Ok, simple question about index notation. If I have this:-
$$
\delta^\mu_\eta (\partial_\mu g_{\eta\nu})
$$
Where $\delta^\mu_\eta$ is the Kronecker delta, does this become:-
$$
\partial_\eta g_{\mu\nu}
$$
Or can I just switch one of the indicies and get this:-
$$
\partial_\mu g_{\mu\nu}
$$
I'm guessing they both can't be correct because in the first all indexes are free and the second $\mu$ is a dummy index.
Best Answer
Even though the Kronecker delta is sometimes treated like an index substitution symbol, it actually defined as $$ \delta_j^k = \begin{cases} 1, & j=k\\ 0, & j\neq k \end{cases} $$ Now consider what happens when summing over one of those indices $$ \sum_\mu \delta^\mu_\eta (\partial_\mu g_{\eta\nu})=\partial_\eta g_{\eta\nu} $$ since this is the only non-zero summand. If we sum over the other index, then $$ \sum_\eta\sum_\mu \delta^\mu_\eta (\partial_\mu g_{\eta\nu})=\sum_\eta \partial_\eta g_{\eta\nu}=\partial_1 g_{1\nu}+\partial_2 g_{2\nu}+\dotsb $$ which is the same result you would get for summing over $\eta$ and then over $\mu$. Observe that in both cases this is different from $\partial_\eta g_{\mu\nu}$.
Finally, as I alluded to in my comment and as Branimir Ćaćić made explicit, in Einstein notation you sum only over those indices that appear both as an upper and a lower index. The other indices are considered constant throughout the summation.