Index Concordia

An identity concerning the Christoffel connection

Posted in Geometry, Relativity by Index Guy on July 11, 2007

One of my summer readings is Sean Carroll’s book on general relativity. It is good to teach the basics of GR; I hope it will serve its purpose when I start the fall with Warren. Anyway, I came upon the following identity regarding the Christoffel connection,

$\Gamma_{\mu \nu}^{\mu} = \displaystyle \tilde{g}^{-1} \partial_{\nu}\tilde{g}$,

where $\tilde{g} = \sqrt{\left|g\right|}$ is the square root of the absolute value of the metric tensor $g_{\mu \nu}$. It is one of those expressions that is preceded by “It is straightforward to show that…”. Since I was a bit doubtful, I decided to try my hand at it.

Recall that the Christoffel connection is metric-compatible and torsion-free. The first statement means that the covariant derivative $\nabla_{\sigma}$ of the metric tensor vanishes. This in turn implies that for the Levi-Civita tensor we have

$\nabla_{\sigma} \epsilon_{\lambda \mu \nu \rho} = 0.$

Recall also that the Levi-Civita tensor $\epsilon$ can be written in terms of the Levi-Civita symbol $\tilde{\epsilon}$, which is a fancy name for the completely antisymmetric symbol. Anyway, we have by definition,

$\epsilon_{I} = \tilde{g} \tilde{\epsilon}_{I}$ and $\epsilon^{I} = \tilde{g}^{-1} \tilde{\epsilon}^{I},$

with $I =\left\{\lambda, \mu, \nu, \rho\right\}$. Recall yet one more thing: the Levi-Civita symbol is the same in every coordinate system, either 1, -1, or 0.

Let us evaluate the covariant derivative of the $\epsilon$ tensor,

$\nabla_{\sigma} \epsilon_{\lambda \mu \nu \rho} = \partial_{\sigma}\epsilon_{\lambda \mu \nu \rho} - \Gamma_{\sigma \lambda}^{\alpha}\epsilon_{\alpha \mu \nu \rho} -\Gamma_{\sigma \mu}^{\alpha}\epsilon_{\lambda \alpha \nu \rho} - \Gamma_{\sigma \nu}^{\alpha}\epsilon_{\lambda \mu \alpha \rho} - \Gamma_{\sigma \rho}^{\alpha}\epsilon_{\lambda \mu \nu \alpha}.$

This is so since for every lower index in the epsilon tensor we need to add a minus Christoffel connection term. Now we know that this covariant derivative vanishes, so we can write

$\partial_{\sigma}\epsilon_{\lambda \mu \nu \rho} = \Gamma_{\sigma \lambda}^{\alpha}\epsilon_{\alpha \mu \nu \rho} +\Gamma_{\sigma \mu}^{\alpha}\epsilon_{\lambda \alpha \nu \rho} + \Gamma_{\sigma \nu}^{\alpha}\epsilon_{\lambda \mu \alpha \rho} + \Gamma_{\sigma \rho}^{\alpha}\epsilon_{\lambda \mu \nu \alpha}.$

The next step is to take a break and rest your hands after writing so many Christoffel connections. Then we use the definition of the Levi-Civita tensor. Since the Levi-Civita symbol is just a bunch of constants, the partial derivative on the left hand side will act only on the $\tilde{g}$ term. If we restore again the tensor, we get

$\epsilon_{\lambda \mu \nu \rho}\tilde{g}^{-1}\partial_{\sigma}\tilde{g} = \Gamma_{\sigma \lambda}^{\alpha}\epsilon_{\alpha \mu \nu \rho} +\Gamma_{\sigma \mu}^{\alpha}\epsilon_{\lambda \alpha \nu \rho} + \Gamma_{\sigma \nu}^{\alpha}\epsilon_{\lambda \mu \alpha \rho} + \Gamma_{\sigma \rho}^{\alpha}\epsilon_{\lambda \mu \nu \alpha}.$

Now we will use another identity of the Levi-Civita tensor, the one that allows us to represent a contraction of a L-C tensor with another L-C tensor in terms of Kronecker deltas,

$\epsilon^{I,M}\epsilon_{I,N} = (-1)^{s}p!(n-p)!\delta^{[M]}_{N}.$

In this expression, the tensors have $n$ components and we contract $p$ of them. The number of negative eigenvalues of the metric tensor is denoted by $s$. Again, $I, M, N$ all represent a string of indices. We want to contract three indices of the left-hand side of our equation with a L-C tensor. We will not write the $(-1)^{s}$ term since it will cancel on both sides. The left-hand side then looks like,

$\epsilon^{\lambda \mu \nu \gamma}\epsilon_{\lambda \mu \nu \rho}\tilde{g}^{-1}\partial_{\sigma}\tilde{g} =12\tilde{g}^{-1}\partial_{\sigma}\tilde{g} \delta^{\gamma}_{\rho}.$

Wait! That 12 looks weird there. Let’s just hope that it will be go away later. Meanwhile, on the right-hand side we had three separate contractions. Two of them will be two-index contractions, and the last one will be a three-index contraction. Let us consider the two-index contractions first. The first thing we should do is re-arrange the indices so to have the contracted indices to the left and the free indices to the right. Lets work with the first term, which both tensors will require two permutations each. There will be no overall sign change. We have,

$\epsilon^{\lambda \mu \nu \gamma}\epsilon_{\alpha \mu \nu \rho} = \epsilon^{\mu \nu \lambda \gamma}\epsilon_{\mu \nu \alpha \rho} = 4 \left(\delta^{\lambda}_{\alpha}\delta^{\gamma}_{\rho} - \delta^{\gamma}_{\alpha}\delta^{\lambda}_{\rho}\right).$

When we combine this with the Christoffel connection, we get something like

$4\Gamma_{\sigma \lambda}^{\lambda}\delta^{\gamma}_{\rho} - 4\Gamma_{\sigma \rho}^{\gamma}.$

The other contraction is exactly the same. The three-index contraction yields a term

$\epsilon^{\lambda \mu \nu \gamma}\epsilon_{\lambda \mu \nu \alpha}\Gamma_{\sigma \rho}^{\alpha} = 12\Gamma_{\sigma \rho}^{\gamma}.$

Finally, we add this three contractions and equate it to the first term we calculated. The result is

$\left[12\tilde{g}^{-1}\partial_{\sigma}\tilde{g}\right]\delta^{\gamma}_{\rho} = 8\Gamma_{\sigma \lambda}^{\lambda}\delta^{\gamma}_{\rho} + 4\Gamma_{\sigma \rho}^{\gamma}.$

Oh no! What now? Well, because of the deltas, this equation will vanish except when $\rho = \gamma$. This condition gives

$12\tilde{g}^{-1}\partial_{\sigma}\tilde{g} = 8\Gamma_{\sigma \lambda}^{\lambda} + 4\Gamma_{\sigma \rho}^{\rho},$

which gives

$\tilde{g}^{-1} \partial_{\sigma}\tilde{g} = \Gamma_{\sigma \mu}^{\mu}.$

Finally, remember that this connection is torsion-free. This means that the connection is symmetric on the two lower indices. Then we can write

$\tilde{g}^{-1} \partial_{\sigma}\tilde{g} = \Gamma_{\mu \sigma }^{\mu},$

which was what we set out to show. So Mr. Carroll was right this time…

3 Responses

1. Psi-String said, on October 26, 2008 at 5:18 am

Hi~ I’m a undergraduate student major in physics, and currently
studying Carroll’s book.

Could you show me how to derive $\nabla_{\sigma} \epsilon_{\lambda \mu \nu \rho} = 0$ ?

Mr. Carroll said it is easy, but I can’t figure it out @@

Thanks!!

2. Psi-String said, on October 26, 2008 at 6:25 am

oops I notice now that you have already derive it in the other topic.

3. Kevin Grosvenor said, on July 29, 2009 at 6:08 pm

I think it’s a heck of a lot easier than that:

$$\Gamma_{\mu\nu}^{\mu} = \tfrac{1}{2} g^{\mu\sigma} ( \partial_\nu g_{\mu\sigma} + \partial_\mu g_{\nu\sigma} – \partial_\sigma g_{\mu\nu} )$$

The last two terms are antisymmetric under the exchange $\mu \leftrightarrow \sigma$ and are multiplied by $g^{\mu\sigma}$ which is symmetric under this exchange. Therefore, these terms vanish and we only have the first term:

$$\Gamma_{\mu\nu}^{\mu} = \tfrac{1}{2} g^{\mu\sigma} \partial_\nu g_{\mu\sigma} = \tr ( g^{-1} \partial_\nu g)$$

Note that this isn’t quite the desired result yet because it is not in terms of the determinants. However, the desired result comes from the well-known result

$$\delta (\det M ) = (\det M) \tr ( M^{-1} \delta M )$$

For any $N \times N$ matrix $M$. This follows directly from the identity $\det M = e^{\tr \log M}$, which is pretty darn straightforward to prove (use the fact that diagonalizing $M$ keeps the determinant and log fixed). Finally use $\det (M^{-1} ) = ( \det M )^{-1}$, which is derived by taking the determinant of $I = MM^{-1}$ and the factorizability $\det (AB) = (\det A ) ( \det B )$. This gives

$$( \det M^{-1} ) \delta (\det M ) = \tr ( M^{-1} \delta M )$$

Replace $M$ by $g$ and $\delta$ by $\partial_\nu$ and you’ve got your desired relation!