## Some (trivial) consequences of metric-compatibility

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

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

and

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

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

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

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)

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. VoilĂ !

notsuresaid, on February 1, 2010 at 1:25 amThe fact than the derivative doesn’t operate on the levi-civita symbol don’t seem obvious for me…

Hanciongsaid, on February 21, 2010 at 1:08 pmhai. could you show me the formula for the determinant of the metric? thanx