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.
Best Answer
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 $$ g^{\mu \rho}\partial_{\rho}g_{\mu\lambda} \to \partial^{\mu}g_{\mu\lambda} $$
this is because you can contract the index of a derivative (only the left-most one, if there are 2 or more derivatives acting on a tensor), while the middle term (the second one) is left untouched.
The first and third term in step 2 are the same , once you see that they differ only in the dummy indices, so they cancel each other and in step 3 we are left only with the second term of step 2.