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\begin{document}
\title{Flat Connections}
\maketitle
In this note we try to prove the following statement. Let $X$ be a smooth manifold and $\pi:E\to X$ be a smooth vector bundle, equipped with a connection 
whose curvature form vanishes. Then we can choose trivialisations of $E$ such that the transition functions are locally constant. The main ingredient is 
the Frobenius theorem on existence of integral submanifolds to involutive distributions. 
\begin{definition}
A connection $D$ is a $\C$ linear map $D:\Gamma(U,E)\to \Gamma(U,E)\otimes \Gamma(U,\Omega_U)$ defined for every open set $U\subset X$ satisfying
Leibniz rule, i.e. for $f\in\Gamma(U,\mathscr{O}_X)$, $D(fs)=s\otimes df + fD(s)$.
\end{definition}
Assume that $rank(E)=n$ and choose a frame over an open subset $U$ over which $E$ has a trivialisation. Say the frame is given by $e_1, \cdots, e_n$. 
$D(e_i)=\sum_{j=1}^{n}e_j\otimes \omega_{ji}^U$. This matrix of 1 forms is called the connection matrix for this trivialisation. Suppose $V$ is another 
open subset over which $E$ is trivial, and we have chosen a frame $f_1,\cdots, f_n$ over $V$, then we may ask what is the relation between $\omega^U$ 
and $\omega^V$. Let $e_i=\sum_{j=1}^{n} g_{ji}f_j$. Then on the one hand, $D(e_i)=\sum_{j=1}^{n}e_j\otimes \omega_{ji}^U=\sum_{j=1}^{n} \sum_{k=1}^{n} 
g_{kj}f_k\otimes \omega_{ji}^U=\sum_{k=1}^{n}(g\omega^U)_{ki}f_k$. On the other hand, $D(e_i)=D(\sum_{j=1}^{n} g_{ji}f_j)=\sum_{j=1}^{n}(dg_{ji}f_j
+\sum_{k=1}^{n}f_k\otimes g_{ji}\omega^V_{kj})=\sum_{k=1}^{n}(dg_{ki}+(\omega^V g)_{ki})f_k$. Equating the coefficients of $f_k$, we get 
$g\omega^U=dg+\omega^V g$, i.e. \[\omega^V=g\omega^Ug^{-1}-dg g^{-1}\]

\vskip 3mm
We can construct a principal $GL_n$ bundle from $E$. Consider the bundle $\underbar{Hom}(\C^n,E)\stackrel{\pi}{\to}X$. The open subset 
$\underbar{Iso}(\C^n,E)$ is the required $GL_n$ bundle. $GL_n$ acts on it naturally on the right, the action being given by $(\phi,g)\mapsto \phi\circ g$. 
Let us call this bundle $P\stackrel{\pi}{\to}U$. If $U$ is an open subset over which $E$ is trivial, once we choose a frame $e_1,\cdots, e_n$, we can 
identify $P\vert_U\cong U\times G$, where $G=GL_n$. This identification is given as follows, let $w_i$ denote the standard basis for $\C^n$. An element 
of $P$ is a pair $(x,\phi)$ with $\phi:\C^n\to E_x$ an isomorphism. Then $\phi(w_i)=\sum_{j=1}^{n}A_{ji}e_j(x)$. We send the element $(x,\phi)$ to the 
pair $(x, A)$. It is clear that $(x,\phi\circ g)\mapsto (x,Ag)$. If $V$ is another open set over which $E$ is trivial with frame $f_1,\cdots, f_n$, then 
$P\vert_V\cong V\times G$. Let $x\in U\cap V$, and let 
$\phi:\C^n\to E_x$ be an isomorphism. Then $\phi(w_i)=\sum_{j=1}^{n}A_{ji}^U e_j(x)=\sum_{j=1}^{n}A_{ji}^U (\sum_{k=1}^{n}g_{kj}(x)f_k(x))=\sum_{k=1}^{n}
(g(x)A^U)_{ki}f_k(x)=\sum_{k=1}^{n}A_{ki}^V f_k(x)$. This shows that $gA^{U}=A^{V}$. Thus, the point $(x,A)\in U\times G$ is identified with the point 
$(x,gA)\in V\times G$.

On the space $P$, we have a natural map of bundles, namely, $TP\to \pi^*TX\to 0$, which comes from the operation of pushing forward tangent vectors. 
Let us describe this map locally. Let $U, e_i$ be a frame as above. Then $P\vert_U\cong U\times G$ and the above map is the 
projection $TU\oplus TG\to TU$. Since $G$ acts on $P$ on the right, it acts on the tangent bundle and this action has the following local description. 
Let $(v_x,w_h)$ be a tangent vector to $P$ at the point $(x,h)$. Then for $g\in G$, this gets mapped to the tangent vector $(v_x, (R_g)_*(w_h))$ at the 
point $(x,hg)$. It is clear that the map $TP\to \pi^*TX$ is $G$ equivariant, where $G$ acts trivially on $\pi^*TX$. 

Let us describe the kernel of the above map. If $G$ is a Lie group acting on a manifold $M$ on the right, then to each element $\xi\in\mathfrak{g}$ 
we can associate a vector field $X_\xi$ on $M$. If $p\in M$, then define $X_\xi(p)=\frac{d}{dt}\big\vert_{t=0}\big(p\cdot\exp(t\xi)\big)$. From 
this, it is clear that $(R_g)_{*}(X_\xi(p))=X_{Ad_{g^{-1}}(\xi)}(p\cdot g)$, where $R_g$ is the action of $g$ on $M$. Since $G$ acts on the right of $P$, 
we can use this construction to produce vector fields, i.e. global sections of $TP$, on $P$. This defines a map from $P\times\mathfrak{g}\to TP$. 
Let us check that this map is $G$ equivariant for the right action on $P\times\mathfrak{g}$ given by $(p,\xi)\mapsto (p\cdot g, Ad_{g^{-1}}(\xi))$. 
But this is clear from the preceding formula.

\begin{comment}
Next we will describe the kernel of this map. Construct a vector bundle on $P$ as follows, over the open set $U\times G$ take the trivial bundle 
$U\times G\times \g$. If $x\in U\cap V$, then identify $(x,h,v)$ with $(x,gh,(L_{g})_*(R_{g^{-1}})_*(v))$. Clearly this gives a bundle $B$ on $P$. 
There is a natural inclusion $B\to TP$ and $B$ is preserved by the action of $G$. We will define this map locally and check that they patch up. We 
identify $\g$ with $T_eG$. Over the open subset $U\times G\times\g$ consider the point $(x,h,v)$ where $v\in T_eG$ . Map this to the tangent vector 
$(0, (R_h)_*(v))$ at the point $(x,h)$. If $x\in U\cap V$, then $(x,h,v)\sim(x,gh,(L_{g})_*(R_{g^{-1}})_*(v))$, which maps to the tangent vector 
$(0,(R_{gh})_*(L_{g})_*(R_{g^{-1}})_*(v))=(0,(R_{h})_*(R_g)_*(L_{g})_*(R_{g^{-1}})_*(v))=(0, (L_g)_*(R_h)_*(v))$. On the other hand, the isomorphism 
$U\cap V\times G\to U\cap V\times G$ given by $(x.h)\mapsto (x,gh)$ sends a tangent vector $(v,w)$ to the tangent vector $(v,(R_h)_*dg(v)+(L_g)_*(w))$, 
where $g:U\cap V\to G$. Thus the tangent vector $(0, (R_h)_*(v))$ at the point $(x,h)$ is identified with the tangent vector $(0, (L_g)_*(R_h)_*(v))$ 
at the point $(x,gh)$. Thus, we have defined a map $B\to TP$. Checking that this map preserves $B$ is again done locally. $(x,h,v)$ gets mapped to 
$(0, (R_h)_*(v))$, which under the action of $g$ gets mapped to $(0, (R_g)_*(R_h)_*(v))=(0, (R_{hg})_*(v))$ which is a tangent vector over the point 
$(x,hg)$, but this is also the image of $(x,hg,v)$. 
\end{comment}

We have shown that there is a $G$ equivariant short exact sequence of bundles 
\[0\to P\times\mathfrak{g}\to TP\to \pi^*TX\to 0\]


\begin{definition}
A connection $D$ on $P$ is a $G$ equivariant splitting of the above short exact sequence.
\end{definition}

Let us analyse this splitting locally. We will freely use the natural identification of $\mathfrak{g}$ with $T_eG$. Over the point $(x,g)\in U\times G$, 
the above short exact sequence is $0\to T_eG\to T_xU\oplus T_gG\to T_xU\to 0$. 
Let $s_U:T_xU\oplus T_gG\to T_eG$ be the splitting. If $w\in T_gG$, then the $G$ equivariance of $s_U$ forces that 
$Ad_{g}(s_U(v,w))=s_U((R_{g^{-1}})_*(v,w))=s_U(v,(R_{g^{-1}})_*(w))=s_U(v,0)+(R_{g^{-1}})_*(w)$. Thus, locally 
giving the splitting is same as giving a map $T_xU\to T_eG$. which is same as giving a matrix of 1-forms, call it $\omega^U$. What is the relation between 
$\omega^U$ and $\omega^V$? Let $W:=U\cap V$. Then $g:W\to G$ is smooth. Let us compute the differential of the map $W\times G\to W\times G$ given by 
$(x,h)\mapsto (x,gh)$. This map is the composite of $(x,h)\mapsto (x,g,h)\mapsto (x,gh)$. If $(v,w)$ is a tangent vector at $(x,h)$, it first gets 
mapped to $(v,dg(v),w)$ and then to $(v,(R_h)_*(dg(v))+(L_g)_*(w))$. This calculation shows that $(v,0)$ at the point $(x,e)$ is mapped to 
the tangent vector $(v,(R_e)_*dg(v))=(v,dg(v))$ at the point $(x,g)$. Thus, 

\begin{align*}
s_U(v,0)=s_V(v,dg(v))&=Ad_{g^{-1}}(s_V(v,(R_{g^{-1}})_*(dg(v)))\\
		&=Ad_{g^{-1}}(s_V(v,0))+Ad_{g^{-1}}((R_{g^{-1}})_*(dg(v)))\\
\end{align*}

In the case of $GL_n$, we may rewrite this as $\omega^V=g(\omega^U)g^{-1}-dgg^{-1}$.
This is exactly the condition for giving a connection on the vector bundle $E$. Thus, both notions are equivalent. 

\begin{definition}
The curvature of a connection $D$ is the composite map $E\to E\otimes\Omega_X\to E\otimes\Omega_X^2\to E\otimes\wedge^2\Omega_X$. It is denoted by $\Theta$.
\end{definition}
\begin{align*}
\Theta(e_i)&=D^2(e_i)=D(\sum_{j=1}^n e_j\otimes\omega_{ji})\\
	&=\sum_{j=1}^n(e_j\otimes d\omega_{ji}+\sum_{k=1}^ne_k\otimes\omega_{kj}\wedge\omega_{ji})\\
	&=\sum_{k=1}^ne_k\otimes (d\omega+\omega\wedge\omega)_{ki}
\end{align*}
\begin{align*}
\Theta(fe_i)&=D^2(fe_i)=D(e_i\otimes df + \sum_{j=1}^n fe_j\otimes\omega_{ji})\\
	&=\sum_{j=1}^n(e_j\otimes \omega_{ji}\wedge df+e_j\otimes df\wedge\omega_{ji}+fD(e_j\otimes\omega_{ji}))\\
	&=f\sum_{k=1}^ne_k\otimes (d\omega+\omega\wedge\omega)_{ki}=f\Theta(e_i)
\end{align*}
The above calculations show that $\Theta\in \Gamma(X,End(E)\otimes\wedge^2\Omega_X)$, and locally after choosing frames, it is given by 
$d\omega+\omega\wedge\omega$.
\begin{theorem}
If $\Theta=0$, then we can choose the transition functions to be locally constant.
\end{theorem}
\begin{proof}
Let us denote by $s$ the splitting $TP\to B$ and restrict our attention to what happens over the open subset $U$. Recall that over the point 
$(x,g)\in U\times G$, the short exact sequence above looks like $0\to T_eG\to T_xU\oplus T_gG\to T_xU\to 0$. If we take the kernel of the splitting, then 
we get $d:=dim(X)$ linearly independent tangent vectors. This happens for every point $(x,g)$, which means that we have a $d$ dimensional distribution on 
the manifold $U\times G$. This distribution occurs as the tangent bundle of a submanifold iff it is involutive, i.e. closed under the Lie bracket. We now 
proceed to check this in steps
\vskip 4mm $\it{1:}$ Let $(v,u)$ be a tangent vector at the point $(x,e)$. Let $\omega$ be defined by $s(v,u)=\omega(v)+u$, where $s$ is the splitting.
\vskip 2mm $\it{2:}$ We could have chosen $U$ to be a coordinate neighbourhood with coordinates $x_1,\cdots, x_d$. Let $X=X_1=\frac{\partial}{\partial{x_1}}$ 
be a vector field on $U$. Using this, we define a tangent vector at points $(x,e)$ by $X_x-\omega(X_x)$. Now using the right action of $G$ define tangent 
vectors on the point $(x,g)$ by $X_x-(R_g)_*(\omega(X_x))$. It is clear that when we do this for all $X_i$, these give a basis for the kernel of $s$ at 
every point $(x,g)$. 
\vskip 2mm $\it{3:}$ Suppose $X_i$ are a bunch of independent vector fields on a manifold $M$, then to check that the distribution generated by them is 
involutive, it is enough to check that $[X_i,X_j]$ is in the distribution. This is because $[fX_i,gX_j]=fX_i(g)X_j-gX_j(f)X_i+fg[X_i,X_j]$. 
\vskip 2mm $\it{4:}$ The vector fields on $U\times G$ given by $X_1-\omega(X_1)$ are invariant under the right action of $G$ by construction. Thus, the 
Lie bracket of any two of these will also be $G$ invariant. Hence, it is enough to compute the value of the Lie bracket at the point $(x,e)$. Let $\eta,\xi$ 
be two left invariant vector fields on $G$. Let $X_\eta, X_\xi$ be right invariant vector fields given by $X_\eta(e)=\eta(e), X_\xi(e)=\xi(e)$. Then the 
vector field $[X_\eta,X_\xi]$ is right invariant and $[X_\eta,X_\xi](e)=[\xi,\eta](e)$. The relevance of this remark here is that $\omega(X_i)$ constructed 
are right invariant.

\begin{align*}
[X_1-\omega(X_1),X_2-\omega(X_2)]&=[X_1,X_2]-[\omega(X_1),X_2]-[X_1,\omega(X_2)]\\
				&\qquad+[\omega(X_1),\omega(X_2)]\\
			&=X_2(\omega(X_1))-X_1(\omega(X_2))+\\
			&\qquad\omega(X_2)\omega(X_1)-\omega(X_1)\omega(X_2)\\
			&=-d\omega(X_1,X_2)+\omega([X_1,X_2])\\
			&\qquad-\omega\wedge\omega(X_1,X_2)\\
			&=-(d\omega+\omega\wedge\omega)(X_1,X_2)=0
\end{align*}
In the above calculation, we have used the following formula
\[d\omega(X_1,X_2)=X_1(\omega(X_2))-X_2(\omega(X_1))+\omega([X_1,X_2])\]

Choose an integral submanifold of this distribution through the point $(x_0,e)$. Call this subset $V$. If $p\in V$, then 
$T_pV\to T_{\pi(p)}U$ is an isomorphism. Let $W:=\pi(V)$. If $w=\pi(v)$, then sending $w\mapsto v$ defines a section to $W\times G\to W$. Use this section 
to give a new trivialisation $\tilde{\pi}:W\times G\to W$. By construction, the connection matrix in this trivialisation is 0. Doing this locally over the 
whole manifold $X$, we would have gotten trivialisations such that the connection matrix is always 0. $\omega^V=g\omega^Ug^{-1}-dgg^{-1}$ shows that 
$dgg^{-1}=0$, which means $dg=0$, which means $g$ is locally constant.  
\end{proof}

\end{document}