19.2 The exponential map

中文

A Lie group homomorphism between two Lie groups $G$ and $H$ is a differentiable map $\phi :G\to H$ that is a group homomorphism, $\phi (gh)=\phi (g)\phi (h)$ for all $g,h\in G$ . In the case where it is a diffeomorphism $\phi$ is said to be a Lie group isomorphism.

Theorem 19.3 · Lie group homomorphisms induce Lie algebra homomorphisms

A Lie group homomorphism $\phi :G\to H$ induces a Lie algebra homomorphism ${\phi }_{\ast }:\mathcal{G}\to \mathcal{H}$ . If$\phi$ is an isomorphism, then ${\phi }_{\ast }$ is a Lie algebra isomorphism.

Proof

The tangent map at the origin, ${\phi }_{\ast }:T_e(G)\to T_e(H)$ , defines a map between Lie algebras $\mathcal{G}$ and $\mathcal{H}$. If $X$ is a left-invariant vector field on $G$ then the left-invariant vector

field ${\phi }_{\ast }X$ on $H$ is defined by

$$({\phi }_{\ast }X{)}_h=L_{h\ast }{\phi }_{\ast }{X}_e.$$

Since $\phi$ is a homomorphism

$$\phi \circ L_a(g)=\phi (ag)=\phi (a)\phi (g)=L_{\phi (a)}\circ \phi (g)$$

for arbitrary $g\in G$ . Thus the vector fields $X$ and ${\phi }_{\ast }X$ are $\phi$-related,

$$\begin{array}{rl}& {\phi }_{\ast }{X}_a={\phi }_{\ast }\circ L_{a\ast }{X}_e\\ & =L_{\phi (a)\ast }\circ {\phi }_{\ast }{X}_e\\ & =({\phi }_{\ast }X{)}_{\phi (a)}.\end{array}$$

It follows by Lemma 19.1 that the Lie brackets [$X$, $Y$] and $[{\phi }_{\ast }X,{\phi }_{\ast }Y]$ are $\phi$-related,

$$[{\phi }_{\ast }X,{\phi }_{\ast }Y]={\phi }_{\ast }[X,Y],$$

so that ${\phi }_{\ast }$ is a Lie algebra homomorphism. In the case where $\phi$ is an isomorphism, ${\phi }_{\ast }$ is one-to-one and onto at $G_e$ -

A one-parameter subgroup of $G$ is a homomorphism $\gamma :\mathbb{R}\to G$

$$\gamma (t+s)=\gamma (t)\gamma (s).$$

Of necessity, $\gamma (0)=e$ . The tangent vector at the origin generates a unique left-invarian vector field $X$ such that $X_e=\dot{\gamma }(0)$ , where we use the self-explanatory notation $\dot{\gamma }(t)$ for ${\dot{\gamma }}_{\gamma (t)}$ . The vector field $X$ is everywhere tangent to the curve, $X_{\gamma (t)}=\dot{\gamma }(t)$ , for if $f:G\to$ $R$ is any differentiable function then

$$\begin{array}{l}{X}_{\gamma (t)}f=(L_{\gamma (t)\ast }\dot{\gamma }(0))f\\ =\dot{\gamma }(0)(f\circ L_{\gamma (t)})\\ =\frac{\mathrm{d}}{\mathrm{d}u}f(\gamma (t)\gamma (u)){|}_{u=0}\\ =\frac{\mathrm{d}}{\mathrm{d}u}f(\gamma (t+u)){|}_{u=0}\\ =\frac{\mathrm{d}}{\mathrm{d}t}f(\gamma (t))=\dot{\gamma }(t)f.\end{array}$$

Conversely, given any left-invariant vector field $X$, there is a unique one-parameter group $\gamma :\mathbb{R}\to G$ that generates it. The proof uses the result of Theorem 15.2 that there exists a local one-parameter group of transformations ${\sigma }_t$ on a neighbourhood $U$ of the identity $e$ such that ${\sigma }_{t+s}={\sigma }_t\circ {\sigma }_s$ for $0\le |t|,|s|,|t+s|<ϵ$ , and for all $h\in U$ and smooth functions $f$ on $U$

$$X_{h}f=\frac{\mathrm{d}f({\sigma }_t(h))}{\mathrm{d}t}{|}_{t=0}.$$

For all $h$, $g\in U$ such that $g^{-1}h\in U$ we have

$$\begin{array}{l}{X}_{h}f=X_{L_g(g^{-1}h)}f\\ =\left(L_{g\ast }{X}_{g^{-1}h}\right)f\\ =X_{g^{-1}h}(f\circ L_g)\\ =\frac{\mathrm{d}f\circ L_g\circ {\sigma }_t(g^{-1}h)}{\mathrm{d}t}{|}_{t=0}\\ =\frac{\mathrm{d}f({\sigma }_t^{\prime }(h))}{\mathrm{d}t}{|}_{t=0}\end{array}$$

where

$${\sigma }_t^{\prime }=L_g\circ {\sigma }_t\circ L_{g^{-1}}.$$

The maps ${\sigma }_t^{\prime }$ form a local one-parameter group of transformations, since

$${\sigma }_t^{\prime }\circ {\sigma }_s^{\prime }=L_g\circ {\sigma }_t\circ {\sigma }_s\circ L_{g^{-1}}=L_g\circ {\sigma }_{t+s}\circ L_{g^{-1}}={\sigma }_{t+s}^{\prime },$$

and generate the same vector field $X$ as generated by ${\sigma }_t$ . Hence ${\sigma }_t^{\prime }={\sigma }_t$ on a neighbourhood $U$ of ${\dot{\theta }}_{e,\theta }$ , so that

$${\sigma }_t\circ L_g=L_g\circ {\sigma }_t.\text{\hspace{1em}(19.6)}$$

Setting $\gamma (t)={\sigma }_t(e)$ , we have

$${\sigma }_t(g)=g{\sigma }_t(e)=g\gamma (t)$$

for all $g\in U$ , and the one-parameter group property follows for $\gamma$ for $0\le |t|,|s|,|t+s|\theta <ϵ$

$$\begin{array}{c}\gamma (t+s)={\sigma }_{t+s}(e)={\sigma }_t\circ {\sigma }_s(e)={\sigma }_t(\gamma (s))\\ =\gamma (s){\sigma }_t(e)=\gamma (s)\gamma (t)=\gamma (t)\gamma (s).\end{array}$$

The local one-parameter group may be extended to all values of $t$ and $s$ by setting

$$\gamma (t)={\left(\gamma (t/n)\right)}^n$$

for $n$ a positive integer chosen such that $|t/n|<ϵ$ . The group property follows for all $t$, $s$ from

$$\gamma (t+s)={\left(\gamma ((t+s)/n)\right)}^n={\left(\gamma (t/n)\gamma (s/n)\right)}^n={\left(\gamma (t/n)\right)}^n{\left(\gamma (s/n)\right)}^n=\gamma (t)\gamma (s).$$

It is straightforward to verify that this one-parameter group is tangent to $X$ for all values of $t$.

Exponential map

The exponential map $\exp$ : $\mathcal{G}\to G$ is defined by

$$\exp X\equiv \exp (X)=\gamma (1)$$

where $\gamma :\mathbb{R}G$ , the one-parameter subgroup generated by the left-invariant vector field $X$. Then

$$\gamma (s)=\exp sX.\text{\hspace{1em}(19.7)}$$

For, let $\alpha$ be the one-parameter subgroup defined by $\alpha (t)=\gamma (st)$

$$\alpha (t+t^{\prime })=\gamma (s(t+t^{\prime }))=\gamma (st)\gamma (s{t}^{\prime })=\alpha (t)\alpha (t^{\prime }).$$

If $f$ is any smooth function on $G_{:}$ , then

$$\dot{\alpha }(0)f=\frac{\mathrm{d}}{\mathrm{d}t}f(\gamma (st)){|}_{t=0}=s\frac{\mathrm{d}}{\mathrm{d}u}f(\gamma (u)){|}_{u=0}=s{X}_{e}f.$$

Thus $\alpha$ is the one-parameter subgroup generated by the left-invariant vector field $sX,$ , and $\exp$ $s$ $X=\alpha (1)=\gamma (s)$

We further have that

$$(Xf)(\exp tX)=X_{\gamma (t)}f=\dot{\gamma }(t)f=\frac{\mathrm{d}}{\mathrm{d}t}f(\gamma (t))$$

so that

$$(Xf)(\exp tX)=\frac{\mathrm{d}}{\mathrm{d}t}f(\exp tX).\text{\hspace{1em}(19.8)}$$

The motivation for the name ‘exponential map’ lies in the identity

$$\exp (sX)\exp (tX)=\gamma (s)\gamma (t)=\gamma (s+t)=\exp (s+t)X.\text{\hspace{1em}(19.9)}$$

Example 19.5

Let

$$X=x_k^i{A}_j^k\frac{\partial }{\partial x_j^i}$$

be a left-invariant vector field on the general linear group $GL(n,\mathbb{R})$ . Setting $f=x_j^i$ we have

$$Xf=X{x}_j^i=x_k^i{A}_j^k$$

and substitution in Eq. (19.8) gives

$$A_j^k{x}_k^i(\exp tX)=\frac{\mathrm{d}}{\mathrm{d}t}{x}_j^i(\exp tX).$$

If $y_k^i=x_k^i(\exp tX)$ are the components of the element $\exp (tX)$, and the matrix $\mathsf{Y}=[y_k^i]$ satisfies the linear differential equation

$$\frac{\mathrm{d}\mathsf{Y}}{\mathrm{d}t}=\mathsf{YA}$$

with the initial condition $\mathsf{Y}(0)=\mathsf{I}$ , having unique solution

$${\rm Y}(t)={\mathrm{e}}^{t\mathsf{A}}=\mathsf{I}+t\mathsf{A}+\frac{t^2{\mathsf{A}}^2}{2!}+\dots ,$$

then if $X_e$ is identified with the matrix $\mathsf{A}\in \mathcal{G}$ , the matrix of components of the elemen $\exp (t{X}_e)\in GL(n,\mathbb{R}){\mathrm{is}\mathrm{e}}^{t\mathsf{A}}$

For any left-invariant vector field $X$ we have, by Eq. (19.8),

$$\begin{array}{rl}\frac{\mathrm{d}}{\mathrm{d}t}f(L_g(\exp tX))& =(X(f\circ L_g))(\exp tX)\\ & =(L_{g\ast }Xf)(\exp tX)\\ & =(Xf)(g\exp tX)\end{array}$$

since $L_{g\ast }X=X$ . The curve $t\mapsto L_g\exp (tX)$ is therefore an integral curve of $X$ through $g$ at $t=0$ . Since $g$ is an arbitrary point of $G_{:}$ , the maps ${\sigma }_t:g\to g\exp tX=R_{\exp tX}g$ form a one-parameter group of transformations that generate $X,$ and we conclude that every left-invariant vector field is complete.

The exponential map is a diffeomorphism of a neighbourhood of the zero vector $0\in \mathcal{G}$ onto a neighbourhood of the identity $e\in G$ . The proof may be found in [6]. If $\phi :H\to G$ is a Lie group homomorphism then

$$\phi \circ \exp =\exp \circ {\phi }_{\ast }\text{\hspace{1em}(19.10)}$$

where ${\phi }_{\ast }:\mathcal{GH}$ is the induced Lie group homomorphism of Theorem 19.3. For, let $\gamma :\mathbb{R}\to G$ be the curve defined by $\gamma (t)=\phi (\exp tX)$ . Since $\phi$ is a homomorphism, this curve is a one-parameter subgroup of $G$

$$\gamma (t)\gamma (s)=\phi (\exp tX\exp sX)=\phi (\exp (t+s)X)=\gamma (t+s).$$

Its tangent vector at $t=0$ is ${\phi }_{\ast }{X}_e$ since

$$\dot{\gamma }(0)f=\frac{\mathrm{d}f\circ \phi (\exp tX)}{\mathrm{d}t}{|}_{t=0}=X_e(f\circ \phi )=({\phi }_{\ast }{X}_e)f,$$

and $\gamma$ is the one-parameter subgroup generated by the left-invariant vector field ${\phi }_{\ast }X$ . Hence $\phi (\exp tX)=\exp t{\phi }_{\ast }X$ and Eq. (19.10) follows on setting $t=1$

Problems

Problem 19.4

A function $f:G\to \mathbb{R}$ is said to be an analytic function on $G$ if it can be expanded as a Taylor series at any point $g\in G$ . Show that if $X$ is a left-invariant vector field and $f$ is an analytic function on $G$ then

$$f(g\exp tX)=\left({\mathrm{e}}^{tX}f\right)(g)$$

where, for any vector field $Y$, we define

$${\mathrm{e}}^{Y}f=f+Yf+\frac{1}{2!}{Y}^{2}f+\frac{1}{3!}{Y}^{3}f+\cdots =\sum _{i=0}^{\infty }\frac{Y^n}{n!}f.$$

The operator $Y^n$ is defined inductively by $Y^{n}f=Y(Y^{n-1}f)$

Problem 19.5

Show that $\exp (tX)$ $\exp$ $tY=\exp t(X+Y)+O(t^2)$