Metric compatibility, Ricci rotation coefficients & non-coordinate bases

Question

I'm currently working through chapter 7 on Riemannian geometry in Nakahara's book "Geometry, topology & physics" and I'm having a bit of trouble reproducing his calculation for the metric compatibility in a non-coordinate basis, using the Ricci rotation coefficients $\Gamma_{\alpha\beta\gamma}\equiv\delta_{\alpha\delta}\Gamma^{\delta}_{\;\beta\gamma}$ that he defines in section 7.8.4 ("Levi-Civita connection in a non-coordinate basis"). Here's his calculation: $$\Gamma_{\alpha\beta\gamma}=\delta_{\alpha\delta}e^{\delta}_{\;\lambda}e_{\beta}^{\;\mu}\nabla_{\mu}e_{\gamma}^{\;\lambda}\\ \qquad\quad=-\delta_{\alpha\delta}e_{\gamma}^{\;\lambda}e_{\beta}^{\;\mu}\nabla_{\mu}e^{\delta}_{\;\lambda}\\ \qquad\quad =-\delta_{\gamma\delta}e^{\delta}_{\;\lambda}e_{\beta}^{\;\mu}\nabla_{\mu}e_{\alpha}^{\;\lambda}\\ =-\Gamma_{\gamma\beta\alpha}\quad$$ where $\Gamma^{\delta}_{\;\beta\gamma}=e^{\delta}_{\;\lambda}e_{\beta}^{\;\mu}\nabla_{\mu}e_{\gamma}^{\;\lambda}$ has been used. He states that this is found using that $\nabla_{\mu}g=0$, but I can't seem to reproduce the result. I assume that between lines 1 and 2 he simply uses that $$e_{\gamma}^{\;\lambda}e_{\beta}^{\;\mu}\nabla_{\mu}e^{\delta}_{\;\lambda}=-\delta_{\gamma\delta}e^{\delta}_{\;\lambda}e_{\beta}^{\;\mu}\nabla_{\mu}e_{\alpha}^{\;\lambda}+\delta_{\gamma\delta}e_{\beta}^{\;\mu}\nabla_{\mu}\left(e_{\alpha}^{\;\lambda}e^{\delta}_{\;\lambda}\right)=-\delta_{\gamma\delta}e^{\delta}_{\;\lambda}e_{\beta}^{\;\mu}\nabla_{\mu}e_{\alpha}^{\;\lambda}$$ since $e_{\alpha}^{\;\lambda}e^{\delta}_{\;\lambda}=\delta_{\alpha\delta}=diag\lbrace 1,1,1,1\rbrace$ and so $\nabla_{\mu}\left(e_{\alpha}^{\;\lambda}e^{\delta}_{\;\lambda}\right)=0$. I don't see how he gets from the second to the third line using $\nabla_{\mu}g=0\Rightarrow \partial_{\mu}g_{\nu\lambda}-g_{\kappa\lambda}\Gamma^{\kappa}_{\;\mu\nu}-g_{\nu\kappa}\Gamma^{\kappa}_{\;\mu\lambda}=0$ though, as when I naively use this result I end up with additional terms and no overall minus sign. Any help would be much appreciated.

Answer 1

I am a bit confused by your notation but the property you are attempting to assert is called the metrillinic property. It is that the metric tensor effectively acts like a constant in covariant differentation. Recall the definitions for the 'christoffel symbols':

$$\Gamma^i_{\ \ jk} = g^{im}.\Gamma_{mjk} = g^{im}.\frac{1}{2}[\partial_k{g_{mj}} + \partial_j{g_{mk}} -\partial_m{g_{jk}}]$$

This relation can be derived upon introduction of the christoffel symbols. Substituting the above relation in your expression for the covariant derivative of the metric you find everything vanishes and hence $\nabla_i{g_{jk}} = 0$

Answer 2

Ok, so I think I may have figured it out (I thought I'd attempt an answer as a comment would be far too long!)

$$\Gamma_{\alpha\beta\gamma}=\delta_{\alpha\delta}e^{\delta}_{\;\lambda}e_{\beta}^{\;\;\mu}\nabla_{\mu}e_{\gamma}^{\;\;\lambda}=-\delta_{\alpha\delta}e_{\gamma}^{\;\;\lambda}e_{\beta}^{\;\;\mu}\nabla_{\mu}e^{\delta}_{\;\lambda}+\delta_{\alpha\delta}e_{\beta}^{\;\;\mu}\nabla_{\mu}\left(e^{\delta}_{\;\lambda}e_{\gamma}^{\;\;\lambda}\right)=-\delta_{\alpha\delta}e_{\gamma}^{\;\;\lambda}e_{\beta}^{\;\;\mu}\nabla_{\mu}e^{\delta}_{\;\lambda}$$ since $\nabla_{\mu}\left(e^{\delta}_{\;\lambda}e_{\gamma}^{\;\;\lambda}\right)=\nabla_{\mu}\left(\delta^{\delta}_{\;\gamma}\right)=0$.

Next, we use that $\nabla_{\mu}g_{\nu\lambda}=0=\nabla_{\mu}\delta^{\alpha\beta}$ to write the following: $$-\delta_{\alpha\delta}e_{\gamma}^{\;\;\lambda}e_{\beta}^{\;\;\mu}\nabla_{\mu}e^{\delta}_{\;\lambda}=-\delta_{\alpha\delta}e_{\gamma}^{\;\;\lambda}e_{\beta}^{\;\;\mu}\nabla_{\mu}\left(g_{\lambda\kappa}\delta^{\delta\sigma}e_{\sigma}^{\;\;\kappa}\right)\\ \qquad\qquad\qquad\;\;\,=-\delta_{\alpha\delta}g_{\lambda\kappa}\delta^{\delta\sigma}e_{\gamma}^{\;\;\lambda}e_{\beta}^{\;\;\mu}\nabla_{\mu}e_{\sigma}^{\;\;\kappa}\\ \qquad\qquad\qquad\qquad\;\;\;\;=-\delta_{\alpha}^{\;\;\sigma}e^{\delta}_{\;\lambda}e^{\xi}_{\;\kappa}\delta_{\delta\xi}e_{\gamma}^{\;\;\lambda}e_{\beta}^{\;\;\mu}\delta_{\delta\xi}\nabla_{\mu}e_{\sigma}^{\;\;\kappa}\\ \qquad\qquad\qquad\quad\;=-\delta^{\delta}_{\;\gamma}\delta_{\delta\xi}e^{\xi}_{\;\kappa}e_{\beta}^{\;\;\mu}\nabla_{\mu}\left(\delta_{\alpha}^{\;\;\sigma}e_{\sigma}^{\;\;\kappa}\right)\\ \qquad\quad\;\;\;=-\delta_{\gamma\xi}e^{\xi}_{\;\kappa}e_{\beta}^{\;\;\mu}\nabla_{\mu}e_{\alpha}^{\;\;\kappa}$$ where we have used that $e^{\delta}_{\;\lambda}=g_{\lambda\kappa}\delta^{\delta\sigma}e_{\sigma}^{\;\;\kappa}$, $\delta_{\alpha\delta}\delta^{\delta\sigma}=\delta_{\alpha}^{\;\;\sigma}$, and that $e^{\delta}_{\;\lambda}e_{\gamma}^{\;\;\lambda}=\delta^{\delta}_{\;\gamma}$.

Upon relabelling indices, we see that $$-\delta_{\alpha\delta}e_{\gamma}^{\;\;\lambda}e_{\beta}^{\;\;\mu}\nabla_{\mu}e^{\delta}_{\;\lambda}=-\delta_{\gamma\delta}e^{\delta}_{\;\lambda}e_{\beta}^{\;\;\mu}\nabla_{\mu}e_{\alpha}^{\;\;\lambda}=-\Gamma_{\gamma\beta\alpha}$$ Hence arriving at the desired result: $$\Gamma_{\alpha\beta\gamma}=-\Gamma_{\gamma\beta\alpha}$$ I think this is the correct way to do it, but I may be wrong.