# Index Concordia

## Some (trivial) consequences of metric-compatibility

Posted in Geometry, Relativity by Index Guy on July 13, 2007

If you were a bit nervous that I used two identities in the previous post without proof, well here I am to make you happy!

Let us start with a connection that is metric compatible. This means that

$\nabla_{\alpha} g_{\mu \nu} = 0.$

We now show that if this is true, then we also have

$\nabla_{\alpha}g^{\mu \nu} = 0$ and $\nabla_{\alpha} \epsilon_{\mu \nu \sigma \rho} = 0.$

We start with the first one. By definition, the inverse metric satisfies

$g^{\alpha \gamma} g_{\gamma \beta} = \delta^{\alpha}_{\beta}.$

If we operate with the covariant derivative on this equation, on the right-hand side we obtain zero, since the Kronecker delta is the same in every coordinate system and to top it all it is just a bunch of constants. Using the product rule we get

$g^{\mu \nu}\nabla_{\alpha} g_{\nu \sigma} + g_{\nu \sigma}\nabla_{\alpha} g^{\mu \nu} = 0.$

Since the connection is metric compatible, we have the first term vanishing. Unless the metric is trivially zero, we have

$\nabla_{\alpha} g^{\mu \nu} = 0.$

So one is in the bag. Now we consider the covariant derivative of the Levi-Civita tensor. By definition, this tensor can be written in terms of the Levi-Civita symbol (which is the same in every coordinate system)

$\displaystyle\epsilon_{\mu \nu \sigma \rho} = \sqrt{\left|g\right|} \tilde{\epsilon}_{\mu \nu \sigma \rho}.$

If we apply the covariant derivative to this equation, it will only operate on the determinant. By the chain rule we will get a term that involves the derivative of the determinant of the metric. This determinant will only depend on the metric entries, which means that the covariant derivative will vanish. Voilà!