# Killing vector contractions along isometric curves

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.

• What is an isometric curve? I, for one, have never heard that qualifier used in reference to just a curve on a manifold. Mar 13 '14 at 15:54
• @joshphysics het.brown.edu/people/danieldf/literary/eric-KKtheories.pdf page 1112 first line Mar 13 '14 at 16:18

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 $$\frac{d}{d\lambda}(\xi_{\nu}\xi^{\nu}) = 2\xi^{\nu}\xi^\mu \nabla_\mu\xi_{\nu} = 2\xi^{(\nu}\xi^{\mu)} \nabla_{(\mu}\xi_{\nu)}=0$$