Geometry and some category theory

Equivalent definitions of connections on vector bundle

Sat, 14 Feb 2026 10:28:30 +0000

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

Vector bundle associated to a principal bundle

Tue, 10 Feb 2026 08:43:42 +0000

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 $G\times F\rightarrow F$.

Connection on vector bundle (Introduction)

Fri, 06 Feb 2026 06:24:52 +0000

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.

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.

computing infimum by an example

Sat, 21 Sep 2024 15:25:30 +0000

Let us consider a problem where you are asked to find infimum of the set

\[\{\int_0^{1}\sqrt{1+f'(x)^2}dx\}_{f\in S}\]

where $S$ is the set of all $f\in C^1(\mathbb{R})$ with the property that $f(0)=10$ and $f(1)=0$.

When we see integral and differential together, that should remind us the famous fundamental theorem of calculus, which says that

limit/limsup/liminf of a sequence (by an example)

Fri, 13 Sep 2024 05:29:07 +0000

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, $\lceil 0.1 \rceil=1, \lceil 0.9 \rceil=1, \lceil -1.2 \rceil=-1, \lceil -2.5 \rceil=-2$

To compute limit (to have a hope of computing), we need to know it converge (which we can check by checking it is Cauchy sequence).

(Alternative description of) Connection on vector bundle

Tue, 09 Jul 2024 18:36:12 +0000

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 $s\in\Gamma(M,E)$

Multilinear algebra : Tensor product

Thu, 16 May 2024 05:40:21 +0000

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 :

Multilinear algebra : an introduction

Sun, 12 May 2024 02:23:31 +0000

In group theory, we mainly study maps that preserve the group structures; which goes by the name of group homomorphisms.

In 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

Universal enveloping algebra

Mon, 29 Apr 2024 17:01:36 +0000

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.

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,

Lie-Rinehart algebras : Introduction and definition of Lie-Rinehart algebra

Wed, 24 Apr 2024 06:42:29 +0000

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 $a,b,c\in A$.

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 $\mathcal{D}$.

Seminar on Geometry/Topology of Principal/fiber bundles

Sun, 11 Aug 2019 14:13:32 +0000

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 :

Books:

YouTube videos :

Fredric Schuller's YouTube channel

MathOverflow/MathStackExchange questions/user pages:

John M. Lee 's MathStackExchange page

Notation

Sat, 02 Feb 2019 17:26:24 +0000

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.

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

Transition maps for principal bundle are smooth

Sat, 26 Jan 2019 08:16:48 +0000

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

$\varphi_\alpha(v)\varphi_\beta(v)^{-1}=\varphi_\alpha(v')\varphi_\beta(v')^{-1}.$

Category theory

Thu, 24 Jan 2019 23:24:08 +0000

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.

Trivializations and sections in Principal bundle

Thu, 24 Jan 2019 19:02:12 +0000

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.

Equivariant maps are Isomorphisms

Wed, 23 Jan 2019 17:49:33 +0000

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

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

Limit of a diagram/functor preserved by Hom functor

Wed, 23 Jan 2019 16:48:25 +0000

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

$\varprojlim_{\mathcal{I}}(\text{Hom}_{\mathcal{C}}(X,F(i))) =\text{Hom}_{\mathcal{C}}(X,\varprojlim_{\mathcal{I}}F(i))$

Definition of gerbe over stack

Sun, 20 Jan 2019 19:12:18 +0000

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

Morphism of Lie groups giving a functor

Fri, 18 Jan 2019 20:46:30 +0000

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 $BG\rightarrow BH$.

Lie groupoids

Wed, 16 Jan 2019 18:37:08 +0000

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 $\phi^{ob}$.

What is a Stack?

Sat, 12 Jan 2019 18:18:08 +0000

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.,

$\mathcal{D}(U)_0=\{V\in \mathcal{D}_0:F(V)=U\}$.

Stacks

Thu, 03 Jan 2019 13:38:28 +0000

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.
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Criterion for a map of stacks to be an atlas

Thu, 03 Jan 2019 12:51:29 +0000

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 $\mathcal{D}$.

Criterion for a stack to be representable

Wed, 02 Jan 2019 15:20:43 +0000

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

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

Invariant polynomials

Sun, 30 Dec 2018 17:32:05 +0000

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

$p=\sum a_{t_1,\cdots,t_r}(e^1)^{t_1}\cdots(e^r)^{t_r}$.

Let $f:V\times V\times \cdots\times V\rightarrow \mathbb{R}$ be a symmetric multilinear mapping. We want to associate

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.

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

Matrix associated to connection/curvature form

Sun, 30 Dec 2018 08:43:58 +0000

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.

Kobayashi and Nomizu's book

Sun, 30 Dec 2018 05:02:27 +0000

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

Sat, 29 Dec 2018 18:01:18 +0000

In this, we see that for a left-invarinat differential form $\omega$ on $G$,

$d\omega(X,Y)=-\frac{1}{2}\omega([X,Y])$

for 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.,

$\omega(g)(v)=\omega(e)((L_{g^{-1}})_{*,g}(v))$

for 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

$(d\omega)(A^*,B^*)=\frac{1}{2}\left[ A^*(\omega(B^*))-B^*(\omega(A^*))-\omega([A^*,B^*])\right]$

Maurer-Cartan form on a Lie group

Sat, 29 Dec 2018 12:40:04 +0000

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

$d\theta(X,Y)=-\frac{1}{2}\theta([X,Y])$

Kernel and cokernel of a Morphism

Wed, 09 Aug 2017 19:22:13 +0000

Additive categories

Wed, 09 Aug 2017 19:21:42 +0000

Pull back and Push forward of two morphisms

Wed, 09 Aug 2017 19:20:53 +0000

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 $g\circ f'=f\circ g'$

Equalizers and Coequalizers

Wed, 09 Aug 2017 19:19:58 +0000

Monomorphisms and epimorphisms

Wed, 09 Aug 2017 19:19:16 +0000

Snake Lemma

Mon, 31 Jul 2017 18:05:04 +0000

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 $\text{Coker}(f)$.

Left exactness of Global section functor

Fri, 28 Jul 2017 17:11:38 +0000

Exact sequences of sheaves

Fri, 28 Jul 2017 17:11:07 +0000

Injective/Surjective Morphisms of Sheaves

Fri, 28 Jul 2017 17:08:34 +0000

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

Group as a category with one object

Wed, 19 Jul 2017 09:30:17 +0000

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

$(g\circ h)\circ f=(g.h)\circ f=(g.h).f=g.(h.f)=g\circ (h\circ f).$

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.

Geometric vector bundle

Mon, 10 Jul 2017 17:10:00 +0000

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.
...
...

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.

About

Mon, 01 Jan 0001 00:00:00 +0000

I am Praphulla Koushik, faculty member in department of Mathematics, NIT Calicut.

I am interested broadly in Differential geometry and Category theory.

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. 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.

Reference Links

Mon, 01 Jan 0001 00:00:00 +0000