almost) all the information in the bilinear map is included in the linear map; in which case we can take some help from the (very little) linear algebra we knew?\nWe need to make it clear : Are we going to associate a linear map \\(P\\rightarrow M\\) for each bilinear map \\(\\varphi:V\\times W\\rightarrow T\\)? It would be useful to remind ourselves the phrase \"less luggage more comfort\". When trying to construct a new structure, it is good to start with least luggage. If we consider \\(V\\times W\\rightarrow T\\) as the data, we will have four components in that: three vector spaces \\(V, W,T\\) and a map \\(V\\times W\\rightarrow T\\). Vector spaces as such was not of issue for us; not in linear algebra. It is the product that is creating a discomfort for us. The product comes with just two vector spaces \\(V,W\\) and \"nothing else\". So, we ask the question again, with a small change:\nCan we associate a single vector space \\(P\\) for the product \\(V\\times W\\) (note that the word product starts with letter p), hoping that, it will help us to construct a linear map out of the bilinear map?\nOne can of course, look at the inclusion map \\(V\\rightarrow V\\times W, W\\rightarrow V\\times W\\) and consider the compositions \\(V\\rightarrow V\\times W\\rightarrow T, W\\rightarrow V\\times W\\rightarrow T\\); giving us two linear maps \\(V\\rightarrow T\\) and \\(W\\rightarrow T\\). This is not what we want. We want one single vector space, one single linear map to gather the information of the bilinear map.\nLet us recall some of the constructions you have seen in your previous courses. Construction of a quotient space \\(X/\\sim\\) from a topological space \\(X\\). We may not even acknowledge, but, the data of quotient space is not just a topological space, it is a topological space along with a map \\(X\\rightarrow X/\\sim\\). This is because the topology on the set \\(X/\\sim\\) is constructed from topology on \\(X\\).\nConstruction of a quotient space \\(V/W\\) from a vector space \\(V\\). As in the above case, the data of quotient space is not just a vector space, it is a vector space along with a map $V\\rightarrow V/W$. This is because the vector space structure on the set \\(V/W\\) is constructed from the vector space structure on \\(V\\). In similar way, in our construction, the data should not be just a vector space \\(P\\), it should be a vector space \\(P\\) along with a map \\(V\\times W\\rightarrow P\\). In case of topological space, any map between two topological spaces is understood to be a continuous map. Similarly, in case of vector spaces, any map between two vector spaces is understood to be a linear map. The last sentence is not quite correct; as in this very setup we are discussing about bilinear maps between two from a product vector space to another vector space. This raises the following question.\nQuestion : Do we want this map \\(V\\times W\\rightarrow P\\) to be a linear map or a bilinear map?\nIt looks like asking for a bilinear map is the opposite of what we should be doing. It is due to bilinear maps we are searching for a comfortable play ground where we can play making use of our previous experience in the play ground of linear algebra. But, it is not what we are after.\nLet us see what was the property of quotients maps mentioned above. The basic property of the map \\(V\\rightarrow V/W\\) is that, every element in \\(W\\) is mapped to the zero element of \\(V/W\\). Not just that. If we take any other linear map \\(V\\rightarrow N\\) in which every element in \\(W\\) is mapped to the zero element in \\(N\\), then, we can get a linear map \\(V/W\\rightarrow N\\) giving a commutative diagram.\nThe same can be taken as a reference point for our construction. In our case, we are looking for one single bilinear map, \\(V\\times W\\rightarrow P\\) that works as a necessary passage for all bilinear maps from \\(V\\times W\\). More precisely, we have the following description.\nA vector space \\(P\\) along with a bilinear map \\(V\\times W\\rightarrow P\\) such that, for any other vector space \\(N\\) and any other bilinear map \\(V\\times W\\rightarrow N\\) there exists a linear map \\(P\\rightarrow N\\) giving\na commutative diagram.\nThere are many questions.\nDoes such space always exists? Is it unique? Is the map \\(P\\rightarrow N\\) unique? Does it have a name? The best we can do in this post is to answer the last question. The name is \"Tensor product\". ","permalink":"https://praphulla-koushik.github.io/2024/05/16/multilinear-algebra-tensor-product/","summary":"\u003c!-- wp:paragraph --\u003e\n\u003cp\u003eLet us look at the first class of multilinear maps; the bilinear maps. \u003c/p\u003e\n\u003c!-- /wp:paragraph --\u003e\n\n\u003c!-- wp:paragraph --\u003e\n\u003cp\u003eWe want to study bilinear maps. The notion of \"study\" will have different meanings as we move forward (or backward) in the course.  \u003c/p\u003e\n\u003c!-- /wp:paragraph --\u003e\n\n\u003c!-- wp:paragraph --\u003e\n\u003cp\u003eLet \\(V,W,T\\) be vector spaces and \\(\\varphi:V\\times W\\rightarrow T\\) be a bilinear map. \u003c/p\u003e\n\u003c!-- /wp:paragraph --\u003e\n\n\u003c!-- wp:paragraph --\u003e\n\u003cp\u003eThe feeling that \"we are good at linear algebra\" suggests us to ask the question :\u003c/p\u003e","tags":["tensor-product"],"title":"Multilinear algebra : Tensor product","type":"wp-import"},{"categories":["Linear algebra"],"content":" In group theory, we mainly study maps that preserve the group structures; which goes by the name of group homomorphisms.\nIn topology, we mainly study maps that preserve the topologies; which goes by the name of continuous functions. In theory of vector spaces, we mainly study maps that preserve the vector space structures; which goes by the name of linear maps. Apart from that, there are many interesting maps that comes up when dealing with vector spaces which are not really linear maps. The very first example that comes to mind is the determinant map \\[\\det:M_n(\\mathbb{R})\\rightarrow \\mathbb{R}\\], in the same lines, \\(\\det:M_n(\\mathbb{C})\\rightarrow \\mathbb{C}\\).\nThere are more such interesting non linear maps that comes up in a linear algebra course. Let \\(V\\) be a vector space over \\(\\mathbb{R}\\). Then, we have the notion of evaluation map \\(V\\times V^*\\rightarrow \\mathbb{R}\\) given by \\((v,f)\\mapsto f(v)\\) for \\(v\\in V, f\\in V^*\\). This map is interesting, but it is not linear. It looks like, this non linearity has something to do with the domains being \"products\" of vector spaces, both in the case of \\(V\\times V^*\\) and the case of \\(M_n(\\mathbb{R})\\). This will become more interesting when we have products of more than two vector spaces. The names that appears, in case of product of two vector spaces is \"bilinear map\", in case of products of three vector spaces is \"trilinear map\"; more generally, a \"multilinear map\" in case of product of vector spaces. For vector spaces \\(V,W,T\\) (over same base field) we say that a set map \\(\\varphi:V\\times W\\rightarrow T\\) is a bilinear map, if, the following conditions are satisfied:\nfor each \\(v\\in V\\) the map \\(\\varphi(v,-):W\\rightarrow T\\) given by \\(w\\mapsto \\varphi(v,w)\\) for \\(w\\in W\\) is a linear map, for each \\(w\\in W\\) the map \\(\\varphi(-,w):V\\rightarrow T\\) given by \\(v\\mapsto \\varphi(v,w)\\) for \\(v\\in V\\) is a linear map. For vector spaces \\(V_1,V_2,\\cdots, V_n, T\\) (over same base filed) we say that a set map \\(\\varphi:V_1\\times V_2\\times\\cdots\\times V_n\\rightarrow T\\) is a multilinear map, if the following conditions are satisfied:\nfor each \\(1\\leq i\\leq n\\) and \\[(v_1,\\cdots,v_{i-1},v_{i+1},\\cdots, v_n)\\in V_1\\times \\cdots \\times V_{i-1}\\times V_{i+1}\\times \\cdots\\times V_n\\], the map \\(\\varphi(v_1,\\cdots,v_{i-1},-,v_{i+1},\\cdots,v_n):V_n\\rightarrow T\\) given by \\[v\\mapsto \\varphi(v_1,\\cdots,v_{i-1},v,v_{i+1},\\cdots, v_n)\\]\nfor \\(v\\in V_i\\) is a linear map.\nIf we define linear algebra as study of vetctor spaces and linear maps, we can say multilinear algebra is study of vector spaces and multilinear maps (at the very least). That is all for now. ","permalink":"https://praphulla-koushik.github.io/2024/05/12/multilinear-algebra-an-introduction/","summary":"\u003c!-- wp:paragraph --\u003e\n\u003cp\u003eIn group theory, we mainly study maps that preserve the group structures; which goes by the name of group homomorphisms.\u003c/p\u003e\n\u003c!-- /wp:paragraph --\u003e\n\n\u003c!-- wp:paragraph --\u003e\n\u003cp\u003eIn topology, we mainly study maps that preserve the topologies; which goes by the name of continuous functions. \u003c/p\u003e\n\u003c!-- /wp:paragraph --\u003e\n\n\u003c!-- wp:paragraph --\u003e\n\u003cp\u003eIn theory of vector spaces, we mainly study maps that preserve the vector space structures; which goes by the name of linear maps. Apart from that, there are many interesting maps that comes up when dealing with vector spaces which are not really linear maps. The very first example that comes to mind is the determinant map \u003c/p\u003e","tags":[],"title":"Multilinear algebra : an introduction","type":"wp-import"},{"categories":["Lie groups"],"content":" In the theory of Lie algebras, we have the notion of representation of a Lie algebra \\(\\mathfrak{g}\\), which consists of a vector space \\(V\\), and a morphism of Lie algebras \\(\\mathfrak{g}\\rightarrow \\mathfrak{gl}(V)\\). We are so used to thinking of \\(\\mathfrak{gl}(V)\\) as a Lie algebra, that, we might not remember that the underlying set \\(End(V)\\) has a structure of an associative algebra, and that we made the underlying set into a Lie algebra by considering the binary operation \\([f,g]=fg-gf\\) for \\(f,g\\in End(V)\\). This is where the notion of enveloping algebra comes into picture. Given a Lie algebra \\(\\mathfrak{g}\\), an enveloping algebra of \\(\\mathfrak{g}\\) consists of\nan associative algebra \\(A\\) a morphism of Lie algebras \\(\\mathfrak{g}\\rightarrow A\\), where we see \\(A\\) as a Lie algebra with the Lie bracket \\([a,b]=ab-ba\\). With that definition, any representation \\((V,\\rho:\\mathfrak{g}\\rightarrow \\mathfrak{gl}(V))\\) of a Lie algebra \\(\\mathfrak{g}\\) gives an enveloping algebra \\(End(V)\\) for \\(\\mathfrak{g}\\).\nOnce we have a notion of a structure, we would ask for a \"universal property\"; the best among the possibilities. We have seen such \"universal\" ideas before, for example universal covering space of a topological space, abelianization of a group, tensor product of two \\(R\\)-modules. In the case of enveloping algebras, one can ask a similar question. Given a Lie algebra \\(\\mathfrak{g}\\), can there be a \"best\" associative algebra \\(A\\), along with a morphism of Lie algebras \\(\\Phi:\\mathfrak{g}\\rightarrow A\\) such that, for any other associative algebra \\(B\\) and a morphism of Lie algebras \\(\\Psi:\\mathfrak{g}\\rightarrow B\\), there exists a unique morphism of associative algebras/Lie algebras \\(\\theta:A\\rightarrow B\\) such that \\(\\theta\\circ \\Phi=\\Psi\\), as in the diagram below,\nSuch enveloping algebra is called the universal enveloping algebra of \\(\\mathfrak{g}\\). In case of representations of Lie algebras, we have seen some notion of \"best\" but it is not unique; the notion of irreducible representation. Given a Lie algebra \\(\\mathfrak{g}\\), there may be many irreducible representations of same/different dimensions.\nQuestion : Does an irreducible representation of a Lie algebra \\(\\mathfrak{g}\\) help as a starting point to understand universal enveloping algebra of \\(\\mathfrak{g}\\)? Given a Lie algebra \\(\\mathfrak{g}\\), is it possible for universal enveloping algebra of \\(\\mathfrak{g}\\) to be of the form \\(End(V)\\) for some vector space \\(V\\)? Are there any other notions in representation theory of Lie algebras that are related to the notion of universal enveloping algebra? ","permalink":"https://praphulla-koushik.github.io/2024/04/29/universal-enveloping-algebra/","summary":"\u003c!-- wp:paragraph --\u003e\n\u003cp\u003eIn the theory of Lie algebras, we have the notion of representation of a Lie algebra \\(\\mathfrak{g}\\), which consists of a vector space \\(V\\), and a morphism of Lie algebras \\(\\mathfrak{g}\\rightarrow \\mathfrak{gl}(V)\\). \u003c/p\u003e\n\u003c!-- /wp:paragraph --\u003e\n\n\u003c!-- wp:paragraph --\u003e\n\u003cp\u003eWe are so used to thinking of \\(\\mathfrak{gl}(V)\\) as a Lie algebra, that, we might not remember that the underlying set \\(End(V)\\) has a structure of an associative algebra, and that we made the underlying set into a Lie algebra by considering the binary operation \\([f,g]=fg-gf\\) for \\(f,g\\in End(V)\\). This is where the notion of enveloping algebra comes into picture. \u003c/p\u003e","tags":["tensor-product"],"title":"Universal enveloping algebra","type":"wp-import"},{"categories":["algebraic-geometry"],"content":" Once we have a reasonably good notion of an object, we would look at a notion of morphisms.\nLet \\((L,A,\\rho,\\tau)\\) to \\((L',A',\\rho',\\tau')\\) be Lie-Rinehart algebras. Our experience suggests that the data of a morphism of Lie-Rinehart algebras from \\((L,A,\\rho,\\tau)\\) to \\((L',A',\\rho',\\tau')\\) should at least have two morphisms, one a morphism of Lie algebras \\(\\Phi:L\\rightarrow L'\\) and a morphism of associative algebras \\(\\Psi:A\\rightarrow A'\\) such that the following diagram commute,\nif someone tells us what the vertical arrows with ?? are defined as. Given a morphism \\(A\\rightarrow A'\\) there is no obvious way to get a morphism of derivations \\(Der(A)\\rightarrow Der(A')\\) and similarly, a morphism \\(L\\rightarrow L'\\) does not suggest a natural choice of morphisms of Endomorphisms \\(End(L)\\rightarrow End(L')\\). So, may be we are going in a wrong direction.\nIf you remember the way the notion of morphism of (affine) schemes is defined, it would give a clue that we are not thinking correctly. Note that, a morphism of ringed spaces from \\((X,\\mathcal{O}_X)\\) to \\((Y,\\mathcal{O}_Y)\\) consists of a morphism of topological spaces \\(f:X\\rightarrow Y\\) and a morphism of sheaves of rings \\(\\mathcal{O}_Y\\rightarrow f_*\\mathcal{O}_X\\). In the same sense, one can think of bring the \"structures\" on same set up. This is where the notion of \"pullback\" comes into picture. This idea is taken from Madeline Jotz Lean's work On ideals in Lie-Rinehart algebras.\nLet \\((L,A,\\rho,\\tau)\\) to \\((L',A',\\rho',\\tau')\\) be Lie-Rinehart algebras. A morphism of Lie-Rinehart algebras from \\((L,A)\\) to \\((L',A')\\) comes with a morphism of algebras \\(\\varphi: A'\\rightarrow A\\), along with certain conditions. Note that this reminds the notion of morphism of schemes; the ring map is in the opposite direction of what we expect. If you are not familiar with the notion of morphism of schemes, one can justify this from the notion of morphism of Lie algebroids. Given Lie algebroids \\((L,M)\\) and \\((L',M')\\) a morphism of Lie algebroids from \\((L,M)\\) to \\((L',M')\\) is a vector bundle morphism \\((\\Phi,f):(L,M)\\rightarrow (L',M')\\) satisfying certain (not so nice) conditions. This data comes with a map \\(f:M\\rightarrow M'\\). This in turn gives a map \\(C^\\infty(M')\\rightarrow C^\\infty (M)\\) given by \\(\\theta:M'\\rightarrow \\mathbb{R}\\) being mapped to \\(f\\circ \\theta:M\\rightarrow \\mathbb{R}\\). So, if we think of Lie-Rinehart algebras associated to the Lie algebroids, we should be having the map \\(A'=C^\\infty(M') \\rightarrow A=C^\\infty(M)\\).\nNote that, this morphism \\(\\varphi:A'\\rightarrow A\\) would make \\(A\\) into an \\(A'\\)-module. One can use this to construct an \\(A\\)-module from an \\(A'\\)-module, in particular, \\(A\\otimes_{A'}L'\\) would be an \\(A\\)-module. Then, the author consider a subset of \\(Der(A)\\times (A\\otimes_{A'}L')\\), denote it as \\(Der(A)\\times_{\\varphi}L'\\), call it the \"pullback\".\nLet \\((L,A,\\rho,\\tau)\\) to \\((L',A',\\rho',\\tau')\\) be Lie-Rinehart algebras. A morphism of Lie-Rinehart algebras from \\((L,A)\\) to \\((L',A')\\) comes with a morphism of algebras \\(\\varphi: A'\\rightarrow A\\), along with a morphism of Lie algebras \\(L\\rightarrow Der(A)\\times_{\\varphi}L'\\) satisfying certain conditions. It is more technical and needs sometime to digest.\nThe construction is natural, but, I am of the opinion that this should be called as push forward and not pull back as we are pushing a structure from \\(A'\\) to get a structure on \\(A\\) through the map \\(\\varphi:A'\\rightarrow A\\). It is not justifiable to expect the definition of morphism of Lie-Rinehart algebras to be straightforward.\nOnce you recall the notion of morphism of Liealgebroids, you would get convinced that, it is not justifiable to expect the definition of morphism of Lie-Rinehart algebras to be any more straightforward than what is mentioned by Madeline Jotz Lean in On ideals in Lie-Rinehart algebras.\nAlong with the idea mentioned above, there is another notion of morphism of Lie-Rinehart algebras, given by Camille Laurent-Gengoux and Ruben Louis in Lie-Rinehart algebras \\(\\simeq\\) acyclic Lie \\(\\infty\\)-algebroids. Consider the same diagram that we mentioned above,\nThere may be no obvious options for the maps \\(End(L)\\rightarrow End(L')\\) and \\(Der(A)\\rightarrow Der(A')\\), but, there is a way to bypass that path. Let \\(a\\in A\\) and \\(l\\in L\\). We have \\(\\Phi(l)\\in L'\\), \\(\\rho'(\\Phi(l))\\in Der(A')\\), \\(\\rho'(\\Phi(l))(\\Psi(a))\\in A'\\). We have \\(\\rho(l)\\in Der(A)\\), \\(\\rho(l)(a)\\in A\\), \\(\\Psi(\\rho(l)(a))\\in A'\\). We ask that \\(\\rho'(\\Phi(l))(\\Psi(a))=\\Psi(\\rho(l)(a))\\).\nWe did not yet mention the condition relating to the maps \\(\\tau, \\tau'\\). Let us do that now. We have \\(\\tau(a)\\in End(L)\\), \\(\\tau(a)(l)\\in L\\), \\(\\Phi(\\tau(a)(l))\\in L'\\). We have\n\\(\\Psi(a)\\in A'\\), \\(\\tau'(\\Psi(a))\\in End(L')\\), \\(\\tau'(\\Psi(a))(\\Phi(l))\\in L'\\). We ask that, \\(\\Phi(\\tau(a)(l))=\\tau'(\\Psi(a))(\\Phi(l))\\). This idea is slightly easier to follow than that of Madeline Jotz Lean's idea of morphism of Lie-Rinehart algebras.\nQuestion : Is the notion of morphism of Lie-Rinehart algebra mentioned in Lie-Rinehart algebras \\(\\simeq\\) acyclic Lie \\(\\infty\\)-algebroids is equivalent to the notion mentioned in On ideals in Lie-Rinehart algebras? For a better understanding of the above two notions of \"morphisms of Lie-Rinehart algebras\" and other interesting ideas about Lie-Rinehart algebras, please see the talk Paths in Lie-Rinehart algebras by Joel Villatoro. The words used are \"morphism\" and \"comorphism.\" ","permalink":"https://praphulla-koushik.github.io/2024/04/26/lie-rinehart-algebras-morphism-of-lie-rinehart-algebras/","summary":"\u003c!-- wp:paragraph --\u003e\n\u003cp\u003eOnce we have a reasonably good notion of an object, we would look at a notion of morphisms.\u003c/p\u003e\n\u003c!-- /wp:paragraph --\u003e\n\n\u003c!-- wp:paragraph --\u003e\n\u003cp\u003eLet \\((L,A,\\rho,\\tau)\\) to \\((L',A',\\rho',\\tau')\\) be Lie-Rinehart algebras. \u003c/p\u003e\n\u003c!-- /wp:paragraph --\u003e\n\n\u003c!-- wp:paragraph --\u003e\n\u003cp\u003eOur experience suggests that the data of a morphism of Lie-Rinehart algebras from \\((L,A,\\rho,\\tau)\\) to \\((L',A',\\rho',\\tau')\\) should at least have two morphisms, one a morphism of Lie algebras \\(\\Phi:L\\rightarrow L'\\) and a morphism of associative algebras \\(\\Psi:A\\rightarrow A'\\) such that the following diagram commute,\u003c/p\u003e","tags":["schemes"],"title":"Lie-Rinehart algebras : Morphism of Lie-Rinehart algebras","type":"wp-import"},{"categories":["Differential geometry"],"content":" Any notion of an \"algebra\" comes with two binary operations:\n\\(A\\times A\\rightarrow A\\), called the addition map, \\(A\\times A\\rightarrow A\\), called the multiplication map. Two properties that are assumed for addition map are that of commutativity and associativity. By the very definition, we would have \\(a+b=b+a\\) and \\(a+(b+c)=(a+b)+c\\) for all \\(a,b,c\\in A\\). These two properties are however not assumed for the multiplicative map. That is where the notion of noncommutative algebras and non associative algebras comes into picture. Note that, \"non commutative\" means not necessarily commutative and \"nonassociative\" means not necessarily associative. One interesting case of nonassociative algebras is that of a Lie algebra, where, the failure of the associativity is controlled by the Jacobi identity. We have \\([x,[y,z]]+[y,[z,x]]+[z,[x,y]]=0\\) for all \\(x,y,z\\in A\\).\nNow comes the notion of Lie-Rinehart algebra. There may be a very intuitive and algebraic way to think about Lie-Rinehart algebra, but, if you are comfortable with the notion of Lie algebroid over a manifold, it is better to approach Lie-Rinehart algebra from Lie algebroid point of view.\nRecall that, given a manifold \\(M\\), a Lie algebroid over \\(M\\) consists of, a vector bundle \\(A\\rightarrow M\\), a morphism of vector bundles \\(\\rho:A\\rightarrow TM\\), a Lie algebra structure on \\(\\Gamma(M,A)\\) satisfying the obvious conditions. Let us seperate out the algebraic ideas here. We have a Lie algebra \\(L=\\Gamma(M,A)\\) an algebra \\(\\mathcal{O}=C^\\infty(M)\\) \\(\\mathcal{O}\\)-module structure on \\(L\\) given by \\((f,s)\\mapsto fs\\) where \\(fs:M\\rightarrow A\\) is given by \\(m\\mapsto f(m)s(m)\\) for \\(m\\in M, f\\in \\mathcal{O}\\), and \\(s\\in L\\) a map \\(L=\\Gamma(M,A)\\rightarrow Der(C^\\infty(M))=\\mathfrak{X}(M)\\) induced from the anchor map \\(A\\rightarrow TM\\) by taking sections \\(\\Gamma(M,A)\\rightarrow \\Gamma(M,TM)=\\mathfrak{X}(M)=Der(C^\\infty(M))\\), satisfying certain conditions.\nWith out mentioning the reference to manifold, writing down the above algebraic data, along with \"certain conditions\" gives the notion of Lie-Rinehart algebra.\nLie-Rinehart algebra A Lie-Rinehart algebra consists of,\na Lie algebra \\(L\\) an associative (some people ask it to be commutative) algebra \\(A\\), a map \\(\\tau : A\\times L\\rightarrow L\\) giving an \\(A\\)-module structure on \\(L\\), a map \\(\\rho : L\\times A\\rightarrow A\\) giving an \\(L\\)-module structure on \\(A\\), such that the following conditions are satisfied:\nthe map \\(\\rho : L\\times A\\rightarrow A\\) gives a morphism of Lie algebras \\(L\\rightarrow Der(A)\\) the map \\(\\tau : A\\times L\\rightarrow L\\) gives a morphism of associative algebras \\(A\\rightarrow End(L)\\), the maps \\(\\rho,\\tau\\) are compatible, in the sense that, \\([u,\\tau(a,v)]=\\tau(a,[u,v])+\\tau(\\rho(u,a),v)\\) for all \\(u,v\\in L\\) and \\(a\\in A\\).\nGiven a smooth manifold \\(M\\), and a Lie algebroid \\(A\\rightarrow M\\), we get a Lie-Rinehart algebra \\((C^\\infty(M), \\Gamma(M,A)\\).\nQuestion : Given a manifold \\(M\\), are there any Lie-Rinehart algebras whose associative algebra is \\(C^\\infty(M)\\) and the Lie algebra is not \\(\\Gamma(M,A)\\) for some Lie algebroid \\(A\\rightarrow M\\)?\n","permalink":"https://praphulla-koushik.github.io/2024/04/24/1483/","summary":"\u003c!-- wp:paragraph --\u003e\n\u003cp\u003eAny notion of an \"algebra\" comes with two binary operations:\u003c/p\u003e\n\u003c!-- /wp:paragraph --\u003e\n\n\u003c!-- wp:list --\u003e\n\u003cul\u003e\u003c!-- wp:list-item --\u003e\n\u003cli\u003e\\(A\\times A\\rightarrow A\\), called the addition map,\u003c/li\u003e\n\u003c!-- /wp:list-item --\u003e\n\n\u003c!-- wp:list-item --\u003e\n\u003cli\u003e\\(A\\times A\\rightarrow A\\), called the multiplication map.\u003c/li\u003e\n\u003c!-- /wp:list-item --\u003e\u003c/ul\u003e\n\u003c!-- /wp:list --\u003e\n\n\u003c!-- wp:paragraph --\u003e\n\u003cp\u003eTwo properties that are assumed for addition map are that of commutativity and associativity. \u003c/p\u003e\n\u003c!-- /wp:paragraph --\u003e\n\n\u003c!-- wp:paragraph --\u003e\n\u003cp\u003eBy the very definition, we would have \\(a+b=b+a\\) and \\(a+(b+c)=(a+b)+c\\) for all \\(a,b,c\\in A\\). \u003c/p\u003e","tags":[],"title":"Lie-Rinehart algebras : Introduction and definition of Lie-Rinehart algebra","type":"wp-import"},{"categories":["Algebraic geometry"],"content":"These are “notes” I have written for myself when reading the book Model Categories by Mark Hovey. This book has some typos, there is an errata by its Author. There might be some more typos. I am assuming some notation and results about topological spaces (fibrations, cofibrations, etc) and homological algebra (chain complexes, etc). Other references for Model categories are : An Introduction to Homotopical categories by Julie Bergner. \u0026nbsp;","permalink":"https://praphulla-koushik.github.io/2020/05/23/model-categories-part-1-motivation/","summary":"These are “notes” I have written for myself when reading the book \u003ca href=\"http://ericmalm.net/ac/projects/symmetric-spectra/hovey--model-cats.pdf\"\u003eModel Categories by Mark Hovey.\u003c/a\u003e  This book has some typos, there \u003ca href=\"https://hopf.math.purdue.edu/Hovey/model-err.pdf\"\u003eis\u003c/a\u003e an errata by its Author. There might be some more typos. I am assuming some notation and results about topological spaces (fibrations, cofibrations, etc) and homological algebra (chain complexes, etc).\n\nOther references for Model categories are :\n\u003col\u003e\n\t\u003cli\u003e\u003ca href=\"https://www.youtube.com/playlist?list=PLN3dwsLfyzcWSIoGOS8Xuh1Ms3Ja10ILd\"\u003eAn Introduction to Homotopical categories by Julie Bergner\u003c/a\u003e.\u003c/li\u003e\n\u003c/ol\u003e\n\u0026nbsp;","tags":[],"title":"Model categories : Part 1 (Motivation)","type":"wp-import"},{"categories":["category-theory","Algebraic geometry"],"content":"Before we move to the notion of \"a model structure on a category\", we need to recall (or introduce) some definitions. Definition : Let \\(\\mathcal{C}\\) be a category. An object \\(X\\) of \\(\\mathcal{C}\\) is said to be a retract of an object \\(\\mathcal{C}\\) if there exists arrows \\(X\\xrightarrow{f} Y\\xrightarrow{g} X\\) such that, the composition \\(g\\circ f:X\\rightarrow X\\) is equal to the identity arrow \\(1_X:X\\rightarrow X\\). Definition : Let \\(\\mathcal{C}\\) be a category. We define the morphism category of \\(\\mathcal{C}\\), denoted by \\(\\text{Map}(\\mathcal{C})\\) whose objects are the arrows of \\(\\mathcal{C}\\), morphisms are commutative diagrams in \\(\\mathcal{C}\\). Definition : Let \\(\\mathcal{C}\\) be a category. A morphism \\(f\\) in \\(\\mathcal{C}\\) is said to be a retract of a morphism \\(g\\) in \\(\\mathcal{C}\\), if, \\(f\\) is a retract of \\(g\\), when both \\(f\\) and \\(g\\) are seen as objects of \\(\\text{Map}(\\mathcal{C})\\). Definition : Let \\(\\mathcal{C}\\) be a category. Let \\(i:A\\rightarrow B\\) and \\(p:X\\rightarrow Y\\) be morphisms in \\(\\mathcal{C}\\). We say that \\(i\\) has the left lifting property with respect to \\(p\\) or \\(p\\) has the right lifting property with respect to \\(i\\) if, for every commutative diagram there exists an arrow \\(h:B\\rightarrow X\\) such that \\(h\\circ i=f\\) and \\(p\\circ h=g\\). Definition : Let \\(\\mathcal{C}\\) be a category. A model structure on \\(\\mathcal{C}\\) consists of the following data : a subcategory of \\(\\mathcal{C}\\) called “weak equivalences”, a subcategory of \\(\\mathcal{C}\\) called “fibrations”, a subcategory of \\(\\mathcal{C}\\) called “cofibrations”, satisfying certain conditions: If \\(f,g\\) are morphisms of \\(\\mathcal{C}\\) such that \\(gf\\) is defined and two of \\(f,g,gf\\) are \"weak equivalences\" then so is the third. A retract of a \"weak equivalece\" is a \"weak equivalence\". A retract of a \"fibration\" is a \"fibration\". A retract of a \"cofibration\" is a \"cofibration\". factorizing property (part \\(1\\)) of arrows of \\(\\mathcal{C}\\) : for each \\(f\\) in \\(\\text{Mor}(\\mathcal{C})\\), there exists a cofibration \\(i\\) and a traivial fibration \\(q\\) such that \\(f=qi\\). factorizing property (part \\(2\\)) of arrows of \\(\\mathcal{C}\\) : for each \\(f\\) in \\(\\text{Mor}(\\mathcal{C})\\), there exists a fibration \\(p\\) and a trivial cofibration \\(j\\) such that \\(f=pj\\). Any commutative diagram of the type has lifting property if either \\(i\\) or \\(p\\) is a \"weak equivalence\". Definition : A model category is defined to be a category that has all small limits, all small colimits, a model structure in \\(\\mathcal{C}\\). Construction of new model categories from old model categories: Let \\(\\mathcal{C}\\) and \\(\\mathcal{D}\\) be model categories. Defining the collection of fibrations (cofibrations, weak equivalences) as pairs \\((f,g)\\) where both \\(f\\) and \\(g\\) are fibrations (cofibrations, weak equivalences) defined a model structure on the product category \\(\\mathcal{C}\\times \\mathcal{D}\\). This model category is called the product model category produced from model categories \\(\\mathcal{C}\\) and \\(\\mathcal{D}\\). ","permalink":"https://praphulla-koushik.github.io/2020/05/21/model-categories-part-1/","summary":"Before we move to the notion of \"a model structure on a category\", we need to recall (or introduce) some definitions.\n\nDefinition : Let \\(\\mathcal{C}\\) be a category. An \u003cem\u003eobject \\(X\\) of \\(\\mathcal{C}\\) is said to be a retract of an object \\(\\mathcal{C}\\)\u003c/em\u003e if there exists arrows \\(X\\xrightarrow{f} Y\\xrightarrow{g} X\\) such that, the composition \\(g\\circ f:X\\rightarrow X\\) is equal to the identity arrow \\(1_X:X\\rightarrow X\\).\n\nDefinition : Let \\(\\mathcal{C}\\) be a category. We define \u003cem\u003ethe morphism category of \\(\\mathcal{C}\\)\u003c/em\u003e, denoted by \\(\\text{Map}(\\mathcal{C})\\) whose\n\u003cul\u003e\n\t\u003cli\u003eobjects are the arrows of \\(\\mathcal{C}\\),\u003c/li\u003e\n\t\u003cli\u003emorphisms are commutative diagrams in \\(\\mathcal{C}\\).\u003c/li\u003e\n\u003c/ul\u003e\nDefinition : Let \\(\\mathcal{C}\\) be a category. A morphism \\(f\\) in \\(\\mathcal{C}\\) is said to be \u003cem\u003ea retract of \u003c/em\u003e a morphism \\(g\\) in \\(\\mathcal{C}\\), if, \\(f\\) is a retract of \\(g\\), when both \\(f\\) and \\(g\\) are seen as objects of \\(\\text{Map}(\\mathcal{C})\\).\n\nDefinition : Let \\(\\mathcal{C}\\) be a category. Let \\(i:A\\rightarrow B\\) and \\(p:X\\rightarrow Y\\) be morphisms in \\(\\mathcal{C}\\). We say that \u003cem\u003e\\(i\\) has the left lifting property with respect to \\(p\\) \u003c/em\u003eor \u003cem\u003e\\(p\\) has the right lifting property with respect to \\(i\\)\u003c/em\u003e if, for every commutative diagram \u003cimg class=\" size-full wp-image-1449 aligncenter\" src=\"/wp-media/2020/05/ae22e72990-screenshot-from-2020-05-23-21-11-33.png\" alt=\"Screenshot from 2020-05-23 21-11-33\" width=\"242\" height=\"213\" /\u003ethere exists an arrow \\(h:B\\rightarrow X\\) such that \\(h\\circ i=f\\) and \\(p\\circ h=g\\).\n\nDefinition : Let \\(\\mathcal{C}\\) be a category. A \u003cem\u003emodel structure\u003c/em\u003e on \\(\\mathcal{C}\\) consists of the following data :\n\u003col\u003e\n\t\u003cli\u003ea subcategory of \\(\\mathcal{C}\\) called “weak equivalences”,\u003c/li\u003e\n\t\u003cli\u003ea subcategory of \\(\\mathcal{C}\\) called “fibrations”,\u003c/li\u003e\n\t\u003cli\u003ea subcategory of \\(\\mathcal{C}\\) called “cofibrations”,\u003c/li\u003e\n\u003c/ol\u003e\nsatisfying certain conditions:\n\u003col\u003e\n\t\u003cli\u003eIf \\(f,g\\) are morphisms of \\(\\mathcal{C}\\) such that \\(gf\\) is defined and two of \\(f,g,gf\\) are \"weak equivalences\" then so is the third.\u003c/li\u003e\n\t\u003cli\u003eA retract of a \"weak equivalece\" is a \"weak equivalence\".\u003c/li\u003e\n\t\u003cli\u003eA retract of a \"fibration\" is a \"fibration\".\u003c/li\u003e\n\t\u003cli\u003eA retract of a \"cofibration\" is a \"cofibration\".\u003c/li\u003e\n\t\u003cli\u003efactorizing property (part \\(1\\)) of arrows of \\(\\mathcal{C}\\) : for each \\(f\\) in \\(\\text{Mor}(\\mathcal{C})\\), there exists a cofibration \\(i\\) and a traivial fibration \\(q\\) such that \\(f=qi\\).\u003c/li\u003e\n\t\u003cli\u003efactorizing property (part \\(2\\)) of arrows of \\(\\mathcal{C}\\) : for each \\(f\\) in \\(\\text{Mor}(\\mathcal{C})\\), there exists a fibration \\(p\\) and a trivial cofibration \\(j\\) such that \\(f=pj\\).\u003c/li\u003e\n\t\u003cli\u003eAny commutative diagram of the type \u003cimg class=\"alignnone size-full wp-image-1455\" src=\"/wp-media/2020/05/789d14d26b-screenshot-from-2020-05-24-09-15-26.png\" alt=\"Screenshot from 2020-05-24 09-15-26\" width=\"616\" height=\"217\" /\u003e has lifting property if either \\(i\\) or \\(p\\) is a \"weak equivalence\".\u003c/li\u003e\n\u003c/ol\u003e\nDefinition : A \u003cem\u003emodel category\u003c/em\u003e is defined to be a category that has\n\u003col\u003e\n\t\u003cli\u003eall small limits,\u003c/li\u003e\n\t\u003cli\u003eall small colimits,\u003c/li\u003e\n\t\u003cli\u003ea model structure in \\(\\mathcal{C}\\).\u003c/li\u003e\n\u003c/ol\u003e\n\u003cb\u003eConstruction of new model categories from old model categories:\u003c/b\u003e\n\u003col\u003e\n\t\u003cli\u003eLet \\(\\mathcal{C}\\) and \\(\\mathcal{D}\\) be model categories. Defining the collection of fibrations (cofibrations, weak equivalences) as pairs \\((f,g)\\) where both \\(f\\) and \\(g\\) are fibrations (cofibrations, weak equivalences) defined a model structure on the product category \\(\\mathcal{C}\\times \\mathcal{D}\\). This model category is called the product model category produced from model categories \\(\\mathcal{C}\\) and \\(\\mathcal{D}\\).\u003c/li\u003e\n\t\u003cli\u003e\u003c/li\u003e\n\u003c/ol\u003e","tags":["model-categories"],"title":"Model categories : Part 2 (Definitions)","type":"wp-import"},{"categories":["differential-geometry","stacks"],"content":"Here I will add notes of the seminar that I am planning to conduct in School of Mathematics, IISER Thiruvananthapuram, India. First lecture is expected to happen on 14 August 2019. --- Some terms which I want to convey the meaning of in this Seminar. Manifold. Differential forms on Manifolds; pullbacks and differential of a Differential form. Lie group. Lie algebra of Lie group. Cohomology of Manifolds / Cohomology of Lie groups. Principal/Vector bundle. Connection (on principal/vector bundle). Curvature (of Connection on principal/vector bundle). Holonomy group. Ambrose-Singer theorem. Characteristic classes (Euler/Chern classes). Lecture notes/Articles : The Topology of Fiber Bundles --- Lecture Notes --- Ralph L. Cohen WHAT IS A CONNECTION? --- TIMOTHY E. GOLDBERG Books: The Topology of Fibre Bundles by Steenrod Foundations of Differentiable Manifolds and Lie Groups by Frank Warner Foundations of Differential Geometry by Kobayashi and Nomizu Introduction to Smooth Manifolds by John Lee Geometry of Differential forms by Shigeyuki Morita Topics in Differential Geometry by Peter W. Michor Differential Geometry - Connections, Curvature, and Characteristic Classes by Loring Tu An Introduction to Manifolds by Loring Tu Differential Geometry, Lie Groups, and Symmetric Spaces by Sigurdur Helgason Differential Forms in Algebraic Topology by Bott and Tu A Geometric Approach to Differential Forms by David Bachman Modern Differential Geometry for Physicists 2nd Edition by Chris J Isham Differential Forms and Connections by R. W. R. Darling Differential Forms - A Heuristic Introduction by M. Schreiber From Calculus to Cohomology by Madsen and Tornehave Manifolds, Sheaves, and Cohomology by Torsten Wedhorn Principal Bundles : The Classical Case by Stephen Bruce Sontz Introduction to the Theory of Lie Groups by Roger Godement Differential Geometry: Bundles, Connections, Metrics and Curvature by Clifford Henry Taubes YouTube videos : Fredric Schuller's YouTube channel MathOverflow/MathStackExchange questions/user pages: John M. Lee 's MathStackExchange page \u0026nbsp;","permalink":"https://praphulla-koushik.github.io/2019/08/11/seminar-on-geometry-topology-of-principal-fiber-bundles/","summary":"Here I will add notes of the seminar that I am planning to conduct in School of Mathematics, IISER Thiruvananthapuram, India.\n\u003col\u003e\n\t\u003cli\u003eFirst lecture is expected to happen on 14 August 2019.\u003c/li\u003e\n\t\u003cli\u003e---\u003c/li\u003e\n\u003c/ol\u003e\nSome terms which I want to convey the meaning of in this Seminar.\n\u003col\u003e\n\t\u003cli\u003eManifold.\u003c/li\u003e\n\t\u003cli\u003eDifferential forms on Manifolds; pullbacks and differential of a Differential form.\u003c/li\u003e\n\t\u003cli\u003eLie group.\u003c/li\u003e\n\t\u003cli\u003eLie algebra of Lie group.\u003c/li\u003e\n\t\u003cli\u003eCohomology of Manifolds / Cohomology of Lie groups.\u003c/li\u003e\n\t\u003cli\u003ePrincipal/Vector bundle.\u003c/li\u003e\n\t\u003cli\u003eConnection (on principal/vector bundle).\u003c/li\u003e\n\t\u003cli\u003eCurvature (of Connection on principal/vector bundle).\u003c/li\u003e\n\t\u003cli\u003eHolonomy group.\u003c/li\u003e\n\t\u003cli\u003eAmbrose-Singer theorem.\u003c/li\u003e\n\t\u003cli\u003eCharacteristic classes (Euler/Chern classes).\u003c/li\u003e\n\u003c/ol\u003e\nLecture notes/Articles :\n\u003col\u003e\n\t\u003cli\u003e \u003ca href=\"http://math.stanford.edu/~ralph/fiber.pdf\"\u003eThe Topology of Fiber Bundles --- Lecture Notes --- Ralph L. Cohen\u003c/a\u003e\u003c/li\u003e\n\t\u003cli\u003e\u003ca style=\"background-color:#ffffff;box-shadow:0 0 0 1px rgba(var(--color-primary-rgb),0.2);\" href=\"http://pi.math.cornell.edu/~goldberg/Notes/AboutConnections.pdf\"\u003eWHAT IS A CONNECTION? --- TIMOTHY E. GOLDBERG\u003c/a\u003e\u003c/li\u003e\n\u003c/ol\u003e\nBooks:\n\u003col\u003e\n\t\u003cli\u003e\u003ca href=\"https://www.jstor.org/stable/j.ctt1bpm9t5\"\u003eThe Topology of Fibre Bundles by Steenrod\u003c/a\u003e\u003c/li\u003e\n\t\u003cli\u003e\u003ca href=\"https://www.springer.com/gp/book/9780387908946\"\u003eFoundations of Differentiable Manifolds and Lie Groups by Frank Warner\u003c/a\u003e\u003c/li\u003e\n\t\u003cli\u003e\u003ca href=\"https://www.amazon.com/dp/0470555580/ref=pd_lpo_sbs_dp_ss_2/133-7323477-4889049\"\u003eFoundations of Differential Geometry by Kobayashi and Nomizu\u003c/a\u003e\u003c/li\u003e\n\t\u003cli\u003e\u003ca href=\"https://www.springer.com/gp/book/9781441999818\"\u003eIntroduction to Smooth Manifolds by John Lee\u003c/a\u003e\u003c/li\u003e\n\t\u003cli\u003e\u003ca href=\"https://bookstore.ams.org/mmono-201\"\u003eGeometry of Differential forms by Shigeyuki Morita\u003c/a\u003e\u003c/li\u003e\n\t\u003cli\u003e\u003ca href=\"https://bookstore.ams.org/gsm-93\"\u003eTopics in Differential Geometry by Peter W. Michor\u003c/a\u003e\u003c/li\u003e\n\t\u003cli\u003e\u003ca href=\"https://www.springer.com/gp/book/9783319550824\"\u003eDifferential Geometry - Connections, Curvature, and Characteristic Classes by Loring Tu\u003c/a\u003e\u003c/li\u003e\n\t\u003cli\u003e\u003ca href=\"https://www.springer.com/gp/book/9781441973993\"\u003eAn Introduction to Manifolds by Loring Tu\u003c/a\u003e\u003c/li\u003e\n\t\u003cli\u003e\u003ca href=\"https://bookstore.ams.org/gsm-34\"\u003eDifferential Geometry, Lie Groups, and Symmetric Spaces by Sigurdur Helgason\u003c/a\u003e\u003c/li\u003e\n\t\u003cli\u003e\u003ca href=\"https://www.springer.com/gp/book/9780387906133\"\u003eDifferential Forms in Algebraic Topology by Bott and Tu\u003c/a\u003e\u003c/li\u003e\n\t\u003cli\u003e\u003ca href=\"https://www.springer.com/gp/book/9780817683030#otherversion=9780817683047\"\u003eA Geometric Approach to Differential Forms by David Bachman\u003c/a\u003e\u003c/li\u003e\n\t\u003cli\u003e\u003ca href=\"https://www.worldscientific.com/worldscibooks/10.1142/3867\"\u003eModern Differential Geometry for Physicists 2nd Edition by Chris J Isham\u003c/a\u003e\u003c/li\u003e\n\t\u003cli\u003e\u003ca href=\"https://www.cambridge.org/core/books/differential-forms-and-connections/767FC792F030D351AF5E65D0434248F5\"\u003eDifferential Forms and Connections by R. W. R. Darling\u003c/a\u003e\u003c/li\u003e\n\t\u003cli\u003e\u003ca href=\"https://www.springer.com/gp/book/9780387902876\"\u003eDifferential Forms - A Heuristic Introduction by M. Schreiber\u003c/a\u003e\u003c/li\u003e\n\t\u003cli\u003e\u003ca href=\"https://www.cambridge.org/us/academic/subjects/mathematics/geometry-and-topology/calculus-cohomology-de-rham-cohomology-and-characteristic-classes?format=PB\u0026amp;isbn=9780521589567\"\u003eFrom Calculus to Cohomology by Madsen and Tornehave\u003c/a\u003e\u003c/li\u003e\n\t\u003cli\u003e\u003ca href=\"https://www.springer.com/gp/book/9783658106324\"\u003e Manifolds, Sheaves, and Cohomology by Torsten Wedhorn\u003c/a\u003e\u003c/li\u003e\n\t\u003cli\u003e\u003ca href=\"https://www.springer.com/gp/book/9783319147642\"\u003e Principal Bundles : The Classical Case by Stephen Bruce Sontz\u003c/a\u003e\u003c/li\u003e\n\t\u003cli\u003e\u003ca href=\"https://www.springer.com/gp/book/9783319543734\"\u003e Introduction to the Theory of Lie Groups by Roger Godement\u003c/a\u003e\u003c/li\u003e\n\t\u003cli\u003e\u003ca href=\"https://www.oxfordscholarship.com/view/10.1093/acprof:oso/9780199605880.001.0001/acprof-9780199605880\"\u003eDifferential Geometry: Bundles, Connections, Metrics and Curvature by Clifford Henry Taubes\u003c/a\u003e\u003c/li\u003e\n\u003c/ol\u003e\nYouTube videos :\n\u003col\u003e\n\t\u003cli\u003e\u003ca style=\"background-color:#ffffff;box-shadow:0 0 0 1px rgba(var(--color-primary-rgb),0.2);\" href=\"https://www.youtube.com/playlist?list=PLPH7f_7ZlzxTi6kS4vCmv4ZKm9u8g5yic\"\u003eFredric Schuller's YouTube channel\u003c/a\u003e\u003c/li\u003e\n\u003c/ol\u003e\nMathOverflow/MathStackExchange questions/user pages:\n\u003col\u003e\n\t\u003cli\u003e\u003ca href=\"https://math.stackexchange.com/users/1421/jack-lee\"\u003eJohn M. Lee 's MathStackExchange page\u003c/a\u003e\u003c/li\u003e\n\u003c/ol\u003e\n\u0026nbsp;","tags":["connections","kobayashi-nomizu","principal-bundles","stacks"],"title":"Seminar on Geometry/Topology of Principal/fiber bundles","type":"wp-import"},{"categories":["stacks"],"content":"Here I add notes about notation I use in this blog. I might use the notion of fibered category and category fibered in groupoids as if there is no difference. I am mostly interested in fibered category \\(\\mathcal{F}\\rightarrow \\mathcal{C}\\) where the fibre \\(\\mathcal{F}(U)\\) is a groupoid for every object \\(U\\) of \\(\\mathcal{C}\\). So, most of the times when I say fibered category, it is most likely that I mean fibred categroy whose fibres are groupoids i.e., category fibered in groupoids. Please let me know if there is some real confusion. ","permalink":"https://praphulla-koushik.github.io/2019/02/02/notation/","summary":"Here I add notes about notation  I use in this blog.\n\u003cul\u003e\n\t\u003cli\u003eI might use the notion of fibered category and category fibered in groupoids as if there is no difference. I am mostly interested in fibered category \\(\\mathcal{F}\\rightarrow \\mathcal{C}\\) where the fibre \\(\\mathcal{F}(U)\\) is a \u003cstrong\u003egroupoid \u003c/strong\u003efor every object \\(U\\) of \\(\\mathcal{C}\\). So, most of the times when I say fibered category, it is most likely that I mean fibred categroy whose fibres are groupoids i.e., category fibered in groupoids. Please let me know if there is some real confusion.\u003c/li\u003e\n\u003c/ul\u003e","tags":[],"title":"Notation","type":"wp-import"},{"categories":["algebraic-geometry","category-theory","stacks"],"content":"I understood most of this from Introduction to the language of stacks and gerbes (section 2) by Ieke Moerdijk and from stacks project Stackification of fibred categories. It is necessary to know what is the sheafification of a presheaf to understand what is the stackification. I studied sheafification from Hartshorne's Algebraic geometry book. You can choose what you are comfortable with. I will mention the result first as in Lemma \\(8.8.1\\). Lemma : Let \\(\\mathcal{C}\\) be a site. Let \\(p:\\mathcal{S}\\rightarrow \\mathcal{C}\\) be a fibred category over \\(\\mathcal{C}\\). There exists a stack \\(p':\\mathcal{S}'\\rightarrow \\mathcal{C}\\) and a morphisms \\(G:\\mathcal{S}\\rightarrow \\mathcal{S}'\\) of fibred categories over \\(\\mathcal{C}\\) such that for every \\(U\\in \\text{Ob}(\\mathcal{C})\\) and \\(x,y\\in \\mathcal{S}(U)\\), the map \\(\\text{Mor}(x,y)\\rightarrow \\text{Mor}(G(x),G(y))\\) induced by \\(G\\) identifies the right hand side with the sheafification of the left hand side. For \\(U\\in \\mathcal{C}_0\\) and \\(x'\\in \\mathcal{S}'(U)\\) there exists a covering \\(\\{U_i\\rightarrow U\\}\\) such that each \\(x'|_{U_i}\\) is in the essential image of the functor \\(G:\\mathcal{S}(U)\\rightarrow \\mathcal{S}'(U)\\). We recall what is \\(\\text{Mor}(a,b)\\). This is a presheaf on \\(U\\) defined as follows. Given an inclusion \\(i : V\\hookrightarrow U\\) we have \\(i^*(a),i^*(b)\\in \\mathcal{S}(V)\\).","permalink":"https://praphulla-koushik.github.io/2019/02/02/stackification-of-fibred-categories/","summary":"I understood most of this from \u003ca href=\"https://arxiv.org/pdf/math/0212266.pdf\"\u003e Introduction to the language of stacks and gerbes\u003c/a\u003e (section 2)  by Ieke Moerdijk and from stacks project \u003ca href=\"https://stacks.math.columbia.edu/tag/02ZM\"\u003eStackification of fibred categories\u003c/a\u003e.\n\nIt is necessary to know what is the sheafification of a presheaf to understand what is the stackification. I studied sheafification from Hartshorne's Algebraic geometry book. You can choose what you are comfortable with.\n\nI will mention the result first as \u003ca href=\"https://stacks.math.columbia.edu/tag/02ZM\"\u003ein\u003c/a\u003e Lemma \\(8.8.1\\).\n\n\u003chr /\u003e\n\n\u003cstrong\u003eLemma\u003c/strong\u003e : Let \\(\\mathcal{C}\\) be a site.  Let \\(p:\\mathcal{S}\\rightarrow \\mathcal{C}\\) be a fibred category over \\(\\mathcal{C}\\). There exists a \u003cstrong\u003estack\u003c/strong\u003e \\(p':\\mathcal{S}'\\rightarrow \\mathcal{C}\\) and a morphisms \\(G:\\mathcal{S}\\rightarrow \\mathcal{S}'\\) of fibred categories over \\(\\mathcal{C}\\) such that\n\u003col\u003e\n\t\u003cli\u003efor every \\(U\\in \\text{Ob}(\\mathcal{C})\\) and \\(x,y\\in \\mathcal{S}(U)\\),  the map \\(\\text{Mor}(x,y)\\rightarrow \\text{Mor}(G(x),G(y))\\) induced by \\(G\\) identifies the right hand side with the sheafification of the left hand side.\u003c/li\u003e\n\t\u003cli\u003eFor \\(U\\in \\mathcal{C}_0\\) and \\(x'\\in \\mathcal{S}'(U)\\) there exists a covering \\(\\{U_i\\rightarrow U\\}\\) such that each \\(x'|_{U_i}\\) is in the essential image of the functor \\(G:\\mathcal{S}(U)\\rightarrow \\mathcal{S}'(U)\\).\u003c/li\u003e\n\u003c/ol\u003e\n\n\u003chr /\u003e\n\nWe recall what is \\(\\text{Mor}(a,b)\\). This is a presheaf on \\(U\\) defined as follows. Given an inclusion \\(i : V\\hookrightarrow U\\) we have \\(i^*(a),i^*(b)\\in \\mathcal{S}(V)\\).","tags":["hartshorne","sheaves","stacks"],"title":"Stackification of fibred categories","type":"wp-import"},{"categories":["Lie groups","Differential geometry"],"content":"Let \\(\\pi:P\\rightarrow M\\) be a principal \\(G\\) bundle. We choose an open covering \\(\\{U_\\alpha\\}\\) of \\(M\\) and trivializations \\(\\psi_\\alpha:\\pi^{-1}(U_\\alpha)\\rightarrow U_\\alpha\\times G\\) defined as \\(\\psi_\\alpha(u)= (\\pi(u),\\varphi_\\alpha(u))\\) such that \\(\\varphi_\\alpha(ua)=\\varphi_\\alpha(u)a\\) for all \\(u\\in \\pi^{-1}(U_\\alpha)\\) and \\(a\\in G\\). Let \\(x\\in U_\\alpha\\cap U_\\beta\\). Given \\(v\\in \\pi^{-1}(x)\\subseteq \\pi^{-1}(U_\\alpha)\\cap \\pi^{-1}(U_\\beta)\\), we have \\(\\varphi_\\alpha(v)\\in G\\) and \\(\\varphi_\\beta(v)\\in G\\). For \\(v'\\in \\pi^{-1}(x)\\) there exists \\(g\\in G\\) such that \\(v'=vg\\). Then, we have \\(\\varphi_\\alpha(v')\\varphi_\\beta(v')^{-1}= \\varphi_\\alpha(vg)\\varphi_\\beta(ua)^{-1} =\\varphi_\\alpha(u)aa^{-1}\\varphi_\\beta(u)^{-1} =\\varphi_\\alpha(u)\\varphi_\\beta(u)^{-1}\\)\nThus, for any \\(v,v'\\in \\pi^{-1}(x)\\), we have \\(\\varphi_\\alpha(v)\\varphi_\\beta(v)^{-1}=\\varphi_\\alpha(v')\\varphi_\\beta(v')^{-1}.\\)\nThis gives a well defined map \\(g_{\\alpha\\beta}:U_\\alpha\\cap U_\\beta\\rightarrow G\\) given by \\(g_{\\alpha\\beta}(x)=\\varphi_\\alpha(u)\\varphi_\\beta(u)^{-1}\\) where \\(u\\in P\\) is such that \\(\\pi(u)=x\\). Now, I have to check that this map \\(g_{\\alpha\\beta}\\) is smooth. As \\(\\pi^{-1}(U_\\alpha)\\rightarrow U_\\alpha\\times G\\) is smooth, so is its projection to \\(G\\) i.e., the map \\(\\pi^{-1}(U_\\alpha)\\rightarrow G\\) given by \\(u\\mapsto \\varphi_\\alpha(u)\\) is smooth. Similarly the map \\(\\pi^{-1}(U_\\beta)\\rightarrow G\\) given by \\(u\\mapsto \\varphi_\\beta(u)\\) is smooth. As inverse map on \\(G\\) is smooth, so is the composition \\(\\pi^{-1}(U_\\beta)\\rightarrow G\\rightarrow G\\) with \\(u\\mapsto \\varphi_{\\beta}(u)^{-1}\\). So, the map \\(\\pi^{-1}(U_\\alpha\\cap U_\\beta)\\rightarrow G\\times G\\) given by \\(u\\mapsto (\\varphi_{\\alpha}(u),\\varphi_{\\beta}^{-1}(u))\\) is smooth. As multiplication map \\(G\\times G\\rightarrow G\\) is smooth, so is the compostion \\(\\pi^{-1}(U_\\alpha\\cap U_\\beta)\\rightarrow G\\times G\\rightarrow G\\) given by \\(u\\mapsto \\varphi_{\\alpha}(u)\\varphi_\\beta^{-1}(u)\\). Thus, the map \\(\\pi^{-1}(U_\\alpha\\cap U_\\beta)\\rightarrow G\\) given by \\(u\\mapsto \\varphi_{\\alpha}(u)\\varphi_{\\alpha}\\beta^{-1}(u)\\) is smooth. Proving \\(g_{\\alpha\\beta}:U_\\alpha\\cap U_\\beta\\rightarrow G\\) is smooth boils down to proving \\(U_\\alpha\\cap U_\\beta\\rightarrow \\pi^{-1}(U_\\alpha\\cap U_\\beta)\\) is smooth. As \\(\\pi^{-1}(U_\\alpha)\\rightarrow U_\\alpha\\times G\\) is diffeomorphism, its inverse \\(U_\\alpha\\times G\\rightarrow \\pi^{-1}(U_\\alpha)\\) is smooth and so is its composition with inclusion \\(U_\\alpha\\rightarrow U_\\alpha\\times G\\) given by \\(x\\mapsto (x,1)\\). Thus, we have smooth map \\(U_\\alpha\\rightarrow \\pi^{-1}(U)\\) given by \\(x\\mapsto (x,1)\\) and so is the map \\(U_\\alpha\\cap U_\\beta\\rightarrow \\pi^{-1}(U_\\alpha\\cap U_\\beta)\\). Thus, the map \\(U_\\alpha\\cap U_\\beta\\rightarrow \\pi^{-1}(U_\\alpha\\cap U_\\beta)\\rightarrow G\\)\nwhich is precisely \\(g_{\\alpha\\beta}\\) is smooth. \u0026nbsp;","permalink":"https://praphulla-koushik.github.io/2019/01/26/transition-maps-for-principal-bundle-are-smooth/","summary":"Let \\(\\pi:P\\rightarrow M\\) be a principal \\(G\\) bundle.\n\nWe choose an open covering \\(\\{U_\\alpha\\}\\) of \\(M\\) and trivializations \\(\\psi_\\alpha:\\pi^{-1}(U_\\alpha)\\rightarrow U_\\alpha\\times G\\) defined as \\(\\psi_\\alpha(u)= (\\pi(u),\\varphi_\\alpha(u))\\) such that \\(\\varphi_\\alpha(ua)=\\varphi_\\alpha(u)a\\) for all \\(u\\in \\pi^{-1}(U_\\alpha)\\) and \\(a\\in G\\).\n\nLet \\(x\\in U_\\alpha\\cap U_\\beta\\). Given \\(v\\in \\pi^{-1}(x)\\subseteq \\pi^{-1}(U_\\alpha)\\cap \\pi^{-1}(U_\\beta)\\), we have \\(\\varphi_\\alpha(v)\\in G\\) and \\(\\varphi_\\beta(v)\\in G\\). For \\(v'\\in \\pi^{-1}(x)\\) there exists \\(g\\in G\\) such that \\(v'=vg\\). Then, we have\n\u003cp style=\"text-align:center;\"\u003e\\(\\varphi_\\alpha(v')\\varphi_\\beta(v')^{-1}= \\varphi_\\alpha(vg)\\varphi_\\beta(ua)^{-1}\n=\\varphi_\\alpha(u)aa^{-1}\\varphi_\\beta(u)^{-1} =\\varphi_\\alpha(u)\\varphi_\\beta(u)^{-1}\\)\u003c/p\u003e\nThus, for any \\(v,v'\\in \\pi^{-1}(x)\\), we have\n\u003cp style=\"text-align:center;\"\u003e\\(\\varphi_\\alpha(v)\\varphi_\\beta(v)^{-1}=\\varphi_\\alpha(v')\\varphi_\\beta(v')^{-1}.\\)\u003c/p\u003e","tags":[],"title":"Transition maps for principal bundle are smooth","type":"wp-import"},{"categories":["category-theory","stacks"],"content":"In this page, I will give links of Category theory posts that I have made here. I learned some category theory from Hilton and Stammbach's book A Course in Homological Algebra. Angelo Vistoli's Descent theory notes. ","permalink":"https://praphulla-koushik.github.io/2019/01/24/category-theory/","summary":"In this page, I will give links of Category theory posts that I have made here.\n\nI learned some category theory from\n\u003cul\u003e\n\t\u003cli\u003eHilton and Stammbach's book A Course in Homological Algebra.\u003c/li\u003e\n\t\u003cli\u003eAngelo Vistoli's Descent theory notes.\u003c/li\u003e\n\u003c/ul\u003e","tags":["stacks"],"title":"Category theory","type":"wp-import"},{"categories":["Differential geometry"],"content":"Given a section \\(\\sigma:N\\rightarrow P\\) we produce a smooth map (trivialization) \\(\\Phi_P:N\\times G\\rightarrow P\\) given by \\((n,g)\\mapsto \\sigma(n)g\\). This is smooth for obvious reasons. The map \\(N\\rightarrow P\\) given by \\(n\\mapsto \\sigma(n)\\) is smooth so is the map \\(N\\times G\\rightarrow P\\times G\\) given by \\((n,g)\\mapsto (\\sigma(n),g)\\). The multiplication map \\(P\\times G\\rightarrow P\\) given by \\((p,g)\\mapsto pg\\) is smooth. Thus the composition \\(N\\times G\\rightarrow P\\times G\\rightarrow P\\)\nis smooth which is simply the map \\(\\Phi_P:N\\times G\\rightarrow P\\) is smooth. We see that this map is a diffeomorphism. What obvious map can you think of \\(P\\rightarrow N\\times G\\)? Given \\(p\\in P\\) we need to associate an element \\((n,g)\\in N\\times G\\). For first coordinate, obvious choice is \\(\\pi(p)\\in N\\). Remember that we are already with a guess that \\(\\Phi\\) is a bijection and this map \\(P\\rightarrow N\\times G\\) has to be inverse of \\(\\Phi:N\\times G\\rightarrow P\\). So, given \\(p\\in P\\) we choose \\(g\\in G\\) such that \\(\\Phi(\\pi(p),g)=p\\) i.e., \\(\\sigma(\\pi(p)).g=p\\). The point is, we can always choose such \\(g\\) and it is unique as action is free. See that \\(\\sigma(\\pi(p))\\in \\pi^{-1}(\\pi(p))\\) and \\(p\\in \\pi^{1}(p)\\). So, as any two elements in fibre are related by an element in \\(G\\) we have \\(g\\in G\\) such that \\(\\sigma(\\pi(p)).g=p\\). Thus, we have an obvious map \\(P\\rightarrow N\\times G\\) given by \\(p\\mapsto (\\pi(p),g)\\) where \\(g\\in G\\) is the unique such \\(g\\) satisfying \\(\\sigma(\\pi(p))g=p\\). It is upto you to see that this map is a smooth map. This is smooth on first projection to \\(N\\) being just the map \\(\\pi\\). It needs some work to see the projectionto \\(G\\) is smooth. It is by definition that this map is actually inverse of \\(\\Phi_P:N\\times G\\rightarrow P\\) and thus we have a diffeomorphism. This diffeomorphism is \\(G\\)-equivariant if you know what it means. Thus, knowing that \\(P\\rightarrow N\\) is a principal \\(G\\) bundle, a section \\(\\sigma:N\\rightarrow P\\) gives a trivialization \\(N\\times G\\rightarrow P\\). Given a trivialization \\(N\\times G\\xrightarrow{\\Phi} P\\), we have a section \\(\\sigma:N\\rightarrow P\\) given by \\(\\sigma(n)=\\Phi(n,1)\\). Thus, giving a section is same thing as giving local trivialization.","permalink":"https://praphulla-koushik.github.io/2019/01/24/trivializations-and-sections-in-principal-bundle/","summary":"Given a section \\(\\sigma:N\\rightarrow P\\) we produce a smooth map (trivialization) \\(\\Phi_P:N\\times G\\rightarrow P\\) given by \\((n,g)\\mapsto \\sigma(n)g\\). This is smooth for obvious reasons. The map \\(N\\rightarrow P\\) given by \\(n\\mapsto \\sigma(n)\\) is smooth so is the map \\(N\\times G\\rightarrow P\\times G\\) given by \\((n,g)\\mapsto (\\sigma(n),g)\\). The multiplication map \\(P\\times G\\rightarrow P\\) given by \\((p,g)\\mapsto pg\\) is smooth. Thus the composition\n\u003cp style=\"text-align:center;\"\u003e\\(N\\times G\\rightarrow P\\times G\\rightarrow P\\)\u003c/p\u003e\nis smooth which is simply the map \\(\\Phi_P:N\\times G\\rightarrow P\\) is smooth. We see that this map is a diffeomorphism. What obvious map can you think of \\(P\\rightarrow N\\times G\\)? Given \\(p\\in P\\) we need to associate an element \\((n,g)\\in N\\times G\\). For first coordinate, obvious choice is  \\(\\pi(p)\\in N\\). Remember that we are already with a \u003cstrong\u003eguess\u003c/strong\u003e that \\(\\Phi\\) is a bijection and this map \\(P\\rightarrow N\\times G\\) has to be inverse of \\(\\Phi:N\\times G\\rightarrow P\\). So, given \\(p\\in P\\) we choose \\(g\\in G\\) such that \\(\\Phi(\\pi(p),g)=p\\) i.e., \\(\\sigma(\\pi(p)).g=p\\). The point is, we can always choose such \\(g\\) and it is unique as action is free.\n\nSee that \\(\\sigma(\\pi(p))\\in \\pi^{-1}(\\pi(p))\\) and \\(p\\in \\pi^{1}(p)\\). So, as any two elements in fibre are related by an element in \\(G\\) we have \\(g\\in G\\) such that \\(\\sigma(\\pi(p)).g=p\\). Thus, we have an obvious map \\(P\\rightarrow N\\times G\\) given by \\(p\\mapsto (\\pi(p),g)\\) where \\(g\\in G\\) is the unique such \\(g\\) satisfying \\(\\sigma(\\pi(p))g=p\\). It is upto you to see that this map is a smooth map. This is smooth on first projection to \\(N\\) being just the map \\(\\pi\\). It needs some work to see the projectionto \\(G\\) is smooth. It is by definition that this map is actually inverse of \\(\\Phi_P:N\\times G\\rightarrow P\\) and thus we have a diffeomorphism. This diffeomorphism is \\(G\\)-equivariant if you know what it means. Thus, knowing that \\(P\\rightarrow N\\) is a principal \\(G\\) bundle,   a section \\(\\sigma:N\\rightarrow P\\) gives a trivialization \\(N\\times G\\rightarrow P\\).\n\nGiven a trivialization \\(N\\times G\\xrightarrow{\\Phi} P\\), we have a section \\(\\sigma:N\\rightarrow P\\) given by \\(\\sigma(n)=\\Phi(n,1)\\).\n\nThus, giving a section is same thing as giving local trivialization.","tags":[],"title":"Trivializations and sections in Principal bundle","type":"wp-import"},{"categories":["Lie groupoids","Lie groups","differential-geometry","stacks"],"content":"Let \\(G\\) be a Lie group and \\(\\pi_P:P\\rightarrow M, \\pi_Q:Q\\rightarrow M\\) be principal \\(G\\) bundles. Then, any \\(G\\)-equivariant map \\(f:P\\rightarrow Q\\) inducing identity on \\(M\\) is a diffeomorphism. The same holds when we have Lie groupoids instead of Lie groups. Let \\(\\mathcal{G}\\) be a Lie groupoid and \\(P\\rightarrow M, Q\\rightarrow M\\) be principal \\(\\mathcal{G}\\) bundles. Then, any \\(\\mathcal{G}\\)-equivariant map \\(f:P\\rightarrow Q\\) inducing identity on \\(M\\) is a diffeomorphism. Above result is very basic thing when defining a stack associated for a Lie groupoid \\(\\mathcal{G}\\). Given a Lie groupoid \\(\\mathcal{G}\\), we define a category fibered in groupoids \\(B\\mathcal{G}\\rightarrow \\text{Man}\\) by associating for each manifold \\(U\\) a category \\(B\\mathcal{G}(U)\\) whose objects are principal \\(\\mathcal{G}\\) bundles whose base space is \\(U\\) i.e., of the form \\(P\\rightarrow U\\) and morphism from an object \\(P\\rightarrow U\\) to another object \\(Q\\rightarrow U\\) is a \\(\\mathcal{G}\\)-equivariant map \\(P\\rightarrow Q\\) that induces \\(Id:U\\rightarrow U\\) on base space of those principal bundles. Thus, to say \\(B\\mathcal{G}(U)\\) is a Lie groupoid, we need to prove that every arrow \\((P\\rightarrow U)\\rightarrow (Q\\rightarrow U)\\) is an isomorphism which is what we are trying to prove. Let us see the proof for the case of Lie groups. See the set up as following diagram. Let \\(p,p'\\in P\\) are such that \\(f(p)=f(p')\\), thus, \\(\\pi_Q(f(p))=\\pi_Q(f(p'))\\). As \\(\\pi_Q\\circ f=\\pi_P\\), we have \\(\\pi_P(p)=\\pi_P(p')\\) i.e., there exists \\(g\\in G\\) such that \\(p'=p.g\\). Thus, \\(f(p')=f(pg)\\). As \\(f\\) is \\(G\\)-equivariant, we have \\(f(pg)=f(p)g\\). Thus, we have \\(f(p')=f(p)g\\). As the action of \\(G\\) on \\(Q\\) is free, \\(f(p')=f(p),f(p')=f(p)g\\) implies \\(g=1\\). Thus, \\(p'=p\\). So, \\(f\\) is one to one mapping. Let \\(q\\in Q\\). We have \\(\\pi_Q(q)\\in M\\). As \\(\\pi_P\\) is surjective, there exists \\(p\\in P\\) such that \\(\\pi_P(p)=\\pi_Q(q)\\). As \\(\\pi_Q\\circ f=\\pi_P\\), we have \\(\\pi_Q(f(p))=\\pi_P(p)=\\pi_Q(q)\\). As \\(\\pi_Q(f(p))=\\pi_Q(q)\\), there exists \\(g\\in G\\) such that \\(f(p)g=q\\). As \\(f\\) is \\(G\\)-equivariant, we have \\(f(p)g=f(pg)\\). Thus, we have \\(q=f(pg)\\) which implies that \\(f\\) is an onto mapping. Suppose that \\(\\pi_P:P\\rightarrow M\\) is trivial \\(G\\) bundle, not for simplicity but because every principal \\(G\\) bundle is locally trivial and diffeomorphism is something that needs to be checked locally. As \\(\\pi_P:P\\rightarrow M\\) is trivial, it has a global section for \\(\\pi_P\\) i.e., a smooth map \\(\\sigma:M\\rightarrow P\\) such that \\(\\pi_P\\circ \\sigma=1\\). This gives a trivialization \\(M\\times G\\xrightarrow{\\Phi} P\\) i.e., an isomorphism. Consider the cimposition \\(f\\circ \\pi:M\\rightarrow Q\\). This is again a smooth map such that \\(\\pi_Q\\circ (f\\circ \\sigma)=(\\pi_Q\\circ f)\\circ \\sigma=\\pi_P\\circ \\sigma=1\\)\ni.e., the composition \\(f\\circ \\sigma:M\\rightarrow Q\\) is a global section for \\(\\pi_Q:Q\\rightarrow M\\). This again gives a trivialization \\(Q\\xrightarrow{\\Psi} M\\times G\\). Check that \\(\\Psi\\circ f\\circ \\Phi=1\\), which is fun (no difficult) to check. As both \\(\\Psi, \\Phi\\) are diffeomorphism, \\(\\Psi\\circ f\\circ \\Phi=1\\) says that \\(f=\\Psi^{-1}\\circ \\Phi^{-1}\\). Being a composition of diffeomorphisms \\(\\Psi^{-1}\\) and \\(\\Phi^{-1}\\), the map \\(f:P\\rightarrow Q\\) is a diffeomorphism. If you consider local trivialization \\(U\\times G\\rightarrow \\pi^{-1}(U)\\), above procedure says that \\(f|_{\\pi^{-1}(U)}\\) is a diffeomorphism. See that \\(P\\) is covered by \\(\\pi^{-1}(U)\\). Thus, \\(f\\) is a diffeomorphism. \u0026nbsp;","permalink":"https://praphulla-koushik.github.io/2019/01/23/equivariant-maps-are-isomorphisms/","summary":"\u003cstrong\u003eLet \\(G\\) be a Lie group and \\(\\pi_P:P\\rightarrow M, \\pi_Q:Q\\rightarrow M\\) be principal \\(G\\) bundles. Then, any \\(G\\)-equivariant map \\(f:P\\rightarrow Q\\) inducing identity on \\(M\\) is a diffeomorphism. \u003c/strong\u003e\n\nThe same holds when we have Lie groupoids instead of Lie groups.\n\n\u003cstrong\u003eLet \\(\\mathcal{G}\\) be a Lie groupoid and \\(P\\rightarrow M, Q\\rightarrow M\\) be principal \\(\\mathcal{G}\\) bundles. Then, any \\(\\mathcal{G}\\)-equivariant map \\(f:P\\rightarrow Q\\) inducing identity on \\(M\\) is a diffeomorphism. \u003c/strong\u003e\n\nAbove result is very basic thing when defining a stack associated for a Lie groupoid \\(\\mathcal{G}\\). Given a Lie groupoid \\(\\mathcal{G}\\), we define a category fibered in \u003cstrong\u003egroupoids \u003c/strong\u003e\\(B\\mathcal{G}\\rightarrow \\text{Man}\\) by associating for each manifold \\(U\\) a category \\(B\\mathcal{G}(U)\\) whose objects are principal \\(\\mathcal{G}\\) bundles whose base space is \\(U\\) i.e., of the form \\(P\\rightarrow U\\) and morphism from an object \\(P\\rightarrow U\\) to another object \\(Q\\rightarrow U\\) is a \\(\\mathcal{G}\\)-equivariant map \\(P\\rightarrow Q\\) that induces \\(Id:U\\rightarrow U\\) on base space of those principal bundles. Thus, to say  \\(B\\mathcal{G}(U)\\) is a Lie groupoid, we need to prove that every arrow \\((P\\rightarrow U)\\rightarrow (Q\\rightarrow U)\\) is an isomorphism which is what we are trying to prove.\n\nLet us see the proof for the case of Lie groups. See the set up as following diagram. \u003cimg class=\" size-full wp-image-1365 aligncenter\" src=\"/wp-media/2019/01/c90b6dbb9e-screenshot-from-2019-01-24-20-14-48.png\" alt=\"screenshot from 2019-01-24 20-14-48\" width=\"272\" height=\"224\" /\u003eLet \\(p,p'\\in P\\) are such that \\(f(p)=f(p')\\), thus, \\(\\pi_Q(f(p))=\\pi_Q(f(p'))\\). As \\(\\pi_Q\\circ f=\\pi_P\\), we have \\(\\pi_P(p)=\\pi_P(p')\\) i.e., there exists \\(g\\in G\\) such that \\(p'=p.g\\). Thus, \\(f(p')=f(pg)\\). As \\(f\\) is \\(G\\)-equivariant, we have \\(f(pg)=f(p)g\\). Thus, we have \\(f(p')=f(p)g\\). As the action of \\(G\\) on \\(Q\\)  is free, \\(f(p')=f(p),f(p')=f(p)g\\) implies \\(g=1\\). Thus, \\(p'=p\\). So, \\(f\\) is one to one mapping.\n\nLet \\(q\\in Q\\). We have \\(\\pi_Q(q)\\in M\\). As \\(\\pi_P\\) is surjective, there exists \\(p\\in P\\) such that \\(\\pi_P(p)=\\pi_Q(q)\\). As \\(\\pi_Q\\circ f=\\pi_P\\), we have \\(\\pi_Q(f(p))=\\pi_P(p)=\\pi_Q(q)\\). As \\(\\pi_Q(f(p))=\\pi_Q(q)\\), there exists \\(g\\in G\\) such that \\(f(p)g=q\\). As \\(f\\) is \\(G\\)-equivariant, we have \\(f(p)g=f(pg)\\). Thus, we have \\(q=f(pg)\\) which implies that \\(f\\) is an onto mapping.\n\nSuppose that \\(\\pi_P:P\\rightarrow M\\) is \u003cstrong\u003etrivial\u003c/strong\u003e \\(G\\) bundle, not for simplicity but because every principal \\(G\\) bundle is locally trivial and diffeomorphism is something that needs to be checked locally.\n\nAs \\(\\pi_P:P\\rightarrow M\\) is \u003cstrong\u003etrivial, \u003c/strong\u003eit has \u003cstrong\u003ea global section\u003c/strong\u003e for \\(\\pi_P\\) i.e., a \u003cstrong\u003esmooth map \u003c/strong\u003e  \\(\\sigma:M\\rightarrow P\\) such that \\(\\pi_P\\circ \\sigma=1\\).  This \u003ca href=\"https://koushik1729.wordpress.com/2019/01/24/trivializations-and-sections-in-principal-bundle/\" target=\"_blank\" rel=\"noopener\"\u003egives a trivialization\u003c/a\u003e \\(M\\times G\\xrightarrow{\\Phi} P\\) i.e., an isomorphism. Consider the cimposition \\(f\\circ \\pi:M\\rightarrow Q\\). This is again a smooth map such that\n\u003cp style=\"text-align:center;\"\u003e\\(\\pi_Q\\circ (f\\circ \\sigma)=(\\pi_Q\\circ f)\\circ \\sigma=\\pi_P\\circ \\sigma=1\\)\u003c/p\u003e","tags":["principal-bundles","stacks"],"title":"Equivariant maps are Isomorphisms","type":"wp-import"},{"categories":["Category theory"],"content":"Let \\(F:\\mathcal{I}\\rightarrow \\mathcal{C}\\) is a functor. This is also called as diagram indexed by \\(\\mathcal{I}\\). By the Limit of this diagram, we mean an object (universal) \\(L\\) of \\(\\mathcal{C}\\) and a collection of arrows (universal again) \\(\\pi_i:L\\rightarrow F(i)\\) such that, for each arrow \\(m:i\\rightarrow j\\) in \\(\\mathcal{I}\\) the following diagram is commutative. This is usually denoted by \\(\\varprojlim_{\\mathcal{I}}F(i)\\) or simply by \\(\\varprojlim_{\\mathcal{I}}F\\). Fixing an object \\(X\\) in \\(\\mathcal{C}\\), I want to prove that \\(\\varprojlim_{\\mathcal{I}}(\\text{Hom}_{\\mathcal{C}}(X,F(i))) =\\text{Hom}_{\\mathcal{C}}(X,\\varprojlim_{\\mathcal{I}}F(i))\\)\ni.e., an isomorphism \\(\\varprojlim_{\\mathcal{I}}(\\text{Hom}_{\\mathcal{C}}(X,F(i))) =\\text{Hom}_{\\mathcal{C}}(X,L)\\)\nLet \\((A,(p_i))\\) be cone for the functor \\(\\mathcal{I}\\rightarrow \\text{Set}\\) given by \\(i\\mapsto \\text{Hom}_{\\mathcal{C}}(X,F(i))\\) i.e., we have following commutative diagrams To prove that \\(\\text{Hom}_{\\mathcal{C}}(X,L)\\) is equal to \\(\\varprojlim_{\\mathcal{I}}(\\text{Hom}_{\\mathcal{C}}(X,F(i)))\\) it suffices to prove that there exists unique arrow \\(p:A\\rightarrow \\text{Hom}_{\\mathcal{C}}(X,L)\\) such that the following diagram is commutative. So, we define an arrow \\(p:A\\rightarrow \\text{Hom}_{\\mathcal{C}}(X,L)\\) i.e., given \\(a\\in A\\) we define arrow \\(p(a):X\\rightarrow L\\) in \\(\\mathcal{C}\\). How to one get such arrow? See above diagram. For each \\(a\\in A\\), we have \\(p_i(a):X\\rightarrow F(i)\\) such that \\(F(m)\\circ p_i(a)=p_j(a)\\) giving following diagram which gives an arrow \\(X\\rightarrow L\\) by universal property This is the \\(p(a):X\\rightarrow L\\) that we associate for each \\(a\\in L\\). This gives the map \\(p:A\\rightarrow \\text{Hom}_{\\mathcal{C}}(X,L)\\) satisfying conditions mentioned above. Thus, we have \\(\\varprojlim_{\\mathcal{I}}(\\text{Hom}_{\\mathcal{C}}(X,F(i))) =\\text{Hom}_{\\mathcal{C}}(X,L)=\\text{Hom}_{\\mathcal{C}}(X,\\varprojlim_{\\mathcal{I}}F(i))\\)\n","permalink":"https://praphulla-koushik.github.io/2019/01/23/limit-of-a-diagram-functor-preserved-by-hom-functor/","summary":"Let \\(F:\\mathcal{I}\\rightarrow \\mathcal{C}\\) is a functor. This is also called as diagram indexed by \\(\\mathcal{I}\\).\n\nBy the Limit of this diagram, we mean an object (universal) \\(L\\) of \\(\\mathcal{C}\\) and a collection of arrows (universal again) \\(\\pi_i:L\\rightarrow F(i)\\) such that, for each arrow \\(m:i\\rightarrow j\\) in \\(\\mathcal{I}\\) the following diagram is commutative.\n\n\u003cimg class=\"alignnone size-full wp-image-1356\" src=\"/wp-media/2019/01/0bfac22033-screenshot-from-2019-01-23-02-04-50.png\" alt=\"screenshot from 2019-01-23 02-04-50\" width=\"398\" height=\"194\" /\u003e\n\nThis is usually denoted by \\(\\varprojlim_{\\mathcal{I}}F(i)\\) or simply by \\(\\varprojlim_{\\mathcal{I}}F\\).\n\nFixing an object \\(X\\) in \\(\\mathcal{C}\\), I want to prove that\n\u003cp style=\"text-align:center;\"\u003e\\(\\varprojlim_{\\mathcal{I}}(\\text{Hom}_{\\mathcal{C}}(X,F(i)))\n=\\text{Hom}_{\\mathcal{C}}(X,\\varprojlim_{\\mathcal{I}}F(i))\\)\u003c/p\u003e","tags":[],"title":"Limit of a diagram/functor preserved by Hom functor","type":"wp-import"},{"categories":["stacks"],"content":"A morphism of stacks \\(F:\\mathcal{D}\\rightarrow \\mathcal{C}\\) is said to be a gerbe over stack if following two conditions hold : Given a manifold \\(U\\) and an object \\(\\xi\\in \\mathcal{C}(U)\\), there exists a covering \\(\\{U_i\\rightarrow U\\}\\) (depending on the Grothendieck topology that we have fixed on the category \\(Man\\) of manifolds) and objects \\(x_i\\in \\mathcal{D}(U_i)\\) with an isomorphism \\(F(x_i)\\rightarrow \\xi|_{U_i}\\) for each \\(i\\). Given a manifold \\(U\\) and an arrow \\(\\xi\\rightarrow \\eta\\) in \\(\\mathcal{C}(U)\\), there exists a covering \\(\\{U_i\\rightarrow U\\}\\) (depending on the Grothendieck topology that we have fixed on the category \\(Man\\) of manifolds) and arrows \\(x_i\\rightarrow y_i\\) in \\(\\mathcal{D}(U_i)\\) such that ","permalink":"https://praphulla-koushik.github.io/2019/01/20/definition-of-gerbe-over-stack/","summary":"A morphism of stacks \\(F:\\mathcal{D}\\rightarrow \\mathcal{C}\\) is said to be a gerbe over stack if following two conditions hold :\n\u003col\u003e\n\t\u003cli\u003eGiven a manifold \\(U\\) and \u003cstrong\u003ean object \\(\\xi\\in \\mathcal{C}(U)\\)\u003c/strong\u003e, there exists a covering \\(\\{U_i\\rightarrow U\\}\\) (depending on the Grothendieck topology that we have fixed on the category \\(Man\\) of manifolds) and objects \\(x_i\\in \\mathcal{D}(U_i)\\) with an isomorphism \\(F(x_i)\\rightarrow \\xi|_{U_i}\\) for each \\(i\\).\u003c/li\u003e\n\t\u003cli\u003eGiven a manifold \\(U\\) and \u003cstrong\u003ean arrow \\(\\xi\\rightarrow \\eta\\) in \\(\\mathcal{C}(U)\\), \u003c/strong\u003ethere exists a covering \\(\\{U_i\\rightarrow U\\}\\) (depending on the Grothendieck topology that we have fixed on the category \\(Man\\) of manifolds) and arrows \\(x_i\\rightarrow y_i\\) in \\(\\mathcal{D}(U_i)\\) such that\u003c/li\u003e\n\u003c/ol\u003e","tags":["stacks"],"title":"Definition of gerbe over stack","type":"wp-import"},{"categories":["category-theory","differential-geometry"],"content":"Given a morphism of Lie groups \\(\\theta:G\\rightarrow H\\) and a principal \\(G\\) bundle \\(\\pi:P\\rightarrow M\\) there are (at least) two ways to assign a principal \\(H\\) bundle. See that the morphism of Lie groups \\(\\theta:G\\rightarrow H\\) gives an action of \\(G\\) on \\(H\\) by \\(g.h=\\theta(g).h\\). Given an action of \\(G\\) on manifold (Lie group in this case) \\(H\\) there is an associated fibre bundle \\(P\\times_G H\\rightarrow M\\) with fibre \\(H\\). This gives a principal \\(H\\) bundle. For principal bundle \\(\\pi:P\\rightarrow M\\), we can find an open cover \\(\\{U_\\alpha\\}\\) of \\(M\\) and (transition) maps \\(g_\\alpha g_\\beta:U_{\\alpha\\beta}\\rightarrow G\\) satifsying the cocycle condition \\(g_{\\alpha\\beta}g_{\\beta\\gamma}=g_{\\alpha\\gamma}\\) on \\(U_\\alpha\\cap U_\\beta\\cap U_\\gamma\\). Then the compositions \\(\\tau_{\\alpha\\beta}=\\theta\\circ g_{\\alpha\\beta}:U_{\\alpha\\beta}\\rightarrow G\\rightarrow H\\) also satifies the cocycle condition \\(\\tau_{\\alpha\\beta}\\tau_{\\beta\\gamma}=\\tau_{\\alpha\\gamma}\\) on \\(U_\\alpha\\cap U_\\beta\\cap U_\\gamma\\). One can then produce a principal \\(H\\) bundle over \\(M\\) given this open cover \\(\\{U_\\alpha\\}\\) of \\(M\\) and smooth maps \\(\\tau_{\\alpha\\beta}:U_\\alpha\\cap U_\\beta\\rightarrow H\\) satisfying the cocycle condition. This gives a principal \\(H\\) bundle. It is a good exercise (that I have not tried) to check that principal \\(H\\) bundles obtained from above two methods are (naturally) isomorphic i.e., one and the same. Given a Lie group \\(G\\), let \\(BG\\) denote the category of principal \\(G\\) bundles. Objects are principal \\(G\\) bundles and morphisms are \\(G\\)-equivariant morphisms. Given a morphism of Lie groups \\(\\theta:G\\rightarrow H\\), above construction gives a functor (at the level of objects) \\(B\\theta:BG\\rightarrow BH\\). It is not difficult to see that, a \\(G\\)-equivarint map induce a \\(H\\)-equivariant map. This gives a functor \\(BG\\rightarrow BH\\).","permalink":"https://praphulla-koushik.github.io/2019/01/18/morphism-of-lie-groups-giving-a-functor/","summary":"Given a morphism of Lie groups \\(\\theta:G\\rightarrow H\\)  and a principal \\(G\\) bundle \\(\\pi:P\\rightarrow M\\) there are (at least) two ways to assign a principal \\(H\\) bundle.\n\u003col\u003e\n\t\u003cli\u003eSee that the morphism of Lie groups \\(\\theta:G\\rightarrow H\\) gives an action of \\(G\\) on \\(H\\) by \\(g.h=\\theta(g).h\\). Given an action of \\(G\\) on manifold (Lie group in this case) \\(H\\) there is an associated fibre bundle \\(P\\times_G H\\rightarrow M\\) with fibre \\(H\\). This gives a principal \\(H\\) bundle.\u003c/li\u003e\n\t\u003cli\u003eFor principal bundle \\(\\pi:P\\rightarrow M\\), we can find an open cover \\(\\{U_\\alpha\\}\\) of \\(M\\) and  (transition) maps \\(g_\\alpha g_\\beta:U_{\\alpha\\beta}\\rightarrow G\\) satifsying the cocycle condition \\(g_{\\alpha\\beta}g_{\\beta\\gamma}=g_{\\alpha\\gamma}\\) on \\(U_\\alpha\\cap U_\\beta\\cap U_\\gamma\\). Then the compositions \\(\\tau_{\\alpha\\beta}=\\theta\\circ g_{\\alpha\\beta}:U_{\\alpha\\beta}\\rightarrow G\\rightarrow H\\) also satifies the cocycle condition \\(\\tau_{\\alpha\\beta}\\tau_{\\beta\\gamma}=\\tau_{\\alpha\\gamma}\\) on \\(U_\\alpha\\cap U_\\beta\\cap U_\\gamma\\). One can then produce a principal \\(H\\) bundle over \\(M\\) given this open cover \\(\\{U_\\alpha\\}\\) of \\(M\\) and smooth maps \\(\\tau_{\\alpha\\beta}:U_\\alpha\\cap U_\\beta\\rightarrow H\\) satisfying the cocycle condition. This gives a principal \\(H\\) bundle.\u003c/li\u003e\n\u003c/ol\u003e\nIt is a good exercise (that I have not tried) to check that principal \\(H\\) bundles obtained from above two methods are (naturally) isomorphic i.e., one and the same.\n\nGiven a Lie group \\(G\\), let \\(BG\\) denote the category of principal \\(G\\) bundles. Objects are principal \\(G\\) bundles and morphisms are \\(G\\)-equivariant morphisms.\n\nGiven a morphism of Lie groups \\(\\theta:G\\rightarrow H\\), above construction gives a functor (at the level of objects) \\(B\\theta:BG\\rightarrow BH\\). It is not difficult to see that, a \\(G\\)-equivarint map induce a \\(H\\)-equivariant map. This gives a functor \\(BG\\rightarrow BH\\).","tags":["principal-bundles"],"title":"Morphism of Lie groups giving a functor","type":"wp-import"},{"categories":["stacks"],"content":"This post is based on (wanted to write after reading) Lie Groupoids and Differentiable stacks by Matias L. del Hoyo. I would suggest this for any one who wants to know about Lie groupoids and Differentiable stacks. This is well written. By a manifold, we always mean a smooth manifold. A Groupoid is a category where every arrow is invertible. A Lie groupoid is a groupoid with additional smooth structures on object set/morphism set and maps between them. Definition : A Lie groupoid consists of a manifold \\(\\mathcal{G}_0\\) of objects, a manifold \\(\\mathcal{G}_1\\) of arrows and following maps : \\(s:\\mathcal{G}_1\\rightarrow \\mathcal{G}_0\\), a submersion, called the source map. \\(t:\\mathcal{G}_1\\rightarrow \\mathcal{G}_0\\), a submersion, called the target map. \\(m:\\mathcal{G}_1\\times_{s,\\mathcal{G}_0,t}\\mathcal{G}_1\\rightarrow \\mathcal{G}_1\\), a smooth map, called the multiplication map. \\(u:\\mathcal{G}_0\\rightarrow \\mathcal{G}_1\\), a smooth map, called the unit map. \\(i:\\mathcal{G}_1\\rightarrow \\mathcal{G}_1\\), a smooth map, called the inverse map. with some compatibility conditions. We denote this Lie groupoid by \\(\\mathcal{G}_1\\rightrightarrows \\mathcal{G}_0\\). Definition : Let \\(\\mathcal{G}_1\\rightrightarrows \\mathcal{G}_0\\) be a Lie groupoid and \\(x\\in \\mathcal{G}_0\\). The set \\(s^{-1}(x)=:G(x,-)\\) is called the \\(s\\)-fibre of \\(x\\) . The set \\(t^{-1}(x)=:G(x,-)\\) is called the \\(s\\)-fibre of \\(x\\) The set \\(s^{-1}(x)\\cap t^{-1}(x)=:G_x\\) is called the Isotropy group of \\(x\\). The set \\(t(s^{-1}(x))=\\{y:x\\rightarrow y\\in \\mathcal{G}_1\\}=:O_x\\) is called the orbit of \\(x\\). Proposition : Given a Lie groupoid \\(\\mathcal{G}_1\\rightrightarrows \\mathcal{G}_0\\) and \\(x,y\\in \\mathcal{G}_0\\), the subset \\(G(y,x)\\subseteq G\\) is an embedded submanifold. In particular, \\(G_x\\) is a Lie group. the subset \\(O_x\\) is a (may not be embedded) submanifold in a canonical way. By a morphism of Lie groupoids \\(\\phi: (\\mathcal{G}_1\\rightrightarrows \\mathcal{G}_0)\\rightarrow (\\mathcal{H}_1\\rightrightarrows \\mathcal{H}_0)\\) we mean a pair of smooth maps \\(\\phi^{ar}:\\mathcal{G}_1\\rightarrow \\mathcal{H}_1\\) and \\(\\phi^{ob}:\\mathcal{G}_0\\rightarrow \\mathcal{H}_0\\) compatible with structure maps \\(s,t,m,u,i\\). We write \\(\\phi\\) for both \\(\\phi^{ar}\\) and \\(\\phi^{ob}\\).","permalink":"https://praphulla-koushik.github.io/2019/01/16/lie-groupoids/","summary":"This post is based on (wanted to write after reading) \u003ca href=\"https://arxiv.org/abs/1212.6714\"\u003eLie Groupoids and Differentiable stacks \u003c/a\u003e by Matias L. del Hoyo. I would suggest this for any one who wants to know about Lie groupoids and Differentiable stacks. This is well written.\n\nBy a manifold, we always mean a smooth manifold. A Groupoid is a category where every arrow is invertible. A Lie groupoid is a groupoid with additional smooth structures on object set/morphism set and maps between them.\n\u003cblockquote\u003e\u003cstrong\u003eDefinition\u003c/strong\u003e : A Lie groupoid consists of a  manifold \\(\\mathcal{G}_0\\) of objects, a manifold \\(\\mathcal{G}_1\\) of arrows and  following maps :\n\u003cul\u003e\n\t\u003cli\u003e\\(s:\\mathcal{G}_1\\rightarrow \\mathcal{G}_0\\), a submersion, called the source map.\u003c/li\u003e\n\t\u003cli\u003e\\(t:\\mathcal{G}_1\\rightarrow \\mathcal{G}_0\\), a submersion, called the target map.\u003c/li\u003e\n\t\u003cli\u003e\\(m:\\mathcal{G}_1\\times_{s,\\mathcal{G}_0,t}\\mathcal{G}_1\\rightarrow \\mathcal{G}_1\\), a smooth map, called the multiplication map.\u003c/li\u003e\n\t\u003cli\u003e\\(u:\\mathcal{G}_0\\rightarrow \\mathcal{G}_1\\), a smooth map, called the unit map.\u003c/li\u003e\n\t\u003cli\u003e\\(i:\\mathcal{G}_1\\rightarrow \\mathcal{G}_1\\), a smooth map, called the inverse map.\u003c/li\u003e\n\u003c/ul\u003e\nwith some compatibility conditions. We denote this Lie groupoid by \\(\\mathcal{G}_1\\rightrightarrows \\mathcal{G}_0\\).\n\n\u003cstrong\u003eDefinition\u003c/strong\u003e : Let \\(\\mathcal{G}_1\\rightrightarrows \\mathcal{G}_0\\) be a Lie groupoid and \\(x\\in \\mathcal{G}_0\\).\n\u003cul\u003e\n\t\u003cli\u003eThe set \\(s^{-1}(x)=:G(x,-)\\) is called  the \\(s\\)-fibre of \\(x\\) .\u003c/li\u003e\n\t\u003cli\u003eThe set \\(t^{-1}(x)=:G(x,-)\\) is called  the \\(s\\)-fibre of \\(x\\)\u003c/li\u003e\n\t\u003cli\u003eThe set \\(s^{-1}(x)\\cap t^{-1}(x)=:G_x\\) is called the Isotropy group of \\(x\\).\u003c/li\u003e\n\t\u003cli\u003eThe set \\(t(s^{-1}(x))=\\{y:x\\rightarrow y\\in \\mathcal{G}_1\\}=:O_x\\) is called the orbit of \\(x\\).\u003c/li\u003e\n\u003c/ul\u003e\n\u003c/blockquote\u003e\n\u003cstrong\u003eProposition\u003c/strong\u003e :  Given a Lie groupoid \\(\\mathcal{G}_1\\rightrightarrows \\mathcal{G}_0\\) and \\(x,y\\in \\mathcal{G}_0\\),\n\u003cul\u003e\n\t\u003cli\u003ethe subset \\(G(y,x)\\subseteq G\\) is \u003cb\u003e an embedded submanifold. \u003c/b\u003eIn particular, \u003cstrong\u003e\\(G_x\\) is a Lie group\u003c/strong\u003e.\u003c/li\u003e\n\t\u003cli\u003ethe subset \\(O_x\\) is a (\u003cstrong\u003emay not be embedded\u003c/strong\u003e) submanifold in   a canonical way.\u003c/li\u003e\n\u003c/ul\u003e\nBy a morphism of Lie groupoids \\(\\phi: (\\mathcal{G}_1\\rightrightarrows \\mathcal{G}_0)\\rightarrow (\\mathcal{H}_1\\rightrightarrows \\mathcal{H}_0)\\) we mean a pair of smooth maps \\(\\phi^{ar}:\\mathcal{G}_1\\rightarrow \\mathcal{H}_1\\) and \\(\\phi^{ob}:\\mathcal{G}_0\\rightarrow \\mathcal{H}_0\\) compatible with structure maps \\(s,t,m,u,i\\). We write \\(\\phi\\) for both \\(\\phi^{ar}\\) and \\(\\phi^{ob}\\).","tags":["stacks"],"title":"Lie groupoids","type":"wp-import"},{"categories":["category-theory","stacks"],"content":"Given a manifold \\(M\\) we have the concept of open cover of \\(M\\). We usually write an open cover of a manifold \\(M\\) as a collection of open subsets \\(\\{U_i\\}\\) (such that \\(\\bigcup U_i=M\\)). In this note we see an open cover of \\(M\\) as a collection of maps (inclusions) \\(\\{U_i\\rightarrow M\\}\\). Some properties of \"open cover\" are. (Pull back exists and gives an open cover) Suppose \\(\\{U_i\\rightarrow M\\}\\) is an open cover for \\(M\\) and \\(\\pi:V\\rightarrow M\\) is a smooth map. Then, \\(\\{\\pi^{-1}(U_i) \\rightarrow V\\}\\) is a cover for \\(V\\). (Diffeomorphisms gives open cover) For any manifold \\(M\\), \\(M\\) itself is considered as an open cover \\(\\{M\\rightarrow M\\}\\). More generally, for any diffeomorphism \\(M'\\rightarrow M\\), \\(\\{M'\\rightarrow M\\}\\) is considered as an open cover. (Open cover of open cover is an open cover) Let \\(\\{U_\\alpha\\rightarrow U\\}\\) be an open cover for \\(U\\) i.e., \\(\\bigcup_{\\alpha} U_\\alpha=U\\). Suppose \\(\\{V_{\\alpha\\beta}\\rightarrow U_\\alpha\\}\\) is an open cover for \\(U_\\alpha\\) for each \\(\\alpha\\) i.e., \\(\\bigcup_{\\beta}V_{\\alpha\\beta}=U_\\alpha\\). Then, \\(\\bigcup_{\\alpha\\beta}V_{\\alpha\\beta}=U\\) i.e., \\(\\{V_{\\alpha\\beta}\\rightarrow U\\}\\) is an open cover for \\(U\\). For a category \\(\\mathcal{C}\\) and an object \\(U\\) of \\(\\mathcal{C}\\), a collection of arrows \\(\\{U_i\\rightarrow U\\}\\) is said to be a cover for \\(U\\). Definition : Let \\(\\mathcal{C}\\) be a category. A Grothendieck topology on \\(\\mathcal{C}\\) is given by a collection of covers \\(\\mathcal{W}=\\{\\{U_i\\rightarrow U\\}: U\\in \\mathcal{C}_0\\}\\) satisfying following conditions. (Pullbacks exists and gives a cover) Suppose \\(\\{U_i\\rightarrow U\\}\\in \\mathcal{W}\\) and \\(\\pi:V\\rightarrow U\\) be an arrow. Then, the pull back \\(U_i\\times_UV\\) exists (as an object in \\(\\mathcal{C}\\)) and \\(\\{U_i\\times_UV \\rightarrow V\\}\\) is a cover for \\(V\\). (Isomorphisms gives an open cover) Suppose \\(V\\in \\mathcal{C}_0\\) and \\(V\\rightarrow U\\) is an isomorphism in \\(\\mathcal{C}\\) then, \\(\\{V\\rightarrow U\\}\\in \\mathcal{W}\\). (cover of a cover is a cover) Suppose \\(\\{U_\\alpha\\rightarrow U\\}\\in \\mathcal{W}\\) and \\(\\{U_{\\alpha\\beta}\\rightarrow U_\\alpha\\}\\in \\mathcal{W}\\) for each \\(\\alpha\\). Then, the collection of compositions \\(\\{U_{\\alpha\\beta}\\rightarrow U_\\alpha\\rightarrow U\\}\\in \\mathcal{W}\\). To talk about a stack over category \\(\\mathcal{C}\\) we fix a Grothendieck topology \\(\\mathcal{W}\\) on \\(\\mathcal{C}\\). When we say cover, we mean it belongs to \\(\\mathcal{W}\\). Let \\(\\mathcal{D}\\) be a category fibered in groupoids over \\(\\mathcal{C}\\) i.e., we have a functor \\(F:\\mathcal{D}\\rightarrow \\mathcal{C}\\) satisfying some conditions. Given an object \\(U\\) of \\(\\mathcal{C}\\) we have what is called fibre of \\(U\\) in \\(\\mathcal{D}\\) usually denoted by \\(\\mathcal{D}(U)\\). Given an object \\(U\\) of \\(\\mathcal{C}\\) and a cover \\(\\{U_i\\rightarrow U\\}\\) (i.e., it belongs to \\(\\mathcal{W}\\)) we have what is called descent category associated to the cover \\(\\{U_i\\rightarrow U\\}\\), usually denoted by \\(\\mathcal{D}(\\{U_i\\rightarrow U\\})\\). There is an obvious functor \\(\\mathcal{D}(U)\\rightarrow \\mathcal{D}(\\{U_i\\rightarrow U\\})\\). Definition : Let \\(\\mathcal{C}\\) be a category with Grothendieck topology \\(\\mathcal{W}\\). A category fibered in groupoids \\(\\mathcal{D}\\rightarrow \\mathcal{C}\\) is said to be a stack over \\(\\mathcal{C}\\) if, for every object \\(U\\) of \\(\\mathcal{C}\\) and every cover \\(\\{U_i\\rightarrow U\\}\\), the functor \\(\\mathcal{D}(U)\\rightarrow \\mathcal{D}(\\{U_i\\rightarrow U\\})\\) is an equivalence of categories. The fibre categroy \\(\\mathcal{D}(U)\\) is a category whose objects are that of \\(\\mathcal{D}\\) which map to \\(U\\) under \\(F\\) i.e., \\(\\mathcal{D}(U)_0=\\{V\\in \\mathcal{D}_0:F(V)=U\\}\\).\nGiven \\(V,V'\\in \\mathcal{D}(U)_0\\), a morphism \\(V\\rightarrow V'\\) in \\(\\mathcal{D}(U)\\) is a morphism in \\(\\mathcal{D}\\) that maps to \\(id:U\\rightarrow U\\) under \\(F\\) i.e., \\(\\mathcal{D}(U)_1=\\{V\\xrightarrow{f} V\\in \\mathcal{D}:F(f:V\\rightarrow V')=id:U\\rightarrow U\\}\\).\nTo define descent category \\(\\mathcal{D}(\\{U_i\\xrightarrow{\\sigma_i} U\\})\\) associated to a cover \\(\\{U_i\\xrightarrow{\\sigma_i}U\\}\\) we need to fix some notations. We have already mentioned that, pull backs exists (in the definition of Grothendieck topology). Thus, \\(U_i\\times_U U_j\\) exists and we denote \\(U_i\\times_U U_j\\) by \\(U_{ij}\\). We have following pull back diagram.We have following diagram for \\(U_{ijk}=U_i\\times_U U_j\\times_U\\times_U U_k\\). Note that functor \\(F:\\mathcal{D}\\rightarrow \\mathcal{C}\\) gives a functor \\(f^*:\\mathcal{D}(V)\\rightarrow \\mathcal{D}(U)\\) for each arrow \\(f:U\\rightarrow V\\) in \\(\\mathcal{C}\\). It is easier to guess what this map has to be than to write down what this map is. So, we skip the description of this functor. Thus, for \\(pr_1:U_{ij}\\rightarrow U_i\\) we have \\(pr_1^*:\\mathcal{D}(U_i)\\rightarrow \\mathcal{D}(U_{ij})\\) and for \\(pr_2:U_{ij}\\rightarrow U_j\\) we have functor \\(pr_2^*:\\mathcal{D}(U_j)\\rightarrow \\mathcal{D}(U_{ij})\\). We have following diagram The descent category \\(\\mathcal{D}(\\{U_i\\xrightarrow{\\sigma_i}U\\})\\) is a category whose objects are a collection \\((\\{\\xi_i\\},\\{\\phi_{ij}\\})\\) where \\(\\xi_i\\in \\mathcal{D}(U_i)\\) and \\(\\phi_{ij}:pr_1^*(\\xi_j)\\rightarrow pr_2^*(\\xi_i)\\) is an isomorphism in \\(\\mathcal{D}(U_{ij})\\) satisfying following cocylce condition on \\(U_{ijk}\\), \\(pr_{13}^*(\\phi_{ik})=pr_{12}^*(\\phi_{ij})\\circ pr_{23}^*(\\phi_{j}):pr_3^*(\\xi_k)\\rightarrow pr_1^*(\\xi_i)\\).\nThis can be seen as following diagram. One must observe that \\(pr_{23}^*(pr_3^*(\\xi_k))=pr_3^*(\\xi_k), pr_{23}^*(pr_2^*(\\xi_j))=pr_2^*(\\xi_j)=pr_{12}^*(pr_2^*(\\xi_j))\\) and \\(pr_{12}^*(pr_1^*(\\xi_k))=pr_1^*(\\xi_i)\\).\nFor \\((\\{\\xi_i\\},\\{\\phi_{ij}\\}),(\\{\\eta_i,\\psi_{ij}\\})\\) in \\(\\mathcal{D}(\\{U_i\\rightarrow U\\})_0\\), an arrow \\((\\{\\xi_i\\},\\{\\phi_{ij}\\})\\xrightarrow{\\alpha} (\\{\\eta_i,\\psi_{ij}\\})\\) is a collection of arrows \\(\\alpha_i:\\xi_i\\rightarrow \\eta_i\\) in \\(\\mathcal{D}(U_i)\\) such that \\(pr_1^*(\\alpha_i):pr_1^*(\\xi_i)\\rightarrow pr_1^*(\\eta_i)\\) and \\(pr_2^*(\\alpha_j):pr_2^*(\\xi_j)\\rightarrow pr_2^*(\\eta_j)\\)\nare compatible with \\(\\phi_{ij}:pr_2^*(\\xi_j)\\rightarrow pr_1^*(\\xi_i)\\) and \\(\\psi_{ij}:pr_2^*(\\eta_j)\\rightarrow pr_1^*(\\eta_i)\\)\ngiving following commutative diagram.Now, we can describe the obvious functor \\(\\mathcal{D}(U)\\rightarrow \\mathcal{D}(\\{U_i\\rightarrow U\\})\\). An object \\(\\xi\\in \\mathcal{D}(U)\\) is mapped to \\((\\{\\xi_i\\},\\{\\phi_{ij}\\})\\) where \\(\\xi_i=\\sigma_i^*(\\xi)\\). For \\(i,j\\) we have \\(pr_1^*:\\mathcal{D}(U_i)\\rightarrow \\mathcal{D}(U_{ij})\\) and \\(pr_2^*:\\mathcal{D}(U_i)\\rightarrow \\mathcal{D}(U_{ij})\\). See that \\(pr_1^*(\\xi_i)=pr_1^*(\\sigma_i^*(\\xi))=(pr_1\\circ \\sigma)^*(\\xi)\\) is same as that of (there exists unique isomorphism) \\(pr_2^*(\\xi_j)=pr_2^*(\\sigma_j^*(\\xi))=(pr_2\\circ \\sigma)^*(\\xi)\\) as \\(pr_2\\circ \\sigma=pr_1\\circ \\sigma\\). Thus, there is a unique isomorphsim \\(pr_2^*(\\xi_j)\\rightarrow pr_1^*(\\xi_i)\\) which we denote by \\(\\phi_{ij}:pr_2^*(\\xi_j)\\rightarrow pr_1^*(\\xi_i)\\). This gives functor \\(\\mathcal{D}(U)\\rightarrow \\mathcal{D}(\\{U_i\\rightarrow U\\})\\) at the level of objects. It is not difficult to see the functor at the level of morphisms. We skip that. The condition for a category fibred in groupoids \\(\\mathcal{D}\\rightarrow \\mathcal{C}\\) to be a stack over \\(\\mathcal{C}\\) is that the functor \\(\\mathcal{D}(U)\\rightarrow \\mathcal{D}(\\{U_i\\rightarrow U\\})\\) is an equivalence of categories i.e., it is essentially surjective and the map \\(\\text{Hom}_{\\mathcal{D}(U)}(\\xi,\\xi')\\rightarrow \\text{Hom}_{\\mathcal{D}(\\{U_i\\rightarrow U\\})}((\\xi_i,\\phi_{ij}),(\\xi_i',\\phi_{ij}'))\\) is a bijection. As \\(\\mathcal{D}(U)\\rightarrow \\mathcal{D}(\\{U_i\\rightarrow U\\})\\) is essentially surjective, it means, given a collection \\(\\xi_i\\in \\mathcal{D}(U_i)\\) together with isomorphisms \\(\\phi_{ij}:pr_2^*(\\xi_j)\\rightarrow pr_1^*(\\xi_i)\\) satisfying cocylce condition mentioned above, there exists an element \\(\\xi\\in \\mathcal{D}(U)\\) that maps to (there is an isomorphism to) \\((\\{\\xi_i\\},\\{\\phi_{ij}\\})\\). This condition is called as gluing objects. The other condition is that for \\(\\xi,\\eta\\in \\mathcal{D}(U)\\) the map \\(\\text{Hom}_{\\mathcal{D}(U)}(\\xi,\\eta)\\rightarrow \\text{Hom}_{\\mathcal{D}(\\{U_i\\rightarrow U\\})}((\\xi_i,\\phi_{ij}),(\\xi_i',\\phi_{ij}'))\\)\nis a bijection. This means that, given an arrow \\(\\alpha_i:\\xi_i\\rightarrow \\eta_i\\) for each \\(i\\) such that they are compatible with \\(\\phi_{ij},\\psi_{ij}\\) as in above commutative diagram i.e., \\(\\psi_{ij}\\circ pr_2^*(\\alpha_j)=pr_1^*(\\alpha_i)\\circ \\phi_{ij}\\), then, there exists unique arrow \\(\\alpha:\\xi\\rightarrow \\eta\\) such that \\(\\sigma_i^*(\\alpha)=\\alpha_i\\) and \\(\\sigma_i^*(\\alpha:\\xi\\rightarrow \\eta)=\\alpha_i:\\xi_i\\rightarrow \\eta_i\\). This condition is called Gluing morphisms. To summarise this, I will write down the definition. Definition : A category fibered in groupoids \\(F:\\mathcal{D}\\rightarrow \\mathcal{C}\\) is said to be a stack over \\(\\mathcal{C}\\) if, for each object \\(U\\) and a cover \\(\\{U_i\\rightarrow U\\}\\) the following conditions holds. Gluing objects : Given \\(\\xi_i \\in \\mathcal{D}(U_i)\\) with compatibility, there exists an object \\(\\xi\\in \\mathcal{D}(U)\\) such that there is an isomorphism \\(\\sigma_i^*(x)\\rightarrow x_i\\) in \\(\\mathcal{D}(U_i)\\) for each \\(i\\). By compatibility, we mean that there exists isomorphisms \\(\\phi_{ij}:pr_2^*(\\xi_j)\\rightarrow pr_1^*(\\xi_i)\\) in \\(\\mathcal{D}(U_{ij})\\) such that \\(\\phi_{ik}=\\phi_{ij}\\circ \\phi_{jk}\\) on \\(U_{ijk}\\). Gluing morphisms : Given \\(\\xi,\\eta\\in \\mathcal{D}(U)\\) and a morphism \\(\\alpha_i:\\sigma_i^*(\\xi)=\\xi_i\\rightarrow \\eta_i=\\sigma_i^*(\\eta)\\) in \\(\\mathcal{D}(U_i)\\) for each \\(i\\) and isomorphisms \\(\\phi_{ij}:pr_2^*(\\xi_j)\\rightarrow pr_1^*(\\xi_i), \\psi_{ij}:pr_2^*(\\eta_j)\\rightarrow pr_1^*(\\eta_i)\\) in \\(\\mathcal{D}(U_{ij})\\) such that \\(\\psi_{ij}\\circ pr_2^*(\\alpha_j)=pr_1^*(\\alpha_i)\\circ \\phi_{ij}\\), there exists unique arrow \\(\\alpha:\\xi\\rightarrow \\eta\\) such that \\(\\sigma_i^*(\\alpha:\\xi\\rightarrow \\eta)=\\alpha_i:\\xi_i\\rightarrow \\eta_i\\). \u0026nbsp;","permalink":"https://praphulla-koushik.github.io/2019/01/12/what-is-a-stack/","summary":"Given a manifold \\(M\\) we have the concept of open cover of \\(M\\). We usually write an open cover of a manifold \\(M\\) \u003cstrong\u003eas a collection of open subsets\u003c/strong\u003e \\(\\{U_i\\}\\) (such that \\(\\bigcup U_i=M\\)). In this note  we see  an open cover of \\(M\\) \u003cstrong\u003eas a collection of maps\u003c/strong\u003e (inclusions)  \\(\\{U_i\\rightarrow M\\}\\).   Some properties of \"open cover\" are.\n\u003col\u003e\n\t\u003cli\u003e(Pull back exists and gives an open cover) Suppose \\(\\{U_i\\rightarrow M\\}\\) is an open cover for \\(M\\) and \\(\\pi:V\\rightarrow M\\) is a smooth map. Then, \\(\\{\\pi^{-1}(U_i) \\rightarrow V\\}\\) is a cover for \\(V\\).\u003c/li\u003e\n\t\u003cli\u003e(Diffeomorphisms gives open cover) For any manifold \\(M\\), \\(M\\) itself is considered as an open cover \\(\\{M\\rightarrow M\\}\\). More generally, for any diffeomorphism \\(M'\\rightarrow M\\), \\(\\{M'\\rightarrow M\\}\\) is considered as an open cover.\u003c/li\u003e\n\t\u003cli\u003e(Open cover of open cover is an open cover) Let \\(\\{U_\\alpha\\rightarrow U\\}\\) be an open cover for \\(U\\) i.e., \\(\\bigcup_{\\alpha} U_\\alpha=U\\). Suppose \\(\\{V_{\\alpha\\beta}\\rightarrow U_\\alpha\\}\\) is an open cover for \\(U_\\alpha\\) for each \\(\\alpha\\) i.e., \\(\\bigcup_{\\beta}V_{\\alpha\\beta}=U_\\alpha\\). Then, \\(\\bigcup_{\\alpha\\beta}V_{\\alpha\\beta}=U\\) i.e., \\(\\{V_{\\alpha\\beta}\\rightarrow U\\}\\) is an open cover for \\(U\\).\u003c/li\u003e\n\u003c/ol\u003e\nFor a category \\(\\mathcal{C}\\) and an object \\(U\\) of  \\(\\mathcal{C}\\), a collection of arrows \\(\\{U_i\\rightarrow U\\}\\) is said to be a cover for \\(U\\).\n\n\u003cstrong\u003eDefinition\u003c/strong\u003e : Let \\(\\mathcal{C}\\) be a category. A Grothendieck topology on \\(\\mathcal{C}\\) is given by a  collection of covers \\(\\mathcal{W}=\\{\\{U_i\\rightarrow U\\}: U\\in \\mathcal{C}_0\\}\\) satisfying following conditions.\n\u003col\u003e\n\t\u003cli\u003e(Pullbacks exists and gives a cover) Suppose  \\(\\{U_i\\rightarrow U\\}\\in \\mathcal{W}\\)  and \\(\\pi:V\\rightarrow U\\) be an arrow. Then, the pull back \\(U_i\\times_UV\\) exists (as an object in \\(\\mathcal{C}\\)) and  \\(\\{U_i\\times_UV \\rightarrow V\\}\\) is a cover for \\(V\\).\u003c/li\u003e\n\t\u003cli\u003e(Isomorphisms  gives an open cover) Suppose \\(V\\in \\mathcal{C}_0\\) and \\(V\\rightarrow U\\) is an isomorphism in \\(\\mathcal{C}\\) then, \\(\\{V\\rightarrow U\\}\\in \\mathcal{W}\\).\u003c/li\u003e\n\t\u003cli\u003e(cover of a cover is a cover) Suppose \\(\\{U_\\alpha\\rightarrow U\\}\\in \\mathcal{W}\\) and \\(\\{U_{\\alpha\\beta}\\rightarrow U_\\alpha\\}\\in \\mathcal{W}\\) for each \\(\\alpha\\). Then, the collection of compositions \\(\\{U_{\\alpha\\beta}\\rightarrow U_\\alpha\\rightarrow U\\}\\in \\mathcal{W}\\).\u003c/li\u003e\n\u003c/ol\u003e\nTo talk about a stack over category \\(\\mathcal{C}\\) we fix a Grothendieck topology \\(\\mathcal{W}\\) on \\(\\mathcal{C}\\). When we say cover, we mean it belongs to \\(\\mathcal{W}\\).\n\nLet \\(\\mathcal{D}\\) be a category fibered in groupoids over \\(\\mathcal{C}\\) i.e., we have a functor \\(F:\\mathcal{D}\\rightarrow \\mathcal{C}\\) satisfying some conditions.\n\u003col\u003e\n\t\u003cli\u003eGiven an object \\(U\\) of \\(\\mathcal{C}\\) we have what is called \u003cstrong\u003efibre of \\(U\\)\u003c/strong\u003e in \\(\\mathcal{D}\\) usually denoted by \\(\\mathcal{D}(U)\\).\u003c/li\u003e\n\t\u003cli\u003eGiven an object \\(U\\) of \\(\\mathcal{C}\\) and a cover \\(\\{U_i\\rightarrow U\\}\\) (i.e., it belongs to \\(\\mathcal{W}\\)) we have what is called \u003cstrong\u003edescent category associated to the cover \\(\\{U_i\\rightarrow U\\}\\),\u003c/strong\u003e usually denoted by \\(\\mathcal{D}(\\{U_i\\rightarrow U\\})\\).\u003c/li\u003e\n\u003c/ol\u003e\nThere is an obvious functor \\(\\mathcal{D}(U)\\rightarrow \\mathcal{D}(\\{U_i\\rightarrow U\\})\\).\n\u003ch4\u003eDefinition : Let \\(\\mathcal{C}\\) be a category with Grothendieck topology \\(\\mathcal{W}\\). A category fibered in groupoids \\(\\mathcal{D}\\rightarrow \\mathcal{C}\\) is said to be \u003cem\u003ea stack over \\(\\mathcal{C}\\) \u003c/em\u003eif, for every object \\(U\\) of \\(\\mathcal{C}\\) and every cover \\(\\{U_i\\rightarrow U\\}\\), the functor \\(\\mathcal{D}(U)\\rightarrow \\mathcal{D}(\\{U_i\\rightarrow U\\})\\) is an equivalence of categories.\u003c/h4\u003e\nThe fibre categroy \\(\\mathcal{D}(U)\\) is a category whose objects are that of \\(\\mathcal{D}\\) which map to \\(U\\) under \\(F\\) i.e.,\n\u003cp style=\"text-align:center;\"\u003e\\(\\mathcal{D}(U)_0=\\{V\\in \\mathcal{D}_0:F(V)=U\\}\\).\u003c/p\u003e","tags":["stacks"],"title":"What is a Stack?","type":"wp-import"},{"categories":["stacks"],"content":"Here, I will add links to WordPress pages where I have written something about Stacks. Papers I am reading are Differentiable Stacks and Gerbes by Kai Behrend and Ping Xu. Orbifolds as Stacks by Eugene Lerman. Non abelian Differentiable Gerbes by Camille, Stienon and Ping Xu. -- -- \u0026nbsp;","permalink":"https://praphulla-koushik.github.io/2019/01/03/stacks/","summary":"Here, I will add links to WordPress pages where I have written something about Stacks.\n\nPapers I am reading are\n\u003col\u003e\n\t\u003cli\u003eDifferentiable Stacks and Gerbes by Kai Behrend and  Ping Xu.\u003c/li\u003e\n\t\u003cli\u003eOrbifolds as Stacks by Eugene Lerman.\u003c/li\u003e\n\t\u003cli\u003eNon abelian Differentiable Gerbes by Camille, Stienon and Ping Xu.\u003c/li\u003e\n\t\u003cli\u003e--\u003c/li\u003e\n\t\u003cli\u003e--\u003c/li\u003e\n\u003c/ol\u003e\n\u0026nbsp;","tags":["stacks"],"title":"Stacks","type":"wp-import"},{"categories":["stacks"],"content":"Definition : A stack \\(\\mathcal{D}\\rightarrow \\text{Man}\\) is differentiable if there exists a manifold \\(X\\) with an atlas \\(p:\\underline{X}\\rightarrow \\mathcal{D}\\) i.e., \\(p\\) is representable surjective submersion. We see a criterion for a map \\(p:\\underline{X}\\rightarrow \\mathcal{D}\\) to be an atlas. By \\(p:\\underline{X}\\rightarrow \\mathcal{D}\\) to be representable surjective submersion, we mean given a map of stacks \\(\\underline{Y}\\rightarrow \\mathcal{D}\\) the fibered product \\(\\underline{X}\\times_{\\mathcal{D}}\\underline{Y}\\) is representable by a manifold and that the map of manifolds \\(\\underline{X}\\times_{\\mathcal{D}}\\underline{Y}\\rightarrow \\underline{Y}\\) is a surjective submersion. As \\(\\underline{X}\\times_{\\mathcal{D}}\\underline{Y}\\) is representable by a manifold for any map of stacks \\(\\underline{Y}\\rightarrow \\mathcal{D}\\), in particular, taking \\(\\underline{Y}\\rightarrow \\mathcal{D}\\) to be the same map \\(\\underline{X}\\rightarrow \\mathcal{D}\\) we see that, in particular \\(\\underline{X}\\times_{\\mathcal{D}}\\underline{X}\\) is representable by a manifold. Remark : If \\(p:\\underline{X}\\rightarrow \\mathcal{D}\\) is an atlas for \\(\\mathcal{D}\\) then \\(\\underline{X}\\times_{\\mathcal{D}}\\underline{X}\\) is representable by a manifold. As \\(\\underline{X}\\times_{\\mathcal{D}}\\underline{Y}\\rightarrow \\underline{Y}\\) is a submersion for any map of stacks \\(\\underline{Y}\\rightarrow \\mathcal{D}\\), in particular, taking \\(\\underline{Y}\\rightarrow \\mathcal{D}\\) to be the same map \\(\\underline{X}\\rightarrow \\mathcal{D}\\) we see that, projecion map \\(\\underline{X}\\times_{\\mathcal{D}}\\underline{X}\\rightarrow \\underline{X}\\) is a submersion. It is not relevant which projection is it as both maps are same. So, both projection maps \\(pr_1:\\underline{X}\\times_{\\mathcal{D}}\\underline{X}\\rightarrow \\underline{X}\\) and \\(pr_2:\\underline{X}\\times_{\\mathcal{D}}\\underline{X}\\rightarrow \\underline{X}\\) are submersions. Remark : If \\(p:\\underline{X}\\rightarrow \\mathcal{D}\\) is an atlas for stack \\(\\mathcal{D}\\) then projection maps \\(pr_1:\\underline{X}\\times_{\\mathcal{D}}\\underline{X}\\rightarrow \\underline{X}\\) and \\(pr_2:\\underline{X}\\times_{\\mathcal{D}}\\underline{X}\\rightarrow \\underline{X}\\) are submersions. As any representable surjective submersion is an epimorphism we have following remark. Remark : If \\(p:\\underline{X}\\rightarrow \\mathcal{D}\\) is an atlas for \\(\\mathcal{D}\\) then \\(p:\\underline{X}\\rightarrow \\mathcal{D}\\) is an epimorphism. Combining all these remarks we have following remark. If \\(p:\\underline{X}\\rightarrow \\mathcal{D}\\) is an atlas for \\(\\mathcal{D}\\) then, \\(\\underline{X}\\times_{\\mathcal{D}}\\underline{X}\\) is representable by a manifold and projection maps \\(pr_1:\\underline{X}\\times_{\\mathcal{D}}\\underline{X}\\rightarrow \\underline{X}\\) and \\(pr_2:\\underline{X}\\times_{\\mathcal{D}}\\underline{X}\\rightarrow \\underline{X}\\) are submersions and \\(p:\\underline{X}\\rightarrow \\mathcal{D}\\) is an epimorphism. It turns out that converse of above remark is true. Proposition : Let \\(p:\\underline{X}\\rightarrow \\mathcal{D}\\) is a morphism of stacks such that \\(\\underline{X}\\times_{\\mathcal{D}}\\underline{X}\\) is representable by a manifold and projection maps \\(pr_1:\\underline{X}\\times_{\\mathcal{D}}\\underline{X}\\rightarrow \\underline{X}\\) and \\(pr_2:\\underline{X}\\times_{\\mathcal{D}}\\underline{X}\\rightarrow \\underline{X}\\) are submersions and that \\(p:\\underline{X}\\rightarrow \\mathcal{D}\\) is an epimorphism. Then, Then, \\(p:\\underline{X}\\rightarrow \\mathcal{D}\\) is a representable surjective submersion i.e., an atlas for \\(\\mathcal{D}\\). Before we give proof of this, we recall a result. Lemma : Let \\(\\mathcal{D}\\rightarrow\\mathcal{C}\\) be a morphism of stacks. Suppose \\(U\\) be a manifold and \\(\\underline{U}\\rightarrow \\mathcal{C}\\) is an epimorphism of stacks such that fiber product \\(\\mathcal{D}\\times_{\\mathcal{C}}\\underline{U}\\) is represented by a manifold and the map of manifolds \\(\\mathcal{D}\\times_{\\mathcal{C}}\\underline{U}\\rightarrow U\\) is a submersion. Then, \\(\\mathcal{D}\\rightarrow \\mathcal{C}\\) is a representable submersion. To prove \\(\\underline{X}\\rightarrow \\mathcal{D}\\) is a representable submersion, consider an epimorphism of stacks, namely \\(\\underline{X}\\rightarrow \\mathcal{D}\\) (it is given to be an epimorphism, condition \\(2\\) above). See that the fibre product \\(\\underline{X}\\times_{\\mathcal{D}}\\underline{X}\\) is representable by a manifold (it is given in condition \\(1\\) above) and that the projection map \\(\\underline{X}\\times_{\\mathcal{D}}\\underline{X}\\rightarrow X\\) is a submersion (it is in condition \\(1\\) above). Thus, by above lemma, \\(p:\\underline{X}\\rightarrow \\mathcal{D}\\) is a representable submersion. Note that, a representable submersion that is an epimorphism is a representable surjective submersion. Thus, \\(p:\\underline{X}\\rightarrow \\mathcal{D}\\) is an atlas for \\(\\mathcal{D}\\). So, we have the following result. Proposition : Let \\(p:\\underline{X}\\rightarrow \\mathcal{D}\\) is a morphism of stacks such that \\(\\underline{X}\\times_{\\mathcal{D}}\\underline{X}\\) is representable by a manifold and projection maps \\(pr_1:\\underline{X}\\times_{\\mathcal{D}}\\underline{X}\\rightarrow \\underline{X}\\) and \\(pr_2:\\underline{X}\\times_{\\mathcal{D}}\\underline{X}\\rightarrow \\underline{X}\\) are submersions and that \\(p:\\underline{X}\\rightarrow \\mathcal{D}\\) is an epimorphism. Then, Then, \\(p:\\underline{X}\\rightarrow \\mathcal{D}\\) is a representable surjective submersion i.e., an atlas for \\(\\mathcal{D}\\).","permalink":"https://praphulla-koushik.github.io/2019/01/03/criterion-for-a-map-of-stacks-to-be-an-atlas/","summary":"\u003cblockquote\u003eDefinition : A stack \\(\\mathcal{D}\\rightarrow \\text{Man}\\) is differentiable if there exists a manifold \\(X\\) with an atlas \\(p:\\underline{X}\\rightarrow \\mathcal{D}\\) i.e., \\(p\\) is representable surjective submersion.\u003c/blockquote\u003e\nWe see a criterion for a map \\(p:\\underline{X}\\rightarrow \\mathcal{D}\\) to be an atlas.\n\nBy \\(p:\\underline{X}\\rightarrow \\mathcal{D}\\) to be representable surjective submersion, we mean given a map of stacks \\(\\underline{Y}\\rightarrow \\mathcal{D}\\) the fibered product \\(\\underline{X}\\times_{\\mathcal{D}}\\underline{Y}\\) is \u003cstrong\u003erepresentable by a manifold\u003c/strong\u003e and that the map of manifolds\n\\(\\underline{X}\\times_{\\mathcal{D}}\\underline{Y}\\rightarrow \\underline{Y}\\) is a surjective submersion.\n\nAs \\(\\underline{X}\\times_{\\mathcal{D}}\\underline{Y}\\) is representable by a manifold for any map of stacks \\(\\underline{Y}\\rightarrow \\mathcal{D}\\), in particular, taking \\(\\underline{Y}\\rightarrow \\mathcal{D}\\) to be the same map \\(\\underline{X}\\rightarrow \\mathcal{D}\\) we see that, in particular \\(\\underline{X}\\times_{\\mathcal{D}}\\underline{X}\\) is \u003cstrong\u003erepresentable by a manifold\u003c/strong\u003e.\n\u003cblockquote\u003eRemark : If \\(p:\\underline{X}\\rightarrow \\mathcal{D}\\) is an atlas for \\(\\mathcal{D}\\) then \\(\\underline{X}\\times_{\\mathcal{D}}\\underline{X}\\) is representable by a manifold.\u003c/blockquote\u003e\nAs \\(\\underline{X}\\times_{\\mathcal{D}}\\underline{Y}\\rightarrow \\underline{Y}\\) is a submersion for any map of stacks \\(\\underline{Y}\\rightarrow \\mathcal{D}\\), in particular, taking \\(\\underline{Y}\\rightarrow \\mathcal{D}\\) to be the same map \\(\\underline{X}\\rightarrow \\mathcal{D}\\) we see that, projecion map \\(\\underline{X}\\times_{\\mathcal{D}}\\underline{X}\\rightarrow \\underline{X}\\) is a submersion. It is not relevant which projection is it as both maps are same. So, both projection maps \\(pr_1:\\underline{X}\\times_{\\mathcal{D}}\\underline{X}\\rightarrow \\underline{X}\\) and \\(pr_2:\\underline{X}\\times_{\\mathcal{D}}\\underline{X}\\rightarrow \\underline{X}\\) are submersions.\n\u003cblockquote\u003eRemark : If \\(p:\\underline{X}\\rightarrow \\mathcal{D}\\) is an atlas for stack \\(\\mathcal{D}\\) then projection maps \\(pr_1:\\underline{X}\\times_{\\mathcal{D}}\\underline{X}\\rightarrow \\underline{X}\\) and \\(pr_2:\\underline{X}\\times_{\\mathcal{D}}\\underline{X}\\rightarrow \\underline{X}\\) are submersions.\u003c/blockquote\u003e\nAs any representable surjective submersion is an epimorphism we have following remark.\n\u003cblockquote\u003eRemark : If \\(p:\\underline{X}\\rightarrow \\mathcal{D}\\) is an atlas for \\(\\mathcal{D}\\) then \\(p:\\underline{X}\\rightarrow \\mathcal{D}\\) is an epimorphism.\u003c/blockquote\u003e\nCombining all these remarks we have following remark.\n\u003cblockquote\u003e\u003cspan style=\"color:#3d596d;background-color:#ffffff;\"\u003eIf \\(p:\\underline{X}\\rightarrow \\mathcal{D}\\) is an atlas for \\(\\mathcal{D}\\) then, \u003c/span\u003e\\(\\underline{X}\\times_{\\mathcal{D}}\\underline{X}\\) is representable by a manifold and projection maps \\(pr_1:\\underline{X}\\times_{\\mathcal{D}}\\underline{X}\\rightarrow \\underline{X}\\) and \\(pr_2:\\underline{X}\\times_{\\mathcal{D}}\\underline{X}\\rightarrow \\underline{X}\\) are submersions and \\(p:\\underline{X}\\rightarrow \\mathcal{D}\\) is an epimorphism.\u003c/blockquote\u003e\nIt turns out that converse of above remark is true.\n\u003cblockquote\u003eProposition : Let \\(p:\\underline{X}\\rightarrow \\mathcal{D}\\) is a morphism of stacks such that \\(\\underline{X}\\times_{\\mathcal{D}}\\underline{X}\\) is representable by a manifold and projection maps \\(pr_1:\\underline{X}\\times_{\\mathcal{D}}\\underline{X}\\rightarrow \\underline{X}\\) and \\(pr_2:\\underline{X}\\times_{\\mathcal{D}}\\underline{X}\\rightarrow \\underline{X}\\) are submersions and that  \\(p:\\underline{X}\\rightarrow \\mathcal{D}\\) is an epimorphism. Then, Then, \\(p:\\underline{X}\\rightarrow \\mathcal{D}\\) is a representable surjective submersion i.e., an atlas for \\(\\mathcal{D}\\).\u003c/blockquote\u003e\nBefore we give proof of this, we recall a \u003ca href=\"https://koushik1729.wordpress.com/2018/12/31/criterion-for-a-map-to-be-representable-submersion/\"\u003eresult\u003c/a\u003e.\n\u003cblockquote\u003eLemma : Let \\(\\mathcal{D}\\rightarrow\\mathcal{C}\\) be a morphism of stacks. Suppose \\(U\\) be a manifold and \\(\\underline{U}\\rightarrow \\mathcal{C}\\) is an epimorphism of stacks such that fiber product \\(\\mathcal{D}\\times_{\\mathcal{C}}\\underline{U}\\) is represented by a manifold and the map of manifolds \\(\\mathcal{D}\\times_{\\mathcal{C}}\\underline{U}\\rightarrow U\\) is a submersion. Then, \\(\\mathcal{D}\\rightarrow \\mathcal{C}\\) is a representable submersion.\u003c/blockquote\u003e\nTo prove \\(\\underline{X}\\rightarrow \\mathcal{D}\\) is a representable submersion, consider an epimorphism of stacks, namely \\(\\underline{X}\\rightarrow \\mathcal{D}\\) (it is given to be an epimorphism, condition \\(2\\) above). See that the fibre product \\(\\underline{X}\\times_{\\mathcal{D}}\\underline{X}\\) is representable by a manifold (it is given in condition \\(1\\) above) and that the projection map \\(\\underline{X}\\times_{\\mathcal{D}}\\underline{X}\\rightarrow X\\) is a submersion (it is in condition \\(1\\) above). Thus, by above lemma, \\(p:\\underline{X}\\rightarrow \\mathcal{D}\\) is a representable submersion. Note that, \u003cstrong\u003ea representable submersion that is an epimorphism is a representable surjective submersion.\u003c/strong\u003e  Thus, \\(p:\\underline{X}\\rightarrow \\mathcal{D}\\) is an atlas for \\(\\mathcal{D}\\).\n\nSo, we have the following result.\n\u003cblockquote\u003eProposition : Let \\(p:\\underline{X}\\rightarrow \\mathcal{D}\\) is a morphism of stacks such that \\(\\underline{X}\\times_{\\mathcal{D}}\\underline{X}\\) is representable by a manifold and projection maps \\(pr_1:\\underline{X}\\times_{\\mathcal{D}}\\underline{X}\\rightarrow \\underline{X}\\) and \\(pr_2:\\underline{X}\\times_{\\mathcal{D}}\\underline{X}\\rightarrow \\underline{X}\\) are submersions and that  \\(p:\\underline{X}\\rightarrow \\mathcal{D}\\) is an epimorphism. Then, Then, \\(p:\\underline{X}\\rightarrow \\mathcal{D}\\) is a representable surjective submersion i.e., an atlas for \\(\\mathcal{D}\\).\u003c/blockquote\u003e","tags":["stacks"],"title":"Criterion for a map of stacks to be an atlas","type":"wp-import"},{"categories":["Stacks"],"content":"","permalink":"https://praphulla-koushik.github.io/2019/01/02/criterion-for-a-stack-to-be-representable/","summary":"","tags":[],"title":"Criterion for a stack to be representable","type":"wp-import"},{"categories":["stacks","Algebraic geometry"],"content":"A morphism of stacks \\(f:\\mathcal{D}\\rightarrow \\mathcal{C}\\) is called a representable submersion if, for every morphism \\(\\underline{M}\\rightarrow \\mathcal{C}\\), the fibred product \\(\\mathcal{D}\\times_{\\mathcal{C}}\\underline{M}\\) is representable i.e., isomorphic to a stack coming from a manifold and that the map of manifolds \\(\\mathcal{D}\\times_{\\mathcal{C}}\\underline{M}\\rightarrow M\\) is a submersion. Following is a criterion for a map of stacks to be representable submersion. The result says it is enough to check for epimorphisms \\(\\underline{M}\\rightarrow \\mathcal{C}\\). Precise statement is as follows. Let \\(f:\\mathcal{D}\\rightarrow \\mathcal{C}\\) be a morphism of stacks. Suppose given a manifold \\(U\\) and a morphism of stacks \\(\\underline{U}\\rightarrow \\mathcal{C}\\) which is an epimorphism. If the fibered product \\(\\mathcal{D}\\times_{\\mathcal{C}}\\underline{U}\\) is representable i.e., isomorphic to a stack coming from a manifold and that the map of manifolds \\(\\mathcal{D}\\times_{\\mathcal{C}}\\underline{U}\\rightarrow U\\) is a submersion, then \\(f\\) is representable submersion. Let us see what this means in the set up of manifolds. Let \\(f:\\mathcal{D}\\rightarrow \\mathcal{C}\\) be a morphism of stacks. Let \\(f:M\\rightarrow N\\) be a morphism of manifolds (which gives a morphism of stacks \\(\\underline{M}\\rightarrow \\underline{N}\\)). Suppose given a manifold \\(U\\) and a morphism of stacks \\(\\underline{U}\\rightarrow \\mathcal{C}\\) which is an epimorphism. A representable surjective submersion is an epimorphism. So, we consider a surjective submersion \\(g:U\\rightarrow N\\) (which gives an epimorphism \\(\\underline{U}\\rightarrow \\underline{N}\\) being a representable surjective submersion). If the fibered product \\(\\mathcal{D}\\times_{\\mathcal{C}}\\underline{U}\\) is representable i.e., isomorphic to a stack coming from a manifold and that the map of manifolds \\(\\mathcal{D}\\times_{\\mathcal{C}}\\underline{U}\\rightarrow U\\) is a submersion. As \\(U\\rightarrow N\\) is submersion, it is anyways true that \\(M\\times_N U\\) is a smooth manifold. What is extra that we have here is that \\(M\\times_NU\\rightarrow U\\) is a submersion. It is anyways true that \\(M\\times_NU\\rightarrow M\\) is a submersion being a pullback of submersion. But it is not true in general that \\(M\\times_NU\\rightarrow U\\) is a submersion. Here, we are given that \\(M\\times_NU\\rightarrow U\\) is a submersion. So, Let \\(f:\\mathcal{D}\\rightarrow \\mathcal{C}\\) be a morphism of stacks. Suppose given a manifold \\(U\\) and a morphism of stacks \\(\\underline{U}\\rightarrow \\mathcal{C}\\) which is an epimorphism. If the fibered product \\(\\mathcal{D}\\times_{\\mathcal{C}}\\underline{U}\\) is representable i.e., isomorphic to a stack coming from a manifold and that the map of manifolds \\(\\mathcal{D}\\times_{\\mathcal{C}}\\underline{U}\\rightarrow U\\) is a submersion, then \\(f\\) is representable submersion. turns to Let \\(f:M\\rightarrow N\\) be a morphism of manifolds be a morphism of manifolds. Suppose given a manifold \\(U\\) and a sujective submersion \\(\\underline{U}\\rightarrow N\\). If the fibered product \\(M\\times_{N}U\\) is a manifold and that the map of manifolds \\(M\\times_{N}U\\rightarrow U\\) is a submersion, then \\(f\\) is submersion. This is more or less obvious. We have following commutative diagram As \\(G:U\\rightarrow N\\) is a submersion (we have started with this) and \\(p_2: M\\times_N U\\rightarrow U\\) is a submersion (we are given this), the composition \\(G\\circ p_2=F\\circ p_1\\) is a submersion which then imply that \\(F:M\\rightarrow N\\) is a submersion. Let \\(m\\in M\\). As \\(G:U\\rightarrow N\\) is surjective, so is \\(p_1\\) (pullback of surjective is surjective) i.e., there exists \\((m,u)\\in M\\times_N U\\) such that \\(p_1(m,u)=m\\). As \\(F\\circ p_1\\) is submersion, \\((F\\circ p_1)_{*,(m,u)}(T_{m,u}(M\\times_N U))=T_{F(m)}N\\). Applying chain rule, we have \\(F_{*,m}((p_1)_{*,(m,u)}(T_{m,u}(M\\times_N U)))=T_{F(m)}N\\), in particular, \\(F_{*,m}(T_mM)=T_{F(m)}N\\). Thus, \\(F\\) is submersion. So, we need both surjectivity and submersion of \\(U\\rightarrow N\\). Now, let us look at more general case. Now, \\(U\\rightarrow N\\) is not a surjective submersion but induces an epimorphism \\(U\\rightarrow N\\). Suppose that the pullback \\(M\\times_N U\\) is a manifold and that the map \\(M\\times_N U\\rightarrow U\\) is a submerson. Let \\(W\\rightarrow N\\) be a map. We need to prove that \\(M\\times_N W\\) is a manifold. We have following diagram As \\(M\\times_N U\\rightarrow U\\) is a submersion, the pullback \\((M\\times_N U)\\times_U W_i=M\\times_N W_i\\) is a manifold. So, we have an open cover \\(\\{W_i\\rightarrow W\\}\\) of \\(W\\) such that the pullbacks \\(M\\times_N W_i\\) are manifolds. I think this should confirm that \\(M\\times_N W\\) is a manifold and just because \\(M\\times_N W\\rightarrow W_i\\) are submersions, so is the map \\(M\\times_N W\\rightarrow W\\). Thus, \\(f:M\\rightarrow N\\) is a representable submersion. The same idea works for an arbitrary map of stacks \\(\\mathcal{D}\\rightarrow \\mathcal{C}\\). Let \\(\\underline{W}\\rightarrow \\mathcal{C}\\) be a map of stacks. We have to prove that \\(\\mathcal{D}\\times_{\\mathcal{C}}\\underline{W}\\) is representable and that the map of manifolds \\(\\mathcal{D}\\times_{\\mathcal{W}}\\underline{W}\\rightarrow W\\) is a submersion. As \\(\\underline{U}\\rightarrow \\mathcal{C}\\) is epimorphism, for \\(\\underline{W}\\rightarrow \\mathcal{C}\\) there exists an open cover \\(\\{W_i\\rightarrow W\\}\\) with commutative diagram as shown below. We have following diagram As \\(\\mathcal{D}\\times_{\\mathcal{C}} \\underline{U}\\rightarrow U\\) is a submersion, the pullback \\((\\mathcal{D}\\times_{\\mathcal{C}} \\underline{U})\\times_U W_i=\\mathcal{D}\\times_{\\mathcal{C}} W_i\\) is a manifold. So, we have an open cover \\(\\{W_i\\rightarrow W\\}\\) of $ W$ such that the \\((\\mathcal{D}\\times_{\\mathcal{C}}W)\\times_W W_i=\\mathcal{D}\\times_{\\mathcal{C}} W_i\\)\nare manifolds. By this question it follows that \\(\\mathcal{D}\\times_{\\mathcal{C}}\\underline{W}\\) is a manifold. For similar reason as mentioned in question, the map of manifolds \\(\\mathcal{D}\\times_{\\mathcal{C}}\\underline{W}\\rightarrow W\\) is a submersion. Thus, \\(\\mathcal{D}\\rightarrow \\mathcal{C}\\) is a representable submersion. Thus, we have following criterion for a map to be a representable submersion. Let \\(f:\\mathcal{D}\\rightarrow \\mathcal{C}\\) be a morphism of stacks. Suppose given a manifold \\(U\\) and a morphism of stacks \\(\\underline{U}\\rightarrow \\mathcal{C}\\) which is an epimorphism. If the fibered product \\(\\mathcal{D}\\times_{\\mathcal{C}}\\underline{U}\\) is representable i.e., isomorphic to a stack coming from a manifold and that the map of manifolds \\(\\mathcal{D}\\times_{\\mathcal{C}}\\underline{U}\\rightarrow U\\) is a submersion, then \\(f\\) is representable submersion.","permalink":"https://praphulla-koushik.github.io/2018/12/31/criterion-for-a-map-to-be-representable-submersion/","summary":"A morphism of stacks \\(f:\\mathcal{D}\\rightarrow \\mathcal{C}\\) is called a  \u003cspan style=\"text-decoration:underline;\"\u003e\u003cem\u003erepresentable  submersion\u003c/em\u003e\u003c/span\u003e if, \u003cstrong\u003efor every morphism \u003c/strong\u003e\\(\\underline{M}\\rightarrow \\mathcal{C}\\), the fibred product \\(\\mathcal{D}\\times_{\\mathcal{C}}\\underline{M}\\) is representable i.e., isomorphic to a stack coming from a manifold and that the map of \u003cstrong\u003emanifolds \u003c/strong\u003e\\(\\mathcal{D}\\times_{\\mathcal{C}}\\underline{M}\\rightarrow M\\) is a submersion.\n\nFollowing is a criterion for a map of stacks to be representable submersion. The result says it is enough to check for epimorphisms \\(\\underline{M}\\rightarrow \\mathcal{C}\\). Precise statement is as follows.\n\u003cblockquote\u003e Let \\(f:\\mathcal{D}\\rightarrow \\mathcal{C}\\) be a morphism of stacks. Suppose given a manifold \\(U\\) and a morphism of stacks \\(\\underline{U}\\rightarrow \\mathcal{C}\\) which is an \u003cstrong\u003eepimorphism. \u003c/strong\u003eIf the  fibered product \\(\\mathcal{D}\\times_{\\mathcal{C}}\\underline{U}\\) is representable i.e., isomorphic to a stack coming from a manifold and that the map of \u003cstrong\u003emanifolds \u003c/strong\u003e\\(\\mathcal{D}\\times_{\\mathcal{C}}\\underline{U}\\rightarrow U\\) is a submersion, then \\(f\\) is representable submersion.\u003c/blockquote\u003e\nLet us see what this means in the set up of manifolds.\n\n\u003cstrong\u003e Let \\(f:\\mathcal{D}\\rightarrow \\mathcal{C}\\) be a morphism of stacks. \u003c/strong\u003e\n\nLet \\(f:M\\rightarrow N\\) be a morphism of manifolds (which gives a morphism of stacks \\(\\underline{M}\\rightarrow \\underline{N}\\)).\n\n\u003cstrong\u003eSuppose given a manifold \\(U\\) and a morphism of stacks \\(\\underline{U}\\rightarrow \\mathcal{C}\\) which is an epimorphism.\u003c/strong\u003e\n\nA representable surjective submersion is an epimorphism. So,  we consider a surjective submersion \\(g:U\\rightarrow N\\) (which gives an epimorphism  \\(\\underline{U}\\rightarrow \\underline{N}\\) being a representable surjective submersion).\n\n\u003cstrong\u003eIf the  fibered product \\(\\mathcal{D}\\times_{\\mathcal{C}}\\underline{U}\\) is representable i.e., isomorphic to a stack coming from a manifold and that the map of manifolds \\(\\mathcal{D}\\times_{\\mathcal{C}}\\underline{U}\\rightarrow U\\) is a submersion. \u003c/strong\u003e\n\nAs \\(U\\rightarrow N\\) is submersion, it is anyways true that \\(M\\times_N U\\) is a smooth manifold. What is \u003cstrong\u003eextra that we have here\u003c/strong\u003e is that \\(M\\times_NU\\rightarrow U\\) is a submersion. It is anyways true that \\(M\\times_NU\\rightarrow M\\) is a submersion being a pullback of submersion. But it is not true in general that \\(M\\times_NU\\rightarrow U\\) is a submersion. Here, we are given that \\(M\\times_NU\\rightarrow U\\) is a submersion.\n\nSo,\n\u003cblockquote\u003e Let \\(f:\\mathcal{D}\\rightarrow \\mathcal{C}\\) be a morphism of stacks. Suppose given a manifold \\(U\\) and a morphism of stacks \\(\\underline{U}\\rightarrow \\mathcal{C}\\) which is an \u003cstrong\u003eepimorphism. \u003c/strong\u003eIf the  fibered product \\(\\mathcal{D}\\times_{\\mathcal{C}}\\underline{U}\\) is representable i.e., isomorphic to a stack coming from a manifold and that the map of \u003cstrong\u003emanifolds \u003c/strong\u003e\\(\\mathcal{D}\\times_{\\mathcal{C}}\\underline{U}\\rightarrow U\\) is a submersion, then \\(f\\) is representable submersion.\u003c/blockquote\u003e\nturns to\n\u003cblockquote\u003e Let \\(f:M\\rightarrow N\\) be a morphism of manifolds be a morphism of manifolds. Suppose given a manifold \\(U\\) and a sujective submersion \\(\\underline{U}\\rightarrow N\\)\u003cstrong\u003e. \u003c/strong\u003eIf the  fibered product \\(M\\times_{N}U\\) is a manifold and that the map of \u003cstrong\u003emanifolds \u003c/strong\u003e\\(M\\times_{N}U\\rightarrow U\\) is a submersion, then \\(f\\) is  submersion.\u003c/blockquote\u003e\nThis is more or less obvious. We have following commutative diagram\n\n\u003cimg class=\"alignnone size-full wp-image-1298\" src=\"/wp-media/2018/12/fada55271c-Screenshot-from-2019-01-02-15-11-38.png\" alt=\"Screenshot from 2019-01-02 15-11-38\" width=\"326\" height=\"213\" /\u003e\n\nAs \\(G:U\\rightarrow N\\) is a submersion (we have started with this) and \\(p_2: M\\times_N U\\rightarrow U\\) is a submersion (we are given this), the composition \\(G\\circ p_2=F\\circ p_1\\) is a submersion which then imply that \\(F:M\\rightarrow N\\) is a submersion.\n\nLet \\(m\\in M\\). As \\(G:U\\rightarrow N\\) is surjective, so is \\(p_1\\) (pullback of surjective is surjective) i.e., there exists \\((m,u)\\in M\\times_N U\\) such that \\(p_1(m,u)=m\\).\n\nAs \\(F\\circ p_1\\) is submersion, \\((F\\circ p_1)_{*,(m,u)}(T_{m,u}(M\\times_N U))=T_{F(m)}N\\). Applying chain rule, we have \\(F_{*,m}((p_1)_{*,(m,u)}(T_{m,u}(M\\times_N U)))=T_{F(m)}N\\), in particular, \\(F_{*,m}(T_mM)=T_{F(m)}N\\). Thus, \\(F\\) is submersion. So, we need both surjectivity and submersion of \\(U\\rightarrow N\\).\n\nNow, let us look at more general case. Now, \\(U\\rightarrow N\\) is not a surjective submersion but induces an epimorphism \\(U\\rightarrow N\\). Suppose that the pullback \\(M\\times_N U\\) is a manifold and that the map \\(M\\times_N U\\rightarrow U\\) is a submerson.\n\nLet \\(W\\rightarrow N\\) be a map. We need to prove that \\(M\\times_N W\\) is a manifold. We have following diagram \u003cimg class=\"alignnone size-full wp-image-1302\" src=\"/wp-media/2018/12/0906ed9c6b-Screenshot-from-2019-01-02-18-11-05.png\" alt=\"Screenshot from 2019-01-02 18-11-05\" width=\"587\" height=\"328\" /\u003e\n\nAs \\(M\\times_N U\\rightarrow U\\) is a submersion, the pullback \\((M\\times_N U)\\times_U W_i=M\\times_N W_i\\) is a manifold. So, we have an open cover \\(\\{W_i\\rightarrow W\\}\\) of \\(W\\) such that the pullbacks \\(M\\times_N W_i\\) are manifolds. I think this should confirm that \\(M\\times_N W\\) is a  manifold and just because \\(M\\times_N W\\rightarrow W_i\\) are submersions, so is the map \\(M\\times_N W\\rightarrow W\\). Thus, \\(f:M\\rightarrow N\\) is a representable submersion.\n\nThe same idea works for an arbitrary map of stacks \\(\\mathcal{D}\\rightarrow \\mathcal{C}\\).\n\nLet \\(\\underline{W}\\rightarrow \\mathcal{C}\\) be a map of stacks. We have to prove that \\(\\mathcal{D}\\times_{\\mathcal{C}}\\underline{W}\\) is representable and that the map of manifolds \\(\\mathcal{D}\\times_{\\mathcal{W}}\\underline{W}\\rightarrow W\\) is a submersion.\n\nAs \\(\\underline{U}\\rightarrow \\mathcal{C}\\) is epimorphism, for \\(\\underline{W}\\rightarrow \\mathcal{C}\\) there exists an open cover \\(\\{W_i\\rightarrow W\\}\\) with commutative diagram as shown below. We have following diagram\n\n\u003cimg class=\"alignnone size-full wp-image-1304\" src=\"/wp-media/2018/12/4003c3ace1-Screenshot-from-2019-01-02-20-24-57.png\" alt=\"Screenshot from 2019-01-02 20-24-57\" width=\"529\" height=\"341\" /\u003e\n\nAs \\(\\mathcal{D}\\times_{\\mathcal{C}} \\underline{U}\\rightarrow U\\) is a submersion, the pullback \\((\\mathcal{D}\\times_{\\mathcal{C}} \\underline{U})\\times_U W_i=\\mathcal{D}\\times_{\\mathcal{C}} W_i\\) is a manifold.\n\nSo, we have an open cover \\(\\{W_i\\rightarrow W\\}\\) of $ W$ such that the\n\u003cp style=\"text-align:center;\"\u003e\n\\((\\mathcal{D}\\times_{\\mathcal{C}}W)\\times_W W_i=\\mathcal{D}\\times_{\\mathcal{C}} W_i\\)\u003c/p\u003e","tags":["sheaves","stacks"],"title":"Criterion for a map to be representable submersion","type":"wp-import"},{"categories":["Linear algebra"],"content":"Let \\(V\\) be a vector space over \\(\\mathbb{R}\\). Let \\(\\{e_1,\\cdots,e_r\\}\\) be a basis of \\(V\\) over \\(\\mathbb{R}\\) and \\(\\{e^1,\\cdots,e^r\\}\\) be the dual basis of \\(V\\). We call \\(e^i:V\\rightarrow \\mathbb{R}\\) to be polynomials over \\(V\\) with values in \\(\\mathbb{R}\\). A map \\(p:V\\rightarrow \\mathbb{R}\\) is said to be a polynomial map if \\(p=\\sum a_{t_1,\\cdots,t_r}(e^1)^{t_1}\\cdots(e^r)^{t_r}\\).\nLet \\(f:V\\times V\\times \\cdots\\times V\\rightarrow \\mathbb{R}\\) be a symmetric multilinear mapping. We want to associate","permalink":"https://praphulla-koushik.github.io/2018/12/30/invariant-polynomials/","summary":"Let \\(V\\) be a vector space over \\(\\mathbb{R}\\). Let \\(\\{e_1,\\cdots,e_r\\}\\) be a basis of \\(V\\) over \\(\\mathbb{R}\\) and \\(\\{e^1,\\cdots,e^r\\}\\) be the dual basis of \\(V\\). We call \\(e^i:V\\rightarrow \\mathbb{R}\\) to be \u003cstrong\u003epolynomials over \\(V\\) with values in \\(\\mathbb{R}\\). \u003c/strong\u003e\n\nA map \\(p:V\\rightarrow \\mathbb{R}\\) is said to be a polynomial map if\n\u003cp style=\"text-align:center;\"\u003e\\(p=\\sum a_{t_1,\\cdots,t_r}(e^1)^{t_1}\\cdots(e^r)^{t_r}\\).\u003c/p\u003e\nLet \\(f:V\\times V\\times \\cdots\\times V\\rightarrow \\mathbb{R}\\) be a symmetric multilinear mapping. We want to associate","tags":[],"title":"Invariant polynomials","type":"wp-import"},{"categories":["differential-geometry","Algebraic geometry"],"content":"Given a principal \\(G\\) bundle \\(P\\rightarrow M\\) we associate what is called a Weil homomorphism \\(I(G)\\rightarrow H^*(M,\\mathbb{R})\\). Given \\(f\\in I^k(G)\\) i.e., \\(f:\\underbrace{\\mathfrak{g}\\times\\cdots\\times\\mathfrak{g}}_{k\\text{ times}}\\rightarrow \\mathbb{R}\\) we associate an element in \\(H^{2k}(M,\\mathbb{R})\\) as follows. This is only an outline. It is useful if you can fill the gaps by your self. Fix a connection \\(\\Gamma\\) on \\(P(M,G)\\) and let \\(\\Omega\\) denote the curvature form associated to \\(\\Gamma\\). The element \\(f\\in I^k(G)\\) gives a \\(2k\\)-form \\(f(\\Omega):P\\rightarrow \\Lambda^{2k}T^*P\\) on \\(P\\) as follows. \\(f(\\Omega)(v_1,\\cdots,v_{2k})=\\frac{1}{(2k)!}\\sum_{\\sigma\\in S_{2k}} f(\\Omega(v_{\\sigma(1)}.v_{\\sigma(2)}),\\cdots\\Omega(v_{\\sigma(2k-1)},v_{\\sigma(2k)}))\\)\nGiven a smooth map \\(\\pi:P\\rightarrow M\\) there is a notion of pull back of forms along \\(\\pi\\) usually denoted by \\(\\pi^*:H^*(M)\\rightarrow H^*(P)\\). A form \\(\\omega:P\\rightarrow \\Lambda^k T^*P\\) on \\(P\\) is called horizantal if \\(\\omega(p)(v_1,\\cdots,v_k)=0\\) if atleast one of \\(v_i\\) is vertical. Check that the right invariant (i.e., \\(R_g^*\\omega=\\omega\\) for every \\(g\\in G\\)) horizantal forms on \\(P\\) with coefficients in \\(\\mathbb{R}\\) are exactly forms in the image of \\(\\pi^*:H^*(M)\\rightarrow H^*(P)\\). Check that \\(f(\\Omega):P\\rightarrow \\Lambda^{2k}T^*P\\) is right invarinat, horizantal form. Thus, \\(f(\\Omega)\\) is in the image of \\(\\pi^*:H^*(M)\\rightarrow H^*(P)\\). So, there exists a \\(2k\\)-form \\(\\tilde{f}(\\Omega)\\) on \\(M\\) such that \\(\\pi^*(\\tilde{f}(\\Omega))=f(\\Omega)\\). Check that \\(f(\\Omega)\\) is closed \\(2k\\)-form on \\(P\\). As \\(\\pi^*(\\tilde{f}(\\Omega))=f(\\Omega)\\), \\(f(\\Omega)\\) being closed \\(2k\\)-form on \\(P\\) implies that \\(\\tilde{f}(\\Omega)\\) is a closed \\(2k\\)-form on \\(M\\) which gives an element in \\(H^{2k}(M,\\mathbb{R})\\). Thus, given a connection \\(\\Gamma\\) on \\(P(M,G)\\) we have \\(I^k(G)\\rightarrow H^{2k}(M,\\mathbb{R})\\) which induces a map \\(I(G)\\rightarrow H^*(M,\\mathbb{R})\\) which we call the Weil homomorphism. We then see that this construction does not depend on the connection \\(\\Gamma\\) that we have chosen. Suppose \\(\\Gamma_0\\) and \\(\\Gamma_1\\) be connections on \\(P(M,G)\\) with curvature forms \\(\\Omega_0\\) and \\(\\Omega_1\\) respectively. We see that \\(\\tilde{f}(\\Omega_0)\\) and \\(\\tilde{f}(\\Omega_1)\\) defines same cohomology class in \\(H^{2k}(M,\\mathbb{R})\\) i.e., \\(\\tilde{f}(\\Omega_0)-\\tilde{f}(\\Omega_1)\\) is an exact \\(2k\\)-form on \\(M\\). This says that the map \\(I(G)\\rightarrow H^*(M,\\mathbb{R})\\) is independent of connection we used to define. We have \\(\\pi^*(\\tilde{f}(\\Omega_0)-\\tilde{f}(\\Omega_1))=\\pi^*(\\tilde{f}(\\Omega_0))-\\pi^*(\\tilde{f}(\\Omega_1))=f(\\Omega_0)-f(\\Omega_1)\\). Suppose \\(f(\\Omega_0)-f(\\Omega_1)=d\\tau\\) for some \\(2k-1\\) form \\(\\tau\\) on \\(P\\) then, \\(\\pi^*(\\tilde{f}(\\Omega_0)-\\tilde{f}(\\Omega_1))=d\\tau\\). Suppose that \\(\\tau=\\pi^*\\tau'\\) for some \\(2k-1\\) form \\(\\tau'\\) on \\(M\\). We the have \\(d\\tau =d(\\pi^*(\\tau'))=\\pi^*(d\\tau')\\). As \\(\\pi^*(\\tilde{f}(\\Omega_0)-\\tilde{f}(\\Omega_1))= d\\tau=\\pi^*(d\\tau')\\) we see that \\(\\tilde{f}(\\Omega_0)-\\tilde{f}(\\Omega_1)=d\\tau'\\) (because \\(\\pi:P\\rightarrow M\\) is a submersion) which proves that \\(\\tilde{f}(\\Omega_0)-\\tilde{f}(\\Omega_1)\\) is exact form on \\(M\\) which further says that, \\([\\tilde{f}(\\Omega_0)]=[\\tilde{f}(\\Omega_1)]\\in H^{2k}(M,\\mathbb{R})\\). Check that \\(f(\\Omega_0)-f(\\Omega_1)=d\\tau\\) and \\(\\tau=\\pi^*\\tau'\\) for some \\(2k-1\\) form \\(\\tau'\\) on \\(M\\). Thus, given a principal \\(G\\) bundle \\(P\\rightarrow M\\) we have Weil homomorphism \\(I(G)\\rightarrow H^*(M,\\mathbb{R})\\).","permalink":"https://praphulla-koushik.github.io/2018/12/30/construction-of-weil-homomorphism/","summary":"Given a principal \\(G\\) bundle \\(P\\rightarrow M\\) we associate what is called a Weil homomorphism \\(I(G)\\rightarrow H^*(M,\\mathbb{R})\\).\n\nGiven \\(f\\in I^k(G)\\) i.e., \\(f:\\underbrace{\\mathfrak{g}\\times\\cdots\\times\\mathfrak{g}}_{k\\text{ times}}\\rightarrow \\mathbb{R}\\) we associate an element in \\(H^{2k}(M,\\mathbb{R})\\) as follows. This is only an outline. It is useful if you can fill the gaps by your self.\n\u003cul\u003e\n\t\u003cli\u003eFix a connection \\(\\Gamma\\) on \\(P(M,G)\\) and let \\(\\Omega\\) denote the curvature form associated to \\(\\Gamma\\).\u003c/li\u003e\n\t\u003cli\u003eThe element \\(f\\in I^k(G)\\) gives a \\(2k\\)-form \\(f(\\Omega):P\\rightarrow \\Lambda^{2k}T^*P\\) on \\(P\\) as follows.\u003c/li\u003e\n\u003c/ul\u003e\n\u003cp style=\"text-align:center;\"\u003e\\(f(\\Omega)(v_1,\\cdots,v_{2k})=\\frac{1}{(2k)!}\\sum_{\\sigma\\in S_{2k}} f(\\Omega(v_{\\sigma(1)}.v_{\\sigma(2)}),\\cdots\\Omega(v_{\\sigma(2k-1)},v_{\\sigma(2k)}))\\)\u003c/p\u003e","tags":["connections"],"title":"Construction of Weil homomorphism","type":"wp-import"},{"categories":["differential-geometry"],"content":"Let \\(G\\) be a Lie group and \\(\\mathfrak{g}\\) be its Lie algebra. Let \\(P\\rightarrow M\\) be a principal \\(G\\) bundle. A connection form on \\(P\\) is a \\(\\mathfrak{g}\\) valued \\(1\\)-form on \\(P\\) satisfying some properties. Suppose \\(G=Gl(n,\\mathbb{R})\\) then \\(\\mathfrak{g}=M(n,\\mathbb{R})\\). A connection is given by \\(\\omega:P\\rightarrow \\Lambda^1_{\\mathfrak{g}}T^*P\\). Given \\(p\\in P\\) we have \\(\\omega(p):T_pP \\rightarrow \\mathfrak{g}\\). Given \\(v\\in T_pP\\), \\(\\omega(p)(v)\\) is a matrix \\((a_{ij})\\in M(n,\\mathbb{R})\\) i.e., given \\(v\\in T_pP\\) we have \\(n^2\\) real numbers \\(a_{ij}\\in \\mathbb{R}\\) associated to it. Varying \\(v\\) over \\(T_pP\\) gives \\(n^2\\) maps \\(a_{ij}:T_pP\\rightarrow \\mathbb{R}\\). So, given \\(p\\in P\\), we have \\(n^2\\) maps \\(\\omega_{ij}(p):T_pP\\rightarrow \\mathbb{R}\\) where \\(\\omega_{ij}(p)(v)\\) is the \\(ij\\) th component of \\(\\omega(p)(v)\\). Fix \\(i,j\\) then, \\(\\omega_{ij}:P\\rightarrow \\Lambda^1 T^*P\\) given by \\(p\\mapsto \\omega_{ij}(p)\\) is a real valued \\(1\\)-form on \\(P\\). Thus, we denote \\(\\omega\\) by \\((\\omega_{ij})\\) where \\(\\omega_{ij}\\) are real valued \\(1\\)-forms on \\(P\\). This is what it means to see connection as a matrix of \\(1\\)-forms. The same can be done for Curvature form also. Curvature form \\(\\Omega:P\\rightarrow \\Lambda^2_{\\mathfrak{g}}TP\\) associates for each \\(p\\in P\\) a map \\(\\Omega(p):T_pP\\times T_pP\\rightarrow \\mathfrak{g}\\). Same explanation as above gives \\(n^2\\) real valued \\(2\\)-forms \\(\\Omega_{ij}:P\\rightarrow \\Lambda^2 TP\\). We denote Curvature form \\(\\Omega\\) by \\((\\Omega_{ij})\\). This is what it means to see curvature as a matrix of \\(2\\)-forms.","permalink":"https://praphulla-koushik.github.io/2018/12/30/matrix-associated-to-connection-curvature-form/","summary":"Let \\(G\\) be a Lie group and \\(\\mathfrak{g}\\) be its Lie algebra.\n\nLet \\(P\\rightarrow M\\) be a principal \\(G\\) bundle. A connection form on \\(P\\) is a \\(\\mathfrak{g}\\) valued \\(1\\)-form on \\(P\\) satisfying some properties.\n\nSuppose \\(G=Gl(n,\\mathbb{R})\\) then \\(\\mathfrak{g}=M(n,\\mathbb{R})\\). A connection is given by \\(\\omega:P\\rightarrow \\Lambda^1_{\\mathfrak{g}}T^*P\\).\n\nGiven \\(p\\in P\\) we have \\(\\omega(p):T_pP \\rightarrow \\mathfrak{g}\\). Given \\(v\\in T_pP\\), \\(\\omega(p)(v)\\) is a matrix \\((a_{ij})\\in M(n,\\mathbb{R})\\) i.e., given \\(v\\in T_pP\\) we have \\(n^2\\) real numbers \\(a_{ij}\\in \\mathbb{R}\\) associated to it. Varying \\(v\\) over \\(T_pP\\) gives \\(n^2\\) maps \\(a_{ij}:T_pP\\rightarrow \\mathbb{R}\\). So, given \\(p\\in P\\), we have \\(n^2\\) maps \\(\\omega_{ij}(p):T_pP\\rightarrow \\mathbb{R}\\) where \\(\\omega_{ij}(p)(v)\\) is the \\(ij\\) th component of \\(\\omega(p)(v)\\).\n\nFix \\(i,j\\) then, \\(\\omega_{ij}:P\\rightarrow \\Lambda^1 T^*P\\) given by \\(p\\mapsto \\omega_{ij}(p)\\) is a  \u003cstrong\u003ereal valued \u003c/strong\u003e\\(1\\)-form  on \\(P\\). Thus, we denote \\(\\omega\\) by \\((\\omega_{ij})\\) where \\(\\omega_{ij}\\) are  \u003cstrong\u003ereal valued \u003c/strong\u003e\\(1\\)-forms  on \\(P\\). This is what it means to see \u003cstrong\u003econnection as a matrix of \\(1\\)-forms\u003c/strong\u003e.\n\nThe same can be done for Curvature form also. Curvature form \\(\\Omega:P\\rightarrow \\Lambda^2_{\\mathfrak{g}}TP\\) associates for each \\(p\\in P\\) a map \\(\\Omega(p):T_pP\\times T_pP\\rightarrow \\mathfrak{g}\\). Same explanation as above gives \\(n^2\\)\u003cstrong\u003e real valued\u003c/strong\u003e \\(2\\)-forms \\(\\Omega_{ij}:P\\rightarrow \\Lambda^2 TP\\). We denote Curvature form \\(\\Omega\\) by \\((\\Omega_{ij})\\). This is what it means to see \u003cstrong\u003ecurvature  as a matrix of \\(2\\)-forms\u003c/strong\u003e.","tags":["connections"],"title":"Matrix associated to connection/curvature form","type":"wp-import"},{"categories":["differential-geometry"],"content":"Here, I will add links for web pages where I have written about concepts from Kobayashi and Nomizu's book Foundations of Differential geometry (Volume \\(1\\) and Volume \\(2\\)). Derivative of Left invariant differential form Maurer-Cartan form on a Lie group Transition maps for principal bundle are smooth Trivializations and sections in Principal bundle Construction of Weil homomorphism Construction of associated bundle Equivariant maps are Isomorphisms Invariant polynomials ","permalink":"https://praphulla-koushik.github.io/2018/12/30/kobayashi-and-nomizus-book/","summary":"Here, I will add links for web pages where I have written about concepts from Kobayashi and Nomizu's book Foundations of Differential geometry (Volume \\(1\\) and Volume \\(2\\)).\n\u003cul\u003e\n\t\u003cli\u003e\u003ca href=\"https://koushik1729.wordpress.com/2018/12/29/derivative-of-left-invariant-differential-form/\"\u003eDerivative of Left invariant differential form\u003c/a\u003e\u003c/li\u003e\n\t\u003cli\u003e\u003ca href=\"https://koushik1729.wordpress.com/2018/12/29/maurer-cartan-form-on-a-lie-group/\"\u003eMaurer-Cartan form on a Lie group\u003c/a\u003e\u003c/li\u003e\n\t\u003cli\u003e\u003ca href=\"https://koushik1729.wordpress.com/2019/01/26/transition-maps-for-principal-bundle-are-smooth/\"\u003eTransition maps for principal bundle are smooth\u003c/a\u003e\u003c/li\u003e\n\t\u003cli\u003e\u003ca href=\"https://koushik1729.wordpress.com/2019/01/24/trivializations-and-sections-in-principal-bundle/\"\u003eTrivializations and sections in Principal bundle\u003c/a\u003e\u003c/li\u003e\n\t\u003cli\u003e\u003ca href=\"https://koushik1729.wordpress.com/2018/12/30/construction-of-weil-homomorphism/\"\u003eConstruction of Weil homomorphism\u003c/a\u003e\u003c/li\u003e\n\t\u003cli\u003e\u003ca href=\"https://koushik1729.wordpress.com/2019/01/07/construction-of-associated-bundle/\"\u003eConstruction of associated bundle\u003c/a\u003e\u003c/li\u003e\n\t\u003cli\u003e\u003ca href=\"https://koushik1729.wordpress.com/2019/01/23/equivariant-maps-are-isomorphisms/\"\u003eEquivariant maps are Isomorphisms\u003c/a\u003e\u003c/li\u003e\n\t\u003cli\u003e\u003ca href=\"https://koushik1729.wordpress.com/2018/12/30/invariant-polynomials/\"\u003eInvariant polynomials\u003c/a\u003e\u003c/li\u003e\n\u003c/ul\u003e","tags":["kobayashi-nomizu","principal-bundles"],"title":"Kobayashi and Nomizu's book","type":"wp-import"},{"categories":["Differential geometry"],"content":"In this, we see that for a left-invarinat differential form \\(\\omega\\) on $G$, \\(d\\omega(X,Y)=-\\frac{1}{2}\\omega([X,Y])\\)\nfor vector fields \\(X, Y\\) in \\(G\\). Let \\(\\omega:G\\rightarrow \\Lambda^1 T^*G\\) be a Left-invariant differential form on \\(G\\) i.e., \\((L_g)^*\\omega=\\omega\\) for all \\(g\\in G\\) i.e., \\(\\omega(g)(v)=\\omega(e)((L_{g^{-1}})_{*,g}(v))\\)\nfor all \\(g\\in G\\) and \\(v\\in T_gG\\). Given \\(A\\in \\mathfrak{g}\\) we have vector field \\(A^*:G\\rightarrow TG\\) defined as \\(A^*(g)=(L_g)_{*,e}(A)\\). Then, \\(\\omega(A^*):G\\rightarrow \\mathbb{R}\\) is constant function, \\(\\omega(A^*)(g)=\\omega(e)(A)\\) for all \\(g\\in G\\). As \\(\\omega(A^*):G\\rightarrow \\mathbb{R}\\) is constant map, \\(X( \\omega(A^*))=0\\) for any vector field \\(X:G\\rightarrow TG\\) on \\(G\\). In particular, \\(B^*(X(\\omega(A^*)))=0\\) for \\(B\\in \\mathfrak{g}\\). Interchanging \\(A\\) and \\(B\\) we have \\(A^*(\\omega(B^*))=0\\) As \\((d\\omega)(A^*,B^*)=\\frac{1}{2}\\left[ A^*(\\omega(B^*))-B^*(\\omega(A^*))-\\omega([A^*,B^*])\\right]\\)\nand \\(A^*(\\omega(B^*))=0, B^*( \\omega(A^*))=0\\) we have \\((d\\omega)(A^*,B^*)=-\\frac{1}{2}\\omega([A^*,B^*])\\). Let \\(X:G\\rightarrow TG\\) be vector fields on \\(G\\). Fix a basis \\(\\{A_1,\\cdots,A_n\\}\\) for the Lie algebra \\(\\mathfrak{g}\\) of \\(G\\). Given \\(g\\in G\\) we have \\(X(g)\\in T_gG\\). Given \\(v\\in T_gG\\), we can find \\(A\\in \\mathfrak{g}\\) such that \\(v=A^*(g)\\). Here, \\(X(g)=A^*(g)\\). As \\(\\{A_1,\\cdots,A_n\\}\\) is basis for \\(\\mathfrak{g}\\) and \\(A\\in \\mathfrak{g}\\), there exists \\(a_i\\in \\mathbb{R}\\) such that \\(A=\\sum_{i=1}^n a_i A_i\\). Then, \\(X(g)=A^*(g)\\) says that \\(X(g)=(\\sum_{i=1}^n a_i A_i)^*(g)=\\sum_{i=1}^n a_i A_i^*(g)\\).\nGiven \\(g\\in G\\) we have \\(a_i\\in \\mathbb{R}\\) such that \\(X(g)=\\sum_{i=1}^n a_iA_i^*(g)\\). Varying \\(g\\) over \\(G\\) gives smooth functions \\(a_i:G\\rightarrow \\mathbb{R}\\) such that \\(X=\\sum a_i A_i^*\\). For a vector field \\(Y:G\\rightarrow TG\\) of \\(G\\), we have \\(b_i:G\\rightarrow \\mathbb{R}\\) such that \\(Y=\\sum b_i A_i^*\\). Then, \\(X(\\omega(Y))=\\sum_{i,j} a_ib_jA_i^*(\\omega(A_j^*))\\)\nAs \\(A_i^*(\\omega(A_j^*))=0\\) for \\(i, j\\), we see that \\(X(\\omega(Y))=0\\). Interchanging \\(X,Y\\) we see that \\(Y(\\omega(X))=0\\). As \\((d\\omega)(X,Y)=\\frac{1}{2}\\left[ X(\\omega(Y))-Y(\\omega(X))-\\omega([X,Y])\\right]\\)\nand \\(X(\\omega(Y))=0=Y(\\omega(X))\\) we see that \\(d\\omega(X,Y)=-\\frac{1}{2}\\omega([X,Y])\\).\nIn particular, for Maurer-Cartan form on a Lie group \\(\\theta:G\\rightarrow TG\\) which is a left invarinat \\(1\\)-form, we see that \\(d\\theta(X,Y)=-\\frac{1}{2}\\theta([X,Y])\\) for vector fields \\(X,Y\\) on \\(G\\). In literature, this property is denoted by \\(d\\theta=-\\frac{1}{2}[\\theta,\\theta]\\).","permalink":"https://praphulla-koushik.github.io/2018/12/29/derivative-of-left-invariant-differential-form/","summary":"In this, we see that for a left-invarinat differential form \\(\\omega\\) on $G$,\n\u003cp style=\"text-align:center;\"\u003e\\(d\\omega(X,Y)=-\\frac{1}{2}\\omega([X,Y])\\)\u003c/p\u003e\nfor vector fields \\(X, Y\\) in \\(G\\).\n\nLet \\(\\omega:G\\rightarrow \\Lambda^1 T^*G\\) be a Left-invariant differential form on \\(G\\) i.e., \\((L_g)^*\\omega=\\omega\\) for all \\(g\\in G\\) i.e.,\n\u003cp style=\"text-align:center;\"\u003e\\(\\omega(g)(v)=\\omega(e)((L_{g^{-1}})_{*,g}(v))\\)\u003c/p\u003e\nfor all \\(g\\in G\\) and \\(v\\in T_gG\\).\n\nGiven \\(A\\in \\mathfrak{g}\\) we have vector field \\(A^*:G\\rightarrow TG\\) defined as \\(A^*(g)=(L_g)_{*,e}(A)\\). Then, \\(\\omega(A^*):G\\rightarrow \\mathbb{R}\\) is constant function, \\(\\omega(A^*)(g)=\\omega(e)(A)\\) for all \\(g\\in G\\).\n\nAs \\(\\omega(A^*):G\\rightarrow \\mathbb{R}\\) is constant map,  \\(X( \\omega(A^*))=0\\) for any vector field \\(X:G\\rightarrow TG\\) on \\(G\\).\n\nIn particular, \\(B^*(X(\\omega(A^*)))=0\\) for \\(B\\in \\mathfrak{g}\\). Interchanging \\(A\\) and \\(B\\) we have \\(A^*(\\omega(B^*))=0\\)\n\nAs\n\u003cp style=\"text-align:center;\"\u003e\\((d\\omega)(A^*,B^*)=\\frac{1}{2}\\left[ A^*(\\omega(B^*))-B^*(\\omega(A^*))-\\omega([A^*,B^*])\\right]\\)\u003c/p\u003e","tags":[],"title":"Derivative of Left invariant differential form","type":"wp-import"},{"categories":["differential-geometry"],"content":"Let \\(G\\) be a Lie group and \\(\\mathfrak{g}\\) be its Lie algebra. We want to associate a \\(\\mathfrak{g}\\) valued \\(1\\) form on \\(G\\). We define \\(\\theta:G\\rightarrow \\Lambda^1_{\\mathfrak{g}}T^*G\\) as follows. For \\(g\\in G\\), we need \\(\\theta(g):T_gG\\rightarrow \\mathfrak{g}=T_eG\\). For manifolds \\(M,N\\), one natural way to get a map between tangent spaces \\(T_mM\\) and \\(T_nN\\) is to think of a smooth map \\(f:M\\rightarrow N\\) such that \\(f(m)=n\\) and take its differential at \\(m\\). We get \\(f_{*,m}:T_mM\\rightarrow T_nN\\). To get \\(\\theta(g):T_gG\\rightarrow \\mathfrak{g}=T_eG\\), we look for a map \\(G\\rightarrow G\\) that takes \\(g\\) to \\(e\\). One such map is multiplication by \\(g^{-1}\\). Consider \\(\\delta_{g^{-1}}:G\\rightarrow G\\) given by \\(h\\mapsto g^{-1}h\\). This map takes \\(g\\) to \\(e\\) and \\(\\delta_{*,g^{-1}}:T_gG\\rightarrow T_eG=\\mathfrak{g}\\). This gives a \\(\\mathfrak{g}\\) valued \\(1\\)-form on \\(G\\) which we call to be the Maurer-Cartan form on \\(G\\) denoted by \\(\\theta\\) defined as \\(\\theta(g)=(\\delta_{g^{-1}})_{*,g}:T_gG\\rightarrow \\mathfrak{g}\\). Kobayashi and Nomizu defines Maurer-Cartan form on \\(G\\) to be \"the left-invariant \\(\\mathfrak{g}\\) valued \\(1\\)-form on \\(G\\) uniquely determined by the condition that \\(\\theta(A)=A\\) for all \\(A\\in \\mathfrak{g}\\). More precisely, this means \\(\\theta (e) :T_eG\\rightarrow T_eG\\) is such that \\(\\theta(e)(A)=A\\) for all \\(A\\in \\mathfrak{g}\\). The condition that \\(\\theta\\) is left-invariant means that \\(\\theta(e):T_eG\\rightarrow \\mathfrak{g}\\) determines what \\(\\theta(g):T_gG\\rightarrow \\mathfrak{g}\\). So, the condition \\(\\theta(e)(A)=A\\) for all \\(A\\in \\mathfrak{g}\\) along with the condition left-invariant gives unique \\(1\\)-form \\(\\theta\\) which is called as the Maurer-Cartan form. Once we unravel how \\(\\theta(e):T_eG\\rightarrow \\mathfrak{g}\\) determines \\(\\theta(g):T_gG\\rightarrow \\mathfrak{g}\\) and that \\(\\theta(e)(A)=A\\) for all \\(A\\in \\mathfrak{g}\\) we see that \\(\\theta(g)(v)=(\\delta_{g^{-1}})_{*,g}(v)\\) for all \\(v\\in T_gG\\) which is precisely what I have written in the first half of this post. Differentiating this Maurer-Cartan form \\(\\theta:G\\rightarrow \\Lambda^1_{\\mathfrak{g}}T^*G\\) we see that \\(d\\theta(X,Y)=-\\frac{1}{2}\\theta([X,Y])\\)\nfor vector fields \\(X,Y\\) on \\(G\\). In literature, this property is denoted by \\(d\\theta==\\frac{1}{2}[\\theta,\\theta]\\).","permalink":"https://praphulla-koushik.github.io/2018/12/29/maurer-cartan-form-on-a-lie-group/","summary":"Let \\(G\\) be a Lie group and \\(\\mathfrak{g}\\) be its Lie algebra. We want to associate a \\(\\mathfrak{g}\\) valued \\(1\\) form on \\(G\\).\n\nWe define \\(\\theta:G\\rightarrow \\Lambda^1_{\\mathfrak{g}}T^*G\\) as follows. For \\(g\\in G\\), we need \\(\\theta(g):T_gG\\rightarrow \\mathfrak{g}=T_eG\\).\n\nFor manifolds \\(M,N\\), one natural way to get a map between tangent spaces \\(T_mM\\) and \\(T_nN\\) is to think of a smooth map \\(f:M\\rightarrow N\\) such that \\(f(m)=n\\) and take its differential at \\(m\\). We get \\(f_{*,m}:T_mM\\rightarrow T_nN\\). To get \\(\\theta(g):T_gG\\rightarrow \\mathfrak{g}=T_eG\\), we look for a map \\(G\\rightarrow G\\) that takes \\(g\\) to \\(e\\). One such map is multiplication by \\(g^{-1}\\). Consider \\(\\delta_{g^{-1}}:G\\rightarrow G\\) given by \\(h\\mapsto g^{-1}h\\). This map takes \\(g\\) to \\(e\\) and \\(\\delta_{*,g^{-1}}:T_gG\\rightarrow T_eG=\\mathfrak{g}\\).  This gives a \\(\\mathfrak{g}\\) valued \\(1\\)-form on \\(G\\) which we call to be the Maurer-Cartan form on \\(G\\)  denoted by \\(\\theta\\) defined as \\(\\theta(g)=(\\delta_{g^{-1}})_{*,g}:T_gG\\rightarrow \\mathfrak{g}\\).\n\nKobayashi and Nomizu defines Maurer-Cartan form on \\(G\\) to be \"the left-invariant \\(\\mathfrak{g}\\) valued \\(1\\)-form on \\(G\\) uniquely determined by the condition that \\(\\theta(A)=A\\) for all \\(A\\in \\mathfrak{g}\\). More precisely, this means \\(\\theta (e) :T_eG\\rightarrow T_eG\\) is such that \\(\\theta(e)(A)=A\\) for all \\(A\\in \\mathfrak{g}\\).  The condition that \\(\\theta\\) is left-invariant means that \\(\\theta(e):T_eG\\rightarrow \\mathfrak{g}\\) determines what \\(\\theta(g):T_gG\\rightarrow \\mathfrak{g}\\).  So, the condition \\(\\theta(e)(A)=A\\) for all \\(A\\in \\mathfrak{g}\\) along with the condition left-invariant gives unique \\(1\\)-form \\(\\theta\\) which is called as the Maurer-Cartan form.\n\nOnce we unravel how \\(\\theta(e):T_eG\\rightarrow \\mathfrak{g}\\) determines  \\(\\theta(g):T_gG\\rightarrow \\mathfrak{g}\\) and that \\(\\theta(e)(A)=A\\) for all \\(A\\in \\mathfrak{g}\\) we see that \\(\\theta(g)(v)=(\\delta_{g^{-1}})_{*,g}(v)\\) for all \\(v\\in T_gG\\) which is precisely what I have written in the first half of this post.\n\n\u003ca href=\"https://koushik1729.wordpress.com/2018/12/29/derivative-of-left-invariant-differential-form/\"\u003eDifferentiating\u003c/a\u003e this Maurer-Cartan form \\(\\theta:G\\rightarrow \\Lambda^1_{\\mathfrak{g}}T^*G\\) we see that\n\u003cp style=\"text-align:center;\"\u003e\\(d\\theta(X,Y)=-\\frac{1}{2}\\theta([X,Y])\\)\u003c/p\u003e","tags":["kobayashi-nomizu"],"title":"Maurer-Cartan form on a Lie group","type":"wp-import"},{"categories":["Homological algebra"],"content":"","permalink":"https://praphulla-koushik.github.io/2017/08/09/kernel-and-cokernel-of-a-morphism/","summary":"","tags":[],"title":"Kernel and cokernel of a Morphism","type":"wp-import"},{"categories":["Category theory"],"content":"","permalink":"https://praphulla-koushik.github.io/2017/08/09/additive-categories/","summary":"","tags":[],"title":"Additive categories","type":"wp-import"},{"categories":["Category theory"],"content":"Definition : Let \\(\\mathcal{C}\\) be a category and \\(f:A\\rightarrow C\\) and \\(g:B\\rightarrow C\\) be morphisms. We define pull back of \\(f,g\\) to be a triple \\((P,f',g')\\) where \\(P\\) is an object of \\(\\mathcal{C}\\) and \\(f':P\\rightarrow B, g':P\\rightarrow A\\) with \\(g\\circ f'=f\\circ g'\\) such that given any object \\(P'\\) of \\(\\mathcal{C}\\) and morphisms \\(g'':P'\\rightarrow A\\) and \\(f'':P'\\rightarrow B\\) with \\(g\\circ f''=f\\circ g''\\), there exists a unique morphism \\(\\eta:P'\\rightarrow P\\) such that \\(g''=g'\\circ \\eta\\) and \\(f''=f'\\circ \\eta\\) as in the following commutative diagram. Definition : Let \\(\\mathcal{C}\\) be a category and \\(f:A\\rightarrow C\\) and \\(g:B\\rightarrow C\\) be morphisms. We define pull back of \\(f,g\\) to be a triple \\((P,f',g')\\) where \\(P\\) is an object of \\(\\mathcal{C}\\) and \\(f':P\\rightarrow B, g':P\\rightarrow A\\) with \\(g\\circ f'=f\\circ g'\\) such that given any object \\(P'\\) of \\(\\mathcal{C}\\) and morphisms \\(g'':P'\\rightarrow A\\) and \\(f'':P'\\rightarrow B\\) with \\(g\\circ f''=f\\circ g''\\), there exists a unique morphism \\(\\eta:P'\\rightarrow P\\) such that \\(g''=g'\\circ \\eta\\) and \\(f''=f'\\circ \\eta\\) as in the following commutative diagram.","permalink":"https://praphulla-koushik.github.io/2017/08/09/pull-back-and-push-forward-of-two-morphisms/","summary":"\u003cstrong\u003eDefinition\u003c/strong\u003e : Let \\(\\mathcal{C}\\) be a category and \\(f:A\\rightarrow C\\) and \\(g:B\\rightarrow C\\) be morphisms. We define pull back of \\(f,g\\) to be a triple \\((P,f',g')\\) where\n\u003col\u003e\n\t\u003cli\u003e\\(P\\) is an object of \\(\\mathcal{C}\\) and\u003c/li\u003e\n\t\u003cli\u003e\\(f':P\\rightarrow B, g':P\\rightarrow A\\) with \\(g\\circ f'=f\\circ g'\\)\u003c/li\u003e\n\u003c/ol\u003e\nsuch that given any object \\(P'\\) of \\(\\mathcal{C}\\) and morphisms \\(g'':P'\\rightarrow A\\) and \\(f'':P'\\rightarrow B\\) with \\(g\\circ f''=f\\circ g''\\), \u003cstrong\u003ethere\u003c/strong\u003e \u003cstrong\u003eexists\u003c/strong\u003e \u003cstrong\u003ea\u003c/strong\u003e \u003cstrong\u003eunique\u003c/strong\u003e morphism \\(\\eta:P'\\rightarrow P\\) such that\n\\(g''=g'\\circ \\eta\\) and \\(f''=f'\\circ \\eta\\) as in the following commutative diagram. \u003cimg class=\" size-full wp-image-1259 aligncenter\" src=\"/wp-media/2017/08/30ce5bb042-ql_9a65ff8c25615e2690891cb3f378db07_l3.png\" alt=\"ql_9a65ff8c25615e2690891cb3f378db07_l3\" width=\"218\" height=\"151\" /\u003e\n\n\u003cstrong\u003eDefinition\u003c/strong\u003e : Let \\(\\mathcal{C}\\) be a category and \\(f:A\\rightarrow C\\) and \\(g:B\\rightarrow C\\) be morphisms. We define pull back of \\(f,g\\) to be a triple \\((P,f',g')\\) where\n\u003col\u003e\n\t\u003cli\u003e\\(P\\) is an object of \\(\\mathcal{C}\\) and\u003c/li\u003e\n\t\u003cli\u003e\\(f':P\\rightarrow B, g':P\\rightarrow A\\) with \\(g\\circ f'=f\\circ g'\\)\u003c/li\u003e\n\u003c/ol\u003e\nsuch that given any object \\(P'\\) of \\(\\mathcal{C}\\) and morphisms \\(g'':P'\\rightarrow A\\) and \\(f'':P'\\rightarrow B\\) with \\(g\\circ f''=f\\circ g''\\), \u003cstrong\u003ethere\u003c/strong\u003e \u003cstrong\u003eexists\u003c/strong\u003e \u003cstrong\u003ea\u003c/strong\u003e \u003cstrong\u003eunique\u003c/strong\u003e morphism \\(\\eta:P'\\rightarrow P\\) such that\n\\(g''=g'\\circ \\eta\\) and \\(f''=f'\\circ \\eta\\) as in the following commutative diagram.","tags":[],"title":"Pull back and Push forward of two morphisms","type":"wp-import"},{"categories":["Category theory"],"content":"","permalink":"https://praphulla-koushik.github.io/2017/08/09/equalizers-and-coequalizers/","summary":"","tags":[],"title":"Equalizers and Coequalizers","type":"wp-import"},{"categories":["Category theory"],"content":"","permalink":"https://praphulla-koushik.github.io/2017/08/09/monomorphisms-and-epimorphisms/","summary":"","tags":[],"title":"Monomorphisms and epimorphisms","type":"wp-import"},{"categories":["Commutative Algebra","homological-algebra"],"content":"In an exercise on Flasque sheaves, I used Snake lemma. So, I thought it is better to mention it separately with proof. Lemma : Given a commutative diagram as below we have exact sequence Proof : This is just question of diagram chasing. It is good if one can prove this on their own with out looking for proof from some other source. Let \\(a\\in \\text{Ker}(f)\\) i.e., \\(f(a)=0\\) which then imply \\(v_1(f(a))=0\\) which is same as saying \\(g(u_1(a))=0\\) (as the first square is commutative) i.e., \\(u_1(a)\\in \\text{Ker}(g)\\). So, we have map \\(\\tilde{u}_1:\\text{Ker}(f)\\rightarrow \\text{Ker}(g)\\) given by \\(a\\mapsto u_1(a)\\). For similar reasons, \\(b\\mapsto u_2(b)\\) gives map \\(\\tilde{u}_2:\\text{Ker}(g)\\rightarrow \\text{Ker}(h)\\). As \\(\\tilde{u}_1\\) and \\(\\tilde{u}_2\\) are just restrictions of \\(u_1\\) and \\(u_2\\), it follows that the sequence \\(0\\rightarrow \\text{Ker(f)}\\xrightarrow{\\tilde{u}_1} \\text{Ker(g)}\\xrightarrow{\\tilde{u}_2} \\text{Ker(h)}\\) is an exact sequence. Define \\(\\tilde{v}_1:\\text{Coker(f)}\\rightarrow \\text{Coker(g)}\\) by \\(a+f(M_1)\\rightarrow v_1(a)+g(M_2)\\). Let \\(a_1+f(M_1)=a_2+f(M_2)\\), then, \\(a_1-a_2\\in f(M_1)\\) i.e., \\(a_1-a_2=f(m)\\) for some \\(m\\in M_1\\). So, we have \\(v_1(a_1-a_2)=v_1(f(m))=g(u_1(m))\\in g(M_2)\\), thus, \\(v_1(a_1)-v_1(a_2)\\in g(M_2)\\) i.e., \\(v_1(a_1)+g(M_2)=v_1(a_2)+g(M_2)\\) i.e., \\(\\tilde{v}_1(a_1+f(M)_1)=\\tilde{v}_1(a_2+f(M_2))\\). Thus, \\(\\tilde{v}_1:\\text{Coker(f)}\\rightarrow\\text{Coker(g)}\\) given by \\(a+f(M_1)\\rightarrow v_1(a)+g(M_2)\\) is well defined. Similarly, \\(\\tilde{v}_2:\\text{Coker(g)}\\rightarrow \\text{Coker(h)}\\) given by \\(b+g(M_2)\\rightarrow v_2(b)+h(M_3)\\) is a well defined map. As these are coming from \\(v_1,v_2\\) the sequence \\(\\text{CoKer}(f) \\xrightarrow{\\tilde{v}_1} \\text{CoKer}(g) \\xrightarrow{\\tilde{v}_2} \\text{CoKer(h)}\\rightarrow 0\\) is an exact sequence. Now, we define (connecting map) \\(d:\\text{Ker(h)}\\rightarrow \\text{Coker(f)}\\) and show that this map connects the two exat sequences \\(0\\rightarrow \\text{Ker(f)}\\xrightarrow{\\tilde{u}_1} \\text{Ker(g)}\\xrightarrow{\\tilde{u}_2} \\text{Ker(h)}\\) and \\(\\text{CoKer}(f) \\xrightarrow{\\tilde{v}_1} \\text{CoKer}(g) \\xrightarrow{\\tilde{v}_2} \\text{CoKer(h)}\\rightarrow 0\\) giving the required exact sequence Let \\(a\\in \\text{Ker(h)}\\) i.e., \\(h(a)=0\\). As \\(u_2\\) is surjective, \\(a=u_2(m_2)\\) for some \\(m_2\\in M_2\\). So, \\(0=h(a)=h(u_2(m_2))=v_2(g(m_2))\\). So, \\(g(m_2)\\in \\text{Ker}(v_2)=\\text{Im}(v_1)\\). So, \\(g(m_2)=v_1(n_1)\\) for some \\(n_1\\in N_1\\). Define \\(d:\\text{Ker(h)}\\rightarrow \\text{Coker(f)}\\) as \\(a\\mapsto n_1+f(M_1)\\) chosen as above. We prove that this is well defined. Let \\(a=u_2(m_2)=u_2(m_2')\\). Then, \\(m_2-m_2'\\in \\text{Ker}(u_2)=\\text{Im}(u_1)\\). So, \\(m_2-m_2'=u_1(m_1)\\). Then, \\(g(m_2)-g(m_2')=g(u_1(m_1))=v_1(f(m_1))\\) i.e., \\(v_1(n_1)-v_1(n_1')=v_1(f(m_1))\\). As \\(v_1\\) is injective, this means \\(n_1-n_1'=f(m_1)\\in f(M_1)\\) i.e., \\(n_1+f(M_1)=n_1'+f(M_1)\\). Thus, \\(d:\\text{Ker(h)}\\rightarrow \\text{Coker(f)}\\) defined as \\(a\\mapsto n_1+f(M_1)\\) is well defined. We have used surjectivity of \\(u_2\\) to define the map \\(d\\) and used injectivity of \\(v_1\\) to prove that it is well defined. I will write proof some other time that the resulting map is exact at \\(\\text{Ker}(h)\\) and \\(\\text{Coker}(f)\\). \u0026nbsp; \u0026nbsp;","permalink":"https://praphulla-koushik.github.io/2017/07/31/snake-lemma/","summary":"In an exercise on Flasque sheaves, I used Snake lemma. So, I thought it is better to mention it separately with proof.\n\n\u003cstrong\u003eLemma : \u003c/strong\u003eGiven a commutative diagram as below\n\n\u003cstrong\u003e\u003cimg class=\" size-full wp-image-1243 aligncenter\" src=\"/wp-media/2017/07/f4f548d154-ql_ee3435353bba304f57df87cb8181252d_l3.png\" alt=\"ql_ee3435353bba304f57df87cb8181252d_l3\" width=\"287\" height=\"85\" /\u003e\u003c/strong\u003e\n\nwe have exact sequence\n\n\u003cimg class=\" size-full wp-image-1245 aligncenter\" src=\"/wp-media/2017/07/03166f5879-ql_853a410f14c77599c146e77c4ca3fded_l3.png\" alt=\"ql_853a410f14c77599c146e77c4ca3fded_l3\" width=\"471\" height=\"81\" /\u003e\n\n\u003cstrong\u003eProof : \u003c/strong\u003eThis is just question of diagram chasing. It is good if one can prove this on their own with out looking for proof from some other source.\n\nLet \\(a\\in \\text{Ker}(f)\\) i.e., \\(f(a)=0\\) which then imply \\(v_1(f(a))=0\\) which is same as saying \\(g(u_1(a))=0\\) (as the first square is commutative) i.e., \\(u_1(a)\\in \\text{Ker}(g)\\). So, we have map \\(\\tilde{u}_1:\\text{Ker}(f)\\rightarrow \\text{Ker}(g)\\) given by \\(a\\mapsto u_1(a)\\). For similar reasons, \\(b\\mapsto u_2(b)\\) gives map \\(\\tilde{u}_2:\\text{Ker}(g)\\rightarrow \\text{Ker}(h)\\). As \\(\\tilde{u}_1\\) and \\(\\tilde{u}_2\\) are just restrictions of \\(u_1\\) and \\(u_2\\), it follows that the sequence  \\(0\\rightarrow \\text{Ker(f)}\\xrightarrow{\\tilde{u}_1} \\text{Ker(g)}\\xrightarrow{\\tilde{u}_2} \\text{Ker(h)}\\) is an exact sequence.\n\nDefine \\(\\tilde{v}_1:\\text{Coker(f)}\\rightarrow \\text{Coker(g)}\\) by \\(a+f(M_1)\\rightarrow v_1(a)+g(M_2)\\). Let \\(a_1+f(M_1)=a_2+f(M_2)\\), then, \\(a_1-a_2\\in f(M_1)\\) i.e., \\(a_1-a_2=f(m)\\) for some \\(m\\in M_1\\). So, we have \\(v_1(a_1-a_2)=v_1(f(m))=g(u_1(m))\\in g(M_2)\\), thus, \\(v_1(a_1)-v_1(a_2)\\in g(M_2)\\) i.e., \\(v_1(a_1)+g(M_2)=v_1(a_2)+g(M_2)\\) i.e., \\(\\tilde{v}_1(a_1+f(M)_1)=\\tilde{v}_1(a_2+f(M_2))\\). Thus, \\(\\tilde{v}_1:\\text{Coker(f)}\\rightarrow\\text{Coker(g)}\\) given by \\(a+f(M_1)\\rightarrow v_1(a)+g(M_2)\\) is well defined. Similarly, \\(\\tilde{v}_2:\\text{Coker(g)}\\rightarrow \\text{Coker(h)}\\) given by \\(b+g(M_2)\\rightarrow v_2(b)+h(M_3)\\) is a well defined map. As these are coming from \\(v_1,v_2\\) the sequence \\(\\text{CoKer}(f) \\xrightarrow{\\tilde{v}_1} \\text{CoKer}(g) \\xrightarrow{\\tilde{v}_2} \\text{CoKer(h)}\\rightarrow 0\\) is an exact sequence.\n\nNow, we define (connecting map) \\(d:\\text{Ker(h)}\\rightarrow \\text{Coker(f)}\\) and show that this map connects the two exat sequences  \\(0\\rightarrow \\text{Ker(f)}\\xrightarrow{\\tilde{u}_1} \\text{Ker(g)}\\xrightarrow{\\tilde{u}_2} \\text{Ker(h)}\\) and\n\\(\\text{CoKer}(f) \\xrightarrow{\\tilde{v}_1} \\text{CoKer}(g) \\xrightarrow{\\tilde{v}_2} \\text{CoKer(h)}\\rightarrow 0\\) giving the required exact sequence\n\n\u003cimg class=\" size-full wp-image-1245 aligncenter\" src=\"/wp-media/2017/07/03166f5879-ql_853a410f14c77599c146e77c4ca3fded_l3.png\" alt=\"ql_853a410f14c77599c146e77c4ca3fded_l3\" width=\"471\" height=\"81\" /\u003e\n\nLet \\(a\\in \\text{Ker(h)}\\) i.e., \\(h(a)=0\\). As \u003cstrong\u003e\\(u_2\\) is surjective, \u003c/strong\u003e\\(a=u_2(m_2)\\) for some \\(m_2\\in M_2\\). So, \\(0=h(a)=h(u_2(m_2))=v_2(g(m_2))\\).\nSo, \\(g(m_2)\\in \\text{Ker}(v_2)=\\text{Im}(v_1)\\). So, \\(g(m_2)=v_1(n_1)\\) for some \\(n_1\\in N_1\\). Define \\(d:\\text{Ker(h)}\\rightarrow \\text{Coker(f)}\\) as \\(a\\mapsto n_1+f(M_1)\\) chosen as above. We prove that this is well defined.\nLet \\(a=u_2(m_2)=u_2(m_2')\\). Then, \\(m_2-m_2'\\in \\text{Ker}(u_2)=\\text{Im}(u_1)\\). So,  \\(m_2-m_2'=u_1(m_1)\\). Then, \\(g(m_2)-g(m_2')=g(u_1(m_1))=v_1(f(m_1))\\) i.e., \\(v_1(n_1)-v_1(n_1')=v_1(f(m_1))\\). As \u003cstrong\u003e\\(v_1\\) is injective,\u003c/strong\u003e this means \\(n_1-n_1'=f(m_1)\\in f(M_1)\\) i.e., \\(n_1+f(M_1)=n_1'+f(M_1)\\). Thus, \\(d:\\text{Ker(h)}\\rightarrow \\text{Coker(f)}\\) defined as \\(a\\mapsto n_1+f(M_1)\\) is well defined.\n\nWe have used surjectivity of \\(u_2\\) to define the map \\(d\\) and used injectivity of \\(v_1\\) to prove that it is well defined. I will write proof  some other time that the resulting map\n\n\u003cimg class=\" size-full wp-image-1245 aligncenter\" src=\"/wp-media/2017/07/03166f5879-ql_853a410f14c77599c146e77c4ca3fded_l3.png\" alt=\"ql_853a410f14c77599c146e77c4ca3fded_l3\" width=\"471\" height=\"81\" /\u003eis exact at \\(\\text{Ker}(h)\\) and \\(\\text{Coker}(f)\\).\n\n\u0026nbsp;\n\n\u0026nbsp;","tags":[],"title":"Snake Lemma","type":"wp-import"},{"categories":["Category theory"],"content":" \u0026nbsp;","permalink":"https://praphulla-koushik.github.io/2017/07/28/left-exactness-of-global-section-functor/","summary":"\u003chr /\u003e\n\n\u0026nbsp;","tags":[],"title":"Left exactness of Global section functor","type":"wp-import"},{"categories":["Homological algebra"],"content":"","permalink":"https://praphulla-koushik.github.io/2017/07/28/exact-sequences-of-sheaves/","summary":"","tags":[],"title":"Exact sequences of sheaves","type":"wp-import"},{"categories":["Sheaves and Schemes"],"content":"","permalink":"https://praphulla-koushik.github.io/2017/07/28/injectivesurjective-morphisms-of-sheaves/","summary":"","tags":[],"title":"Injective/Surjective Morphisms of Sheaves","type":"wp-import"},{"categories":["Algebraic geometry"],"content":"","permalink":"https://praphulla-koushik.github.io/2017/07/28/sheafification-of-a-presheaf/","summary":"","tags":[],"title":"Sheafification of a presheaf","type":"wp-import"},{"categories":["Sheaves and Schemes","Algebraic geometry"],"content":"Definition : Let \\(X\\) be a topological space, \\(\\mathcal{F},\\mathcal{G}\\) be presheaves on \\(X\\). A morphism \\(\\varphi:\\mathcal{F}\\rightarrow \\mathcal{G}\\) is a collection \\(\\{\\varphi(U):\\mathcal{F}(U)\\rightarrow \\mathcal{G}(U)\\}\\) indexing over all open \\(U\\subseteq X\\) such that the following diagram is commutative for open \\(U\\subseteq V\\subseteq X\\). Morphism of sheaves inducing Morphism of stalks : We see that any morphism of sheaves \\(\\varphi:\\mathcal{F}\\rightarrow \\mathcal{G}\\) induces morphism of stalks \\(\\varphi_p:\\mathcal{F}_p\\rightarrow \\mathcal{G}_p\\) for each \\(p\\in X\\). Fix \\(p\\in X\\). Let us define \\(\\varphi_p:\\mathcal{F}_p\\rightarrow \\mathcal{G}_p\\) i.e., for \\((U,s)\\in \\mathcal{F}_p\\) we give an open set \\(V\\) containing \\(p\\) and a section \\(t\\in \\mathcal{G}(V)\\) giving an element \\((V,t)\\in \\mathcal{G}_p\\). One obvious choice of an open set containing \\(p\\) is \\(U\\). For this \\(U\\), we have \\(\\varphi(U):\\mathcal{F}(U)\\rightarrow \\mathcal{G}(U)\\) sending \\(s\\) to \\(\\varphi(U)(s)\\in \\mathcal{G}(U)\\). Define \\(\\varphi_p:\\mathcal{F}_p\\rightarrow \\mathcal{G}_p\\) as \\((U,s)\\mapsto (U,\\varphi(U)(s))\\). Map is well defined : We prove \\((U,s) \\sim (V,t)\\) implies \\((U,\\varphi(U)(s))\\sim (V,\\varphi(V)(t))\\). As \\((U,s) \\sim (V,t)\\) there exists an open subset \\(W\\subseteq U\\cap V\\) containing \\(p\\) such that \\(s|_W=t|_W\\). We prove that \\(\\varphi(U)(s)|_W=\\varphi(V)(t)|_W\\) which implies \\((U,\\varphi(U)(s))\\sim (V,\\varphi(V)(t))\\). The commutative diagram gives \\(\\varphi(U)(s)|_W=\\varphi(W)(s|_W)\\). Similar diagram in which \\(U\\) is replaced by \\(V\\) gives that \\(\\varphi(V)(t)|_W=\\varphi(W)(t|_W)\\). As \\(s|_W=t|_W\\) we have \\(\\varphi(W)(s|_W)=\\varphi(W)(t|_W)\\), concluding that \\(\\varphi(U)(s)|_W=\\varphi(V)(t)|_W\\). So, given a morphism of sheaves \\(\\varphi:\\mathcal{F}\\rightarrow \\mathcal{G}\\) we have well defined morphism \\(\\varphi_p:\\mathcal{F}_p\\rightarrow\\mathcal{G}_p\\) for each \\(p\\in X\\). Theorem : A morphism of sheaves \\(\\varphi:\\mathcal{F}\\rightarrow \\mathcal{G}\\) is an isomorphism of sheaves iff the induced map \\(\\varphi_p:\\mathcal{F}_p\\rightarrow\\mathcal{G}_p\\) is an isomorphism for each \\(p\\in X\\). Proof : Let \\(\\varphi:\\mathcal{F}\\rightarrow \\mathcal{G}\\) is an isomorphism of sheaves i.e., \\(\\varphi(U):\\mathcal{F}(U)\\rightarrow \\mathcal{G}(U)\\) is an isomorphism of groups for each open \\(U\\subseteq X\\). Fixing \\(p\\in X\\) we prove that \\(\\varphi_p:\\mathcal{F}_p\\rightarrow\\mathcal{G}_p\\) is an isomorphism. Let \\((U,s)\\in \\mathcal{F}_p\\) be such hat \\((U,\\varphi(U)(s))=0\\in \\mathcal{G}_p\\) i.e., \\(\\varphi(U)(s)|_W=0\\) for some open \\(W\\subseteq U\\). We thus have \\(\\varphi(W)(s|_W)=\\varphi(U)(s)|_W=0\\). As \\(\\varphi(W):\\mathcal{F}(W)\\rightarrow \\mathcal{G}(W)\\) is injective, this means \\(s|_W=0\\). Thus, \\(\\varphi_p:\\mathcal{F}_p\\rightarrow\\mathcal{G}_p\\) is an injective map. Let \\((V,t)\\in \\mathcal{G}_p\\) i.e., \\(p\\in V\\) and \\(t\\in \\mathcal{F}(V)\\). As \\(\\varphi(V):\\mathcal{F}(V)\\rightarrow \\mathcal{G}(V)\\) is surjective, there exists \\(s\\in \\mathcal{F}(V)\\) such that \\(\\varphi(V)(s)=t\\). So, \\(\\varphi_p((V,s))=(V,\\varphi(V)(s))=(V,t)\\). Thus, \\(\\varphi_p\\) is surjective. So, \\(\\varphi:\\mathcal{F}\\rightarrow \\mathcal{G}\\) is an isomorphism of sheaves implies \\(\\varphi_p:\\mathcal{F}_p\\rightarrow\\mathcal{G}_p\\) is an isomorphism for each \\(p\\in X\\). Conversely, suppose that \\(\\varphi_p:\\mathcal{F}_p\\rightarrow\\mathcal{G}_p\\) is an isomorphism for each \\(p\\in X\\). We prove \\(\\varphi(U):\\mathcal{F}(U)\\rightarrow \\mathcal{G}(U)\\) is an isomorphism for each open \\(U\\subseteq X\\). Fix \\(U\\subseteq X\\) and consider \\(\\varphi(U):\\mathcal{F}(U)\\rightarrow\\mathcal{G}(U)\\). Let \\(s\\in \\mathcal{F}(U)\\) be such that \\(\\varphi(U)(s)=0\\). Fix \\(p\\in U\\) and consider \\(\\varphi_p:\\mathcal{F}_p\\rightarrow \\mathcal{G}_p\\). We have \\((U,s)\\in \\mathcal{F}_p\\) with \\(\\varphi_p((U,s))=(U,\\varphi(U)(s))=0\\). As \\(\\varphi_p\\) is injective, this means that \\(s|_{W_p}=0\\) for some \\(W_p\\subseteq U\\) containing \\(p\\). This is true for all \\(p\\in U\\). So, we have an open cover \\(\\{W_p\\}_{p\\in U}\\) of \\(U\\) and \\(s\\in \\mathcal{F}(U)\\) such that \\(s|_{W_p}=0\\). Identity axiom of sheaf implies that \\(s=0\\). So, \\(\\varphi(U):\\mathcal{F}(U)\\rightarrow \\mathcal{G}(U)\\) is injective. Let \\(s\\in \\mathcal{G}(U)\\). Fix \\(p\\in U\\) and consider \\(\\varphi_p:\\mathcal{F}_p\\rightarrow \\mathcal{G}_p\\). As \\((U,s)\\in \\mathcal{G}_p\\) and \\(\\varphi_p\\) is surjective, there exists \\((V,t_p)\\in \\mathcal{F}_p\\) such that \\(\\varphi_p((V,t_p))=(U,s)\\) i.e., \\((V,\\varphi(V)(t_p))=(U,s)\\in \\mathcal{F}_p\\) i.e., \\(s|_{W_p}=\\varphi(V)(t_p)|_{W_p}\\) for some \\(p\\in W_p\\subseteq U\\cap V\\). Idea is to glue the sections \\(t_p|_{W_p}\\in \\mathcal{F}(W_p)\\) to get a section \\(t\\in \\mathcal{F}(U)\\). For that we show that \\(t_p|_{W_p\\cap W_q}=t_q|_{W_p\\cap W_q}\\). We have the following commuative diagram,which says that \\(\\varphi(W_p)(t_p|_{W_p})|_{W_p\\cap W_q}=\\varphi(W_p\\cap W_q)(t_p|_{W_p\\cap W_q}).\\)\nSimilarly, we have \\(\\varphi(W_q)(t_q|_{W_q})|_{W_p\\cap W_q}=\\varphi(W_p\\cap W_q)(t_q|_{W_p\\cap W_q}).\\)\nAs \\(s|_{W_p}=\\varphi(V)(t_p)|_{W_p}=\\varphi(W_p)(t_p|_{W_p})\\), we have \\(s|_{W_p\\cap W_q}=(s|_{W_p})|_{W_p\\cap W_q}=\\varphi(W_p)(t_p|_{W_p})|_{W_p\\cap W_q}=\\varphi(W_p\\cap W_q)(t_p|_{W_p\\cap W_q})\\)\nand \\(s|_{W_p\\cap W_q}=(s|_{W_q})|_{W_p\\cap W_q}=\\varphi(W_q)(t_q|_{W_q})|_{W_p\\cap W_q}=\\varphi(W_p\\cap W_q)(t_q|_{W_p\\cap W_q}).\\)\nSo, we have \\(\\varphi(W_p\\cap W_q)(t_p|_{W_p\\cap W_q})=s|_{W_p\\cap W_q}=\\varphi(W_p\\cap W_q)(t_q|_{W_p\\cap W_q}).\\)\nAs \\(\\varphi(W_p\\cap W_q):\\mathcal{F}(W_p\\cap W_q)\\rightarrow \\mathcal{G}(W_p\\cap W_q)\\) is injective we have \\(t_p|_{W_p\\cap W_q}=t_q|_{W_p\\cap W_q}\\). So, sections, \\(t_p|_{W_p}\\in \\mathcal{F}(W_p)\\) glue together and gives a section \\(t\\in \\mathcal{F}(U)\\) such that \\(t|_{W_p}=t_p|_{W_p}\\). We have \\(\\varphi(U)(t)|_{W_p}=\\varphi(W_p)(t|_{W_p})=\\varphi(W_p)(t_p|_{W_p})=s|_{W_p}.\\)\nAs \\(\\varphi(U)(t)|_{W_p}=s|_{W_p}\\) for all $p\\in U$ and as \\(\\{W_p\\}_{p\\in U}\\) is an open cover for \\(U\\), identitiy axiom of sheaves says that \\(\\varphi(U)(t)=s\\). Thus, \\(\\varphi(U):\\mathcal{F}(U)\\rightarrow \\mathcal{G}(U)\\) is surjective. So, \\(\\varphi_p:\\mathcal{F}_p\\rightarrow\\mathcal{G}_p\\) is an isomorphism for each \\(p\\in X\\) implies that \\(\\varphi:\\mathcal{F}\\rightarrow \\mathcal{G}\\) is an isomorphism. \u0026nbsp; \u0026nbsp; \u0026nbsp;","permalink":"https://praphulla-koushik.github.io/2017/07/28/morphism-of-sheaves-morphism-of-stalks/","summary":"\u003cstrong\u003eDefinition : \u003c/strong\u003eLet \\(X\\) be a topological space, \\(\\mathcal{F},\\mathcal{G}\\) be presheaves on \\(X\\). A morphism  \\(\\varphi:\\mathcal{F}\\rightarrow \\mathcal{G}\\) is a collection \\(\\{\\varphi(U):\\mathcal{F}(U)\\rightarrow \\mathcal{G}(U)\\}\\) indexing over all open \\(U\\subseteq X\\) such that the following diagram is commutative for open \\(U\\subseteq V\\subseteq X\\).\n\n\u003cimg class=\" size-full wp-image-1226 aligncenter\" src=\"/wp-media/2017/07/9d75150cbd-ql_9e6a39093afc407e385bb7497c06f5ea_l3.png\" alt=\"ql_9e6a39093afc407e385bb7497c06f5ea_l3\" width=\"137\" height=\"94\" /\u003e\n\n\u003cstrong\u003eMorphism of sheaves inducing Morphism of stalks : \u003c/strong\u003eWe  see that any morphism of sheaves \\(\\varphi:\\mathcal{F}\\rightarrow \\mathcal{G}\\) induces morphism of stalks \\(\\varphi_p:\\mathcal{F}_p\\rightarrow \\mathcal{G}_p\\) for each \\(p\\in X\\).\n\nFix \\(p\\in X\\). Let us define \\(\\varphi_p:\\mathcal{F}_p\\rightarrow \\mathcal{G}_p\\) i.e., for \\((U,s)\\in \\mathcal{F}_p\\) we  give an open set \\(V\\) containing \\(p\\) and a section \\(t\\in \\mathcal{G}(V)\\) giving an element \\((V,t)\\in \\mathcal{G}_p\\). One obvious choice of an open set containing \\(p\\) is \\(U\\). For this \\(U\\), we have \\(\\varphi(U):\\mathcal{F}(U)\\rightarrow \\mathcal{G}(U)\\) sending \\(s\\) to \\(\\varphi(U)(s)\\in \\mathcal{G}(U)\\). Define \\(\\varphi_p:\\mathcal{F}_p\\rightarrow \\mathcal{G}_p\\)  as \\((U,s)\\mapsto (U,\\varphi(U)(s))\\).\n\n\u003cstrong\u003eMap is well defined : \u003c/strong\u003eWe prove \\((U,s) \\sim (V,t)\\) implies \\((U,\\varphi(U)(s))\\sim (V,\\varphi(V)(t))\\).\n\nAs \\((U,s) \\sim (V,t)\\) there exists an open subset \\(W\\subseteq U\\cap V\\) containing \\(p\\) such that \\(s|_W=t|_W\\). We prove that \\(\\varphi(U)(s)|_W=\\varphi(V)(t)|_W\\) which implies \\((U,\\varphi(U)(s))\\sim (V,\\varphi(V)(t))\\).\n\nThe commutative diagram\u003cimg class=\" size-full wp-image-1227 aligncenter\" src=\"/wp-media/2017/07/744a42c726-ql_7816294aa49414ddfd513a4b26251a36_l3.png\" alt=\"ql_7816294aa49414ddfd513a4b26251a36_l3\" width=\"144\" height=\"94\" /\u003e\n\ngives \\(\\varphi(U)(s)|_W=\\varphi(W)(s|_W)\\).\n\nSimilar diagram in which \\(U\\) is replaced by \\(V\\) gives that \\(\\varphi(V)(t)|_W=\\varphi(W)(t|_W)\\).\n\nAs \\(s|_W=t|_W\\) we have \\(\\varphi(W)(s|_W)=\\varphi(W)(t|_W)\\), concluding that \\(\\varphi(U)(s)|_W=\\varphi(V)(t)|_W\\).\n\nSo, given a morphism of sheaves \\(\\varphi:\\mathcal{F}\\rightarrow \\mathcal{G}\\) we have well defined morphism \\(\\varphi_p:\\mathcal{F}_p\\rightarrow\\mathcal{G}_p\\) for each \\(p\\in X\\).\n\n\u003cstrong\u003eTheorem : A morphism of sheaves \\(\\varphi:\\mathcal{F}\\rightarrow \\mathcal{G}\\) is an isomorphism of sheaves iff the induced map \\(\\varphi_p:\\mathcal{F}_p\\rightarrow\\mathcal{G}_p\\) is an isomorphism for each \\(p\\in X\\).\u003c/strong\u003e\n\nProof : Let \\(\\varphi:\\mathcal{F}\\rightarrow \\mathcal{G}\\) is an isomorphism of sheaves i.e., \\(\\varphi(U):\\mathcal{F}(U)\\rightarrow \\mathcal{G}(U)\\) is an isomorphism of groups for each open  \\(U\\subseteq X\\). Fixing \\(p\\in X\\) we prove that \\(\\varphi_p:\\mathcal{F}_p\\rightarrow\\mathcal{G}_p\\) is an isomorphism.\n\nLet \\((U,s)\\in \\mathcal{F}_p\\) be such hat \\((U,\\varphi(U)(s))=0\\in \\mathcal{G}_p\\) i.e., \\(\\varphi(U)(s)|_W=0\\) for some open \\(W\\subseteq U\\). We thus have \\(\\varphi(W)(s|_W)=\\varphi(U)(s)|_W=0\\). As \\(\\varphi(W):\\mathcal{F}(W)\\rightarrow \\mathcal{G}(W)\\) is injective, this means \\(s|_W=0\\). Thus, \\(\\varphi_p:\\mathcal{F}_p\\rightarrow\\mathcal{G}_p\\) is an injective map.\n\nLet \\((V,t)\\in \\mathcal{G}_p\\) i.e., \\(p\\in V\\) and \\(t\\in \\mathcal{F}(V)\\). As \\(\\varphi(V):\\mathcal{F}(V)\\rightarrow \\mathcal{G}(V)\\) is surjective, there exists \\(s\\in \\mathcal{F}(V)\\) such that \\(\\varphi(V)(s)=t\\). So, \\(\\varphi_p((V,s))=(V,\\varphi(V)(s))=(V,t)\\). Thus, \\(\\varphi_p\\) is surjective.\n\nSo, \\(\\varphi:\\mathcal{F}\\rightarrow \\mathcal{G}\\) is an isomorphism of sheaves implies \\(\\varphi_p:\\mathcal{F}_p\\rightarrow\\mathcal{G}_p\\) is an isomorphism for each \\(p\\in X\\).\n\nConversely, suppose that \\(\\varphi_p:\\mathcal{F}_p\\rightarrow\\mathcal{G}_p\\) is an isomorphism for each \\(p\\in X\\). We prove \\(\\varphi(U):\\mathcal{F}(U)\\rightarrow \\mathcal{G}(U)\\) is an isomorphism for each open \\(U\\subseteq X\\).\n\nFix \\(U\\subseteq X\\) and consider \\(\\varphi(U):\\mathcal{F}(U)\\rightarrow\\mathcal{G}(U)\\). Let \\(s\\in \\mathcal{F}(U)\\) be such that \\(\\varphi(U)(s)=0\\). Fix \\(p\\in U\\) and consider \\(\\varphi_p:\\mathcal{F}_p\\rightarrow \\mathcal{G}_p\\). We have \\((U,s)\\in \\mathcal{F}_p\\) with \\(\\varphi_p((U,s))=(U,\\varphi(U)(s))=0\\). As \\(\\varphi_p\\) is injective, this means that \\(s|_{W_p}=0\\) for some  \\(W_p\\subseteq U\\) containing \\(p\\). This is true for all \\(p\\in U\\). So, we have an open cover \\(\\{W_p\\}_{p\\in U}\\) of \\(U\\) and \\(s\\in \\mathcal{F}(U)\\) such that \\(s|_{W_p}=0\\). Identity axiom of sheaf implies that \\(s=0\\). So, \\(\\varphi(U):\\mathcal{F}(U)\\rightarrow \\mathcal{G}(U)\\) is injective.\n\nLet \\(s\\in \\mathcal{G}(U)\\). Fix \\(p\\in U\\) and consider \\(\\varphi_p:\\mathcal{F}_p\\rightarrow \\mathcal{G}_p\\). As \\((U,s)\\in \\mathcal{G}_p\\) and \\(\\varphi_p\\) is surjective, there exists \\((V,t_p)\\in \\mathcal{F}_p\\) such that \\(\\varphi_p((V,t_p))=(U,s)\\) i.e., \\((V,\\varphi(V)(t_p))=(U,s)\\in \\mathcal{F}_p\\) i.e., \\(s|_{W_p}=\\varphi(V)(t_p)|_{W_p}\\) for some \\(p\\in W_p\\subseteq U\\cap V\\). Idea is to glue the sections \\(t_p|_{W_p}\\in \\mathcal{F}(W_p)\\) to get a section \\(t\\in \\mathcal{F}(U)\\). For that we show that \\(t_p|_{W_p\\cap W_q}=t_q|_{W_p\\cap W_q}\\).\n\nWe have the following commuative diagram,\u003cimg class=\" size-full wp-image-1229 aligncenter\" src=\"/wp-media/2017/07/9fa15c49b4-ql_a5b08531149147e84f9fe543a6476896_l3.png\" alt=\"ql_a5b08531149147e84f9fe543a6476896_l3\" width=\"238\" height=\"96\" /\u003ewhich says that\n\u003cp style=\"text-align:center;\"\u003e\\(\\varphi(W_p)(t_p|_{W_p})|_{W_p\\cap W_q}=\\varphi(W_p\\cap W_q)(t_p|_{W_p\\cap W_q}).\\)\u003c/p\u003e","tags":["sheaves"],"title":"Morphism of Sheaves - Morphism of Stalks","type":"wp-import"},{"categories":["Sheaves and Schemes","algebraic-geometry"],"content":"Let \\(A\\) be a ring. We have corresponding topological space \\(X=\\text{Spec}(A)\\), the collection of all prime ideals of \\(A\\) with Zariski Topology. We now define a sheaf on \\(X\\) called the structure sheaf, denoted by \\(\\mathcal{O}_X\\). This \\(X\\) with this structure sheaf \\(\\mathcal{O}_X\\) is called an affine scheme, These affine schemes are building blocks of what is called an arbitrary scheme. To define a sheaf on \\(X\\) we need to associate a ring for each \\(U\\) open in \\(X\\). We do that as follows :where the condition \\(\\dagger\\) says that given \\(p\\in U\\) we have \\(s(p)\\in A_p\\) and that \\(s\\) is locally a fraction i.e., given \\(p\\in U\\) there exists an open set \\(U(p)\\subseteq U\\) and \\(a\\in A, f\\in A\\) such that \\(s(q)=\\frac{a}{f}\\in A_q\\) for all \\(q\\in U(p)\\). The verification that this gives a sheaf on \\(X\\) is same as that of the verification that sheafification of a sheaf is a sheaf. We can see the similarity between the definitions. More details can be found here about the similarity. So, \\((X,\\mathcal{O}_X)\\) forms a ringed space, which we call an affine scheme. We will now see results about the global sections, stalks of structure sheaf and what does structure sheaf give on basic open subsets of \\(X=\\text{Spec}(A)\\). Proposition : Let \\(A\\) be a ring, and \\((\\text{Spec A},\\mathcal{O})\\) its spectrum. For any \\(\\mathfrak{p}\\in \\text{Spec A}\\), the stalk \\(\\mathcal{O}_{\\mathfrak{p}}\\) of the sheaf \\(\\mathcal{O}_{}\\) is isomorphic to the local ring \\(A_{\\mathfrak{p}}\\) i.e., \\(\\mathcal{O}_{\\mathfrak{p}}\\cong A_{\\mathfrak{p}}\\). For any element \\(f\\in A\\), the ring \\(\\mathcal{O}(D(f))\\) is isomorphic to the localized ring \\(A_f\\) i.e., \\(\\mathcal{O}(D(f))\\cong A_f\\). In particular, \\(\\Gamma(\\text{Spec A}, \\mathcal{O})\\cong A\\). Proof : Let \\(\\mathfrak{p}\\in X\\). We define a map \\(\\mathcal{O}_{\\mathfrak{p}}\\rightarrow A_{\\mathfrak{p}}\\) and show that this is a bijection. Defining the map - Let \\([(U,s)]\\in \\mathcal{O}_{\\mathfrak{p}}\\) i.e., \\(U\\) is an open set in \\(X\\) containing \\(p\\) and \\(s\\in \\mathcal{O}(U)\\). By definition, \\(s:U\\rightarrow \\bigsqcup_{\\mathfrak{q}\\in U}A_{\\mathfrak{q}}\\). To get an element in \\(A_{\\mathfrak{p}}\\) given \\(s\\), its only natural to consider image of \\(\\mathfrak{p}\\) under \\(s\\) namely \\(s(\\mathfrak{p})\\in A\\). Defining \\(s\\mapsto s(\\mathfrak{p})\\) gives a map \\(\\mathcal{O}_{\\mathfrak{p}}\\rightarrow A_{\\mathfrak{p}}\\). Showing that the map is well defined - Suppose \\([(U,s)]=[(V,t)]\\in \\mathcal{O}_{\\mathfrak{p}}\\) i.e., there is an open set \\(W\\subset U\\cap V\\) containing \\(\\mathfrak{p}\\) such that \\(s|_{W}=t|_W\\). As \\(\\mathfrak{p}\\in W\\), we have in particular \\(s(\\mathfrak{p})=t(\\mathfrak{p})\\). So, there is a well defined map \\(\\mathcal{O}_{\\mathfrak{p}}\\rightarrow A_{\\mathfrak{p}}\\). Showing that the map is Injective - For \\([(U,s)],[(V,t)]\\in \\mathcal{O}_{\\mathfrak{p}}\\) with \\(s(\\mathfrak{p})=t(\\mathfrak{p})\\), we show that \\([(U,s)]=[(V,t)]\\in \\mathcal{O}_{\\mathfrak{p}}\\). As \\(\\mathfrak{p}\\in U\\), for \\(s: U\\rightarrow\\bigsqcup_{\\mathfrak{q}\\in U}A_{\\mathfrak{q}}\\) there exists open \\(U(\\mathfrak{p})\\subset U\\) containing \\(\\mathfrak{p}\\) and \\(a,f\\in A\\) such that \\(s(\\mathfrak{q})=\\frac{a}{f}\\) for all \\(\\mathfrak{q}\\in U(\\mathfrak{p})\\). Similarly, for \\(t:V\\rightarrow\\bigsqcup_{\\mathfrak{q}\\in V}A_{\\mathfrak{q}}\\) there exists open \\(V(\\mathfrak{q})\\subseteq V\\) and \\(b,g\\in A\\) such that \\(t(\\mathfrak{q})=\\frac{b}{g}\\) for all \\(\\mathfrak{q}\\in V(\\mathfrak{p})\\). In particular, \\(\\frac{a} {f}=s(\\mathfrak{p})=t(\\mathfrak{p})=\\frac{b}{g}\\). Let \\(\\mathfrak{q}\\in U(\\mathfrak{p})\\cap V(\\mathfrak{p})\\). Then, \\(s(q)=\\frac{a}{f}=\\frac{b}{g}=t(q)\\). Thus, we have \\(s|_{U(\\mathfrak{p})\\cap V(\\mathfrak{q})}=t|_{U(\\mathfrak{p})\\cap V(\\mathfrak{q})}\\). Thus, \\([(U,s)]=[(V,t)]\\). So, \\(s\\mapsto s(\\mathfrak{p})\\) is injective. Showing that the map is surjective - Let \\(\\frac{a}{f}\\in A_{\\mathfrak{p}}\\), we want to choose an open set \\(U\\) containing \\(\\mathfrak{p}\\) and \\(s:U\\rightarrow \\bigsqcup_{\\mathfrak{q}\\in U}A_{\\mathfrak{q}}\\) such that \\(s(\\mathfrak{p})=\\frac{a}{f}\\). One choice for \\(s\\) is sending \\(q\\) to image of \\(\\frac{a}{f}\\) in \\(A_{\\mathfrak{q}}\\). For this, we need \\(f\\notin\\mathfrak{q}\\) i.e., \\(\\mathfrak{q}\\in D(f)\\). Let \\(U=D(f)\\) and consider \\(s:U\\rightarrow \\bigsqcup_{\\mathfrak{q}\\in U}A_{\\mathfrak{q}}\\) sending \\(\\mathfrak{q}\\) to image of \\(\\frac{a}{f}\\) in \\(A_{\\mathfrak{q}}\\). We then have \\(s(\\mathfrak{p})=\\frac{a}{f}\\in A_{\\mathfrak{p}}\\). Thus, the map is surjective. So, we have isomorphism \\(\\mathcal{O}_{\\mathfrak{p}}\\rightarrow A_{\\mathfrak{p}}\\) given by \\(s\\mapsto s(\\mathfrak{p})\\). Let \\(f\\in A\\). We define a map \\(A_f\\rightarrow \\mathcal{O}(D(f))\\) and show that this is a bijection. Defining the map - Given \\(\\frac{a}{f}\\in A_f\\) we assign \\(s\\in \\mathcal{O}(D(f))\\) where \\(s:D(f)\\rightarrow \\bigsqcup_{q\\in D(f)}A_q\\). Let \\(q\\in D(f)\\) then, \\(f\\notin q\\). So, \\(\\frac{a}{f}\\) is defined in \\(A_q\\). So, define \\(s(q)\\) to be the image of \\(\\frac{a}{f}\\) in \\(A_q\\) for each \\(q\\in D(f)\\). It is clearly a well defined function. Similarly we define for \\(\\frac{a}{f^n}\\in A_f\\) a map \\(s:D(f^n)=D(f)\\rightarrow \\bigsqcup_{q\\in D(f)}A_q\\) as \\(q\\mapsto \\frac{a}{f^n}\\in A_q\\). Showing that the map is injective - Suppose \\(\\frac{a}{f^n},\\frac{b}{f^m}\\in A_f\\) is such that the corresponding maps \\(s,t\\) are equal i.e., \\(\\frac{a}{f^n}=\\frac{b}{f^m}\\in A_q~\\forall q\\in D(f)\\) i.e., given \\(q\\in D(f)\\) there exists \\(t_q\\notin q\\) such that \\(t_q(af^m-bf^n)=0\\). Consider the case when \\(D(f)=\\{q\\}\\). As \\(t\\notin q\\), we have \\(q\\in D(t)\\) i.e., \\(D(f)\\subseteq D(t)\\) i.e., \\(V(t)\\subseteq V(f)\\) i.e., \\(\\sqrt{(f)}\\subseteq \\sqrt{(t)}\\). As \\(f\\in \\sqrt{(f)}\\) we have \\(f^l=td\\) for some \\(d\\in A\\). We have \\(t(af^m-bf^n)=0\\) which implies \\(td(af^m-bf^n)=0\\) i.e., \\(f^l(af^m-bf^n)=0\\) i.e., \\(\\frac{a}{f^n}=\\frac{b}{f^m}\\in A_f\\) and we are done. Suppose \\(D(f)=\\{q_i\\}_{i\\in \\Lambda}\\). As \\(t_i\\notin q_i\\) we have \\(q_i\\in D(t_i)\\) i.e., \\(D(f)\\subseteq \\bigcup_{i\\in \\Lambda} D(t_i)\\). As in previous observation, this means \\(f^l\\) is in the ideal generated by \\(\\{t_i\\}\\) for some \\(l\\in \\mathbb{N}\\). So, we have (after rearranging indices in \\(\\Lambda\\)) \\(f^l=a_1t_1+\\cdots+a_nt_n\\) for some \\(a_i\\in A\\). As \\(t_i(af^m-bf^n)=0\\), we have \\(a_it_i(af^m-bf^n)=0\\) for all \\(i\\). So, \\(\\sum_{i=1}^na_it_i(af^m-bf^n)=0\\) i.e., \\(f^l(af^m-bf^n)=0\\). Thus, \\(\\frac{a}{f^n}=\\frac{b}{f^m}\\in A_f\\). Thus, the map \\(A_f\\rightarrow \\mathcal{O}(D(f))\\) is injective. \u0026nbsp;","permalink":"https://praphulla-koushik.github.io/2017/07/25/structure-sheaf-on-spectrum-of-a-ring/","summary":"Let \\(A\\) be a ring. We have corresponding topological space \\(X=\\text{Spec}(A)\\), the collection of all prime ideals of \\(A\\) with Zariski Topology. We now define a sheaf on \\(X\\) called the structure sheaf, denoted by \\(\\mathcal{O}_X\\). This \\(X\\) with this structure sheaf \\(\\mathcal{O}_X\\) is called an affine scheme,  These affine schemes  are building blocks of what is called an arbitrary scheme.\n\nTo define a sheaf on \\(X\\) we need to associate a ring for each \\(U\\) open in \\(X\\). We do that as follows :\u003cimg class=\" size-full wp-image-621 aligncenter\" src=\"/wp-media/2017/07/85383b043a-ql_95ea6db253a445248fdc006fdfabc129_l3.png\" alt=\"ql_95ea6db253a445248fdc006fdfabc129_l3\" width=\"392\" height=\"64\" /\u003ewhere the condition \\(\\dagger\\) says that given \\(p\\in U\\) we have \\(s(p)\\in A_p\\) and that \\(s\\) is locally a fraction i.e., given \\(p\\in U\\) there exists an open set \\(U(p)\\subseteq U\\) and \\(a\\in A, f\\in A\\) such that \\(s(q)=\\frac{a}{f}\\in A_q\\) for all \\(q\\in U(p)\\).\n\nThe verification that this gives a sheaf on \\(X\\) is same as that of the verification that sheafification of a sheaf is a sheaf. We can see the similarity between the definitions. \u003ca href=\"https://mathoverflow.net/questions/80548/affine-scheme-on-speca-of-a-ring-a-as-the-sheafification-of-a-pre-sheave-on-sp\"\u003eMore details can be found here about the similarity\u003c/a\u003e. So, \\((X,\\mathcal{O}_X)\\) forms a ringed space, which we call an affine scheme.\n\nWe will now see results about the global sections, stalks of structure sheaf and what does structure sheaf give on basic open subsets of \\(X=\\text{Spec}(A)\\).\n\n\u003cstrong\u003eProposition : \u003c/strong\u003eLet \\(A\\) be a ring, and \\((\\text{Spec A},\\mathcal{O})\\) its spectrum.\n\u003col\u003e\n\t\u003cli\u003eFor any \\(\\mathfrak{p}\\in \\text{Spec A}\\), the stalk \\(\\mathcal{O}_{\\mathfrak{p}}\\) of the sheaf \\(\\mathcal{O}_{}\\) is isomorphic to the local ring \\(A_{\\mathfrak{p}}\\) i.e., \\(\\mathcal{O}_{\\mathfrak{p}}\\cong A_{\\mathfrak{p}}\\).\u003c/li\u003e\n\t\u003cli\u003eFor any element \\(f\\in A\\), the ring \\(\\mathcal{O}(D(f))\\) is isomorphic to the localized ring \\(A_f\\) i.e., \\(\\mathcal{O}(D(f))\\cong A_f\\).\u003c/li\u003e\n\t\u003cli\u003eIn particular, \\(\\Gamma(\\text{Spec A}, \\mathcal{O})\\cong A\\).\u003c/li\u003e\n\u003c/ol\u003e\n\u003cstrong\u003eProof : \u003c/strong\u003eLet \\(\\mathfrak{p}\\in X\\). We define a map \\(\\mathcal{O}_{\\mathfrak{p}}\\rightarrow A_{\\mathfrak{p}}\\) and show that this is a bijection.\n\n\u003cstrong\u003eDefining the map - \u003c/strong\u003eLet \\([(U,s)]\\in \\mathcal{O}_{\\mathfrak{p}}\\) i.e., \\(U\\) is an open set in \\(X\\) containing \\(p\\) and \\(s\\in \\mathcal{O}(U)\\). By definition, \\(s:U\\rightarrow \\bigsqcup_{\\mathfrak{q}\\in U}A_{\\mathfrak{q}}\\). To get an element in \\(A_{\\mathfrak{p}}\\) given \\(s\\), its only natural to consider image of \\(\\mathfrak{p}\\) under \\(s\\) namely \\(s(\\mathfrak{p})\\in A\\). Defining \\(s\\mapsto s(\\mathfrak{p})\\) gives a map \\(\\mathcal{O}_{\\mathfrak{p}}\\rightarrow A_{\\mathfrak{p}}\\).\n\n\u003cstrong\u003eShowing that the map is well defined\u003c/strong\u003e -  Suppose \\([(U,s)]=[(V,t)]\\in \\mathcal{O}_{\\mathfrak{p}}\\) i.e., there is an open set \\(W\\subset U\\cap V\\) containing \\(\\mathfrak{p}\\) such that \\(s|_{W}=t|_W\\). As \\(\\mathfrak{p}\\in W\\), we have in particular \\(s(\\mathfrak{p})=t(\\mathfrak{p})\\). So, there is a well defined map \\(\\mathcal{O}_{\\mathfrak{p}}\\rightarrow A_{\\mathfrak{p}}\\).\n\n\u003cstrong\u003eShowing that the map is Injective -\u003c/strong\u003e For \\([(U,s)],[(V,t)]\\in \\mathcal{O}_{\\mathfrak{p}}\\) with \\(s(\\mathfrak{p})=t(\\mathfrak{p})\\), we show that \\([(U,s)]=[(V,t)]\\in \\mathcal{O}_{\\mathfrak{p}}\\).\n\nAs \\(\\mathfrak{p}\\in U\\), for \\(s: U\\rightarrow\\bigsqcup_{\\mathfrak{q}\\in U}A_{\\mathfrak{q}}\\) there exists open \\(U(\\mathfrak{p})\\subset U\\) containing \\(\\mathfrak{p}\\) and \\(a,f\\in A\\) such that  \\(s(\\mathfrak{q})=\\frac{a}{f}\\) for all \\(\\mathfrak{q}\\in U(\\mathfrak{p})\\). Similarly, for \\(t:V\\rightarrow\\bigsqcup_{\\mathfrak{q}\\in V}A_{\\mathfrak{q}}\\) there exists open \\(V(\\mathfrak{q})\\subseteq V\\) and  \\(b,g\\in A\\) such that \\(t(\\mathfrak{q})=\\frac{b}{g}\\) for all \\(\\mathfrak{q}\\in V(\\mathfrak{p})\\).  In particular, \\(\\frac{a} {f}=s(\\mathfrak{p})=t(\\mathfrak{p})=\\frac{b}{g}\\).\n\nLet \\(\\mathfrak{q}\\in U(\\mathfrak{p})\\cap V(\\mathfrak{p})\\). Then, \\(s(q)=\\frac{a}{f}=\\frac{b}{g}=t(q)\\). Thus,  we have \\(s|_{U(\\mathfrak{p})\\cap V(\\mathfrak{q})}=t|_{U(\\mathfrak{p})\\cap V(\\mathfrak{q})}\\). Thus, \\([(U,s)]=[(V,t)]\\). So, \\(s\\mapsto s(\\mathfrak{p})\\) is injective.\n\n\u003cstrong\u003eShowing that the map is surjective - \u003c/strong\u003eLet \\(\\frac{a}{f}\\in A_{\\mathfrak{p}}\\), we want to choose an open set \\(U\\) containing \\(\\mathfrak{p}\\) and \\(s:U\\rightarrow \\bigsqcup_{\\mathfrak{q}\\in U}A_{\\mathfrak{q}}\\) such that \\(s(\\mathfrak{p})=\\frac{a}{f}\\). One choice for \\(s\\) is sending \\(q\\) to image of \\(\\frac{a}{f}\\) in \\(A_{\\mathfrak{q}}\\). For this, we need \\(f\\notin\\mathfrak{q}\\) i.e., \\(\\mathfrak{q}\\in D(f)\\). Let \\(U=D(f)\\) and consider \\(s:U\\rightarrow \\bigsqcup_{\\mathfrak{q}\\in U}A_{\\mathfrak{q}}\\) sending \\(\\mathfrak{q}\\) to image of \\(\\frac{a}{f}\\) in \\(A_{\\mathfrak{q}}\\). We then have \\(s(\\mathfrak{p})=\\frac{a}{f}\\in A_{\\mathfrak{p}}\\). Thus, the map is surjective.\n\nSo, we have isomorphism \\(\\mathcal{O}_{\\mathfrak{p}}\\rightarrow A_{\\mathfrak{p}}\\) given by \\(s\\mapsto s(\\mathfrak{p})\\).\n\nLet \\(f\\in A\\). We define a map \\(A_f\\rightarrow \\mathcal{O}(D(f))\\) and show that this is a bijection.\n\n\u003cstrong\u003eDefining the map -\u003c/strong\u003e Given \\(\\frac{a}{f}\\in A_f\\) we assign  \\(s\\in \\mathcal{O}(D(f))\\) where \\(s:D(f)\\rightarrow \\bigsqcup_{q\\in D(f)}A_q\\). Let \\(q\\in D(f)\\) then, \\(f\\notin q\\). So, \\(\\frac{a}{f}\\) is defined in \\(A_q\\). So, define \\(s(q)\\) to be the image of \\(\\frac{a}{f}\\) in \\(A_q\\) for each \\(q\\in D(f)\\). It is clearly a well defined function. Similarly we define for \\(\\frac{a}{f^n}\\in A_f\\) a map \\(s:D(f^n)=D(f)\\rightarrow \\bigsqcup_{q\\in D(f)}A_q\\) as \\(q\\mapsto \\frac{a}{f^n}\\in A_q\\).\n\n\u003cstrong\u003eShowing that the map is injective - \u003c/strong\u003eSuppose \\(\\frac{a}{f^n},\\frac{b}{f^m}\\in A_f\\) is such that the corresponding maps \\(s,t\\) are equal i.e., \\(\\frac{a}{f^n}=\\frac{b}{f^m}\\in A_q~\\forall q\\in D(f)\\) i.e., given \\(q\\in D(f)\\) there exists \\(t_q\\notin q\\) such that \\(t_q(af^m-bf^n)=0\\).\n\nConsider the case when \\(D(f)=\\{q\\}\\). As \\(t\\notin q\\), we have \\(q\\in D(t)\\) i.e., \\(D(f)\\subseteq D(t)\\)  i.e., \\(V(t)\\subseteq V(f)\\) i.e., \\(\\sqrt{(f)}\\subseteq \\sqrt{(t)}\\). As \\(f\\in \\sqrt{(f)}\\) we have \\(f^l=td\\) for some \\(d\\in A\\). We have \\(t(af^m-bf^n)=0\\) which implies \\(td(af^m-bf^n)=0\\) i.e., \\(f^l(af^m-bf^n)=0\\) i.e., \\(\\frac{a}{f^n}=\\frac{b}{f^m}\\in A_f\\) and we are done.\n\nSuppose \\(D(f)=\\{q_i\\}_{i\\in \\Lambda}\\). As \\(t_i\\notin q_i\\) we have \\(q_i\\in D(t_i)\\) i.e., \\(D(f)\\subseteq \\bigcup_{i\\in \\Lambda} D(t_i)\\). As in previous observation, this means \\(f^l\\) is in the ideal generated by \\(\\{t_i\\}\\) for some \\(l\\in \\mathbb{N}\\). So,  we have (after rearranging indices in \\(\\Lambda\\)) \\(f^l=a_1t_1+\\cdots+a_nt_n\\)  for some \\(a_i\\in A\\). As \\(t_i(af^m-bf^n)=0\\), we have \\(a_it_i(af^m-bf^n)=0\\) for all \\(i\\). So, \\(\\sum_{i=1}^na_it_i(af^m-bf^n)=0\\) i.e., \\(f^l(af^m-bf^n)=0\\). Thus, \\(\\frac{a}{f^n}=\\frac{b}{f^m}\\in A_f\\). Thus, the map \\(A_f\\rightarrow \\mathcal{O}(D(f))\\) is injective.\n\n\u0026nbsp;","tags":["schemes","sheaves"],"title":"Structure sheaf on spectrum of a ring","type":"wp-import"},{"categories":["Sheaves and Schemes","algebraic-geometry"],"content":"This is an exercise from Hartshorne's Algebraic Geometry book. A part of this exercise is called QcQs lemma in Ravi Vakil's Foundations of Algebraic Geometry notes.\n\n\u0026nbsp;","permalink":"https://praphulla-koushik.github.io/2017/07/23/qcqs-lemma/","summary":"This is an exercise from Hartshorne's Algebraic Geometry book. A part of this exercise is called QcQs lemma in Ravi Vakil's Foundations of Algebraic Geometry notes.\n\n\u0026nbsp;","tags":["hartshorne"],"title":"QcQs lemma","type":"wp-import"},{"categories":["algebraic-geometry","category-theory"],"content":"Let \\(A\\) be a ring and let \\((X,\\mathcal{O}_X)\\) be a scheme. Given a morphism \\(f:X\\rightarrow \\text{Spec}(A)\\) we have an associated map on sheaves \\(f^{\\#}:\\mathcal{O}_{\\text{Spec}(A)}\\rightarrow f_* \\mathcal{O}_X\\). Taking global sections, we obtain a homomorphism \\(A\\rightarrow \\mathcal{O}_X(X)\\). Thus there is a natural map \\(\\alpha: \\text{Hom}_{\\text{Schemes}}(X,\\text{Spec}(A))\\rightarrow \\text{Hom}_{\\text{Rings}}(A,\\mathcal{O}_X(X)).\\)\nThen, \\(\\alpha\\) is bijective. We will try to understand this adjointness of Global section functor and Spec functor. Suppose we are given a scheme \\((X,\\mathcal{O}_X)\\) and a ring homomoprhism \\(\\varphi: A\\rightarrow \\mathcal{O}_X(X)\\). We construct a morphism of schemes \\((f,f^{\\#}):(X,\\mathcal{O}_X)\\rightarrow (\\text{Spec}(A),\\mathcal{O}_{\\text{Spec}(A)})\\). We first define morphism of topological spaces \\(f:X\\rightarrow \\text{Spec}(A)\\). Let \\(x\\in X\\), we want to assign a prime ideal \\(P\\) in \\(A\\). Let \\(X=\\text{Spec}(B)\\) and \\(x=\\mathfrak{P}\\in X=\\text{Spec}(B)\\), as we have a ring homomorphism \\(\\varphi : A\\rightarrow \\mathcal{O}_X(X)=O_{\\text{Spec}(B)}(\\text{Spec}(B))=B\\)\n\\(\\varphi^{-1}(\\mathfrak{P})\\) is a prime ideal in \\(A\\) and we define \\(f(x)=\\varphi^{-1}(\\mathfrak{P})\\). This defines a morphism of topological spaces \\(\\text{Spec}(B)\\rightarrow \\text{Spec}(A)\\). Suppose \\(X\\) is an arbitrary scheme, given \\(x\\in X\\) there is no natural choice of prime ideal in \\(\\mathcal{O}_X(X)\\) whose inverse in \\(A\\) defines a function \\(X\\rightarrow \\text{Spec}(A)\\). Given \\(x\\in X\\) we have local ring \\(\\mathcal{O}_x\\) which has unique maximal ideal \\(\\mathfrak{m}_x\\) which is in particular a prime ideal. We have canonical ring homomorphism \\(O_X(X)\\xrightarrow{\\pi} \\mathcal{O}_x\\) with \\(s\\mapsto s_x\\). So, \\(\\pi^{-1}(\\mathfrak{m}_x)\\) is a prime ideal in \\(\\mathcal{O}_X(X)\\), and its inverse image under \\(\\varphi\\) namely \\(\\varphi^{-1}(\\pi^{-1}(\\mathfrak{m}_x))\\) is a prime ideal in \\(A\\). We thus have a map \\(f:X\\rightarrow \\text{Spec}(A)\\) with \\(x\\rightarrow \\varphi^{-1}(\\pi^{-1}(\\mathfrak{m}_x))\\). We prove that \\(f:X\\rightarrow \\text{Spec}(A)\\) is a continuous map. It suffices to prove \\(f^{-1}(D(a))\\) is an open set in \\(X\\) for each \\(a\\in A\\) as \\(\\{D(a)\\}_{a\\in A}\\) is a basis for topology on \\(\\text{Spec}(A)\\). We have Result : Let \\(Y\\) be a scheme, and \\(f\\in \\mathcal{O}_Y(Y)\\). Then \\(Y_f=\\{x\\in Y : f_x\\notin \\mathfrak{m}_x\\}\\) is open in \\(Y\\). Here, \\(\\varphi(a)\\in \\mathcal{O}_X(X)\\). Thus, \\(X_{\\varphi(a)}\\) is an open subset of \\(X\\) i.e., \\(f^{-1}(D(a))\\) is an open subset of \\(X\\). So, \\(f\\) is a continuous function. We now construct morphism of schemes \\(f^{\\#}:\\mathcal{O}_{\\text{Spec}(A)}\\rightarrow f_*\\mathcal{O}_X\\). It suffices to define morphisms \\(f^{\\#}(D(a)):\\mathcal{O}_{\\text{Spec}(A)}(D(a))\\rightarrow f_*\\mathcal{O}_X(D(a))=\\mathcal{O}_X(f^{-1}(D(a)))=\\mathcal{O}_X(X_{\\varphi(a)}).\\)\nAs \\(\\mathcal{O}_{\\text{Spec}(A)}(D(a))=A_a\\), it boils down to defining morphism \\(f^{\\#}(D(a)):A_a\\rightarrow \\mathcal{O}_X(X_{\\varphi(a)})\\). Considering the isomorphism \\(\\mathcal{O}_X(X_{\\varphi(a)})\\cong (\\mathcal{O}_X)_{\\varphi(a)}\\) (which is natural to expect and happens most of the times, in particular when \\(X\\) is quasi compact and quasi separated) where the right side component is localization of the ring \\(\\mathcal{O}_X(X)\\) at \\(\\varphi(a)\\), it boils down to defining morphism \\(f^{\\#}(D(a)):A_a\\rightarrow \\mathcal{O}_X(X)_{\\varphi(a)}\\). Given \\(\\varphi: A\\rightarrow \\mathcal{O}_X(X)\\), we have induced map \\(A_a\\rightarrow \\mathcal{O}_X(X)_{\\varphi(a)}\\) for each \\(a\\in A\\). Set \\(f^{\\#}(D(a)): A_a\\rightarrow \\mathcal{O}_X(X)_{\\varphi(a)}\\) to be the composition \\(A_a\\rightarrow \\mathcal{O}_X(X)_{\\varphi(a)} \\xrightarrow{\\cong} (\\mathcal{O}_X)_{\\varphi(a)}\\)\nWhile proving isomorphism \\(\\mathcal{O}_X(X_{\\varphi(a)})\\cong (\\mathcal{O}_X)_{\\varphi(a)}\\), we need the condition that \\(X\\) is quasi compact and quasi separated only to show that the map is surjective. We always have natural map \\(\\mathcal{O}_X(X_{\\varphi(a)})\\rightarrow (\\mathcal{O}_X)_{\\varphi(a)}\\). So, composition \\(A_a\\rightarrow \\mathcal{O}_X(X)_{\\varphi(a)} \\rightarrow (\\mathcal{O}_X)_{\\varphi(a)}\\) gives maps \\(f^{\\#}(D(a))\\) for each \\(a\\in A\\). These maps glue to give morphism of schemes \\((f,f^{\\#}):(X,\\mathcal{O}_X)\\rightarrow (\\text{Spec}(A),\\mathcal{O}_{\\text{Spec}(A)})\\).\nThus, \\(\\alpha\\) is surjective and thus an isomorphism. \u0026nbsp; \u0026nbsp; \u0026nbsp; \u0026nbsp;","permalink":"https://praphulla-koushik.github.io/2017/07/22/adjointness-of-the-global-section-functor-and-the-spec-functor/","summary":"Let \\(A\\) be a ring and let \\((X,\\mathcal{O}_X)\\) be a scheme. Given a morphism \\(f:X\\rightarrow \\text{Spec}(A)\\) we have an associated map on sheaves \\(f^{\\#}:\\mathcal{O}_{\\text{Spec}(A)}\\rightarrow f_* \\mathcal{O}_X\\). Taking global sections, we obtain a homomorphism \\(A\\rightarrow \\mathcal{O}_X(X)\\). Thus there is a natural map\n\u003cp style=\"text-align:center;\"\u003e\\(\\alpha: \\text{Hom}_{\\text{Schemes}}(X,\\text{Spec}(A))\\rightarrow \\text{Hom}_{\\text{Rings}}(A,\\mathcal{O}_X(X)).\\)\u003c/p\u003e\nThen, \\(\\alpha\\) is bijective.\n\nWe will try to understand this adjointness of Global section functor and Spec functor.\n\nSuppose we are given a scheme \\((X,\\mathcal{O}_X)\\) and a ring homomoprhism \\(\\varphi: A\\rightarrow \\mathcal{O}_X(X)\\). We construct a morphism of schemes \\((f,f^{\\#}):(X,\\mathcal{O}_X)\\rightarrow (\\text{Spec}(A),\\mathcal{O}_{\\text{Spec}(A)})\\).\n\nWe first define morphism of topological spaces \\(f:X\\rightarrow \\text{Spec}(A)\\). Let \\(x\\in X\\), we want to assign a prime ideal \\(P\\) in \\(A\\).\n\nLet \\(X=\\text{Spec}(B)\\) and \\(x=\\mathfrak{P}\\in X=\\text{Spec}(B)\\), as we have a ring homomorphism\n\u003cp style=\"text-align:center;\"\u003e\\(\\varphi : A\\rightarrow \\mathcal{O}_X(X)=O_{\\text{Spec}(B)}(\\text{Spec}(B))=B\\)\u003c/p\u003e","tags":["schemes"],"title":"Adjointness of the global section functor and the Spec functor","type":"wp-import"},{"categories":["algebraic-geometry"],"content":"Definition : Let \\(S\\) be a scheme. An \\(S\\) scheme is a scheme \\(X\\) together with a morphism \\(p:X\\rightarrow S\\). A morphism of \\(S\\) schemes \\((X,p:X\\rightarrow S)\\) and \\((Y,q:Y\\rightarrow S)\\) is a morphism of schemes \\(f:X\\rightarrow Y\\) such that \\(q\\circ f=p\\). \u0026nbsp;","permalink":"https://praphulla-koushik.github.io/2017/07/22/fiber-product-of-schemes/","summary":"\u003cstrong\u003eDefinition : \u003c/strong\u003eLet \\(S\\) be a scheme. An \\(S\\) scheme is a scheme \\(X\\) together with a morphism \\(p:X\\rightarrow S\\).  A morphism of \\(S\\) schemes \\((X,p:X\\rightarrow S)\\) and \\((Y,q:Y\\rightarrow S)\\) is a morphism of schemes \\(f:X\\rightarrow Y\\) such that \\(q\\circ f=p\\).\n\n\u0026nbsp;","tags":["schemes"],"title":"Fiber product of Schemes","type":"wp-import"},{"categories":["Algebraic geometry"],"content":"let \\(\\mathcal{F}\\) be a presheaf. We define associated presheaf of \\(\\mathcal{F}\\) to be the sheaf given by map \\(U\\mapsto \\widetilde{F}(U)\\) where where the condition \\(^\\dagger\\) says that \\(s(p)\\in \\mathcal{F}_p\\) for every \\(p\\in U\\) and there exists an open subset \\(U(p)\\subset U\\) containing \\(p\\) and a section \\(t\\in \\mathcal{F}(U(p))\\) such that \\(t_q=s(q)\\) for every \\(q\\in U(p)\\). Suppose \\(\\mathcal{F}\\) is actually a sheaf then we will see that \\(\\widetilde{F}(U)\\cong \\mathcal{F}(U)\\) for every open \\(U\\subseteq X\\). Let \\(s\\in \\widetilde{F}(U)\\) i.e., \\(s:U\\rightarrow \\bigsqcup_{i\\in \\Lambda}\\mathcal{F}_{p_i}\\) (this notation is just for my comfort, we have \\(U=\\bigcup_{i\\in \\Lambda}\\{p_i\\}\\)) satisfying some conditions given above. Given such \\(s\\) we want to assign an element in \\(\\mathcal{F}(U)\\).\nGiven \\(s\\) we have \\(t^{p_i}\\in \\mathcal{F}(U(p_i))\\) for each \\(i\\in \\Lambda\\). If we can show that these sections agree on intersections, as \\(\\{U(p_i)\\}_{i\\in \\Lambda}\\) cover \\(U\\) and as \\(\\mathcal{F}\\) is a sheaf, we get a section \\(t\\in \\mathcal{F}(U)\\). It remains to prove that \\(t^{p_i}|_{U(p_i)\\bigcap U(q_i)}=t^{p_j}|_{U(p_i)\\bigcap U(p_j)}\\) for each \\(i,j\\in \\Lambda\\). It is equivalent to proving that \\(t^{p_i}_q=t^{p_j}_q\\) for every \\(q\\in U(p_i)\\cap U(q_j)\\). By definition, as \\(q\\in U(p_i)\\) we have \\(t^{p_i}_q=s(q)\\), as \\(q\\in U(p_j)\\) we have \\(t^{p_j}_q=s(q)\\). So, we have \\(t^{p_i}_q=t^{p_j}_q\\) for every \\(q\\in U(p_i)\\cap U(q_j)\\) and this is true for all \\(i,j\\in \\Lambda\\). Thus, we can get a section \\(t\\in \\mathcal{F}(U)\\) such that \\(t|_{U(p_i)}=t^{p_i}\\). So, we have a map \\(\\widetilde{F}(U)\\rightarrow F(U)\\). It remains to prove that it is one one and onto. Let \\(s_1,s_2\\in \\mathcal{F}(U)\\) be such that \\(t_1=t_2\\in \\mathcal{F}(U)\\) i.e., \\((t_1)_q=(t_2)_q\\) for all \\(q\\in U\\). As \\((t_1)_q=s_1(q)\\) and \\((t_2)_q=s_2(q)\\) we have \\(s_1(q)=s_2(q)\\) for all \\(q\\in U\\). Thus, \\(s_1=s_2\\), the map is injective. It remains to prove that it is surjective. Given \\(t\\in \\mathcal{F}(U)\\) define \\(s:U\\rightarrow \\bigsqcup_{p\\in U}\\mathcal{F}_p\\) as \\(p\\mapsto t_p\\). By the very definition, it is an element of \\(\\mathcal{F}(U)\\). It goes to \\(t\\) under the above map. So, \\(\\widetilde{\\mathcal{F}}(U)\\rightarrow \\mathcal{F}(U)\\) defined above is an isomorphism. Thus, sheafification of a presheaf that is already a sheaf is itself.","permalink":"https://praphulla-koushik.github.io/2017/07/21/sheafification-of-a-presheaf-that-is-already-a-sheaf-is-itself-reality-check/","summary":"let \\(\\mathcal{F}\\) be a presheaf. We define associated presheaf of \\(\\mathcal{F}\\) to be the sheaf given by map \\(U\\mapsto \\widetilde{F}(U)\\) where\u003cimg class=\" size-full wp-image-1064 aligncenter\" src=\"/wp-media/2017/07/595b0383e3-ql_05c03babc4c7f7ad0bae3f0ff48eda81_l3.png\" alt=\"ql_05c03babc4c7f7ad0bae3f0ff48eda81_l3\" width=\"409\" height=\"64\" /\u003e\n\nwhere the condition \\(^\\dagger\\) says that \\(s(p)\\in \\mathcal{F}_p\\) for every \\(p\\in U\\) and there exists an open subset \\(U(p)\\subset U\\)  containing \\(p\\) and a section \\(t\\in \\mathcal{F}(U(p))\\) such that \\(t_q=s(q)\\) for every \\(q\\in U(p)\\).\n\nSuppose \\(\\mathcal{F}\\) is actually a sheaf then we will see that \\(\\widetilde{F}(U)\\cong \\mathcal{F}(U)\\) for every open  \\(U\\subseteq X\\).\n\u003cp style=\"text-align:justify;\"\u003eLet \\(s\\in \\widetilde{F}(U)\\) i.e., \\(s:U\\rightarrow \\bigsqcup_{i\\in \\Lambda}\\mathcal{F}_{p_i}\\) (this notation is just for my comfort, we have \\(U=\\bigcup_{i\\in \\Lambda}\\{p_i\\}\\)) satisfying some conditions given above. Given such \\(s\\) we want to assign an element in \\(\\mathcal{F}(U)\\).\u003c/p\u003e","tags":["sheaves"],"title":"Sheafification of a presheaf that is already a sheaf is itself - Reality check","type":"wp-import"},{"categories":["algebraic-geometry"],"content":"I will add links of blogs of topics in sequence as in Hartshorne. Morphism of Sheaves – Morphism of Stalks Sheafification of a presheaf that is already a sheaf is itself – Reality check Structure sheaf on spectrum of a ring \u0026nbsp; \u0026nbsp; \u0026nbsp; \u0026nbsp; \u0026nbsp;","permalink":"https://praphulla-koushik.github.io/2017/07/20/hartshornes-algebraic-geometry-solutions/","summary":"I will add links of blogs of topics in sequence as in Hartshorne.\n\n\u003ca href=\"https://koushik1729.wordpress.com/2017/07/28/morphism-of-sheaves-morphism-of-stalks/\"\u003eMorphism of Sheaves – Morphism of Stalks\u003c/a\u003e\n\n\u003ca href=\"https://koushik1729.wordpress.com/2017/07/21/sheafification-of-a-presheaf-that-is-already-a-sheaf-is-itself-reality-check/\"\u003eSheafification of a presheaf that is already a sheaf is itself – Reality check\u003c/a\u003e\n\n\u003ca href=\"https://koushik1729.wordpress.com/2017/07/25/structure-sheaf-on-spectrum-of-a-ring/\"\u003eStructure sheaf on spectrum of a ring\u003c/a\u003e\n\n\u0026nbsp;\n\n\u0026nbsp;\n\n\u0026nbsp;\n\n\u0026nbsp;\n\n\u0026nbsp;","tags":["hartshorne","sheaves"],"title":"Hartshorne's Algebraic Geometry","type":"wp-import"},{"categories":["Category theory"],"content":"Let \\(G\\) be a group. We are going to construct a category with one element whose morphisms are elements of this group \\(G\\). Definition : A category \\(\\mathcal{C}\\) consists of a collection of objects, \\(\\text{Ob}(\\mathcal{C})\\), for each \\(A,B\\in \\text{Ob}(\\mathcal{C})\\) a collection of maps from \\(A\\) to \\(B\\) denoted by \\(\\mathcal{C}(A,B)\\). for each \\(A,B,C \\in \\text{Ob}(\\mathcal{C})\\) a function \\(\\mathcal{C}(B,C)\\times \\mathcal{C}(A,B)\\rightarrow \\mathcal{C}(A,C)\\) with \\((g,f)\\mapsto g\\circ f\\) called the composition for each \\(A\\in \\text{Ob}(\\mathcal{C})\\) an element \\(1_A\\in \\mathcal{C}(A,A)\\) called the identity on \\(A\\) satisfying the following conditions Associativity : for each \\(f\\in \\mathcal{C}(A,B), g\\in \\mathcal{C}(B,C), h\\in \\mathcal{C}(C,D)\\) we have \\((h\\circ g)\\circ f=h\\circ (g\\circ f)\\). Identity law : for each \\(f\\in \\mathcal{C}(A,B)\\) we have \\(f\\circ 1_A=f=1_B\\circ f\\). We are not constructing a categroy \\(\\mathcal{C}\\) with \\(\\text{Ob}(\\mathcal{C})=G\\), we are constructing a category \\(\\mathcal{C}\\) with \\(\\text{Ob}(\\mathcal{C})=\\{A\\}\\)(one point set) and \\(\\mathcal{C}(A,A)=G\\). We have defined objects and morphisms collections (here there is only one object so there is only one morphisms collection). We have to define composition \\(\\mathcal{C}(A,A)\\times \\mathcal{C}(A,A)\\rightarrow \\mathcal{C}(A,A)\\) i.e., we have to give a map \\(G\\times G\\rightarrow G\\). There are two natural ways to give this map \\((g,h)\\mapsto g.h\\) or \\((g,h)\\mapsto h.g\\). We consider the map \\((g,h)\\mapsto g\\circ h:=g.h\\) to give composition. Obvious choice of an identity element in \\(\\mathcal{C}\\) is $llatex 1_A=e$ identity element of the group. Associativity of group implies associativity of composition \\((g\\circ h)\\circ f=(g.h)\\circ f=(g.h).f=g.(h.f)=g\\circ (h\\circ f).\\)\nWe have \\(1_A\\circ g=e.g=g\\) and \\(g\\circ 1_A=g.e=g\\).\nSo, we have a category whose sets of objects contains exactly one element and whose set of morphisms is exactly one set equal to given group \\(G\\). Given a category \\(\\mathcal{C}\\) and an element \\(A\\in\\text{Ob}(\\mathcal{C})\\) we have a functor \\(h_A:\\mathcal{C}^{\\text{op}}\\rightarrow \\text{Sets}\\) defined by \\(B\\mapsto \\mathcal{C}(B,A)\\) and given an element \\(f\\in \\mathcal{C}(B,C)\\) we have \\(h_A(f)\\in \\text{Mor}((h_A(C),h_A(B))\\) given by \\(h_A(f)(g)=g\\circ f\\). We will see explicitly what is \\(h_A\\) in case of group as a category with exactly one object as defined above. We have \\(h_A(A)=\\mathcal{C}(A,A)=G\\). Let \\(h\\in \\mathcal{C}(A,A)=G\\). Then, \\(h_A(h):G\\rightarrow G\\) given by \\(g\\mapsto g\\circ h=g.h\\). Let us see explicitly what are the natural transformations \\(\\eta:h_A\\rightarrow h_A\\). This should come up with \\(\\eta(A):h_A(A)\\rightarrow h_A(A)\\) i.e., with a map \\(\\eta(A)=\\sigma:G\\rightarrow G\\) satisfying the naturality condition i.e., for every \\(h\\in G=\\mathcal{C}(A,A)\\) the following diagram commutes i.e., we want \\(\\sigma:G\\rightarrow G\\) to be such that \\((h_A(h)\\circ \\sigma)(g)=(\\sigma\\circ h_A(h))(g)\\) for all \\(g\\in h_A(A)=G\\) i.e., \\(h_A(h)(\\sigma(g))=\\sigma(h_A(h)(g))\\) ie., \\(\\sigma(g)h=\\sigma(gh)\\) for all $g\\in G$. So, the set of all natural transformations from \\(h_A\\) to \\(h_A\\) is given by \\(\\text{Nat}(h_A,h_A)=\\{\\sigma:G\\rightarrow G : \\sigma(gh)=\\sigma(g)h \\forall g,h\\in G\\}\\).\nSee that every element in \\(\\text{Nat}(h_A,h_A)\\) is an isomorphism, as a set. It is not even necessary that \\(\\sigma\\in \\text{Nat}(h_A,h_A)\\) is a group homomorphism. See that \\(\\sigma(h)=\\sigma(1.h)=\\sigma(1).h ~~\\forall h\\in G\\). Given \\(\\sigma\\in \\text{Nat}(h_A,h_A)\\) define \\(\\tau:G\\rightarrow G\\) as \\(\\tau(h)=(\\sigma(1))^{-1}h\\). Then, \\(\\sigma(\\tau(h))=\\sigma(\\sigma(1)^{-1}h)=\\sigma(1).\\sigma(1)^{-1}.h=h\\). \\(\\tau(\\sigma(h))=\\tau(\\sigma(1)h)=\\sigma(1)^{-1}.\\sigma(1).h=h\\). So, \\(\\tau\\) is inverse of \\(\\sigma\\). Thus, every element in \\(\\text{Nat}(h_A,h_A)\\) is an isomorphism of sets. Thus, we have \\(\\text{Nat}(h_A,h_A)=\\{\\sigma\\in \\text{Sym}(G) : \\sigma(gh)=\\sigma(g)h \\forall g,h\\in G\\}\\leq \\text{Sym}(G)\\).\nYoneda lemma states that \\(\\mathcal{C}(A,A)\\cong \\text{Nat}(h_A,h_A)\\) i.e., \\(G\\cong H\\leq \\text{Sym}(G)\\) which is exactly the statement of Cayley's theorem. So, yoneda lemma does say that any group can be identified with a subgroup of its symmetric group. \u0026nbsp;","permalink":"https://praphulla-koushik.github.io/2017/07/19/group-as-a-category-with-one-object/","summary":"Let \\(G\\) be a group. We are going to construct a category with one element whose morphisms are elements of this group \\(G\\).\n\n\u003cstrong\u003eDefinition : \u003c/strong\u003eA category \\(\\mathcal{C}\\) consists of\n\u003cul\u003e\n\t\u003cli\u003ea collection of objects, \\(\\text{Ob}(\\mathcal{C})\\),\u003c/li\u003e\n\t\u003cli\u003efor each \\(A,B\\in \\text{Ob}(\\mathcal{C})\\) a collection of maps from \\(A\\) to \\(B\\) denoted by \\(\\mathcal{C}(A,B)\\).\u003c/li\u003e\n\t\u003cli\u003efor each \\(A,B,C \\in \\text{Ob}(\\mathcal{C})\\) a function \\(\\mathcal{C}(B,C)\\times \\mathcal{C}(A,B)\\rightarrow \\mathcal{C}(A,C)\\) with \\((g,f)\\mapsto g\\circ f\\) called the composition\u003c/li\u003e\n\t\u003cli\u003efor each \\(A\\in \\text{Ob}(\\mathcal{C})\\) an element \\(1_A\\in \\mathcal{C}(A,A)\\) called the identity on \\(A\\)\u003c/li\u003e\n\u003c/ul\u003e\nsatisfying the following conditions\n\u003cul\u003e\n\t\u003cli\u003eAssociativity : for each \\(f\\in \\mathcal{C}(A,B), g\\in \\mathcal{C}(B,C), h\\in \\mathcal{C}(C,D)\\) we have \\((h\\circ g)\\circ f=h\\circ (g\\circ f)\\).\u003c/li\u003e\n\t\u003cli\u003eIdentity law : for each \\(f\\in \\mathcal{C}(A,B)\\) we have \\(f\\circ 1_A=f=1_B\\circ f\\).\u003c/li\u003e\n\u003c/ul\u003e\nWe are not  constructing a categroy \\(\\mathcal{C}\\) with \\(\\text{Ob}(\\mathcal{C})=G\\), we are  constructing a category \\(\\mathcal{C}\\) with \\(\\text{Ob}(\\mathcal{C})=\\{A\\}\\)(one point set) and \\(\\mathcal{C}(A,A)=G\\).\n\nWe have defined objects and morphisms collections (here there is only one object so there is only one morphisms collection).\n\nWe have to define composition \\(\\mathcal{C}(A,A)\\times \\mathcal{C}(A,A)\\rightarrow \\mathcal{C}(A,A)\\) i.e., we have to give a map \\(G\\times G\\rightarrow G\\).  There are two natural ways to give this map \\((g,h)\\mapsto g.h\\) or \\((g,h)\\mapsto h.g\\). \u003cstrong\u003eWe consider the map \\((g,h)\\mapsto g\\circ h:=g.h\\) to give composition. \u003c/strong\u003eObvious choice of an identity element in \\(\\mathcal{C}\\) is $llatex 1_A=e$ identity element of the group.\n\nAssociativity of group implies associativity of composition\n\u003cp style=\"text-align:center;\"\u003e\\((g\\circ h)\\circ f=(g.h)\\circ f=(g.h).f=g.(h.f)=g\\circ (h\\circ f).\\)\u003c/p\u003e","tags":["yoneda-lemma"],"title":"Group as a category with one object","type":"wp-import"},{"categories":["algebraic-geometry"],"content":"Let \\(f:X\\rightarrow Y\\) be an affine morphism and \\(\\mathcal{F}\\) is a quasi coherent sheaf of \\(\\mathcal{O}_X\\) modules. Then, \\(f_*\\mathcal{F}\\) is a quasi coherent sheaf of \\(\\mathcal{O}_Y\\) modules. We have the following result : Let \\(X\\) be a scheme. Then, an \\(\\mathcal{O}_X\\) module \\(\\mathcal{F}\\) is quasi coherent iff for every open affine subset \\(U=\\text{Spec}(A)\\) of \\(X\\), there is an \\(A\\) module \\(M\\) such that \\(\\mathcal{F}|_U=\\tilde{M}\\). Let \\(U\\subseteq Y\\) be an open affine subset say \\(U=\\text{Spec}(A)\\). As \\(f\\) is affine, \\(f^{-1}(U)\\) is affine open, say \\(f^{-1}(U)=\\text{Spec}(B)\\subseteq X\\). As \\(\\mathcal{F}\\) is quasi coherent sheaf of \\(\\mathcal{O}_X\\) modules and \\(f^{-1}(U)=\\text{Spec}(B)\\) is open affine subset of \\(X\\), there exists a \\(B\\) module \\(M\\) such that \\(\\mathcal{F}|_{f^{-1}(U)}\\cong \\widetilde{M}\\). As \\(f^{-1}(U)=\\text{Spec}(B)\\) we have \\(f:\\text{Spec}(B)\\rightarrow \\text{Spec}(A)\\) which also gives a ring morphsim \\(A\\rightarrow B\\). The isomorphism \\(\\mathcal{F}|_{f^{-1}(U)}\\cong \\widetilde{M}\\) implies \\(f_*(\\mathcal{F}|_{f^{-1}(U)})\\cong f_*\\widetilde{M}\\).\nAs \\(f_*(\\mathcal{F}|_{f^{-1}(U)})\\cong f_*\\mathcal{F}|_U\\) we have \\(f_*\\mathcal{F}|_U\\cong f_*\\widetilde{M}\\).\nFrom previous observation we see that \\(f_*\\widetilde{M}\\cong (_A M)\\) where \\(_A M\\) is \\(M\\) considered as a \\(A\\) module under ring morphism \\(A\\rightarrow B\\) defined above. So, we have \\(f_*\\mathcal{F}|_U\\cong \\widetilde{_A M}\\).\nSo, given an open affine \\(U=\\text{Spec}(A)\\subseteq X\\) there exists an \\(A\\) module \\(_A M\\) such that \\((f_*\\mathcal{F})|_U\\cong \\widetilde{_A M}\\).\nThus, \\(f_*\\mathcal{F}\\) is a quasi coherent sheaf of \\(\\mathcal{O}_Y\\) modules.","permalink":"https://praphulla-koushik.github.io/2017/07/14/push-forward-of-quasi-coherent-sheaf-of-modules/","summary":"Let \\(f:X\\rightarrow Y\\) be an affine morphism and \\(\\mathcal{F}\\) is a quasi coherent sheaf of \\(\\mathcal{O}_X\\) modules. Then, \\(f_*\\mathcal{F}\\) is a quasi coherent sheaf of \\(\\mathcal{O}_Y\\) modules.\n\nWe have the following result :\n\n\u003cstrong\u003eLet \\(X\\) be a scheme. Then, an \\(\\mathcal{O}_X\\) module \\(\\mathcal{F}\\) is quasi coherent iff for every open affine subset \\(U=\\text{Spec}(A)\\) of \\(X\\), there is an \\(A\\) module \\(M\\)  such that \\(\\mathcal{F}|_U=\\tilde{M}\\).\u003c/strong\u003e\n\nLet \\(U\\subseteq Y\\) be an open affine subset say \\(U=\\text{Spec}(A)\\). As \\(f\\) is affine, \\(f^{-1}(U)\\) is affine open, say  \\(f^{-1}(U)=\\text{Spec}(B)\\subseteq X\\).\n\nAs \\(\\mathcal{F}\\) is quasi coherent sheaf of \\(\\mathcal{O}_X\\) modules and \\(f^{-1}(U)=\\text{Spec}(B)\\) is open affine subset of \\(X\\), there exists a \\(B\\) module \\(M\\) such that \\(\\mathcal{F}|_{f^{-1}(U)}\\cong \\widetilde{M}\\). As \\(f^{-1}(U)=\\text{Spec}(B)\\) we have \\(f:\\text{Spec}(B)\\rightarrow \\text{Spec}(A)\\) which also gives a ring morphsim \\(A\\rightarrow B\\). The isomorphism \\(\\mathcal{F}|_{f^{-1}(U)}\\cong \\widetilde{M}\\) implies\n\u003cp style=\"text-align:center;\"\u003e\\(f_*(\\mathcal{F}|_{f^{-1}(U)})\\cong f_*\\widetilde{M}\\).\u003c/p\u003e","tags":["schemes","sheaves"],"title":"Push forward of quasi coherent sheaf of modules","type":"wp-import"},{"categories":["Sheaves of Modules","algebraic-geometry"],"content":"Let \\(X=\\text{Spec}(R)\\) be an affine scheme.\nLet \\(X'\\) be an affine \\(X\\) scheme i.e., \\(X'=\\text{Spec}(R')\\) for some ring \\(R'\\) with a morphism of schemes \\(\\pi: X'\\rightarrow X\\). This \\(\\pi\\) comes with morphism of global sections \\(R=\\Gamma(X,\\mathcal{O}_X)\\rightarrow \\Gamma(X',\\mathcal{O}_X')=R'\\)\ngiving \\(R'\\), structure of an \\(R\\) algebra. So, any affine scheme over \\(X=\\text{Spec}(R)\\) is simply the specturm of an \\(R\\) algebra. Conversely, given an \\(R\\) algebra say \\(R'\\), we have an affine scheme \\(X'=\\text{Spec}(R')\\) over \\(X\\) with morphism \\(\\pi:X'\\rightarrow X\\). Being a morphism of affine schemes, \\(\\pi: X\\rightarrow X\\) is an affine morphism. Let \\(X\\) be an arbitrary scheme. We want to associate an \\(X\\) scheme \\(X'\\) such that the structure morphism \\(\\pi:X'\\rightarrow X\\) is an affine morphism. To do this in case of \\(X=\\text{Spec}(R)\\) we have fixed an \\(R\\) algebra and then associated an affine scheme for this. In case of an arbitrary scheme \\(X\\) unlike the case of affine scheme \\(X=\\text{Spec}(R)\\) there is no single ring that has all information about the scheme \\(X\\). It is only natural to consider the collection \\(\\{\\mathcal{O}_X(U): U\\subseteq X\\}\\) varying over all open subsets of \\(X\\) to get information about the scheme \\(X\\). Choosing an \\(\\mathcal{O}_X(U)\\) algebra \\(\\mathcal{F}(U)\\) for each open \\(U\\subseteq X\\) we associate an \\(X\\) scheme \\(X'\\) for this collection \\(\\{\\mathcal{F}(U)\\}\\) of \\(\\mathcal{O}_X(U)\\) algebras. It is only natural to put a condition that this collection \\(\\{\\mathcal{F}(U)\\}\\) to be compatible with structure sheaf \\(\\mathcal{O}_X\\) i.e., we want \\(U\\mapsto \\mathcal{F}(U)\\) to give a structure of sheaf of \\(\\mathcal{O}_X\\) algebras on \\(X\\). So, given an arbitrary scheme \\(X\\) and a sheaf \\(\\mathcal{F}\\) of \\(\\mathcal{O}_X\\) algebras we associate an \\(X\\) scheme \\(X'\\) such that the structure map \\(X'\\rightarrow X\\) is an affine morphism. It is not obvious at this point but we also want \\(\\mathcal{F}\\) to be a quasicoherent sheaf of \\(\\mathcal{O}_X\\) modules. We call this \\(X'\\), Global spec or Relative spec of sheaf \\(\\mathcal{F}\\) of \\(\\mathcal{O}_X\\) algebras over \\(X\\) denoted by \\(\\textbf{Spec} (\\mathcal{F})\\). Here we make two important remarks : The \\(\\textbf{Spec}\\) construction gives an important way to understand affine morphisms. Note that \\(\\textbf{Spec}(\\mathcal{F})\\rightarrow X\\) is an affine morphism. The converse is also true. If \\(f:X\\rightarrow Y\\) is an affine morphism then \\(\\mathcal{A}=f_*\\mathcal{O}_X\\) is a quasi coherent sheaf of \\(\\mathcal{O}_Y\\) algebras and \\(X\\cong \\text{Spec} (\\mathcal{A})\\). The \\(\\textbf{Spec}\\) construction is used to assign a geometric vector bundle on a scheme \\(Y\\) to each locally free sheaf \\(\\mathcal{E}\\) of rank \\(n\\) on a scheme \\(Y\\) which gives a bijection between isomorphism classes of locall free sheaves of rank \\(n\\) on \\(Y\\), and isomorphism classes of vector bundles of rank \\(n\\) on \\(Y\\). To define \\(\\textbf{Spec}(\\mathcal{F})\\) we do not need \\(\\mathcal{F}\\) to be quasi coherent, but \\(\\mathcal{F}\\) quasi coherent implies the structure map \\(\\textbf{Spec}(\\mathcal{F})\\rightarrow X\\) is an affine morphism. Now, we try to construct \\(\\textbf{Spec}(\\mathcal{F})\\) and structure map \\(\\textbf{Spec}(\\mathcal{F})\\rightarrow X\\). One way to do this is gluing the schemes \\(\\text{Spec}(\\mathcal{F}(U))\\) over all open subsets \\(U\\subseteq X\\). Another way is to use universal property of \\(\\text{Spec}\\) of a ring. We have following result : Let \\(A\\) be a ring and let \\((X,\\mathcal{O}_X)\\) be a scheme. Given a morphism \\(f:X\\rightarrow \\text{Spec}(A)\\) we have an associated map on sheaves \\(f^{\\#}:\\mathcal{O}_{\\text{Spec}(A)}\\rightarrow f_*\\mathcal{O}_X\\). Taking global sections, we obtain a ring homomorphism \\(A\\rightarrow \\mathcal{O}_X(X)\\). Thus there is a natural map \\(\\alpha: \\text{Hom}_{\\text{Schemes}}(X,\\text{Spec}(A))\\rightarrow\\text{Hom}_{\\text{Rings}}(A,\\mathcal{O}_X(X)).\\)\nThis map is bijective. So, we have \\(\\text{Hom}_{\\text{Schemes}}(X,\\text{Spec}(A))\\cong\\text{Hom}_{\\text{Rings}}(A,\\mathcal{O}_X(X)).\\)\nWe have the similar result for relative schemes over an affine scheme Let \\(A\\) be a ring and \\(B\\) be an \\(A\\) algebra. let \\((X,\\mathcal{O}_X)\\) be an \\(S=\\text{Spec}(A)\\) scheme. Given a morphism \\(f:X\\rightarrow \\text{Spec}(B)\\) of \\(S\\) schemes we have an associated map on sheaves \\(f^{\\#}:\\mathcal{O}_{\\text{Spec}(B)}\\rightarrow f_*\\mathcal{O}_X\\). Taking global sections, we obtain an \\(A\\) algebra homomorphism \\(B\\rightarrow \\mathcal{O}_X(X)\\). Thus there is a natural map \\(\\alpha: \\text{Hom}_{S - \\text{Schemes}}(X,\\text{Spec}(B))\\rightarrow\\text{Hom}_{A \\text{algebra}}(B,\\mathcal{O}_X(X)).\\)\nThis map is bijective. So, we have \\(\\text{Hom}_{S-\\text{Schemes}}(X,\\text{Spec}(B))\\cong\\text{Hom}_{A \\text{algebra}}(B,\\mathcal{O}_X(X)).\\)\nIn terms of commutative diagram we have the following : Now, for an arbitrary scheme \\(X\\) (which is generalization of \\(S\\) above), and a quasi coherent sheaf of \\(\\mathcal{O}_X\\) algebras \\(\\mathcal{F}\\) (which is generalization of \\(B\\) above), we want to define \\(\\text{Spec}(\\mathcal{F})\\) that comes with a structure map \\(\\text{Spec}(\\mathcal{F})\\rightarrow X\\) such that given any \\(X\\) scheme \\(f:Y\\rightarrow X\\) and a morphism of \\(X\\) schemes \\(\\alpha : Y \\rightarrow \\text{Spec}(\\mathcal{F})\\) something similar to that of the above situation happens. In terms of commutative diagram, we have As we are generalizing the case of \\(S=\\text{Spec}(R)\\) we expect to have similar commutative diagram as in the case of \\(S\\) except that in this case commutative diagram not be of ring homomorphism but it would be of sheaves of algebras. We expect to have something like As \\(\\mathcal{F}\\) is a sheaf of \\(\\mathcal{O}_X\\) modules and \\(\\mathcal{O}_Y\\) is a sheaf of \\(\\mathcal{O}_Y\\) modules, it does not make sense to talk about morphism between these two, we make a slight change by considering morphsim \\(\\mathcal{F}\\rightarrow f_*\\mathcal{O}_Y\\) where \\(f^*\\mathcal{O}_Y\\) is the push forward of $\\mathcal{O}_Y$ under \\(f:Y\\rightarrow X\\), thus a sheaf of $\\mathcal{O}_X$ modules. So, we want the following commutative diagram So, our definition of \\(\\textbf{Spec}(\\mathcal{F})\\) would be such that it satisfy the universal property We will discuss properties of this construction in next blog post. There are still some loose ends here, we will fix that soon.","permalink":"https://praphulla-koushik.github.io/2017/07/13/global-spec-or-relative-spec-of-a-scheme/","summary":"\u003cp style=\"text-align:justify;\"\u003eLet \\(X=\\text{Spec}(R)\\) be an affine scheme.\u003c/p\u003e\nLet \\(X'\\) be an affine \\(X\\) scheme i.e., \\(X'=\\text{Spec}(R')\\) for some ring \\(R'\\) with a morphism of schemes \\(\\pi: X'\\rightarrow X\\).  This \\(\\pi\\) comes with morphism of global sections\n\u003cp style=\"text-align:center;\"\u003e\\(R=\\Gamma(X,\\mathcal{O}_X)\\rightarrow \\Gamma(X',\\mathcal{O}_X')=R'\\)\u003c/p\u003e\ngiving \\(R'\\), structure of an \\(R\\) algebra. So, any affine scheme over \\(X=\\text{Spec}(R)\\) is simply the specturm of an \\(R\\) algebra. Conversely, given an \\(R\\) algebra say \\(R'\\), we have an affine scheme \\(X'=\\text{Spec}(R')\\) over \\(X\\) with morphism \\(\\pi:X'\\rightarrow X\\). Being a morphism of affine schemes, \\(\\pi: X\\rightarrow X\\) is an affine morphism.\n\nLet \\(X\\) be an arbitrary scheme. We want to associate an \\(X\\) scheme \\(X'\\) such that the structure morphism \\(\\pi:X'\\rightarrow X\\) is an affine morphism.\n\nTo do this in case of \\(X=\\text{Spec}(R)\\) we have fixed an \\(R\\) algebra and then associated an affine scheme for this.\n\nIn case of an arbitrary scheme \\(X\\) unlike the case of affine scheme \\(X=\\text{Spec}(R)\\) there is no single ring that has all information about  the scheme \\(X\\). It is only natural to consider the collection \\(\\{\\mathcal{O}_X(U): U\\subseteq X\\}\\) varying over all open subsets of \\(X\\) to get information about the scheme \\(X\\). Choosing an \\(\\mathcal{O}_X(U)\\) algebra \\(\\mathcal{F}(U)\\) for each open \\(U\\subseteq X\\) we associate an \\(X\\) scheme \\(X'\\) for this collection \\(\\{\\mathcal{F}(U)\\}\\) of \\(\\mathcal{O}_X(U)\\) algebras. It is only natural to put a condition that this collection \\(\\{\\mathcal{F}(U)\\}\\) to be compatible with structure sheaf \\(\\mathcal{O}_X\\) i.e., we want \\(U\\mapsto \\mathcal{F}(U)\\) to give a structure of  sheaf of \\(\\mathcal{O}_X\\) algebras on \\(X\\).\n\nSo, given an arbitrary scheme \\(X\\) and a sheaf \\(\\mathcal{F}\\) of \\(\\mathcal{O}_X\\) algebras we associate an \\(X\\) scheme \\(X'\\) such that the structure map \\(X'\\rightarrow X\\) is an affine morphism. It is not obvious at this point but we also want \\(\\mathcal{F}\\) to be a quasicoherent sheaf of \\(\\mathcal{O}_X\\) modules. We call this \\(X'\\),  Global spec or Relative spec of sheaf \\(\\mathcal{F}\\) of \\(\\mathcal{O}_X\\) algebras over \\(X\\) denoted by \\(\\textbf{Spec} (\\mathcal{F})\\).\n\nHere we make two important remarks :\n\u003col\u003e\n\t\u003cli\u003eThe \\(\\textbf{Spec}\\) construction gives an important way to understand affine morphisms. Note that \\(\\textbf{Spec}(\\mathcal{F})\\rightarrow X\\) is an affine morphism. The converse is also true. \u003cstrong\u003eIf \\(f:X\\rightarrow Y\\) is an affine morphism then \\(\\mathcal{A}=f_*\\mathcal{O}_X\\) is a quasi coherent sheaf of \\(\\mathcal{O}_Y\\) algebras and \\(X\\cong \\text{Spec} (\\mathcal{A})\\).\u003c/strong\u003e\u003c/li\u003e\n\t\u003cli\u003eThe \\(\\textbf{Spec}\\) construction is used to assign a geometric vector bundle on a scheme \\(Y\\) to each locally free sheaf \\(\\mathcal{E}\\) of rank \\(n\\) on a scheme \\(Y\\) which gives a bijection between \u003cstrong\u003eisomorphism classes of locall free sheaves of rank \\(n\\) on \\(Y\\), \u003c/strong\u003eand \u003cstrong\u003eisomorphism classes of vector bundles of rank \\(n\\) on \\(Y\\)\u003c/strong\u003e.\u003c/li\u003e\n\t\u003cli\u003eTo define \\(\\textbf{Spec}(\\mathcal{F})\\) we do not need \\(\\mathcal{F}\\) to be quasi coherent, but \\(\\mathcal{F}\\) quasi coherent implies the structure map \\(\\textbf{Spec}(\\mathcal{F})\\rightarrow X\\) is an affine morphism.\u003c/li\u003e\n\u003c/ol\u003e\nNow, we try to construct \\(\\textbf{Spec}(\\mathcal{F})\\) and structure map \\(\\textbf{Spec}(\\mathcal{F})\\rightarrow X\\). One way to do this is gluing the schemes \\(\\text{Spec}(\\mathcal{F}(U))\\) over all open subsets \\(U\\subseteq X\\). Another way is to use universal property of \\(\\text{Spec}\\) of a ring.\n\nWe have following result :\n\n\u003cstrong\u003eLet \\(A\\) be a ring and let \\((X,\\mathcal{O}_X)\\) be a scheme. Given a morphism \\(f:X\\rightarrow \\text{Spec}(A)\\) we have an associated map on sheaves \\(f^{\\#}:\\mathcal{O}_{\\text{Spec}(A)}\\rightarrow f_*\\mathcal{O}_X\\). Taking global sections, we obtain a ring homomorphism \\(A\\rightarrow \\mathcal{O}_X(X)\\). Thus there is a natural map\u003c/strong\u003e\n\u003cp style=\"text-align:center;\"\u003e\u003cstrong\u003e \\(\\alpha: \\text{Hom}_{\\text{Schemes}}(X,\\text{Spec}(A))\\rightarrow\\text{Hom}_{\\text{Rings}}(A,\\mathcal{O}_X(X)).\\)\u003c/strong\u003e\u003c/p\u003e","tags":["schemes","sheaves"],"title":"Global Spec Or Relative Spec of a Scheme","type":"wp-import"},{"categories":["algebraic-geometry","stacks"],"content":"Here I will add links of pages of interesting questions/answers from Math Stack Exchange and Math over flow. Learning Algebraic Geometry 1 transitive Lie groupoid is Morita equivalent to the isotropy group ","permalink":"https://praphulla-koushik.github.io/2017/07/11/mathstack-exchange-stack-overflow-pages/","summary":"Here I will add links of pages of interesting questions/answers from \u003ca href=\"https://math.stackexchange.com/\"\u003eMath Stack Exchange\u003c/a\u003e and \u003ca href=\"https://mathoverflow.net/\"\u003eMath over flow\u003c/a\u003e.\n\u003cul\u003e\n\t\u003cli\u003e\u003ca href=\"https://math.stackexchange.com/questions/285201/path-to-basics-in-algebraic-geometry-from-hs-algebra-and-calculus/285355#285355\"\u003eLearning Algebraic Geometry 1\u003c/a\u003e\u003c/li\u003e\n\t\u003cli\u003e\u003ca href=\"https://math.stackexchange.com/questions/3097017/transitive-lie-groupoid-is-morita-equivalent-to-the-isotropy-group\"\u003etransitive Lie groupoid is Morita equivalent to the isotropy group\u003c/a\u003e\u003c/li\u003e\n\u003c/ul\u003e","tags":["stacks"],"title":"Math stack exchange/ stack overflow questions","type":"wp-import"},{"categories":["Differential geometry"],"content":"","permalink":"https://praphulla-koushik.github.io/2017/07/10/geometric-vector-bundle/","summary":"","tags":[],"title":"Geometric vector bundle","type":"wp-import"},{"categories":["Algebraic geometry"],"content":"","permalink":"https://praphulla-koushik.github.io/2017/07/10/tensor-algebra-symmetric-algebra-and-exterior-algebra-of-a-sheaf/","summary":"","tags":[],"title":"Tensor algebra, symmetric algebra and exterior algebra of a sheaf","type":"wp-import"},{"categories":["Algebraic geometry"],"content":"","permalink":"https://praphulla-koushik.github.io/2017/07/10/associated-sheaf-and-global-section-functors-are-adjoint/","summary":"","tags":[],"title":"Associated sheaf and global section functors are adjoint","type":"wp-import"},{"categories":["algebraic-geometry"],"content":"Andreas Gathmann Algebraic Geometry Foundations of Algebraic Geometry Ravi Vakil Kiran Kedlaya Algebraic Geometry Lecture Notes ","permalink":"https://praphulla-koushik.github.io/2017/07/10/algebraic-geometry-lecture-notes-books/","summary":"\u003ca title=\"Andreas Gathmann Algebraic Geometry\" href=\"https://koushik1729.wordpress.com/wp-content/uploads/2015/06/andreas-gathmann-algebraic-geometry.pdf\"\u003eAndreas Gathmann Algebraic Geometry\u003c/a\u003e\n\n\u003ca title=\"Foundations of Algebraic Geometry Ravi Vakil\" href=\"https://koushik1729.wordpress.com/wp-content/uploads/2017/07/foundations-of-algebraic-geometry-ravi-vakil.pdf\"\u003eFoundations of Algebraic Geometry Ravi Vakil\u003c/a\u003e\n\n\u003ca href=\"https://ocw.mit.edu/courses/mathematics/18-726-algebraic-geometry-spring-2009/lecture-notes/\"\u003eKiran Kedlaya Algebraic Geometry Lecture Notes \u003c/a\u003e","tags":[],"title":"Algebraic Geometry Lecture Notes/ Books","type":"wp-import"},{"categories":["Category theory","Algebraic geometry"],"content":"Yoneda lemma : Let \\(\\mathcal{C}\\) be a (locally) small category. Then naturally in \\(A\\in \\mathcal{C}\\) and \\(X\\in [\\mathcal{C}^{\\rm{op}},\\rm{Set}]\\). Terminology : \\(\\mathcal{C}\\) is a category mentioned in the lemma, \\(\\mathcal{C}^{\\rm{op}}\\) is the opposite category associated to \\(\\mathcal{C}\\). \\(\\rm{Set}\\) is the category with elements as sets and morphisms as functions. \\(X:\\mathcal{C}^{op}\\rightarrow \\rm{Set}\\) is a functor. Given \\(A\\in \\mathcal{C}\\), \\(H_A\\) is the functor \\(H_A:\\mathcal{C}^{\\rm{op}}\\rightarrow \\rm{Set}\\) given by $B\\mapsto \\mathcal{C}(B,A)$. \\([\\mathcal{C}^{\\rm{op}},\\rm{Set}]\\) is the category with functors from \\(\\mathcal{C}^{\\rm{op}}\\) to $\\rm{Set}$ as elements and natural transformations between these functors as morphisms. \\([\\mathcal{C},\\rm{Set}]\\) is the category with functors from \\(\\mathcal{C}\\) to \\(rm{Set}\\) as elements and natural transformations between these functors as morphisms. Given \\(X,A\\) as above, \\(X(A)\\) is a set and \\([\\mathcal{C}^{\\rm{op}},\\rm{Set}](H_A,X)\\) is a set. Yoneda lemma says that there is a bijection between these sets, natural bijection in both \\(A\\) and \\(X\\). Natural transformation : Let \\(\\mathcal{A},\\mathcal{B}\\) be two categories and \\(F,G:\\mathcal{A}\\rightarrow \\mathcal{B}\\) be both contravariant or both covariant functors. A natural transformation \\(\\eta:F\\rightarrow G\\) is a family of arrows (morphisms) \\(F(A)\\xrightarrow{\\eta(A)}G(A)\\) such that for each \\(A\\xrightarrow{f}A'\\) in \\(\\mathcal{A}\\) the following appropriate diagram commutes. Now that we have defined terminology used in the statement, we will now prove the statement. We will first give a bijectionand then prove that it is natural in \\(A\\) and \\(X\\). Construction of Bijective map : Let \\(\\eta\\in [\\mathcal{C}^{\\rm{op}},\\rm{Set}](H_A,X)\\) i.e., \\(\\eta:H_A\\rightarrow X\\) is a natural transformation. We want to assign an element in \\(X(A)\\) with this \\(\\eta\\). It is only natural to consider the map \\(\\eta(A):H_A(A)\\rightarrow X(A)\\). The set \\(H_A(A)\\) has a special element namely \\(1_A\\in H_A(A)\\), its image \\(\\eta(A)(1_A)\\in X(A)\\). Define \\(\\Phi(\\eta)=\\eta(A)(1_A)\\). This give a map \\(\\Phi:[\\mathcal{C}^{\\rm{op}},\\rm{Set}](H_A,X)\\rightarrow X(A)\\).\nWe will now prove that this map is injective and surjective. Surjectivity of \\(\\Phi\\) : Let \\(a\\in X(A)\\). We want to assign a natural trasformation \\(\\eta:H_A\\rightarrow X\\) with this \\(a\\) such that \\(\\Phi(\\eta)=a\\). We need to define \\(\\eta(B):H_A(B)\\rightarrow X(B)\\) for each \\(B\\in \\rm{Ob}(\\mathcal{C})\\). Let \\(B\\in \\rm{Ob}(\\mathcal{C})\\) be fixed and \\(f\\in H_A(B)\\) i.e., \\(f:B\\rightarrow A\\). The functor \\(X:\\mathcal{C}^{\\rm{op}}\\rightarrow \\text{Set}\\) induces \\(X(f):X(A)\\rightarrow X(B)\\). We have \\(X(f)(a)\\in X(B)\\). Define \\(\\eta(B)(f)=X(f)(a)\\). This gives a natural transformation \\(\\eta:H_A\\rightarrow X\\) and \\(\\eta(A)(1_A)=X(1_A)(a)=1_{X(A)}(a)=a\\).\nSo, \\(\\Phi:[\\mathcal{C}^{\\rm{op}},\\rm{Set}](H_A,X)\\rightarrow X(A)\\) is surjective.\nInjectivity of \\(\\Phi\\) : Suppose \\(\\eta_1,\\eta_2\\in [\\mathcal{C}^{\\rm{op}},\\rm{Set}](H_A,X)\\) be such that \\(\\Phi(\\eta_1)=\\Phi(\\eta_2)\\) i.e., \\(\\eta_1(A)(1_A)=\\eta_2(A)(1_A)\\). We prove that $\\eta_1=\\eta_2$ i.e., \\(\\eta_1(B)=\\eta_2(B)\\) for every \\(B\\in \\mathcal{C}\\) i.e., \\(\\eta_1(B)(f)=\\eta_2(B)(f)\\) for every \\(f\\in H_A(B)\\).\nFix \\(B\\in \\mathcal{C}\\) and \\(f\\in H_A(B)\\) i.e., \\(f:B\\rightarrow A\\) in \\(\\mathcal{C}\\). We then have following commutative diagrams We have. So, Thus, \\(\\Phi:[\\mathcal{C}^{\\rm{op}},\\rm{Set}](H_A,X)\\rightarrow X(A)\\) is injective. So, we have bijective correspondence Bijection is natural in \\(A\\) : We prove that the above bijection is natural in \\(A\\) i.e., given \\(A,B\\in \\mathcal{C}\\) with \\(f:B\\rightarrow A\\) the following diagram is commutativewhere, for \\(\\eta\\in [\\mathcal{C}^{\\rm{op}},\\rm{Set}](H_A,X)\\), \\(\\varphi(f)\\) sends \\(\\eta\\) to the composition \\(\\eta\\circ f^*: H_B\\xrightarrow{f^*} H_A\\xrightarrow{\\eta} X\\). For \\(\\eta\\in [\\mathcal{C}^{\\rm{op}},\\rm{Set}](H_A,X)\\) we have \\(\\varphi(f)(\\eta) :H_B\\rightarrow X\\). Let \\(C\\in \\mathcal{C}\\) then, we have We now prove thatWe haveThe following commutative diagramsays that \\(X(f)(\\eta(A)(1_A))=\\eta(B)(H_A(f)(1_A))\\). As \\(H_A(f)(1_A)=f\\), we have Now,We have defined \\(\\varphi(f)\\) as \\(\\varphi(f)(\\eta)(C)(g)=\\eta(C)(f\\circ g)\\). So,So, we see that So, the bijection \\(\\Phi:[\\mathcal{C}^{\\rm{op}},\\rm{Set}](H_A,X)\\rightarrow X(A)\\) is natural in \\(A\\). Bijection is natural in \\(X\\) : We prove that the above bijection is natural in \\(X\\) i.e., for functors \\(X,Y:\\mathcal{C}^{\\rm{op}}\\rightarrow \\rm{Set}\\) and a natural transformation \\(\\tilde{\\eta}:X\\rightarrow Y\\) we prove that the following diagram is commutativeLet \\(\\eta:H_A\\rightarrow X\\) be a natural transformation. Then, \\(\\varphi(\\eta)=\\tilde{\\eta}\\circ \\eta:H_A\\rightarrow Y\\). We haveSo,So, the bijection \\(\\Phi:[\\mathcal{C}^{\\rm{op}},\\rm{Set}](H_A,X)\\rightarrow X(A)\\) is natural in \\(X\\). Special case: We have seen that \\(\\Phi:[\\mathcal{C}^{\\rm{op}},\\rm{Set}](H_A,X)\\rightarrow X(A)\\) is natural in \\(A\\) and \\(X\\). Let \\(X=H_B\\) for some \\(B\\in \\mathcal{C}\\) then, we have \\(\\Phi:[\\mathcal{C}^{\\rm{op}},\\rm{Set}](H_A,H_B)\\rightarrow H_B(A)=\\mathcal{C}(A,B)\\). So, let \\(\\mathcal{D}\\) denote the category \\([\\mathcal{C}^{\\rm{op}},\\rm{Set}]\\). Then, Here \\(\\mathcal{C}(A,B)\\) denotes the collection of morphisms and \\(\\mathcal{D}(H_A,H_B)\\) denotes the collection of all natural transformations. We have Covariant version of Yoneda lemma : Let \\(\\mathcal{C}\\) be a (locally) small category. Then naturally in \\(A\\in \\mathcal{C}\\) and \\(X\\in [\\mathcal{C},\\rm{Set}]\\), where \\(H^A:\\mathcal{C}\\rightarrow \\rm{Set}\\) is given by \\(B\\mapsto \\mathcal{C}(A,B)\\). As a special case when \\(X=H^B\\) for some \\(B\\in \\mathcal{C}\\) we have Here \\(\\mathcal{C}(B,A)\\) denotes the collection of morphisms and \\(\\mathcal{D}(H^A,H^B)\\) denotes the collection of all natural transformations. We have ","permalink":"https://praphulla-koushik.github.io/2017/07/09/yoneda-lemma/","summary":"\u003cstrong\u003eYoneda lemma : \u003c/strong\u003eLet \\(\\mathcal{C}\\) be a (locally) small category. Then\n\n\u003cimg class=\" size-full wp-image-688 aligncenter\" src=\"/wp-media/2017/07/e54e03b561-ql_5d7a52d15b0bc85c34de3666351c92e9_l3.png\" alt=\"ql_5d7a52d15b0bc85c34de3666351c92e9_l3\" width=\"192\" height=\"19\" /\u003e\n\nnaturally in \\(A\\in \\mathcal{C}\\) and \\(X\\in [\\mathcal{C}^{\\rm{op}},\\rm{Set}]\\).\n\n\u003cstrong\u003eTerminology\u003c/strong\u003e :\n\u003col\u003e\n\t\u003cli\u003e \\(\\mathcal{C}\\) is a category mentioned in the lemma, \\(\\mathcal{C}^{\\rm{op}}\\) is the opposite category associated to \\(\\mathcal{C}\\).\u003c/li\u003e\n\t\u003cli\u003e \\(\\rm{Set}\\) is the category with elements as sets and morphisms as functions.\u003c/li\u003e\n\t\u003cli\u003e \\(X:\\mathcal{C}^{op}\\rightarrow \\rm{Set}\\) is a functor.\u003c/li\u003e\n\t\u003cli\u003e Given \\(A\\in \\mathcal{C}\\), \\(H_A\\) is the functor \\(H_A:\\mathcal{C}^{\\rm{op}}\\rightarrow \\rm{Set}\\) given by $B\\mapsto \\mathcal{C}(B,A)$.\u003c/li\u003e\n\t\u003cli\u003e \\([\\mathcal{C}^{\\rm{op}},\\rm{Set}]\\) is the category with functors from \\(\\mathcal{C}^{\\rm{op}}\\) to $\\rm{Set}$ as elements and natural transformations between these functors as morphisms.\u003c/li\u003e\n\t\u003cli\u003e \\([\\mathcal{C},\\rm{Set}]\\) is the category with functors from \\(\\mathcal{C}\\) to \\(rm{Set}\\) as elements and natural transformations between these functors as morphisms.\u003c/li\u003e\n\u003c/ol\u003e\nGiven \\(X,A\\) as above, \\(X(A)\\) is a set and \\([\\mathcal{C}^{\\rm{op}},\\rm{Set}](H_A,X)\\) is a set. Yoneda lemma says that there is a bijection between these sets, natural bijection in both \\(A\\) and \\(X\\).\n\n\u003cstrong\u003eNatural transformation : \u003c/strong\u003eLet \\(\\mathcal{A},\\mathcal{B}\\) be two categories and \\(F,G:\\mathcal{A}\\rightarrow \\mathcal{B}\\) be both contravariant or both covariant functors. A natural transformation \\(\\eta:F\\rightarrow G\\) is a family of arrows (morphisms) \\(F(A)\\xrightarrow{\\eta(A)}G(A)\\) such that for each \\(A\\xrightarrow{f}A'\\) in \\(\\mathcal{A}\\) the following appropriate diagram commutes.\u003cimg class=\" size-full wp-image-678 aligncenter\" src=\"/wp-media/2017/07/22e01cc582-ql_f5f1b45cf5b9428b42e7bcfd77134d7e_l3.png\" alt=\"ql_f5f1b45cf5b9428b42e7bcfd77134d7e_l3\" width=\"311\" height=\"94\" /\u003e\n\nNow that we have defined terminology used in the statement, we will now prove the statement. We will first give a bijection\u003cimg class=\" size-full wp-image-769 aligncenter\" src=\"/wp-media/2017/07/b08c0f55b6-ql_d75b467242e412c976d496142aba4b20_l3.png\" alt=\"ql_d75b467242e412c976d496142aba4b20_l3\" width=\"225\" height=\"19\" /\u003eand then prove that it is natural in \\(A\\) and \\(X\\).\n\n\u003cstrong\u003eConstruction of Bijective map : \u003c/strong\u003eLet \\(\\eta\\in [\\mathcal{C}^{\\rm{op}},\\rm{Set}](H_A,X)\\) i.e., \\(\\eta:H_A\\rightarrow X\\) is a natural transformation. We want to assign an element in \\(X(A)\\) with this \\(\\eta\\). It is only natural to consider the map \\(\\eta(A):H_A(A)\\rightarrow X(A)\\). The set \\(H_A(A)\\) has a special element namely \\(1_A\\in H_A(A)\\), its image \\(\\eta(A)(1_A)\\in X(A)\\). Define \\(\\Phi(\\eta)=\\eta(A)(1_A)\\). This give a map\n\u003cp style=\"text-align:center;\"\u003e\\(\\Phi:[\\mathcal{C}^{\\rm{op}},\\rm{Set}](H_A,X)\\rightarrow X(A)\\).\u003c/p\u003e","tags":["yoneda-lemma"],"title":"Yoneda Lemma","type":"wp-import"},{"categories":["Sheaves of Modules","Algebraic geometry"],"content":"","permalink":"https://praphulla-koushik.github.io/2017/07/08/quasi-coherentcoherent-sheaf-of-modules/","summary":"","tags":[],"title":"Quasi coherent/coherent sheaf of Modules","type":"wp-import"},{"categories":["Algebraic geometry"],"content":"","permalink":"https://praphulla-koushik.github.io/2017/07/08/sheaf-associated-to-a-graded-module-over-a-graded-ring/","summary":"","tags":[],"title":"Sheaf associated to a graded module over a graded ring","type":"wp-import"},{"categories":["Sheaves of Modules","algebraic-geometry","category-theory"],"content":"Let \\(A\\) be a ring and \\(M\\) be an \\(A\\) module. We associate a sheaf of modules \\(\\widetilde{M}\\) on \\(X= \\text{Spec(A)}\\) with this module \\(M\\). These modules are our models for quasi-coherent sheaves. For an open subset \\(U\\subseteq \\text{Spec(A)}\\) we definewhere the condition \\(\\dagger\\) says that given \\(p\\in U\\) we have \\(s(p)\\in M_p\\) and that \\(s\\) is locally a fraction i.e., given \\(p\\in U\\) there exists an open set \\(U(p)\\subseteq U\\) and \\(m\\in M, f\\in A\\) such that \\(s(q)=\\frac{m}{f}\\in M_q\\) for all \\(q\\in U(p)\\). With obvious restriction maps this defines a sheaf \\(\\widetilde{M}\\) on \\(X= \\text{Spec(A)}\\) called the sheaf associated with \\(M\\). This should remind you something similar we have done before. We have defined structure sheaf on \\(X=\\text{Spec(A)}\\) in exactly same way where \\(M\\) in this definition is replaced by ring \\(A\\). Just to confirm, we havewhere the condition \\(\\dagger\\) says that given \\(p\\in U\\) we have \\(s(p)\\in A_p\\) and that \\(s\\) is locally a fraction i.e., given \\(p\\in U\\) there exists an open set \\(U(p)\\subseteq U\\) and \\(a\\in A, f\\in A\\) such that \\(s(q)=\\frac{a}{f}\\in A_q\\) for all \\(q\\in U(p)\\). This should suggest some relation between \\(\\widetilde{M}(U)\\) and \\(\\mathcal{O}_X(U)\\). It turns out that \\(\\widetilde{M}(U)\\) is an \\(\\mathcal{O}_X(U)\\) module for every open \\(U\\subseteq \\text{Spec(A)}\\). So, \\(\\widetilde{M}\\) is a sheaf of \\(\\mathcal{O}_X\\) modules. In case of structure sheaf \\(\\mathcal{O}_X\\) on \\(X=\\text{Spec (A)}\\) we have following proposition. Proposition : Let \\(A\\) be a ring, and \\((\\text{Spec A},\\mathcal{O})\\) its spectrum. For any \\(\\mathfrak{p}\\in \\text{Spec A}\\), the stalk \\(\\mathcal{O}_{\\mathfrak{p}}\\) of the sheaf \\(\\mathcal{O}_{}\\) is isomorphic to the local ring \\(A_{\\mathfrak{p}}\\) i.e., \\(\\mathcal{O}_{\\mathfrak{p}}\\cong A_{\\mathfrak{p}}\\). For any element \\(f\\in A\\), the ring \\(\\mathcal{O}(D(f))\\) is isomorphic to the localized ring \\(A_f\\) i.e., \\(\\mathcal{O}(D(f))\\cong A_f\\). In particular, \\(\\Gamma(\\text{Spec A}, \\mathcal{O})\\cong A\\). In case of associated sheaf \\(\\widetilde{M}\\) on \\(X=\\text{Spec A}\\) we have same results with \\(A\\) replaced by \\(M\\) and \\(\\mathcal{O}\\) replaced by \\(\\widetilde{M}\\). Precisely, we have following proposition. Proposition : Let \\(A\\) be a ring, \\(M\\) be an \\(A\\) module and and \\(\\widetilde{M}\\) be the associated sheaf on \\(X=\\text{Spec A}\\). For any \\(\\mathfrak{p}\\in \\text{Spec A}\\), the stalk \\(\\widetilde{M}_{\\mathfrak{p}}\\) of the sheaf \\(\\widetilde{\u0026lt;}\\) is isomorphic to the localization \\(M_{\\mathfrak{p}}\\) i.e., \\(\\widetilde{M}_{\\mathfrak{p}}\\cong M_{\\mathfrak{p}}\\). For any element $f\\in A$, the ring \\(\\widetilde{M}(D(f))\\) is isomorphic to the localization \\(M_f\\) i.e., \\(\\widetilde{M}(D(f))\\cong M_f\\). In particular, \\(\\Gamma(\\text{Spec A}, \\widetilde{M})\\cong M\\). Proposition : Let \\(A\\) be a ring and let \\(X=\\text{Spec}(A)\\). Also let \\(A\\rightarrow B\\) be a ring homomorphism, and let \\(f:\\text{Spec}(B)\\rightarrow \\text{Spec}(A)\\) be the correpsonding morphism of spectra. Then : the map \\(M\\mapsto \\widetilde{M}\\) gives an exact, fully faithful functor from category of \\(A\\) modules to the category of \\(\\mathcal{O}_X\\) modules. \\(\\widetilde{M\\otimes_A N}=\\widetilde{M}\\otimes_{\\mathcal{O}_X}\\widetilde{N}\\). \\(\\widetilde{\\bigoplus M_i}=\\bigoplus \\widetilde{M_i}\\). For a \\(B\\) module \\(N\\), we have \\(f_*(\\widetilde{N})=\\widetilde{~_A N}\\) where \\(~_A N\\) is \\(N\\) considered as an \\(A\\) module. For a \\(A\\) module \\(M\\), we have \\(f^*(\\widetilde{M})=\\widetilde{M\\otimes_A B}\\). ","permalink":"https://praphulla-koushik.github.io/2017/07/08/sheaf-associated-to-a-module-over-a-ring/","summary":"Let \\(A\\) be a ring and \\(M\\) be an \\(A\\) module.\n\nWe associate a sheaf of modules \\(\\widetilde{M}\\)  on \\(X= \\text{Spec(A)}\\) with this module \\(M\\). These modules are our models for quasi-coherent sheaves.\n\nFor an open subset \\(U\\subseteq \\text{Spec(A)}\\) we define\u003cimg class=\" size-full wp-image-620 aligncenter\" src=\"/wp-media/2017/07/8c822dc158-ql_9405d0bdb2b92f4edcef95865429d34f_l3.png\" alt=\"ql_9405d0bdb2b92f4edcef95865429d34f_l3\" width=\"389\" height=\"64\" /\u003ewhere the condition \\(\\dagger\\) says that given \\(p\\in U\\) we have \\(s(p)\\in M_p\\) and that \\(s\\) is locally a fraction i.e., given \\(p\\in U\\) there exists an open set \\(U(p)\\subseteq U\\) and \\(m\\in M, f\\in A\\) such that \\(s(q)=\\frac{m}{f}\\in M_q\\) for all \\(q\\in U(p)\\). With obvious restriction maps this defines a sheaf \\(\\widetilde{M}\\) on \\(X= \\text{Spec(A)}\\) called the sheaf associated with \\(M\\).\n\nThis should remind you something similar we have done before. We have defined structure sheaf on \\(X=\\text{Spec(A)}\\) in exactly same way where \\(M\\) in this definition is replaced by ring \\(A\\). Just to confirm, we have\u003cimg class=\" size-full wp-image-621 aligncenter\" src=\"/wp-media/2017/07/85383b043a-ql_95ea6db253a445248fdc006fdfabc129_l3.png\" alt=\"ql_95ea6db253a445248fdc006fdfabc129_l3\" width=\"392\" height=\"64\" /\u003ewhere the condition \\(\\dagger\\) says that given \\(p\\in U\\) we have \\(s(p)\\in A_p\\) and that \\(s\\) is locally a fraction i.e., given \\(p\\in U\\) there exists an open set \\(U(p)\\subseteq U\\) and \\(a\\in A, f\\in A\\) such that \\(s(q)=\\frac{a}{f}\\in A_q\\) for all \\(q\\in U(p)\\).\n\nThis should suggest some relation between \\(\\widetilde{M}(U)\\) and \\(\\mathcal{O}_X(U)\\).  It turns out that \\(\\widetilde{M}(U)\\) is an \\(\\mathcal{O}_X(U)\\) module for every open \\(U\\subseteq \\text{Spec(A)}\\). So, \\(\\widetilde{M}\\) is a sheaf of \\(\\mathcal{O}_X\\) modules.\n\n\u003cstrong\u003eIn case of structure sheaf \\(\\mathcal{O}_X\\) on \\(X=\\text{Spec (A)}\\) we have following proposition.\u003c/strong\u003e\n\n\u003cstrong\u003eProposition : \u003c/strong\u003eLet \\(A\\) be a ring, and \\((\\text{Spec A},\\mathcal{O})\\) its spectrum.\n\u003col\u003e\n\t\u003cli\u003eFor any \\(\\mathfrak{p}\\in \\text{Spec A}\\), the stalk \\(\\mathcal{O}_{\\mathfrak{p}}\\) of the sheaf \\(\\mathcal{O}_{}\\) is isomorphic to the local ring \\(A_{\\mathfrak{p}}\\) i.e., \\(\\mathcal{O}_{\\mathfrak{p}}\\cong A_{\\mathfrak{p}}\\).\u003c/li\u003e\n\t\u003cli\u003eFor any element \\(f\\in A\\), the ring \\(\\mathcal{O}(D(f))\\) is isomorphic to the localized ring \\(A_f\\) i.e., \\(\\mathcal{O}(D(f))\\cong A_f\\).\u003c/li\u003e\n\t\u003cli\u003eIn particular, \\(\\Gamma(\\text{Spec A}, \\mathcal{O})\\cong A\\).\u003c/li\u003e\n\u003c/ol\u003e\n\u003cstrong\u003eIn case of associated sheaf \\(\\widetilde{M}\\) on \\(X=\\text{Spec A}\\) we have same results with \\(A\\) replaced by \\(M\\) and \\(\\mathcal{O}\\) replaced by \\(\\widetilde{M}\\). Precisely, we have following proposition.\u003c/strong\u003e\n\n\u003cstrong\u003eProposition : \u003c/strong\u003eLet \\(A\\) be a ring, \\(M\\) be an \\(A\\) module and and \\(\\widetilde{M}\\) be the associated sheaf on \\(X=\\text{Spec A}\\).\n\u003col\u003e\n\t\u003cli\u003eFor any \\(\\mathfrak{p}\\in \\text{Spec A}\\), the stalk \\(\\widetilde{M}_{\\mathfrak{p}}\\) of the sheaf \\(\\widetilde{\u0026lt;}\\) is isomorphic to the localization \\(M_{\\mathfrak{p}}\\) i.e., \\(\\widetilde{M}_{\\mathfrak{p}}\\cong M_{\\mathfrak{p}}\\).\u003c/li\u003e\n\t\u003cli\u003eFor any element $f\\in A$, the ring \\(\\widetilde{M}(D(f))\\) is isomorphic to the localization \\(M_f\\) i.e., \\(\\widetilde{M}(D(f))\\cong M_f\\).\u003c/li\u003e\n\t\u003cli\u003eIn particular, \\(\\Gamma(\\text{Spec A}, \\widetilde{M})\\cong M\\).\u003c/li\u003e\n\u003c/ol\u003e\n\u003cstrong\u003eProposition : \u003c/strong\u003eLet \\(A\\) be a ring and let \\(X=\\text{Spec}(A)\\). Also let \\(A\\rightarrow B\\) be a ring homomorphism, and let \\(f:\\text{Spec}(B)\\rightarrow \\text{Spec}(A)\\) be the correpsonding morphism of spectra. Then :\n\u003col\u003e\n\t\u003cli\u003ethe map \\(M\\mapsto \\widetilde{M}\\) gives an exact, fully faithful functor from category of \\(A\\) modules to the category of \\(\\mathcal{O}_X\\) modules.\u003c/li\u003e\n\t\u003cli\u003e\\(\\widetilde{M\\otimes_A N}=\\widetilde{M}\\otimes_{\\mathcal{O}_X}\\widetilde{N}\\).\u003c/li\u003e\n\t\u003cli\u003e\\(\\widetilde{\\bigoplus M_i}=\\bigoplus \\widetilde{M_i}\\).\u003c/li\u003e\n\t\u003cli\u003eFor a \\(B\\) module \\(N\\), we have \\(f_*(\\widetilde{N})=\\widetilde{~_A N}\\) where \\(~_A N\\) is \\(N\\) considered as an \\(A\\) module.\u003c/li\u003e\n\t\u003cli\u003eFor a \\(A\\) module \\(M\\), we have \\(f^*(\\widetilde{M})=\\widetilde{M\\otimes_A B}\\).\u003c/li\u003e\n\u003c/ol\u003e","tags":["schemes","sheaves"],"title":"Sheaf associated to a Module over a ring","type":"wp-import"},{"categories":["Algebraic geometry"],"content":"In this post we will see definitions of the following terms sheaf of \\(\\mathcal{O}_X\\) module. Tensor product of two sheaves. Direct image sheaf \\(\\mathcal{O}_X\\) module. Inverse image sheaf \\(\\mathcal{O}_X\\) module. Definition : Let \\((X,\\mathcal{O}_X)\\) be a ringed space. A sheaf of \\(\\mathcal{O}_X\\) modules is a sheaf \\(\\mathcal{F}\\) on \\(X\\) such that for each open \\(U\\subseteq X\\), \\(\\mathcal{F}(U)\\) is an \\(\\mathcal{O}_X(U)\\) module and for each inclusion \\(V\\subseteq U\\) we have compatibility of restriction maps with module structure i.e., following diagram is commuatative","permalink":"https://praphulla-koushik.github.io/2017/07/08/sheaves-of-modules-introducton/","summary":"In this post we will see definitions of the following terms\n\u003cul\u003e\n\t\u003cli\u003esheaf of \\(\\mathcal{O}_X\\) module.\u003c/li\u003e\n\t\u003cli\u003eTensor product of two sheaves.\u003c/li\u003e\n\t\u003cli\u003eDirect image sheaf \\(\\mathcal{O}_X\\) module.\u003c/li\u003e\n\t\u003cli\u003eInverse image sheaf \\(\\mathcal{O}_X\\) module.\u003c/li\u003e\n\u003c/ul\u003e\n\u003cstrong\u003eDefinition\u003c/strong\u003e : Let \\((X,\\mathcal{O}_X)\\) be a ringed space. A sheaf of \\(\\mathcal{O}_X\\) modules is a sheaf \\(\\mathcal{F}\\) on \\(X\\) such that for each open \\(U\\subseteq X\\), \\(\\mathcal{F}(U)\\) is an \\(\\mathcal{O}_X(U)\\) module and for each inclusion \\(V\\subseteq U\\) we have compatibility of restriction maps with module structure i.e., following diagram is commuatative\u003cimg class=\" size-full wp-image-604 aligncenter\" src=\"/wp-media/2017/07/b68c66846b-ql_b9327d0c981bbce4965f8a77497f2f86_l3.png\" alt=\"ql_b9327d0c981bbce4965f8a77497f2f86_l3\" width=\"209\" height=\"84\" /\u003e","tags":["sheaves","tensor-product"],"title":"Sheaves of Modules - Introducton","type":"wp-import"},{"categories":["algebraic-geometry","category-theory","differential-geometry"],"content":"I am Praphulla Koushik, faculty member in department of Mathematics, NIT Calicut. I am interested broadly in Differential geometry and Category theory. I want to share my thoughts as I read the following books. Hartshorne's Algebraic geometry. ... ... I wish to add some articles that seems to be interesting for me on different topics. Most of the times, I start writing a post and leave it in the middle.. I usually write in usual latex before copying it here. It is time consuming to put latex in between dollars. I have to take screen shot of figures, crop into image and then add here which is really time consuming. So, If I feel I am wasting time, I will leave it like that. If you find any of my posts interesting and incomplete, do leave a message. I will continue it.","permalink":"https://praphulla-koushik.github.io/2015/06/02/about/","summary":"I am Praphulla Koushik, faculty member in department of Mathematics, NIT Calicut.\n\nI am interested broadly in Differential geometry and Category theory.\n\nI want to share my thoughts as I read the following books.\n\u003col\u003e\n \t\u003cli\u003eHartshorne's Algebraic geometry.\u003c/li\u003e\n \t\u003cli\u003e...\u003c/li\u003e\n \t\u003cli\u003e...\u003c/li\u003e\n\u003c/ol\u003e\nI wish to add some articles that seems to be interesting for me on different topics.\n\nMost of the times, I start writing a post and leave it in the middle.. I usually write in usual latex before copying it here.  It is time consuming to put latex in between dollars. I have to take screen shot of figures, crop into image and then add here which is really time consuming. So, If I feel I am wasting time, I will leave it like that. If you find any of my posts interesting and incomplete, do leave a message. I will continue it.","tags":["hartshorne"],"title":"About","type":"wp-import"},{"categories":[],"content":"I am Praphulla Koushik, faculty member in department of Mathematics, NIT Calicut.\nI am interested broadly in Differential geometry and Category theory.\nMost of the times, I start writing a post and leave it in the middle. I usually write in usual $ \\LaTeX $ before copying it here. I have to take screen shot of figures, crop into image and then add here which is really time consuming. So, If I feel I am wasting time, I will leave it like that. If you find any of my posts interesting and incomplete, do leave a message. I will continue it. This is my blog page. One can look at my webpage here.\n","permalink":"https://praphulla-koushik.github.io/about/","summary":"\u003cp\u003eI am Praphulla Koushik, faculty member in department of Mathematics, NIT Calicut.\u003c/p\u003e\n\u003cp\u003eI am interested broadly in Differential geometry and Category theory.\u003c/p\u003e\n\u003cp\u003eMost of the times, I start writing a post and leave it in the middle. I usually write in usual $ \\LaTeX $ before copying it here. I have to take screen shot of figures, crop into image and then add here which is really time consuming. So, If I feel I am wasting time, I will leave it like that. If you find any of my posts interesting and incomplete, do leave a message. I will continue it. This is my blog page. One can look at my webpage \u003ca href=\"https://sites.google.com/view/praphulla-koushik\"\u003ehere.\u003c/a\u003e\u003c/p\u003e","tags":[],"title":"About","type":"page"},{"categories":[],"content":" Algebraic Geometry Lecture Notes / Books Hartshorne\u0026rsquo;s Algebraic Geometry Kobayashi and Nomizu\u0026rsquo;s Book Math StackExchange / Stack Overflow Pages Notation Seminar on Geometry/Topology of Principal/Fiber Bundles Stacks ","permalink":"https://praphulla-koushik.github.io/library/","summary":"\u003col\u003e\n\u003cli\u003e\u003ca href=\"/2017/07/10/algebraic-geometry-lecture-notes-books/\"\u003eAlgebraic Geometry Lecture Notes / Books\u003c/a\u003e\u003c/li\u003e\n\u003cli\u003e\u003ca href=\"/2017/07/20/hartshornes-algebraic-geometry-solutions/\"\u003eHartshorne\u0026rsquo;s Algebraic Geometry\u003c/a\u003e\u003c/li\u003e\n\u003cli\u003e\u003ca href=\"/2018/12/30/kobayashi-and-nomizus-book/\"\u003eKobayashi and Nomizu\u0026rsquo;s Book\u003c/a\u003e\u003c/li\u003e\n\u003cli\u003e\u003ca href=\"/2017/07/11/mathstack-exchange-stack-overflow-pages/\"\u003eMath StackExchange / Stack Overflow Pages\u003c/a\u003e\u003c/li\u003e\n\u003cli\u003e\u003ca href=\"/2019/02/02/notation/\"\u003eNotation\u003c/a\u003e\u003c/li\u003e\n\u003cli\u003e\u003ca href=\"/2019/08/11/seminar-on-geometry-topology-of-principal-fiber-bundles/\"\u003eSeminar on Geometry/Topology of Principal/Fiber Bundles\u003c/a\u003e\u003c/li\u003e\n\u003cli\u003e\u003ca href=\"/2019/01/03/stacks/\"\u003eStacks\u003c/a\u003e\u003c/li\u003e\n\u003c/ol\u003e","tags":[],"title":"Reference Links","type":"page"}]