5.6 Lie groups and Lie algebras

A Lie group is a manifold on which the group manipulations, product and inverse, are defined. Lie groups play an extremely important role in the theory of fibre bundles and also find vast applications in physics. Here we will work out the geometrical aspects of Lie groups and Lie algebras.

5.6.1 Lie groups

Definition 5.8

A Lie group $G$ is a differentiable manifold which is endowed with a group structure such that the group operations

(i) $\cdot :G\times G\to G$ , $(g_1,g_2)\mapsto g_1\cdot g_2$
(ii) ${}^{-1}:G\to G$ , $g\mapsto g^{-1}$

are differentiable. [Remark: It can be shown that $G$ has a unique analytic structure with which the product and the inverse operations are written as convergent power series.]

The unit element of a Lie group is written as $e$ . The dimension of a Lie group $G$ is defined to be the dimension of $G$ as a manifold. The product symbol may be omitted and $g_1\cdot g_2$ is usually written as $g_1{g}_2$ . For example, let ${\mathbb{R}}^{\ast }\equiv \mathbb{R}-\{0\}$ . Take three elements $x,y,z\in {\mathbb{R}}^{\ast }$ such that $xy=z$ . Obviously if we multiply a number close to $x$ by a number close to $y$ , we have a number close to $z$ . Similarly, an inverse of a number close to $x$ is close to $1/x$ . In fact, we can differentiate these maps with respect to the relevant arguments and ${\mathbb{R}}^{\ast }$ is made into a Lie group with these group operations. If the product is commutative, namely $g_1{g}_2=g_2{g}_1$ , we often use the additive symbol + instead of the product symbol.

(a) Show that ${\mathbb{R}}^{+}=\{x\in \mathbb{R}|x>0\}$ is a Lie group with respect to multiplication.
(b) Show that $\mathbb{R}$ is a Lie group with respect to addition.
(c) Show that ${\mathbb{R}}^2$ is a Lie group with respect to addition defined by $(x_1,y_1)+(x_2,y_2)=(x_1+x_2,y_1+y_2)$ .

Example 5.14

Let $S^1$ be the unit circle on the complex plane,

$$S^1=\{{\mathrm{e}}^{\mathrm{i}\theta }|\theta \in \mathbb{R}(\mathrm{mod}2\pi )\}.$$

The group operations defined by ${\mathrm{e}}^{\mathrm{i}\theta }{\mathrm{e}}^{\mathrm{i}\phi }={\mathrm{e}}^{\mathrm{i}(\theta +\phi )}$ and $({\mathrm{e}}^{\mathrm{i}\theta }{)}^{-1}={\mathrm{e}}^{-\mathrm{i}\theta }$ are differentiable and $S^1$ is made into a Lie group, which we call $\mathrm{U}(1)$ . It is easy to see that the group operations are the same as those in exercise 5.19(b) modulo $2\pi$ .

Of particular interest in physical applications are the matrix groups which are subgroups of general linear groups $\mathrm{GL}(n,\mathbb{R})$ or $\mathrm{GL}(n,\mathbb{C})$ . The product of

elements is simply the matrix multiplication and the inverse is given by the matrix inverse. The coordinates of $\mathrm{GL}(n,\mathbb{R})$ are given by $n^2$ entries of $M=\{x_{ij}\}$ . $\mathrm{GL}(n,\mathbb{R})$ is a non-compact manifold of real dimension $n^2$ .

Interesting subgroups of $\mathrm{GL}(n,\mathbb{R})$ are the orthogonal group $\mathrm{O}(n)$ , the special linear group $\mathrm{SL}(n,\mathbb{R})$ and the special orthogonal group $\mathrm{SO}(n)$ :

$$\mathrm{O}(n)=\{M\in \mathrm{GL}(n,\mathbb{R})|M{M}^{\mathrm{t}}=M^{\mathrm{t}}M=I_n\}\text{\hspace{1em}(5.103)}$$

$$\mathrm{SL}(n,\mathbb{R})=\{M\in \mathrm{GL}(n,\mathbb{R})|\det M=1\}\text{\hspace{1em}(5.104)}$$

$$\mathrm{SO}(n)=\mathrm{O}(n)\cap \mathrm{SL}(n,\mathbb{R})\text{\hspace{1em}(5.105)}$$

where ${}^{\mathrm{t}}$ denotes the transpose of a matrix. In special relativity, we are familiar with the Lorentz group

$$\mathrm{O}(1,3)=\{M\in \mathrm{GL}(4,\mathbb{R})|M\eta M^t=\eta \}$$

where $\eta$ is the Minkowski metric, $\eta =\mathrm{diag}(-1,1,1,1)$ . Extension to higher-dimensional spacetime is trivial.

Exercise 5.20

Show that the group $\mathrm{O}(1,3)$ is non-compact and has four connected components according to the sign of the determinant and the sign of the $(0,0)$ entry. The component that contains the unit matrix is denoted by ${\mathrm{O}}_{+}^{\uparrow }(1,3)$ .

The group $\mathrm{GL}(n,\mathbb{C})$ is the set of non-singular linear transformations in ${\mathbb{C}}^n$ , which are represented by $n\times n$ non-singular matrices with complex entries. The unitary group $\mathrm{U}(n)$ , the special linear group $\mathrm{SL}(n,\mathbb{C})$ and the special unitary group $\mathrm{SU}(n)$ are defined by

$$\mathrm{U}(n)=\{M\in \mathrm{GL}(n,\mathbb{C})|M{M}^{†}=M^{†}M=1\}\text{\hspace{1em}(5.106)}$$

$$\mathrm{SL}(n,\mathbb{C})=\{M\in \mathrm{GL}(n,\mathbb{C})|\det M=1\}\text{\hspace{1em}(5.107)}$$

$$\mathrm{SU}(n)=\mathrm{U}(n)\cap \mathrm{SL}(n,\mathbb{C})\text{\hspace{1em}(5.108)}$$

where $†$ is the Hermitian conjugate.

So far we have just mentioned that the matrix groups are subgroups of a Lie group $\mathrm{GL}(n,\mathbb{R})$ (or $\mathrm{GL}(n,\mathbb{C})$ ). The following theorem guarantees that they are Lie subgroups, that is, these subgroups are Lie groups by themselves. We accept this important (and difficult to prove) theorem without proof.

Theorem 5.2

Every closed subgroup $H$ of a Lie group $G$ is a Lie subgroup.

For example, $\mathrm{O}(n)$ , $\mathrm{SL}(n,\mathbb{R})$ and $\mathrm{SO}(n)$ are Lie subgroups of $\mathrm{GL}(n,\mathbb{R})$ . To see why $\mathrm{SL}(n,\mathbb{R})$ is a closed subgroup, consider a map $f:\mathrm{GL}(n,\mathbb{R})\to \mathbb{R}$ defined by $A\mapsto \det A$ . Obviously $f$ is a continuous map and $f^{-1}(1)=\mathrm{SL}(n,\mathbb{R})$ . A point $\{1\}$ is a closed subset of $\mathbb{R}$ , hence $f^{-1}(1)$ is closed in $\mathrm{GL}(n,\mathbb{R})$ . Then theorem 5.2 states that $\mathrm{SL}(n,\mathbb{R})$ is a Lie subgroup. The reader should verify that $\mathrm{O}(n)$ and $\mathrm{SO}(n)$ are also Lie subgroups of $\mathrm{GL}(n,\mathbb{R})$ .

Let $G$ be a Lie group and $H$ a Lie subgroup of $G$ . Define an equivalence relation $\sim$ by $g\sim g^{\prime }$ if there exists an element $h\in H$ such that $g^{\prime }=gh$ . An equivalence class $[g]$ is a set $\{gh|h\in H\}$ . The coset space $G/H$ is a manifold (not necessarily a Lie group) with $\dim G/H=\dim G-\dim H$ . $G/H$ is a Lie group if $H$ is a normal subgroup of $G$ , that is, if $gh{g}^{-1}\in H$ for any $g\in G$ and $h\in H$ . In fact, take equivalence classes $[g]$ , $[g^{\prime }]\in G/H$ and construct the product $[g][g^{\prime }]$ . If the group structure is well defined in $G/H$ , the product must be independent of the choice of the representatives. Let $gh$ and $g^{\prime }{h}^{\prime }$ be the representatives of $[g]$ and $[g^{\prime }]$ respectively. Then $gh{g}^{\prime }{h}^{\prime }=g{g}^{\prime }{h}^{\prime \prime }{h}^{\prime }\in [g{g}^{\prime }]$ where the equality follows since there exists $h^{\prime \prime }\in H$ such that $h{g}^{\prime }=g^{\prime }{h}^{\prime \prime }$ . It is left as an exercise to the reader to show that $[g{]}^{-1}$ is also a well defined operation and $[g{]}^{-1}=[g^{-1}]$ .

5.6.2 Lie algebras

Definition 5.9

Let $a$ and $g$ be elements of a Lie group $G$ . The right-translation $R_a:G\to G$ and the left-translation $L_a:G\to G$ of $g$ by $a$ are defined by

$$R_{a}g=ga\text{\hspace{1em}(5.109a)}$$

$$L_{a}g=ag.\text{\hspace{1em}(5.109b)}$$

By definition, $R_a$ and $L_a$ are diffeomorphisms from $G$ to $G$ . Hence, the maps $L_a:G\to G$ and $R_a:G\to G$ induce $L_{a\ast }:T_{g}G\to T_{ag}G$ and $R_{a\ast }:T_{g}G\to T_{ga}G$ ; see section 5.2. Since these translations give equivalent theories, we are concerned mainly with the left-translation in the following. The analysis based on the right-translation can be carried out in a similar manner.

Given a Lie group $G$ , there exists a special class of vector fields characterized by an invariance under group action. [On the usual manifold there is no canonical way of discriminating some vector fields from the others.]

Definition 5.10

Let $X$ be a vector field on a Lie group $G$ . $X$ is said to be a left-invariant vector field if $L_{a\ast }X{|}_g=X{|}_{ag}$ .

Exercise 5.21

Verify that a left-invariant vector field $X$ satisfies

$${L_{a\ast }X|}_g=X^{\mu }(g)\frac{\partial x^{\nu }(ag)}{\partial x^{\mu }(g)}{\frac{\partial }{\partial x^{\nu }}|}_{ag}=X^{\nu }(ag){\frac{\partial }{\partial x^{\nu }}|}_{ag}\text{\hspace{1em}(5.110)}$$

where $x^{\mu }(g)$ and $x^{\mu }(ag)$ are coordinates of $g$ and $ag$ , respectively.

A vector $V\in T_{e}G$ defines a unique left-invariant vector field $X_V$ throughout $G$ by

$${X_V|}_g=L_{g\ast }Vg\in G.\text{\hspace{1em}(5.111)}$$

In fact, we verify from (5.34) that $X_V{|}_{ag}=L_{ag\ast }V=(L_a{L}_g{)}_{\ast }V=L_{a\ast }{L}_{g\ast }V=L_{a\ast }{X}_V{|}_g$ . Conversely, a left-invariant vector field $X$ defines a unique vector $V=X{|}_e\in T_{e}G$ . Let us denote the set of left-invariant vector fields on $G$ by

$\mathfrak{g}$ . The map $T_{e}G\to \mathfrak{g}$ defined by $V\mapsto X_V$ is an isomorphism and it follows that the set of left-invariant vector fields is a vector space isomorphic to $T_{e}G$ . In particular, $\dim \mathfrak{g}=\dim G$ .

Since $\mathfrak{g}$ is a set of vector fields, it is a subset of $\mathcal{X}(G)$ and the Lie bracket defined in section 5.3 is also defined on $\mathfrak{g}$ . We show that $\mathfrak{g}$ is closed under the Lie bracket. Take two points $g$ and $ag=L_{a}g$ in $G$ . If we apply $L_{a\ast }$ to the Lie bracket $[X,Y]$ of $X,Y\in \mathfrak{g}$ , we have

$${L_{a\ast }[X,Y]|}_g={\left[L_{a\ast }X\right|}_g,{L_{a\ast }Y|}_g={[X,Y]|}_{ag}\text{\hspace{1em}(5.112)}$$

where the left-invariances of $X$ and $Y$ and (5.52) have been used. Thus, $[X,Y]\in \mathfrak{g}$ , that is $\mathfrak{g}$ is closed under the Lie bracket.

It is instructive to work out the left-invariant vector field of $\mathrm{GL}(n,\mathbb{R})$ . The coordinates of $\mathrm{GL}(n,\mathbb{R})$ are given by $n^2$ entries $x^{ij}$ of the matrix. The unit element is $e=I_n=({\delta }^{ij})$ . Let $g=\{x^{ij}(g)\}$ and $a=\{x^{ij}(a)\}$ be elements of $\mathrm{GL}(n,\mathbb{R})$ . The left-translation is

$$L_{a}g=ag=\sum x^{ik}(a)x^{kj}(g).$$

Take a vector $V=\sum V^{ij}\partial /\partial x^{ij}{|}_e\in T_{e}G$ where the $V^{ij}$ are the entries of $V$ . The left-invariant vector field generated by $V$ is

$$\begin{array}{rl}{X}_V{|}_g& =L_{g\ast }V=\sum _{ijklm}{V}^{ij}\frac{\partial }{\partial x^{ij}}{|}_e{x}^{kl}(g)x^{lm}(e)\frac{\partial }{\partial x^{km}}{|}_g\\ & =\sum V^{ij}{x}^{kl}(g){\delta }_i^l{\delta }_j^m{\frac{\partial }{\partial x^{km}}|}_g\\ & =\sum x^{ki}(g)V^{ij}{\frac{\partial }{\partial x^{kj}}|}_g=\sum (gV{)}^{kj}\frac{\partial }{\partial x^{kj}}{|}_g\end{array}\text{\hspace{1em}(5.113)}$$

where $gV$ is the usual matrix multiplication of $g$ and $V$ . The vector $X_V{|}_g$ is often abbreviated as $gV$ since it gives the components of the vector.

The Lie bracket of $X_V$ and $X_W$ generated by $V=V^{ij}\partial /\partial x^{ij}{|}_e$ and $W=W^{ij}\partial /\partial x^{ij}{|}_e$ is

$$\begin{array}{rl}{\left[X_V,X_W\right]|}_g& =\sum x^{ki}(g)V^{ij}\frac{\partial }{\partial x^{kj}}{|}_g{x}^{ca}(g)W^{ab}\frac{\partial }{\partial x^{cb}}{|}_g-(V\leftrightarrow W)\\ & =\sum x^{ij}(g)[V^{jk}{W}^{kl}-W^{jk}{V}^{kl}]{\frac{\partial }{\partial x^{il}}|}_g\\ & =\sum (g[V,W]{)}^{ij}\frac{\partial }{\partial x^{ij}}{|}_g.\end{array}\text{\hspace{1em}(5.114)}$$

Clearly, (5.113) and (5.114) remain true for any matrix group and we establish that

$$\begin{array}{rl}{L}_{g\ast }V& =gV(5.115)\\ & {\left[X_V,X_W\right]|}_g=L_{g\ast }[V,W]=g[V,W].(5.116)\end{array}$$

Now a Lie algebra is defined as the set of left-invariant vector fields $\mathfrak{g}$ with the Lie bracket.

Definition 5.11

The set of left-invariant vector fields $\mathfrak{g}$ with the Lie bracket $[,]:\mathfrak{g}\times \mathfrak{g}\to \mathfrak{g}$ is called the Lie algebra of a Lie group $G$

We denote the Lie algebra of a Lie group by the corresponding lower-case German gothic letter. For example $\mathfrak{so}(n)$ is the Lie algebra of $\mathrm{SO}(n)$ .

(a) Take $G=\mathbb{R}$ as in exercise 5.19(b). If we define the left translation $L_a$ by $x\mapsto x+a$ , the left-invariant vector field is given by $X=\partial /\partial x$ . In fact,

$$L_{a\ast }X{|}_x=\frac{\partial (a+x)}{\partial x}\frac{\partial }{\partial (a+x)}=\frac{\partial }{\partial (x+a)}=X{|}_{x+a}.$$

Clearly this is the only left-invariant vector field on $\mathbb{R}$ . We also find that $X=\partial /\partial \theta$ is the unique left-invariant vector field on $G=\mathrm{SO}(2)=\{{\mathrm{e}}^{\mathrm{i}\theta }|0\le \theta \le 2\pi \}$ . Thus, the Lie groups $\mathbb{R}$ and $\mathrm{SO}(2)$ share the common Lie algebra.

(b) Let $\mathfrak{gl}(n,\mathbb{R})$ be the Lie algebra of $\mathrm{GL}(n,\mathbb{R})$ and $c:(-\epsilon ,\epsilon )\to \mathrm{GL}(n,\mathbb{R})$ be a curve with $c(0)=I_n$ . The curve is approximated by $c(s)=I_n+sA+O(s^2)$ near $s=0$ , where $A$ is an $n\times n$ matrix of real entries. Note that for small enough $s$ , $\det c(s)$ cannot vanish and $c(s)$ is, indeed, in $\mathrm{GL}(n,\mathbb{R})$ . The tangent vector to $c(s)$ at $I_n$ is $c^{\prime }(s){|}_{s=0}=A$ . This shows that $\mathfrak{gl}(n,\mathbb{R})$ is the set of $n\times n$ matrices. Clearly $\dim \mathfrak{gl}(n,\mathbb{R})=n^2=\dim \mathrm{GL}(n,\mathbb{R})$ . Subgroups of $\mathrm{GL}(n,\mathbb{R})$ are more interesting.

(c) Let us find the Lie algebra $\mathfrak{sl}(n,\mathbb{R})$ of $\mathrm{SL}(n,\mathbb{R})$ . Following this prescription, we approximate a curve through $I_n$ by $c(s)=I_n+sA+O(s^2)$ . The tangent vector to $c(s)$ at $I_n$ is ${c^{\prime }(s)|}_{s=0}=A$ . Now, for the curve $c(s)$ to be in $\mathrm{SL}(n,\mathbb{R})$ , $c(s)$ has to satisfy $\det c(s)=1+\mathrm{str}A=1$ , namely $\mathrm{tr}A=0$ . Thus, $\mathfrak{sl}(n,\mathbb{R})$ is the set of $n\times n$ traceless matrices and $\dim \mathfrak{sl}(n,\mathbb{R})=n^2-1$ .

(d) Let $c(s)=I_n+sA+O(s^2)$ be a curve in $\mathrm{SO}(n)$ through $I_n$ . Since $c(s)$ is a curve in $\mathrm{SO}(n)$ , it satisfies $c(s{)}^{\mathfrak{t}}c(s)=I_n$ . Differentiating this identity, we obtain $c^{\prime }(s{)}^{\mathfrak{t}}c(s)+c(s{)}^{\mathfrak{t}}{c}^{\prime }(s)=0$ . At $s=0$ , this becomes $A^{\mathfrak{t}}+A=0$ . Hence, $\mathfrak{so}(n)$ is the set of skew-symmetric matrices. Since we are interested only in the vicinity of the unit element, the Lie algebra of $\mathrm{O}(n)$ is the same as that of $\mathrm{SO}(n)$ : $\mathfrak{o}(n)=\mathfrak{so}(n)$ . It is easy to see that $\dim \mathfrak{o}(n)=\dim \mathfrak{so}(n)=n(n-1)/2$ .

(e) A similar analysis can be carried out for matrix groups of $\mathrm{GL}(n,\mathbb{C})$ . $\mathfrak{gl}(n,\mathbb{C})$ is the set of $n\times n$ matrices with complex entries and $\dim \mathfrak{gl}(n,\mathbb{C})=2n^2$ (the dimension here is a real dimension). $\mathfrak{sl}(n,\mathbb{C})$ is the set of traceless matrices with real dimension $2(n^2-1)$ . To find $\mathfrak{u}(n)$ , we consider a curve $c(s)=I_n+sA+O(s^2)$ in $\mathrm{U}(n)$ . Since $c(s{)}^{†}c(s)=I_n$ , we have $c^{\prime }(s{)}^{†}c(s)+c(s{)}^{†}{c}^{\prime }(s)=0$ . At $s=0$ , we have $A^{†}+A=0$ .

Hence, $\mathfrak{u}(n)$ is the set of skew-Hermitian matrices with $\dim \mathfrak{u}(n)=n^2$ . $\mathfrak{su}(n)=\mathfrak{u}(n)\cap \mathfrak{sl}(n)$ is the set of traceless skew-Hermitian matrices with $\dim \mathfrak{su}(n)=n^2-1$ .

Exercise 5.22

Let

$$c(s)=\left(\begin{array}{ccc}\cos s& -\sin s& 0\\ \sin s& \cos s& 0\\ 0& 0& 1\end{array}\right)$$

be a curve in SO(3). Find the tangent vector to this curve at $I_3$ .

5.6.3 The one-parameter subgroup

A vector field $X\in \mathcal{X}(M)$ generates a flow in $M$ (section 5.3). Here we are interested in the flow generated by a left-invariant vector field.

Definition 5.12

A curve $\varphi :\mathbb{R}\to G$ is called a one-parameter subgroup of $G$ if it satisfies the condition

$$\varphi (t)\varphi (s)=\varphi (t+s).\text{\hspace{1em}(5.117)}$$

It is easy to see that $\varphi (0)=e$ and ${\varphi }^{-1}(t)=\varphi (-t)$ . Note that the curve $\varphi$ thus defined is a homomorphism from $\mathbb{R}$ to $G$ . Although $G$ may be non-Abelian, a one-parameter subgroup is an Abelian subgroup: $\varphi (t)\varphi (s)=\varphi (t+s)=\varphi (s+t)=\varphi (s)\varphi (t)$ .

Given a one-parameter subgroup $\varphi :\mathbb{R}\to G$ , there exists a vector field $X$ such that

$$\frac{\mathrm{d}{\varphi }^{\mu }(t)}{\mathrm{d}t}=X^{\mu }(\varphi (t)).\text{\hspace{1em}(5.118)}$$

We now show that the vector field $X$ is left-invariant. First note that the vector field $\mathrm{d}/\mathrm{d}t$ is left-invariant on $\mathbb{R}$ , see example 5.15(a). Thus, we have

$${{\left(L_t\right)}_{\ast }\frac{\mathrm{d}}{\mathrm{d}t}|}_0={\frac{\mathrm{d}}{\mathrm{d}t}|}_t.\text{\hspace{1em}(5.119)}$$

Next, we apply the induced map ${\varphi }_{\ast }:T_t\mathbb{R}\to T_{\varphi (t)}G$ on the vectors $\mathrm{d}/\mathrm{d}t{|}_0$ and $\mathrm{d}/\mathrm{d}t{|}_t$ ,

$${{\varphi }_{\ast }\frac{\mathrm{d}}{\mathrm{d}t}|}_0={\frac{\mathrm{d}{\varphi }^{\mu }(t)}{\mathrm{d}t}|}_0{\frac{\partial }{\partial g^{\mu }}|}_e=X{|}_e\text{\hspace{1em}(5.120a)}$$

$${{\varphi }_{\ast }\frac{\mathrm{d}}{\mathrm{d}t}|}_t={\frac{\mathrm{d}{\varphi }^{\mu }(t)}{\mathrm{d}t}|}_t{\frac{\partial }{\partial g^{\mu }}|}_g=X{|}_g\text{\hspace{1em}(5.120b)}$$

where we put $\varphi (t)=g$ . From (5.119) and (5.120b), we have

$${{\left(\varphi L_t\right)}_{\ast }\frac{\mathrm{d}}{\mathrm{d}t}|}_0={{\varphi }_{\ast }{L}_{t\ast }\frac{\mathrm{d}}{\mathrm{d}t}|}_0=X{|}_g.\text{\hspace{1em}(5.121a)}$$

It follows from the commutativity $\varphi L_t=L_g\varphi$ that ${\varphi }_{\ast }{L}_{t\ast }=L_{g\ast }{\varphi }_{\ast }$ . Then (5.121a) becomes

$${{\varphi }_{\ast }{L}_{t\ast }\frac{\mathrm{d}}{\mathrm{d}t}|}_0={L_{g\ast }{\varphi }_{\ast }\frac{\mathrm{d}}{\mathrm{d}t}|}_0=L_{g\ast }X{|}_e.\text{\hspace{1em}(5.121b)}$$

From (5.121), we conclude that

$$L_{g\ast }X{|}_e=X{|}_g.\text{\hspace{1em}(5.122)}$$

Thus, given a flow $\varphi (t)$ , there exists an associated left-invariant vector field $X\in \mathfrak{g}$ .

Conversely, a left-invariant vector field $X$ defines a one-parameter group of transformations $\sigma (t,g)$ such that $\mathrm{d}\sigma (t,g)/\mathrm{d}t=X$ and $\sigma (0,g)=g$ . If we define $\varphi :\mathbb{R}\to G$ by $\varphi (t)\equiv \sigma (t,e)$ , the curve $\varphi (t)$ becomes a one-parameter subgroup of $G$ . To prove this, we have to show $\varphi (s+t)=\varphi (s)\varphi (t)$ . By definition, $\sigma$ satisfies

$$\frac{\mathrm{d}}{\mathrm{d}t}\sigma (t,\sigma (s,e))=X(\sigma (t,\sigma (s,e))).\text{\hspace{1em}(5.123)}$$

[We have omitted the coordinate indices for notational simplicity. If readers feel uneasy, they may supplement the indices as in (5.118).] If the parameter $s$ is fixed, $\bar{\sigma }(t,\varphi (s))\equiv \varphi (s)\varphi (t)$ is a curve $\mathbb{R}\to G$ at $\varphi (s)\varphi (0)=\varphi (s)$ . Clearly $\sigma$ and $\bar{\sigma }$ satisfy the same initial condition,

$$\sigma (0,\sigma (s,e))=\bar{\sigma }(0,\varphi (s))=\varphi (s).\text{\hspace{1em}(5.124)}$$

$\bar{\sigma }$ also satisfies the same differential equation as $\sigma$ ..

$$\begin{array}{rl}\frac{\mathrm{d}}{\mathrm{d}t}\bar{\sigma }(t,\varphi (t))& =\frac{\mathrm{d}}{\mathrm{d}t}\varphi (s)\varphi (t)={\left(L_{\varphi (s)}\right)}_{\ast }\frac{\mathrm{d}}{\mathrm{d}t}\varphi (t)\\ & =\left(L_{\varphi (s)}\right)\ast X(\varphi (t))\\ & =X(\varphi (s)\varphi (t))\text{(l e f t - i n v a r i a n c e)}\\ & =X(\bar{\sigma }(t,\varphi (s))).\end{array}\text{\hspace{1em}(5.125)}$$

From the uniqueness theorem of ODEs, we conclude that

$$\varphi (s+t)=\varphi (s)\varphi (t).\text{\hspace{1em}(5.126)}$$

We have found that there is a one-to-one correspondence between a one-parameter subgroup of $G$ and a left-invariant vector field. This correspondence becomes manifest if we define the exponential map as follows.

Definition 5.13

Let $G$ be a Lie group and $V\in T_{e}G$ . The exponential map $\exp :T_{e}G\to G$ is defined by

$$\exp V\equiv {\varphi }_V(1)\text{\hspace{1em}(5.127)}$$

where ${\varphi }_V$ is a one-parameter subgroup of $G$ generated by the left-invariant vector field $X_V{|}_g=L_{g^{\ast }}V$ .

Proposition 5.2

Let $V\in T_{e}G$ and let $t\in \mathbb{R}$ . Then

$$\exp (tV)={\varphi }_V(t)\text{\hspace{1em}(5.128)}$$

where ${\varphi }_V(t)$ is a one-parameter subgroup generated by $X_V{|}_g=L_{g\ast }V$ .

Proof

Let $a\ne 0$ be a constant. Then ${\varphi }_V(at)$ satisfies

$${\frac{\mathrm{d}}{\mathrm{d}t}{\varphi }_V(at)|}_{t=0}={a\frac{\mathrm{d}}{\mathrm{d}t}{\varphi }_V(t)|}_{t=0}=aV$$

which shows that ${\varphi }_V(at)$ is a one-parameter subgroup generated by $L_{g\ast }aV$ . The left-invariant vector field $L_{g\ast }aV$ also generates ${\varphi }_{aV}(t)$ and, from the uniqueness of the solution, we find that ${\varphi }_V(at)={\varphi }_{aV}(t)$ . From definition 5.13, we have

$$\exp (aV)={\varphi }_{aV}(1)={\varphi }_V(a).$$

The proof is completed if $a$ is replaced by $t$ .

For a matrix group, the exponential map is given by the exponential of a matrix. Take $G=\mathrm{GL}(n,\mathbb{R})$ and $A\in \mathfrak{gl}(n,\mathbb{R})$ . Let us define a one-parameter subgroup ${\varphi }_A:\mathbb{R}\to \mathrm{GL}(n,\mathbb{R})$ by

$${\varphi }_A(t)=\exp (tA)=I_n+tA+\frac{t^2}{2!}{A}^2+\cdots +\frac{t^n}{n!}{A}^n+\dots .\text{\hspace{1em}(5.129)}$$

In fact, ${\varphi }_A(t)\in \mathrm{GL}(n,\mathbb{R})$ since $[{\varphi }_A(t){]}^{-1}={\varphi }_A(-t)$ exists. It is also easy to see ${\varphi }_A(t){\varphi }_A(s)=\varphi (t+s)$ . Now the exponential map is given by

$${\varphi }_A(1)=\exp (A)=I_n+A+\frac{1}{2!}{A}^2+\cdots +\frac{1}{n!}{A}^n+\dots .\text{\hspace{1em}(5.130)}$$

The curve $g\exp (tA)$ is a flow through $g\in G$ . We find that

$$\frac{\mathrm{d}}{\mathrm{d}t}g\exp (tA){|}_{t=0}=L_{g\ast }A=X_A{|}_g$$

where $X_A$ is a left-invariant vector field generated by $A$ . From (5.115), we find, for a matrix group $G$ , that

$$L_{g\ast }A=X_A{|}_g=gA.\text{\hspace{1em}(5.131)}$$

The curve $g\exp (tA)$ defines a map ${\sigma }_t:G\to G$ by ${\sigma }_t(g)\equiv g\exp (tA)$ which is also expressed as a right-translation,

$${\sigma }_t=R_{\exp (tA)}.\text{\hspace{1em}(5.132)}$$

5.6.4 Frames and structure equation

Let the set of $n$ vectors $\{V_1,V_2,\dots ,V_n\}$ be a basis of $T_{e}G$ where $n=\dim G$ . [We assume throughout this book that $n$ is finite.] The basis defines the set of $n$ linearly independent left-invariant vector fields $\{X_1,X_2,\dots ,X_n\}$ at each point $g$ in $G$ by $X_{\mu }{|}_g=L_{g\ast }{V}_{\mu }$ . Note that the set $\{X_{\mu }\}$ is a frame of a basis defined throughout $G$ . Since $[X_{\mu },X_{\nu }]{|}_g$ is again an element of $\mathfrak{g}$ at $g$ , it can be expanded in terms of $\{X_{\mu }\}$ as

$$\left[X_{\mu },X_v\right]=c_{\mu v}^{\lambda }{X}_{\lambda }\text{\hspace{1em}(5.133)}$$

where $c_{\mu \nu }^{\lambda }$ are called the structure constants of the Lie group $G$ . If $G$ is a matrix group, the LHS of (5.133) at $g=e$ is precisely the commutator of matrices $V_{\mu }$ and $V_{\nu }$ ; see (5.116). We show that the $c_{\mu \nu }^{\lambda }$ are, indeed, constants independent of $g$ . Let $c_{\mu \nu }^{\lambda }(e)$ be the structure constants at the unit element. If $L_{g\ast }$ is applied to the Lie bracket, we have

$$\left[X_{\mu },X_{\nu }\right]{|}_g=c_{\mu \nu }^{\lambda }(e)X_{\lambda }{|}_g$$

which shows the $g$ -independence of the structure constants. In a sense, the structure constants determine a Lie group completely (Lie’s theorem).

Exercise 5.23

Show that the structure constants satisfy

(a) skew-symmetry

$$c_{\mu \nu }^{\lambda }=-c_{\nu \mu }^{\lambda }\text{\hspace{1em}(5.134)}$$

(b) Jacobi identity

$$c_{\mu \nu }^{\tau }{c}_{\tau \rho }^{\lambda }+c_{\rho \mu }^{\tau }{c}_{\tau \nu }^{\lambda }+c_{\nu \rho }^{\tau }{c}_{\tau \mu }^{\lambda }=0.\text{\hspace{1em}(5.135)}$$

Let us introduce a dual basis to $\{X_{\mu }\}$ and denote it by $\{{\theta }^{\mu }\}$ ; $⟨{\theta }^{\mu },X_{\nu }⟩={\delta }_{\nu }^{\mu }$ . $\{{\theta }^{\mu }\}$ is a basis for the left-invariant one-forms. We will show that the dual basis satisfies Maurer-Cartan’s structure equation,

$$\mathrm{d}{\theta }^{\mu }=-\frac{1}{2}{c}_{\nu \lambda }^{\mu }{\theta }^{\nu }\wedge {\theta }^{\lambda }.\text{\hspace{1em}(5.136)}$$

This can be seen by making use of (5.70):

$$\begin{array}{rl}\mathrm{d}{\theta }^{\mu }\left(X_{\nu },X_{\lambda }\right)& =X_{\nu }\left[{\theta }^{\mu }\left(X_{\lambda }\right)\right]-X_{\lambda }\left[{\theta }^{\mu }\left(X_{\nu }\right)\right]-{\theta }^{\mu }\left(\left[X_{\nu },X_{\lambda }\right]\right)\\ & =X_{\nu }\left[{\delta }_{\lambda }^{\mu }\right]-X_{\lambda }\left[{\delta }_{\nu }^{\mu }\right]-{\theta }^{\mu }\left(c_{\nu \lambda }^{\kappa }{X}_{\kappa }\right)=-c_{\nu \lambda }^{\mu }\end{array}$$

which proves (5.136).

We define a Lie-algebra-valued one-form $\theta :T_{g}G\to T_{e}G$ by

$$\theta :X\mapsto {\left(L_{g^{-1}}\right)}_{\ast }X={\left(L_g\right)}_{\ast }^{-1}XX\in T_{g}G.\text{\hspace{1em}(5.137)}$$

$\theta$ is called the canonical one-form or Maurer-Cartan form on $G$ .

Theorem 5.3

(a) The canonical one-form $\theta$ is expanded as

$$\theta =V_{\mu }\otimes {\theta }^{\mu }\text{\hspace{1em}(5.138)}$$

where $\{V_{\mu }\}$ is the basis of $T_{e}G$ and $\{{\theta }^{\mu }\}$ the dual basis of $T_e^{\ast }G$ .

(b) The canonical one-form $\theta$ satisfies

$$\mathrm{d}\theta +\frac{1}{2}[\theta \wedge \theta ]=0\text{\hspace{1em}(5.139)}$$

where $\mathrm{d}\theta \equiv V_{\mu }\otimes \mathrm{d}{\theta }^{\mu }$ and

$$[\theta \wedge \theta ]\equiv [V_{\mu },V_{\nu }]\otimes {\theta }^{\mu }\wedge {\theta }^{\nu }.\text{\hspace{1em}(5.140)}$$

Proof

(a) Take any vector $Y=Y^{\mu }{X}_{\mu }\in T_{g}G$ , where $\{X_{\mu }\}$ is the set of frame vectors generated by $\{V_{\mu }\}$ ; $X_{\mu }{|}_g=L_{g\ast }{V}_{\mu }$ . From (5.137), we find

$$\theta (Y)=Y^{\mu }\theta (X_{\mu })=Y^{\mu }(L_{g\ast }{)}^{-1}[L_{g\ast }{V}_{\mu }]=Y^{\mu }{V}_{\mu }.$$

However,

$$(V_{\mu }\otimes {\theta }^{\mu })(Y)=Y^v{V}_{\mu }{\theta }^{\mu }(X_v)=Y^v{V}_{\mu }{\delta }_v^{\mu }=Y^{\mu }{V}_{\mu }.$$

Since $Y$ is arbitrary, we have $\theta =V_{\mu }\otimes {\theta }^{\mu }$

(b) We use the Maurer-Cartan structure equation (5.136):

$$\mathrm{d}\theta +\frac{1}{2}[\theta \wedge \theta ]=-\frac{1}{2}{V}_{\mu }\otimes c_{\nu \lambda }^{\mu }{\theta }^{\nu }\wedge {\theta }^{\lambda }+\frac{1}{2}{c}_{\nu \lambda }^{\mu }{V}_{\mu }\otimes {\theta }^{\nu }\wedge {\theta }^{\lambda }=0$$

where the $c_{\nu \lambda }^{\mu }$ are the structure constants of $G$

来源与版本

正文：英文 · OCR 机器稿 · 待校对

核对状态：OCR 机器稿 · 待校对

原书 PDF

来源版本：second-edition-local

来源 PDF SHA-256：fcd24c3371d092a8c699ccf5cf0765462924b6c48412804231076e8084b1940d

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