2
$\begingroup$

Imagine $\xi_{\nu}$ is a Killing vector field on a manifold. Does $\xi_{\nu}\xi^{\nu}$ remain constant along any isometric curve defined by the Killing vector field?

My guess is that yes since as you move along an isometric curve every point "around" you looks pretty much as it did a differential step ahead, but I am don't have a full grasp on this and would thank some clarification.

$\endgroup$
4
$\begingroup$

Let $\lambda$ be an affine parameter of the integral curves of $\xi^{\nu}$ then you question translates as $$ \frac{d}{d\lambda}(\xi_{\nu}\xi^{\nu}) = \xi^\mu \nabla_\mu(\xi_{\nu}\xi^{\nu}) = (\xi^\mu \nabla_\mu\xi_{\nu})\xi^{\nu} + \xi_{\nu}(\xi^\mu \nabla_\mu\xi^{\nu}) $$ if the connection is Levi-Civita (i.e metric compatible) $$ \frac{d}{d\lambda}(\xi_{\nu}\xi^{\nu}) = 2\xi^{\nu}\xi^\mu \nabla_\mu\xi_{\nu} $$ but now since $\xi^\nu$ is a Killing vector and it satisfies Killing's equation $$ \nabla_{(\nu}\xi_{\mu)}=\nabla_\mu\xi_{\nu} + \nabla_\nu\xi_{\mu} = 0 $$ and since $\xi^{\nu}\xi^\mu = \xi^{(\nu}\xi^{\mu)}$ is symmetric we have that $$ \frac{d}{d\lambda}(\xi_{\nu}\xi^{\nu}) = 2\xi^{\nu}\xi^\mu \nabla_\mu\xi_{\nu} = 2\xi^{(\nu}\xi^{\mu)} \nabla_{(\mu}\xi_{\nu)}=0 $$

$\endgroup$

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.