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

Question

Given the following equation

\begin{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) \end{equation}

Suppose I have the following metric

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

$\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 $g_{\phi\phi}$?

Answer 1

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.