How Is the Christoffel Symbol Related to the Metric Tensor's Determinant?

[bdforbes]

Homework Statement

Prove that
$$\Gamma^\mu_{\mu\lambda}=\frac{1}{\sqrt{-g}}\partial_\lambda(\sqrt{-g})$$

where g is the determinant of the metric, and $\Gamma$ are the Christoffel connection coefficients.

The Attempt at a Solution

From the general definition of the coefficients I got:
$$\Gamma^\mu_{\mu\lambda}=(1/2)g^{\mu\rho}\partial_\lambda g_{\rho\mu}$$

But I have no idea how to work with the determinant of the metric. I'm not sure if I'm allowed to use this:

det(g)=exp[Tr ln G]

And if I did, would I have to use the GR definition of the trace?

$$Tr R = R^\mu_\mu$$

I cleaned it up a little bit with the chain rule:

$$\frac{1}{\sqrt{-g}}\partial_\lambda(\sqrt{-g})$$
$$=\frac{1}{2g}\partial_\lambda(g)$$

 

[dextercioby]

The last part you wrote can't be right.

I'd start with

$$\delta |g| = \delta \[\exp \(\mbox{Tr} \ln |g|\)\]$$

and then see what the 'delta' of the RHS brings me.

 

[Hurkyl]

Isn't there an index-notation expression for the determinant?

 

[bdforbes]

Thanks for the tips. Turns out it's a bit simpler:

1. From Jacobi's formula we get:
$$dg=A^{ik}dg_{ik}$$
where $A^{ik}$ are elements of the adjugate of the metric.

2. Standard differential expansion:
$$dg=\frac{\partial g}{\partial g_{ik}}dg_{ik}$$
$$\Rightarrow A^{ik}=\frac{\partial g}{\partial g_{ik}}$$

3. From Laplace's formula for the determinant we get:
$$g^{ik}=(1/g)A^{ik}$$Putting this all together we have:
$$g^{ik}=\frac{1}{g}\frac{\partial g}{\partial g_{ik}}$$

Substitute this into my expression for $\Gamma^\mu_{\mu\lambda}$ above and the result follows.

I'm still working on the proofs to steps 1 and 3 above, but essentially the problem is solved. Feels good to brush up on my linear algebra, haven't used it for years.