Differential Geometry on 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.

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

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 $a,b,c\in A$.

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.
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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

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

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

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

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

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 $BG\rightarrow BH$.

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

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

Geometric vector bundle

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

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.