Christoffel symbol metric connection Suppose that you are given an arbitrary metric $g_{\mu\nu}$ such that you want to calculate all of the Christoffel symbol $\Gamma^{\lambda}_{\mu\nu}$. The equation for a Christoffel symbols are $$\Gamma^{\lambda}_{\mu\nu}=\frac{1}{2}g^{\lambda\sigma}\left(\partial_{\nu}g_{\sigma\mu}+\partial_{\mu}g_{\sigma\nu}-\partial_{\sigma}g_{\mu\nu}\right).$$ It is pretty simple to calculate all 40 of the unique Christoffel symbols, just plug in values of the metric. For instance, given an arbitary metric $$ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}$$ where $ds^2$ is arbitary. For arbitrary metrics, the Christoffel symbols $$\Gamma^{\lambda}_{\mu\nu}=\frac{1}{2}g^{\lambda\sigma}\left(\partial_{\nu}g_{\sigma\mu}+\partial_{\mu}g_{\sigma\nu}-\partial_{\sigma}g_{\mu\nu}\right)$$ dummy index $\sigma$ can be $$x^0,x^1,x^2,x^3.$$ How do you determine what value the dummy takes? For instance $\Gamma^{3}_{22}$; this would obviously be $$\Gamma^{3}_{22}=\frac{1}{2}g^{3\sigma}\left(\partial_2g_{\sigma 2}+\partial_2g_{\sigma 2}-\partial_{\sigma}g_{22}\right).$$ How would  I choose what coordinate $\sigma$ takes? Would it take all $x^0,x^1,x^2,x^3$ and sum those metric terms up such that it becomes $$\Gamma^{3}_{22}=\frac{1}{2}g^{3 \sigma}\partial_2 g_{\sigma2}+\frac{1}{2}g^{3 \sigma}\partial_2 g_{\sigma2}-\frac{1}{2}g^{3 \sigma}\partial_{\sigma}g_{22}=(\frac{1}{2}g^{3 0}\partial_2 g_{02}+\frac{1}{2}g^{3 1}\partial_2 g_{12}+\frac{1}{2}g^{3 2}\partial_2 g_{22}+\frac{1}{2}g^{3 3}\partial_2 g_{32})+(\frac{1}{2}g^{3 0}\partial_2 g_{02}+\frac{1}{2}g^{3 1}\partial_2 g_{12}+\frac{1}{2}g^{3 2}\partial_2 g_{22}+\frac{1}{2}g^{3 3}\partial_2 g_{32})-(\frac{1}{2}g^{3 0}\partial_{0}g_{22}+\frac{1}{2}g^{3 1}\partial_{1}g_{11}+\frac{1}{2}g^{3 2}\partial_{2}g_{22}+\frac{1}{2}g^{3 3}\partial_{3}g_{33})?$$ If that is the case is there any way to simplify such that the simplification holds for any combination of $x^0,x^1,x^2,x^3?$ Furthermore what is the simplification for diagonal metrics and the naming of the dummy index?
 A: You sum over all $\sigma$, because it's a repeated index. It's called contraction. For example,$$\begin{align}\Gamma_{22}^3&=\frac12g^{30}(\partial_2g_{02}+\partial_2g_{02}-\partial_0g_{22})\\&+\frac12g^{31}(\partial_2g_{12}+\partial_2g_{12}-\partial_1g_{22})\\&+\frac12g^{32}(\partial_2g_{22}+\partial_2g_{22}-\partial_2g_{22})\\&+\frac12g^{33}(\partial_2g_{32}+\partial_2g_{32}-\partial_3g_{22})\end{align}.$$Luckily, when you do these calculations you often find lots of terms vanish, usually due to symmetries. This happens to an especially large number of terms with a symmetry-respecting choice of coordinate system. For diagonal metrics, we unfortunately need to deactivate contraction in our notation to understand the simplification:$$\Gamma_{\mu\nu}^\lambda=\frac12g^{\lambda\lambda}(\color{red}{\partial_\mu g_{\lambda\nu}}+\color{blue}{\partial_\nu g_{\lambda\mu}}-\color{limegreen}{\partial_\lambda g_{\mu\nu}})$$does not sum over $\lambda$. The red term vanishes if $\lambda\ne\nu$; the blue term vanishes if $\lambda\ne\mu$; the green term vanishes if $\mu\ne\nu$. Therefore, cases where $\mu,\,\nu,\,\lambda$ are all different give $\Gamma_{\mu\nu}^\lambda=0$. Terms may also vanish due to a derivative's effect.
