\classheader{2013-03-13}

Let $G$ be a Lie group. A tangent vector at the identity can be extended uniquely to a left-invariant vector field on $G$, i.e. an element of $\Lie(G)$. Similarly, it can be extended in a unique way to a right-invariant vector field on $G$. Let us call the collection of these $\Lie^r(G)$. If $i:G\to G$ is the inverse map $i(g)=g^{-1}$, then we clearly see that $X\in\Lie(G)$ if and only if $i^*X\in\Lie^r(G)$. Because $i$ induces $-1$ on the tangent space $T_eG$, we have a commutative diagram
\begin{center}
\begin{tikzcd}
\Lie(G)\ar{r}{i^*} \ar{d}[swap]{\ev_g} & \Lie^r(G) \ar{d}{\ev_g}\\
T_eG \ar{r}[swap]{-1} & T_eG
\end{tikzcd}
\end{center}
where $\ev_e$ is evaluation at the identity (we've previously called this map $I_G$).

Given a $\Cinfty$ homomorphism $\sigma:G\to\Diffeo(M)$, we get a $\Cinfty$ group action $G\times M\to M$ given by $(g,m)\mapsto \sigma(g)m$.

Let $m\in M$. Define $f_m:G\to M$ by $f_m(g)=\sigma(g)m$. We have the derivative $f_m'(e):T_eG\to T_mM$. For all $v\in T_eG$, we get a vector field $\widetilde{v}$ on $M$ defined by $\widetilde{m}=f_m'(e)v$. This is a map from $T_eG$ to $\Cinfty$ vector fields on $M$.

For all $v\in T_eG$, let $[v]$ be the unique right-invariant vector field with $[v](e)=v$. Then $[v](g)=r_g'(e)v$ for all $g\in G$, where $r_g$ is right multiplication by $g$.

\begin{lemma}
$\text{}$

\begin{enumerate}
\item Let $m\in M$. Let $v\in T_eG$. Then for $f_m:G\to M$, $[v]$ is $f_m$-related to $\widetilde{v}$.
\item If $v,w\in T_eG$, then $[[v],[w]]$ is $f_m$-related to $[\widetilde{v},\widetilde{w}]$.
\item To rephrase the above, the map $\Lie^r(G)\to$ vector fields on $M$ defined by $[v]\mapsto \widetilde{v}$ is a Lie algebra homomorphism.
\end{enumerate}
\end{lemma}
\begin{proof}
Let's prove 1 first. Let $g\in G$. We have $\widetilde{v}(gm)=f_{gm}'(e)v=f_m'(g)r_g'(e)v$ (we are now neglecting to write the $\sigma$). But also
\[f_{gm}(g)=hgm=f_m(hg)=(f_m\circ r_g)(h)=f_m'(g)[v](g),\]
which is the same.
\end{proof}
\begin{proposition}[Converse]
There is also a converse: a Lie algebra homomorphism $h:\Lie^r(G)\to$ $\Cinfty$ vector fields on $M$ yields an action of $G$ on $M$, if we assume that (a) $G$ is simply connected and (b) $M$ is compact.
\end{proposition}
\begin{proof}
Each $w\in\Lie^r(G)$ produces a vector field $\theta(w)$ on $G\times M$ defined by
\[\theta(w)(g,m)=(w(g),h(w)(m)).\]
We claim that $w\mapsto \theta(w)$ is a homomorphism of Lie algebras. Note that the map $\mathrm{Vect}(M)\times\mathrm{Vect}(N)\to\mathrm{Vect}(M\times N)$ defined by $(v,w)\mapsto (p_1^*v,p_2^*w)$ is a Lie algebra homomorphism. Thus, we have an involutive subbundle $S(g,m)=\{\theta(w)(g,m)\mid w\in\Lie^r(G)\}$ (observation 1). Observe (2) that
\begin{center}
\begin{tikzcd}
S(g,m) \ar[hook]{r} \ar[bend right]{rr}[swap]{\cong} & T_{(g,m)}(G\times M) \ar{r} & T_gG
\end{tikzcd}
\end{center}
It follows that a leaf is, locally, the graph of a function $U\to M$ for some $U\hookrightarrow G$. We get the following observation (3): for all $m\in M$, there is a neighborhood $U$ of $e\in G$, a neighborhood $U_m$ of $m\in M$, and a map $f:U\times U_m\to M$ such that for all $m'\in U_m$, the image of $g\mapsto (g,f(g,m'))$ has tangent space $S$, and $f(e,m')=m'$ for all $m'\in U_m$.

Observation 4: Assuming that $M$ is compact, we get a neighborhood $U$ of $e\in G$ and $f:U\times M\to M$ with $f(e,m)=m$ for all $m\in M$, and a leaf $g\mapsto (g,f(g,m))$ of $S$, for all $m\in M$.

Observation 5: Giving $G\times M$ the topology it gets as a disjoint union of leaves, we have a commutative diagram
\begin{center}
\begin{tikzcd}
(U\times M)^\new \ar[hook]{r}{\text{open}} \ar{d}[swap]{\substack{\text{covering}\\\text{map}}} & (G\times M)^\new\ar{d}\\
U \ar{r} & G
\end{tikzcd}
\end{center}
where we know this map is a covering map by 4. 

\end{proof}


\subsection*{The adjoint representation}

Let $G$ be a Lie group. For any $g\in G$, there is the conjugation map $C_g:G\to G$ defined by $C_g(h)=ghg^{-1}$. Clearly $C_{g_1}\circ C_{g_2}=C_{g_1g_2}$. The derivative $C_g'(e):T_eG\to T_eG$ is denoted by $\mathrm{Ad}(g):T_eG\to T_eG$. We have that $\mathrm{Ad}$ is an action of $G$ on $\Lie(G)$. As an exercise, compute $d(\mathrm{Ad})$, which is an action of $\Lie(G)$ on itself.

Let $G$ be a Lie group and $H$ a closed Lie subgroup. We say that $\pi:G\to G/H$ is a homogeneous manifold. A homogeneous vector bundle on $G/H$ is just a vector bundle $W$ on $G$ equipped with an action of $G$ such that the diagram
\begin{center}
\begin{tikzcd}
W \ar{r}{\sigma(g)} \ar{d}[swap]{p} & W\ar{d}{p}\\
G/H\ar{r}[swap]{\ell_g} & G/H
\end{tikzcd}
\end{center}
commutes, and such that $\sigma(g):W(x)\to W(x)$ is a linear transformation for all $g\in G$ and $x\in G$.

Observe that $\ell_h\overline{e}=\overline{e}$ for all $h\in H$, where $\overline{e}=eH\in G/H$. It follows that $\sigma(h)(W_{\overline{e}})=W_{\overline{e}}$. We see that $h\mapsto \sigma(h)|_{W_{\overline{e}}}$ is a representation on $H$ on $W(\overline{e})$. Conversely, given $M$ and a representation of $H$, we can construct $W$ as the quotient of $G\times M$ by $(gh,m)\sim (g,hm)$ for all $g\in G$, $h\in H$, $m\in M$.










