What's the 2nd covariant derivative of guv?

[jmc]

Hi, I recently joined www.scienceforums.net and I have a question that may sound stupid:

 

The first covariant derivative of the metric tensor is equal to zero. Fine. But does anyone know if the covariant derivative of THAT is equal to zero? Part of me says yes because the first covariant derivative is zero but part of me says maybe not because it's an entirely different operator acting on the metric tensor. I'm not an expert on tensors, just learning them for enjoyment, so any help will be greatly appreciated!

[ajb]

The first covariant derivative of the metric tensor is equal to zero. Fine.

 

This is not true for a given connection in general. If a connection satisfies

 

[math]\nabla_{\mu}g_{\nu \rho} = 0[/math],

 

then it is said to be metric compilable. The Levi-Civita connection is metric compatible.

 

The fundamental theorem of Riemannian geometry states that the Levi-Civita connection is the unique torsion-free metric compatible connection on a Riemannian manifold. (This generalises to pseudo-Riemannian manifolds)

 

 

But does anyone know if the covariant derivative of THAT is equal to zero? Part of me says yes because the first covariant derivative is zero but part of me says maybe not because it's an entirely different operator acting on the metric tensor. I'm not an expert on tensors, just learning them for enjoyment, so any help will be greatly appreciated!

 

So, if the connection is metric compatible, then at all points the covariant derivative annihilates the metric. Recall that everything is tensorial. That is, if the components are zero at a point in one set of coordinates they are zero in all admissible coordinates.