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

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!

]]>At this point I have not outlined his lectures properly. ðŸ˜‰ I should start doing that soon since the topics he discussed were among the most non-trivial.

]]>My bad.

]]>studying Carroll’s book.

Could you show me how to derive ?

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

Thanks!!

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