Various texts make this claim, but no proof is given.

Explicitly, let $L$ denote the Lie derivative. Suppose $L_X g_{ab} = 0$ for some vector field $X$, called a Killing vector field. Suppose that the stress-energy tensor $T_{ab}$ satisfies the Einstein field equations. Then, show that $L_X T_{ab} = 0$.

  • 3
    $\begingroup$ I can give you the hint that this is equivalent to showing that $\pound_{X}R_{ab}=0$ $\endgroup$ Dec 2, 2013 at 3:06
  • $\begingroup$ Yes, I figured out how to prove it given that. However, I can't see how to proceed any further. To write out $L_X R_{ab}$ in terms of the metric tensor would be long and tedious and I'm sure there must be a nicer proof. $\endgroup$
    – Brian Bi
    Dec 2, 2013 at 3:07
  • $\begingroup$ see also mathoverflow.net/q/47332 $\endgroup$
    – Christoph
    Dec 2, 2013 at 16:14

3 Answers 3


Edit: Note: I have posted another proof of this in another question, here. Those who prefer coordinates may find it slightly more palatable.

I gather from your comments that you can do this if you have $\mathcal{L}_X\text{Ric} = 0$. Thus I will outline a somewhat more general result, assuming a certain identity connecting Killing vectors and Riemann curvature that's an exercise in many textbooks.

A Killing vector field $X$ has $R^a{}_{bcd}X^d = \nabla_c\nabla_bX^a$, i.e., $$\nabla_Z\nabla_YX - \nabla_{\nabla_ZY}X= R(Z,X)Y \equiv_{\text{def}} \nabla_Z\nabla_XY - \nabla_X\nabla_ZY - \nabla_{[Z,X]}Y\text{.}$$ Whenever torsion vanishes, substitution gives $$\begin{eqnarray*} \mathcal{L}_X(\nabla_ZY) &=& \nabla_X\nabla_ZY - \nabla_{\nabla_ZY}X\\ &=&\nabla_Z\nabla_XY - \nabla_Z\nabla_YX + \nabla_{[X,Z]}Y\\ &=&\nabla_Z(\mathcal{L}_XY) + \nabla_{\mathcal{L}_XZ}Y\text{.}\end{eqnarray*} $$ You should be able to apply this formula to the covariant-derivative form of the Riemann tensor to show that $$\mathcal{L}_X[R(Y,Z)W] = R(\mathcal{L}_XY,Z)W + R(Y,\mathcal{L}_XZ)W + R(Y,Z)\mathcal{L}_XW\text{,}$$ i.e., the Lie derivative of the Riemann tensor along a Killing vector vanishes. The corresponding result for the Ricci tensor follows trivially.

  • 1
    $\begingroup$ Thanks for the answer! Unfortunately, I'm not familiar with the notation $R(Z,X)Y$ and I'm not sure how to interpret the expression $\nabla_{\nabla_Z Y} X$. Any idea how to do it in index notation? $\endgroup$
    – Brian Bi
    Dec 2, 2013 at 12:11
  • $\begingroup$ I am afraid I missed some points too... 1) I don't see from where comes the term $\nabla_{\nabla_Z Y} X$, in the LHS of the second equation... 2) I am not able to make the link between the expression of $\mathcal{L}_X(\nabla_ZY)$ and the expression of $\mathcal{L}_X[R(Y,Z)W]$... 3) I don't see why the last expression $\mathcal{L}_X[R(Y,Z)W]$ is zero (unless $\mathcal{L}_X Y, \mathcal{L}_X Z, \mathcal{L}_X W$ are zero, but I don't see why !) $\endgroup$
    – Trimok
    Dec 2, 2013 at 12:34
  • 1
    $\begingroup$ @Trimok: see physics.stackexchange.com/a/88705/6389 $\endgroup$
    – Christoph
    Dec 2, 2013 at 13:45
  • 2
    $\begingroup$ Standard notation: $[R(Z,X)Y]^a = R^a{}_{bcd}Y^bZ^cX^d$. Also, the following is a definition in some books and an exercise in others (e.g., d'Inverno 6.10): $$\nabla_Z\nabla_XY^a-\nabla_X\nabla_ZY^a-\nabla_{[Z,X]}Y^a = R^a{}_{bcd}Y^bZ^cX^d\text{.}$$ Interpret $\nabla_{\nabla_YZ}X$ literally: $\nabla_YZ$ is itself a vector field, so it's just $\nabla_WX$ with $W = \nabla_YZ$. I think everything else was elaborated by @Christoph in-depth. $\endgroup$
    – Stan Liou
    Dec 2, 2013 at 14:26
  • 1
    $\begingroup$ +1 : OK, with your answer and Christoph's answer, I think I have checked all the details... $\endgroup$
    – Trimok
    Dec 2, 2013 at 21:02

This is not a complete answer, but fills in some of the missing pieces Trimok asked about in the comments to Stan's answer:

Note that I did not verify that Stan's proof actually works.

Re 1)

$$ \begin{align} \nabla_Z \nabla_Y X &= Z^\lambda(Y^\mu X^\nu_{;\mu})_{;\lambda} \partial_\nu \\&= Z^\lambda(Y^\mu_{;\lambda} X^\nu_{;\mu} + Y^\mu X^\nu_{;\mu;\lambda}) \partial_\nu \\&= \nabla_{\nabla_ZY}X + Z^\lambda Y^\mu\nabla_{\partial_\lambda}\nabla_{\partial_\mu} X \\&= \nabla_{\nabla_ZY}X + R(Z,X)Y \end{align} $$ where the last equality was assumed and is proven over there.

Now, there's another step that might not be obvious to all readers: $$ \mathcal L_X(\nabla_YZ) = [X,\nabla_YZ] = \nabla_X\nabla_YZ - \nabla_{\nabla_YZ}X $$ where the second equality is due to zero torsion, ie $$ \nabla_AB - \nabla_BA - [A,B] = 0 $$ for arbitrary $A,B$ and in particular $A=X, B=\nabla_YZ$.

Re 2)

This step was left as an exercise for the reader - just compute the left-hand side of the last equation. The expression $\mathcal L_X(\nabla_YZ)$ holds for generic $Y,Z$ - that's just a bit of unfortunate naming.

Re 3)

That's the Leibniz rule for tensors with one term $\mathcal L_XR$ missing.

  • 1
    $\begingroup$ +1 : OK, I lost my last neurone, but I think that I checked all the steps in details... $\endgroup$
    – Trimok
    Dec 2, 2013 at 21:01

A slightly different and, perhaps, more simple proof follows. $\def\lie{\mathit{£}}$ For $K^a$ a Killing vector, we have (Kostant formula), $$\nabla_a \nabla_b K^c = -R_{bca}{}^d K^d.$$ Using this, we may prove that the covariant and the Lie derivatives commute: $$\lie_K \nabla_a X^b = \nabla_a \lie_K X^b,$$ for arbitrary $X^a$. We have: $$\lie_K(R_{abcd}X^d )= 2\nabla_{[a}\nabla_{b]} \lie_K X_c = R_{abcd}\lie_K X^d,$$ and $$\lie_K (R_{abcd}X^d) = \lie_K R_{abcd} X^d + R_{abcd} \lie_K X^d,$$ and these two combined yield $\lie_K R_{abcd} X^d = 0$ for arbitrary $X^a$, therefore $\lie_K R_{abcd} = 0$. Obviously, the Lie derivatives of Riemann tensor's contractions also vanish, therefore $\lie_K T_{ab} = 0$ by Einstein's equation.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.