I am reading Landau & Lifshitz's The Classical Theory of Fields. There is a derivation about metric tensor and Christoffel symbol I cannot get. On page 261, section 86, first the authors introduce the relation between the differential of the metric tensor and the differential of its determinant: $$dg=gg^{ik}dg_{ik}=-gg_{ik}dg^{ik}.\tag{86.4}$$ Later, they compute a term:$$g^{kl}\Gamma^{i}_{kl}=\frac{1}{2}g^{kl}g^{im}(\frac{\partial g_{mk}}{\partial x^l}+\frac{\partial g_{lm}}{\partial x^k}-\frac{\partial g_{kl}}{\partial x^m})=g^{kl}g^{im}(\frac{\partial g_{mk}}{\partial x^l}-\frac{1}{2}\frac{\partial g_{kl}}{\partial x^m})$$and say, with the help of $(86.4)$, this can be transformed to $$g^{kl}\Gamma^{i}_{kl}=-\frac{1}{\sqrt{-g}}\frac{\partial (\sqrt{-g}g^{ik})}{\partial x^k}.\tag{86.6}$$ I cannot get the result $(86.6)$. Can someone help?
2 Answers
The piece you may be missing is that $$ 0 = \frac{\partial \delta^i {}_k}{\partial x^l} = \frac{\partial (g^{im} g_{mk})}{\partial x^l} = g^{im} \frac{\partial g_{mk}}{\partial x^l} + g_{mk}\frac{\partial g^{im} }{\partial x^l} $$ where I have applied the product rule in the last step. So we have $$ g^{kl}g^{im}(\frac{\partial g_{mk}}{\partial x^l}-\frac{1}{2}\frac{\partial g_{kl}}{\partial x^m}) = -g^{kl} g_{mk} \frac{\partial g^{im}}{\partial x^l} - \frac{1}{2} g^{im} g^{kl} \frac{\partial g_{kl}}{\partial x^m}. $$ From there, it is mainly a matter of applying the product rule to (86.6) and using the identity in (86.4).
Another trick that is often very useful to use in proofs about the determinant of the metric is to leverage the levi-civita symbol:
$$g = \frac{1}{d!}\epsilon^{abc...}\epsilon^{ijk...}(g_{ai}g_{bj}g_{ck}...)$$
Then, taking a derivative is trivial:
$$\partial_{\alpha}g = \partial_{\alpha}g_{ai}\left[\frac{1}{(d-1)!}\epsilon^{abc...}\epsilon^{ijk...}(g_{bj}g_{ck}...)\right]$$
Where the factor of $d$ comes about by having the derivative act on each term, and permuting the two episilon indices in pairs to the front (I keep the ai indices, but they are dummy indices). By inspection [hint: multiply that factor by $g_{ai}$], it should be clear that the term in brackets is equal to $ g^{ai}g$, and you have: $\partial_{c}g = g^{ab}\partial_{c}g_{ab}$
With this ansatz, proving the result is much easier.
