\classheader{2013-03-08}

Let us start with the following data: $M$ is a $\Cinfty$ manifold, $S\subset TM$ is an involutive subbundle, and
\[B=\{N\subseteq M\mid N\text{ is a locally closed submanifold of }M,\text{ and }T_xN=S(x)\text{ for all }x\in N\}.\]
The set $B$ forms a basis for a topology on $M$, which we'll denote $M^\new$. Then $M^\new$ is a $\Cinfty$ manifold. A leaf of $S$ is defined to be a connected component of $M^\new$.

\begin{exercise}
If $M$ is second-countable and Hausdorff, then every leaf is second-countable. Note that because $M^\new\to M$ is continuous and $M$ is Hausdorff, we have that $M^\new$ is Hausdorff.
\end{exercise}

Here is an application of this.

Let $G$ be a Lie group, and let $W\subset \Lie(G)$ be a Lie subalgebra (i.e. $W$ is closed under taking Lie brackets). Let $\langle W\rangle$ be the subbundle spanned by $W$, so that if $X_1,\ldots,X_k$ are an $\R$-basis of $W$, then $\langle W\rangle(g)$ is the span of $X_1(g),\ldots,X_k(g)$ for all $g\in G$. Note that $\langle W\rangle$ is involutive; this is because any section of $W$ is of the form $\sum_{i=1}^i f_iX_i$, and just using the rule $[X,fY]=X(f)Y+f[X,Y]$.

Let $H$ be the connected component of $G^\new$ containing the identity $e\in G$. From the definition of $G^\new$ (which, recall, depends on the $W$ we initially chose), it is clear that the tangent space of $H$ at $e$ is
\[T_eH=\{X(e)\in T_eG\mid X\in W\}.\]
Because $\ell_g^*X=X$ for all $X\in\Lie(G)$ and $g\in G$, we see that for any linear subspace $W\subset\Lie(G)$, we have $\ell_g^*\langle W\rangle=\langle W\rangle$. It follows that the leaves are permuted amongst each other, and thus $\ell_g:G^\new\to G^\new$ is continuous. Thus, for any $a\in H$, it follows that $a^{-1}H$ is also a leaf; but $e\in a^{-1}H$, and $a^{-1}H$ is a connected component, so we must have $H=a^{-1}H$. Thus $H$ is a subgroup.

\begin{corollary}
Let $W$ and $G$ be as above. Then there is a Lie group $H$ and Lie group homomorphism $f:H\to G$ such that
\begin{center}
\begin{tikzcd}
\Lie(H) \ar{rr}{df} \ar{rd}[swap]{\cong} & & \Lie(G)\\
& W\ar[hook]{ru}
\end{tikzcd}
\end{center}
\end{corollary}

\begin{lemma}
Let $G$ be a connected Lie group. Let $p:\widetilde{G}\to G$ be a universal covering space. Let $\widetilde{e}\in G$ be an element of $p^{-1}(e)$. Then $\widetilde{G}$ with its natural $\Cinfty$ structure has the structure of a Lie group with $\widetilde{e}$ as its identity and $p:\widetilde{G}\to G$ a homomorphism of Lie groups.
\end{lemma}
\begin{proof}
$\widetilde{G}\times\widetilde{G}$ is simply connected and $p$ is a covering map, so by covering space theory, in the following diagram
\begin{center}
\begin{tikzcd}
\widetilde{G}\times\widetilde{G} \ar{d}[swap]{p\times p} & G\ar{d}{p}\\
G\times G\ar{r}[swap]{B}& G
\end{tikzcd}\hspace{1in}
\begin{tikzcd}
(\widetilde{e},\widetilde{e}) \ar[mapsto]{d} & \widetilde{e} \ar[mapsto]{d}\\
(e,e) \ar[mapsto]{r} & e
\end{tikzcd}
\end{center}
there is a unique lift $\widetilde{B}:\widetilde{G}\times\widetilde{G}\to\widetilde{G}$ such that $\widetilde{B}(\widetilde{e},\widetilde{e})=\widetilde{e}$ and $p\circ \widetilde{B}=B\circ(p\times p)$. The binary operation that $B$ defines on $G$ will just be written as juxtaposition.

We can define maps $P$ and $Q$
\begin{center}
\begin{tikzcd}[column sep=1in]
\widetilde{G}\times\widetilde{G}\times\widetilde{G} \ar[bend left=5]{r}{P} \ar[bend right=5]{r}[swap]{Q} \ar{rd}& \widetilde{G} \ar{d}{p}\\
& G
\end{tikzcd}
\end{center}
by $P(a,b,c)=a(bc)$ and $Q(a,b,c)=(ab)c$, and the diagonal is just $p(a)p(b)p(c)$. It follows taht there is a covering transformation $\gamma:\widetilde{G}\to\widetilde{G}$ of $p$ such that $Q=\gamma\circ P$. But $P(\widetilde{e},\widetilde{e},\widetilde{e})=Q(\widetilde{e},\widetilde{e},\widetilde{e})$. The rest of the arguments are similar and are omitted.
\end{proof}


\begin{theorem}
The functor $G\mapsto \Lie(G)$ from the category of simply connected Lie groups and Lie group homomorphisms to the category of Lie algebras and Lie algebra homomorphisms is an equivalence of categories.
\end{theorem}

In particular, given Lie groups $G_1$ and $G_2$, then $\Mor(G_1,G_2)\to\Mor(\Lie(G_1),\Lie(G_2))$ is a bijection. To see that it is injective, note that we know this is true when $G_1=\R$, i.e. that $\Mor(\R,G)\to\Lie(G)=\Mor(\R,\Lie(G))$ is injective, so when given $f,g:G_1\to G_2$ such that $df=dg:\Lie(G_1)\to\Lie(G_2)$, let $X\in\Lie(G_1)$; the maps $\gamma:t\mapsto f(\exp(tX))$ and $\delta:t\mapsto g(\exp(tX))$ satisfy
\[\gamma'(0)=df(X)(e)=dg(X)(e)=\delta'(0),\]
so that $f(\exp(tX))=g(\exp(tX))$ for all $X\in\Lie(G_1)$ and $t\in\R$.

Now note that $M=\{z\in G_1\mid f(z)=g(z)\}$ is clearly a subgroup of $G_1$. The set $\{\exp(X)\mid X\in\Lie(G)\}$ contains a neighborhood $U$ of $e\in G$. Because $G$ is connected, we therefore must have that $M=G$. Thus $f=g$.

Now we prove that the map $\Mor(G_1,G_2)\to\Mor(\Lie(G_1),\Lie(G_2))$ is surjective.

Given a homomorphism $L:\Lie(G_1)\to\Lie(G_2)$ of Lie algebras, we want to show it comes from a Lie group homomorphism. Let $G=G_1\times G_2$, so that $\Lie(G)=\Lie(G_1)\times \Lie(G_2)$. If $p_1,p_2$ are the projections, then for any vector fields $v_1,v_2$ on $G_1,G_2$, we have that $p_i^*v_i$ is a section of $p_i^*TG_i$, and there is a natural inclusion $p_i^*TG_i\hookrightarrow TG$, so we may regard each $p_i^*v_i$ as a section of $TG$.

We claim that $[p_1^*v_1,p_2^*v_2]=0$. Let $W=\{(\alpha,L(\alpha))\mid \alpha\in\Lie(G)\}\subset \Lie(G)$. Because $L$ is a Lie algebra homomorphism, $W$ is a Lie subalgebra. It follows that there is a connected $H$ and map $\rho:H\to G$ such that the image of $d\rho:\Lie(H)\to\Lie(G)$ is $W$. We claim that the map $\sigma$, in the diagram
\begin{center}
\begin{tikzcd}
H\ar{r}{\rho} \ar{rd}[swap]{\sigma} & G_1\times G_2\ar{d}{p_1}\\ & G_1
\end{tikzcd}
\end{center}
is an isomorphism. Note that, if this is so, then $p_2\circ \rho\circ \sigma^{-1}$ is the desired map $G_1\to G_2$.

Let's prove the claim. We know that $d\sigma:\Lie(H)\to\Lie(G_1)$ is an isomorphism, so it suffices to show that $\sigma$ is a covering map (because $G_1$ is simply connected). We make the following observations:
\begin{enumerate}
\item The image $\im(\sigma)$ is a subgroup, $\im(\sigma)$ contains a non-empty open subset, and $G_1$ connected $\implies$ $\im(\sigma)=G_1$
\item The map $\sigma:H\to G_1$ is a group homomorphism and surjective. It now suffices to check that there is a neighborhood $U$ of $e\in G_1$ that is evenly covered.
\end{enumerate}

We have a neighborhood $P$ of $e\in H$ such that $\sigma|_P:P\to\sigma(P)$ is a homeomorphism. By continuity, there is a neighborhood $Q$ of $e$ such that $Q\cdot Q\subset P$. Assume that $Q=Q^{-1}$ in addition. Let $\Gamma=\ker(\sigma)$; we claim that $\Gamma\times Q\to\sigma^{-1}(\sigma(Q))$, $(\gamma,q)\mapsto \gamma q$ is a homeomorphism. 

Next time we'll finish this proof.









