19.1 Lie groups

中文

In Section 10.8 we defined a topological group as a group $G$ that is also a topological space such that the map $\psi :(g,h)\mapsto g{h}^{-1}$ is continuous. If $G$ has the structure of a differentiable manifold and $\psi$ is a smooth map it is said to be a Lie group. The arguments given in Section 10.8 to show that the maps $\varphi :(g,h)\mapsto gh$ and $\tau :g\mapsto g^{-1}$ are both continuous are easily extended to show that both are differentiable in the case of a Lie group. The map $\tau$ is clearly a diffeomorphism of $G$ since ${\tau }^{-1}=\tau$ . Details of proofs in Lie group theory can sometimes become rather technical. We will often resort to outline proofs, when the full proof is not overly instructive. Details can be found in a number of texts, such as [1–7]. Warner [6] is particularly useful in this respect.

Example 19.1

The additive group ${\mathbb{R}}^n$ , described in Example 10.20, is an $n$-dimensiona abelian Lie group, as is the $n$-torus $T^n={\mathbb{R}}^n/{\mathbb{Z}}^n$

Example 19.2

The set $M_n(\mathbb{R})$ of $n\times n$ real matrices is a differentiable manifold, diffeo morphic to ${\mathbb{R}}^{n^2}$ , with global coordinates $x^i{}_j(\mathsf{A})=A^i{}_j$ , where $A$ is the matrix $[A^i{}_j]$ . The general linear group $GL(n,\mathbb{R})$ is an $n^2$-dimensional Lie group, which is an open submani fold of $M_n(\mathbb{R})$ and a Lie group since the function $\psi$ is given by

$$x_j^i(\psi (\mathsf{A},\mathsf{B}))=x_k^i(\mathsf{A})x_j^k({\mathsf{B}}^{-1}),$$

which is differentiable since $x_j^k({\mathsf{B}}^{-1})$ are rational polynomial functions of the components $B_j^i$ with non-vanishing denominator on $\det$ ${}^{-1}(\dot{\mathbb{R}})$ . In a similar way, $GL(n,\mathbb{C})$ is a Lie group, since it is an open submanifold of $M_n(\mathbb{C})\cong {\mathbb{C}}^{n^2}\cong {\mathbb{R}}^{2n^2}$

Left-invariant vector fields

If $G$ is a Lie group, the operation of left translation $L_g:G\to G$ defined by $L_{g}h\equiv L_g(h)=gh$ is a diffeomorphism of $G$ onto itself. Similarly, the operation of right trans lation $R_g:G\to G$ defined by $R_{g}h=h_g$ is a diffeomorphism.

A left translation $L_g$ induces a map on the module of vector fields, $L_{g\ast }:T(M)\to T(M)$ by setting

$$(L_{g\ast }X{)}_a=L_{g\ast }{X}_{g^{-1}a}\text{\hspace{1em}(19.1)}$$

where $X$ is any smooth vector field. On the right-hand side of Eq. (19.1) $L_{g\ast }$ is the tangent map at the point $g^{-1}a$ . A vector field $X$ on $G$ is said to be left-invariant if $L_{g\ast }X=X$ for all $g\in G$

Given a tangent vector $A$ at the identity, $A\in T_e(G)$ , define the vector field $X$ on $G$ by $X_g=L_{g\ast }A$ . This vector field is left-invariant for, by Eq. (15.17),

$$L_{g\ast }{X}_h=L_{g\ast }\circ L_{h\ast }A=(L_g\circ L_h{)}_{\ast }A=L_{gh\ast }A=X_{gh}.$$

It is clearly the unique left-invariant vector field on $G$ such that $X_e=A$ . We must show, however, that $X$ is a differentiable vector field. In a local coordinate chart $(U,\phi ;x^i)$ at the identity $e$ let the composition law be represented by $n$ differentiable functions ${\psi }^i$ : $\phi (U)\times \phi (U)\to \phi (U)$

$$x^i(g{h}^{-1})={\psi }^i(x^1(g),\dots ,x^n(g),x^1(h),\dots ,x^n(h))\text{ where }{\psi }^i=x^i\circ \psi .$$

For any smooth function $f:G\to \mathbb{R}$

$$\begin{array}{l}Xf=(L_{g\ast }A)f=A(f\circ L_g)\\ =A^i\frac{\partial f\circ L_g}{\partial x^i}{|}_{x^i(e)}\\ =A^i\frac{\partial }{\partial y^i}\hat{f}({\psi }^1(x^1(g),\dots ,x^n(g),\mathbf{y}),\dots ,{\psi }^n(x^1(g),\dots ,x^n(g),\mathbf{y})){|}_{y^i=x^i(e)}\end{array}$$

where $\hat{f}=f\circ {\phi }^{-1}$ . Hence $Xf$ is differentiable at $e$ since it is differentiable on the neighbourhood $U$. If $g$ is an arbitrary point of $G$ then $gU$ is an open neighbourhood of $g$ and every point $h\in gU$ can be written $h=g{g}^{\prime }$ where $g^{\prime }\in U$ , so that

$$\begin{array}{l}(Xf)(h)=A(f\circ L_h)\\ =A(f\circ L_g\circ L_{g^{\prime }})\\ =X(f\circ L_g)(g^{\prime })\\ =X(f\circ L_g)\circ L_{g^{-1}}(h).\end{array}$$

Thus $Xf=X(f\circ L_g)\circ L_{g^{-1}}$ on $gU$ , and it follows that $Xf$ is differentiable at $g.$ .

Hence $X$ is the unique differentiable left-invariant vector field everywhere on $G$ such that $X_e=A$ . Left-invariant vector fields on $G$ are therefore in one-to-one correspondence with tangent vectors at $e,$ and form a vector space of dimension $n$, denoted $\mathcal{G}$ .

Lie algebra of a Lie group

Given a smooth map $\phi :M\to N$ between manifolds $M$ and $M^{\prime }$ , we will say vector fields $X$ on $M$ and $X^{\prime }$ on $M^{\prime }$ are $\phi$-related if ${\phi }_{\ast }{X}_p=X_{\phi (p)}^{\prime }$ for every $p\in M$ . In general there does not exist a vector field on $M$ that is $\phi$-related to a given vector field $X$ on $M$ unless $\phi$ is a diffeomorphism (see Section 15.4).

Lemma 19.1 · Lie brackets of related vector fields

If $\phi :M\to M^{\prime }$ is a smooth map and $X$ and $Y$ are two vectorfields on $M$, $\phi$-related respectively to $X^{\prime }$ and $Y^{\prime }$ on $M^{\prime }$ , then their Lie brackets $[X,Y]$ and $[X^{\prime },Y^{\prime }]$ are $\phi$-related.

Proof

If $f:M^{\prime }\to \mathbb{R}$ is a smooth map on $M^{\prime }$ then for any $p\in M$

$$(X^{\prime }f)\circ \phi =X(f\circ \phi )$$

since $X_{\phi (p)}^{\prime }f=({\phi }_{\ast }{X}_p)f=X_p(f\circ \phi )$ . Hence

$$\begin{array}{rl}[X^{\prime },Y^{\prime }{]}_{\phi (p)}f& =X_{\phi (p)}^{\prime }(Y^{\prime }f)-Y_{\phi (p)}^{\prime }(X^{\prime }f)\\ & =X_p((Y^{\prime }f)\circ \phi )-Y_p((X^{\prime }f)\circ \phi )\\ & =X_p(Y(f\circ \phi ))-Y_p(X(f\circ \phi ))\\ & =[X,Y{]}_p(f\circ \phi )\\ & ={\phi }_{\ast }[X,Y{]}_{p}f,\end{array}$$

as required.

Exercise

Show that if $X$ is a left-invariant vector field then $(Xf)\circ L_g=X(f\circ L_g)$

If $X$ and $Y$ are left-invariant vector fields on a Lie group $G$ it follows from Lemma 19.1 that

$$L_{g\ast }[X,Y]=[X,Y].\text{\hspace{1em}(19.2)}$$

The vector space $\mathcal{G}$ therefore forms an $n$-dimensional Lie algebra called the Lie algebra of the Lie group $G.$ . Because of the one-to-one correspondence between $\mathcal{G}$ and $T_e(G)$ ) it is meaningful to write $[A,B]$ for any pair $A$, $B\in T_e(G)$ , and the Lie algebra structure can be thought of as being imposed on the tangent space at the identity $e.$

Let $A_1,\dots ,A_n$ be a basis of the tangent space at the identity $T_e(G)$ , and $X_1,\dots ,X_n$ the associated set of left-invariant vector fields forming a basis of $\mathcal{G}.$ . As in Section 6.5 define the structure constants $C_{ij}^k=-C_{ji}^k$ by

$$[X_i,X_j]=C_{ij}^k{X}_k.$$

Exercise

Show that the Jacobi identities (15.24) are equivalent to

$$C_{mi}^k{C}_{jl}^m+C_{mj}^k{C}_{li}^m+C_{ml}^k{C}_{ij}^m=0.\text{\hspace{1em}(19.3)}$$

Example 19.3

Let ${\mathbb{R}}^n$ be the additive abelian Lie group ofExample 19.1. The vector field $X$ generated by a tangent vector $A=A^i{\left({\partial }_{x^i}\right)}_{ϵ}$ has components

$$X^i(g)=(X{x}^i)(g)=(L_{g\ast }A)x^i=A(x^i\circ L_g).$$

Now $x^i\circ L_g=g^i+x^i$ where $g^i=x^i(g)$ , whence

$$X^i(g)=A^j\frac{\partial }{\partial x^j}(g^i+x^i){|}_{x^i=0}=A^j{\delta }_j^i=A^i.$$

If $X=A^i{\partial }_{x^i}$ $i$ and $Y=B^j{\partial }_{x^j}$ are left-invariant vector fields, then for any function $f$

$$\begin{array}{c}[X,Y]f=A^i\frac{\partial }{\partial x^i}\left(B^j\frac{\partial f}{\partial x^j}\right)-B^j\frac{\partial }{\partial x^j}\left(A^i\frac{\partial f}{\partial x^i}\right)\\ =A^i{B}^j(f_{,ji}-f_{,ij})=0.\end{array}$$

Hence $[X,Y]=0$ for all left-invariant vector fields. The Lie algebra of the abelian Lie group ${\mathbb{R}}^n$ is commutative.

Example 19.4

Let $A$ be a tangent vector at the identity element of $GL(n,\mathbb{R})$

$$A=A_j^i{\left(\frac{\partial }{\partial x_j^i}\right)}_{\mathsf{X}=\mathsf{I}}.$$

The tangent space at $e=1$ is thus isomorphic with the vector space of $n\times n$ real matrices $M_n(\mathbb{R})$ . The left-invariant vector field $X$ generated by this tangent vector is

$$X=X_j^i\frac{\partial }{\partial x_j^i}$$

with components

$$\begin{array}{l}{X}_j^i(\mathsf{G})=(L_{\mathsf{G}\ast }A)x_j^i\\ =A(x_j^i\circ L_{\mathsf{G}})\\ =A_q^p\frac{\partial }{\partial x_q^p}\left(G_k^i{x}_j^k\right){|}_{\mathsf{X}=\mathsf{I}}\text{where}{G}_j^i=x_j^i(\mathsf{G})\\ =A_q^p{G}_k^i{\delta }_p^k{\delta }_j^q\\ =x_k^i(\mathsf{G})A_j^k.\end{array}$$

Hence

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

If $X$ and $Y$ are left-invariant vector fields such that $X_e=A$ and $Y_e=B$ , then their Lie bracket has components

$$\begin{array}{c}[X,Y{]}_j^i=[X,Y]x_j^i=x_m^p{A}_q^m\frac{\partial }{\partial x_q^p}(x_k^i{B}_j^k)-x_m^p{B}_q^m\frac{\partial }{\partial x_q^p}(x_k^i{A}_j^k)\\ =x_m^i(A_k^m{B}_j^k-B_k^m{A}_j^k).\end{array}$$

At the identity element $e=1$ the components of $[X,Y]$ are therefore formed by taking the matrix commutator product $\mathsf{AB}-\mathsf{BA}$ where $\mathsf{A}=[A^i{}_j]$ and $\mathsf{B}=[B_j^i]$ , and the Lie algebra of $GL(n,\mathbb{R})$ is isomorphic to the Lie algebra formed by taking commutators of $n\times n$ matrices from $M_n(\mathbb{R})$ , known as $\mathcal{G}\mathcal{L}(n,\mathbb{R})$

Maurer–Cartan relations

We say that a differential form $\alpha$ is left-invariant if $L_g^{\ast }\alpha =\alpha$ for all $g\in G$ . Its exterior derivative $d\alpha$ is also left-invariant, fo $L_g^{\ast }\mathrm{d}\alpha =\mathrm{d}{L}_g^{\ast }\alpha =\mathrm{d}\alpha$ . If $\omega$ is a left-invariant 1-form and $X$ a left-invariant vector field then $⟨\omega ,X⟩$ is constant over $G_{:}$ , for

$$⟨{\omega }_g,X_g⟩=⟨{\omega }_g,L_{g\ast }{X}_e⟩=⟨L_g^{\ast }{\omega }_g,X_e⟩=⟨{\omega }_e,X_e⟩.$$

By the Cartan identity, Eq. (16.14), we therefore have

$$\begin{array}{rl}& \mathrm{d}\omega (X,Y)=\frac{1}{2}[X(⟨Y,\omega ⟩)-Y(⟨X,\omega ⟩)-⟨\omega ,[X,Y]⟩]\\ & =-\frac{1}{2}⟨\omega ,[X,Y]⟩.\end{array}\text{\hspace{1em}(19.4)}$$

Let $E_1,\dots ,E_n$ be a left-invariant set of vector fields, forming a basis of the Lie algebra $\mathcal{G}$ and ${\epsilon }^1,\dots ,{\epsilon }^n$ the dual basis of differential 1-forms such that

$$⟨{\epsilon }^i,E_i⟩={\delta }_j^i.$$

These 1-forms are left-invariant, for $L_g^{\ast }{\epsilon }^i={\epsilon }^i$ for $i=1,\dots ,n$ as

$$⟨L_g^{\ast }{\epsilon }^i,E_j⟩=⟨{\epsilon }^i,L_{g\ast }{E}_j⟩=⟨{\epsilon }^i,E_j⟩={\delta }_j^i.$$

Hence, by Eq. (19.4),

$$\mathrm{d}{\epsilon }^i(E_j,E_k)=-\frac{1}{2}⟨{\epsilon }^i,[E_j,E_k]⟩=-\frac{1}{2}⟨{\epsilon }^i,C_{jk}^l{E}_l⟩=-\frac{1}{2}{C}_{jk}^i,$$

from which we can deduce the Maurer–Cartan relations

$$\mathrm{d}{\epsilon }^i=-\frac{1}{2}{C}_{jk}^i{\epsilon }^j\wedge {\epsilon }^k.\text{\hspace{1em}(19.5)}$$

Exercise

Show that the Jacobi identities Eq. (19.3) follow by taking the exterior derivative of the Maurer– Cartan relations.

Theorem 19.2 · Vanishing structure constants and local commutativity

A Lie group $G$ has vanishing structure constants if and only if it is iso morphic to the abelian group ${\mathbb{R}}^n$ in some neighbourhood of the identity.

Proof

Ifthere exists a coordinate neighbourhood of the identity such that $(gh{)}^i=g^i+h^i$ then from Example $19.3[X,Y]=0$ throughout this neighbourhood. Thu $C_{jk}^i=0_{:}$ , since the Lie algebra structure is only required in a neighbourhood of ${\mathbf{\Pi }}_{e.}$

Conversely, if all structure constants vanish then the Maurer–Cartan relations imply $\mathrm{d}{\epsilon }^i=0$ . By the Poincaré lemma 17.5, there exist functions $y^i$ in a neighbourhood of the identity such that ${\epsilon }^i=\mathrm{d}{\mathbf{y}}^i$ . Using these as local coordinates at $e$, we may assume that the identity is at the origin of these coordinates, $y^i(e)=0$ . For any $a,g$ in the domain of these coordinates

$$L_g^{\ast }({\epsilon }^i{)}_{L_{g}a}=({\epsilon }^i{)}_a=\mathrm{d}{a}^i\text{ where }{a}^i=y^i(a),$$

and

$$L_g^{\ast }({\epsilon }^i{)}_{L_{g}a}=L_g^{\ast }(\mathrm{d}{y}^i{)}_{ga}={\left(\mathrm{d}(y^i\circ L_g)\right)}_{ga}.$$

Writing $\varphi (g,h)=gh$ , we have for a fixed $g$

$$\mathrm{d}{a}^i=\mathrm{d}\left(y^i\circ \varphi (g,a)\right)=\frac{\partial {\varphi }^i\left(g^1,\dots ,g^n,a^1,\dots ,a^n\right)}{\partial a^j}\mathrm{d}{a}^j$$

where ${\varphi }^i=y^i\circ \varphi$ . Thus

$$\frac{\partial {\varphi }^i}{\partial a^j}={\delta }_j^i,$$

equations that are easily integrated to give

$${\varphi }^i(g^1,\dots ,g^n,a^1,\dots ,a^n)=f^i(g^1,\dots ,g^n)+a^i.$$

If $a=e,$ , so that $a^i=0$ , we have ${\varphi }^i(g^1,\dots ,g^n,0,\dots ,0)=f^i(g^1,\dots ,g^n)=g^i$ . The $\mathrm{\Delta }\mathrm{re}-$ quired local isomorphism with ${\mathbb{R}}^n$ follows immediately,

$${\varphi }^i(g^1,\dots ,g^n,a^1,\dots ,a^n)=g^i+a^i.$$

Problems

Problem 19.1

Let $E_j^i$ be the matrix whose $(i,j)$ th component is 1 and all other components vanish. Show that these matrices form a basis of $\mathcal{G}\mathcal{L}(n,\mathbb{R})$ , and have the commutator relations

$$[E_j^i,E_l^k]={\delta }_l^i{E}_j^k-{\delta }_j^k{E}_l^i.$$

Write out the structure constants with respect to this algebra in this basis

Problem 19.2

Let $E_j^i$ be the matrix defined as in the previous problem, and $F_j^i=i{E}_j^i$ where $i=\sqrt{-1}$ . Show that these matrices form a basis of $\mathcal{G}\mathcal{L}(n,\mathbb{C})$ , and write all the commutator relations between these generators of $\mathcal{G}\mathcal{L}(n,\mathbb{C})$

Problem 19.3

Define the $T_e(G)$ -valued 1-form $\theta$ on a Lie group $G,$ , by setting

$${\theta }_g(X_g)=L_{g^{-1}\ast }{X}_g$$

for any vector field $X$ on $G$ (not necessarily left-invariant). Show tha $\theta$ is left-invariant, $L_a^{\ast }{\theta }_g={\theta }_{a^{-1}g}$ for all $a,g\in G$

With respect to a basis $E_i$ of left-invariant vector fields and its dual basis ${\epsilon }^i$ , show that

$$\theta =\sum _{i=1}^n{\left(E_i\right)}_e{\epsilon }^i.$$