I'm going over an assignment for my general relativity course. My solution to the question below strikes me as too short, considering that it appeared in the "longer questions" section of the assignment.


A Killing vector $V^\mu$ is one that satisfies the equation $\nabla_{\mu} V_{\nu}+\nabla_{\nu} V_{\mu}=0$. The commutator of two vectors $U^\mu$ and $V^\mu$ is defined as $[U,V]^\mu=U^\nu \nabla_\nu V^\mu-U^\nu \nabla_\nu U^\mu$. Show that the commutator of two Killing vectors is itself a Killing vector.

My solution could be (and is likely to be) incorrect, but I'm just using the definitions that I've seen so far.

My solution: To get from $[U,V]^\mu$ to $[U,V]_\mu$ we have to use the metric i.e. $[U,V]_\nu=g_{\nu \lambda}[U,V]^\lambda$. Now I just plugged this in to the Killing equation to get (for the first term)

$\nabla_{\mu}[U,V]_\nu=\nabla_{\mu}g_{\nu \lambda}[U,V]^\lambda$.

Using that the metric is covariantly constant, we have $\nabla_{\mu}g_{\nu \lambda}=0$, hence the whole term is equal to zero. The same thing happens with the second term, so that the whole expression is zero. I can't see that I made a mistake, but it seems way too short a solution. When I looked at the solutions provided in the course, they are about a page long, and begin

$$\nabla_{\mu}[U,V]_\nu + \nabla_\nu[U,V]_\mu=\nabla_\mu (U^\tau \nabla_\tau V_\nu-V^\tau \nabla_\tau U_\nu ) + \nabla_\nu (U^\tau \nabla_\tau V_\mu-V^\tau \nabla_\tau U_\mu )=\dots$$

i.e. they do things explicitly. Actually, having written this out, I can see that what I've done can't be correct otherwise every vector would be a killing vector just by writing $V_\mu$ as $V_\mu=g_{\mu \nu}V^\nu$. However I don't know what I've done wrong, so any help would be very much appreciated!


1 Answer 1


$$\nabla_{\mu}\left[U,V\right]_{\nu} = \nabla_{\mu}\left(g_{\nu\lambda}\left[U,V\right]^{\lambda}\right)$$

You have two terms inside the parentheses, and you have to apply the derivative to both of them. Myself, I'd just remember that:

$$\left[U,V\right]^{a} = U^{b}\nabla_{b}V^{a} - V^{b}\nabla_{b}U^{a}$$,

so, we have:

$$\begin{align} \left[U,V\right]_{a} &= g_{ab} \left[U,V\right]^{b}\\ &=g_{ab}\left( U^{c}\nabla_{c}V^{b} - V^{c}\nabla_{c}U^{b}\right)\\ &=U^{c}\nabla_{c}V_{a} - V^{c}\nabla_{c}U_{a} \end{align}$$

  • 1
    $\begingroup$ Oh wow, in getting tangled up in the index notation I completely ignored the product rule, doh! Ok thanks a lot $\endgroup$
    – Lammey
    Jul 29, 2014 at 16:26
  • $\begingroup$ Instead of covariant derivatives, shouldn't we have partial derivatives in the expression for the components of the commutator $[U,V]$? Because this expression comes easily from the fact that $\{ \partial_\mu \}$ is a basis for the tangent space $\endgroup$
    – soap
    Nov 8, 2017 at 11:35
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    $\begingroup$ @Soap : it doesn't matter which one you use, the Christoffel symbols cancel. I like using the covariant derivatives in the derivation to avoid taking derivatives of the metric tensor, but the Lie derivative is (torsion-free) connection independent. $\endgroup$ Nov 8, 2017 at 16:38

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