7.6 Isometries and conformal transformations

7.6.1 Isometries

Definition 7.4

Let $(M,g)$ be a (pseudo-)Riemannian manifold. A diffeomorphism $f:M\to M$ is an isometry if it preserves the metric

$$f^{\ast }{g}_{f(p)}=g_p\text{\hspace{1em}(7.96a)}$$

that is, if $g_{f(p)}(f_{\ast }X,f_{\ast }Y)=g_p(X,Y)$ for $X,Y\in T_{p}M$

In components, the condition (7.96a) becomes

$$\frac{\partial y^{\alpha }}{\partial x^{\mu }}\frac{\partial y^{\beta }}{\partial x^{\nu }}{g}_{\alpha \beta }(f(p))=g_{\mu \nu }(p)\text{\hspace{1em}(7.96b)}$$

where $x$ and $y$ are the coordinates of $p$ and $f(p)$ , respectively. The identity map, the composition of the isometries and the inverse of an isometry are isometries; all these isometries form a group. Since an isometry preserves the length of a vector, in particular that of an infinitesimal displacement vector, it may be regarded as a rigid motion. For example, in ${\mathbb{R}}^n$ , the Euclidean group $E^n$ , that is the set of maps $f:x\mapsto Ax+T$ ( $A\in \mathrm{SO}(n)$ , $T\in {\mathbb{R}}^n$ ), is the isometry group.

7.6.2 Conformal transformations

Definition 7.5

Let $(M,g)$ be a (pseudo-)Riemannian manifold. A diffeomorphism $f:M\to M$ is called a conformal transformation if it preserves the metric up to a scale,

$$f^{\ast }{g}_{f(p)}={\mathrm{e}}^{2\sigma }{g}_p\sigma \in \mathcal{F}(M)\text{\hspace{1em}(7.97a)}$$

namely, $g_{f(p)}(f_X,f_Y)={\mathrm{e}}^{2\sigma }{g}_p(X,Y)$ for $X,Y\in T_{p}M$ .

In components, the condition (7.97a) becomes

$$\frac{\partial y^{\alpha }}{\partial x^{\mu }}\frac{\partial y^{\beta }}{\partial x^{\nu }}{g}_{\alpha \beta }(f(p))={\mathrm{e}}^{2\sigma (p)}{g}_{\mu \nu }(p).\text{\hspace{1em}(7.97b)}$$

The set of conformal transformations on $M$ is a group, the conformal group denoted by $\mathrm{Conf}(M)$ . Let us define the angle $\theta$ between two vectors $X=X^{\mu }{\partial }_{\mu }$ , $Y=Y^{\mu }{\partial }_{\mu }\in T_{p}M$ by

$$\cos \theta =\frac{g_p(X,Y)}{\sqrt{g_p(X,X)g_p(Y,Y)}}=\frac{g_{\mu \nu }{X}^{\mu }{Y}^{\nu }}{\sqrt{g_{\zeta \eta }{X}^{\zeta }{X}^{\eta }{g}_{\kappa \lambda }{Y}^{\kappa }{Y}^{\lambda }}}.\text{\hspace{1em}(7.98)}$$

If $f$ is a conformal transformation, the angle ${\theta }^{\prime }$ between $f_{\ast }X$ and $f_{\ast }Y$ is given by

$$\cos {\theta }^{\prime }=\frac{{\mathrm{e}}^{2\sigma }{g}_{\mu \nu }{X}^{\mu }{Y}^{\nu }}{\sqrt{{\mathrm{e}}^{2\sigma }{g}_{\zeta \eta }{X}^{\zeta }{X}^{\eta }\cdot {\mathrm{e}}^{2\sigma }{g}_{\kappa \lambda }{Y}^{\kappa }{Y}^{\lambda }}}=\cos \theta$$

hence $f$ preserves the angle. In other words, $f$ changes the scale but not the shape.

A concept related to conformal transformations is Weyl rescaling. Let $g$ and $\bar{g}$ be metrics on a manifold $M$ . $\bar{g}$ is said to be conformally related to $g$ if

$${\bar{g}}_p={\mathrm{e}}^{2\sigma (p)}{g}_p.\text{\hspace{1em}(7.99)}$$

Clearly this is an equivalence relation among the set of metrics on $M$ . The equivalence class is called the conformal structure. The transformation $g\to {\mathrm{e}}^{2\sigma }g$ is called a Weyl rescaling. The set of Weyl rescalings on $M$ is a group denoted by $\mathrm{Weyl}(M)$ .

Example 7.8

Let $w=f(z)$ be a holomorphic function defined on the complex plane $\mathbb{C}$ . [A $C^{\infty }$ -function regarded as a function of $z=x+\mathrm{i}y$ and $\bar{z}=x-\mathrm{i}y$ is holomorphic if ${\partial }_{\bar{z}}f(z,\bar{z})=0$ .] We write the real part and the imaginary part of the respective variables as $z=x+\mathrm{i}y$ and $w=u+\mathrm{i}v$ . The map $f:(x,y)\mapsto (u,v)$ is conformal since

$$\begin{array}{rl}\mathrm{d}{u}^2+\mathrm{d}{v}^2& ={\left(\frac{\partial u}{\partial x}\mathrm{d}x+\frac{\partial u}{\partial y}\mathrm{d}y\right)}^2+{\left(\frac{\partial v}{\partial x}\mathrm{d}x+\frac{\partial v}{\partial y}\mathrm{d}y\right)}^2\\ & =\left[{\left(\frac{\partial u}{\partial x}\right)}^2+{\left(\frac{\partial u}{\partial y}\right)}^2\right](\mathrm{d}{x}^2+\mathrm{d}{y}^2)\end{array}\text{\hspace{1em}(7.100)}$$

where use has been made of the Cauchy-Riemann relations

$$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}.$$

Exercise 7.15

Let $f:M\to M$ be a conformal transformation on a Lorentz manifold $(M,g)$ . Show that $f_{\ast }:T_{p}M\to T_{f(p)}M$ preserves the local light cone structure, namely

$$f_{\ast }:\{\begin{array}{ccc}\text{t i m e l i k e v e c t o r}& \mapsto & \text{t i m e l i k e v e c t o r}\\ \text{n u l l v e c t o r}& \mapsto & \text{n u l l v e c t o r}\\ \text{s p a c e l i k e v e c t o r}& \mapsto & \text{s p a c e l i k e v e c t o r .}\end{array}\text{\hspace{1em}(7.101)}$$

Let $\bar{g}$ be a metric on $M$ , which is conformally related to $g$ as $\bar{g}={\mathrm{e}}^{2\sigma (p)}g$ . Let us compute the Riemann tensor of $\bar{g}$ . We could simply substitute $\bar{g}$ into the defining equation (7.42). However, we follow the elegant coordinate-free derivation of Nomizu (1981). Let $K$ be the difference of the covariant derivatives $\bar{\nabla }$ with respect to $\bar{g}$ and $\nabla$ with respect to $g$ ,

$$K(X,Y)\equiv {\bar{\nabla }}_{X}Y-{\nabla }_{X}Y.\text{\hspace{1em}(7.102)}$$

Proposition 7.1

Let $U$ be a vector field which corresponds to the one-form $\mathrm{d}\sigma$ : $Z[\sigma ]=⟨\mathrm{d}\sigma ,Z⟩=g(U,Z)$ . Then

$$K(X,Y)=X[\sigma ]Y+Y[\sigma ]X-g(X,Y)U.\text{\hspace{1em}(7.103)}$$

Proof

It follows from the torsion-free condition that $K(X,Y)=K(Y,X)$ . Since ${\bar{\nabla }}_X\bar{g}={\nabla }_{X}g=0$ , we have

$$X[\bar{g}(Y,Z)]={\bar{\nabla }}_X[\bar{g}(Y,Z)]=\bar{g}({\bar{\nabla }}_X,Z)+\bar{g}(Y,{\bar{\nabla }}_{X}Z)$$

and also

$$\begin{array}{rl}X[\bar{g}(Y,Z)]& ={\nabla }_X[{\mathrm{e}}^{2\sigma }g(Y,Z)]\\ & =2X[\sigma ]{\mathrm{e}}^{2\sigma }g(Y,Z)+{\mathrm{e}}^{2\sigma }[g({\nabla }_X,Z)+g(Y,{\nabla }_{X}Z)].\end{array}$$

Taking the difference between these two expressions, we have

$$g(K(X,Y),Z)+g(Y,K(X,Z))=2X[\sigma ]g(Y,Z).\text{\hspace{1em}(7.104a)}$$

Permutations of $(X,Y,Z)$ yield

$$\begin{array}{rl}& g(K(Y,X),Z)+g(X,K(Y,Z))=2Y[\sigma ]g(X,Z)(7.104b)\\ & g(K(Z,X),Y)+g(X,K(Z,Y))=2Z[\sigma ]g(X,Y).(7.104c)\end{array}$$

The combination (7.104a) + (7.104b) - (7.104c) yields

$$g(K(X,Y),Z)=X[\sigma ]g(Y,Z)+Y[\sigma ]g(X,Z)-Z[\sigma ]g(X,Y).\text{\hspace{1em}(7.105)}$$

The last term is modified as

$$Z[\sigma ]g(X,Y)=g(U,Z)g(X,Y)=g(g(Y,X)U,Z).$$

Substituting this into (7.105), we find

$$g(K(X,Y)-X[\sigma ]Y-Y[\sigma ]X+g(X,Y)U,Z)=0.$$

Since this is true for any $Z$ , we have (7.103).

The component expression for $K$ is

$$\begin{array}{rl}K\left(e_{\mu },e_v\right)& ={\bar{\nabla }}_{\mu }{e}_v-{\nabla }_{\mu }{e}_v=\left({\bar{\Gamma }}_{\mu v}^{\lambda }-{\Gamma }_{\mu v}^{\lambda }\right)e_{\lambda }\\ & =e_{\mu }[\sigma ]e_v+e_v[\sigma ]e_{\mu }-g\left(e_{\mu },e_v\right)g^{\kappa \lambda }{\partial }_{\kappa }\sigma e_{\lambda }\end{array}$$

from which it is readily seen that

$${\bar{\Gamma }}_{\mu v}^{\lambda }={\Gamma }_{\mu v}^{\lambda }+{\delta }_v^{\lambda }{\partial }_{\mu }\sigma +{\delta }_{\mu }^{\lambda }{\partial }_v\sigma -g_{\mu v}{g}^{\kappa \lambda }{\partial }_{\kappa }\sigma .\text{\hspace{1em}(7.106)}$$

To find the Riemann curvature tensor, we start from the definition,

$$\begin{array}{rl}& \bar{R}(X,Y)Z={\bar{\nabla }}_X{\bar{\nabla }}_{Y}Z-{\bar{\nabla }}_Y{\bar{\nabla }}_{X}Z-{\bar{\nabla }}_{[X,Y]}Z\\ & ={\bar{\nabla }}_X[{\nabla }_{Y}Z+K(Y,Z)]-{\bar{\nabla }}_Y[{\nabla }_{X}Z+K(X,Z)]\\ & -\left\{{\nabla }_{[X,Y]}Z+K([X,Y],Z)\right\}\\ & ={\nabla }_X\left\{{\nabla }_{Y}Z+K(Y,Z)\right\}+K(X,{\nabla }_{Y}Z+K(Y,Z))\\ & -{\nabla }_Y\left\{{\nabla }_{X}Z+K(X,Z)\right\}-K(Y,{\nabla }_{X}Z+K(X,Z))\\ & -\left\{{\nabla }_{[X,Y]}Z+K([X,Y],Z)\right\}.\end{array}\text{\hspace{1em}(7.107)}$$

After a straightforward but tedious calculation, we find that

$$\begin{array}{rl}& \bar{R}(X,Y)Z=R(X,Y)Z+⟨{\nabla }_X\mathrm{d}\sigma ,Z⟩Y-⟨{\nabla }_Y\mathrm{d}\sigma ,Z⟩X\\ & -g(Y,Z){\nabla }_{X}U+Y[\sigma ]Z[\sigma ]X\\ & -g(Y,Z)U[\sigma ]X+X[\sigma ]g(Y,Z)U\\ & +g(X,Z){\nabla }_{Y}U-X[\sigma ]Z[\sigma ]Y\\ & +g(X,Z)U[\sigma ]Y-Y[\sigma ]g(X,Z)U.\end{array}\text{\hspace{1em}(7.108)}$$

Let us define a type $(1,1)$ tensor field $B$ by

$$BX\equiv -X[\sigma ]U+{\nabla }_{X}U+\frac{1}{2}U[\sigma ]X.\text{\hspace{1em}(7.109)}$$

Since $g({\nabla }_{Y}U,Z)=⟨{\nabla }_Y\mathrm{d}\sigma ,Z⟩$ (exercise 7.8(c)), (7.108) becomes

$$\begin{array}{rl}& \bar{R}(X,Y)Z=R(X,Y)Z-[g(Y,Z)BX-g(BX,Z)Y\\ & +g(BY,Z)X-g(X,Z)BY].\end{array}\text{\hspace{1em}(7.110)}$$

In components, this becomes

$${\bar{R}}_{\lambda \mu v}^{\kappa }=R_{\lambda \mu v}^{\kappa }-g_{v\lambda }{B}_{\mu }^{\kappa }+g_{\xi \lambda }{B}_{\mu }^{\xi }{\delta }_v^{\kappa }-g_{\xi \lambda }{B}_v^{\xi }{\delta }_{\mu }^{\kappa }+g_{\mu \lambda }{B}_v^{\kappa }\text{\hspace{1em}(7.111)}$$

where the components of the tensor $B$ are

$$\begin{array}{rl}{B}_{\mu }^{\kappa }& =-{\partial }_{\mu }\sigma U^{\kappa }+({\nabla }_{\mu }U{)}^{\kappa }+\frac{1}{2}U[\sigma ]{\delta }_{\mu }^{\kappa }\\ & =-{\partial }_{\mu }\sigma g^{\kappa \lambda }{\partial }_{\lambda }\sigma +g^{\kappa \lambda }\left({\partial }_{\mu }{\partial }_{\lambda }\sigma -{\Gamma }_{\mu \lambda }^{\xi }{\partial }_{\xi }\sigma \right)+\frac{1}{2}{g}^{\lambda \xi }{\partial }_{\lambda }\sigma {\partial }_{\xi }\sigma {\delta }_{\mu }^{\kappa }.\end{array}\text{\hspace{1em}(7.112)}$$

Note that $B_{\mu \nu }\equiv g_{\nu \lambda }{B}_{\mu }^{\lambda }=B_{\nu \mu }$

By contracting the indices in (7.111), we obtain

$${\overline{Ric}}_{\mu \nu }=Ri{c}_{\mu \nu }-g_{\mu \nu }{B}_{\lambda }^{\lambda }-(m-2)B_{\nu \mu }\text{\hspace{1em}(7.113)}$$

$${\mathrm{e}}^{2\sigma }\bar{\mathcal{R}}=\mathcal{R}-2(m-1)B_{\lambda }^{\lambda }\text{\hspace{1em}(7.114a)}$$

where $m=\dim M$ . Equation (7.114a) is also written as

$${\bar{g}}_{\mu \nu }\bar{\mathcal{R}}=[\mathcal{R}-2(m-1)B_{\lambda }^{\lambda }]g_{\mu \nu }.\text{\hspace{1em}(7.114b)}$$

If we eliminate $g_{\mu \nu }{B}_{\lambda }^{\lambda }$ and $B_{\mu \nu }$ in ${\bar{R}}_{\lambda \mu \nu }^{\kappa }$ in favour of $\overline{Ric}$ and $\bar{\mathcal{R}}$ and separate barred and unbarred terms, we find a combination which is independent of $\sigma$ ,

$$\begin{array}{rl}& C_{\kappa \lambda \mu \nu }=R_{\kappa \lambda \mu \nu }+\frac{1}{m-2}\left(Ri{c}_{\kappa \mu }{g}_{\lambda \nu }-Ri{c}_{\lambda \mu }{g}_{\kappa \nu }+Ri{c}_{\lambda \nu }{g}_{\kappa \mu }-Ri{c}_{\kappa \nu }{g}_{\lambda \mu }\right)\\ & +\frac{\mathcal{R}}{(m-2)(m-1)}\left(g_{\kappa \mu }{g}_{\lambda \nu }-g_{\kappa \nu }{g}_{\lambda \mu }\right)\end{array}\text{\hspace{1em}(7.115)}$$

where $m\ge 4$ (see problem 7.2 for $m=3$ ). The tensor $C$ is called the Weyl tensor. The reader should verify that $C_{\kappa \lambda \mu \nu }={\bar{C}}_{\kappa \lambda \mu \nu }$ .

If every point $p$ of a (pseudo-)Riemannian manifold $(M,g)$ has a chart $(U,\phi )$ containing $p$ such that $g_{\mu \nu }={\mathrm{e}}^{2\sigma }{\delta }_{\mu \nu }$ , then $(M,g)$ is said to be conformally flat. Since the Weyl tensor vanishes for a flat metric, it also vanishes for a conformally flat metric. If $\dim M\ge 4$ , then $C=0$ is the necessary and sufficient condition for conformal flatness (Weyl-Schouten). If $\dim M=3$ , the Weyl tensor vanishes identically; see problem 7.2. If $\dim M=2$ , $M$ is always conformally flat; see the next example.

Example 7.9

Any two-dimensional Riemannian manifold $(M,g)$ is conformally flat. Let $(x,y)$ be the original local coordinates with which the metric takes the form

$$\mathrm{d}{s}^2=g_{xx}\mathrm{d}{x}^2+2g_{xy}\mathrm{d}x\mathrm{d}y+g_{yy}\mathrm{d}{y}^2.\text{\hspace{1em}(7.116)}$$

Let $g\equiv g_{xx}{g}_{yy}-g_{xy}^2$ and write (7.116) as

$$\mathrm{d}{s}^2=\left(\sqrt{g_{xx}}\mathrm{d}x+\frac{g_{xy}+\mathrm{i}\sqrt{g}}{\sqrt{g_{xx}}}\mathrm{d}y\right)\left(\sqrt{g_{yy}}\mathrm{d}x+\frac{g_{xy}-\mathrm{i}\sqrt{g}}{\sqrt{g_{xx}}}\mathrm{d}y\right).$$

According to the theory of differential equations, there exists an integrating factor $\lambda (x,y)={\lambda }_1(x,y)+\mathrm{i}{\lambda }_2(x,y)$ such that

$$\lambda \left(\sqrt{g_{xx}}dx+\frac{g_{xy}+i\sqrt{g}}{\sqrt{g_{xx}}}dy\right)=du+idv\text{\hspace{1em}(7.117a)}$$

$$\bar{\lambda }\left(\sqrt{g_{yy}}dx+\frac{g_{xy}-i\sqrt{g}}{\sqrt{g_{xx}}}dy\right)=du-idv.\text{\hspace{1em}(7.117b)}$$

Then ${\mathrm{ds}}^2=({\mathrm{du}}^2+{\mathrm{dv}}^2)/|\lambda {|}^2$ and by setting $|\lambda {|}^{-2}={\mathrm{e}}^{2\sigma }$ , we have the desired coordinate system. The coordinates $(u,v)$ are called the isothermal coordinates. [Remark: If the curve $u=a$ constant is regarded as an isothermal curve, $v=a$ constant corresponds to the line of heat flow.]

For example, let ${\mathrm{ds}}^2=\mathrm{d}{\theta }^2+{\sin }^2\theta \mathrm{d}{\varphi }^2$ be the standard metric of $S^2$ . Noting that

$$\frac{\mathrm{d}}{\mathrm{d}\theta }\log \left|\tan \frac{\theta }{2}\right|=\frac{1}{\sin \theta }$$

we find that $f:(\theta ,\varphi )\mapsto (u,v)$ defined by $u=\log |\tan \frac{1}{2}\theta |$ and $v=\varphi$ yields a conformally flat metric. In fact,

$$\mathrm{d}{s}^2={\sin }^2\theta \left(\frac{\mathrm{d}{\theta }^2}{{\sin }^2\theta }+\mathrm{d}{\varphi }^2\right)={\sin }^2\theta (\mathrm{d}{u}^2+\mathrm{d}{v}^2).$$

If $(M,g)$ is a Lorentz manifold, we have integrating factors $\lambda (x,y)$ and $\mu (x,y)$ such that

$$\lambda \left(\sqrt{g_{xx}}dx+\frac{g_{xy}+\sqrt{-g}}{\sqrt{g_{xx}}}dy\right)=du+dv\text{\hspace{1em}(7.118a)}$$

$$\mu \left(\sqrt{g_{xx}}\mathrm{d}x+\frac{g_{xy}-\sqrt{-g}}{\sqrt{g_{xx}}}\mathrm{d}y\right)=\mathrm{d}u-\mathrm{d}v.\text{\hspace{1em}(7.118b)}$$

In terms of the coordinates $(u,v)$ the metric takes the form ${\mathrm{ds}}^2={\lambda }^{-1}{\mu }^{-1}({\mathrm{du}}^2-{\mathrm{dv}}^2)$ . The product $\lambda \mu$ is either positive definite or negative definite and we may set $1/|\lambda \mu |={\mathrm{e}}^{2\sigma }$ to obtain the form

$$\mathrm{d}{s}^2=\pm {\mathrm{e}}^{2\sigma }(\mathrm{d}{u}^2-\mathrm{d}{v}^2).\text{\hspace{1em}(7.119)}$$

Exercise 7.16

Let $(M,g)$ be a two-dimensional Lorentz manifold with $g=-\mathrm{d}t\otimes \mathrm{d}t+t^2\mathrm{d}x\otimes \mathrm{d}x$ (the Milne universe). Use the transformation $|t|\mapsto {\mathfrak{e}}^{\eta }$ to show that $g$ is conformally flat. In fact, it is further simplified by $(\eta ,x)\mapsto (u={\mathrm{e}}^{\eta }\sinh x,v={\mathrm{e}}^{\eta }\cosh x)$ . What is the resulting metric?

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