Calculating divergence using covariant derivative

[v2536]

Homework Statement

Using the definition of divergence $d(i_{X}dV) = (div X)dV$ where $X:M\rightarrow TM$ is a vector field, $dV$ is a volume element and $i_X$ is a contraction operator e.g. $i_{X}T = X^{k}T^{i_{1}...i_{r}}_{kj_{2}...j_{s}}$, prove that if we use Levi-Civita connection then the divergence can also be written as
$div X = X^{i}_{;i}$

2. The attempt at a solution

This is what i tried:
since $dV = dx^{1} \wedge ... \wedge dx^{n}$
after some calculation i conclude that $i_{X}dV = \sum_{i=1}^{n}(-1)^{i}X_{i}dx^{1} \wedge ... \wedge dx^{i-1} \wedge dx^{i+1} \wedge ... \wedge dx^{n}$
so $d(i_{X} dV) = (\partial _{i}X^{i})dV$
Then i attempt the use the fact that $\Gamma^{i}_{jk} = \Gamma^{i}_{kj}$ to get a lot of cancellation and show that $\partial _{i}X^{i} = X^{i}_{;i}$
but i couldn't.
So can anyone please help? Thx in advanced :)

 

[chiro]

Hey v2536 and welcome to the forums.

I don't know much about differential geometry, but I do understand in essence what you are trying to do.

Maybe you should look at this website (http://en.wikipedia.org/wiki/Divergence#Generalizations). A lot of the results you need are found in the coordinate free form you are trying to prove.

 

[Ben Niehoff]

Your first step is wrong. You should have

$$dV = \sqrt{g} \, dx^1 \wedge \ldots \wedge dx^n$$
where $\sqrt{g}$ is the square root of the determinant of the Riemannian metric.

 

[v2536]

Thank you chiro and Ben for your reply :)
I see that i forgot about $\sqrt{g}$ in front but can't i always find an orthonormal basis and work in that?
and by doing that g should become 1 right?

 

[v2536]

ok i think i understand now, if i use orthonormal basis then the connection may no be Levis-Civita.

 

[Ben Niehoff]

If you use an orthonormal basis, then your basis 1-forms are not necessarily closed (so it would be incorrect to write them as dx).

 

[v2536]

yes. that make sense.
Now i got the solution, thanks for your help.