2
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ Suggestion to the question (v1): Replace the word Killing form with Killing vector field. $\endgroup$ – Qmechanic Dec 29 '15 at 8:59
  • 1
    $\begingroup$ In the very last expression if you change $\mu$ and $\nu$ the expression will be the same (since there is a summation over them). On the other hand $\xi$ is Killing so it will change sign. Thus the expression is equal to its negative, so it must be zero. $\endgroup$ – MBN Dec 29 '15 at 9:20
  • $\begingroup$ And all the calculation, from where you expanded the covarient derivative till the end, are unnecessary. $\endgroup$ – MBN Dec 29 '15 at 9:21
  • $\begingroup$ @Qmechanic, the reason to call it a form is that symmetrization of indices with the covariant index of a derivative ought to be a covariant index as well. But I agree that it might lead to confusion with the other Killing form used in Lie algebras $\endgroup$ – lurscher Dec 29 '15 at 13:48
  • $\begingroup$ @MBN, no, you are missing the $\frac{d^2 P^{\mu}}{d \lambda^2} \xi_{\mu}$ term $\endgroup$ – lurscher Dec 29 '15 at 14:14
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ A few observations: 1) your expression has the index $\mu$ repeated four times, which might be confusing $\endgroup$ – lurscher Dec 29 '15 at 13:52
  • $\begingroup$ 2) What is the point of using two notations for the covariant derivative? $\nabla_{\mu}=\partial_{;\mu}$ $\endgroup$ – lurscher Dec 29 '15 at 13:52
  • $\begingroup$ 3) $\partial_{;\nu} \frac{d P^{\mu}}{d\lambda}$ is not meaningful since $P^{\mu}$ is not a function of coordinates but of $\lambda$, so there is no point in complicating things here; $\frac{d}{d\lambda}( \frac{d P^{\mu}}{d \lambda})$ is just $\frac{d^2 P^{\mu}}{d \lambda^2}$ $\endgroup$ – lurscher Dec 29 '15 at 13:54
  • $\begingroup$ I copy pasted your answer and made clear the part where I think your notation is confusing you. By all means change the indices or use semicolons, I am just trying to communicate where the error is to you. It is not just a complication, to use the product rule over a contraction like you did you need to use the covariant derivative. Since you are dotting the index with the tangent vector it does not matter if $dP^\mu/d\lambda$ is only defined on the path. $\endgroup$ – octonion Dec 29 '15 at 17:08
  • $\begingroup$ How do you evaluate the expression $\partial_{;\nu} \frac{dP^\mu}{d\lambda}$? (or $\nabla_{\nu} \frac{dP^\mu}{d\lambda}$ in your previous notation) $\endgroup$ – lurscher Dec 29 '15 at 17:30

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.