6.5 Lie algebra - Lie group correspondence

TLDR

The correspondence between Lie groups and Lie algebras includes the following three main results:

• ^a98121
• ^d9ae7e
• ^3b51d3

All of them start at Lie algebra/Lie algebra homomorphism and get a unique Lie group or Lie group homomorphism.

Motivation. How to handle a Lie group homomorphism using the associate Lie algebra homomorphism?

Some consequences of Cartan’s closed subgroup theorem

Proposition

Suppose $G$, $H$ are smooth Lie groups and $\phi :G\to H$ is a Lie group homomorphism. Then $\ker \phi$ is a closed Lie subgroup with $$ \mathrm{Lie}\space(\ker\varphi)=\ker d\varphi.

\begin{proof}Note $\ker \phi$ is closed because $\phi$ is continuous. Due to Cartan’s closed subgroup theorem, $\ker \phi$ is an embedded Lie subgroup.

For $X\in \mathrm{Lie}(\ker \phi )$, there is $e^{tX}\in \ker \phi$ and $1_H=\phi (e^{tX})$. By the commute diagram, there is $\phi (e^{tX})=e^{td\phi (X)}$ yields $d\phi (X)=0$. Thus $X\in \ker d\phi$. Conversely, if $X\in \ker d\phi$, then $\phi (e^{tX})=e^{td\phi (X)}=1_H$. Hence $e^{tX}\in \ker \phi$. \end{proof}

continuous group homomorphism is smooth

Suppose $G,H$ are smooth Lie groups. If $\phi :G\to H$ is a group homomorphism and continuous, then $\phi$ is smooth.

\begin{proof}Consider the following diagram. Define a graph ${\Gamma }_{\phi }=\{(g,\phi (g)):g\in G\}$. Note that by Cartan’s closed subgroup theorem, ${\Gamma }_{\phi }$ is a closed Lie subgroup of $G$. Whence ${\tilde{p}}_1=p_1{|}_{{\Gamma }_{\phi }}$ is a bijective smooth Lie group homomorphism. Also we have the following proposition.

Proposition

If $\phi :H_1\to H_2$ is bijective smooth Lie group homomorphism, then ${\phi }^{-1}$ is also smooth.

\begin{proof}Since $\ker \phi =1$, $d\phi$ is injective by ^b18c72. Also we need to show $d\phi$ is surjective. If it is true, then $\phi$ is local diffeomorphism near $1_G$ and a local diffeomorphism at every $g\in G$, thus ${\phi }^{-1}$ is smooth. We will use Haar measure or Sard’s theorem to prove it.

By Haar measure: If $d\phi {|}_e$ is not surjective, then a small open neighborhood $U$ of $1_G$ has an image of measure $0$ in $H$. On the other hand, choose a compact neighborhood $V$ of $1_G$ contained in $U$, then $G$ is covered by countably many translates of $V$. It follows that $\phi (G)$ has measure $0$ in $H$. It contradicts $\phi (G)=H$.

By Sard’s theorem: If $d\phi {|}_e$ is not surjective, then $d\phi {|}_g$ is not surjective at every $g\in G$. So the critical point set at $\phi$ is $G$, which contradicts with Sard’s theorem. \end{proof}

about the above proposition

I think there is no need to take a compact neighborhood contained in the open neighborhood in the argument using Haar measure, as Lie group is second countable, and Lebesgue measure, Haar measure can be identified, and this property.

With the proposition above, we know ${\tilde{p}}_1^{-1}$ is smooth. Then $p_2\circ {\tilde{p}}_1^{-1}=\phi$ is smooth. \end{proof}

Questions

• Suppose $g$ a finite dimensional Lie algebra over $\mathbb{R}$. Does there exists a connected Lie subgroup $G$ such that $\mathrm{Lie}G=g$?
• $G$ a connected Lie group with Lie algebra $g$. If $h$ is a Lie subalgebra of $g$, does there exist a Lie subgroup $H$ of $G$ such that $\mathrm{Lie}H=h$?
• What’s the relation between Lie groups with the same Lie algebras?
• Let $G,H$ be connected Lie groups with Lie algebras $g,h$, if $\rho :g\to h$ is a Lie algebra homomorphism, does there exists a Lie group homomorphism $\phi :G\to H$ such that $d\phi =\rho$?

Answer of question 3

existence of universal covering Lie group

Let $G$ be a connected Lie group with Lie algebra $g$. Let $\tilde{G}$ be the universal covering of $G$, then $\tilde{G}$ has a Lie group structure and the projection map $\tilde{G}\to G$ is a smooth Lie group homomorphism.

\begin{proof}By the theorem of existence of the universal covering space, $G$ has a universal covering $\tilde{G}$. And by the lifting criterion, define multiplication and inverse operations on $\tilde{G}$. By the uniqueness of lifting, these operations do make $\tilde{G}$ a group. And the preimage of projection map gives charts of $\tilde{G}$. Thus $\tilde{G}$ is a Lie group. Also $\pi :\tilde{G}\to G$ is smooth because on every chart the map equals identity. The notes on the class is attached here. \end{proof}

Proposition

Let $G,H$ be connected Lie groups and $\phi :G\to H$ is a Lie group homomorphism with $d\phi :g\to h$ bijective. Then:

• $\phi$ is a covering map.
• If $H$ is further simply connected, then $\phi$ is an isomorphism.

\begin{proof}Note that $\phi$ is a local homeomorphism near $e_G$, and a connected Lie group is generated by an arbitrary neighborhood of the identity element by exercise. Thus $\phi$ is surjective and so $\phi$ is a covering map. Furthermore, when $H$ is simply connected, ${\phi }^{-1}$ must be a homeomorphism and hence smooth by ^7e0dad. \end{proof}

Answer of question 2

The subgroups-subalgebras theorem

Let $G$ be a connected Lie group and $g$ be the Lie algebra of $G$. If $h$ is a Lie subalgebra of $g$, then there exists a unique connected Lie subgroup $H$ of $G$ such that $\mathrm{Lie}H=h$.

\begin{proof}It is just a sketch of proof. There exists a maximal connected integral manifold $H$ of $G$ such that for any $h\in H$, $T_{h}H=d{L}_{h}h$. Then equip $H$ with a group structure. \end{proof}

Remark

I cannot complete the sketch of proof and I think it is not important :) maybe later I will finish it.

maximal integral manifold $(M,\psi )$ of $N$ is an integral submanifold of a distribution $\mathcal{D}$ on $N$ if $d\psi (M_m)=\mathcal{D}(\psi (m))$for each $m\in M$. A maximal integral manifold $(M,\psi )$ of a distribution $\mathcal{D}$ on a manifold $N$ is a connected integral manifold of $\mathcal{D}$ whose image in $N$ is not a proper subset of any other connected integral manifold of $\mathcal{D}$.

A submanifold 

Answer of question 4

The homomorphisms theorem

Let $G,H$ be connected Lie group. Let $\rho :g\to h$ be a Lie algebra homomorphism. If $G$ is simply connected, then there exists unique Lie group homomorphism $\phi :G\to H$ such that $d\phi =\rho$.

\begin{proof}Firstly, the uniqueness is clear, because $\phi$‘s behaviour near $e_G$ is determined by $\rho$ and $G$ is generated by an open neighborhood of $e_G$ by exercise.

Now we consider the existence. Consider the Lie algebra $k=\{(X,\rho (X)):X\in g\}\subseteq g\oplus h.$Let $K$ be the unique connected subgroup of $G\times H$ whose Lie algebra is $k$. Put ${\tilde{p}}_1=p_1\circ i$. Then $d{\tilde{p}}_1=d{p}_1\circ di$ is a map sending $(X,\rho (X))$ to $X$. So it is a bijection map. Hence ${\tilde{p}}_1$ is a covering map. Because $G$ is simply connected, ${\tilde{p}}_1$ is a Lie group homomorphism by ^103f0c. Put $\phi =p_2\circ {\tilde{p}}_1^{-1}$, then $\phi$ meets the purpose. \end{proof}

Corollary

If $G,H$ are simply connected Lie groups with isomorphic Lie algebras, then $G$ and $H$ are isomorphism.

Answer of question 1

Lie's third theorem

Every finite-dimensional real Lie algebra is the Lie algebra of some simply connected Lie group.

\begin{proof}See here. The key point is to use Ado’s theorem and there are some references. I am too lazy to read it. \end{proof}

Connection to the representation theory

Rmk. Combining the theory of finite dimensional representations of semi-simple Lie algebra and Lie algebra-Lie group correspondence, we have:

• the theory of finite dimensional representations of simply connected complex semi-simple groups, such as ${\mathrm{SL}}_n(\mathbb{C})$.
• the theory of finite dimensional representations of connected complex semi-simple groups.

Rmk. If $G$ is a connected Lie group and $\phi :G\to {\mathrm{GL}}_n(\mathbb{C})$ is a smooth Lie group homomorphism, then $d\phi :g\to g{l}_n(\mathbb{C})$ is a Lie algebra homomorphism with $g\cong {\mathbb{R}}^m$, which induces a Lie algebra homomorphism

$$(d\phi {)}_{\mathbb{C}}:g{\otimes }_{\mathbb{R}}\mathbb{C}\to g{l}_n(\mathbb{C}).$$

Rmk. If $g$ is a complex semi-simple Lie algebra, then there exists a real Lie subalgebra $g_{\mathbb{R}}$ such that $g=g_{\mathbb{R}}\otimes \mathbb{C}$, and $g_{\mathbb{R}}=\mathrm{Lie}K$ for some compact Lie group $K$.

Compact representation

Definition

Let $G$ be a topological group. A unitary representation of $G$ on a Hilbert space $V$ is a continuous group homomorphism $\rho :G\to U(V)$. And such a representation is called irreducible if $V$ has no closed invariant subspace.

Theorem

Let $G$ be a compact, Hausdorff, second-countable topological group. Then:

• Every irreducible continuous Hilbert space representation is finite dimensional.
• As a representation of $G\times G$,

L^2(G)=\oplus_{\pi\in\hat G}\pi\times\pi^{\vee} $$where $\hat{G}=\{\text{finite dimensional irreducible representation of }G\}$.