Algebraic-Geometry on Geometry and some category theory

Is it true that eigenvalues of skew-symmetric matrices are always zero?

Sun, 18 May 2025 10:20:47 +0000

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 $-v^TMv=\lambda v^Tv$.

non-abelian simple group of order less than 100

Mon, 18 Nov 2024 15:24:18 +0000

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.

Lie-Rinehart algebras : Morphism of Lie-Rinehart algebras

Fri, 26 Apr 2024 17:37:25 +0000

Once we have a reasonably good notion of an object, we would look at a notion of morphisms.

Let $(L,A,\rho,\tau)$ 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,

Model categories : Part 1 (Motivation)

Sat, 23 May 2020 14:57:32 +0000

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.

Model categories : Part 2 (Definitions)

Thu, 21 May 2020 17:59:45 +0000

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 $\mathcal{D}$.

Stackification of fibred categories

Sat, 02 Feb 2019 13:50:32 +0000

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 $i^*(a),i^*(b)\in \mathcal{S}(V)$.

Criterion for a map to be representable submersion

Mon, 31 Dec 2018 17:53:25 +0000

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

$(\mathcal{D}\times_{\mathcal{C}}W)\times_W W_i=\mathcal{D}\times_{\mathcal{C}} W_i$

Construction of Weil homomorphism

Sun, 30 Dec 2018 14:31:22 +0000

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.

$f(\Omega)(v_1,\cdots,v_{2k})=\frac{1}{(2k)!}\sum_{\sigma\in S_{2k}} f(\Omega(v_{\sigma(1)}.v_{\sigma(2)}),\cdots\Omega(v_{\sigma(2k-1)},v_{\sigma(2k)}))$

Sheafification of a presheaf

Fri, 28 Jul 2017 17:07:43 +0000

Morphism of Sheaves - Morphism of Stalks

Fri, 28 Jul 2017 11:55:19 +0000

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

$\varphi(W_p)(t_p|_{W_p})|_{W_p\cap W_q}=\varphi(W_p\cap W_q)(t_p|_{W_p\cap W_q}).$

Structure sheaf on spectrum of a ring

Tue, 25 Jul 2017 20:04:23 +0000

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.

QcQs lemma

Sun, 23 Jul 2017 09:16:19 +0000

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.

Adjointness of the global section functor and the Spec functor

Sat, 22 Jul 2017 17:38:23 +0000

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

$\alpha: \text{Hom}_{\text{Schemes}}(X,\text{Spec}(A))\rightarrow \text{Hom}_{\text{Rings}}(A,\mathcal{O}_X(X)).$

Then, $\alpha$ 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

$\varphi : A\rightarrow \mathcal{O}_X(X)=O_{\text{Spec}(B)}(\text{Spec}(B))=B$

Fiber product of Schemes

Sat, 22 Jul 2017 11:01:40 +0000

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 $q\circ f=p$.

Sheafification of a presheaf that is already a sheaf is itself - Reality check

Fri, 21 Jul 2017 05:48:15 +0000

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 $\mathcal{F}(U)$.

Hartshorne's Algebraic Geometry

Thu, 20 Jul 2017 20:02:30 +0000

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

Push forward of quasi coherent sheaf of modules

Fri, 14 Jul 2017 21:07:28 +0000

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

$f_*(\mathcal{F}|_{f^{-1}(U)})\cong f_*\widetilde{M}$.

Global Spec Or Relative Spec of a Scheme

Thu, 13 Jul 2017 09:03:44 +0000

Let $X=\text{Spec}(R)$ be an affine scheme.

Let $X'$ 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

$R=\Gamma(X,\mathcal{O}_X)\rightarrow \Gamma(X',\mathcal{O}_X')=R'$

giving $R'$, 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

$\alpha: \text{Hom}_{\text{Schemes}}(X,\text{Spec}(A))\rightarrow\text{Hom}_{\text{Rings}}(A,\mathcal{O}_X(X)).$

Math stack exchange/ stack overflow questions

Tue, 11 Jul 2017 14:33:19 +0000

Here I will add links of pages of interesting questions/answers from Math Stack Exchange and Math over flow.

Tensor algebra, symmetric algebra and exterior algebra of a sheaf

Mon, 10 Jul 2017 17:09:20 +0000

Associated sheaf and global section functors are adjoint

Mon, 10 Jul 2017 17:05:52 +0000

Algebraic Geometry Lecture Notes/ Books

Mon, 10 Jul 2017 14:05:39 +0000

Andreas Gathmann Algebraic Geometry Foundations of Algebraic Geometry Ravi Vakil Kiran Kedlaya Algebraic Geometry Lecture Notes

Yoneda Lemma

Sun, 09 Jul 2017 08:06:02 +0000

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

$\Phi:[\mathcal{C}^{\rm{op}},\rm{Set}](H_A,X)\rightarrow X(A)$.

Quasi coherent/coherent sheaf of Modules

Sat, 08 Jul 2017 03:55:50 +0000

Sheaf associated to a graded module over a graded ring

Sat, 08 Jul 2017 03:53:18 +0000

Sheaf associated to a Module over a ring

Sat, 08 Jul 2017 03:46:15 +0000

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 $\widetilde{<}$ 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 $f^*(\widetilde{M})=\widetilde{M\otimes_A B}$.

Sheaves of Modules - Introducton

Sat, 08 Jul 2017 02:48:46 +0000

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

About

Tue, 02 Jun 2015 08:24:56 +0000

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.
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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.