# Can I simply find the Christoffel symbols by dividing by $g$?

Given the following equation

$$g_{\alpha\delta} \Gamma^{\delta}_{\beta\gamma} = \frac{1}{2} \left(\partial_\gamma g_{\alpha\beta} + \partial_\beta g_{\alpha\gamma} - \partial_\alpha g_{\beta\gamma} \right)$$

Suppose I have the following metric

$$g_{\alpha\beta} = \begin{bmatrix} a^2 & 0 \\ 0 & a^2\sin ^2 \theta \\ \end{bmatrix}$$

$\alpha,\beta \in \lbrace \theta, \phi \rbrace$

So suppose I want to find all the Christoffel symbols for this metric.

Work I Did:

\begin{align*} g_{\theta\theta} \Gamma^\theta_{\theta\theta} & = 0\\ g_{\theta\theta} \Gamma^\theta_{\theta\phi} & = 0\\ g_{\theta\theta} \Gamma^\theta_{\phi\phi} & = -\sin\theta\cos\theta\\ g_{\phi\phi} \Gamma^\phi_{\theta\theta} & = 0\\ g_{\phi\phi} \Gamma^\phi_{\theta\phi} & = \cot \theta \\ g_{\phi\phi} \Gamma^\phi_{\phi\phi} & = 0\\ \end{align*}

But here's the problem, suppose I just want to find $\Gamma^\phi_{\theta\phi}$ by itself, can I just divide $\cot \theta$ by $g_{\phi\phi}$?

• Out of curiosity, why did you think you couldn't do the division? If you have a plausible reason for thinking so, and you had included that in the question, I think it would have been a great question. (Don't get me wrong; as is, it's perfectly fine.) Jan 21, 2015 at 5:54
• Well, I am learning GR and Riemannian geometry. And I am always careful about performing simple operations on mathematical objects because it's easy to get the symbols mixed up. If I'm not careful, I might divide by accident without realizing a summation is implicit and screw up the whole calculation. So I like to check with more knowledgeable people to make sure I am on the right track before spending tons of time on nonsensical calculations. I expect as I get the hang of it, I won't have as many questions over small things. But in the beginning they can be hard. Jan 21, 2015 at 5:59
• I find the summation convention is bittersweet. I cant stand it because i always forget it. But when I actually try to write out the sums, i see why it is needed. The sums become so overfilled with symbols including the sigma would just make it worse. Jan 21, 2015 at 6:01

Yes, you can do the division as planned. This is because the equation $$g_{\phi\phi} \Gamma^\phi_{\theta\phi} = \cot\theta$$ is just a scalar equation with two factors on the left and one on the right. None of the indices are being summed over by this point. The problem only occurs when there is an implicit summation of terms indicated with repeated dummy indices.
But: I think you should check your equations. In particular, the usual definition of Christoffel symbols is $$\Gamma^\alpha_{\beta\gamma} = \frac{1}{2} g^{\alpha\lambda} \left(\partial_\beta g_{\gamma\lambda} + \partial_\gamma g_{\beta\lambda} - \partial_\lambda g_{\beta\gamma}\right).$$ Contracting both sides with $g_{\alpha\delta}$ gives $$g_{\alpha\delta} \Gamma^\alpha_{\beta\gamma} = \frac{1}{2} \delta^\lambda_\delta \left(\partial_\beta g_{\gamma\lambda} + \partial_\gamma g_{\beta\lambda} - \partial_\lambda g_{\beta\gamma}\right) = \frac{1}{2} \left(\partial_\beta g_{\gamma\delta} + \partial_\gamma g_{\beta\delta} - \partial_\delta g_{\beta\gamma}\right).$$ This latter equation is what you have, and it often isn't useful, since the left-hand side is a sum of metric-component-weighted Christoffel symbols. In your lucky case one of the metric components is $0$: $$g_{\phi\delta} \Gamma^\delta_{\theta\phi} = g_{\phi\theta} \Gamma^\theta_{\theta\phi} + g_{\phi\phi} \Gamma^\phi_{\theta\phi} = a^2 \sin^2\!\theta\ \Gamma^\phi_{\theta\phi}.$$ This won't happen in general. Moreover, I suspect you used the right-hand side of the former definition with the left-hand side of the latter, so your scalar equations in which you seek to divide things aren't right.
• Since there are $3$ indices, there are $d^3$ Christoffel symbols in $d$ dimensions. However, the symbols are always symmetric in the lower two indices, so we usually only bother writing one of the pair, meaning you should check for $d^2(d+1)/2$ symbols, none of which should be obtained from the others by merely switching the lower indices. In your case of $d=2$, you do indeed list $6$ symbols, and they satisfy the no-reversed-indices condition, so you are good.