5.7 The action of Lie groups on manifolds

In physics, a Lie group often appears as the set of transformations acting on a manifold. For example, SO(3) is the group of rotations in ${\mathbb{R}}^3$ , while the Poincaré group is the set of transformations acting on the Minkowski spacetime. To study more general cases, we abstract the action of a Lie group $G$ on a manifold $M$ . We have already encountered this interaction between a group and geometry. In section 5.3 we defined a flow in a manifold $M$ as a map $\sigma :\mathbb{R}\times M\to M$ , in which $\mathbb{R}$ acts as an additive group. We abstract this idea as follows.

5.7.1 Definitions

Definition 5.14

Let $G$ be a Lie group and $M$ be a manifold. The action of $G$ on $M$ is a differentiable map $\sigma :G\times M\to M$ which satisfies the conditions

(i) $\sigma (e,p)=p$ for any $p\in M$ (5.141a)
(ii) $\sigma (g_1,\sigma (g_2,p))=\sigma (g_1{g}_2,p).$ (5.141b)

[Remark: We often use the notation $gp$ instead of $\sigma (g,p)$ . The second condition in this notation is $g_1(g_{2}p)=(g_1{g}_2)p$ .]

Example 5.16

(a) A flow is an action of $\mathbb{R}$ on a manifold $M$ . If a flow is periodic with a period $T$ , it may be regarded as an action of $\mathrm{U}(1)$ or $\mathrm{SO}(2)$ on $M$ . Given a periodic flow $\sigma (t,x)$ with period $T$ , we construct a new action $\bar{\sigma }(\exp (2\pi \mathrm{i}t/T),x)\equiv \sigma (t,x)$ whose group $G$ is $\mathrm{U}(1)$ .

(b) Let $M\in \mathrm{GL}(n,\mathbb{R})$ and let $x\in {\mathbb{R}}^n$ . The action of $\mathrm{GL}(n,\mathbb{R})$ on ${\mathbb{R}}^n$ is defined by the usual matrix action on a vector:

$$\sigma (M,x)=M\cdot x.\text{\hspace{1em}(5.142)}$$

The action of the subgroups of $\mathrm{GL}(n,\mathbb{R})$ is defined similarly. They may also act on a smaller space. For example, $\mathrm{O}(n)$ acts on $S^{n-1}(r)$ , an $(n-1)$ -sphere of radius $r$ ,

$$\sigma :\mathrm{O}(n)\times S^{n-1}(r)\to S^{n-1}(r).\text{\hspace{1em}(5.143)}$$

(c) It is known that $\mathrm{SL}(2,\mathbb{C})$ acts on a four-dimensional Minkowski space $M_4$ in a special manner. For $x=(x^0,x^1,x^2,x^3)\in M_4$ , define a Hermitian matrix,

$$X(x)\equiv x^{\mu }{\sigma }_{\mu }=\left(\begin{array}{cc}{x}^0+x^3& x^1-\mathrm{i}{x}^2\\ x^1+\mathrm{i}{x}^2& x^0-x^3\end{array}\right)\text{\hspace{1em}(5.144)}$$

where ${\sigma }_{\mu }=(I_2,{\sigma }_1,{\sigma }_2,{\sigma }_3)$ , ${\sigma }_i$ ( $i=1,2,3$ ) being the Pauli matrices. Conversely, given a Hermitian matrix $X$ , a unique vector $(x^{\mu })\in M_4$ is defined as

$$x^{\mu }=\frac{1}{2}\mathrm{tr}\left({\sigma }_{\mu }X\right)\text{\hspace{1em}(5.130)}$$

where $\mathrm{tr}$ is over the $2\times 2$ matrix indices. Thus, there is an isomorphism between $M_4$ and the set of $2\times 2$ Hermitian matrices. It is interesting to note that $\det X(x)=(x^0{)}^2-(x^1{)}^2-(x^2{)}^2-(x^3{)}^2=-X^{\mathsf{t}}\eta X=-(\mathrm{Minkowski}\mathrm{norm}{)}^2$ . Accordingly

$$\begin{array}{rl}& \det X(x)>0\text{i f}x\text{i s a t i m e l i k e v e c t o r}\\ & =0\text{i f}x\text{i s o n t h e l i g h t c o n e}\\ & <0\text{i f}x\text{i s a s p a c e l i k e v e c t o r}.\end{array}$$

Take $A\in \mathrm{SL}(2,\mathbb{C})$ and define an action of $\mathrm{SL}(2,\mathbb{C})$ on $M_4$ by

$$\sigma (A,x)\equiv AX(x)A^{†}.\text{\hspace{1em}(5.145)}$$

The reader should verify that this action, in fact, satisfies the axioms of definition 5.14. The action of $\mathrm{SL}(2,\mathbb{C})$ on $M_4$ represents the Lorentz transformation $\mathrm{O}(1,3)$ . First we note that the action preserves the Minkowski norm,

$$\det \sigma (A,x)=\det [AX(x)A^{†}]=\det X(x)$$

since $\det A=\det A^{†}=1$ . Moreover, there is a homomorphism $\phi :\mathrm{SL}(2,\mathbb{C})\to \mathrm{O}(1,3)$ since

$$A(BX{B}^{†})A^{†}=(AB)X(AB{)}^{†}.$$

However, this homomorphism cannot be one to one, since $A\in \mathrm{SL}(2,\mathbb{C})$ and $-A$ give the same element of $\mathrm{O}(1,3)$ ; see (5.145). We verify (exercise 5.24) that the following matrix is an explicit form of a rotation about the unit vector $\hat{n}$ by an angle $\theta$ ,

$$A=\exp \left[-\mathrm{i}\frac{\theta }{2}(\hat{\mathbf{n}}\cdot \mathbf{\sigma })\right]=\cos \frac{\theta }{2}{I}_2-\mathrm{i}(\hat{\mathbf{n}}\cdot \mathbf{\sigma })\sin \frac{\theta }{2}.\text{\hspace{1em}(5.146a)}$$

The appearance of $\theta /2$ ensures that the homomorphism between $\mathrm{SL}(2,\mathbb{C})$ and the O(3) subgroup of $\mathrm{O}(1,3)$ is indeed two to one. In fact, rotations about $\hat{n}$ by $\theta$ and by $2\pi +\theta$ should be the same O(3) rotation, but $A(2\pi +\theta )=-A(\theta )$ in $\mathrm{SL}(2,\mathbb{C})$ . This leads to the existence of spinors. [See Misner et al (1973) and Wald (1984).] A boost along the direction $\hat{n}$ with the velocity $v=\tanh \alpha$ is given by

$$A=\exp \left[\frac{\alpha }{2}(\hat{\mathbf{n}}\cdot \mathbf{\sigma })\right]=\cosh \frac{\alpha }{2}{I}_2+(\hat{\mathbf{n}}\cdot \mathbf{\sigma })\sinh \frac{\alpha }{2}.\text{\hspace{1em}(5.146b)}$$

We show that $\phi$ maps $\mathrm{SL}(2,\mathbb{C})$ onto the proper orthochronous Lorentz group ${\mathrm{O}}_{+}^{\uparrow }(1,3)=\{\Lambda \in \mathrm{O}(1,3)|\det \Lambda =+1,{\Lambda }_{00}>0\}$ . Take any

$$A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in \mathrm{SL}(2,\mathbb{C})$$

and suppose $x^{\mu }=(1,0,0,0)$ is mapped to $x^{\prime \mu }$ . If we write $\phi (A)=\Lambda$ , we have

$$\begin{array}{rl}{x}^{\prime 0}& =\frac{1}{2}\mathrm{tr}(AX{A}^{†})=\frac{1}{2}\mathrm{tr}\left[\left(\begin{array}{ll}a& b\\ c& d\end{array}\right)\left(\begin{array}{ll}\bar{a}& \overline{cc}\\ \bar{b}& \bar{d}\end{array}\right)\right]\\ & =\frac{1}{2}(|a{|}^2+|b{|}^2+|c{|}^2+|d{|}^2)>0\end{array}$$

hence ${\Lambda }_{00}>0$ . To show $\det A=+1$ , we note that any element of $\mathrm{SL}(2,\mathbb{C})$ may be written as

$$\begin{array}{rl}A& =\left(\begin{array}{cc}{\mathrm{e}}^{\mathrm{i}\alpha }& 0\\ 0& {\mathrm{e}}^{-\mathrm{i}\alpha }\end{array}\right)\left(\begin{array}{cc}\cos \beta & \sin \beta {\mathrm{e}}^{\mathrm{i}\gamma }\\ -\sin \beta {\mathrm{e}}^{-\mathrm{i}\gamma }& \cos \beta \end{array}\right)B\\ & ={\left(\begin{array}{cc}{\mathrm{e}}^{\mathrm{i}\alpha /2}& 0\\ 0& {\mathrm{e}}^{\mathrm{i}\alpha /2}\end{array}\right)}^2{\left(\begin{array}{cc}\cos (\beta /2)& \sin (\beta /2){\mathrm{e}}^{\mathrm{i}\gamma }\\ -\sin (\beta /2){\mathrm{e}}^{-\mathrm{i}\gamma }& \cos (\beta /2)\end{array}\right)}^{2}B\\ & \equiv M^2{N}^2{B}_0^2\end{array}$$

where $B\equiv B_0^2$ is a positive-definite matrix. This shows that $\phi (A)$ is positive definite:

$$\det \phi (A)=(\det \phi (M){)}^2(\det \phi (N){)}^2(\det \phi (B_0){)}^2>0.$$

Now we have established that $\phi (\mathrm{SL}(2,\mathbb{C}))\subset {\mathrm{O}}_{+}^{\uparrow }(1,3)$ . Equations (5.146a) and (5.146b) show that for any element of ${\mathrm{O}}_{+}^{\uparrow }(1,3)$ , there is a corresponding matrix $A\in \mathrm{SL}(2,\mathbb{C})$ , hence $\phi$ is onto. Thus, we have established that

$$\phi (\mathrm{SL}(2,\mathbb{C}))={\mathrm{O}}_{+}^{\uparrow }(1,3).\text{\hspace{1em}(5.147)}$$

It can be shown that $\mathrm{SL}(2,\mathbb{C})$ is simply connected and is the universal covering group $\mathrm{Spin}(1,3)$ of ${\mathrm{O}}_{+}^{\uparrow }(1,3)$ , see section 4.6.

Exercise 5.24

Verify by explicit calculations that

(a)

$$A=\left(\begin{array}{cc}{\mathrm{e}}^{-\mathrm{i}\theta /2}& 0\\ 0& {\mathrm{e}}^{\mathrm{i}\theta /2}\end{array}\right)$$

represents a rotation about the $z$ -axis by $\theta$

(b)

$$A=\left(\begin{array}{cc}\cosh (\alpha /2)+\sinh (\alpha /2)& 0\\ 0& \cosh (\alpha /2)-\sinh (\alpha /2)\end{array}\right)$$

represents a boost along the $z$ -axis with the velocity $v=\tanh \alpha$ .

Definition 5.15

Let $G$ be a Lie group that acts on a manifold $M$ by $\sigma :G\times M\to M$ . The action $\sigma$ is said to be

(a) transitive if, for any $p_1,p_2\in M$ , there exists an element $g\in G$ such that $\sigma (g,p_1)=p_2$ ;
(b) free if every non-trivial element $g\ne e$ of $G$ has no fixed points in $M$ , that is, if there exists an element $p\in M$ such that $\sigma (g,p)=p$ , then $g$ must be the unit element $e$ ; and
(c) effective if the unit element $e\in G$ is the unique element that defines the trivial action on $M$ , i.e. if $\sigma (g,p)=p$ for all $p\in M$ , then $g$ must be the unit element $e$ .

Exercise 5.25

Show that the right translation $R:(a,g)\mapsto R_{a}g$ and left translation $L:(a,g)\mapsto L_{a}g$ of a Lie group are free and transitive.

5.7.2 Orbits and isotropy groups

Given a point $p\in M$ , the action of $G$ on $p$ takes $p$ to various points in $M$ . The orbit of $p$ under the action $\sigma$ is the subset of $M$ defined by

$$Gp=\{\sigma (g,p)|g\in G\}.\text{\hspace{1em}(5.148)}$$

If the action of $G$ on $M$ is transitive, the orbit of any $p\in M$ is $M$ itself. Clearly the action of $G$ on any orbit $Gp$ is transitive.

Example 5.17

(a) A flow $\sigma$ generated by a vector field $X=-y\partial /\partial x+x\partial /\partial y$ is periodic with period $2\pi$ , see example 5.9. The action $\sigma :\mathbb{R}\times {\mathbb{R}}^2\to {\mathbb{R}}^2$ defined by $(t,(x,y))\to \sigma (t,(x,y))$ is not effective since $\sigma (2\pi n,(x,y))=(x,y)$ for all $(x,y)\in {\mathbb{R}}^2$ . For the same reason, this flow is not free either. The orbit through $(x,y)\ne (0,0)$ is a circle $S^1$ centred at the origin.

(b) The action of $\mathrm{O}(n)$ on ${\mathbb{R}}^n$ is not transitive since if $|x|\ne |x^{\prime }|$ , no element of $\mathrm{O}(n)$ takes $x$ to $x^{\prime }$ . However, the action of $\mathrm{O}(n)$ on $S^{n-1}$ is obviously transitive. The orbit through $x$ is the sphere $S^{n-1}$ of radius $|x|$ . Accordingly, given an action $\sigma :\mathrm{O}(n)\times {\mathbb{R}}^n\to {\mathbb{R}}^n$ , the orbits divide ${\mathbb{R}}^n$ into mutually disjoint spheres of different radii. Introduce a relation by $x\sim y$ if $y=\sigma (g,x)$ for some $g\in G$ . It is easily verified that $\sim$ is an equivalence relation. The equivalence class $[x]$ is an orbit through $x$ . The coset space ${\mathbb{R}}^n/\mathrm{O}(n)$ is $[0,\infty )$ since each equivalence class is parametrized by the radius.

Definition 5.16

Let $G$ be a Lie group that acts on a manifold $M$ . The isotropy group of $p\in M$ is a subgroup of $G$ defined by

$$H(p)=\{g\in G|\sigma (g,p)=p\}.\text{\hspace{1em}(5.149)}$$

$H(p)$ is also called the little group or stabilizer of $p$ .

It is easy to see that $H(p)$ is indeed a subgroup. Let $g_1,g_2\in H(p)$ , then $g_1{g}_2\in H(p)$ since $\sigma (g_1{g}_2,p)=\sigma (g_1,\sigma (g_2,p))=\sigma (g_1,p)=p$ . Clearly $e\in H(p)$ since $\sigma (e,p)=p$ by definition. If $g\in H(p)$ , then $g^{-1}\in H(p)$ since $p=\sigma (e,p)=\sigma (g^{-1}g,p)=\sigma (g^{-1},\sigma (g,p))=\sigma (g^{-1},p)$ .

Exercise 5.26

Suppose a Lie group $G$ acts on a manifold $M$ freely. Show that $H(p)=\{e\}$ for any $p\in M$ .

Theorem 5.4

Let $G$ be a Lie group which acts on a manifold $M$ . Then the isotropy group $H(p)$ for any $p\in M$ is a Lie subgroup.

Proof

For fixed $p\in M$ , we define a map ${\phi }_p:G\to M$ by ${\phi }_p(g)\equiv gp$ . Then $H(p)$ is the inverse image ${\phi }_p^{-1}(p)$ of a point $p$ , and hence a closed set. The group properties have been shown already. It follows from theorem 5.2 that $H(p)$ is a Lie subgroup.

For example, let $M={\mathbb{R}}^3$ and $G=\mathrm{SO}(3)$ and take a point $p=(0,0,1)\in {\mathbb{R}}^3$ . The isotropy group $H(p)$ is the set of rotations about the $z$ -axis, which is isomorphic to SO(2).

Let $G$ be a Lie group and $H$ any subgroup of $G$ . The coset space $G/H$ admits a differentiable structure and $G/H$ becomes a manifold, called a homogeneous space. Note that $\dim G/H=\dim G-\dim H$ . Let $G$ be a Lie group which acts on a manifold $M$ transitively and let $H(p)$ be an isotropy group of $p\in M$ . $H(p)$ is a Lie subgroup and the coset space $G/H(p)$ is a homogeneous space.

In fact, if $G,H(p)$ and $M$ satisfy certain technical requirements (for example, $G/H(p)$ compact) is, it can be shown that $G/H(p)$ is homeomorphic to $M$ , see example 5.18.

Example 5.18

(a) Let $G=\mathrm{SO}(3)$ be a group acting on ${\mathbb{R}}^3$ and $H=\mathrm{SO}(2)$ be the isotropy group of $x\in {\mathbb{R}}^3$ . The group $\mathrm{SO}(3)$ acts on $S^2$ transitively and we have $\mathrm{SO}(3)/\mathrm{SO}(2)\cong S^2$ . What is the geometrical picture of this? Let $g^{\prime }=gh$ where $g,g^{\prime }\in G$ and $h\in H$ . Since $H$ is the set of rotations in a plane, $g$ and $g^{\prime }$ must be rotations about the common axis. Then the equivalence class $[g]$ is specified by the polar angles $(\theta ,\varphi )$ . Thus, we again find that $G/H=S^2$ . Since $\mathrm{SO}(2)$ is not a normal subgroup of $\mathrm{SO}(3)$ , $S^2$ does not admit a group structure.

It is easy to generalize this result to higher-dimensional rotation groups and we have the useful result

$$\mathrm{SO}(n+1)/\mathrm{SO}(n)=S^n.\text{\hspace{1em}(5.150)}$$

$\mathrm{O}(n+1)$ also acts on $S^n$ transitively and we have

$$\mathrm{O}(n+1)/\mathrm{O}(n)=S^n.\text{\hspace{1em}(5.151)}$$

Similar relations hold for $\mathrm{U}(n)$ and $\mathrm{SU}(n)$ :

$$\mathrm{U}(n+1)/\mathrm{U}(n)=\mathrm{SU}(n+1)/\mathrm{SU}(n)=S^{2n+1}.\text{\hspace{1em}(5.152)}$$

(b) The group $\mathrm{O}(n+1)$ acts on $\mathbb{R}{P}^n$ transitively from the left. Note, first, that $\mathrm{O}(n+1)$ acts on ${\mathbb{R}}^{n+1}$ in the usual manner and preserves the equivalence relation employed to define $\mathbb{R}{P}^n$ (see example 5.12). In fact, take $x,x^{\prime }\in {\mathbb{R}}^{n+1}$ and $g\in \mathrm{O}(n+1)$ . If $x\sim x^{\prime }$ (that is if $x^{\prime }=ax$ for some $a\in \mathbb{R}-\{0\}$ ), then it follows that $gx\sim g{x}^{\prime }$ ( $g{x}^{\prime }=agx$ ). Accordingly, this action of $\mathrm{O}(n+1)$ on ${\mathbb{R}}^{n+1}$ induces the natural action of $\mathrm{O}(n+1)$ on $\mathbb{R}{P}^n$ . Clearly this action is transitive on $\mathbb{R}{P}^n$ . (Look at two representatives with the same norm.) If we take a point $p$ in $\mathbb{R}{P}^n$ , which corresponds to a point $(1,0,\dots ,0)\in {\mathbb{R}}^{n+1}$ , the isotropy group $H(p)$ is

$$H(p)=\left(\begin{array}{ccccc}\pm 1& 0& 0& \dots & 0\\ 0& & & & \\ 0& & & & \\ ⋮& & \mathrm{O}(n)& & \\ 0& & & & \end{array}\right)=\mathrm{O}(1)\times \mathrm{O}(n)\text{\hspace{1em}(5.153)}$$

where $\mathrm{O}(1)$ is the set $\{-1,+1\}={\mathbb{Z}}_2$ . Now we find that

$$\mathrm{O}(n+1)/[\mathrm{O}(1)\times \mathrm{O}(n)]\cong S^n/{\mathbb{Z}}_2\cong \mathbb{R}{P}^n.\text{\hspace{1em}(5.154)}$$

(c) This result is easily generalized to the Grassmann manifolds: $G_{k,n}(\mathbb{R})=\mathrm{O}(n)/[\mathrm{O}(k)\times \mathrm{O}(n-k)]$ . We first show that $\mathrm{O}(n)$ acts on $G_{k,n}(\mathbb{R})$ transitively.

Let $A$ be an element of $G_{k,n}(\mathbb{R})$ , then $A$ is a $k$ -dimensional plane in ${\mathbb{R}}^n$ . Define an $n\times n$ matrix $P_A$ which projects a vector $v\in {\mathbb{R}}^n$ to the plane $A$ . Let us introduce an orthonormal basis $\{e_1,\dots ,e_n\}$ in ${\mathbb{R}}^n$ and another orthonormal basis $\{f_1,\dots ,f_k\}$ in the plane $A$ , where the orthonormality is defined with respect to the Euclidean metric in ${\mathbb{R}}^n$ . In terms of $\{e_i\}$ , $f_a$ is expanded as $f_a={\sum }_i{f}_{ai}{e}_i$ and the projected vector is

$$\begin{array}{rl}{P}_{A}v& =\left(v{f}_1\right)f_1+\cdots +\left(v{f}_k\right)f_k\\ & =\sum _{i,j}\left(v_i{f}_{1i}{f}_{1j}+\cdots +v_i{f}_{ki}{f}_{kj}\right)e_j=\sum _{i,a,j}{v}_i{f}_{ai}{f}_{aj}{e}_j.\end{array}$$

Thus, $P_A$ is represented by a matrix

$${\left(P_A\right)}_{ij}=\sum f_{ai}{f}_{aj}.\text{\hspace{1em}(5.155)}$$

Note that $P_A^2=P_A$ , $P_A^{\mathrm{t}}=P_A$ and $\mathrm{tr}{P}_A=k$ . [The last relation holds since it is always possible to choose a coordinate system such that

$$P_A=\mathrm{diag}(\underset{k}{\underbrace{1,1,\dots ,1}},\underset{n-k}{\underbrace{0,\dots ,0}}).$$

This guarantees that $A$ is, indeed, a $k$ -dimensional plane.] Conversely any matrix $P$ that satisfies these three conditions determines a unique $k$ -dimensional plane in ${\mathbb{R}}^n$ , that is a unique element of $G_{k,n}(\mathbb{R})$ .

We now show that $\mathrm{O}(n)$ acts on $G_{k,n}(\mathbb{R})$ transitively. Take $A\in G_{k,n}(\mathbb{R})$ and $g\in \mathrm{O}(n)$ and construct $P_B\equiv g{P}_A{g}^{-1}$ . The matrix $P_B$ determines an element $B\in G_{k,n}(\mathbb{R})$ since $P_B^2=P_B$ , $P_B^{\mathrm{t}}=P_B$ and $\mathrm{tr}{P}_B=k$ . Let us denote this action by $B=\sigma (g,A)$ . Clearly this action is transitive since given a standard $k$ -dimensional basis of $A$ , $\{f_1,\dots ,f_k\}$ for example, any $k$ -dimensional basis $\{{\tilde{f}}_1,\dots ,{\tilde{f}}_k\}$ can be reached by an action of $\mathrm{O}(n)$ on this basis.

Let us take a special plane $C_0$ which is spanned by the standard basis $\{f_1,\dots ,f_k\}$ . Then an element of the isotropy group $H(C_0)$ is of the form

$$\begin{array}{rl}& kn-k\\ & M=\left(\begin{array}{cc}{g}_1& 0\\ 0& g_2\end{array}\right)\begin{array}{l}k\\ n-k\end{array}\end{array}\text{\hspace{1em}(5.156)}$$

where $g_1\in \mathrm{O}(k)$ . Since $M\in \mathrm{O}(n)$ , an $(n-k)\times (n-k)$ matrix $g_2$ must be an element of $\mathrm{O}(n-k)$ . Thus, the isotropy group is isomorphic to $\mathrm{O}(k)\times \mathrm{O}(n-k)$ . Finally we verified that

$$G_{k,n}(\mathbb{R})\cong \mathrm{O}(n)/[\mathrm{O}(k)\times \mathrm{O}(n-k)].\text{\hspace{1em}(5.157)}$$

The dimension of $G_{k,n}(\mathbb{R})$ is obtained from the general formula as

$$\begin{array}{rl}\dim G_{k,n}(\mathbb{R})& =\dim \mathrm{O}(n)-\dim [\mathrm{O}(k)\times \mathrm{O}(n-k)]\\ & =\frac{1}{2}n(n-1)-\left[\frac{1}{2}k(k-1)+\frac{1}{2}(n-k)(n-k-1)\right]\\ & =k(n-k)\end{array}\text{\hspace{1em}(5.158)}$$

in agreement with the result of example 5.5. Equation (5.157) also shows that the Grassmann manifold is compact.

5.7.3 Induced vector fields

Let $G$ be a Lie group which acts on $M$ as $(g,x)\mapsto gx$ . A left-invariant vector field $X_V$ generated by $V\in T_{e}G$ naturally induces a vector field in $M$ . Define a flow in $M$ by

$$\sigma (t,x)=\exp (tV)x,\text{\hspace{1em}(5.159)}$$

$\sigma (t,x)$ is a one-parameter group of transformations, and define a vector field called the induced vector field denoted by $V^{♯}$ ,

$${V^{♯}|}_x={\frac{\mathrm{d}}{\mathrm{d}t}\exp (tV)x|}_{t=0}.\text{\hspace{1em}(5.160)}$$

Thus, we have obtained a map $♯:T_{e}G\to \mathcal{X}(M)$ defined by $V\mapsto V^{♯}$ .

Exercise 5.27

The Lie group $\mathrm{SO}(2)$ acts on $M={\mathbb{R}}^2$ in the usual way. Let

$$V=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$$

be an element of $\mathfrak{so}(2)$ .

(a) Show that

$$\exp (tV)=\left(\begin{array}{cc}\cos t& -\sin t\\ \sin t& \cos t\end{array}\right)$$

and find the induced flow through

$$\mathbf{x}=\left(\begin{array}{c}x\\ y\end{array}\right)\in {\mathbb{R}}^2.$$

(b) Show that $V^{♯}{|}_x=-y\partial /\partial x+x\partial /\partial y$ .

Example 5.19

Let us take $G=\mathrm{SO}(3)$ and $M={\mathbb{R}}^3$ . The basis vectors of $T_{e}G$ are generated by rotations about the $x,y$ and $z$ axes. We denote them by $X_x,X_y$ and $X_z$ , respectively (see exercise 5.22),

$$X_x=\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& -1\\ 0& 1& 0\end{array}\right),X_y=\left(\begin{array}{ccc}0& 0& 1\\ 0& 0& 0\\ -1& 0& 0\end{array}\right),X_z=\left(\begin{array}{ccc}0& -1& 0\\ 1& 0& 0\\ 0& 0& 0\end{array}\right).$$

Repeating a similar analysis to the previous one, we obtain the corresponding induced vectors,

$$X_x^{♯}=-z\frac{\partial }{\partial y}+y\frac{\partial }{\partial z},X_y^{♯}=-x\frac{\partial }{\partial z}+z\frac{\partial }{\partial x},X_z^{♯}=-y\frac{\partial }{\partial x}+x\frac{\partial }{\partial y}.$$

5.7.4 The adjoint representation

A Lie group $G$ acts on $G$ itself in a special way.

Definition 5.17

Take any $a\in G$ and define a homomorphism ${\mathrm{ad}}_a:G\to G$ by the conjugation,

$${\mathrm{ad}}_a:g\mapsto ag{a}^{-1}.\text{\hspace{1em}(5.161)}$$

This homomorphism is called the adjoint representation of $G$ .

Exercise 5.28

Show that ${\mathrm{ad}}_a$ is a homomorphism. Define a map $\sigma :G\times G\to G$ by $\sigma (a,g)\equiv {\mathrm{ad}}_{a}g$ . Show that $\sigma (a,g)$ is an action of $G$ on itself.

Noting that ${\mathrm{ad}}_{a}e=e$ , we restrict the induced map ${\mathrm{ad}}_{a\ast }:T_{g}G\to T_{{\mathrm{ad}}_{a}g}G$ to $g=e$ ,

$${\mathrm{Ad}}_a:T_{e}G\to T_{e}G\text{\hspace{1em}(5.162)}$$

where ${\mathrm{Ad}}_a\equiv {\mathrm{ad}}_{a\ast }{|}_{T_{e}G}$ . If we identify $T_{e}G$ with the Lie algebra $\mathfrak{g}$ , we have obtained a map $\mathrm{Ad}:G\times \mathfrak{g}\to \mathfrak{g}$ called the adjoint map of $G$ . Since ${\mathrm{ad}}_{a\ast }{\mathrm{ad}}_{b\ast }={\mathrm{ad}}_{ab\ast }$ , it follows that ${\mathrm{Ad}}_a{\mathrm{Ad}}_b={\mathrm{Ad}}_{ab}$ . Similarly, ${\mathrm{Ad}}_{a^{-1}}={\mathrm{Ad}}_a^{-1}$ follows from ${\mathrm{ad}}_{a^{-1}\ast }{\mathrm{ad}}_{a\ast }{|}_{T_{e}G}={\mathrm{id}}_{T_{e}G}$ .

If $G$ is a matrix group, the adjoint representation becomes a simple matrix operation. Let $g\in G$ and $X_V\in \mathfrak{g}$ , and let ${\sigma }_V(t)=\exp (tV)$ be a one-parameter subgroup generated by $V\in T_{e}G$ . Then ${\mathrm{ad}}_g$ acting on ${\sigma }_V(t)$ yields $g\exp (tV)g^{-1}=\exp (tgV{g}^{-1})$ . As for ${\mathrm{Ad}}_g$ we have ${\mathrm{Ad}}_g:V\mapsto gV{g}^{-1}$ since

$$\begin{array}{rl}{\mathrm{Ad}}_{g}V& ={\frac{\mathrm{d}}{\mathrm{d}t}[{\mathrm{ad}}_g\exp (tV)]|}_{t=0}\\ & ={\frac{\mathrm{d}}{\mathrm{d}t}\exp (tgV{g}^{-1})|}_{t=0}=gV{g}^{-1}.\end{array}\text{\hspace{1em}(5.163)}$$

5.1 The Stiefel manifold $V(m,r)$ is the set of orthonormal vectors $\{e_i\}$ ( $1\le i\le r$ ) in ${\mathbb{R}}^m$ ( $r\le m$ ). We may express an element $A$ of $V(m,r)$ by an $m\times r$ matrix $(e_1,\dots ,e_r)$ . Show that $\mathrm{SO}(m)$ acts transitively on $V(m,r)$ . Let

$$A_0\equiv \left(\begin{array}{cccc}1& 0& \dots & 0\\ 0& 1& \dots & 0\\ \dots & \dots & \dots & \dots \\ 0& 0& \dots & 1\\ 0& 0& \dots & 0\\ 0& 0& \dots & 0\end{array}\right)$$

be an element of $V(m,r)$ . Show that the isotropy group of $A_0$ is $\mathrm{SO}(m-r)$ . Verify that $V(m,r)=\mathrm{SO}(m)/\mathrm{SO}(m-r)$ and $\dim V(m,r)=[r(r-1)]/2+r(m-r)$ . [Remark: The Stiefel manifold is, in a sense, a generalization of a sphere. Observe that $V(m,1)=S^{m-1}$ .]

5.2 Let $M$ be the Minkowski four-spacetime. Define the action of a linear operator $\ast :{\Omega }^r(M)\to {\Omega }^{4-r}(M)$ by

$$\begin{array}{rl}& r=0:\ast 1=-\mathrm{d}{x}^0\wedge \mathrm{d}{x}^1\wedge \mathrm{d}{x}^2\wedge \mathrm{d}{x}^3;\\ & r=1:\ast \mathrm{d}{x}^i=-\mathrm{d}{x}^j\wedge \mathrm{d}{x}^k\wedge \mathrm{d}{x}^0\ast \mathrm{d}{x}^0=-\mathrm{d}{x}^1\wedge \mathrm{d}{x}^2\wedge \mathrm{d}{x}^3;\\ & r=2:\ast \mathrm{d}{x}^i\wedge \mathrm{d}{x}^j=\mathrm{d}{x}^k\wedge \mathrm{d}{x}^0\ast \mathrm{d}{x}^i\wedge \mathrm{d}{x}^0=-\mathrm{d}{x}^j\wedge \mathrm{d}{x}^k;\\ & r=3:\ast \mathrm{d}{x}^1\wedge \mathrm{d}{x}^2\wedge \mathrm{d}{x}^3=-\mathrm{d}{x}^0\ast \mathrm{d}{x}^i\wedge \mathrm{d}{x}^j\wedge \mathrm{d}{x}^0=-\mathrm{d}{x}^k;\\ & r=4:\ast \mathrm{d}{x}^0\wedge \mathrm{d}{x}^1\wedge \mathrm{d}{x}^2\wedge \mathrm{d}{x}^3=1;\end{array}$$

where $(i,j,k)$ is an even permutation of $(1,2,3)$ . The vector potential $A$ and the electromagnetic tensor $F$ are defined as in example 5.11. $J=J_{\mu }\mathrm{d}{x}^{\mu }=\rho \mathrm{d}{x}^0+j_k\mathrm{d}{x}^k$ is the current one-form.

(a) Write down the equation $\mathrm{d}\ast F=\ast J$ and verify that it reduces to two of the Maxwell equations $\nabla \cdot \mathbf{E}=\rho$ and $\nabla \times \mathbf{B}-\partial \mathbf{E}/\partial t=\mathbf{j}$ .
(b) Show that the identity $0=\mathrm{d}(\mathrm{d}\ast F)=\mathrm{d}\ast J$ reduces to the charge conservation equation

$${\partial }_{\mu }{J}^{\mu }=\frac{\partial \rho }{\partial t}+\nabla \cdot \mathbf{j}=0.$$

(c) Show that the Lorentz condition ${\partial }_{\mu }{A}^{\mu }=0$ is expressed as $\mathrm{d}\ast A=0$

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

OCR 来源 SHA-256：0b6324f01172a3548cd34382e82fd8bfae574602b6b2cfaa357ba96bc2c35478

OCR 产物 SHA-256：7d6a5d49d35abf37dd6d957ae7a4caf172480186ab3f994a97bbd4240d0ba984