Differential Geometry

In this note we collect some references that discuss the notion of connection on vector bundle Differential 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 \(\pi:P\rightarrow M\) be a principal \(G\) bundle. Let \(F\) 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 \(G\times F\rightarrow F\). We 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 \(G\times F\rightarrow F\). ...

We will understand the notion of a connection on a vector bundle in the following steps: Give the definition of a connection Explain the objects appearing in the definition (sections and their algebraic structure) Study the trivial bundle case, which motivates the axioms Examine the tangent bundle case and test familiar operations Explain why the usual differential of a section does not give what we want Let \(E\rightarrow M\) be a vector bundle. ...

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 \[\nabla : \Gamma(M,TM)\times \Gamma(M,E)\rightarrow \Gamma(M,E)\] satisfying the following conditions: \(\nabla\) 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 \(s\in\Gamma(M,E)\) ...

Any notion of an "algebra" comes with two binary operations: \(A\times A\rightarrow A\), 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 \(a,b,c\in A\). ...

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  

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 \(\varphi_\alpha(v')\varphi_\beta(v')^{-1}= \varphi_\alpha(vg)\varphi_\beta(ua)^{-1} =\varphi_\alpha(u)aa^{-1}\varphi_\beta(u)^{-1} =\varphi_\alpha(u)\varphi_\beta(u)^{-1}\) Thus, for any \(v,v'\in \pi^{-1}(x)\), we have \(\varphi_\alpha(v)\varphi_\beta(v)^{-1}=\varphi_\alpha(v')\varphi_\beta(v')^{-1}.\) ...

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 \(N\times G\rightarrow P\times G\rightarrow P\) is 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.

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 \(\pi_Q\circ (f\circ \sigma)=(\pi_Q\circ f)\circ \sigma=\pi_P\circ \sigma=1\) ...

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 \(BG\rightarrow BH\).