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I'm attempting to follow a proof that the commutator of two Killing vectors is itself a Killing vector. The source that I've posted is from my course notes.

Question

I've highlighted the part I'm stuck on. I've been working on this question for quite a while, and I've understood everything else in the solution. I'm struggling to see how the 4 terms in the first row inside the red box cancel out.

Thanks for any help!

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1 Answer 1

Notice that $(\nabla_\alpha A^\mu) (\nabla_\beta B_\mu) = (\nabla_\alpha A_\mu) (\nabla_\beta B^\mu) $ and hence (looking only at the first row), the first term reads $$-(\nabla_\mu U^\tau)(\nabla_\nu V_\tau) = -(\nabla_\nu V_\tau)(\nabla_\mu U^\tau)= -(\nabla_\nu V^\tau)(\nabla_\mu U_\tau),$$ which is the symmetric of the forth and thus cancels with it. Same thing with the second term: $$(\nabla_\mu V^\tau)(\nabla_\nu U_\tau) = (\nabla_\nu U_\tau)(\nabla_\mu V^\tau)= (\nabla_\nu U^\tau)(\nabla_\mu V_\tau),$$ which is the symmetric of (and cancels with) the third.

The source of confusion seems to concern the "swapping indices up and down". This works even when covariant derivatives are present as long as one works with the metric connection (see, for instance, chapter 6 of Ray D'Inverno's "Introducing Einstein's Relativity", namely section 6.10 and others leading up to it). The metric tensor then satisfies $\nabla_\alpha \, g_{\mu\nu} = 0$, and one indeed has:

$$\nabla_\alpha A^\mu = \nabla_\alpha (g^{\mu\nu} A_\nu) = g^{\mu\nu} (\nabla_\alpha A_\nu), $$ so that

$$(\nabla_\alpha A^\mu) (\nabla_\beta B_\mu) = (\nabla_\alpha A_\nu) (\nabla_\beta (g^{\mu\nu}B_\mu)) = (\nabla_\alpha A_\nu) (\nabla_\beta B^\nu) = (\nabla_\alpha A_\mu) (\nabla_\beta B^\mu),$$ where in the last step one has changed with impunity the letter representing repeated indices.

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