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\begin{document}
\begin{preamble}
\subject{Vorlesung aus dem Sommersemester 2012}
\title{Mathematical Gauge Theory 1}
\author{Prof.\,Dr\kern-.1em.~Dieter Kotschick}%Prof. Dr. Dieter Kotschick
\date{}
\publishers{\small ge\TeX{}t von Florian Stecker}
\maketitle
\thispagestyle{empty}
\tableofcontents
\clearpage
\end{preamble}

\section{Lie groups and Lie algebras}
\begin{definition}
A {\em smooth/differentiable manifold} $M$ is a topological space with the property that
\begin{enumerate}[(1)]
\item $M$ is Hausdorff
\item $M$ is second countable, i.e. there is a countable basis of its topology
\item There exists a covering of $M$ by open $U_i$, $i \in I$, and homeomorphisms $\varphi_i \colon U_i \to \mathbb R^n$ such that when $U_i \cap U_j \neq \varnothing$, then $\varphi_j \circ \varphi_i^{-1} \colon \varphi_i(U_i \cap U_j) \to \varphi_j(U_i \cap U_j)$ is a diffeomorphism.
\end{enumerate}
\end{definition}

If $f \colon M \to N$ is a smooth map between smooth manifolds, then there is a derivative $Df \colon TM \to TN$. $Df$ is linear on the fibers of $TM$, i.e. $D_pf \colon T_pM \to T_{f(p)}N$ is linear, and the following diagram commutes:
\[\xymatrix{
TM \ar[r]^{Df}\ar[d]_\pi & TN \ar[d]_\pi \\ M \ar[r]^f & N
}\]

$\mathfrak X(M) \coloneqq \{X \colon M \to TM \mid \text{$X$ is smooth} \land \pi \circ X = \id_M\}$ is the set of smooth sections of $TM$, i.e. vector fields on $M$. For $f \in C^\infty(M)$ we have $L_X f \in C^\infty(M)$, the Lie derivative of $f$ in the direction of $X$. $[X,Y]$ is the unique vector field with the property $L_{[X,Y]}f = L_XL_Y f - L_YL_Xf$.

\begin{definition}
A (real) {\em Lie algebra} is an $\mathbb R$-vector space $\mathfrak g$ together with a bilinear map $[-,-] \colon \mathfrak g \times \mathfrak g \to \mathfrak g$ satisfying:
\begin{enumerate}[(1)]
\item $[v,w] = -[w,v]$
\item $[[x,y],z] + [[y,z],x] + [[z,x],y] = 0$
\end{enumerate}
\end{definition}

\begin{example}
\ \vspace{-.5em}
\begin{enumerate}[(1)]
\item $\mathfrak X(M)$ with $[-,-]$ defined by the Lie bracket is an $\infty$-dimensional Lie algebra.
\item Let $V$ be any $\mathbb R$-vector space. Then $[-,-] = 0$ defines a Lie algebra. These are abelian Lie algebras.
\item $\Mat(n \times n, \mathbb R) = \mathfrak {gl}(n, \mathbb R)$, $[A,B] = AB - BA$.
\end{enumerate}
\end{example}

\begin{definition}
A {\em Lie group} is a group which is also a smooth manifold and has the property that $G \times G \to G, (a,b) \mapsto ab^{-1}$ is smooth.
\end{definition}

\begin{example}\
\vspace{-.5em}
\begin{enumerate}[(1)]
\item Any finite dimensional $\mathbb R$-vector space $V$ with the group structure given by $+$.
\item $\GL(n, \mathbb R)$
\item subgroups of $\GL(n,\mathbb R)$: $\O(n)$, $\SO(n)$, $\GL(n, \mathbb C)$, $\U(n)$, $\SU(n)$.
\end{enumerate}
\end{example}

A Lie group $G$ acts on itself by left multiplication. For any $g \in G$, $\ell_g \colon G \to G, a \mapsto ga$ is a smooth map. $\ell_g$ is a diffeomorphism with inverse $(\ell_g)^{-1} = \ell_{g^{-1}}$.

\begin{definition}
A vector field $X \in \mathfrak X(G)$ is called {\em left--invariant} if $X_{ga} = X_{\ell_g(a)} = (D_a\ell_g)X_a$.
\end{definition}

\begin{proposition}
The subset of left--invariant vector fields in $\mathfrak X(G)$ is a linear subspace closed under the Lie bracket $[-,-]$. Thus it is a Lie algebra.
\end{proposition}
\begin{proof}
$X$ left invariant means $(D\ell_g)X = X \circ \ell_g$ for all $g \in G$. Assume that $X,Y$ are both left invariant. Then $[X,Y] = [(D \ell_g)X, (D \ell_g)Y] = (D \ell_g)[X,Y]$ for all $g \in G$. So $[X,Y]$ is also left--invariant.
\end{proof}

\begin{definition}
$\mathfrak X(G)$ with the Lie bracket $[-,-]$ is the {\em Lie algebra $\mathfrak g = \L(G)$ of the Lie group $G$}.
\end{definition}

\begin{convention}
From now on, $G$, $H$ are always Lie groups and $\mathfrak g = \L(G)$, $\mathfrak h = \L(H)$ its Lie algebras.
\end{convention}

\begin{definition}
Let $G$ be a Lie group with neutral element $e \in G$. Define $\ev \colon \mathfrak g \to T_eG, X \mapsto X_e$.
\end{definition}

\begin{proposition}
$\ev$ is an isomorphism of $\mathbb R$--vector spaces.
\end{proposition}
\begin{proof}\
\vspace{-.5em}
\begin{enumerate}[(1)]
\item $\ev$ is clearly linear.
\item $\ev$ is injective: Suppose $X,Y \in \mathfrak g$ and $\ev(X) = \ev(Y)$. This means $X_e = Y_e \Rightarrow (D_e \ell_g)X_e = (D_e \ell_g)Y_e \Rightarrow X_g = Y_g \Rightarrow X = Y$.
\item $\ev$ is surjective: Take $v \in T_eG$. Define $X \in \mathfrak X(G)$ by $X_g \coloneqq (D_e \ell_g)(v)$. This is a smooth vector field and $X_{ga} = (D_e \ell_{ga})(v) = (D_e(\ell_g \circ \ell_a))(v) = (D_a\ell_g \circ D_e \ell_a)(v) = (D_a\ell_g)(X_a)$.\qedhere
\end{enumerate}
\end{proof}

\begin{corollary}
The dimension of $G$ is constant and equals the dimension of $\mathfrak g$ as $\mathbb R$--vector space.
\end{corollary}
\begin{proof}
Denoting by $G_g$ the connected component of $G$ containing $g \in G$, $\ell_g \colon G_e \to G_g$ is a diffeomorphism for every $g \in G$. Thus all connected components have the same dimension and $\dim G_0 = \dim T_eG = \dim \mathfrak g$.
\end{proof}

\begin{corollary}
The tangent bundle of $G$ is globally trivial, i.e. $G$ is parallelizable.
\end{corollary}
\begin{proof}
Consider $G \times \mathfrak g \to TG, (g, X) \mapsto X_g$. This sends the fiber of $G \times \mathfrak g$ over $g \in G$ to $T_gG$ linearly. At the point $e$ we have $(e,X) \mapsto X_e$, which is an isomorphism by the result about the evaluation map. For arbitrary $g \in G$ we have:
\[\xymatrix{
\{g\} \times \mathfrak g \ar[r] & T_gG \\ \{e\} \times \mathfrak g \ar[u]^\simeq_{\ell_g \times \id_{\mathfrak g}} \ar[r] & T_eG \ar[u]^\simeq_{D_e \ell_g}
}\]
This implies that the top horizontal map is an isomorphism, as claimed.
\end{proof}

\begin{definition}
A {\em one--parameter subgroup} of $G$ is a smooth map $s \colon \mathbb R \to G$ with $s(0) = e$ and $s(t_1 + t_2) = s(t_1)s(t_2)$.
\end{definition}

\begin{example}
Consider $G = U(1) = \{z \in \mathbb C \mid |z| = 1\}$. Then $s \colon \mathbb R \to U(1), t \mapsto e^{2\pi i t}$ is a (not injective) one--parameter subgroup.
\end{example}

\begin{proposition}\
\vspace{-.5em}
\begin{enumerate}[(1)]
\item Every left--invariant vector field on $G$ is complete, i.e. it generates a global flow on $G$.
\item For every $X \in \mathfrak g$ there is a unique one--parameter subgroup $s_X \colon \mathbb R \to G$ such that $\dot s_X(0) \coloneqq D_0s_X(\partial_t) = X_e$. The flow of $X$ is given by $\varphi \colon \mathbb R \times G \to G, (t,g) \mapsto gs_X(t) = \ell_g(s_X(t))$.
\end{enumerate}
\end{proposition}

\begin{notation}
We write $\varphi_t(g) \coloneqq \varphi(t,g)$.
\end{notation}

\begin{proof}
Given a left--invariant vector field, there is a local flow at $e \in G$:
\[\varphi \colon (-\varepsilon, \varepsilon) \times U \longrightarrow G\]
where $U$ is an open neighbourhood of $e$ in $G$, $\varepsilon > 0$. For any $g \in G$ consider
\[\psi \colon (-\varepsilon, \varepsilon) \times \ell_g(U) \longrightarrow G \qquad \psi \coloneqq \ell_g \circ \varphi \circ (\id_{\mathbb R} \times \ell_g^{-1}) \quad\Rightarrow\quad \psi_t = \ell_g \circ \varphi_t \circ \ell_g^{-1}\]
Claim: $\psi$ is a local flow around $g$ for $X$. Proof: $\ell_g(U)$ is an open neighbourhood of $g$ in $G$. First we check that $\psi$ defines a local flow:
\[\psi_0(h) = g\varphi_0(g^{-1}h) = gg^{-1}h = h\]
\[\psi_{t_1+t_2} = \ell_g \circ \varphi_{t_1+t_2} \circ \ell_g^{-1} = \ell_g \circ \varphi_{t_1} \circ \varphi_{t_2} \circ \ell_g^{-1} = \ell_g \circ \varphi_{t_1} \circ \ell_g^{-1} \circ \ell_g \circ \varphi_{t_2} \circ \ell_g^{-1} = \psi_{t_1} \circ \psi_{t_2}\]
So $\psi$ is a flow generated by some vector field, which we can calculate by differentiating the flow lines of $\psi$. To do this, consider the flow lines defined by
\[s_h(t) \coloneqq \psi_t(h) = g \varphi_t(g^{-1}h) = (\ell_g \circ s_{g^{-1}h})(t)\]
Then
\begin{align*}
\dot s_g(0) &= D_0 s_g(\partial_t) = D_0(\ell_g \circ s_e)(\partial_t) = (D_{s_e(0)}\ell_g \circ D_0 s_e)(\partial_t) =(D_{\varphi_0(e)}\ell_g \circ D_0 s_e)(\partial_t) = \\
&= D_e\ell_g(D_0s_e(\partial_t)) = D_e\ell_g(\dot s_e(0)) = D_e\ell_g(X_e) = X_g
\end{align*}
This proves the claim.

These local flows defined at different points in $G$ are all defined for the same time interval $(-\varepsilon, \varepsilon)$, and so define a flow $\overline\varphi \colon (-\varepsilon, \varepsilon) \times G \to G$ for $X$. $\overline\varphi$ can be extended to all $t \in \mathbb R$, so $X$ is complete.

To prove part (2) fix $X \in \mathfrak g$. By (1) we have a global flow $\varphi \colon \mathbb R \times G \to G$ for $X$. Define $s_X(t) \coloneqq \varphi_t(e)$. Since
\[\varphi_t(g) = (\ell_g \circ \varphi_t \circ \ell_g^{-1})(g) = \ell_g(\varphi_t(e)) = g\varphi_t(e)\]
we have $s_X(0) = \varphi_0(e) = e$ and
\[s_X(t_1 + t_2) = \varphi_{t_1+t_2}(e) = \varphi_{t_2+t_1}(e) = \varphi_{t_2}(\varphi_{t_1}(e)) = \varphi_{t_1}(e)\varphi_{t_2}(e) = s_X(t_1)s_X(t_2)\]
Also $\dot s_X(0) = X_e$, since $s_X$ is the flow line at $e$, and the formula $\varphi_t(g) = g s_X(t)$ follows from the claim above. One can easily check that $\varphi$ defined by this formula is a global flow for any one--parameter subgroup $s_X$ and thus $s_X$ is unique by the uniqueness of global flows.
\end{proof}

\begin{definition}
The map $\exp \colon \mathfrak g \to G, X \mapsto s_X(1)$ is the {\em exponential map} of $G$.
\end{definition}

\begin{example}
Let $G = \GL(n,\mathbb R)$ and $\mathfrak g = \mathfrak{gl}(n, \mathbb R) = \Mat(n \times n, \mathbb R)$. Then $\exp$ is given by the usual exponential function.
\end{example}

\begin{lemma}
$s_X(t) = \exp(tX)$.
\end{lemma}
\begin{proof}
Clear.\phantom{\qedhere}
\end{proof}

\begin{definition}
A homomorphism of Lie groups is a smooth map $f \colon G \to H$ which is also a group homomorphism.
\end{definition}

\begin{proposition}
Any homomorphism $f \colon G \to H$ as above induces an $\mathbb R$--linear homomorphism $f_* \colon \mathfrak g \to \mathfrak h$ such that $f_*[X,Y] = [f_*X,f_*Y]$.
\end{proposition}

\begin{proof}[Proof of proposition]
Since $f$ is a smooth group homomorphism, it has a derivative at $e$ and $f(e) = e$.
\[\xymatrix{
T_eG \ar[r]^{D_ef} & T_eH \\
\mathfrak g \ar[u]_\ev^\simeq \ar[r]^{f_*} & \mathfrak h \ar[u]_\ev^\simeq
}\]
Under the homomorphism provided by $\ev$, $D_ef$ corresponds to a unique linear $f_*$.

\begin{lemma}
For any $X \in \mathfrak g$ and $g \in G$ we have $(D_g f)(X_g) = (f_*X)_{f(g)}$.
\end{lemma}
\begin{proof}
Direct calculation using the left--invariance of $X$ and $f(e) = e$.
\begin{align*}
(D_g f)(X_g) &= D_gf \circ D_e \ell_g(X_e) = D_e(f \circ \ell_g)(X_e) = D_e(\ell_{f(g)} \circ f)(X_e) = \\
&= D_e \ell_{f(g)} \circ D_ef(X_e) = D_e \ell_{f(g)}((f_*X)_e) = (f_*X)_{f(g)} \qedhere
\end{align*}
\end{proof}

% \begin{lemma*}
% If $f \colon M \to N$ is a smooth map between smooth manifolds, then
% \[Df[X,Y] = [Df(X), Df(Y)]\]
% \end{lemma*}
% \begin{proof}
% Let $h \colon N \to \mathbb R$ be a $C^\infty$ function, $X \in \mathfrak X(M)$. Then ???
% \end{proof}

% If $f \colon G \to H$ is a homomorphism, then $X,Y \in \mathfrak g$ define vector fields $f_* X, f_* Y \in \mathfrak h$.
% \[f_*[X,Y] = Df[X,Y] = [Df(X),Df(Y)] = [f_*X, f_*X]\]
% This completes the proof of the proposition.

Using this lemma, we can show that $f_*X(h) \circ f = X(h \circ f)$ for any $X \in \mathfrak X(M)$ and $h \in C^\infty(H)$:
\[(f_*X)(h)(f(g)) = (f_*X)_{f(g)}[h]_{f(g)} = (D_gf)(X_g)[h]_{f(g)} = X_g[h \circ f]_g = X(h \circ f)(g)\]
Thus, applying this successively:
\begin{align*}
[f_*X,f_*Y](h) \circ f &= f_*X(f_*Y(h)) \circ f - f_*Y(f_*X(h)) \circ f = \\
&= X(f_*Y(h) \circ f) - Y(f_*X(h) \circ f) = \\
&= X(Y(h \circ f)) - Y(X(h \circ f)) = \\
&= [X,Y](h \circ f) = f_*[X,Y](h) \circ f
\end{align*}
Since this holds for any $h \in C^\infty(H)$, we have $[f_*X,f_*Y] \circ f = f_*[X,Y] \circ f$, so in particular $[f_*X,f_*Y]$ and $f_*[X,Y]$ coincide on $e \in H$, hence they are equal by left--invariance, which proves the proposition.
\end{proof}

\subsection{Digression on integrability and the Frobenius theorem}

\begin{theorem}
Let $M$ be a smooth manifold and $X_1, \dots, X_k \in \mathfrak X(M)$ with $[X_i,X_j] = 0$ for all $i,j$. If $X_1(p),\dots,X_k(p)$ are linearly independent for some $p \in M$, then there is a chart $(U,\varphi)$ for $M$ with $p \in U$ and $D\varphi(X_i|U) = \partial_i$ for all $i = 1,\dots,k$.
\end{theorem}
\begin{proof}
The problem is local, so we may assume $M$ is an open neighborhood of $0$ in $\mathbb R^n$ and $p = 0$. We may choose $U$ open around $0$ with $X_1,\dots,X_k$ linearly independent throughout $U$. After a linear change of coordinates we may assume that
\[X_1(0), \dots, X_k(0), \partial_k(0), \dots \partial_n(0)\]
form a basis of $\mathbb R^n$. We may assume that the local flow $\varphi^i$ for $X_i$ is defined for all $t \in (-\varepsilon, \varepsilon)$, $i = 1,\dots,k$.
\[f \colon U \to \mathbb R^n \qquad f(x_1,\dots,x_n) = \varphi_{x_1}^1 \circ \dots \circ \varphi_{x_k}^k (0,\dots,0,x_{k+1},\dots,x_k)\]
is a smooth map. Moreover $f(0) = 0$ and $D_0f(\partial_i) = \partial_i$ for $i = k+1,\dots,n$. For all $x \in U$ we have, since the flows $\varphi^i$ commute:
\[D_xf(\partial_i) = X_i(f(x)) \qquad i = 1,\dots,k\]
For $x = 0$ we see that $D_0f$ is an isomorphism, so $f$ is a local diffeomorphism around $0$ by the inverse function theorem. Set $\varphi = f^{-1}$ after possibly shrinking $U$:
\[D\varphi(X_i|f(U)) = \partial_i\qedhere\]
\end{proof}

Let $M$ be a smooth manifold of dimension $n$.

\begin{definition}
A {\em rank $k$ distribution} on $M$ is a rank $k$ subbundle $E \subset TM$.
\end{definition}

What this means is that around every point $p \in M$ there exists an open set $U$ and $X_1,\dots,X_k \in \mathfrak X(M)$ such that
\[E_x = \{X_1(x),\dots,X_k(x)\}\]

\begin{definition}
An {\em integral submanifold} for $E$ is a $k$--dimensional submanifold $N \subset M$ with $TN = E|N$.
\end{definition}

\begin{definition}
$E$ is called {\em integrable} if for all $p \in M$ there exists an integrable submanifold $N$ with $p \in N$.
\end{definition}

\begin{definition}
$E$ is {\em involutive} if $[X,Y] \in \Gamma(E)$ whenever $X,Y \in \Gamma(E)$.
\end{definition}

\newpage
\begin{theorem}[Frobenius theorem]
For a distribution $E$ of rank $k$ on $M$, the following are equivalent:
\begin{enumerate}[(1)]
\item $E$ is integrable.
\item $E$ is involutive.
\item There is a covering of $M$ by domains of charts $(U, \varphi)$ with the property that
\[(D\varphi)(E) \ni \partial_i \qquad \forall i = 1,\dots,k\]
\end{enumerate}
\end{theorem}
%TODO: check this proof
\begin{proof}\
\vspace{-.5em}
\begin{itemize}
\item[3 $\Rightarrow$ 1]
If $E = \Span\{\partial_1,\dots,\partial_k\}$, then the equations
\begin{align*}
x_{k+1} &= c_{k+1} \\
&\hspace{.5em}\vdots \\
x_n &= c_n
\end{align*}
define $k$--dimensional submanifolds for $E$.
\item[1 $\Rightarrow$ 2]
  Take $X,Y \in \Gamma(E)$ and $p \in M$. By (1) we have a submanifold $i \colon N \hookrightarrow M$ with $p \in N$ and $E|N = TN$. The restrictions of $X,Y$ to $N$ are vector fields on $N$. Furthermore, $[X,Y]_p \in E_p$.
\item[2 $\Rightarrow$ 3]
Everything is local, so we work at $0 \in \mathbb R^n$.
\begin{enumerate}[Step 1:]
\item Consider the projection
\[\pi \colon \mathbb R^n \to \mathbb R^k, (x_1,\dots,x_n) \mapsto (x_1,\dots,x_k)\]
If for some point $p$, $D_p\pi$ is injective on $E_p$, then the same is true for all $x$ in an open neighbourhood of $p$.
\item At every point $p$ there is a chart so that w.r.t. the coordinates given by the chart $D_x\pi$ is an isomorphism from $E_x$ to $\mathbb R^n$ for all $x$ in the domain of the chart. To prove this, by step 1 it is enough to ensure $D_p\pi$ is injective. We can choose local coordinates $(x_1,\dots,x_k)$ in such a way that if
\[X_1(p),\dots,X_k(p)\]
 is a basis for $E_p$, then
\[X_1(p),\dots,X_k(p),\partial_{k+1},\dots,\partial_n\]
is a basis of $T_p\mathbb R^n$.
\item Let $E, p, U, \pi$ be as above. Let $Z_i \in \Gamma(E)$ be the unique section such that $D_x\pi(Z_i(x)) = \partial_i(x)$ for all $x \in U$, $i = 1,\dots,k$. So $Z_1,\dots,Z_k$ span $E$ throughout $U$.
\item \ \vspace{-.5em}
\[D\pi[Z_i,Z_j] = [D\pi Z_i, D\pi Z_j] = [\partial_i,\partial_j] = 0\]
By involutivity, $[Z_i,Z_j] \in \Gamma(E|U)$. But $D\pi$ is an isomorphism on $E$ so $[Z_i, Z_j] = 0$. By the previous theorem, we can find a chart in which $D\varphi(Z_i) = \partial_i$ for $i = 1,\dots,k$. This gives (3) in the Frobenius theorem.\qedhere
\end{enumerate}
\end{itemize}
\end{proof}

%TODO: "connected" necessary ?
\begin{definition}
If $E$ satisfies the conditions in the theorem above and $p \in M$, let $L_p$ be the maximal connected submanifold of $E$ with $p \in L_p$. This is called the {\em leaf} through $p$. The collection of all leaves formes a {\em foliation} of $M$.
\end{definition}

\begin{remark}
The leaves of a foliation $\mathcal F$ are generally not closed subsets of $M$ and the subspace topology is not the same as the manifold topology of a leaf.
\end{remark}

\begin{definition}
Let $G$ be a Lie group, $H \subset G$ a subset. $H$ is a {\em Lie subgroup} of $G$ if $H$ has a Lie group structure such that the inclusion $i \colon H \hookrightarrow G$ is a homomorphism of Lie groups and an injective immersion.
\end{definition}

\begin{theorem}
For any Lie group $G$, there is a bijection between connected Lie subgroups $H \subset G$ and Lie subalgebras $\mathfrak h \subset \mathfrak g$.
\end{theorem}
\begin{proof}
Suppose $H \subset G$ is a Lie subgroup. Then, since $i$ is an immersion,
\[D_e i \colon T_e H \to T_eG \quad \text{and} \quad i_* \colon \mathfrak h \to \mathfrak g,\]
which are essentially the same maps, are injective. So $i_*(\mathfrak h)$ is a Lie subalgebra, which can be identified with $\mathfrak h$.

Conversely, let $\mathfrak h \subset \mathfrak g$ be a Lie subalgebra. Let $E_g \coloneqq D_e\ell_g(\ev(\mathfrak h)) \subset T_gG$ for all $g \in G$. This is in fact the evaluation of $\mathfrak h$ at $g$. For every $g \in G$, $E_g$ is a $k$--dimensional subspace of $T_gG$ with $k = \dim \mathfrak h$. The collection of all $E_g$ is a smooth rank $k$ distribution $E \subset TG$.
\begin{enumerate}[Step 1:]
\item $E$ is involutive, and thus integrable by the Frobenius theorem. To see this, let $X_1,\dots,X_k$ be a basis for $\mathfrak h$. Then all $X,Y \in \Gamma(E)$ are of the form
\[X = \sum_{i=1}^k f_i X_i \qquad Y = \sum_{i=1}^k h_i X_i \qquad f_i,h_i \in C^\infty(G)\]
Then $[X,Y]$ is a linear combination of the $X_i$ and the $[X_i,X_j]$. Since $\mathfrak h$ is a Lie subalgebra, $[X_i,X_j] \in \mathfrak h$ and so $[X,Y] \in \Gamma(E)$.
\item Let $\mathcal F$ be the foliation of $G$ defined by the integral submanifolds of $E$, and $H \coloneqq L_e$. Then $L_g = \ell_g(H)$. Proof: Both sides are connected subsets containing $g$. $L_g$ is a leaf of $\mathcal F$ by definition. Once we prove that $\ell_g(H)$ is a leaf, we have the conclusion by the uniqueness of leaves. For any $a \in G$, $b \in H$ we have
\[T_{ab}\ell_a(H) \!=\! D_b \ell_a (T_bH) = D_b \ell_a(E_b) \!=\! (D_b \ell_a \circ D_e \ell_b)(\ev(\mathfrak h)) \!=\! D_e \ell_{ab}(\ev(\mathfrak h)) \!=\! E_{ab}\]
so $\ell_a(H)$ is an integral submanifold of $E$ and thus $g \cdot H \coloneqq \ell_g(H)$ is the leaf of $\mathcal F$ through $g$.
\item Let $a,b \in H$. Then $aH = H \ni b$, since $aH$ is the leaf through $a$. But $b \in aH$, implies $a^{-1}b \in H$, so $H$ is a subgroup of $G$.
\item The inclusion $i \colon H \hookrightarrow G$ makes $H$ into a Lie subgroup. Since $D_g i(T_gH) = E_g$, $i$ is an injective immersion, with the manifold structure of $H$ given by its construction as an integral submanifold for $E$.
\item This $H$ is the only connected Lie subgroup of $G$ with Lie algebra $\mathfrak h$. To prove this, suppose $\overline H$ is another Lie subgroup with Lie algebra $\mathfrak h$. Both $H, \overline H$ are injectively immersed in $G$ with the same tangent space $\ev(\mathfrak h)$ at $e$. $\overline H$ will then be an integral submanifold for $E$ through $e$. By uniqueness of the leaf through $e$, we must have $H = \overline H$.\qedhere
\end{enumerate}
\end{proof}

\begin{example}
The 2--torus $G = T^2 = \mathbb R^2/\mathbb Z^2$ is a connected Lie group with $e = [(0,0)]$. The Lie algebra is $\mathfrak g = \mathbb R^2$ with $[X,Y] = 0$ for all $X,Y \in \mathfrak g$. Every vector subspace $\mathfrak h \subset \mathfrak g$ is a Lie subalgebra in this case, giving rise to a Lie subgroup. If $\mathfrak h = \Span(1,\lambda)$, $\lambda \in \mathbb R \setminus \mathbb Q$, then the corresponding connected Lie subgroup is
\[H = \{\exp(t(1,\lambda)) \mid t \in \mathbb R\}\]
This is densely immersed in $T^2$, in particular it is not a closed subgroup.
\end{example}

\subsection{Actions of Lie groups on manifolds}

\begin{definition}
A {\em (left) action} of a Lie group $G$ on a smooth manifold $M$ is a smooth map
\[\mu \colon G \times M \to M \qquad \mu(g,p) = g \cdot p = \ell_g(p)\]
such that for any $p \in M$ and $g,h \in G$
\[e \cdot p = p \qquad g \cdot (h \cdot p) = (gh) \cdot p\]
A {\em right action} is a smooth map $\mu \colon G \times M \to M$ such that for any $p \in M$ and $g,h \in G$
\[\mu(e,p) = p \qquad \mu(g,\mu(h,p)) = \mu(hg,p)\]
For a right action, we write $\mu(g,p) = p \cdot g = r_g(p)$. Then the axioms become
\[p \cdot e = p \qquad (p \cdot h) \cdot g = p \cdot (hg)\]
\end{definition}

\begin{remark}
If $\mu \colon G \times M \to M$ is a left action, we can define $\overline \mu(g,p) = \mu(g^{-1}, p)$. This is a right action.
\end{remark}

\begin{definition}
Let $\mu \colon G \times M \to M$ be a an action of a Lie group on a smooth manifold.
\begin{enumerate}[(1)]
\item $\mu$ is {\em effective} if for every $g \in G \setminus \{e\}$, there exists $p \in M$ such that $\mu(g,p) \neq p$.
\item For $p \in M$, the subset
\[G(p) \coloneqq \{\mu(g,p) \mid g \in G\}\]
is the {\em orbit} of $p$ under the action.
\item The action is {\em transitive} if $G(p) = M$ for some $p \in M$ (and thus for all $p \in M$).
\item The {\em isotropy group} of $p \in M$ is
\[G_p \coloneqq \{g \in G \mid \mu(g,p) = p\}\]
\end{enumerate}
\end{definition}

\begin{proposition}
Let $f \colon G \to H$ be a homomorphism of Lie groups and $f_* \colon \mathfrak g \to \mathfrak h$ the induced Lie algebra homomorphism. Then the following diagram commutes:
\[\xymatrix{
G \ar[r]^f & H \\
\mathfrak g \ar[u]^\exp \ar[d]_\ev^\simeq \ar[r]^{f_*} & \mathfrak h \ar[u]_\exp \ar[d]^\ev_\simeq \\
T_eG \ar[r]_{D_ef} & T_eH.
}\]
\end{proposition}

\begin{proof}
$f(\exp(tX))$ is a $C^\infty$ curve in $H$ passing through $e$ at time $t = 0$. Since $f$ is a homomorphism, this is also a 1--parameter subgroup whose tangent vector at $e$ is $D_ef(\ev(X)) = \ev(f_*(X))$. So $f(\exp(tX))$ is the unique 1--parameter subgroup of $H$ generated by $f_*X = Y$. We know that the 1--parameter subgroup generated by $Y$ is $\exp(tY)$. Therefore
\[f(\exp(tX)) = \exp(tY) = \exp(tf_*X) \quad \xRightarrow{\ t=1\ } \quad f(\exp(X)) = \exp(f_*X)\qedhere\]
\end{proof}

The isotropy group $G_p$ is a closed subgroup of $G$. This means that $G_p$ is actually a Lie subgroup of $G$ (not proved). If we restrict $\mu$ from $G$ to $G_p$, then $p$ is a fixed point for the action of $G_p$ on $M$.

Under an action $\mu \colon G \times M \to M$, every $g \in G$ gives a diffeomorphism
\[\ell_g \colon M \to M, p \mapsto g \cdot p\]
with inverse $(\ell_g)^{-1} = \ell_{g^{-1}}$.

\begin{lemma}
If $p$ is a fixed point of the action $\mu \colon G \times M \to M$, then $G$ acts linearly on $T_pM$, so we have a representation $G \to \GL(T_pM)$. This is called the {\em isotropy representation} of $G$ at $p$.
\end{lemma}

\begin{proof}
Since $p$ is a fixed point, we have $\ell_g(p) = p$ for all $g \in G$, so $D_p\ell_g \colon T_pM \to T_pM$ is a linear isomorphism since $\ell_g$ is a diffeomorphism. We obtain a map
\[G \to \GL(T_pM), g \mapsto D_p\ell_g\]
which is smooth since $\mu$ is smooth. It is also a homomorphism because of the chain rule:
\[D_p\ell_{g_1g_2} = D_p(\ell_{g_1} \circ \ell_{g_2}) = D_p\ell_{g_1} \circ D_p\ell_{g_2}\qedhere\]
\end{proof}

\begin{example}
The action
\[\mu \colon G \times G \to G, (g,p) \mapsto g \cdot p = \ell_g(p)\]
is effective, transitive, and $G_p = \{e\}$ for all $p \in G$.
\end{example}

Let $G$ act on itself by conjugation:
\[a \colon G \times G \to G, (g,p) \mapsto g \cdot p \cdot g^{-1} \eqqcolon a_g(p)\]
Note that $g$ and $p$ commute in $G$ if and only if $a_g(p) = p$. The isotropy group $G_p$ of a point $p \in G$ under the conjugation action $a$ is the centraliser of $p$ in $G$. If $G$ has a non--trivial center, then the conjugation action $a$ is not effective. One has $G_e = G$ since $g e g^{-1} = e$ for all $g \in G$. By the Lemma, we obtain the isotropy representation of $a$ at $p = e$:
\[\Ad \colon G \to \GL(T_eG).\]
This is the {\em adjoint representation} of $G$ (on $\mathfrak g$), whereas the map $\ad$ defined by
\[\xymatrix@C=3em{
T_eG \ar[r]^-{D_e\Ad} & T_e\GL(T_eG) \\
\mathfrak g \ar[u]^\ev_\simeq \ar[r]_-\ad & \End(\mathfrak g) \ar[u]_\ev^\simeq
}\]
with the identification $\End(\mathfrak g) = \mathfrak{gl}(T_eG)$ is the {\em adjoint representation} of $\mathfrak g$. Since $\ad = \Ad_*$, it is a homomorhism of Lie algebras. By the proposition, we have the commutative diagram:
\[\xymatrix{
G \ar[r]^-\Ad & \GL(T_eG) \\
\mathfrak g \ar[u]^\exp \ar[r]^-\ad & \End(\mathfrak g) \ar[u]_\exp
}\]
Also $a_g$ is a homomorphism of $G$ to itself, so the following diagram commutes:
\[\xymatrix@C=3em{
G \ar[r]^{a_g} & G \\
\mathfrak g \ar[u]^\exp \ar[d]_\ev^\simeq \ar[r]^{(a_g)_*} & \mathfrak g \ar[u]_\exp \ar[d]^\ev_\simeq \\
T_eG \ar[r]_{D_ea_g} & T_eG
}\]
\begin{notation}
Define $\ad_X \colon \mathfrak g \to \mathfrak g$ by $\ad_X(Y) = \ad(X)(Y)$ for any $X,Y \in \mathfrak g$ and $\Ad_g \colon T_eG \to T_eG$ by $\Ad_g(X) = \Ad(g)(X)$ for any $g \in G$ and $X \in T_eG$.
\end{notation}

Then $(a_g)_* = \Ad_g = \Ad(g)$, since $(a_g)_*$ is defined by $D_e a_g = \Ad(g)$ by the definition of $\Ad$. Take a vector space $V$ and $G = \GL(V)$. Then the above diagrams become:
\[\xymatrix{
\GL(V) \ar[r]^-\Ad & \GL(\End(V)) & & \GL(V) \ar[r]^{a_g} & \GL(V) \\
\End(V) \ar[u]^\exp \ar[r]_-\ad & \End(\End(V)) \ar[u]_\exp & & \End(V) \ar[u]^\exp \ar[r]_{\Ad_g} & \End(V) \ar[u]^\exp
}\]
We claim that $\Ad_g(M) = gMg^{-1}$. This can be seen by
\[\Ad_g(M) = D_ea_g(M) = \]
%TODO: rest der VL

{\em (something is missing here)}

Consider $r_{g^{-1}} \colon G \to G$, the right multiplication by $g^{-1}$. Recall that $TG \cong G \times \mathfrak g$ is a trivialization of $TG$ given by
\[G \times \mathfrak g \to TG \quad (p,X) \mapsto (p,X_p)\]
\begin{lemma}
$\Ad_g \in \GL(T_eG)$ is given by the composition of $D_e r_{g^{-1}}$ with the identification of $T_{g^{-1}}G$ with $T_eG$ via the trivialization of the tangent bundle by left--invariant vector fields.
\end{lemma}
\[\xymatrix@C=4em{
T_eG \ar[r]_{D_er_{g^{-1}}} \ar@/^1.5pc/[rr]^{\Ad_g} & T_{g^{-1}}G \ar[r]_{D_{g^{-1}}\ell_g} & T_eG \\
& & \mathfrak g \ar[ul]^/-.7em/{X \mapsto X_{g^{-1}}\!\!}\ar[u]_\ev
}\]
\begin{proof}
Any tangent vector $v \in T_{g^{-1}}G$ can be identified with $X_e \in T_eG$ for the unique $X \in \mathfrak g$ such that $X_{g^{-1}} = v$. This identification is via $D_{g^{-1}}\ell_g$:
\[D_{g^{-1}} \ell_g(v) = D_{g^{-1}}\ell_g(X_{g^{-1}}) = D_{g^{-1}}\ell_g \circ D_e\ell_{g^{-1}}(X_e) = X_e\]
Using this, $\Ad_G$ we get the claim:
\[D_{g^{-1}}\ell_g \circ D_e r_{g^{-1}} = D_ea_g = \Ad_g\qedhere\]
\end{proof}

\begin{definition}
Let $G$ be a Lie group and $\mathfrak g$ its Lie algebra.
\[C(G) \coloneqq \{g \in G \mid gh = hg \, \forall h \in G\}\]
is the {\em center} of $G$ and
\[C(\mathfrak g) \coloneqq \{X \in \mathfrak g \mid [X,Y] = 0 \, \forall Y \in \mathfrak g\}\]
is the {\em center} of $\mathfrak g$.
\end{definition}

\begin{lemma}
Let $G$ be a connected Lie group. Then $\ker \Ad = C(G)$.
\end{lemma}

\begin{proof}
If $g \in C(G)$, then $a_g = \id_g$, so $\Ad_g = D_ea_g = \id_{T_eG}$, so $g \in \ker \Ad$. Conversely, suppose that $g \in \ker \Ad$. Then
\[g \exp(tX) g^{-1} = a_g(\exp(tX)) = \exp(\Ad_g(tX)) = \exp(tX)\]
for all $X \in T_eG$. So $g$ commutes with all $h \in G$ containted in a small enough neighbourhood in $G$, because $\exp$ is a local diffeomorphism at $0 \in T_eG$. Every open neighbourhood of $e$ in $G$ generates the connected component of $e$ in $G$ by taking products. Therefore, if $G$ is connected, then $g \in C(G)$.
\end{proof}

\begin{corollary}
Let $G$ be a connected Lie group. Then $C(G)$ is a closed Lie subgroup whose Lie algebra is the center of $\mathfrak g$.
\end{corollary}

\begin{proof}
The center $C(G)$ is a closed subgroup of $G$. So it is a Lie subgroup. The Lie algebra of $C(G)$ is $C(\mathfrak g)$.
\end{proof}

\begin{corollary}
If $G$ is a connected Lie group, then $G$ is abelian iff $\mathfrak g$ has trivial Lie brackets.
\end{corollary}

\subsection{Homogeneous spaces}
Let $G$ be a Lie group, $H \subset G$ a closed Lie subgroup. Consider
\[G/H \coloneqq \{aH \mid a \in G\},\]
the set of left cosets of $H$. We denote by
\[\ell_g \colon G/H \to G/H \quad aH \mapsto (ga)H\]
the action induced by left multiplication. For any two $a,b \in G$, there exists $g \in G$ such that $\ell_g(aH) = bH$.

\begin{theorem}
$G/H$ has a natural structure as a smooth manifold of dimension $\dim G - \dim H$, such that
\[\pi \colon G \to G/H \quad a \mapsto aH\]
is a smooth map that admits local smooth sections. $\pi$ will actually be a submersion.
\end{theorem}
\begin{proof}
This proof will be added later \phantom{\qedhere}
% $G/H$ is given the quotient topology defined by
% \[U \in G/H\ \text{open} \Leftrightarrow \pi^{-1}(U) \in G\ \text{open}.\]
% $G/H$ then has a countable basis of the topology, because $G$ does.

% To see that $G/H$ is Hausdorff, suppose that $aH \neq bH$. Consider
% \[f \colon G \times G \to G \quad (g_1,g_2) \mapsto g_1^{-1}g_2\]
% By continuity, $R \coloneqq f^{-1}(H)$ is closed, since $H \subset G$ is closed. Since $aH \neq bH$ implies $b^{-1}a \not\in H$, we have $(b,a) \not\in R$. Since $R$ is closed, there exist open subsets $U,V \subset G$ such that $a \in U$, $b \in V$ and $(V \times U) \cap R = \varnothing$. This implies $\pi(V)$ and $\pi(U)$ are disjoint in $G/H$ and $bH \in \pi(V)$, $aH \in \pi(U)$. In fact, $\pi(U)$ and $\pi(V)$ are disjoint open neighbourhoods of $aH$ ad $bH$ in $G/H$, so $G/H$ is Hausdorff.
%TODO: looooooooong proof
\end{proof}

\begin{corollary}
The action
\[\mu \colon G \times G/H \to G/H \quad (g, aH) \mapsto (ga)H\]
defines a transitive smooth action of $G$ on $G/H$. The isotropy group of $H = eH$ is $H$.
\end{corollary}
\begin{proof}
$\mu$ is a smooth map by the construction of the smooth structure on $G/H$. $\mu$ is a left action of $G$. The action is transitive, and
\[G_{eH} = \{g \in G \mid \mu(g, eH) = eH\} = H\qedhere\]
\end{proof}

\section{Principal bundles}
\begin{definition}
A {\em principal $G$--bundle} over a smooth manifold $M$ is a smooth manifold $P$, a smooth projection $\pi \colon P \to M$ and a right $G$--action $\cdot \colon P \times G \to P$ satisfying:
\begin{enumerate}[(a)]
\item There is a covering of $M$ by open sets $U_\alpha$, together with diffeomorphisms
\[\pi^{-1}(U_\alpha) \to U_\alpha \times G \qquad p \mapsto (\pi(p), \varphi_\alpha(p))\]
\item For all $g \in G$, $p \in \pi^{-1}(U_\alpha)$, we have $\varphi_\alpha(r_g(p)) = r_g(\varphi_\alpha(p))$, i.e. $\varphi_\alpha(pg) = \varphi_\alpha(p)g$ ($G$--equivariance)
\end{enumerate}
$G$ is called the {\em structure group} of $P$.
\end{definition}

\begin{remark}\
\vspace{-.5em}
\begin{enumerate}[(1)]
\item $\pi$ is a submersion because $\pi^{-1}(U_\alpha) \cong U_\alpha \times G$ and $\pi_1$ is a submersion.
\item $\pi^{-1}(m) \cong G$ for all $m \in M$.
\item The action of $G$ on $P$ maps $\pi^{-1}(m)$ to itself for all $m \in M$.
\item On each fiber $\pi^{-1}(m)$, the $G$--action is simply transitive, i.e. transitive and has trivial stabilizers. It follows also that $G$ is free on the whole principal bundle $P$.
\end{enumerate}
\end{remark}

Let $X \in \mathfrak g$, then $\exp(tX)$ is a one--parameter subgroup of $G$. By restricting the right action $P \times G \to P$ we obtain a flow on $P$, which is generated by some vector field $X^* \in \mathfrak X(P)$.

\begin{definition}
$X^*$ is the {\em fundamental vector field} generated by $X$. $X^*$ is tangent to the fiber of $\pi$.
\end{definition}

\begin{lemma}
For any $g \in G$, $X \in \mathfrak g$ and $p \in P$, we have
\[D_pr_g(X^*_p) = (\Ad_{g^{-1}}(X))^*_{pg}\]
\end{lemma}
\begin{proof}
We have the commutative diagram
\[\xymatrix{
G \ar[r]^{a_{g^{-1}}} & G \\
\mathfrak g \ar[u]^\exp \ar[r]_{\Ad_{g^{-1}}} & \mathfrak g \ar[u]_\exp
}\]
so, defining $Y \coloneqq \Ad_{g^{-1}}(X)$:
$\exp(tY) = \exp(t\Ad_{g^{-1}}(X)) = \exp(\Ad_{g^{-1}}(tX)) = g^{-1}\exp(tX)g$
Next, we can define two smooth maps
\[s, s' \colon \mathbb R \to P \qquad s(t) = p \exp(tX) \qquad s'(t) = pg\exp(tY) = p\exp(tX)g\]
Then
\[D_pr_g(X_p^*) = (D_pr_g \circ D_0s)(\partial) = D_0(r_g \circ s)(\partial) = D_0s'(\partial) = Y_{pg}^* = (\Ad_{g^{-1}}(X))_{pg}^* \qedhere\]
\end{proof}

\begin{remark}\
\vspace{-.5em}
\begin{enumerate}[(1)]
\item The map
\[\mathfrak g \to \mathfrak X(P) \qquad X \mapsto X^*\]
is an injective homomorphism of Lie algebras, because $G$ acts freely.
\item For all $p \in P$, $\ker D_p\pi$ is spanned by the values of the fundamental vector fields at $p$: The map
\[\mathfrak g \to T_pP \qquad X \mapsto X_p^*\]
is a linear map of $\mathbb R$-vector spaces. This map is injective because $X^*$ has no zeroes, and its image is in $\ker D_p\pi$. So by dimensional reasons, it is $\ker D_p\pi$.
%TODO: end of remark 2
\end{enumerate}
\end{remark}

\begin{lemma}
A principal $G$--bundle $P$ admits a global smooth section $s \colon M \to P$ if and only if it is isomorphic to the product bundle $M \times G \xrightarrow{\pi_1} M$.
\end{lemma}

\begin{proof}
The product bundle has a smooth section
\[s \colon M \to M \times G \qquad m \mapsto (m,e)\]
If $f \colon P \to M \times G$ is an isomorphism, then $f^{-1} \circ s$ is a smooth section of $P$.

Conversely, suppose $P$ admits a smooth section $s \colon M \to P$. Then define
\[f \colon M \times G \to P \qquad (m,g) \mapsto s(m)g\]
Clearly $f$ is a smooth map. For any $h \in G$,
\[(f \circ r_h)(m,g) = f(m,gh) = s(m)gh = r_h(s(m)g) = (r_h \circ f)(m,g)\]
so $f$ maps $\{m\} \times G$ to $\pi^{-1}(m)$. This map $f \colon \{m\} \times G \to \pi^{-1}(m)$ is bijective. To show injectivity, assume $s(m)g_1 = s(m)g_2$, then $g_1g_2^{-1} \in \Stab(s(m)) = \{e\}$, so $g_1 = g_2$. It is also surjective, since $G$ acts transitively on $\pi^{-1}(m)$. So $f^{-1}$. One can check smoothness of $f^{-1}$ in the local trivialization for $p$.
\end{proof}

\begin{example}\
\vspace{-.5em}
\begin{enumerate}[(1)]
\setcounter{enumi}{-1}
\item For any smooth manifold $M$ and any Lie group $G$, the trivial bundle $M \times G \xrightarrow{\pi_1} M$ is a principal bundle.
\item Let $H \subset G$ be a closed Lie subgroup. Then $P = G$ is a principal $H$--bundle over $G/H$ with the action
\[G \times H \to G \qquad (g,h) \mapsto gh\]
The projection $\pi \colon G \to G/H$ admits local smooth sections. Let $U \subset G$ be open such that there is a smooth $s \colon U \to G$ with $\pi \circ s = \id_U$ and define
\[f \colon U \times H \to G \qquad (m,h) \mapsto s(m)h\]
This is a diffeomorphism between $U \times H$ and $\pi^{-1}(U)$ and for any $h' \in H$, we have
\[(r_{h'} \circ f)(m,h) = r_{h'}(s(m)h) = s(m)hh' = f(m,hh')\]
So all the requirements are satisfied such that $G$ is a principal $H$--bundle.
\item Let $M$ be a smooth manifold and $P$ the set of bases for tangent spaces of $M$. $P$ has a $C^\infty$ manifold structure such that $\pi$ is smooth and $P$ is the total space of a principal $\GL_n(\mathbb R)$--bundle over $M$, $n = \dim(M)$.
\end{enumerate}
\end{example}

Let $P \xrightarrow{\pi} M$ be a principal $G$--bundle and $M = \bigcup_\alpha U_\alpha$ a covering by local trivialisations. Suppose $U_\alpha \cap U_\beta \neq \varnothing$. Then the composition
\[(U_\alpha \cap U_\beta) \times G \xrightarrow{(\pi \times \varphi_\beta)^{-1}} \pi^{-1}(U_\alpha \cap U_\beta) \xrightarrow{\pi \times \varphi_\alpha} (U_\alpha \cap U_\beta) \times G\]
forms a diffeomorphism from $(U_\alpha \cap U_\beta) \times G$ to itself, which is the identity on the first factor. We denote this map by $(m,g) \mapsto (m,\overline{\psi_{\alpha\beta}}(m,g))$. By $G$--equivariance of the local trivializations, $\overline{\psi_{\alpha\beta}}(m,gh) = \overline{\psi_{\alpha\beta}}(m,g)h$ holds for any $m \in M$, $g,h \in G$, so we have $\overline{\psi_{\alpha\beta}}(m,g) = \overline{\psi_{\alpha\beta}}(m,e)g \eqqcolon \psi_{\alpha\beta}(m)g$. These {\em transition maps} $\psi_{\alpha\beta}$ have the following properties:
\begin{enumerate}[(1)]
\item If $U_\alpha \cap U_\beta \neq \varnothing$, then $\psi_{\alpha\beta} \colon U_\alpha \cap U_\beta \to G$ is a smooth map.
\item $\psi_{\alpha\alpha}(m) = e$ for all $m \in U_\alpha$.
\item $\psi_{\alpha\beta}(m) = \psi_{\beta\alpha}(m)^{-1}$ for all $m \in U_\alpha \cap U_\beta$.
\item For all $m \in U_\alpha \cap U_\beta \cap U_\gamma$ we have the following:
\[\psi_{\alpha\beta}(m)\psi_{\beta\gamma}(m) = \psi_{\alpha\gamma}(m)\]
\end{enumerate}
The properties $(2)-(4)$ are summarized by saying that the maps $\psi_{\alpha\beta}$ satisfy the {\em cocycle conditions}. Property (3) follows directly from (2) and (4).

Now suppose we are given a smooth manifold $M$, an open covering $M = \bigcup_\alpha U_\alpha$ and smooth maps $\psi_{\alpha\beta} \colon U_\alpha \cap U_\beta \to G$ satisfying the cocycle conditions. Then we can construct a principal $G$--bundle $P \xrightarrow{\pi} M$ trivial over each $U_\alpha$ such that $\psi_{\alpha\beta}$ are the transition maps of $P$:
\[P = \coprod_\alpha (U_\alpha \times G) \bigg/\!\!\sim\]
The equivalence relation $\sim$ is given as follows:
\[U_\alpha \times G \ni (m,g) \sim (m, \psi_{\alpha\beta}(m)g) \in U_\beta \times G \quad\Longleftrightarrow\quad m \in U_\alpha \cap U_\beta\]
This really is an equivalence relation because $\psi_{\alpha\beta}$ satisfy the cocycle conditions. $P$ is a smooth manifold that each $U_\alpha \times G$ projects to an open submanifold of $P$.

Now define a projection $\pi \colon P \to M$ by $\pi([(m,g)]) = m$. In the chart given by $U_\alpha \times G$ this is $\pi_1$ and so it is smooth. Also define an action
\[\mu \colon P \times G \to P \qquad ([(m,g)],h) \mapsto [(m,gh)]\]
It is well--defined and smooth. This definition of $P,\pi,\mu$ satisfies the properties (a) and (b) in the definition of a principal bundle. So we do indeed have a principal $G$--bundle defined from the $\psi_{\alpha\beta}$.

\begin{example}
Let $M$ be a smooth manifold, $(U_\alpha,f_\alpha)$ an atlas for $M$ and $n = \dim M$. If $U_\alpha \cap U_\beta \neq \varnothing$, then we have the transition map
\[f_{\alpha\beta} \coloneqq f_\alpha \circ f_\beta^{-1} \colon f_\beta(U_\alpha \cap U_\beta) \to f_\alpha(U_\alpha \cap U_\beta)\]
between open sets in $\mathbb R^n$. This $f_{\alpha\beta}$ is a diffeomorphism. Let $\psi_{\alpha\beta} \colon U_\alpha \cap U_\beta \to \GL_n(\mathbb R)$ be defined as follows:
\[\psi_{\alpha\beta}(x) = D_{f_\beta(x)}f_{\alpha\beta} \in \GL_n(\mathbb R)\]
The $\psi_{\alpha\beta}$ so defined are smooth and satisfy
\[\psi_{\alpha\alpha}(x) = D_{f_\alpha(x)} f_{\alpha\alpha} = D_{f_\alpha(x)} \id_{f_\alpha(U_\alpha)} = e \in \GL_n(\mathbb R)\]
and, for any $x \in U_\alpha \cap U_\beta \cap U_\gamma$:
\[\psi_{\alpha\gamma}(x) = D_{f_\gamma(x)} f_{\alpha\gamma} = D_{f_\gamma(x)} (f_{\alpha\beta} \circ f_{\beta\gamma}) = D_{f_\beta(x)} f_{\alpha\beta} \circ D_{f_\gamma(x)} f_{\beta\gamma} = \psi_{\alpha\beta}(x) \psi_{\beta\gamma}(x)\]
We have checked that $\psi_{\alpha\beta}$ satisfy (2) and (4) of the cocycle conditions and (3) follows. Therefore $\psi_{\alpha\beta}$ define a principal $\GL_n(\mathbb R)$--bundle over $\mathbb R^n$. This is the {\em bundle of bases/frames for tangent spaces} to $M$.
\end{example}

\begin{definition}
Let $P \xrightarrow{\pi} M$ and $P' \xrightarrow{\pi'} M'$ be principal $G$-- resp. $G'$--bundles. A homomorphism $f$ from $P$ to $P'$ is a pair of smooth maps
\[f' \colon P \to P' \qquad f'' \colon G \to G'\]
such that $f''$ is a homomorphism of Lie groups and
\[f'(pg) = f'(p)f''(g) \qquad \forall p \in P, g \in G\]
\end{definition}

\begin{notation}
For a homomorphism $f$ of principal bundles $P$ and $P'$, we usually denote both $f'$ and $f''$ by $f$. We write
\[f \colon P \to P' \qquad f \colon G \to G'\]
and the equivariance is written
\[f(pg) = f(p)f(g)\]
\end{notation}

Note that a homomorphism $P \to P'$ sends the fibers of $P$ to the fibers of $P'$. There is a well--defined smooth $\overline f \colon M \to M'$ such that the following diagram commutes:
\[\xymatrix{
P \ar[d]_\pi \ar[r]^f & P' \ar[d]^{\pi'} \\
M \ar[r]_{\overline f} & M'
}\]

\begin{definition}
Let $P \xrightarrow{\pi} M$ be a principal $G$--bundle and $H \subset G$ a Lie subgroup. A {\em reduction of the structure group} of $P$ from $G$ to $H$ is an injective homomorphism $f$ of a principal $H$--bundle $Q \to M$ into $P$ such that $\overline f = \id_M$.
\end{definition}

\begin{example}
Let $P = M \times G$ be the product bundle. Let $H = \{e\} \subset G$, $Q = M \times H$. Define
\[f \colon H \to G, e \mapsto e \qquad f \colon Q \to P, (m,e) \mapsto (m,e)\]
This $f$ is a homomorphism and defines a reduction of the structure group of $P$ to $\{e\}$.
\end{example}

\begin{proposition}
Let $P \xrightarrow{\pi} M$ be a principal $G$--bundle and $H \subset G$ a Lie subgroup. The structure group of $P$ can be reduced to $H$ if and only if there is a system of local trivializations for $P$ such that the corresponding transition maps $\psi_{\alpha\beta}$ take values in $H$.
\end{proposition}

\begin{proof}
Assume there is a reduction $f \colon Q \to P$, where $Q$ is a principal $H$-bundle. We may assume $f \colon H \to G$ is the inclusion. The corresponding transition maps take values in $H$.
%TODO: understand this part (one line is missing)
Conversely, suppose $P$ admits local trivializations $U_\alpha \times G$ such that all transition maps $\psi_{\alpha\beta}$ take values in $H$. Then we can construct a principal $H$--bundle $Q \xrightarrow{\pi} M$ from the $\psi_{\alpha\beta}$. In each trivialization $U_\alpha \times G$ for $P$ we have $U_\alpha \times H \subset U_\alpha \times G$. These inclusions induce an injective homomorphism $f \colon Q \to P$ giving a reduction of the structure group of $P$ from $G$ to $H$.
\end{proof}

\subsection{Associated bundles}

\begin{definition}
Let $P \xrightarrow{\pi} M$ be a principal $G$--bundle, $F$ a smooth manifold and $\mu \colon G \times F \to F$ a left action of $G$ on $F$. The {\em associated bundle} $E \xrightarrow{\pi_E} M$ is defined as follows:
\[E := P \times_G F := (P \times F) / \!\sim\]
where
\[(p,f) \sim (pg, g^{-1}f) \qquad \forall p \in P, f \in F, g \in G\]
and
\[\pi_E([(p,f)]) \coloneqq \pi(p)\]
\end{definition}

Let $U \subset M$ be an open set over which $P$ is trivial and let
\[\pi^{-1}(U) \to U \times G \qquad p \mapsto (\pi(p),\varphi(p))\]
be a local trivialization. Performing the construction of $E$ with $\pi^{-1}(U)$ in place of $P$, we obtain
\[\pi_E^{-1}(U) = (\pi^{-1}(U) \times F)/\!\sim\  \cong (U \times G \times F)/\!\simeq\]
with $\simeq$ defined as follows:
\[(m,h,f) \simeq (m,hg,g^{-1}f)\]
We claim that
\[(U \times G \times F)/\!\simeq \ \cong U \times F\]
To prove this, define $\psi_1 \colon (U \times G \times F)/\!\simeq\ \to U \times F$ by $\psi_1([(m,h,f)]) = (m,hf)$ and $\psi_2 \colon U \times F \to (U \times G \times F)/\simeq$ by $\psi_2(m,f) = [(m,e,f)]$. Then $\psi_1$ is well--defined and $\psi_1, \psi_2$ are mutually inverse:
\[\psi_2 \circ \psi_1([(m,h,f)]) = \psi_2(m,hf) = [(m,e,hf)] = [(m,h,f)]\]
\[\psi_1 \circ \psi_2(m,f) = \psi_1([(m,e,f)]) = (m,ef) = (m,f)\]
The associated bundle $E$ has a unique differentiable structure in which the open subsets $\pi_E^{-1}(U)$ are open smooth submanifolds diffeomorphic to $U \times F$. This shows that $E$ is a locally trivial smooth fiber bundle with fiber $F$ and structure group $G$.

\begin{example}\
\vspace{-.5em}
\begin{enumerate}[(1)]
\item If $\mu \colon G \times F \to F$ is the terminal action $(g,f) \mapsto f$ then $E = P \times_G F$ is diffeomorphic to $M \times F$ such that $\pi_E$ corresponds to $\pi_1$.
\item Let $\rho \colon G \to \GL_n(\mathbb R)$ be a homomorphism of Lie groups. Then $G$ acts on $\mathbb R^n$ via $\rho$:
\[\mu \colon G \times \mathbb R^n \to \mathbb R^n \qquad (g,v) \mapsto \rho(g)v\]
In this case, $E = P \times_G \mathbb R^n \eqqcolon P \times_\rho \mathbb R^n$ is a vector bundle over $M$.
\item[(2')] Suppose $V \to M$ is a vector bundle of rank $k$. The basis for fibers of $V$ form a principal $GL_k(\mathbb R)$--bundle $P \xrightarrow{\pi} M$. Take $\id \colon \GL_k(\mathbb R) \to \GL_k(\mathbb R)$. Then $E = P \times_\rho \mathbb R^n$ is isomorphic to $V$.
\item Let $P \xrightarrow{\pi} M$ be a principal $G$--bundle and $H \subset G$ a closed subgroup. Using the action
\[\mu \colon G \times G/H \to G/H \qquad (g,aH) \mapsto (ga)H\]
we can form the associated bundle $E$ with fiber $G/H$.
\end{enumerate}
\end{example}

\begin{lemma}
In example (3), the associated bundle $E$ with fiber $G/H$ is diffeomorphic to the orbit space $P/H$, where $H$ acts on $P$ by restricting the $G$--action.
\end{lemma}

\begin{proof}
We define to mutually inverse smooth maps $\psi_1$ and $\psi_2$ between $E$ and $P/H$.
\[\psi_2 \colon E \to P/H, [(p,aH)] \mapsto H(pa) \qquad \psi_1 \colon P/H \to E, H(p) \mapsto [(p,H)]\]
These are indeed well--defined and smooth and $\psi_1 \circ \psi_2 = \psi_2 \circ \psi_1 = \id$.
\end{proof}

\begin{proposition}
Let $P \xrightarrow{\pi} M$ be a principal $G$--bundle and $H \subset G$ a closed Lie subgroup. The structure group of $P$ can be reduced to $H$ if and only if the associated bundle $E$ with fiber $G/H$ has a section.
\end{proposition}

\begin{proof}
Suppose the structure group of $P$ can be reduced to $H$, so that there is a principal $H$--bundle $Q \to M$ and an injective homomorphism $f \colon Q \to P$. We claim that the composition $Q \xrightarrow{f} P \to P/H$ is constant on every fiber of $Q$. To see this, let $\alpha, \beta \in Q$ be in the same fiber of $Q$. Then there exists $h \in H$ such that $\alpha h = \beta$, so $f(\alpha)h = f(\beta)$ and thus $[f(\alpha)] = [f(\beta)]$, i.e. the images in $P/H$ agree.
\[\xymatrix{
Q \ar[d] \ar[r]^f & P \ar[r] & P/H = E \\
M \ar[rru]
}\]
By the claim, this map factors through the projection $Q \to M$, and so gives a section $s \colon M \to E$. Conversely, suppose $E \xrightarrow{\pi_E} M$ admits a section $s \colon M \to E$. Define $Q$ as the preimage of $s(M)$ under the map $P \to P/H = E$. The restriction to $H$ of the $G$--action on $P$ preserves $Q \subset P$ and is simply transitive on the fibers of $Q \to M$. $Q$ is a principal $H$--bundle and the inclusion $Q \subset P$ is a reduction of the structure group of $P$ to $H$.
\end{proof}

\begin{definition}
Suppose $P \xrightarrow{\pi} M$ is a principal $G$--bundle and $f \colon N \to M$ is a smooth map. Then define
\[f^* P \coloneqq \{(n,p) \in N \times P \mid f(n) = \pi(p)\]
\[\xymatrix{
f^* P \ar[d]_{\pi_1} \ar[r]^{\pi_2} & P \ar[d]^\pi \\
N \ar[r]_f & M}\]
$f^*P$ is a principal $G$--bundle. It is called the {\em pullback bundle} obtained by pulling back $P \xrightarrow{\pi} M$ via $f$.
\end{definition}

\subsection{Connections}

Let $P \xrightarrow{\pi} M$ be a principal $G$--bundle. Then $\pi$ is a submersion and $D_p\pi \colon T_pP \to T_{\pi(p)}M$ has as kernel the tangent space at $p$ to the fiber $\pi^{-1}(\pi(p))$. Moreover, $\ker(D_p\pi)$ is spanned by the fundamental vector fields $X^*$ generated by the $G$--action on $P$. We call $\ker(D_p\pi) \eqqcolon V_p$ the {\em vertical tangent space} at $p$.

\begin{definition}
A {\em connection} on $P$ is a choice of a complement $H_p$ for $V_p$ in $T_pP$ for all $p \in P$ such that
\begin{enumerate}[(1)]
\item $H_p$ depends smoothly on $p$.
\item $D_p r_g(H_p) = H_{pg}$ for all $p \in P$, $g \in G$.
\end{enumerate}
\end{definition}

\begin{remark}
Property (1) is equivalent to saying that $\bigcup_{p\in P} H_p$ is a smooth subbundle $H$ of $TP$. If $V = \bigcup_{p\in P} V_p$ is the vertical subbundle in $TP$ with $V = \ker(D\pi)$, then $H$ has to be a complement to $V$ in $TP$, so that $TP = V \oplus H$. A connection $H$ on $P$ is a $G$--invariant smooth complement to $V$.

If $H$ is a connection on $P$, then
\[D_p\pi \colon H_p \to T_{\pi(p)}M\]
is an isomorphism for all $p \in P$. Under this isomorphism, vector fields on $M$ can be lifted to horizontal vector fields on $P$.
\end{remark}

If $\alpha \in \Omega^1(P)$ is a 1--form on $P$, then at every point $p \in P$ with $\alpha_p \neq 0$, $\ker \alpha_p \subset T_pP$ is a hyperplane. If $\alpha$ is a 1--form with values in $\mathbb R^k$, then at every point $p \in P$ it defines a linear map
\[\alpha_p \colon T_pP \to \mathbb R^k\]
If $\alpha_p$ is surjective onto $\mathbb R^k$, then $\ker(\alpha_p) \subset T_pP$ is a subspace of codimension $k$.

\begin{definition}
Given a connection $H$ on $P$, we define a 1--form $\omega$ on $P$ with values in $\mathfrak g$ as follows:
\[\omega_p(X) = \begin{cases} 0 & \text{if $X \in H_p$} \\ A & \text{if $X = A^*_p$ for any $A \in \mathfrak g$}\end{cases}\]
where $A^*$ is the fundamental vector field on $P$ generated by the right action of $\exp(tA)$. Then $\omega$ is the {\em connection 1--form} corresponding to $H$.
\end{definition}

At every point $p \in P$, we have $T_pP = V_p \oplus H_p$. $H_p$ is in the kernel of $\omega_p$ and $V_p$ contains only elements of the form $A^*_p$ and for those $\omega(A^*_p) = A$. So $\omega$ is well--defined and $\ker \omega_p = H_p$, because $\omega_p \colon T_pP \to \mathfrak g$ is surjective. $\omega$ is also smooth, since $H$ and $A^*$ are smooth.

\begin{lemma}
For $g \in G$, we have
\[r_g^*\omega = \Ad_{g^{-1}} \omega\]
where $\Ad_{g^{-1}} \omega$ is the composition of $\omega \colon TP \to \mathfrak g$ and $\Ad_{g^{-1}} \colon \mathfrak g \to \mathfrak g$.
\end{lemma}

\begin{proof}
Let $X \in T_pP$. Since $(r_g^* \omega)_p(X) = \omega(D_pr_g(X))$, the claim of the lemma is equivalent to
\[\omega(Dr_g(X)) = \Ad_{g^{-1}}(\omega(X))\]
Both sides are linar in $X$, therefore it is enough to check the claim for $X \in V$ and $X \in H$.

If $X \in H_p$, then $\omega(Dr_g(X)) \in \omega(H_{pg}) = \{0\}$ by the $G$--invariance of $H$. Also $\Ad_{g^{-1}}(\omega(X)) = \Ad_{g^{-1}}(0) = 0$.

Now let $X \in V_p$. Since the fundamental vector fields span $V$, we can choose $Y \in \mathfrak g$ with $Y^*_p = X$. Then
\[\omega(Dr_g(X)) = \omega(Dr_g(Y^*_p)) = \omega((\Ad_{g^{-1}}(Y))^*_{pg}) = (\Ad_{g^{-1}}(Y))\]
and
\[\Ad_{g^{-1}}(\omega(X)) = \Ad_{g^{-1}}(\omega(Y^*_p)) = \Ad_{g^{-1}}(Y)\qedhere\]
\end{proof}

\begin{proposition}
Suppose $\omega$ is a $\mathfrak g$--valued 1--form on $P$ with the property that $r_g^* \omega = \Ad_{g^{-1}} \omega$. Assume also that for fundamental vector fields $A^*$, we have $\omega(A^*) = A$. Then $H \coloneqq \ker \omega$ is a connection on $P$.
\end{proposition}

\begin{proof}
The two requirements on $\omega$ are consistent. If $A^*$ is the fundamental vector field generated by $A$, then $Dr_g(A^*_p) = (\Ad_{g{-1}}(A))^*_{pg}$. This implies
\[(\Ad_{g^{-1}}(A))_{pg} = \omega((\Ad_{g^{-1}}(A))^*_{pg}) = \omega(Dr_g(A^*_p)) = r_g^*\omega(A^*_p)\]
For all $p \in P$, the map
\[\omega_p \colon T_pP \to \mathfrak g\]
is surjective, so $\ker \omega_p = H_p$ is a subspace of $T_pP$ whose codimension is $\dim \mathfrak g$. The requirement $\omega(A^*) = A$ means that $\omega_p|_{V_p} \colon V_p \to \mathfrak g$ is an isomorphism for all $p \in P$. So $H_p$ is a complement for $V_p$ in $T_pP$.

Suppose $X \in H_p$. To prove $G$--invariance of $H$, we have to show $D_pr_g(X) \in H_{pg}$. Now $X \in H_p$ means that $\omega(X) = 0$. We have to prove $\omega(Dr_g(X)) = 0$. Indeed,
\[\omega(Dr_g(X)) = r_g^*\omega(X) = \Ad_{g^{-1}}(\omega(X)) = \Ad_{g^{-1}}(0) = 0\qedhere\]
\end{proof}

\begin{proposition}
Every principal $G$--bundle $P \xrightarrow{\pi} M$ admits a connection.
\end{proposition}

\begin{proof}
Let $\{U_i\}_{i\in I}$ be an open cover of $M$ with the property that $P$ restricted to each $U_i$ is trivial:
\[\pi^{-1}(U_i) \xrightarrow{\ \ \sim\ \ } U_i \times G\]
On the product bundle $U_i \times G$, there is a connection with
\[H_{(m,g)} = T_mU_i \times \{0\} \subset T_mU_i \oplus T_gG = T_{(m,g)}(U_i \times G)\]
Let $\omega_i$ be the connection 1--form on $\pi^{-1}(U_i)$, whose kernel corresponds to $H$ under the trivialization $\pi^{-1}(U_i) \to U_i \times G$. Let $\{\rho_j\}_{j\in J}$ be a smooth partition of unity on $M$ subordinate to the covering by the $U_i$, i.e. $\rho_j \colon M \to \mathbb R$ are smooth non-negative functions such that $\supp \rho_j$ are locally finite in $M$ and $\sum_j \rho_j = 1$ and for all $j \in J$ there exists an $i \in I$ such that $\supp \rho_j \subset U_i$. Then define
\[\omega \coloneqq \sum_{j\in J} \pi^* \rho_j \cdot \omega_j = \sum_{j\in J} (\rho_j \circ \pi) \cdot \omega_j\]
where for all $j \in J$, $\omega_j \coloneqq \omega_i$ for some $i \in I$ such that $\supp \rho_j \subset U_i$, and the summands, which are supported only inside $\pi^{-1}(U_i)$ are being extended by $0$ to all of $P$.

We claim that $\omega$ is a connection 1--form on $P$. To see this, we need to check that $r_g^*\omega = \Ad_{g^{-1}} \omega$ for all $g \in G$ and $\omega(A^*_p) = A_p$ for all $A \in \mathfrak g$, $p \in P$. Since $\pi^*\rho_j$ is constant under right $G$--action, the first equality follows by
\begin{align*}
r_g^* \omega &= r_g^*\left(\sum_{j\in J} \pi^*\rho_j \cdot \omega_j \right) = \sum_{j\in J} \pi^* \rho_j \cdot r_g^* \omega_j = \sum_{j\in J} \pi^* \rho_j \cdot \Ad_{g^{-1}} \omega_j = \\
&= \Ad_{g^{-1}}\left(\sum_{j\in J} \pi^* \rho_j \cdot \omega_j \right) = \Ad_{g^{-1}} \omega
\end{align*}
and the second by
\[\omega(A^*_p) = \left(\sum_{j\in J} \pi^*\rho_j \cdot \omega_j\right)(A^*_p) = \sum_{j\in J} \pi^*\rho_j(p) \cdot \omega_j(A^*_p) = \sum_{j\in J} \pi^*\rho_j(p) \cdot A_p = A_p\qedhere\]
\end{proof}

\begin{proposition}
The set of connections on a principal $G$--bundle $P \xrightarrow{\pi} M$ is naturally an affine space whose vector space of translations is the space of 1--forms on $M$ with values in the vector bundle $P \times_{\Ad} \mathfrak g \longrightarrow M$.

More precisely, for any difference of connection 1--forms $\widetilde\omega = \omega_1 - \omega_2$ on $P$, there is a unique $\omega \in \Gamma((P \times_{\Ad} \mathfrak g) \otimes T^*\!M)$ such that
\[\pi^*\omega(Y) = [(p,\widetilde\omega(Y))] \qquad \forall Y \in T_pP\]
\end{proposition}

\begin{proof}
By the previous proposition, the set of connections $P$ is non--empty, so we can choose a reference connection 1--form $\omega_0$. For any connection 1--form $\omega_1$, let $\widetilde\omega \coloneqq \omega_1 - \omega_0$. For all $p \in P$, $V_p$ is spanned by the values of the fundamental vector fields $A^*_p$ with $A \in \mathfrak g$, but
\[\widetilde\omega(A^*_p) = \omega_1(A^*_p) - \omega_0(A^*_p) = A - A = 0\]
so $\widetilde\omega$ vanishes on $V$. Let $E \coloneqq P \times_{\Ad} \mathfrak g$, then $\Omega^1(M,E) = \Gamma(T^*\!M \otimes E)$ is the vector space of 1--forms on $M$ with values in the vector bundle $E$. We want to define $\omega \in \Omega^1(M,E)$ by
\[\omega(X) = [(p, \widetilde \omega(Y))]\]
for all $X \in T_mM$, where $Y \in T_pP$ is a lift of $X$ at some $p \in \pi^{-1}(m)$, i.e. $D\pi(Y) = X$. We can always choose such a lift since $D\pi$ is surjective. For $\omega$ to be well--defined, we have to check that
\[[(p,\widetilde\omega(Y))] = [(q,\widetilde\omega(Z))]\]
for any lift $Z \in T_qP$ at $q \in \pi^{-1}(m)$. To see this, first let $q = p$ and $Z \in T_pP$ be a lift of $X$ at $p$. Then
\[D\pi(Z-Y) = D\pi(Z) - D\pi(Y) = X - X = 0\]
so $Z-Y \in V$ and therefore
\[\widetilde\omega(Z) = \widetilde\omega(Y) - \widetilde\omega(Z-Y) = \widetilde\omega(Y)\]
Now let $q \in \pi^{-1}(m)$ be arbitrary and $Z \in T_qP$ a lift of $X$ at $q$. There is a unique $g \in G$ such that $q = pg$. Since
\[D\pi(Dr_g(Y)) = D(\pi \circ r_g)(Y) = D\pi(Y) = X,\]
$Dr_g(Y) \in T_qP$ is a lift of $X$ at $q$, so $\widetilde\omega(Z) = \widetilde\omega(Dr_g(Y))$. But then we have
\[[(q,\widetilde\omega(Z))] = [(q, \widetilde\omega(Dr_g(Y)))] = [(q, r_g^*\widetilde\omega(Y))] = [(pg, \Ad_{g^{-1}}\widetilde\omega(Y))] = [(p, \widetilde\omega(Y))].\]
This shows that although $\omega$ is not well--defined as an ordinary $\mathfrak g$--valued 1--form on $M$, it is well--defined as a 1--form on $M$ with values in $E$.

Conversely, let $\omega \in \Omega^1(M,E)$. We have to check that with $\widetilde\omega$ defined by
\[\pi^*\omega(Y) = [(p,\widetilde\omega(Y))] \qquad \forall Y \in T_pP\]
the $\mathfrak g$--valued 1--form $\omega_1 = \omega_0 + \widetilde \omega$ is a connection 1--form. First, we show $\omega_1(X^*_p) = X$ for all $X \in \mathfrak g$ and $p \in P$: Since
\[[(p,\widetilde\omega(X^*_p))] = \pi^*\omega(X^*_p) = \omega(D\pi(X^*_p)) = \omega_{\pi(p)}(0) = \omega_{\pi(p)}(D_p\pi(0)) = (\pi^*\omega)_p(0) \!=\! [(p,0)]\]
and the first element uniquely determines the representative of the equivalence class, we have
\[\omega_1(X^*_p) = \omega_0(X^*_p) + \widetilde\omega(X^*_p) = X + 0 = X.\]
Left to show is $r_g^*\omega_1 = \Ad_{g^{-1}}\omega_1$ for $g \in G$, but for $Y \in T_pP$,
\begin{align*}
[(pg,r_g^*\widetilde\omega(Y))] &= [(pg,\widetilde\omega(Dr_g(Y)))] = \pi^*\omega(Dr_g(Y)) = \omega(D\pi(Dr_g(Y))) = \\
&= \omega(D(\pi \circ r_g)(Y)) = \omega(D\pi(Y)) = \pi^*\omega(Y) = [(p,\widetilde\omega(Y))] = \\
&= [(pg, \Ad_{g^{-1}} \widetilde\omega(Y))],
\end{align*}
so $r_g^*\widetilde\omega = \Ad_{g^{-1}} \widetilde \omega$ and thus by linearity $r_g^*\omega_1 = \Ad_{g^{-1}}\omega_1$.
\end{proof}

Let $P \xrightarrow{\pi} M$ be a principal $G$--bundle, $\omega$ a connection 1--form on $P$, $U_i,U_j \subset M$ open sets over which $P$ is trivial. The trivializations
\[\psi_i \colon \pi^{-1}(U_i) \to U_i \times G\]
correspond to sections
\[s_i \colon U_i \to \pi^{-1}(U_i) \qquad m \mapsto \psi_i^{-1}(m,e)\]
$\omega_i \coloneqq s_i^* \omega$ is a $\mathfrak g$--valued 1--form on $U_i \subset M$. Suppose $U_i \cap U_j \neq \varnothing$. Then on $U_i \cap U_j$ both $\omega_i$ and $\omega_j$ are defined. We have the smooth transition maps $\psi_{ij} \colon U_i \cap U_j \to G$ defined by
\[\psi_i \circ \psi_j^{-1} \colon (U_i \cap U_j) \times G \to (U_i \cap U_j) \times G \qquad (m,g) \mapsto (m,\psi_{ij}(m)g)\]
and want to use them to find a formula for transition between $\omega_j$ and $\omega_i$.

On $G$ we have a canonical 1--form $\theta$ with values in $\mathfrak g$ defined by
\[\theta(A_g) = A \qquad \forall A \in \mathfrak g, g \in G\]
This is well--defined since it is equivalent to $\theta(X) = D\ell_{g^{-1}}(X)$ for $X \in T_gG$.

\begin{lemma}
We have the following translation from $\omega_i$ to $\omega_j$:
\[\omega_j(X) = \Ad_{\psi_{ij}(m)}^{-1}\omega_i(X) + \psi_{ij}^*\theta(X) \qquad \forall X \in T_mM\]
\end{lemma}

\begin{proof}
Differentiating the function
\[s_j(m) = \psi_j^{-1}(m,e) = \psi_i^{-1} \circ \psi_i \circ \psi_j^{-1}(m,e) = \psi_i^{-1}(m,\psi_{ij}(m))  = s_i(m) \cdot \psi_{ij}(m)\]
gives for $X \in T_mM$
\begin{align*}
Ds_j(X) &= D(\mu \circ (s_i \times \psi_{ij}) \circ \Delta)(X) = (D\mu \circ D(s_i \times \psi_{ij}) \circ D\Delta)(X) = \\
&= D\mu (Ds_i(X), D\psi_{ij}(X)) = Dr_{\psi_{ij}(m)}(Ds_i(X)) + D\ell_{s_i(m)}(D\psi_{ij}(X))
\end{align*}
where we can transform the last summand using $A_{pg}^* = D\ell_p(A_g)$ like
\begin{align*}
D\ell_{s_i(m)}(D\psi_{ij}(X)) &= (D\ell_{s_i(m)} \circ D\ell_{\psi_{ij}(m)} \circ D\ell_{\psi_{ij}(m)^{-1}} \circ D\psi_{ij})(X) = \\
&= D\ell_{s_i(m)}((D\ell_{\psi_{ij}(m)^{-1}}(D\psi_{ij}(X)))_{\psi_{ij}(m)}) = \\
&= (D\ell_{\psi_{ij}(m)^{-1}}(D\psi_{ij}(X)))^*_{s_i(m)\psi_{ij}(m)} = \\
&= (\theta(D\psi_{ij}(X)))^*_{s_j(m)} = \\
&= (\psi_{ij}^*\theta(X))^*_{s_j(m)}
\end{align*}
Putting the above identities together we get the desired result
\begin{align*}
\omega_j(X) &= \omega(Ds_j(X)) = \omega(Dr_{\psi_{ij}(m)}(Ds_i(X)) + \omega((\psi_{ij}^*\theta(X))^*_{s_j(m)}) = \\
&= r_{\psi_{ij}(m)}^*\omega(Ds_i(X)) + \psi_{ij}^*\theta(X) = \Ad_{\psi_{ij}(m)}^{-1}\omega_i(X) + \psi_{ij}^*\theta(X)\qedhere
\end{align*}
\end{proof}

\subsection{Parallel transport}
Let $P \xrightarrow{\pi} M$ be a principal $G$--bundle, $H = \ker \omega$ a connection on $P$. For every $p \in P$, $D_p\pi|_H$ is an isomorphism $H_p \to T_{\pi(p)}M$. Every $X \in T_{\pi(p)}M$ has a unique preimage $X^* \in H_p$ under this isomorphism. Every vector field $X \in \mathfrak X(M)$ gives rise to a unique vector field $X^*$ on $P$ such that
\[X^*_p = (D_p\pi)^{-1} (X_{\pi(p)})\]
This construction has the following simple properties: $(fX)^* = \pi^*f \cdot X^*$ for any $f \in C^\infty(M,\mathbb R)$ and $(X+Y)^* = X^* + Y^*$. But $[X,Y]^* \neq [X^*,Y^*]$:
Although $[X,Y]_{\pi(p)} = D\pi([X^*,Y^*]_p)$ holds, $[X^*,Y^*]$ is not neccessarily horizontal. Denoting the projections from to the horizontal and vertical subbundles of $TP$ by
\[\mathscr V \colon TP \to V \qquad \mathscr H \colon TP \to H\]
the above equality shows that
\[\mathscr H([X^*,Y^*]) = [X,Y]^*\]
\begin{remark}
In general $\mathscr V([X^*,Y^*]) \neq 0$, and is related to the curvature of $H$.
\end{remark}

\begin{definition}
A smooth curve $c \colon [0,1] \to P$ is {\em horizontal} (wrt H) if $\dot c(t) \in H_{c(t)}$ for all $t \in [0,1]$.
\end{definition}

\begin{proposition}
Let $c \colon [0,1] \to M$ be a smooth curve and $p \in \pi^{-1}(c(0))$. Then there is a unique horizontal curve $\overline c \colon [0,1] \to P$ with $\overline c(0) = p$ and $\pi \circ \overline c = c$. $\overline c$ is called a {\em horizontal lift} of $c$.
\end{proposition}

\begin{proof}
Given $c$ and $p \in \pi^{-1}(c(0))$, there exists some smooth $\overline c \colon [0,1] \to P$ with $\overline c(0) = p$ and $\pi \circ \overline c = c$ (by local triviality). Any other lift of $c$ to $P$ with starting point $p$ is of the form $\overline c \cdot g$ where $g \colon [0,1] \to G$ is a smooth map with $g(0) = e$. We need to find a $g$ such that $\overline c \cdot g$ is horizontal. This is the case iff $\frac{\mathrm d}{\mathrm d t}(\overline c(t) \cdot g(t)) \in H$, i.e. $\omega(\frac{\mathrm d}{\mathrm d t}(\overline c \cdot g)) = 0$ for all $t \in [0,1]$. As in the lemma above, we get
\[\frac{\mathrm d}{\mathrm d t}(\overline c \cdot g) = D_t(\overline c \cdot g)(\partial) = (D\mu \circ D(\overline c \cdot g) \circ D_t\Delta)(\partial) = D_{\overline c}r_{g}(D_t\overline c(\partial)) + D_{g}\ell_{\overline c}(D_tg(\partial))\]
and
\begin{align*}
D_g\ell_{\overline c}(D_tg(\partial)) &= (D_g\ell_{\overline c} \circ D_e\ell_g \circ D_g \ell_{g^{-1}} \circ D_t g)(\partial) = D_g\ell_{\overline c}(((D_g \ell_{g^{-1}} \circ D_t g)(\partial))_g) = \\
&= ((D_g \ell_{g^{-1}} \circ D_t g)(\partial))^*_{\overline c \cdot g} = (D_g\ell_{g^{-1}} (\dot g))^*_{\overline c \cdot g}
\end{align*}
so
\begin{align*}
\omega\left(\frac{\mathrm d}{\mathrm d t}(\overline c \cdot g)\right) &= \omega(D_{\overline c}r_g(\dot{\overline c})) + D_g\ell_{g^{-1}}(\dot g) = r_g^*\omega(\dot{\overline c}) + D\ell_{g^{-1}}(\dot g) = \\
&= \Ad_{g^{-1}}\omega(\dot{\overline c}) + D\ell_{g^{-1}}(\dot g) = D\ell_{g^{-1}}(Dr_g(\omega(\dot{\overline c}))) + D\ell_{g^{-1}}(\dot g) = \\
&= (D\ell_{g^{-1}} \circ Dr_g)(\omega(\dot{\overline c}) + Dr_{g^{-1}}(\dot g))
\end{align*}
Since $D\ell_{g^{-1}} \circ Dr_g$ is an isomorphism, this is 0 if and only if
\[Dr_{g(t)^{-1}}(\dot g(t)) = - \omega(\dot{\overline c}(t))\]
So the statement of the proposition is just that this differential equation has a unique solution with $g(0) = e$. This is shows by the following lemma.
\end{proof}

\begin{lemma}
Let $X \colon [0,1] \to \mathfrak g$ be smooth. There exists a unique $g \colon [0,1] \to G$ with $g(0) = e$ and
\[Dr_{g(t)^{-1}}(\dot g(t)) = X(t)\]
\end{lemma}

\begin{proof}
On $G \times [0,1]$, $X$ defines a time--independent vector field $\mathbb X$ with
\[\mathbb X_{(g,t)} = (X_g(t), \partial)\]
The flow of $\mathbb X$ is defined for all t (see the proof of completeness of left--invariant vector fields on G). Under the flow $\varphi$ of $\mathbb X$ we have
\[\varphi_t(e,0) = (g(t),t)\]
for a $g(t)$ which solves our equation. This is the only solution with $g(0) = e$.
\end{proof}

Let $c \colon [0,1] \to M$ be a smooth curve. Define
\[P_c \colon \pi^{-1}(c(0)) \to \pi^{-1}(c(1)) \qquad p \mapsto \overline c(1)\]
where $\overline c$ is the unique horizontal lift of $c$ with $\overline c(0) = p$.
%TODO: picture

The map $P_c$ is the {\em parallel transport map} defined by $c$. $P_c$ is invertible by running back along $c$. Except from change of direction, it is independent of the parametrization of $c$. The parallel transport map can also be defined for a piecewise smooth curve $c$ by concatenating the $P_{c_i}$ for $c_i$ obtained by restricting $c$ to subintervals of $[0,1]$ where it is smooth.

Fix a basepoint $m_0 \in M$. For every closed piecewise smooth curve $c \colon [0,1] \to M$ with $c(0) = c(1) = m_0$ we have
\[P_c \colon \pi^{-1}(m_0) \to \pi^{-1}(m_0)\]
\begin{claim}
The set of all these $P_c$ is a group with composition as group operation.
\end{claim}

\begin{proof}
$\{P_c\}$ has $\id_{\pi^{-1}(m_0)}$ as an element obtained as $P_c$ for $c$ the constant path at $m_0$. If $c_1$ and $c_2$ are two closed paths beginning and ending at $m_0$, then $P_{c_2} \circ P_{c_1} = P_{c_1c_2}$, where $c_1c_2$ denotes the concatenation of $c_1$ and $c_2$. This is associative and $P_c^{-1} = P_{\overline c}$ where $\overline c$ is $c$ parametrized backwards.
\end{proof}

Fix a basepoint $p_0 \in P$ with $\pi(p_0) = m_0$. If $P_c$ is one of the parallel transport maps defined above, then $P_c(p_0) \in \pi^{-1}(m_0)$. Since $G$ acts simply transitively on the fiber, there exists a unique $g(c) \in G$ such that $P_c(p_0) = p_0 g(c)$.

\begin{claim}
The map
\[h \colon \{P_c\} \to G \qquad P_c \mapsto g(c)\]
is an injective homomorphism of groups.
\end{claim}

\begin{proof}
Since $h(P_{c_1} \circ P_{c_2}) = h(P_{c_2c_1}) = g(c_2c_1)$ and $h(P_{c_1})h(P_{c_2}) = g(c_1)g(c_2)$ need to prove that $g(c_2c_1) = g(c_1)g(c_2)$.
%TODO: picture

For any curve $c \colon [0,1] \to M$, the curve $\overline c$ from $p \in P$ to $P_c(p)$ is horizontal, since it is a horizontal lift of $c$. Since $H$ is $G$--invariant, the $G$--action maps horizontal curves to horizontal curves, so for any $g \in G$,  $\overline c \cdot g$ is a horizontal curve from $pg$ to $P_c(p) g$. This means $P_c(pg) = P_c(p)g$, i.e. $P_c$ is $G$--equivariant. So
\[p_0 g(c_2c_1) = P_{c_2c_1}(p_0) = P_{c_1} \circ P_{c_2}(p_0) = P_{c_1}(p_0 g(c_2)) = P_{c_1}(p_0) g(c_2) = p_0 g(c_1) g(c_2)\]
and thus $g(c_2c_1) = g(c_1)g(c_2)$. We have proved that $h$ is a homomorphism. Now suppose $P_c \in \ker h$, i.e. $g(c) = h(P_c) = e$. Then $P_c(p_0) = p_0$. Every $p \in \pi^{-1}(p_0)$ is of the form $p = p_0 g$ for some $g \in G$. But $P_c(p) = P_c(p_0g) = P_c(p_0)g = p_0g = p$ by the $G$--equivariance of $P_c$, so $P_c = \id_{\pi^{-1}(p_0)}$. Thus $h$ is injective.
\end{proof}

\begin{definition}
The {\em holonomy group} $\Hol(H,p_0) = \Hol(p_0)$ of the connection $H$ wrt $p_0 \in P$ is the subgroup of $G$ obtained by parallel transport along closed loops based at $p_0$, i.e. $\Hol(H,p_0) = \im h$.

The {\em restricted holonomy group} $\Hol_0(H,p_0) = \Hol_0(p_0)$ is the subgroup obtained by considering only parallel transports $P_c$ for closed loops $c$ which are contractible or null--homotopic.
\end{definition}

\begin{properties}\
\begin{enumerate}[(1)]
\item $\Hol(H,p_1) = g^{-1} \Hol(H,p_0) g$ if $p_1 = p_0 g$.
\item $\Hol(H,p_1) = \Hol(H,p_0)$ if $p_1$ is obtained from $p_0$ by parallel transport.
\item If $M$ is connected, $\Hol(H,p_0)$ and $\Hol(H,p_1)$ are conjugate in $G$ for any $p_0,p_1 \in H$.
\end{enumerate}
These properties also hold for $\Hol_0$.
\end{properties}

\begin{proof}\
\begin{enumerate}[(1)]
\item Consider $h_0 \in \Hol(H,p_0)$ and $h_1 \in \Hol(H,p_1)$ defined by $P_c(p_0) = p_0 h_0$ and $P_c(p_1) = p_1 h_1$ for the same curve $c$ on $M$. Then
\[p_0 g h_1 = p_1 h_1 = P_c(p_1) = P_c(p_0g) = P_c(p_0) g = p_0 h_0 g\]
so $g h_1 = h_0 g$ and $h_1 = g^{-1} h_0 g$.
\item Let $c$ be a smooth curve on $M$ such that $p_1 = P_c(p_0)$ and let $h_1 \in \Hol(H,p_1) = \Hol(H,P_c(p_0))$. There exists a closed curve $c_1$ on $M$ such that $P_{c_1}(P_c(p_0)) = P_c(p_0)h_1$. The curve $cc_1\overline c$ is a closed curve in $M$ based at $\pi(p_0)$, so there exists $h_0 \in \Hol(H,p_0)$ such that $P_{cc_1\overline c}(p_0) = p_0 h_0$. But
\[P_c(p_0)h_1 = P_c \circ P_c^{-1} \circ P_{c_1} \circ P_c(p_0) = P_c(P_{cc_1\overline c}(p_0)) = P_c(p_0h_0) = P_c(p_0)h_0\]
so $h_1 = h_0$ and thus $h_1 \in \Hol(H,p_0)$. By using $\overline c$ instead of $c$, we get the other inclusion.
\item Since $M$ is connected, there is a smooth curve $c$ from $\pi(p_0)$ to $\pi(p_1)$, which has a horizontal lift from $p_0$ to $P_c(p_0) \in \pi^{-1}(\pi(p_1))$, $\Hol(H,P_c(p_0)) = \Hol(H,p_0)$ by (2). By (1), this is conjugate to $\Hol(H,p_1)$.\qedhere
\end{enumerate}
\end{proof}

% \begin{properties}\
% \begin{enumerate}[(1)]
% \item $\Hol(H, p_1) = g^{-1} \Hol(H, p_0) g$ if $p_1 = p_0g$.
% \item If $p_1 \in \pi^{-1}(p_0)$ is obtained from $p_0$ by parallel transport along some $c$ based at $m_0$, then $\Hol(H,p_1) = \Hol(H,p_0)$.
% \item If $M$ is connected, $\Hol(H,p_1)$ and $\Hol(H,p_0)$ are conjugate in $G$, even if $\pi(p_0) \neq \pi(p_1)$.
% \end{enumerate}
% These properties also hold for $\Hol_0$.
% \end{properties}

% \begin{proof}\
% \begin{enumerate}[(1)]
% \item Consider $h_0 \in \Hol(H,p_0)$ and $h_1 \in \Hol(H,p_1)$ defined by $P_c(p_0) = p_0 h_0$ and $P_c(p_1) = p_1 h_1$ for the same curve $c$ on $M$. Then
% \[p_0 h_1 = p_1 h_1 = P_c(p_1) = P_c(p_0g) = P_c(p_0) g = p_0 h_0 g\]
% so $g h_1 = h_0 g$ and $h_1 = g^{-1} h_0 g$.
% \item That $p_1$ is obtained from $p_0$ by parallel transport means $p_1 = P_c(p_0) = p_0 h_0$ for some $h_0 \in \Hol(H,p_0)$. Using (1), we get $\Hol(H,p_1) = h_0^{-1} \Hol(H,p_0) h_0 = \Hol(H,p_0)$.
% \item Let $c$ be a smooth curve connecting $m_0 = \pi(p_0)$ to $m_1 = \pi(p_1)$ in $M$, and let $\overline c$ be the horizontal lift of $c$ with starting point $p_0$. We show $\Hol(H,p_0) = \Hol(H,P_c(p_0))$. Let $h_0 \in \Hol(H,p_0)$, then there is a closed curve $c_0$ on $M$ with starting point $m_0$ such that $P_{c_0}(p_0) = p_0 h_0$. The concatenation $\overline c c_0 c$ is a closed curve on $M$ with starting point $m_1$, so there is a $h_1 \in \Hol(H,P_c(p_0))$ such that $P_{\overline c c_0 c}(P_c(p_0)) = P_c(p_0) h_1$. Now $h_0 = h_1$, since
% \[P_c(p_0) h_1 \!=\! P_{\overline c c_0 c}(P_c(p_0)) \!=\! P_c \circ P_{c_0} \circ P_c^{-1} \circ P_c(p_0) \!=\! P_c(P_{c_0}(p_0)) \!=\! P_c(p_0h_0) \!=\! P_c(p_0) h_0\]
% by $G$--equivariance of $P_c$. By using $\overline c$ instead of $c$, we get the other inclusion. So $\Hol(H,p_0) = \Hol(H,P_c(p_0))$. But $\Hol(H,P_c(p_0))$ is conjugate to $\Hol(H,p_1)$ by (1), so $\Hol(H,p_0)$ and $\Hol(H,p_1)$ are conjugate in $G$.\qedhere
% \end{enumerate}
% \end{proof}

\begin{theorem}
The restricted holonomy group $\Hol_0(H,p_0)$ is a connected Lie subgroup of $G$.
\end{theorem}

\begin{proof}
By definition, $\Hol_0(H,p_0) \subset G$ is a subgroup. We claim that $\Hol_0(H,p_0)$ is connected, more precisely for any $g \in \Hol_0(H,p_0)$ there is a piecewise smooth curve $\widetilde g \colon [0,1] \to G$ with $\widetilde g(0) = e$, $\widetilde g(1) = g$ and $\widetilde g(s) \in \Hol_0(H,p_0)$ for all $s \in [0,1]$. The property $g \in \Hol_0(H,p_0)$ means that there is a piecewise smooth curve $c \colon [0,1] \to M$ with $c(0) = c(1) = m_0$ such that $P_c(p_0) = p_0 g$ and $c$ is contractible as a curve based at $m_0$. There exists a piecewise smooth map
\[H \colon [0,1] \times [0,1] \to M\]
such that
\[H(t,0) = m_0, \quad H(t,1) = c(t), \quad H(0,s) = H(1,s) = m_0 \qquad \forall s,t \in [0,1]\]
For every $s \in [0,1]$,
\[c_s \colon [0,1] \to M \qquad t \mapsto H(t,s)\]
is a piecewise smooth curve in $M$ based at $m_0$. Define $\widetilde g$ by $p_0\widetilde g(s) = P_{c_s}(p_0)$. This is piecewise smooth in $s$. We have $\widetilde g(0) = e$ since $c_0$ is constant and $\widetilde g(1) = g$ since $c_1 = c$. $\widetilde g(s) \in \Hol_0(H,p_0)$ because each $c_s$ is a closed loop based at $m_0$ and is contractible. The proof of the theorem is completed by the following proposition.
\end{proof}

\begin{proposition}
Let $G$ be a Lie group and $H \subset G$ a subgroup with the property that every $g \in H$ can be connected to $e$ by a piecewise smooth curve in $H$. Then $H$ is a Lie subgroup of $G$.
\end{proposition}

%TODO: 11.6. 14.6. 18.6. 21.6. 25.6. 28.6. 2.7. (one proof missing) | 5.7.

%Vorlesung vom 11.6.:

\begin{proof}
Let
\[\mathfrak h = \left\{\dot c(0) \mid c \colon [0,1] \to G\;\text{piecewise smooth},\; c(0) = e,\; \forall t \colon c(t) \in H\right\}.\]
We claim that $\mathfrak h$ is a Lie subalgebra of $\mathfrak g = L(G)$. If $X \in \mathfrak g$ is in $\mathfrak h$, then so is $\lambda X$ by reparametrizing $c$ with $\dot c(0) = E$. If $c_1$ and $c_2$ are two curves with $\dot{c_1}(0) = X \in \mathfrak h$ and $\dot{c_2} = Y \in \mathfrak h$, then, with
\[c \colon [0,1] \to G, \quad t \mapsto c_1(t) c_2(t),\]
we have $\dot c(0) = X+Y$, so $\mathfrak h \in \mathfrak g$ is a linear subspace. It remains to prove that for $X,Y \in \mathfrak h$ we have $[X,Y] \in \mathfrak h$. Let $c_1, c_2$ be as above. Consider $c(t^2) = c_1(t)c_2(t)c_1(-t)c_2(-t)$. Without loss of generalit, we may take $c_1,c_2$ defined on $(-\varepsilon, 1]$ for some small $\varepsilon > 0$. The defining formula for $c$ makes sense for small positive $t$. We have $\dot c(0) = [X,Y]$ and $c$ satisfies $c(t) \in H$ for all $t$, so $[X,Y] \in \mathfrak h$.

Now we define a subbundle $E \subset TG$ by $E_g \coloneqq D\ell_g(\mathfrak h) \subset T_gG$ forall $g \in G$. We prove that $E$ is integrable: Since $\mathfrak h$ is closed under $[-,-]$, $E$ is involutive. By the Frobenius theorem, $E$ is integrable, i.e. there exists an integral submanifold through every point. Let $K \subset G$ be the maximal connected integral submanifold of $E$ through $e \in G$. $K$ is a connected Lie subgroup with Lie algebra $\mathfrak h$.

Next, we show that $H \subset K$. By assumption, every $h \in H$ is connected to $e \in G$ by a piecewise smooth $c \colon [0,1] \to G$ with $c(t) \in H$ for all $t$. Suppose $c$ is smooth. For a fixed $t_0$ define $\overline c(t) = c(t_0)^{-1}c(t) \in H$. We have $\overline c(t_0) = e$ and $\dot{\overline c}(t_0) = D_{c(t_0)}\ell_{c(t_0)^{-1}}(\dot c(t_0)) \in \mathfrak h$, so $\dot c(t_0) = D_e \ell_{c(t_0)}(\dot{\overline c}(t_0)) \in E_{c(t_0)}$ for all $t_0$. Hence $c$ is contained in $K$ and in particular $h = c(1) \in K$. For piecewise smooth $c$ we repeat this argument for each subinterval of $[0,1]$ where $c$ is smooth. So $H \subset K$.

To prove that $K \subset H$, pick a basis $X_1,\dots, X_k$ for $\mathfrak h$. For each $i$ we can choose a curve $c_i$ with $c_i(0)=e$, $c_i(t) \in H \,\forall t$ and $\dot{c_i}(0) = X_i$. Define
\[f \colon (-\varepsilon,\varepsilon)^k \to G, \qquad (t_1,\dots,t_k) \mapsto c_1(t_1) c_2(t_2)\cdots c_k(t_k).\]
Since each $c_i$ is contained in $H$ and $H \subset G$ is a subgroup, $f(t_1,\dots,t_k) \in H$ for all $t_i$ and $f(0,\dots,0) = e$. The linear map $D_{(0,\dots,0)} f$ is an isomorphism between $\mathbb R^k$ and $\mathfrak h$. So, for $\varepsilon > 0$ small enough, $f$ is an immersion. We obtain an immersed submanifold through $e$ which is an integral submanifold for $E$, since $\im (Df)$ is contained in $E$ at every point. The maximal integral submanifold of $E$ through $e$ is $K$, so $\im f \subset K$. In fact, $\im f$ and $K$ have the same dimension, so $\im f$ is an open neighbourhood of $e$ in $K$. Since $\im f \subset H$ by definition of $f$, $K \subset H$ is true locally near $e$. Since $K$ is a connected Lie subgroup, each element of $K$ is a product of finitely many elements of $\im f \subset H$. since $H$ is also a subgroup, $H \subset K$.
\end{proof}

\subsection{Curvature}

\begin{definition}
The curvature of $H$ is the following $\mathfrak g$--valued 2--form on $P$:
\[\Omega(X,Y) = \dd \omega(\mathscr H X, \mathscr H Y) \qquad \forall X,Y \in T_pP, p \in P\]
\end{definition}

For $X,Y \in T_pP$, $p \in P$, we have
\begin{align*}
(r_g^* \Omega)(X,Y) &= \Omega(Dr_g(X),Dr_g(Y)) = \dd \omega(\mathscr H Dr_g(X), \mathscr H Dr_g(Y)) = \\
&= \dd \omega (Dr_g(\mathscr HX), Dr_g(\mathscr HY)) = (r_g^* \dd \omega)(\mathscr H X, \mathscr H Y) = \\
&= \dd (r_g^*\omega) (\mathscr HX, \mathscr HY) = (\dd \Ad_{g^{-1}}\omega)(\mathscr HX, \mathscr HY) = \\
&= \Ad_{g^{-1}} \dd \omega(\mathscr HX, \mathscr HY) = \Ad_{g^{-1}} \Omega(X,Y),
\end{align*}
so $r_g^*\Omega = \Ad_{g^{-1}}\Omega$.

\begin{proposition}[Structure equation]
$\Omega(X,Y) = \dd \omega(X,Y) + [\omega (X), \omega (Y)]$.
\end{proposition}

%TODO: überarbeiten, Lie ableitungen statt derivationen? Funktioniert das mit Tangentialvektoren?

\begin{proof}
Both sides of the claim are bilinear and skew--symmetric, so we may assume that each $X,Y$ is either in $V$ or in $H$. In the case $X,Y \in H$, we have $\omega(X) = \omega(Y) = 0$, and $\Omega(X,Y) = \dd \omega(X,Y)$ by definition of $\Omega$. If instead $X,Y \in V_p$ with $p \in P$, we may assume $X = A_p^*$ and $Y = B_p^*$ for $A,B \in \mathfrak g$. Since $\mathscr HX = \mathscr HY = 0$, we have $\Omega(X,Y) = 0$. The structure equation then holds since
\begin{align*}
\dd \omega(A^*,B^*) &= A^*(\omega(B^*)) - B^*(\omega(A^*)) - \omega([A^*,B^*]) = \\
&= A^*(B) - B^*(A) - \omega([A,B]^*) = 0 - 0 - [A,B] = \\
&= -[\omega(A^*),\omega(B^*)]
\end{align*}
so in particular $\dd \omega(X,Y) + [\omega(X),\omega(Y)] = 0 = \Omega(X,Y)$. In the third case, $X_p \in H_p$ and $Y_p \in V$, where $X$ is a vector field and $Y_p = B_p^*$ for some $B \in \mathfrak g$. Again,
\[\Omega(X,Y) = \dd\omega(\mathscr HX, \mathscr HY) = \dd \omega(X,0) = 0.\]
Then we have
\begin{align*}
\dd \omega(X,Y) &= X(\omega(Y)) - Y(\omega(X)) - \omega([X,Y]) = \\
&= X(B) - 0 - \omega([X,B^*]) = -\omega([X,B^*]) = 0
\end{align*}
by the following lemma. Since also $[\omega(X),\omega(Y)] = [0,B] = 0$, this completes the proof of case 3 and of the proposition.
\end{proof}

\begin{lemma}
Let $X$ be a vector field with $X_p \in H_p$. Then $[X,B_p^*] \in H_p$.
\end{lemma}

\begin{proof}
For any two vector fields $X$ and $Y$ we have
\[[X,Y] = L_X Y = - \frac{\dd}{\dd t} D\varphi_t(Y) \Big|_{t=0}\]
With this we have for $X$ horizontal and $Y = B^*_p$:
\[[X,B^*_p] = -[B^*_p,X] = \frac{\dd}{\dd t} Dr_{\exp(tB)}(X) \Big|_{t=0} \in H\]
since $H$ is invariant under right $G$--action.
\end{proof}

\begin{definition}
Let $\alpha$ be a differential $k$--form on $P$. Then $D$ is defined by
\[D\alpha = \dd \alpha \circ \mathscr H,\quad\text{i.e.}\;D\alpha(X_1,\dots,X_{k+1}) = \dd \alpha(\mathscr H X_1,\dots,\mathscr H X_{k+1}).\]
$D$ depends on the connection $H$ and is called the {\em covariant derivative} defined by $H$.
\end{definition}

\begin{proposition}[Bianchi identity]
$D\Omega = 0$, in other words $\dd \Omega(\mathscr HX, \mathscr HY, \mathscr HZ) = 0$ for all $X,Y,Z \in T_pP$.
\end{proposition}

\begin{proof}
Let $X,Y,Z \in H_p$, then
\begin{align*}
\dd\Omega(X,Y,Z) &= L_X(\Omega(Y,Z)) + L_Y(\Omega(Z,X)) + L_Z(\Omega(X,Y)) \\
&- \Omega([X,Y],Z) - \Omega([Z,X],Y) - \Omega([Y,Z],X) \\
&= L_X(\dd\omega(Y,Z)) + L_Y(\dd\omega(Z,X)) + L_Z(\dd\omega(X,Y) \\
&- \dd\omega([X,Y],Z) - \dd\omega([Z,X],Y) - \dd\omega([Y,Z],X) \\
&= \dd\dd\omega(X,Y,Z) = 0 \qedhere
\end{align*}
\end{proof}

\begin{proposition}
Let $P \xrightarrow{\pi} M$ be a principal $G$--bundle an $H$ a connection of $P$ with curvature $\Omega$. The following conditions are equivalent:
\begin{enumerate}[(1)]
\item $\Omega = 0$.
\item $H$ is involutive, i.e. closed under $[-,-]$.
\item $H$ is integrable.
\end{enumerate}
\end{proposition}

\begin{proof}
Conditions (2) and (3) are equivalent by the Frobenius theorem. Assume $X,Y \in H$. Then
\[\Omega(X,Y) = \dd \omega(X, Y) = L_X(\omega(Y)) - L_Y(\omega(X)) - \omega([X,Y]) = - \omega([X,Y]),\]
so $\Omega = 0$ iff $[X,Y] \in H$ whenever $X,Y \in H$. This shows the equivalence of (1) and (2).
\end{proof}

Let $L \subset P$ be the maximal connected integral submanifold for $H$ with $p_0 \in L$. If $c$ is any loop in $M$ based at $m_0 = \pi(p_0)$, then the horizontal lift $\overline c$ of $c$ with initial value $p_0$ is contained in $L$. Note that $L$ is a covering space of $M$.
\[\pi|_L \colon L \to M \qquad D(\pi|_L) \colon T_pL = H_p \to T_{\pi(p)}M\]
$D(\pi|_L)$ is an isomorphism, so $\pi|_L$ is a local diffeomorphism.

\begin{definition}
The connection $H = \ker \omega$ is {\em flat} if $\Omega = 0$.
\end{definition}

Now let $H$ be arbitrary, not neccessarily flat. For $p_0 \in P$ let
\[H(p_0) = \{p \in P \mid\;\text{$p$ is obtained from $p_0$ by parallel transport}\}\]

Note that $\Hol_0(H,p_0)$ is a connected Lie subgroup of $G$. Moreover, $\Hol_0(H,p_0)$ is the connected compontent of $e$ in $\Hol(H,p_0)$. $\Hol(H,p_0)$ has at most countable many connected components, each diffeomorphis to $\Hol_0(H,p_0)$. So $\Hol(H,p_0) \subset G$ is a Lie subgroup as well.

\begin{proposition}
If $M$ is connected, then $H(p_0)$ is a principal $\Hol(H,p_0)$--bundle.
\end{proposition}

\begin{proof}
$H(p_0)$ is a subset of $P$, so we have a projection $\pi \colon H(p_0) \to M$ by restricting $\pi \colon P \to M$. Consider
\[H(p_0) \cap \pi^{-1}(m_0) = \{p_0 g(c) \mid \text{$c$ is a closed loop based at $m_0$}\}.\]
Since $g(c) \in \Hol(H,p_0)$, this shows that $\Hol(H,p_0)$ acts simply transitively on the fiber of $H(p_0)$ over $m_0$. $G$ acts on $P$ on the right, and we restrict this action to the subgroup $\Hol(H,p_0) \in G$.

We claim that the restricted action maps $H(p_0)$ to itself. To prove this, let $p \in H(p_0)$ and $\overline c$ be a horizontal curve connecting $p_0$ to $p$. Assume $g \in \Hol(H,p_0)$. Then $r_g \circ \overline c$ is a horizontal curve connecting $p_0 g$ to $pg$. Because $g \in \Hol(H,p_0)$, there is a horizontal curve from $p_0$ to $p_0g$. Concatenating with $r_g \circ \overline c$ gives a piecewise smooth horizontal curve from $p_0$ to $pg$, so $pg \in H(p_0)$.

Next, we want to show that the action of $\Hol(H,p_0)$ on $H(p_0)$ is simply transitive on every fiber of $H(p_0) \xrightarrow{\pi} M$. Let $p \in H(p_0)$ and $\overline c$ be a horizontal curve from $p_0$ to $p$. For every $g \in \Hol(H,p_0)$, $pg \in H(p_0)$ by the previous paragraph. Suppose $p' \in H(p_0) \cap \pi^{-1}(\pi(p))$. There exists a horizontal curve $\overline c'$ from $p_0$ to $p'$. Since both $p$ and $p'$ are connected to $p_0$ by horizontal curves, there is a horizontal curve from $p$ to $p'$. So there exists $g' \in \Hol(H,p)$ such that $p' = pg'$. Since there is a horizontal curve between $p_0$ and $p$, we have $\Hol(H,p)= \Hol(H,p_0)$. We have shown that $\Hol(H,p_0)$ acts transitively on $H(p_0) \cap \pi^{-1}(\pi(p))$ for all $p \in H(p_0)$. Since the action of $G$ on $P$ has trivial stabilizers, the restricted action of $\Hol(H,p_0)$ on $H(p_0)$ has trivial stabilizers.

By connectedness of $M$, $H(p_0)$ intersects every fiber of $P$. Take local trivializations for $P$:
\[\psi \colon \pi^{-1}(U) \to U \times G\]
Since $\Hol_0(H,p_0) \subset G$ is a Lie subgroup, $U \times \Hol_0(H,p_0)$ is a smooth manifold and so is $U \times \Hol(H,p_0)$. Using the restriction of $\psi$ to $\pi^{-1}(U) \cap H(p_0)$, we can identify $\pi^{-1}(U) \cap H(p_0)$ with $U \times \Hol(H,p_0)$. $H(p_0)$ has a unique smooth structure for which these identifications are diffeomorphisms. With respect to this smooth structure on $H(p_0)$, the action of $\Hol(H,p_0)$ on $H(p_0)$ discussed before is smooth.
\end{proof}

\begin{definition}
Let $p_0 \in P$ and $H$ be a connection on $P$. The principal $\Hol(H,p_0)$--bundle $H(p_0)$ is the {\em holonomy bundle} of $H$ through $p_0$.
\end{definition}

\begin{remark}
Consider $H(p_0) \hookrightarrow P$. This is a homomorphism of principal bundles where the corresponding homomorphism of Lie groups is $\Hol(H,p_0) \hookrightarrow G$.
\end{remark}

A reduction of the structure group of $P$ to $H$ is a principal $H$--bundle $Q \to M$ together with a homomorphism $Q \to P$ with respect to the inclusion $H \hookrightarrow G$ such that
\[\xymatrix{
Q \ar[r]\ar[d] & P\ar[ld] \\
M
}\]
commutes.

The existence of such a reduction is equivalent to a section of the bundle
\[\xymatrix{
G/H \ar[r] & P/H \ar[d] \\
& H
}\]
Extending this definition slightly, we have proved that if $H$ is a connection of $P$ and $Q \coloneqq H(p_0)$ is the holonomy bundle through a basepoint $p_0 \in P$, then $Q \hookrightarrow P$ defines a reduction of the structure group of $P$ to $\Hol(H,p_0) \subset G$.

Suppose $Q$ is a reduction of the structure group $G$ of $P$ to a subgroup $G' \subset G$. Let $H$ be a connection on $Q$.

\begin{claim}
$H$ extends uniquely to $P$.
\end{claim}

\begin{proof}
For $p \in P$ consider some $p' \in Q \cap \pi^{-1}(\pi(p))$. The connection $H$ on $Q$ defines $H_{p'} \subset T_{p'}Q$. $H_{p'}$ is a complement to the vertical subspace at $p'$ in $Q$ and also in $P$. There exists $g \in g$ such that $p' g = p$. Define $H_p \coloneqq Dr_g(H_{p'})$. This is a horizontal subspace at $p$ in $P$.

This defines a connection on $P$! $H$ is clearly horizontal. It is also smooth, so we only have to check $G$--invariance. Let $\overline p \in \pi^{-1}(\pi(p))$. There exists $\overline g \in G$ such that $\overline p = p \overline g$. We have to check that
\[Dr_{\overline g}(H_p) = H_{\overline p}\]
There exist $p' \in Q \cap \pi^{-1}(\pi(p))$ and $g' \in G$ such that $p' = pg'$. Then $p = p' g'^{-1}$ and $\overline p = p \overline g = p' g'^{-1} \overline g$, so
\[H_{\overline p} = Dr_{g'^{-1}\overline g}(H_{p'}) = Dr_{\overline g}(Dr_{g'^{-1}}(H_{p'})) = Dr_{\overline g}(H_p).\]
Suppose $\overline H$ is a connection on $P$ which restricts to the given $H$ on $Q$, i.e. $\overline{H_p} = H_p$ if $p \in Q$. Then $\overline H$ is the connection defined above, because every fiber of $P$ contains a point in $Q$ and the connection $\overline H$ on $P$ is completely determined by $G$--invariance and $\overline H$ at a single point in every fiber.
\end{proof}

\begin{definition}
Let $\overline H$ be a connection on a principal $G$--bundle $P \xrightarrow{\pi} M$ and $G' \subset G$ a Lie subgroup. $\overline H$ is {\em reducible to the subgroup $G'$} if there is a reduction $Q \to M$ of $P$ to a principal $G'$--bundle and a connection $H$ on $Q$ whose extension to $P$ is $\overline H$.
\end{definition}

\begin{proposition}
Every connection $\overline H$ on $P$ is reducible to $\Hol(\overline H, p_0)$.
\end{proposition}

\begin{proof}
Set $Q = H(p_0)$ and $H$ the restriction of $\overline H$ to $Q$.
\end{proof}

\begin{theorem}[Ambrose--Singer]
Le $P \xrightarrow{\pi} M$ be a principal $G$--bundle with a connection $H = \ker \omega$ and curvature $\Omega$, where $M$ is connected, and let $p \in P$. Define
\[\mathfrak g' \coloneqq \{\Omega(X_p,Y_p) \mid X_p, Y_p \in T_pP\} \subset \mathfrak g\]
Then $\mathfrak g'$ is the Lie algebra of $\Hol(H,p)$.
\end{theorem}

\begin{proof}
We replace $P$ by the hononomy bundle $Q = H(p)$ through $P$. We work on $Q$ whose structure group is $G' = \Hol(H,p)$. We have to prove that the values of $\Omega$ span the full Lie algebra of $G'$.

Take a basis $A_1,\dots,A_k$ for the subspace $\mathfrak g'$. The $A_i$ induce fundamental vector fields $A_i^*$ on $Q$. Let $X_1,\dots,X_n$ be a basis for $H_p$ and $X_1^*, \dots, X_n^*$ extensions of the $X_i$ to horizontal vector fields. Let $S \subset T_pQ$ be the span of the $(A_i^*)_p$ and $(X_j^*)_p = X_j$. We extend $S$ to a smooth distribution on $Q$ by setting
\[S_q \coloneqq \Span\{(A_1^*)_q,\dots,(A_k^*)_q\} \oplus H_q.\]

We claim that $S$ is integrable as a distribution on $Q$. To check that $S$ is closed under $[-,-]$, we can check at $p$: First, $[A_i^*, A_j^*] = [A_i,A_j]^*$, so we need to check that $\mathfrak g' = \Span \{A_1,\dots,A_k\}$ is a Lie subalgebra. Second, $[A_i^*, X_j^*]$ is horizontal by a previous Lemma, so it is in $H \subset S$. And third,
\[[X_i^*,X_j^*] = \mathscr H([X_i^*,X_j^*]) + \mathscr V([X_i^*,X_j^*]).\]
Since the horizontal part is in $S$, we only need to prove that $\mathscr V([X_i^*,X_j^*]) \in S$. Since the fundamental vector fields span the vertical subspace, we can write
\[\mathscr V([X_i^*,X_j^*](p)) = B_p^*\quad\text{for some $B \in \mathfrak g$,}\]
so $\omega(\mathscr V([X_i^*,X_j^*](p))) = B$. Then
\begin{align*}
\Omega(X_i^*(p),X_j^*(p)) &= \dd \omega(X_i^*(p),X_j^*(p)) = L_{X_i^*}(\omega(X_j^*)) - L_{X_j^*}(\omega(X_i^*)) - \omega([X_i^*,X_j^*]) = \\
&= - \omega([X_i^*,X_j^*](p)) = - \omega(\mathscr V([X_i^*,X_j^*](p))) = -B
\end{align*}
so $B \in \Span\{A_1,\dots,A_k\}$. The equation $\mathscr V([X_i^*,X_j^*](p)) = B_p^*$ then shows that $\mathscr V([X_i^*,X_j^*](p)) \in \Span\{(A_1^*)_p,\dots,(A_k^*)_p\} \subset S$.

We have proved that $S$ is closed under $[-,-]$, so it is integral by the Frobenius theorem.

Let $L$ be the maximal connected integral submanifold of $S$ through the point $p \in Q$. Every point in $Q$ can be reached from $p$ by parallel transport of $p$. The correspoinding horizontal curve is tangent to $H$ and therefore tangent to $S$. Therefore the whole curve is contained in $L$ and so $L = Q$. Since
\[k + \dim M = \rank S = \dim L = \dim Q = \dim G' + \dim M,\]
we have $\dim \mathfrak g' = k = \dim G' = \dim L(G')$, so $\mathfrak g' = L(G')$.
\end{proof}

By definition, $H$ is $G$--invariant. So, since $r_g^* \omega = \Ad_{g^{-1}}\omega$, we also have $r_g^*\Omega = \Ad_{g^{-1}}\Omega$, i.e. $\Omega(Dr_g(X), Dr_g(Y)) = \Ad_{g^{-1}}\Omega(X,Y)$ for all $X,Y \in T_pP$. Since $\Ad_{g^{-1}}$ is a linear isomorphism, there is as much curvature at $pg$ as there is at $p$ for all $g \in G$.

With $E = P \times_{\Ad} \mathfrak g = P \times \mathfrak g / \simeq$ where $(p,A) \simeq (pg, \Ad_{g^{-1}}A)$ for all $g \in G$, the bundle $\pi_E \colon E \to M, [(p,A)] \to \pi(p)$ is a vector bundle on $M$ whose fiber at every point $m \in M$ is isomorphic to $\mathfrak g$. We interpret the curvature $\Omega$ of a connection $H$ on $P$ as a 2--form on $M$ with values in $E$, i.e.
\[\Omega \in \Omega^2(M) \otimes E = \Gamma(\Lambda^2 T^* M \otimes E)\]
For $X,Y \in T_mM$ we should have $\Omega(X,Y) \in E_m = \pi_E^{-1}(m)$. Given $X,Y \in T_mM$, choose preimages $\widetilde X, \widetilde Y \in T_pP$ with $\pi(p) = m$ and $D_p\pi(\widetilde X) = X$, $D_p\pi(\widetilde Y) = Y$. Then
\[\Omega(X,Y) = [(p,\Omega(\widetilde X, \widetilde Y))] \in E_{\pi(p)} = E_m\]
If we replace $\widetilde X$ by $\widetilde X' \in T_pP$ with $D_p\pi(\widetilde X') = X$, then $\widetilde X - \widetilde X' \in \ker D_p\pi = V_p$, so
\[\Omega(\widetilde X', \widetilde Y) = \Omega(\widetilde X' - \widetilde X, \widetilde Y) + \Omega(\widetilde X, \widetilde Y) = \Omega(\widetilde X, \widetilde Y)\]
This shows that $\Omega(X,Y)$ is independent of the choice of $\widetilde X$ at $p$ and similarily for $\widetilde Y$ at $p$.

Any $p' \in P$ with $\pi(p') = m$ is of the form $p' = pg$ for some $g \in G$. Then
\[(p, \Omega(\widetilde X, \widetilde Y)) \simeq (pg, \Ad_{g^{-1}}\Omega(\widetilde X, \widetilde Y)) = (p', r_g^*\Omega(\widetilde X, \widetilde Y)) = (p',\Omega(Dr_g(\widetilde X), Dr_g(\widetilde Y)))\]
and $D\pi(Dr_g(\widetilde X)) = X$, $D\pi(Dr_g(\widetilde Y)) = Y$. This shows that $\Omega(X,Y) \coloneqq [(p,\Omega(\widetilde X, \widetilde Y))]$ is well--defined as an element of $E_m$.

\subsection{Global gauge transformations}

\begin{definition}
An {\em automorphism} (or {\em global gauge transformation}) of $P$ is a diffeomorphism $\phi \colon P \to P$ such that $\pi \circ \phi = \pi$ ad $\phi(pg) = \phi(p)g$ for all $g \in G$.
\end{definition}

Every such $\phi$ has an inverse $\phi^{-1}$ which is a diffeomorphism. Moreover, $\pi \circ \phi^{-1} = \pi$ and $\phi^{-1}(pg) = \phi^{-1}(p)g$ for all $g \in G$. So the automorphisms of $P$ form a group.

\begin{definition}
The group of automorphisms of $P$ is called the {\em gauge group} $\mathscr G$ of $P$.
\end{definition}

\begin{proposition}
$\mathscr G$ is the space of sections of the bundle $F \to M$ with fiber $G$ associated to $P$ by the conjugation action of $G$ on itself. %Make more precise
\end{proposition}

\begin{proof}
$F$ is defined by $F = P \times G/ \simeq$ where $(p,h) \simeq (pg,g^{-1}hg)$ for all $g \in G$. Let $\phi \in \mathscr G$. Then $\phi(p) = p u(p)$ for some smooth $u \colon P \to G$. The definition of automorphisms gives
\[(pg)u(pg) = pu(p)g \Rightarrow gu(pg) = u(p)g \Rightarrow u(pg) = g^{-1}u(p)g\]
Define a section $s \colon M \to F$ by $s(m) = [(p,u(p))]$ for any $p \in \pi^{-1}(m)$. Since
\[(p,u(p)) \simeq (pg, g^{-1}u(p)g) = (pg, u(pg)),\]
$s$ is well--defined and so it is a smooth section of $F$.

Conversely, suppose $s \colon M \to F$ is a smooth section of $F$. Define $u \colon P \to G$ by $p \mapsto g$ if $[(p,g)] = s(\pi(p))$. Then $\phi(p) = pu(p)$ is a gauge transformation of $P$.
\end{proof}

The gauge group $\mathscr G$ acts on connections. If $\omega$ is a connection 1--from defining a connection $H$ on $P$, then $\phi^*\omega$ is also a connection 1--form defining the pulled--back connection:
\[(\phi^*H)_p = (D_p\phi)^{-1} H_{\phi(p)}\]
For every $H$ we can think of its curvature $\Omega$ as a section of $\Lambda^2 T^*M \otimes E$, equivalently a 2--form on $M$ with values in $E$, where $E = P \times_{\Ad} \mathfrak g$. The action of $\mathscr G$ on $P$ by automorphisms induces an action on $E$:
\[\mathscr G \times E \to E, \quad (\phi, [p,A]) \mapsto [\phi(p),A)]\]
If $[q,B] = [p,A]$, then there exists $g \in G$ such that $q = pg$ and $B = \Ad_{g^{-1}}(A)$, so
\[[\phi(q), B] = [\phi(pg), B] = [\phi(p), \Ad_g (B)] = [\phi(p), A].\]
This shows that the action of $\mathscr G$ on $E$ is well--defined. $\phi$ maps $\phi^*H$ to $H$ and so it maps the curvature $\widetilde \Omega$ of $\phi^*H$ to the curvature $\Omega$ of $H$, i.e.
\[\phi(\widetilde \Omega(X,Y)) = \Omega(X,Y) \qquad \forall X,Y \in T_mM\]
Choose$p \in \pi^{-1}(m)$ and write $\widetilde \Omega(X,Y) = [p,A]$ for some $A \in \mathfrak g$. Then
\[\phi\widetilde\Omega(X,Y) = [\phi(p),A] = [pu(p),A] = [p, \Ad_{u(p)}(A)]\]
So if $A \in \mathfrak g$ represents the curvature of $\phi^*H$, then $\Ad_{u}(A)$ represents the curvature of $H$ itself. If $\phi(p) = pu(p)$, then the curvature $\widetilde\Omega$ of $\phi^*H$ is $\widetilde \Omega = \Ad_{u^{-1}}\Omega$ where $\Omega$ is the curvature of $H$. (The same formula holds for the curvature as a 2--form).

\begin{corollary}
If $\phi \in \mathscr G$ and $H$ is a connection on $P$, then $\phi^*H$ is flat if and only if $H$ is flat.
\end{corollary}

\begin{definition}
Two connections $H_1, H_2$ on $P$ are called gauge equivalent if there is a $\phi \in \mathscr G$ with $\phi^*H_1 = H_2$.
\end{definition}

The corollary says that if $H_1, H_2$ are gauge equivalent, then $H_1$ is flat if and only if $H_2$ is flat.

Let $\langle - , - \rangle$ be a Riemannian metric on $M$. If $M$ is oriented, then $\langle - , - \rangle$ induces a volume form $\dvol$ on $M$ characterized by
\[\dvol(e_1,\dots,e_n) = 1\]
if $e_1,\dots,e_n$ is a positively oriented orthonormal basis for $(T_mM, \langle -, - \rangle)$. Assume that $\mathfrak g$ is equipped with an $\Ad$--invariant, positive definite scalar product. $\langle -, - \rangle$ together with the $\Ad$--invariant scalar product in $\mathfrak g$ induces a smooth fiber--wise metric on $\Lambda^2 T^*M \otimes E$.

\begin{definition}
If $H$ is a connection on $P \to M$ with curvature $\Omega$, where $M$ is an oriented compact manifold, we define the {\em Yang--Mills--functional}
\[\YM(H) \coloneqq \int_M \|\Omega\|^2 \,\dvol\]
\end{definition}

\begin{lemma}
$\YM(\phi^*H) = \YM(H)$, i.e. $\YM$ is $\mathscr G$--invariant.
\end{lemma}

\begin{proof}
Let $\widetilde \Omega$ be the curvature of $\phi^*H$. Since the scalar product on the fiber of $E$ is $\Ad$--invariant, we have
\[\|\widetilde\Omega\|^2 = \|\Omega\|^2\qedhere\]
\end{proof}

\begin{remark}
We have $\YM(H) \geq 0$ with equality if and only if $H$ is flat.
\end{remark}

\begin{theorem}
There is a 1:1 correspondence between the flat connections on all possible principal $G$--bundles $P \to M$ up to gauge equivalence and the set $\Hom(\pi_1(M),G) / G$ where $G$ acts on homomorphisms by conjugation.
\end{theorem}

\begin{lemma}
Let $P \xrightarrow{\pi} M$ be a principal $G$--bundle and $H$ a connection on $P$. If $H$ is flat, then $P_c$ (with respect to $H$) depends only on the homotopy class of $c$.
\end{lemma}

\begin{proof}
If $H$ is flat, then by the Ambrose--Singer theorem, $\Hol_0$ is trivial, so $P_c$ is the identity if $c$ is a closed loop which is null--homotopic through closed loops.

Let $c_1,c_2$ be paths from $m_1$ to $m_2$. Then $c_1$ travelled backwards followed by $c_2$ is a closed loop based at $m_2$. Parallel transport along this loop maps $P_{c_1}(p)$ to $P_{c_2}(p)$ for all $p \in \pi^{-1}(m_2)$. If $c_1$ and $c_2$ are homotopic with fixed endpoints, then this loop at $m_2$ is null--homotopic as a loop. Since $H$ is flat, this implies $P_{c_1}(p) = P_{c_2}(p)$.
\end{proof}

If $H$ is a flat connection on $P \xrightarrow{\pi} M$ and $p_0 \in \pi^{-1}(m_0)$, define the {\em holonomy representation}
\[\hol \colon \pi_i(M,m_0) \to G, \quad [\gamma] \mapsto g(\gamma)^{-1}\]
where $P_\gamma(p_0) = p_0 g(\gamma)$. This is well--defined by the lemma and is a group homomorphism $g(\gamma)g(\gamma') = g(\gamma'\gamma)$ since
\[p_0 g(\gamma)^{-1}g(\gamma')^{-1} = P_{\gamma'}(P_\gamma(p_0)) = P_{\gamma \gamma'}(p_0) = p_0 g(\gamma \gamma')^{-1}.\]
Suppose we use $p_1 \in \pi^{-1}(m_0)$ instead of $p_0$ to define $\hol$. Then we get $g_1$ defined by $P_\gamma(p_1) = p_1g_1(\gamma)$. There exists a unique $h \in G$ such that $p_1 = p_0h$, so
\[p_0 g(\gamma) h = P_\gamma(p_0)h = P_\gamma(p_0h) = p_0 h g_1(\gamma).\]
This implies $g_1(\gamma)^{-1} = h^{-1}g(\gamma)^{-1} h$, so the conjugacy class of $\hol$ is independent of the choice of basepoint $p_0 \in \pi^{-1}(m_0)$.

\begin{lemma}
If $H_1$, $H_2$ are gauge equivalent flat connections on $P \xrightarrow{\pi} M$, then their holonomy representations are conjugate.
\end{lemma}

\begin{proof}
Let $\phi \colon P \to P$ be a gauge transformation with $D\phi(H_1) = H_2$. Pick a basepoint $p_1 \in \pi^{-1}(m_0)$ to define $\hol_1$, the holonomy representation of $H_1$. To define $\hol_2$, the holonomy representation of $H_2$, use the basepoint $p_2 = \phi(p_1)$. Then $\hol_1([\gamma]) = g_1(\gamma)^{-1}$ and $\hol_2([\gamma]) = g_2(\gamma)^{-1}$ where $g_1$ and $g_2$ are defined by $P_\gamma^1(p_1) = p_1\gamma_1(\gamma)$ and $P_\gamma^2 = p_2\gamma_2(\gamma)$ where $P^1$ is the parallel transport with respect to $H_1$ and $P^2$ is the parallel transport with respect to $H_2$. We have
\[p_2 g_2(\gamma) = P_\gamma^2(p_2) = \phi(p_1g_1(\gamma)) = \phi(p_1)g_1(\gamma) = p_2 g_1(\gamma)\]
so $g_1(\gamma) = g_2(\gamma)$, meaning that $\hol_1 = \hol_2$ if we use $p_1$ respectively $p_2 = \phi(p_1)$ as basepoints for the definition of $\hol$.
\end{proof}

Let $\rho \colon \pi_1(M,m_0) \to G$ be a representation. Consider the universal covering $\widetilde M \to M$. Definie $P \colon \widetilde M \times_\rho G = \widetilde M \times G / \simeq$ where
\[(x,g) \simeq (\gamma x, \rho(\gamma) g) \qquad \forall \gamma \in \pi_1(M,m_0)\]
$P$ is the quotient of $\widetilde M \times G$ by the action of $\pi_1(M,m_0)$ given by
\[\pi_1(M,m_0) \times \widetilde M \times G \to \widetilde M \times G \quad (\gamma,x,g) \mapsto (\gamma x, \rho(\gamma) g)\]
The equivalence class of $(x,g)$ is denoted by $[x,g]$. Define $\pi \colon P \to M, [x,g] \mapsto [x]$ where $\widetilde M \to M, x \mapsto [x]$ comes from the covering. This is well--defined and smooth. $G$ acts on $P$ on the right as follows:
\[P \times G \to P, \quad ([x,h],g) \mapsto [x,hg] \eqqcolon [x,h]g\]
For any $\gamma \in \pi_1(M,m_0)$, we have $[x,h] = [\gamma x, \rho(\gamma)h]$. Since also
\[[\gamma x, \rho(\gamma) h]g = [\gamma x, \rho(\gamma)hg] = [x,hg]\]
this right $G$--action on $P$ is well--defined.

The $P$ constructed in this way from $\rho$ is a principal $G$--bundle over $M$. $\widetilde M \times G$ has a natural connection whose horizontal subspaces are tangent spaces to $\widetilde M \times \{g\}$. This distribution is tautologically integrable, so the connection is flat. $\pi_1(M,m_0)$ acts on $\widetilde M \times G$ preserving the flat product connection, which therefore descends to $P$ as a flat connection.

Suppose $\overline \rho$ is defined by $\overline \rho(\gamma) = \alpha \rho(\gamma) \alpha^{-1}$ for all $\gamma$ and some fixed $\alpha \in G$. Then $\overline \rho$ gives rise to a principal $G$--bundle $\overline P$ with a flat connection $\overline H$.

\begin{lemma}
There is an isomorphism of principal bundles $\phi \colon \overline P \to P$ with $D\phi(\overline H) = H$.
\end{lemma}

\begin{proof}
We have $\overline P = \widetilde M \times G / \simeq$ with
\[(x,g) \simeq (\gamma x, \overline \rho(\gamma)g) = (\gamma x, \alpha \rho(\gamma) \alpha^{-1} g) \qquad \forall \gamma \in \pi_1(M,m_0)\]
Now define
\[\phi \colon \overline P \to P \quad [x,g]_{\overline \rho} \mapsto [x, \alpha^{-1}g]_\rho\]
This $\phi$ is well--defined since $[x,g]_{\overline \rho} = [\gamma x, \alpha \rho(\gamma) \alpha^{-1}g]_{\overline \rho}$ is mapped to $[x,\alpha^{-1}g]_\rho = [\gamma x, \rho(\gamma) \alpha^{-1} g]_\rho$. $\phi$ is also smooth. We have $\pi_P \circ \phi = \pi_{\overline P}$ and
\[\phi([x,g]_{\overline \rho} h) = \phi([x,gh]_{\overline \rho}) = [x,\alpha^{-1}gh]_\rho = [x, \alpha^{-1}g]_\rho h = \phi([x,g]_{\overline \rho})h.\]
This shows that $\phi$ is an automorphism of principal $G$--bundles. $\phi$ preserves the local product structures in which $H$, $\overline H$ are given by the tangent spaces to the first factor in $U \times G$, $U \subset M$ open. So $D\phi(\overline H) = H$.
\end{proof}

% TODO: picture of bijection

For the theorem, it remains to prove that the composition of these two maps. in both directions is the identity. Start with a representation $\rho \colon \pi_1(M,m_0) \to G$ and consider the corresponding principal $G$--bundle $P$ and the flat connection $H$ on $P$. Choose a basepoint $x_0 \in \overline M$ with $[x_0] = m_0$. Let $\gamma$ be a loop in $M$ based at $m_0$ Then $\gamma$ has a unique lift $\widetilde \gamma$ to $\overline M$ with starting point $x_0$ and endpoint $[\gamma]x_0$.

Define $\overline \gamma(t) \coloneqq [\widetilde \gamma(t),e]$, where $e \in G$ is the neutral element. If $\pi \colon P \to M$ is the projection, then
\[\pi(\overline \gamma) = \pi([\widetilde \gamma(t),e]) = [\widetilde \gamma(t)] = \gamma(t)\]
so $\overline \gamma$ is a lift of $\gamma$ from $M$ to $P$. The starting point of $\overline \gamma$ is $[x_0,e]$. This lift is horizontal for the flat connection $H$ because the curve $(\widetilde \gamma(t),e)$ in $\widetilde M \times G$ has tangent vector tangent to the first factors.

We use $p_0 = [x_0, e]$ as a basepoint in $P$ to define $\hol$.
\[\hol \colon \pi_1(M,m_0) \to G, \quad [\gamma] \mapsto g(\gamma)^{-1}\]
where $P_\gamma(p_0) = p_0 g(\gamma)$. Since $\overline \gamma$ is the unique horizontal lift of $\gamma$,
\[p_0 g(\gamma) = P_\gamma(p_0) = \overline \gamma(1) = [\widetilde \gamma(1), e] = [[\gamma]x_0,e] = [x_0, \rho([\gamma])^{-1}] = [x_0,e]\rho([\gamma])^{-1} = p_0 g(\gamma)^{-1}\]
so $\hol = \rho$.

Finally, start with a flat connection $H$ on some principal $G$--bundle $P \xrightarrow{\pi} M$. Fix $p_0 \in \pi^{-1}(m_0)$ and define $\hol \colon \pi_1(M,m_0) \to G$ using the basepoint $p_0$. Define $\overline P$ by $\widetilde M \times G / \simeq$ where $(x,g) \simeq (\gamma x, \hol(\gamma)g)$ for all $\gamma \in \pi_1(M,m_0)$. $\overline P$ has an obvious flat connection $\overline H$. We need to find an isomorphism $\phi \colon P \to \overline P$ with $D\phi(H) = \overline H$.

Let $H(p)$ be the holonomy bundle of $p \in P$. So $H(p) = \widetilde M / \Gamma$ where $\Gamma \subset \pi_1(M,m_0)$ is a subgroup. In fact, $\Gamma = \ker (\hol)$. Define
\[\phi \colon H(p) \to \overline P, \quad [x] \mapsto [x,e]\]
If $\gamma \in \Gamma$, then $[x] = [\gamma x]$. Since
\[[\gamma x, e] = [x,\hol(\gamma)^{-1}] = [x,e],\]
$\phi$ is well--defined and smooth. If $q \in P \setminus H(p)$, there exists $g$ such that $qg \in H(p)$. Define $\phi(q) \coloneqq \phi(qg)g^{-1}$. We will leave out the check that this is well--defined. This $\phi$ defined on all of $P$ is then an isomorphism $P \to \overline P$ mapping $H$ to $\overline H$.

\begin{proposition}
Let $\varphi$ be a $\mathfrak g$--valued 1--form on $P$ satisfying $r_g^*\varphi = \Ad_{g^{-1}}\varphi$ and $\varphi(X) = 0$ if $X$ is vertical. Then
\[D\varphi(X,Y) = \dd \varphi(X,Y) + [\omega(X), \varphi(Y)] + [\varphi(X),\omega(Y)] \qquad \forall X,Y \in TP\]
\end{proposition}

\begin{proof}
Both sides of the equation are bilinear and skew--symmetric. It suffices to check the three cases that $X,Y$ are both horizontal, both vertical or one horizontal and the other vertical.

If both $X$ and $Y$ are horizontal, $\omega(X) = 0 = \omega(Y)$ and $D\varphi(X,Y) = \dd \varphi(X,Y)$ by the definition of the covariant deriviative.

If $X$ and $Y$ are vertical, $\varphi(X) = 0 = \varphi(Y)$ and $D\varphi(X,Y) = 0$ since $\mathscr HX = 0 = \mathscr HY$. Extend $X,Y$ to fundamental vector fields $A_p^* = X$, $B_p^* = Y$, then
\[\dd \varphi(A^*,B^*) = L_{A^*}(\varphi(B^*)) - L_{B^*}(\varphi(A^*)) - \varphi([A^*,B^*]) = 0\]
since $[A^*,B^*]$ is vertical and $\varphi$ vanishes on fundamental vector fields, so $\dd \varphi(X,Y) = (\dd\varphi(A^*,B^*))(p) = 0$.

Given $X \in V_p$ and $Y \in H_p$ we choose extensions to vector fields on $P$ as follows: $X$ is extended by $A^*$ with $A \in \mathfrak g$, such that $A^*_p = X$. $Y$ is extended to a $G$--invariant horizontal vector field $\widetilde Y$ on $P$. This is possible since: $D_p \pi \colon H_p \to T_{\pi(p)}M$ is an isomorphism. We extend $D_p\pi(Y)$ to a vector field on $M$ with support in a neighbourhood of $\pi(p)$ over which $P$ is trivial. Choosing a section $s \colon U \to P$, $U$ containing the support of the vector field in $M$, we can an isomorphism $D\pi \colon H_{s(U)} \to TU$. We lift the vector field on $M$ under this isomorphism and use the $G$--action to extend it to a $G$--invariant horizontal vector field on $P$ extending the original $Y$.

Now $X$ is vertical, so $\varphi(X) = 0$ and $D\varphi(X,Y) = 0$ since $\mathscr HX = 0$. To check the claim in this case, we have to prove $\dd \varphi(X,Y) = -[\omega(X),\varphi(Y)]$.
%TODO: Beweis zu Ende führen (voodoo!)
\end{proof}

Let $\omega_0$ be a connection 1--form on $P$ and $\omega$ is a 1--form on $P$ with values in $\mathfrak g$ satisfying $r_g^*\omega = \Ad_{g^{-1}}\omega$ and $\omega|_V = 0$. Let $\omega_t = \omega_0 + t \omega$ with $t \in \mathbb R$. This is a smoothly varying familiy of connection 1--forms defining $H_t = \ker \omega_t$. Let $\Omega_t$ be the curvature of $H_t$. Then
\begin{align*}
\Omega &= \dd\omega_t + [\omega_t, \omega_t] = \dd \omega_0 + t \dd \omega + [\omega_0 + t \omega, \omega_0 + t \omega] = \\
&= \dd \omega_0 + [\omega_0, \omega_0] + t ( \dd \omega + [\omega_0, \omega] + [\omega, \omega_0]) + t^2 [\omega, \omega] = \\
&= \Omega_0 + t D_0\, \omega + t^2 [\omega, \omega]
\end{align*}
where $D_0$ is the derivative with respect to $\omega_0$ or $H_0$.
\[\frac{\dd}{\dd t} \Omega_t \Big|_{t=0} = D_0 \omega\]
With the Yang--Mills--functional
\[\YM \colon \operatorname{Conn}(P) \to R, \quad H \mapsto \int_M \|\Omega\|^2 \,\dvol\]
where we think of $\Omega$ as a section of $\Lambda^2 T^*M \otimes (P \times_{\Ad} \mathfrak g)$ and choose a Riemannian metric on $M$ and an $\Ad$--invariant scalar product on $\mathfrak g$, we have
\begin{align*}
\frac{\dd}{\dd t} \YM(H_t)\Big|_{t=0} &= \frac{\dd}{\dd t} \int_M \langle \Omega_t, \Omega_t \rangle \,\dvol \\
&= \frac{\dd}{\dd t} \int_M \langle \Omega_0 + t D_0 \omega + t^2[\omega, \omega], \Omega_0 + t D_0 \omega + t^2 [\omega, \omega] \rangle \,\dvol \\
&= 2 \int_M \langle \Omega_0, D_0 \omega \rangle \,\dvol = 2 \int_M \langle D_0^* \Omega_0, \omega \rangle \,\dvol
\end{align*}

\begin{proposition}
$H_0 = \ker \omega_0$ is a critical point of $\YM$ if and only if $D_0^* \Omega_0 = 0$.
\end{proposition}

\begin{remark}
We always have $D_0 \Omega_0 = 0$ by the Bianchi identity.
\end{remark}

\subsection{Principal $S^1$--bundles}

We consider principal $S^1$--bundles over $M$. $\mathfrak g$--valued forms on $P$ or $M$ are ordinary forms. Because $G$ is Abelian, $\Ad \colon G \to \Aut(\mathfrak g)$ sends $G$ to $\id_{\mathfrak g}$. A connection 1--form $\omega$ on a principal $S^1$--bundle $P \xrightarrow{\pi} M$ is an ordinary $S^1$--invariant 1--form on $P$, since
\[r_g^*\omega = \Ad_{g^{-1}}\omega = \omega.\]
The curvature $\Omega$ is an $S^1$--invariant ordinary 2--form on $P$ which vanishes on vertical vectors. We can think of $\Omega$ as an ordinary 2--form on $M$. By the structure equation
\[D\omega = \Omega = \dd \omega + [\omega, \omega] = \dd \omega\]
on $P$ since $G$ is Abelian. On $M$, $\Omega$ is closed but not neccessarily exact, because $\omega$ is not defined on $M$, only on $P$. In this case $D = \dd$. The Yang--Mills--equation $D_0^* \Omega_0 = 0$ becomes $\dd^* \Omega_0 = 0$. Since $\Omega_0$ is closed, $H_0$ is a Yang--Mills--connection if and only if $\Omega_0$ is a harmonic 2--form. Let $\omega_0, \omega_1$ be 2 different connection 1--forms on a principal $S^1$--bundle $P \to M$ and $\omega \coloneqq \omega_1 - \omega_0$. Then
\[\Omega_1 = \dd \omega_1 = \dd (\omega_0 + \omega) = \dd \omega_0 + \dd \omega = \Omega_1 + \dd \omega\]
with $\omega$ defined on $M$. So $[\Omega] \in H^2_{\mathrm{dR}}(M)$ is independent of the connection whose curvature we take.

\begin{definition}
$[\Omega] \in H^2_{\mathrm{dR}}(M)$ is the {\em Euler class} of $P \to M$ (or {\em first Chern class} if $G = U(1)$).
\end{definition}

Given a principal $S^1$--bundle $P \xrightarrow{\pi} M$, let $C(P) \in H^2_{\mathrm{dR}}(M)$ be its Euler class and $\mathscr C_P$ the space of closed 2--forms on $M$ whose cohomology class is $C(P)$.

\begin{lemma}
Every $\alpha \in \mathscr C_P$ is the curvature of some connection on $P$.
\end{lemma}

\begin{proof}
Choose some connection $\omega_0$ on $P$ with curvature $\Omega_0$. Then $\Omega_0 - \alpha$ is exact and we can write $\alpha = \Omega_0 + \dd \omega$ for some 1--form $\omega$ on $M$. Then $\omega_0 + \pi^* \omega$ is a connection 1--form with curvature $\alpha$.
\end{proof}

Let $\mathcal A$ be the affine space of connections on $P$ and define the map
\[c \colon \mathcal A \to \mathscr C_P, \quad \ker \omega = H \mapsto \Omega.\]
Then by the above lemma, $c$ is surjective.

\begin{lemma}
For every $\Omega \in \mathscr C_P$, the preimage $c^{-1}(\Omega)$ can be identified with $\mathcal C^1$, the space of closed forms on $M$.
\end{lemma}

\begin{proof}
Let $H_0 = \ker \omega_0 \in c^{-1}(\Omega)$. Every other connection 1--form $\omega_1$ on $P$ is defined by $\omega_1 = \omega_0 + \pi^*\omega$ for some 1--form $\omega$ on $M$. The curvature $\Omega_1$ of $H_1 = \ker \omega_1$ is
\[\Omega_1 = \dd \omega_1 = \dd \omega_0 + \dd \pi^* \omega = \Omega + \pi^* \dd \omega\]
so $H_1 \in c^{-1}(\Omega)$ if and only if $\pi^* \dd \omega = 0$, which is equivalent to $\dd \omega = 0$, i.e. $\omega \in \mathcal C^1$.
\end{proof}

The gauge group $\mathscr G$ is, since $S^1$ is Abelian,
\begin{align*}
\mathscr G &= \Aut(P) = \{u \colon P \to S^1 \mid u(pg) = g^{-1}u(p)g\} = \{u \colon P \to S^1 \mid u(pg) = u(p)\} \\
&= \{\overline u \colon M \to S^1\}.
\end{align*}
$\mathscr G$ acts on $A$ by
\[\phi^* \omega_0 = \Ad_{u^{-1}}\omega_0 + u^* \theta = \omega_0 + u^*\theta = \omega_0 + \pi^*(\overline u^* \theta).\]
The curvature of $\phi^*\omega_0$ is
\[\dd \phi^*\omega_0 = \dd \omega_0 + \dd \pi^*(\overline u^*\theta) = \dd \omega_0 + \pi^*\overline u^* \dd \theta = \dd \omega_0.\]
The map $c \colon \mathcal A \to \mathscr C_P$ descends to
\[\mathcal A / \mathscr G \xrightarrow{\quad\overline c\quad} \mathscr C_P\]
which is again surjective and $\mathcal C^1$ surjects onto $\overline c^{-1}(\Omega)$.

Let $\mathscr G H_0 = \mathscr G \ker \omega_0 \in \overline c^{-1}(\Omega)$. Then every other gauge equivalence class in $\overline c^{-1}(\Omega)$ is represented by $\omega_0 + \pi^* \omega$ for some closed $\omega$. What is the condition on $\omega$ and $\omega'$ to ensure that $\omega_0 + \pi^* \omega$ and $\omega_0 + \pi^* \omega'$ are gauge equivalent?
\[\omega_0 + \pi^* \omega' = \phi^*(\omega_0 + \pi^*\omega) = \omega_0 + \pi^*\omega + \pi^*(\overline u^*\theta) \Longleftrightarrow \omega' = \omega + \overline u^*\theta\]
The map $\exp \colon \mathbb R \to S^1, t \mapsto e^{2\pi i t}$ is a universal cover of $S^1$. We lift $\overline u \colon M \to S^1$ to $\widetilde u \colon M \to \mathbb R$ such that
\[\xymatrix{
& \mathbb R \ar[d]^{e^{2\pi i t}} \\
M \ar[ur]^{\widetilde u} \ar[r]^{\overline u} & S^1
}\]
commutes. So $\overline u^* \theta = \widetilde u^* \exp^* \theta = \widetilde u^*(\dd t) = \dd \widetilde u$, i.e. if $\overline u = \exp \circ \,\widetilde u$, then $\overline u^* \theta = \dd \widetilde u$. Conversely, for every exact 1--form $\alpha$ on $M$, we can choose a $\widetilde u \in C^\infty(M)$ such that $\alpha = \dd \widetilde u$ and consider $\exp \circ\,\widetilde u = \overline u$ as a gauge transformation of $P$. So $\mathcal C^1 / \mathcal E^1 = H_{\mathrm{dR}}^1(M)$ surjects onto $\overline c^{-1}(\Omega)$.
\[[M, S^1] \to H^1(M,\mathbb Z), \quad [\overline u] \mapsto [\overline u^* \theta] = \overline u^* [\theta]\]

\begin{lemma}
For any $\Omega \in \mathscr C_P$, the preimage $\overline c^{-1}(\Omega) \in \mathcal A / \mathscr G$ can be parametrized by the quotient
\[H^1(M,\mathbb R) / H^1(M, \mathbb Z),\]
i.e. the sequence $\mathbb C^1 \to \mathcal A \to \mathscr C_P$ induces
\[H^1(M,\mathbb R)/H^1(M, \mathbb Z) \longrightarrow \mathcal A/\mathscr G \longrightarrow \mathscr C_P.\]
\end{lemma}

A connection $H_0 = \ker \omega_0$ is a critical point of the Yang--Mills--functional if and only if $\dd^*\Omega_0 = 0$. We know that $\dd \Omega_0 = 0$. If $M$ is compact without boundary, the pair of equations
\[\dd \Omega_0 = 0 \qquad \dd^* \Omega_0 = 0\]
Is equivalent to $\Delta \Omega_0 = 0$, where $\Delta = \dd \dd^* + \dd^* \dd$. By Hodge theory, there is a unique harmonic 2--form in $\mathscr C_P$. All Yang--Mills--connections on $P$ map to the unique harmonic 2--form in $\mathscr C_P$.

The gauge equivalence classes of Yang--Mills--connections on $P$ are parametrized by $H^1(M,\mathbb R)/H^1(M, \mathbb Z)$. If $H^1(M, \mathbb R) = 0$, then there is a unique gauge equivalence class of Yang--Mills--connections on every principal $S^1$--bundle $P \to M$.

\subsection{The Yang--Mills--equations}

Fix a Lie group $G$ with the property that $\mathfrak g$ as a positive definite scalar product (e.g. $G$ is compact). We fix once and for all such a $\langle -,- \rangle$. Let $(M,g)$ be a compact Riemannian manifold, oriented without boundary.

If $P \xrightarrow{\pi} M$ is a principal $G$--bundle, then the space of connections $\mathcal A$ on $P$ is an affince space for $\Omega^1(M,E) = \Gamma(T^*M \otimes E)$ where $E = P \times_{\Ad} \mathfrak g$. We have metrics on $T^*M$ and $E$.

More generally, we can look at $k$--forms on $M$ with values in $E$.
\[\Omega^k(M,E) = \Gamma(\Lambda^k T^*M \otimes E)\]
\begin{example}
The curvature form of a connection on $P$ is an element of $\Omega^1(M,E)$.
\end{example}

Elements of $\Omega^k(M,E)$ correspond to $k$--forms $\alpha$ on $P$ with the following properties:
\begin{enumerate}[(1)]
\item $\alpha(X_1,\dots,X_k) = 0$ if one of the $X_1,\dots,X_k$ is vertical.
\item $r_g^*\alpha = \Ad_{g^{-1}}\alpha$ for all $g \in G$.
\end{enumerate}

We define the following operations on $\Omega^k(M,E)$:
\[[-,-] \colon \Omega^k(M,E) \times \Omega^l(M,E) \to \Omega^{k+l}(M,E), \quad (\alpha \otimes v, \beta \otimes w) \mapsto (\alpha \wedge \beta) \otimes [v,w]_{\mathfrak g}\]
\[\wedge \colon \Omega^k(M,E) \times \Omega^l(M,E) \to \Omega^{k+l}(M), \quad (\alpha \otimes v, \beta \otimes w) \mapsto \langle v, w \rangle\, \alpha \wedge \beta\]
%TODO: make array

A connection $H = \ker \omega$ on $P$ defines a covariant derivative $D$ on $\mathfrak g$--valued $k$--forms on $P$ by
\[D\alpha = \dd \alpha \circ \mathscr H.\]
If $\alpha$ has values in $\mathfrak g$ and satisfies (1) and (2), then so does $D\alpha$. Therefore, $D$ can be thought of as
\[D \colon \Omega^k(M,E) \to \Omega^{k+1}(M,E)\]
$D$ is compatible with the metric and with $[-,-]$ and $\wedge$. That means: Let $V \to M$ be a vector boundle with scalar product $\langle -, - \rangle$. A covariant derivative $D \colon \Omega^k(M,V) \to \Omega^{k+1}(M,V)$ is compatible with $\langle -, - \rangle$ if
\[\dd \langle s_1, s_2 \rangle = \langle Ds_1, s_2 \rangle + \langle s_1, Ds_2 \rangle.\]
For $\omega^k \in \Omega^{k}(M,E)$ and $\omega^l \in \Omega^l(M,E)$, we have
\[\dd(\omega^k \wedge \omega^l) = D\omega^k \wedge \omega^l \pm \omega^k \wedge D\omega^l\]

Let $V$ be a real vector space with an orientation, and $\langle - , - \rangle$ a positive definit scalar product on $V$. The volume form $\dvol \in \Lambda^n V^*$ where $n = \dim V$ is defined by $\dvol(e_1,\dots,e_n) = 1$ if $e_1,\dots,e_n$ is a positively oriented orthonormal basis of $V$. The scalar product induces a scalar product $V^*$ by the requirement that the isomorphism $V \to V^*, v \mapsto \langle v, - \rangle$ should be isometric. This also gives rise to a scalar product $\langle -, - \rangle$ on $\Lambda^k V^*$ where vectors of the form $\lambda_{i_1} \wedge \dots \wedge \lambda_{i_k}$ with $\lambda_1,\dots,\lambda_n$ an orthonormal basis of $V^*$ and $i_1 < \dots < i_k$ have length 1.

If $\langle -, - \rangle_0$ and $\langle -, - \rangle_1$ are two different scalar products on $V$, such that $\langle - , - \rangle_1 = \lambda^2 \langle -, - \rangle_0$ with $\lambda > 0$, then $\langle -, - \rangle_1 = \lambda^{-2k}\langle -, - \rangle_0$ on $\Lambda^k V^*$ and $\dvol_1 = \lambda^n \dvol_0$.

We define the Hodge operator
\[\star \colon \Lambda^k V^* \to \Lambda^{n-k} V^*\]
by $\alpha \wedge \star \beta = \langle \alpha, \beta \rangle \dvol$ for all $\alpha,\beta \in \Lambda^k V^*$. One can check that $\star\star = (-1)^{k(n-k)}$.

If again $\langle -, - \rangle_1 = \lambda^2 \langle -, - \rangle_0$ on $V$, then
\[\alpha \wedge \star_1 \beta = \langle \alpha, \beta \rangle_1 \,\dvol_1 = \lambda^{-2k} \langle \alpha,\beta \rangle_0 \lambda^n \,\dvol_0 = \lambda^{n-2k} \alpha \wedge \star_0 \beta\]
for all $\alpha, \beta \in \Lambda^k V^*$, so $\star_1 = \lambda^{n-2k} \star_0$ on $\Lambda^k V^*$.

Let $M$ be an oriented Riemannian manifold. Then the Hodge star
\[\star \colon \Omega^k(M) \to \Omega^{n-k}(M)\]
where $n = \dim M$ is defined fiberwise as above, i.e.
\[\alpha \wedge \star \beta = \langle \alpha, \beta \rangle \, \dvol \qquad \star\star = (-1)^{n(n-k)} \qquad \forall \alpha,\beta \in \Omega^k(M)\]
Let $P \to M$ be a principal $G$--bundle, fix an $\Ad$--invariant positive definit scalar product on the Lie algebra $\mathfrak g$ and define $E \coloneqq P \times_{\Ad} \mathfrak g$. We can extend $\star$ to a map $\Omega^k(M,E) \to \Omega^{n-k}(M,E)$ using the same formula with the above defined operation $\wedge \colon \Omega^k(M,E) \times \Omega^{n-k}(M,E) \to \Omega^n(M)$.

\begin{lemma}
Let $M$ be a compact oriented Riemannian manifold without boundary. On $\Omega^{k}(M,E)$, $k \geq 1$, we have $D^* = (-1)^{nk-n+1}\star D \star$, where $D$ is the covariant derivative defined b a connection on $P$.
\end{lemma}

\begin{proof}
Let $\alpha \in \Omega^{k-1}(M,E)$ and $\beta \in \Omega^k(M,E)$. Then by Stokes' theorem,
\begin{align*}
0 &= \int_M \dd(\alpha \wedge \star \beta) = \int_M D\alpha \wedge \star\beta + (-1)^{k-1} \int_M \alpha \wedge D \star \beta \\
&= \int_M \langle D\alpha, \beta \rangle \,\dvol + (-1)^{k-1} \int_M \langle \alpha, \star^{-1}D\star\beta\rangle\,\dvol
\end{align*}
So we have, for $\beta \in \Omega^k(M,E)$:
\[D^* \beta = -(-1)^{k-1}(-1)^{(k-1)(n-k+1)} \star D \star \beta = (-1)^{nk + n + 1} \star D \star \beta \qedhere\]
\end{proof}

Now for $\omega_t = \omega_0 + t \omega$ with $t \in \mathbb R$ and $\omega \in \Omega^1(M,E)$, if $\omega_0$ is a critical point of the Yang--Mills--functional, we have
\[0 = \frac{\dd}{\dd t} \YM (\omega_t) \Big|_{t=0} = 2 \int_M \langle \Omega_0, D_0 \omega \rangle\,\dvol = (-1)^{n+1}2 \int_M \langle \star D_0 \star \Omega_0, \omega \rangle \,\dvol.\]
So
\[D_0 \star \Omega_0 = 0\]
which is the {\em Yang--Mills--equation} for $\omega_0$. The Yang--Mills--equation is a second order differential equation.

Assume $n = \dim M = 4$. Then $\star \Omega_0$ is a 2--form just like $\Omega_0$ itself. By the Bianchi identity, $D_0 \Omega_0 = 0$. We want to know when $\Omega_0 = \star \Omega_0$ holds. In the case $n = 4$, $k = 2$, $\star$ is an endomorphism of the 6--dimensional vector space $\Lambda^2 V$. It has eigenvalues $\pm 1$ and this splits $\Lambda^2 V$:
\[\Lambda^2 V = \Lambda^2_+ V \oplus \Lambda^2_-V\]
where $\star$ acts as $\pm$ on $\Lambda^2_\pm V$. $\Lambda^2_+V$ is called the {\em self--dual (SD)} and $\Lambda^2_-V$ the {\em anti--self--dual (ASD)} part of $\Lambda^2V$.

If $\alpha_1,\dots,\alpha_4$ is an orthonormal basis for $V^*$, then we have a basis for $\Lambda^2_\pm$ given by
\begin{align*}
\alpha_1 \wedge \alpha_2 &\pm \alpha_3 \wedge \alpha_4 \\
\alpha_1 \wedge \alpha_3 &\pm \alpha_4 \wedge \alpha_2 \\
\alpha_1 \wedge \alpha_4 &\pm \alpha_2 \wedge \alpha_3
\end{align*}
If $\omega \in \Lambda^2_+V$ and $\eta \in \Lambda^2_-V$, then
\[\omega \wedge \eta = \star \omega \wedge \eta = \eta \wedge \star \omega = \langle \eta, \omega \rangle \, \dvol = 0\]
$\Lambda^2_\pm V$ are orthogonal for $\langle - , - \rangle$ and for $\wedge$. If a connection $\omega_0$ has self--dual or anti--self--dual curvature, then $\star \Omega_0 = \pm \Omega_0$ and $D_0 \star \Omega_0 = 0$ because of the Bianchi identity.

The equation $\star \Omega_0 = \pm \Omega_0$ is the {\em (anti--)self--duality equation} for $\omega_0$. It implies the Yang--Mills equation.

\begin{lemma}
On a 4--manifold $M$, the Yang--Mills--equation and the (anti--)self--duality equation are conformally invariant.
\end{lemma}

\begin{proof}
Two metrics $\langle -, - \rangle_1$ and $\langle -, - \rangle_0$ on $M$ are conformally invariant if $\langle -, - \rangle_1 = \lambda^2 \langle -, - \rangle_0$ for some $\lambda \neq 0$, $\lambda \in C^\infty(M)$.

We calculated that on $k$--forms, $\star_1 = \lambda^{n-2k}\star_0$, so in our case $\star_1 = \star_0$ for 2--forms and so the (A)SD equations for 2--forms with respect to the two metrics agree. The Yang--Mills equation is $D_0 \star \Omega_0 = 0$. Since the metric enters only in $\star$, this is the same for both metrics.
\end{proof}
\end{document}
