19.3 Lie subgroups

中文

A Lie subgroup $H$ of a Lie group $G$ is a subgroup that is a Lie group, and such that the natural injection map $i:H\to G$ defined by $i(g)=g$ makes it into an embedded submanifold of $G$. It is called a closed subgroup if in addition $H$ is a closed subset of $G$. In this case the embedding is regular and its topology is that induced by the topology of $G$ (see Example 15.12). The injection $i$ induces a Lie algebra homomorphism $i_{\ast }:\mathcal{H}\to (G)$ , which is clearly an isomorphism of $\mathcal{H}$ with a Lie subalgebra of $\mathcal{G}$. We may therefore regard the Lie algebra of the Lie subgroup as being a Lie subalgebra of $\mathcal{G}$.

Example 19.6

Let $T^2=S^1\times S^1$ be the 2-torus, where $S^1$ is the one-dimensional Lie group where composition is addition modulo 1. This is evidently a Lie group whose elements can be written as pairs of complex numbers $({\mathrm{e}}^{i\theta },{\mathrm{e}}^{i\varphi })$ , where

$$({\mathrm{e}}^{i\theta },{\mathrm{e}}^{i\varphi })({\mathrm{e}}^{i{\theta }^{\prime }},{\mathrm{e}}^{i{\varphi }^{\prime }})=({\mathrm{e}}^{i(\theta +{\theta }^{\prime })},{\mathrm{e}}^{i(\varphi +{\varphi }^{\prime })}).$$

The subset

$$H=\{({\mathsf{e}}^{iat},{\mathsf{e}}^{ibt})|-\infty <t<\infty \}$$

is a Lie subgroup for arbitrary values of $a$ and $b$. If $a/b$ is rational it is isomorphic with $S^1$ , the embedding is regular and it is a closed submanifold. If $a/b$ is irrational then the subgroup winds around the torus an infinite number of times and is arbitrarily close to itself everywhere. In this case the embedding is not regular and the induced topology does no correspond to the submanifold topology. It is still referred to as a Lie subgroup. This is done so that all Lie subalgebras correspond to Lie subgroups.

The following theorem shows that there is a one-to-one correspondence between Lie subgroups of $G$ and Lie subalgebras of $\mathcal{G}$. The details of the proof are a little technical and the interested reader is referred to the cited literature for a complete proof.

Theorem 19.4 · Unique connected Lie subgroup for a Lie subalgebra

Let $G$ be a Lie group with Lie algebra $\mathcal{G}$. For every Lie subalgebra $\mathcal{H}$ of $\mathcal{G}$, there exists a unique connected Lie subgroup $H\subseteq G$ with Lie algebra $\mathcal{H}.$

Proof outline

The Lie subalgebra $\mathcal{H}$ defines a distribution $D^k$ on $G$, by

$$D^k(g)=\{X_g|X\in \mathcal{H}\}.$$

Let the left-invariant vector fields $E_1,\dots ,E_k$ be a basis of the Lie algebra $\mathcal{H}$, so that a vector field $X$ belongs to $D^k$ if and only if it has the form $X=X^i{E}_i$ where $X^i$ are real-valued functions on $G.$ . The distribution $D^k$ is involutive, for if $X$ and $Y$ belong to $D^k$ , then so does their Lie bracket [$X$, $Y$]:

$$[X,Y{]}_g=X^i(g)Y^j(g)C_{ij}^k{E}_k+X^i(g)E_i{Y}^j(g)E_j-Y^j(g)E_j{X}^i(g)E_i.$$

By the Frobenius theorem 15.4, every point $g\in G$ has an open neighbourhood $U$ such that every $h\in U$ lies in an embedded submanifold $N_h$ of $G$ whose tangent space spans $\mathcal{H}$ at all points $h^{\prime }\in N_h$ . More specifically, it can be proved that through any point of $G$ there exists a unique maximal connected integral submanifold – see [2, $p$. 92] or $[6,\mathsf{p}.48]$ . Let $H$ be the maximal connected integral submanifold through the identity $e\in G$ . Since $D^k$ is invariant under left translations $g^{-1}H=L_{g^{-1}}H$ is also an integral submanifold of $D^k$ By maximality, we must have $g^{-1}\subseteq H$ . Hence, $\mathrm{if}g,h\in H$ then $g^{-1}h\in H$ , so that $H$ is a subgroup of $G$. It remains to show that $(g,h)\mapsto g^{-1}h$ is a smooth function with respect to the differentiable structure on $H$, and that $H$ is the unique subgroup having $\mathcal{H}$ as its Lie algebra. Further details may be found in [6, $p$. 94]. -

If the Lie subalgebra is set to be $\mathcal{H}=\mathcal{G}$ , namely the Lie algebra of $G$ itself, then the unique Lie subgroup corresponding to $\mathcal{G}$ is the connected component of the identity, often denoted $G_0$

Matrix Lie groups

All the groups discussed in Examples 2.10–2.15 of Section 2.3 are instances of matrix Lie groups; that is, they are all Lie subgroups of the general linear group $GL(n,\mathbb{R})$ . Their Lie algebras were discussed heuristically in Section 6.5.

Example 19.7

As seen in Example 19.4 the Lie algebra of $GL(n,\mathbb{R})$ is isomorphic to the Lie algebra of all $n\times n$ matrices with respect to commutator products $[\mathsf{A},\mathsf{B}]=\mathsf{AB}-\mathsf{BA}$. The set of all trace-free matrices $\mathcal{H}=\{\mathsf{A}\in M_n(\mathbb{R})|\mathrm{tr}\mathsf{A}=0\}$ is a Lie subalgebra of $\mathcal{G}\mathcal{L}(n,\mathbb{R})$ since it is clearly a vector subspace of $M_n(\mathbb{R})$ ,

$$\mathrm{tr}\mathsf{A}=0,\mathrm{tr}\mathsf{B}=0⟹\mathrm{tr}(\mathsf{A}+a\mathsf{B})=0,$$

and is closed with respect to taking commutators,

$$\mathrm{tr}[\mathsf{A},\mathsf{B}]=\mathrm{tr}(\mathsf{AB})-\mathrm{tr}(\mathsf{BA})=0.$$

It therefore generates a unique connected Lie subalgebra $H\subset GL(n,\mathbb{R})$ (see Theorem 19.4). To show that this Lie subgroup is the unimodular group $SL(n,\mathbb{R})$ , we use the wellknown identity

$$\det {\mathrm{e}}^{\mathsf{A}}={\mathrm{e}}^{\mathrm{tr}\mathsf{A}}.\text{\hspace{1em}(19.11)}$$

Thus, for all $\mathsf{A}\in \mathcal{H}$

$$\det {\mathrm{e}}^{t\mathsf{A}}={\mathrm{e}}^{t\mathrm{tr}\mathsf{A}}={\mathrm{e}}^0=1,$$

and by Example 19.5 the entire one-parameter subgroup $\exp (tA)$ lies in the unimodular group $SL(n,\mathbb{R})$

Since the map $\exp$ is a diffeomorphism from an open neighbourhood $U$ of0 in $\mathcal{G}\mathcal{L}(n,\mathbb{R})$ onto $\exp$($U$), every non-singular matrix $X$ in a connected neighbourhood of $I$ is uniquely expressible as an exponential, $\mathsf{X}={\mathsf{e}}^{\mathsf{A}}$ . Note the importance of connectedness here: the set of non-singular matrices has two connected components, being the inverse images of the two components of $\dot{\mathbb{R}}$ under the continuous map $\det$ : $GL(n,\mathbb{R})\to \mathbb{R}$ . The matrices of negative determinant clearly cannot be connected to the identity matrix by a smooth curve in $GL(n,\mathbb{R})$ since the determinant would need to vanish somewhere along such a curve. In particular the subgroup $SL(n,\mathbb{R})$ is connected since it is the inverse image of the connected set 1 under the determinant map. Every $\mathsf{X}\in SL(n,\mathbb{R})$ has $\det$ $\mathsf{X}=1$ and is therefore of the form $\mathsf{X}={\mathsf{e}}^{\mathsf{A}}$ where, by Eq. (19.11), $\mathsf{A}\in \mathcal{H}$ . Let $H$ be the unique connected

Lie subgroup $H$ whose Lie algebra is $\mathcal{H},$ , according to Theorem 19.4. In a neighbourhood of the identity every $\mathsf{X}={\mathsf{e}}^{\mathsf{A}}$ for some $\mathsf{A}\in \mathcal{H}.$ , and every matrix of the form ${\mathrm{e}}^{\mathsf{A}}$ belongs to $H$. Hence $SL(n,\mathbb{R})=H$ is the connected Lie subgroup of $GL(n,\mathbb{R})$ with Lie algebra $\begin{array}{r}S\mathcal{L}(n,\mathbb{R})=\mathcal{H}\end{array}$ . Since a Lie group and its Lie algebra are of equal dimension, the Lie group $\dim$ $SL(n,\mathbb{R})=\dim \mathcal{H}=n^2-1$

The Lie group $GL(n,\mathbb{C})$ has Lie algebra isomorphic with $M_n(\mathbb{C})$ , with bracket [$A$, $B$] again the commutator of the complex matrices $A$ and $B$. As discussed in Chapter 6 this complex Lie algebra must be regarded as the complexification of a real Lie algebra by restricting the field of scalars to the real numbers. In this way any complex Lie algebra of dimension $n$ can be considered as being a real Lie algebra of dimension 2$n$. As a real Lie group $GL(n,\mathbb{C})$ has dimension $2n^2$ , as does its Lie algebra $\mathcal{G}\mathcal{L}(n,\mathbb{C})$ ). A similar discussion to that above can be used to show that the unimodular group $SL(n,\mathbb{C})$ of complex matrices of determinant 1 has Lie algebra consisting of trace-free complex $n\times n$ matrices. Both $SL(n,\mathbb{C})$ and $\mathcal{SL}(n,\mathbb{C})$ have (real) dimension $2n^2-2$

Example 19.8

The orthogonal group $O(n)$ consists of real $n\times n$ matrices $R$ such that

$${\mathsf{RR}}^T=\mathsf{I}.$$

A one-parameter group of orthogonal transformations has the form $\mathsf{R}(t)=\exp (t\mathsf{A})={\mathsf{e}}^{t\mathsf{A}}$ , whence

$${\mathrm{e}}^{t\mathsf{A}}{\left({\mathrm{e}}^{t\mathsf{A}}\right)}^T={\mathrm{e}}^{t\mathsf{A}}{\mathrm{e}}^{t{\mathsf{A}}^T}=1.$$

Performing the derivative with respect to $t$ of this matrix equation results in

$${\mathsf{Ae}}^{t\mathsf{A}}{\mathsf{e}}^{t{\mathsf{A}}^T}+{\mathsf{e}}^{t\mathsf{A}}{\mathsf{e}}^{t{\mathsf{A}}^T}{\mathsf{A}}^T=\mathsf{A}+{\mathsf{A}}^T=\mathsf{O},$$

so that $A$ is a skew-symmetric matrix.

The set of skew-symmetric $n\times n$ matrices $\mathcal{O}(n)$ forms a Lie algebra since it is a vecto subspace of $M_n(\mathbb{R})$ and is closed with respect to commutator products,

$$\begin{array}{c}[\mathsf{A},\mathsf{B}{]}^T=(\mathsf{AB}-\mathsf{BA}{)}^T\\ ={\mathsf{B}}^T{\mathsf{A}}^T-{\mathsf{A}}^T{\mathsf{B}}^T\\ =-[{\mathsf{A}}^T,{\mathsf{B}}^T]=-[\mathsf{A},\mathsf{B}].\end{array}$$

Since every matrix ${\mathrm{e}}^{\mathsf{A}}$ is orthogonal for a skew-symmetric matrix $A$,

$$\mathsf{A}=-{\mathsf{A}}^T⟹{\mathrm{e}}^{\mathsf{A}}({\mathrm{e}}^{\mathsf{A}}{)}^T={\mathrm{e}}^{\mathsf{A}}{\mathrm{e}}^{{\mathsf{A}}^T}={\mathrm{e}}^{\mathsf{A}}{\mathrm{e}}^{-\mathsf{A}}=\mathsf{I},$$

$\mathcal{O}(n)$ is the Lie algebra corresponding to the connected Lie subgroup $SO(n)=O(n)\cap SL(n,\mathbb{R})$ . The dimensions of this Lie group and Lie algebra are clearly $\frac{1}{2}n(n-1)$

Similar arguments show that the unitary group $U(n)$ of complex matrices such that

$${\mathsf{UU}}^{†}\equiv \mathsf{U}{\overline{\mathsf{U}}}^T=\mathsf{I}$$

is a Lie group with Lie algebra $\mathcal{U}(n)$ consisting ofskew-hermitian matrices, $\mathsf{A}=-{\mathsf{A}}^{†}$ . The dimensions of $U(n)$ and $\mathcal{U}(n)$ are both $n^2$ . The group $SU(n)=U(n)\cap SL(n,\mathbb{C})$ has Lie algebra consisting of trace-free skew-hermitian matrices and has dimension $n^2-1$

Problems

Problem 19.6

For any $n\times n$ matrix $A$, show that

$$\frac{\mathrm{d}}{\mathrm{d}t}\det {{\mathrm{e}}^{t\mathsf{A}}|}_{t=0}=\mathrm{tr}\mathsf{A}.$$

Problem 19.7

Prove Eq. (19.11). One method is to find a matrix $S$ that transforms $A$ to upper triangular Jordan form by a similarity transformation as in Section 4.2, and use the fact that both determinant and trace are invariant under such transformations.

Problem 19.8

Show that $GL(n,\mathbb{C})$ and $SL(n,\mathbb{C})$ are connected Lie groups. Is $U(n)$ a connected group?

Problem 19.9

Show that the groups $SL(n,\mathbb{R})$ and $SO(n)$ are closed subgroups of $GL(N,\mathbb{R})$ , and that $U(n)$ and $SU(n)$ are closed subgroups of $GL(n,\mathbb{C})$ . Show furthermore that $SO(n)$ and $U(n)$ are compact Lie subgroups.

Problem 19.10

As in Example 2.13 let the symplectic group $Sp(n)$ consist of $2n\times 2n$ matrices $S$ such that

$${\mathsf{S}}^T\mathsf{JS}=\mathsf{J},\mathsf{J}=\left(\begin{array}{cc}\mathsf{O}& \mathsf{I}\\ -\mathsf{I}& \mathsf{O}\end{array}\right),$$

where $O$ is the $n\times n$ zero matrix and $L$ is the $n\times n$ unit matrix. Show that the Lie algebra $S_{\mathcal{P}}(n)$ consists of matrices $A$ satisfying

$${\mathsf{A}}^T\mathsf{J}+\mathsf{JA}=\mathsf{O}.$$

Verify that these matrices form a Lie algebra and generate the symplectic group. What is the dimension of the symplectic group? Is it a closed subgroup of $GL(2n,\mathbb{R})?$ Is it compact?