6.2 Relate Lie algebra with Lie group

We want to relate $\mathrm{Lie}G$ with $G$ and we do so via the “exponential map” which follows naturally from the tool of flow.

Integral curve

In general, let $M$ be a (smooth) manifold and $X$ be a (smooth) vector field. An integral curve through $p\in M$ is a smooth map $\gamma :I\to M$ where $t_0\in I$ is an interval such that $\gamma (t_0)=p$ and ${\gamma }^{\prime }(t)=X_{\gamma (t)}$.

Local existence and uniqueness

Lemma

Integral curve locally exists and is unique.

\begin{proof} To solve ${\gamma }^{\prime }(t)=X_{\gamma (t)}$ with the initial condition $\gamma (0)=p$, it is enough to show $d\Gamma /dt=d\phi (X)$ where $\Gamma =\phi \circ \gamma$ and $\phi :U\to {\mathbb{R}}^n$ is a chart containing $p$.

Note $d\phi (X):=\sum F_i(x)\partial /\partial x_i$ is a vector field on $\phi (U)\subseteq {\mathbb{R}}^n$. Since $T{\mathbb{R}}^n$ is trivial and $\partial /{\partial }_i$ form a basis everywhere, the above equation is $d\vec{x}/dt=(F_1(\vec{x}),\cdots ,F_n(\vec{x}))=\vec{F}(\vec{x})$where $\Gamma =(x_1,\cdots ,x_n)=\vec{x}$. Then by the uniqueness and existence of ODE, the integral curve locally exists and is unique. \end{proof}

Corollary

Let ${\gamma }_1:I_1\to M$ and ${\gamma }_2:I_2\to M$ be two integral curves of a vector field $X$. If ${\gamma }_1(t_0)={\gamma }_2(t_0)$ for some $t_0\in I_1\cap I_2$, then ${\gamma }_1$ and ${\gamma }_2$ agree on $I_1\cap I_2$.

Complete vector field

complete vector field

A vector field $X$ is said to be complete if on a manifold $M$, for any $p\in M$ there exists an integral curve $\gamma :\mathbb{R}\to M$ such that $\gamma (0)=p$.

Lemma

If $M$ is compact, then every vector field is complete.

\begin{proof} It is a well-known conclusion in differential manifold. See here.

Lemma

If $M=G$ is a Lie group, then left-invariant vector fields are complete.

\begin{proof} I think the following proof is clear enough, and there is no need to write it again in markdown.

\end{proof}

Define exponential map

Flow and 1-parameter group

In general, if $X$ is a complete vector field on $M$, then for any $p\in M$, there exists a global integral curve ${\gamma }_p:\mathbb{R}\to M$ with ${\gamma }_p(0)=p$. Put \begin{align} \Phi_X:\mathbb{R}\times M&\to M, \\ (t,p)&\mapsto\gamma_p(t) \end{align}and ${\varphi }_t=\Phi (t,\cdot )$. Then ${\varphi }_t\circ {\varphi }_s={\varphi }_{t+s}$.

Definition

${\Phi }_X$ is called the flow associated to the complete vector field $X$.

Definition

$\{{\varphi }_t{\}}_{t\in \mathbb{R}}$ is called a $1$-parameter group.

$h_X$

Now let $G$ be a Lie group. For $X\in \mathrm{Lie}G$, define$h_X(t):={\gamma }_{e,X}(t)$as the integral curve w.r.t. $X$ and passing through the identity element at $t=0$.

$h_X:\mathbb{R}\to G$ is a group homomorphism and is the unique smooth group homomorphism $h:\mathbb{R}\to G$ such that $h^{\prime }(0)=X_e$.

\begin{proof} Firstly, we need to show $h_X(s+t)=h_X(s)h_X(t)$. Since both of them are integral curves going through $h_X(s)$ at $t=0$, they agree on $\mathbb{R}$ by previous lemma. To show the uniqueness, suppose $h:\mathbb{R}\to G$ is a smooth group homomorphism with $h^{\prime }(0)=X_e$. Then \begin{align} h'(s)&=d/dt\space h(t+s)|_{t=0} \\ &=d/dt\space h(t)h(s)|_{t=0} \\ &=dL_{h(s)}(h'(0)) \\ &=dL_{h(s)}(X_e) \\ &=X_{h(s)}. \end{align}In the other word, $h(s)$ is an integral curve w.r.t $X$ and going through $e$ at $s=0$. So $h(s)={\gamma }_{e,X}(s)=h_X(s)$. \end{proof}

Exponential map

Now put $\exp (X)=h_X(1)={\gamma }_{e,X}(1)$.

Since ${\gamma }_{e,X}(t)$ depends smoothly on $X\in$ Lie $G\cong T_{e}G\cong {\mathbb{R}}^n$. After choosing a basis $e_1,\cdots ,e_n$ of ${\mathbb{R}}^n$, ${\gamma }_{e,X}(t)$ depends smoothly on $e_1,\cdots ,e_n$. Then $\exp :\text{Lie }G\to G$ is a smooth map.

Properties of exponential map

Commute diagram

Proposition

• Let $G_1,G_2$ be two connected Lie groups and $\phi :G_1\to G_2$ a Lie group homomorphism. Then the diagram is commutative. It is easy to prove by the uniqueness in the above lemma.
• Let $G$ be a Lie group and its associated Lie algebra $\mathfrak{g}$. For any $g\in G,x\in \mathfrak{g}$,\begin{align} \mathrm{Ad}_{\exp x}&=\exp(\mathrm{ad}x), \\ g(\exp x)g^{-1}&=\exp((\mathrm{Ad}_g)x), \end{align}where ${\mathrm{Ad}}_g:G\to G,x\mapsto gx{g}^{-1}$ which induces ${\mathrm{Ad}}_g:\mathfrak{g}\to \mathfrak{g}$ denoted also by ${\mathrm{Ad}}_g$. On the other hand, $\mathrm{ad}x\in \mathrm{End}(\mathfrak{g})$ is in the Lie algebra of $\mathrm{GL}(\mathfrak{g})$. So $\exp (\mathrm{ad}x)\in \mathrm{GL}(\mathfrak{g})$.

Rmk. (exercise) Show that $\mathrm{Lie}{\mathrm{GL}}_n(\mathbb{R})\cong M_{n\times n}(\mathbb{R})$ and $\exp :M_{n\times n}(\mathbb{R})\to {\mathrm{GL}}_n(\mathbb{R})$ is given by $\exp (x)={\sum }_{n=0}^{\infty }{x}^n/n!$, where $\mathrm{Lie}G$ denotes the Lie algebra of a Lie group $G$.

Local diffeomorphism

Lemma

$(d\exp {)}_e=\mathrm{id}$. As a consequence of inversion function theorem, $\exp$ is a diffeomorphism near $0\in \text{Lie }G$.

\begin{proof} For $X\in \text{Lie }G\simeq T_{e}G$, there is \begin{align} (d\exp)_e(X)&=d/dt\exp(tX)|_{t=0} \\ &=d/dt \space h_{tX}(1)|_{t=0}\\ &=d/dt \space h_X(t)|_{t=0} \\ &=X. \end{align}It remains to show $h_{tX}(s)=h_X(ts)$. For any fixed $t$, both sides are smooth group homomorphisms from $\mathbb{R}\to G$ whose derivations at $s=0$ are both $tX$. By the previous lemma, they must agree. \end{proof}

Dynkin’s formula

Motivation. When $X,Y\in g$ are close to $0\in g$, $\exp X,\exp Y$ are close to $e$. Therefore, there exists a $\mu (X,Y)\in g$ is close to $0$ such that $\exp \mu (X,Y)=\exp X\cdot \exp Y$. The question is, is there a formula expressing $\mu (X,Y)$ in terms of $X,Y$? The answer is Yes: Dynkin formula.

Dynkin's formula

$\mu (X,Y)=X+Y+\frac{1}{2}[X,Y]+\mathcal{O}(|(X,Y){|}^3).$ More explicitly,

• what’s the meaning of $O(|(X,Y){|}^3)$?

Rmk. Dynkin’s formula tells that the product of $G$ near $e$ is determined by the bracket structure of the Lie algebra.

Trivial but useful properties

Proposition

For any $X\in \mathrm{Lie}G$, $\exp (X)\exp (-X)={\mathrm{id}}_G$.

\begin{proof} Since $h_{tX}(s)=h_X(ts)$, there is $h_X(t)=h_{-X}(-t)$. Since $h_{-X}$ is a group homomorphism, $h_{-X}(t)h_{-X}(-t)={\mathrm{id}}_G$. Then $h_X(t)=(h_{-X}(t){)}^{-1}$ and so $h_X(1)h_{-X}(1)={\mathrm{id}}_G$. That is, $\exp (X)\exp (-X)={\mathrm{id}}_G$. \end{proof}

Proposition

For any $X\in \mathrm{Lie}G$ and $n\in \mathbb{N}$, $\exp (nX)=(\exp (X){)}^n$.

\begin{proof}Note that $$ \begin{aligned} \exp(nX)&=h_{nX}(1)=h_X(n)\ &=h_X(1)\cdots h_X(1)\ &=h_X(1)^n=(\exp(X))^n. \end{aligned}

**Rmk.** When you compute something about $\exp(tX)$, it seems a good idea to consider $h_X$.