Is this derivation correct? $$ R^{ab}_{;a}=0 $$ $$ g_{ac}g_{bd}R^{ab}_{;a}=0 $$ $$ (g_{ac}g_{bd}R^{ab})_{;a}=0 $$ $$ R_{cd;a}=0 $$
And does that mean I now have $n^3$ equation as opposed to $n$?
2026-04-01 16:28:56.1775060936
Riemannian geometry algebra
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No. In the second line, you've incorrectly used the summation convention. The upper $a$ in the first line is a dummy index, already contracted with a lower $a$, so you can't introduce another $a$ into the formula.