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](https://en.wikipedia.org/wiki/Picard–Lindelöf_theorem#Theorem), 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 [!definition] 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 [[Pasted image 20240102193901.png|here]]. [!lemma] If $M=G$ is a Lie group, then left-invariant vector fields are complete. ^ff5f02 `\begin{proof}` I think the following proof is clear enough, and there is no need to write it again in markdown. ![[Pasted image 20231223230929.png]] `\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 $\phi_t=\Phi(t,\cdot)$. Then $\phi_t\circ\phi_s=\phi_{t+s}$. [!definition] $\Phi_X$ is called the flow associated to the complete vector field $X$. [!definition] $\{\phi_t\}_{t\in\mathbb{R}}$ is called a $1$-parameter group. ## $h_X$ Now let $G$ be a Lie group. For $X\in\mathrm{Lie}\space G$, define

h_X(t):=\gamma_{e,X}(t)

as the integral curve w.r.t. $X$ and passing through the identity element at $t=0$. [!proposition] $h_X:\mathbb{R}\to G$ is a group homomorphism and is the unique smooth group homomorphism $h:\mathbb{R}\to G$ such that $h'(0)=X_e$. ^3c468f `\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 $h'(0)=X_e$. 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 $G\cong T_eG\cong\mathbb{R}^n$. 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 $\varphi:G_1\to G_2$ a Lie group homomorphism. Then the [[Pasted image 20240103150136.png|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 $\mathrm{Ad}_g:G\to G,x\mapsto gxg^{-1}$ 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.** *([[0.0 Exercise#week-12|exercise]])* Show that $\mathrm{Lie}\space\mathrm{GL}_n(\mathbb{R})\cong M_{n\times n}(\mathbb{R})$ 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 $\mathrm{Lie}\space G$ 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 $X\in \text{Lie }G\simeq T_eG$, 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. [!proposition] Dynkin's formula $\mu(X,Y)=X+Y+\dfrac{1}{2}[X,Y]+\mathcal{O}(|(X,Y)|^3).$ More explicitly, ![[Pasted image 20231224160445.png]] - [ ] 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 $X\in\mathrm{Lie}\space G$, $\exp(X)\exp(-X)=\mathrm{id}_G$. `\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 $X\in\mathrm{Lie}\space G$ 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}

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