6.3 Basic theorems in the Lie theory

About Lie subgroups

Theorem

Let $G$ be a (smooth) Lie group and $H$ a Lie subgroup of $G$. Then

$$h:=\{X\in g:{\exp }_G(tX)\in H,\forall t\in \mathbb{R}\}.$$

Heuristics. If $H\subseteq G$ is an embedded Lie subgroup, then $H$ is a regular sub-manifold. In this situation, if

$$X\in \{X\in g:\exp (tX)\in H,\forall t\in \mathbb{R}\},$$

then for $t$ sufficiently small, $\exp (tX)\in U$ where $U$ is a neighborhood of $e\in G$. So $\exp (tX)\in U\cap H$ and $d\exp (tX)/dt\in T_{e}H=h$. On the other hand, if $H$ is not an embedding Lie subgroup, then $U\cap H$ may not a regular sub-manifold. Recall the example, that is the place where the argument fails.

Example. It is possible that there exists $X\in g$ such that ${\exp }_G(tX)\in H$ for some $t\in \mathbb{R}$. Let

$$G:=\{x_0+x_{1}i+x_{2}j+x_{3}k:\sum x_i^2=1\}$$

be a Lie group where $i^2=j^2=k^2=-1$, $ij=-ji=k$. Then

$$\begin{array}{r}{H}_1:=\{x_0+x_{1}i:x_0^2+x_1^2=1\}\\ H_2:=\{x_0+x_{2}i:x_0^2+x_2^2=1\}\\ H_3:=\{x_0+x_{3}i:x_0^2+x_3^2=1\}\end{array}$$

are a Lie subgroups and Lie $H_i=h_i\cong \mathbb{R}$. Since $\ker {\exp }_G\cong \mathbb{Z}\oplus \mathbb{Z}\oplus \mathbb{Z}$, there do exist $X\in g\backslash h_i$ such that ${\exp }_G(X)=e\in h_i$. Also, note that the “bad points” are countable, which gives us an idea to finish the proof.

\begin{proof}First, one direction is easy. If $X\in h$, then ${\exp }_G(tX)=i({\exp }_H(tX))\in i(H)=H$ by diagram. Second, it remains to show that if $X\notin h$ then $\exp (tX)\in H$ does not hold for all $t\in \mathbb{R}$. Actually, one can show for a fixed $X\notin H$, $\{t\in \mathbb{R}:\exp (tX)\in H\}$ is countable.

To figure out $\{t\in \mathbb{R}:\exp (tX)\in H\}$, we need to do it “locally”. In other words, choosing an open neighborhood $U$ of $0\in h$ and $\epsilon \ge 0$, ask about

$$\{t\in (-\epsilon ,\epsilon ):\exp (tX)\in {\exp }_H(U)\}.$$

It is possible that the above set is $\{0\}$. With such an answer, we can quickly pass to the global using translation.

Local assertion.

Consider

\begin{align} \varphi:\mathbb{R}\oplus h&\to G, \\

(tX,Y)&\mapsto \exp(-tX)\exp(Y).

\end{align}

Because $d\varphi|_{e}(1)=-X$ and $d\varphi|_{e}(Y)=Y$, $d\varphi$ is injective. As a conclusion, $\varphi$ is locally an embedding and specifically injective by inverse function theorem. So there exists small $\varepsilon$ and small neighborhood $U$ of $0$ in $h$ such that $\varphi|_{(-\epsilon,\epsilon)\times U}$ is injective, that is, $\exp(tX)\in H$ iff $\varphi(t,Y)=e$ iff $t=0$ for some $Y\in h$. Whence

{t\in(-\varepsilon,\varepsilon):\exp(tX)\in \exp_H(U)}={0}.

Choose a smaller open neighborhood $V\subseteq U$ such that $\exp (V)\exp (-V)\subseteq \exp (U)$.

With the local assertion at hand, we proceed to the global question as follows. Since $\exp (V)$ is a neighborhood of $e$ in $H$ and $H$ is second countable, one can write $H={\cup }_{i\in I}{h}_i\exp (V)$ where $I$ is an countable set. If $t_1,t_2\in \{t\in \mathbb{R}:\exp (tX)\in h_i\exp (V)\}$, then

$$\exp (-t_{1}X)\exp (t_{2}X)\in \exp (-V)h_i^{-1}{h}_i\exp (V)\subseteq U.$$

Thus if $t_1\ne t_2$ then $|t_2-t_1|>\epsilon$. Therefore $\{t\in \mathbb{R}:\exp (tX)\in H\}$ is countable for any $X\notin H$. \end{proof}

Embedding Lie subgroup

Embedded Lie subgroup is closed

Lemma

Let $G$ be a smooth Lie group and $H$ be an embedded Lie subgroup. Then $H$ is closed in $G$.

\begin{proof}It remains to show $H=\overline{H}$, where $\overline{H}$ is the closure of $H$ in $G$. First, there exists an open neighborhood $U$ of $e$ in $G$ such that $U\cap H$ is closed in $U$ by the definition of embedding Lie subgroup. In the other word, $H$ is locally closed in $G$.

Claim: $U\cap H=U\cap \overline{H}$. One direction is easy. For $h^{\prime }\in U\cap \overline{H}$, there exists $\{h_i\}\subseteq H$ such that $h_n\to h^{\prime }$. When $n$ is big enough, $h_n\in U\cap H$. Since $U\cap H$ is closed in $U$, $h_n\to h^{\prime }\in U\cap H$.

Second, we argue for $\overline{H}=H$. Choose a smaller $V\subseteq U$ such that $V^{-1}V\subseteq U$ and $V\subseteq U$. If $h^{\prime }\in \overline{H}$, there exists $\{h_n\}\subseteq H$ with $h_n\to h^{\prime }$. So there exists $N$ such that $h_n\in h^{\prime }V$ for any $n\ge N$. For any $n,m\ge N$, there is $h_n^{-1}{h}_m\in V^{-1}V\subseteq U$. Also $h_n^{-1}{h}_m\in H$ yields $h_n^{-1}{h}_m\in U\cap H$. Fix $n$ and let $m\to \infty$, $h_n^{-1}{h}^{\prime }\to h_n^{-1}{h}^{\prime }\in V^{-1}\subseteq U$. Since $U\cap H$ is closed in $U$, we have $h_n^{-1}{h}^{\prime }\in U\cap H$. Thus $h^{\prime }\in H$. \end{proof}

A closed subgroup is an embedded Lie subgroup

Cartan's closed subgroup theorem

Let $G$ be a smooth Lie group. If $H\subseteq G$ is a closed subgroup, then $H$ has an embedded submanifold structure such that $H$ is an embedded Lie subgroup.

Heuristics. The key is to give a suitable local chart of $G$ whose restriction to $H$ give local charts of $H$. For this purpose, one needs to work out how to do it as $e\in G$.

Candidate of local charts

Lemma

Suppose $G$ is a smooth Lie group and $H$ is a closed subgroup of $G$. Then $h:=\{X\in g:\exp (tX)\in H,\forall t\in \mathbb{R}\}$ is a vector space.

\begin{proof}Obviously, $h$ is closed under scalar multiplication. We now argue that $X,Y\in h$ yields $X+Y\in h$. Actually,

$$\begin{array}{rl}(\exp \frac{tX}{n}\exp \frac{tY}{n}{)}^n& =(\exp (\frac{tX}{n}+\frac{tY}{n}+O(\frac{t^2|(X,Y){|}^2}{n^2})){)}^n\\ & =\exp (tX+tY+O(\frac{t^2|(X,Y){|}^2}{n}))\end{array}$$

by Dynkin’s formula and $s:\mathbb{R}\to G,s\mapsto \exp (sz)$ is a group homomorphism for a fixed $z$. Let $n\to \infty$, then

$$\exp (t(X+Y))=\underset{n\to \infty }{\lim }(\exp \frac{tX}{n}\exp \frac{tY}{n}{)}^n\in H.$$

Now we finish the proof. \end{proof}

The key proposition

Proposition

Suppose $H$ is a closed subgroup of a Lie group $G$. There exists a small open neighborhood $U$ of $G$ such that $\exp (U)\cap H=\exp (U\cap h)$ and $\exp {|}_U$ is a diffeomorphism.

To finish the proof, we will first establish the statement for an approximation of exponential.

Approximation of exponential

Observation. Choose a complementary subspace $h^{\prime }$ of $h$ in $g$. Define

$$\begin{array}{rl}\Phi :g=h\oplus h^{\prime }& \to G,\\ (X,Y)& \mapsto \exp (X)\exp (Y).\end{array}$$

Then $d{\Phi }_e=\mathrm{id}$ and hence $\Phi$ is a local diffeomorphism.

Claim: There exists a sufficiently small open neighborhood $V$ of $0\in g$ such that $\Phi (V)\cap H=\Phi (V\cap h)$ and $\Phi {|}_V$ is a diffeomorphism. To prove the claim, we need the following lemma.

Lemma. $H$ is a closed subgroup of a Lie group $G$. Suppose $\{X_n\}\subseteq g$ satisfying

• $X_n\to 0$;
• $\exp (X_n)\in H$;
• ${\lim }_{n\to \infty }\frac{X_n}{|X_n|}=X$ exists in $g$. Then $X\in h$.

Proof. Note that

$$\begin{array}{rl}\exp (tX)& =\exp (t\underset{n\to \infty }{\lim }\frac{X_n}{|X_n|})\\ & =\exp (\underset{n\to \infty }{\lim }[\frac{t}{|X_n|}]X_n)\\ & =\underset{n\to \infty }{\lim }\exp (X_n{)}^{[t/|X_n|]}\\ & \in H.\end{array}$$

So we finish the proof. $■$

Proof of the claim. "$\supseteq$" is easy. We prove "$\subseteq$" by contradiction. First, choose a small open neighborhood $V_0$ of $0$ in $g$ such that $\Phi {|}_{V_0}:V_0\to \Phi (V_0)$ is a diffeomorphism. If the assertion is false, then for any small open neighborhood $V\subseteq V_0$, there is $\Phi (V)\cap H⊋\Phi (V\cap h)$.

Let $V_n$ run over a local neighborhood basis of $0$ in $g$, then there exists $p_n\in (\Phi (V_n)\cap H)\backslash \Phi (V_n\cap h)$. Suppose $p_n=\Phi (X_n+Y_n)=\exp (X_n)\exp (Y_n)$ with $X_n,Y_n\to 0$ and $\exp (Y_n)\in H$. Thus there exists a subsequence $\{Y_{n_k}\}$ such that $Y_{n_k}\to 0$, $\exp (Y_{n_k})\in H$ and $Y:={\lim }_{k\to \infty }\frac{Y_{n_k}}{|Y_{n_k}|}$ exists. By Lemma, we know $Y\in h$. Also $Y\in h^{\prime }$. Thus $Y=0$, which is impossible as $|Y|=1$. Contradiction. $■$

Proof of the proposition

The choice of $V$ in the lemma can be arbitrary small. Choose a small open neighborhood $U_0$ of $0$ in $g$ such that $\exp {|}_{U_0}$ is a diffeomorphism. Choose $V$ so that $\Phi (V)\subseteq \exp (U_0)$. Put $U={\exp }^{-1}(\Phi (V))$, then $U$ meets the requirement by ${\exp }^{-1}(\Phi (V))\cap {\exp }^{-1}(H)={\exp }^{-1}(\Phi (V)\cap H)$. $■$

Proof of Cartan’s theorem

It follows from the key proposition.

Example of $O(n)$

Define

$$O(n):=\{g\in \mathrm{GL}(n,\mathbb{R}):g^{T}g=I_n\}\subseteq \mathrm{GL}(n,\mathbb{R})\subseteq M_{n\times n}(\mathbb{R}).$$

Since $\mathrm{GL}(n,\mathbb{R})$ is a Lie group and $O(n)$ is an closed subgroup, $O(n)$ is an embedded Lie subgroup. The question is, what is the Lie algebra of $O(n)$?

Solution. The Lie algebra of $\mathrm{GL}(n,\mathbb{R})$ is $gl(n,\mathbb{R})=M_{n\times n}(\mathbb{R})$. The exponential map $gl(n,\mathbb{R})\to \mathrm{GL}(n,\mathbb{R})$ is given by $\exp (X)={\sum }_{n=0}^{\infty }{X}^n/n!$ because $\sum X^n/n!$ is a group homomorphism from $gl(n,\mathbb{R})$ to $\mathrm{GL}(n,\mathbb{R})$ whose derivation at $0$ is $X$. Thus the Lie algebra $o(n)$ of $O(n)$ consists of $X\in gl(n,\mathbb{R})$ such that $\exp (tX)\in O(n)$, that is, $\exp (tX)\exp (tX{)}^T=I_n$. Since $\exp (tX)\exp (tX{)}^T=\exp (tX)\exp ((tX{)}^T)=I_n$, there is

$$d/dt\exp (tX)\exp ((tX{)}^T){|}_{t=0}=X+X^T=0.$$

On the other hand, if $X+X^T=0$, then

$$\exp (tX)\exp ((tX{)}^T)=\exp (tX)\exp (-tX)=\exp (0)=I_n$$

and so $X\in o(n)$. Therefore, $X\in o(n)$ iff $X+X^T=0$ and

$$o(n)=\{x\in gl(n,\mathbb{R}):X+X^T=0\}.$$

Exercise. What about $\mathrm{Sp}(n)$ and $\mathrm{U}(n)$? See here.

A remark

Lie subgroups: about injective Lie group homomorphisms

Closed Lie subgroups: related to surjective Lie group homomorphism