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 $L_X T_{ab} = 0$.

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    I can give you the hint that this is equivalent to showing that $\pound_{X}R_{ab}=0$ – Jerry Schirmer Dec 2 '13 at 3:06
  • 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. – Brian Bi Dec 2 '13 at 3:07
  • see also – Christoph Dec 2 '13 at 16:14

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.

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    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? – Brian Bi Dec 2 '13 at 12:11
  • 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 !) – Trimok Dec 2 '13 at 12:34
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    @Trimok: see – Christoph Dec 2 '13 at 13:45
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    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. – Stan Liou Dec 2 '13 at 14:26
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    +1 : OK, with your answer and Christoph's answer, I think I have checked all the details... – Trimok Dec 2 '13 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.

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    +1 : OK, I lost my last neurone, but I think that I checked all the steps in details... – Trimok Dec 2 '13 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.

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