Proving that Killing form contractions with geodesics are constants of motion

Question

I want to prove the fundamental theorem of Killing forms, namely that

$$\frac{d}{d \lambda} \Big( \frac{d P^{\mu}}{d \lambda} \xi_{\mu}(P(\lambda)) \Big) = 0 $$

If $P(\lambda)$ is a Geodesic curve, which implies that $\dot{P}^{\mu} \xi_{\mu}(P(\lambda))$ are constants of geodesic motion

This should be straightforward to prove, basically expanding the derivative expression

\begin{align*} \frac{d}{d \lambda} \Big( \frac{d P^{\mu}}{d \lambda} \xi_{\mu}(P(\lambda)) \Big) &= \\ &=\frac{d^2 P^{\mu}}{d \lambda^2}\xi_{\mu}(P(\lambda))+\frac{d P^{\mu}}{d \lambda}\partial_{;\nu} \xi_{\mu}(P(\lambda)) \frac{d P^{\nu}}{d \lambda} \\ \end{align*}

We now use the fact that $\xi_{\nu}$ is a Killing form, that is:

$$\partial_{;\nu} \xi_{\mu}(P(\lambda)) = - \partial_{;\mu} \xi_{\nu}(P(\lambda)) $$

And we expand the covariant derivative:

\begin{align*} \frac{d}{d \lambda} \Big( \frac{d P^{\mu}}{d \lambda} \xi_{\mu}(P(\lambda)) \Big) &= \\ &=\frac{d^2 P^{\mu}}{d \lambda^2}\xi_{\mu}(P(\lambda))-\frac{d P^{\mu}}{d \lambda}\partial_{;\mu} \xi_{\nu}(P(\lambda)) \frac{d P^{\nu}}{d \lambda} \\ &= \frac{d^2 P^{\mu}}{d \lambda^2}\xi_{\mu}(P(\lambda))-\frac{d P^{\mu}}{d \lambda} \Big[ \partial_{\mu} \xi_{\nu}(P(\lambda)) - \Gamma^{\theta}_{\mu \nu} \xi_{\theta}(P(\lambda)) \Big] \frac{d P^{\nu}}{d \lambda} \\ &= \frac{d^2 P^{\mu}}{d \lambda^2}\xi_{\mu}(P(\lambda))+ \Gamma^{\theta}_{\mu \nu} \xi_{\theta}(P(\lambda))\frac{d P^{\mu}}{d \lambda}\frac{d P^{\nu}}{d \lambda} - \partial_{\mu} \xi_{\nu}(P(\lambda)) \frac{d P^{\mu}}{d \lambda} \frac{d P^{\nu}}{d \lambda} \end{align*}

But

$$\Gamma^{\theta}_{\mu \nu}\frac{d P^{\mu}}{d \lambda}\frac{d P^{\nu}}{d \lambda}= - \frac{d^2 P^{\theta}}{d \lambda^2} $$

Since the curve is a geodesic, which means that the expression simplifies:

\begin{align*} \frac{d}{d \lambda} \Big( \frac{d P^{\mu}}{d \lambda} \xi_{\mu}(P(\lambda)) \Big) &= \\ &= \frac{d^2 P^{\mu}}{d \lambda^2}\xi_{\mu}(P(\lambda))- \frac{d^2 P^{\theta}}{d \lambda^2} \xi_{\theta}(P(\lambda)) - \partial_{\mu} \xi_{\nu}(P(\lambda)) \frac{d P^{\mu}}{d \lambda} \frac{d P^{\nu}}{d \lambda} \\ &= - \partial_{\mu} \xi_{\nu}(P(\lambda)) \frac{d P^{\mu}}{d \lambda} \frac{d P^{\nu}}{d \lambda} \end{align*}

So, I am able to get rid of those two terms, but there is still an uncancelled term with the coordinate derivative of the $\xi$ form. I don't know how to proceed next

Suggestions?

Answer 1

Your $d/d\lambda$ should be understood in terms of parallel transport $$\frac{d}{d\lambda}\equiv \frac{dP^\mu}{d\lambda}\partial_{;\mu}$$ It is not an ordinary derivative when it acts on a vector.

So writing your third line in more transparent notation: $$\frac{d}{d \lambda} \Big( \frac{d P^{\mu}}{d \lambda} \xi_{\mu} \Big) =\left(\frac{dP^\nu}{d\lambda}\partial_{;\nu} \frac{dP^\mu}{d\lambda}\right)\xi_{\mu}+\left(\frac{dP^\nu}{d\lambda}\partial_{;\nu} \,\xi_{\mu}\right)\frac{dP^\mu}{d\lambda} $$ The second term vanishes since $\partial_{;\mu}\xi_\nu$ is antisymmetric in the indices as MBN points out in his comment (although it only applies to the covariant derivative). The first term vanishes by the geodesic equation, as should be clear in this notation.