# Contracted Christoffel symbol

On page 3 of this document:

it shows how to calculate the contracted Christoffel symbol by

\begin{align*} \Gamma^{\mu}_{\mu \lambda} &= \frac{1}{2}g^{\mu \rho}(\partial_{\mu}g_{\rho\lambda} + \partial_{\lambda}g_{\mu\rho} - \partial_{\rho}g_{\mu\lambda} ) \\ &= \frac{1}{2}(\partial^{\rho}g_{\rho\lambda} + g^{\mu \rho}\partial_{\lambda}g_{\mu\rho} - \partial^{\mu}g_{\mu\lambda} ) \\ &= \frac{1}{2}g^{\mu \rho}\partial_{\lambda}g_{\mu\rho} \end{align*}

Can someone kindly please help and explain what's happening in step $2$ and $3$? I just don't get it.

• Nothing much is happening there. From line one to two $g^{\mu\rho}$ is pulled into the bracket and indices are raised. From step two to three the summation index in the first term is renamed to $\mu$ (or equivalently in the third to $\rho$). Then the first and last terms cancel. If that is not clear to you I would suggest reintroducing the suppressed sums explicitly. – N0va Sep 21 '17 at 11:17

The tensor product in step 1 is expanded and the 3 terms in step 1 are transformed into 3 terms in step 2 according to these rules: $$g^{\mu \rho}\partial_{\mu}g_{\rho\lambda} \to \partial^{\rho}g_{\rho\lambda}$$ and $$g^{\mu \rho}\partial_{\rho}g_{\mu\lambda} \to \partial^{\mu}g_{\mu\lambda}$$