7.7 Killing vector fields and conformal Killing vector fields

7.7.1 Killing vector fields

Let $(M,g)$ be a Riemannian manifold and $X\in \mathfrak{X}(M)$ . If a displacement $\epsilon X$ , $\epsilon$ being infinitesimal, generates an isometry, the vector field $X$ is called a Killing vector field. The coordinates $x^{\mu }$ of a point $p\in M$ change to $x^{\mu }+\epsilon X^{\mu }(p)$ under this displacement, see (5.42). If $f:x^{\mu }\mapsto x^{\mu }+\epsilon X^{\mu }$ is an isometry, it satisfies (7.96b),

$$\frac{\partial \left(x^{\kappa }+\epsilon X^{\kappa }\right)}{\partial x^{\mu }}\frac{\partial \left(x^{\lambda }+\epsilon X^{\lambda }\right)}{\partial x^{\nu }}{g}_{\kappa \lambda }(x+\epsilon X)=g_{\mu \nu }(x).$$

After a simple calculation, we find that $g_{\mu \nu }$ and $X^{\mu }$ satisfy the Killing equation

$$X^{\xi }{\partial }_{\xi }{g}_{\mu v}+{\partial }_{\mu }{X}^{\kappa }{g}_{\kappa v}+{\partial }_v{X}^{\lambda }{g}_{\mu \lambda }=0.\text{\hspace{1em}(7.120a)}$$

From the definition of the Lie derivative, this is written in a compact form as

$${\left({\mathcal{L}}_{X}g\right)}_{\mu v}=0.\text{\hspace{1em}(7.120b)}$$

Let ${\varphi }_t:M\to M$ be a one-parameter group of transformations which generates the Killing vector field $X$ . Equation (7.120b) then shows that the local geometry does not change as we move along ${\varphi }_t$ . In this sense, the Killing vector fields represent the direction of the symmetry of a manifold.

A set of Killing vector fields are defined to be dependent if one of them is expressed as a linear combination of others with constant coefficients. Thus, there may be more Killing vector fields than the dimension of the manifold. [The number of independent symmetries has no direct connection with $\dim M$ . The maximum number, however, has; see example 7.10.]

Exercise 7.17

Let $\nabla$ be the Levi-Civita connection. Show that the Killing equation is written as

$$({\nabla }_{\mu }X{)}_v+({\nabla }_{v}X{)}_{\mu }={\partial }_{\mu }{X}_v+{\partial }_v{X}_{\mu }-2{\Gamma }_{\mu v}^{\lambda }{X}_{\lambda }=0.\text{\hspace{1em}(7.121)}$$

Exercise 7.18

Find three Killing vector fields of $({\mathbb{R}}^2,\delta )$ . Show that two of them correspond to translations while the third corresponds to a rotation; cf next example.

Example 7.10

Let us work out the Killing vector fields of the Minkowski spacetime $({\mathbb{R}}^4,\eta )$ , for which all the Levi-Civita connection coefficients vanish. The Killing equation becomes

$${\partial }_{\mu }{X}_{\nu }+{\partial }_{\nu }{X}_{\mu }=0.\text{\hspace{1em}(7.122)}$$

It is easy to see that $X_{\mu }$ is, at most, of the first order in $x$ . The constant solutions

$$X_{(i)}^{\mu }={\delta }_i^{\mu }(0\le i\le 3)\text{\hspace{1em}(7.123a)}$$

correspond to spacetime translations. Next, let $X_{\mu }=a_{\mu \nu }{x}^{\mu }$ , $a_{\mu \nu }$ being constant. Equation (7.122) implies that $a_{\mu \nu }$ is anti-symmetric with respect to $\mu \leftrightarrow \nu$ . Since $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$ , there are six independent solutions of this form, three of which

$$X_{(j)0}=0X_{(j)m}={\epsilon }_{jmn}{x}^n(1\le j,m,n\le 3)\text{\hspace{1em}(7.123b)}$$

correspond to spatial rotations about the $x^j$ -axis, while the others

$$X_{(k)0}=x^k{X}_{(k)m}=-{\delta }_{km}{x}^0(1\le k,m\le 3)\text{\hspace{1em}(7.123c)}$$

correspond to Lorentz boosts along the $x^k$ -axis.

In $m$ -dimensional Minkowski spacetime ( $m\ge 2$ ), there are $m(m+1)/2$ Killing vector fields, $m$ of which generate translations, ( $m-1$ ), boosts and ( $m-1$ )(m - 2) / 2, space rotations. Those spaces (or spacetimes) which admit $m(m+1)/2$ Killing vector fields are called maximally symmetric spaces.

Let $X$ and $Y$ be two Killing vector fields. We easily verify that

(i) a linear combination $aX+bY$ $(a,b\in \mathbb{R})$ is a Killing vector field; and
(ii) the Lie bracket $[X,Y]$ is a Killing vector field.

(i) is obvious from the linearity of the covariant derivative. To prove (ii), we use (5.58). We have ${\mathcal{L}}_{[X,Y]}g={\mathcal{L}}_X{\mathcal{L}}_{Y}g-{\mathcal{L}}_Y{\mathcal{L}}_{X}g=0$ , since ${\mathcal{L}}_{X}g={\mathcal{L}}_{Y}g=0$ . Thus, all the Killing vector fields form a Lie algebra of the symmetric operations on the manifold $M$ ; see the next example.

Example 7.11

Let $g=d\theta \otimes \mathrm{d}\theta +{\sin }^2\theta \mathrm{d}\varphi \otimes \mathrm{d}\varphi$ be the standard metric of $S^2$ . The Killing equations (7.121) are:

$${\partial }_{\theta }{X}_{\theta }+{\partial }_{\theta }{X}_{\theta }=0\text{\hspace{1em}(7.124a)}$$

$${\partial }_{\varphi }{X}_{\varphi }+{\partial }_{\varphi }{X}_{\varphi }+2\sin \theta \cos \theta X_{\theta }=0\text{\hspace{1em}(7.124b)}$$

$${\partial }_{\theta }{X}_{\varphi }+{\partial }_{\varphi }{X}_{\theta }-2\cot \theta X_{\varphi }=0.\text{\hspace{1em}(7.124c)}$$

It follows from (7.124a) that $X_{\theta }$ is independent of $\theta$ : $X_{\theta }(\theta ,\varphi )=f(\varphi )$ . Substituting this into (7.124b), we have

$$X_{\varphi }=-F(\varphi )\sin \theta \cos \theta +g(\theta )\text{\hspace{1em}(7.125)}$$

where $F(\varphi )={\int }^{\varphi }f(\varphi )\mathrm{d}\varphi$ . Substitution of (7.125) into (7.124c) yields

$$-F(\varphi )({\cos }^2\theta -{\sin }^2\theta )+\frac{\mathrm{d}g}{\mathrm{d}\theta }+\frac{\mathrm{d}f}{\mathrm{d}\varphi }+2\cot \theta (F(\varphi )\sin \theta \cos \theta -g(\theta ))=0.$$

This equation may be separated into

$$\frac{\mathrm{d}g}{\mathrm{d}\theta }-2\cot \theta g(\theta )=-\frac{\mathrm{d}f}{\mathrm{d}\varphi }-F(\varphi ).$$

Since both sides must be separately constant $(\equiv C)$ , we have

$$\frac{\mathrm{d}g}{\mathrm{d}\theta }-2\cot \theta g(\theta )=C\text{\hspace{1em}(7.126a)}$$

$$\frac{\mathrm{d}f}{\mathrm{d}\varphi }+F(\varphi )=-C.\text{\hspace{1em}(7.126b)}$$

Equation (7.126a) is solved if we multiply both sides by $\exp (-\int \mathrm{d}\theta 2\cot \theta )={\sin }^{-2}\theta$ to make the LHS a total derivative,

$$\frac{\mathrm{d}}{\mathrm{d}\theta }\left(\frac{g(\theta )}{{\sin }^2\theta }\right)=\frac{C}{{\sin }^2\theta }.$$

The solution is easily found to be

$$g(\theta )=\left(C_1-C\cot \theta \right){\sin }^2\theta .$$

Differentiating (7.126b) again, we find that $f$ is harmonic,

$$X_{\theta }(\varphi )=f(\varphi )=A\sin \varphi +B\cos \varphi$$

$$F(\varphi )=-A\cos \varphi +B\sin \varphi -C.$$

Substituting these results into (7.125), we have

$$\begin{array}{rl}{X}_{\varphi }(\theta ,\varphi )& =-(-A\cos \varphi +B\sin \varphi -C)\sin \theta \cos \theta +(C_1-C\cot \theta ){\sin }^2\theta \\ & =(A\cos \varphi -B\sin \varphi )\sin \theta \cos \theta +C_1{\sin }^2\theta .\end{array}$$

A general Killing vector is given by

$$\begin{array}{rl}& X=X^{\theta }\frac{\partial }{\partial \theta }+X^{\varphi }\frac{\partial }{\partial \varphi }\\ & =A\left(\sin \varphi \frac{\partial }{\partial \theta }+\cos \varphi \cot \theta \frac{\partial }{\partial \varphi }\right)\\ & +B\left(\cos \varphi \frac{\partial }{\partial \theta }-\sin \varphi \cot \theta \frac{\partial }{\partial \varphi }\right)+C_1\frac{\partial }{\partial \varphi }.\end{array}\text{\hspace{1em}(7.127)}$$

The basis vectors

$$L_x=-\cos \varphi \frac{\partial }{\partial \theta }+\cot \theta \sin \varphi \frac{\partial }{\partial \varphi }\text{\hspace{1em}(7.128a)}$$

$$L_y=\sin \varphi \frac{\partial }{\partial \theta }+\cot \theta \cos \varphi \frac{\partial }{\partial \varphi }\text{\hspace{1em}(7.128b)}$$

$$L_z=\frac{\partial }{\partial \varphi }\text{\hspace{1em}(7.128c)}$$

generate rotations round the $x$ , $y$ and $z$ axes respectively.

These vectors generate the Lie algebra $\mathfrak{so}(3)$ . This reflects the fact that $S^2$ is the homogeneous space $\mathrm{SO}(3)/\mathrm{SO}(2)$ and the metric on $S^2$ retains this SO(3) symmetry (see example 5.18(a)). In general $S^n=\mathrm{SO}(n+1)/\mathrm{SO}(n)$ with the usual metric has $\dim \mathrm{SO}(n+1)=n(n+1)/2$ Killing vectors and they form the Lie algebra $\mathfrak{so}(n+1)$ . The sphere $S^n$ with the usual metric is a maximally symmetric space. We may squash $S^n$ so that it has fewer symmetries. For example, if $S^2$ considered here is squashed along the $z$ -axis it has a rotational symmetry around the $z$ -axis only and there exists one Killing vector field $L_z=\partial /\partial \varphi$ .

7.7.2 Conformal Killing vector fields

Let $(M,g)$ be a Riemannian manifold and let $X\in \mathfrak{X}(M)$ . If an infinitesimal displacement given by $\epsilon X$ generates a conformal transformation, the vector field $X$ is called a conformal Killing vector field (CKV). Under the displacement $x^{\mu }\to x^{\mu }+\epsilon X^{\mu }$ , this condition is written as

$$\frac{\partial (x^{\kappa }+\epsilon X^{\kappa })}{\partial x^{\mu }}\frac{\partial (x^{\lambda }+\epsilon X^{\lambda })}{\partial x^{\nu }}{g}_{\kappa \lambda }(x+\epsilon X)={\mathrm{e}}^{2\sigma }{g}_{\mu \nu }(x).$$

Noting that $\sigma \propto \epsilon$ , we set $\sigma =\epsilon \psi /2$ , where $\psi \in \mathcal{F}(M)$ . Then we find that $g_{\mu \nu }$ and $X^{\mu }$ satisfy

$${\mathcal{L}}_X{g}_{\mu v}=X^{\xi }{\partial }_{\xi }{g}_{\mu v}+{\partial }_{\mu }{X}^{\kappa }{g}_{\kappa v}+{\partial }_v{X}^{\lambda }{g}_{\mu \lambda }=\psi g_{\mu v}.\text{\hspace{1em}(7.129a)}$$

Equation (7.129a) is easily solved for $\psi$ to yield

$$\psi =\frac{X^{\xi }{g}^{\mu v}{\partial }_{\xi }{g}_{\mu v}+2{\partial }_{\mu }{X}^{\mu }}{m}\text{\hspace{1em}(7.129b)}$$

where $m=\dim M$ . We verify that

(i) a linear combination of CKVs is a CKV: $({\mathcal{L}}_{aX+bY}g{)}_{\mu \nu }=(a\phi +b\psi )g_{\mu \nu }$ where $a,b\in \mathbb{R},{\mathcal{L}}_X{g}_{\mu \nu }=\phi g_{\mu \nu }$ and ${\mathcal{L}}_Y{g}_{\mu \nu }=\psi g_{\mu \nu }$
(ii) the Lie bracket $[X,Y]$ of a CKV is again a CKV: ${\mathcal{L}}_{[X,Y]}{g}_{\mu \nu }=(X[\psi ]-Y[\phi ])g_{\mu \nu }$ .

Example 7.12

Let $x^{\mu }$ be the coordinates of $({\mathbb{R}}^m,\delta )$ . The vector

$$D\equiv x^{\mu }\frac{\partial }{\partial x^{\mu }}\text{\hspace{1em}(7.130)}$$

(dilatation vector) is a CKV. In fact,

$${\mathcal{L}}_D{\delta }_{\mu \nu }={\partial }_{\mu }{x}^{\kappa }{\delta }_{\kappa \nu }+{\partial }_{\nu }{x}^{\lambda }{\delta }_{\mu \lambda }=2{\delta }_{\mu \nu }.$$

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