# Ricci rotation coefficients and non-coordinate bases

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1. Sep 21, 2015

### "Don't panic!"

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.

Last edited: Sep 21, 2015
2. Sep 21, 2015

### fzero

I think it's just raising and lowering indices across the covariant derivative:
$$A^\lambda \nabla_\mu B_\lambda = A^\lambda \nabla_\mu (g_{\lambda\sigma} B^\sigma) = A^\lambda g_{\lambda\sigma} \nabla_\mu B^\sigma.$$

3. Sep 21, 2015

### "Don't panic!"

I thought it was $\left(\nabla_{\mu}g\right)_{\nu\lambda}=0$, and not $\nabla_{\mu}g_{\nu\lambda}=0$ though?

4. Sep 21, 2015

### fzero

The Levi-Civita connection satisfies $\nabla_{\mu}g_{\nu\lambda}=0$. In index-free notation, it would be $\nabla_X g(Y,Z)=0$. I'm not sure what you mean by the first expression, if not this.

5. Sep 21, 2015

### "Don't panic!"

What I meant was that $(\nabla_{\mu}g)_{\nu\lambda}=\partial_{\mu} g_{\nu\lambda}-g_{\kappa\lambda}\Gamma^{\kappa}_{\;\mu\nu}-g_{\nu\kappa}\Gamma^{\kappa}_{\;\mu\lambda}=0$ are the components of the covariant derivative of the metric tensor, whereas $\nabla_{\mu}g_{\nu\lambda}=e_{\mu}[g_{\nu\lambda}]=\partial_{\mu}g_{\nu\lambda}$ is the covariant derivative acting on the components of the metric tensor.

6. Sep 21, 2015

### fzero

I don't have time to look at Nakahara at the moment, but $\nabla_{\mu}g_{\nu\lambda}=e_{\mu}[g_{\nu\lambda}]=\partial_{\mu}g_{\nu\lambda}$ doesn't make any sense to me. You can write an expression for the covariant derivative in terms of a spin connection, but that is not what you've done here. In any case, you can't simply lose the Christoffel symbols at the end of this expression, so I would suggest that you unlearn it.

It is true that if we have a function $f$, then the covariant derivative is the same as the ordinary derivative $\nabla_\mu f = \partial_\mu f$, or in index-free notation $\nabla_X f = X f$. Could you have been confused by such an expression, perhaps in which $g$ was the determinant of the metric tensor?

7. Sep 21, 2015

### "Don't panic!"

Maybe I've misunderstood thinks, but I thought that the expression for the covariant derivative of the metric is derived as follows $$\nabla_{\mu}g=\nabla_{\mu}\left(g_{\alpha\nu}dx^{\alpha}\otimes dx^{\nu}\right)=(\nabla_{\mu}g_{\alpha\nu})dx^{\alpha}\otimes dx^{\nu}+g_{\alpha\nu}(\nabla_{\mu}dx^{\alpha})\otimes dx^{\nu}+g_{\alpha\nu}dx^{\alpha}\otimes (\nabla_{\mu} dx^{\nu})\\=\partial_{\mu}g_{\alpha\nu}dx^{\alpha}\otimes dx^{\nu}-g_{\alpha\nu}\Gamma^{\nu}_{\;\mu\beta}dx^{\alpha}\otimes dx^{\beta}-g_{\alpha\nu}\Gamma^{\alpha}_{\;\mu\beta}dx^{\beta}\otimes dx^{\nu}\\=\left(\partial_{\mu}g_{\alpha\nu}-g_{\alpha\beta}\Gamma^{\beta}_{\;\mu\nu}-g_{\beta\nu}\Gamma^{\beta}_{\;\mu\alpha}\right)dx^{\alpha}\otimes dx^{\nu}\\=(\nabla_{\mu}g)_{\alpha\nu}dx^{\alpha}\otimes dx^{\nu}$$ Is this correct at all?

8. Sep 21, 2015

### fzero

Yes that calculation is correct (in going from the 1st line to the 2nd you've flipped the 2nd and 3rd terms around). I see where we are confusing each other. You are using the rigorous formulation where a tensor is always expressed as a mixed form
$$T = {T^{\mu_1\cdots \mu_p}}_{\nu_1\cdots\nu_q} \mathbf{e}_{\mu_1}\otimes \cdots \mathbf{e}_{\mu_p} \otimes \mathbf{e}^{\nu_1}\otimes \cdots \mathbf{e}^{\mu_q}.$$
Then when we take the the covariant derivative, we view the coefficients ${T^{\mu_1\cdots \mu_p}}_{\nu_1\cdots\nu_q}$ as functions and take ordinary derivatives, while the Christoffel symbols come from the covariant derivative of the basis vectors and forms.

Physicists then abuse the notation and take the coefficient of the covariant derivative of the tensor and call this the covariant derivative of the coefficient of the original tensor, i.e., for a vector field $V = V^a \mathbf{e}_a$, we would write
$$\nabla_b A^a = \partial_b V^a + {\gamma^a}_{bc} V^c.$$
In this spirit, we say that the covariant derivative of the metric is
$$\nabla_\mu g_{\alpha\nu} = \partial_{\mu}g_{\alpha\nu}-g_{\alpha\beta}\Gamma^{\beta}_{\;\mu\nu}-g_{\beta\nu}\Gamma^{\beta}_{\;\mu\alpha},$$
when we really mean the full calculation that you did above.

In this context, rather than use the abusive notation to get the result, let us consider
$$\nabla_\mu ( \delta_{\alpha\delta} {e^\delta}_\lambda - g_{\lambda\nu} {e_\alpha}^\nu ) \mathbf{e}^\alpha \otimes \mathbf{e}^\lambda .$$
The first term is what appears on the 2nd line of the equation in your OP and the second term is the expression in the 3rd line that we want to get. My expression above is that for the coefficients of the covariant derivative
$$\nabla_\mu ( \delta_{\alpha\delta} {e^\delta}_\lambda\mathbf{e}^\alpha \otimes \mathbf{e}^\lambda - g_{\lambda\nu} {e_\alpha}^\nu \mathbf{e}^\alpha \otimes \mathbf{e}^\lambda) = \nabla_\mu ( \delta_{\alpha\delta}\mathbf{e}^\alpha \otimes \mathbf{e}^\delta - g_{\lambda\nu} \mathbf{e}^\nu \otimes \mathbf{e}^\lambda ).$$
This vanishes by metric compatibility so we understand that the manipulation of indices in the first expression was a valid identity.

9. Sep 22, 2015

### "Don't panic!"

Oh good, I'm glad I haven't misunderstood that (apologies about the term flipping, an accident on my part).

Thanks for your help, I think I understand it now.
Just out of curiosity, would the following be correct? (using the standard notation $\nabla_{\mu}g_{\nu\lambda}=0$ for the metric compatibility condition):

Starting from $$-\delta_{\alpha\delta}e_{\gamma}^{\;\;\lambda}e_{\beta}^{\;\;\mu}\nabla_{\mu}e^{\delta}_{\;\lambda}$$
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: $$\Gamma_{\alpha\beta\gamma}=-\Gamma_{\gamma\beta\alpha}$$

10. Sep 22, 2015

### fzero

Yes, that was the type of calculation I was originally suggesting.

11. Sep 23, 2015

### lautaaf

There's another (somewhat shorter) way, in which you always work in the non-coordinate frames.
Noting that $$\nabla_{\hat e_{a}} \hat \theta^{b} = - \Gamma^b{}_{ac} \hat \theta^{c}$$ and replacing this relation into the metric compatibility
$$\nabla_{X}g=0 \Rightarrow X^{a} [\nabla_{\hat e_{a}} (\delta_{bc} \hat \theta^{b} \otimes \hat \theta^{c})]=0 \Rightarrow \nabla_{\hat e_{a}} (\delta_{bc} \hat \theta^{b} \otimes \hat \theta^{c})=0$$ the result follows immediately.
($a$,$b$,$c$,.... super(sub)-scripts stand for quantities in the non-coordinate frames $\{\hat \theta^{a}\}$,$\{\hat e_{a}\}$)