# Metric compatibility, Ricci rotation coefficients & non-coordinate bases

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.

• @Dr.MV this seems like your specialty? It has tied a knot in my brain! Sep 22, 2015 at 9:06

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 $\nabla_i{g_{jk}} = 0$

• I appreciate your answer, but I'm actually asking how to show that $\Gamma_{\alpha\beta\gamma}=-\Gamma_{\gamma\beta\alpha}$ in the non-coordinate basis, deriving it from the vierbeins, and using the property that $(\nabla_{\mu}g)_{\nu\lambda}=0$. I have just paraphrased what's written in Nakahara's book. In particular I don't understand how he gets from line 2 to line 3 using that $(\nabla_{\mu}g)_{\nu\lambda}=0$?
– Will
Sep 21, 2015 at 15:24
• @Will, Ah I understand. I will look at the book to see what he did. Although the terms you use are new to me so I am trying not to say something stupid :P Sep 21, 2015 at 15:37
• No problem. I appreciate your help!
– Will
Sep 21, 2015 at 15:45

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.