# GR Lie Derivative of metric vanish <=> metric is independent

1. Jul 18, 2017

### binbagsss

1. The problem statement, all variables and given/known data

How to show that lie deriviaitve of metric vanish $(L_v g)_{uv}=0$ <=> metric is independent of this coordinate, for example if $v=\partial_z$ then $g_{uv}$ is independent of $z$ (and vice versa)

2. Relevant equation

I am wanting to show this for the levi-civita symbol as the connection i.e. metric compatability were we have $\nabla_{a}g^{uv} =0$ so the first term of $(L_v g)_{uv}=0$ vanishes trivially.

3. The attempt at a solution

$(L_ug)_{uv} = U^{\alpha}\nabla_{\alpha}g_{uv}+g_{u\alpha}\nabla_vU^{\alpha}+g_{\alpha v}\nabla_u U^{\alpha}$ (1)

If the metric is independent of some coordinate, e.g $z$ then we have $\partial_z g_{ab} =0$ but , assuming metric compatibility $\nabla_{\alpha}g_{uv}=0$ anyway, so there's no need to split it up into the partial and connection term, which seemed to me the most obvious way and only way I can see to substitute the information of metric independence of (say) $z$ into (1)

So we have $(L_ug)_{uv} = g_{u\alpha}\nabla_vU^{\alpha}+g_{\alpha v}\nabla_u U^{\alpha}$, where there are no derivatives on the metric in these two terms, so I'm pretty stuck..

Many thanks

2. Jul 18, 2017

### davidge

This is not so in general. You can have metric invariance along a vector which is not in one of the coordinate directions. The metric of an ordinary sphere is a example.
So, you should try to show that vanishing of the Lie derivative of the metric along a vector $\Rightarrow$ metric is invariant along that vector.

3. Jul 18, 2017

### stevendaryl

Staff Emeritus
First of all, the Lie derivative does not involve the connection, so it's actually the case that:

$(L_U g)_{\mu \nu} = U^\alpha \partial_\alpha g_{\mu \nu} + g_{\mu \alpha} \partial_\nu U^\alpha + g_{\alpha \nu} \partial_\mu U^\alpha$

The first term doesn't vanish in general (although the corresponding expression involving the covariant derivative does: $U^\alpha \nabla_\alpha g_{\mu \nu} = 0$)

Now, let's look at the special case where $U$ is a coordinate basis vector, $e_\lambda$, that means that $U^\lambda = 1$ and $U^\alpha = 0$ if $\alpha \neq \lambda$. So for this particular choice of $U$, $\partial_\mu U^\alpha = \partial_\nu U^\alpha = 0$. So we have:

$(L_U g)_{\mu \nu} = \partial_\lambda g_{\mu \nu} + 0 + 0$

4. Jul 20, 2017

### binbagsss

perfect, thank you