Deriving Riemann tensor with four lower indices

Question

In terms of the connection coefficients

$$R_{abc}^d = \partial_a \Gamma_{bc}^d-\partial_b \Gamma_{ac}^d - \Gamma_{be}^d\Gamma_{ac}^e+\Gamma_{ae}^d\Gamma_{bc}^e$$

Pick an event $A$ and choose coordinates such that $\partial_a g_{bc} = 0$ at $A$... So at the event $A$, but not elsehwere in general,

$$R_{abcd} = g_{de}\partial_a(\Gamma_{bc}^e) - g_{de}\partial_b(\Gamma_{ac}^e)$$

Where has this last equation (RHS) come from?

Answer 1

Because of the hypothesis at event A (which is called Riemann normal coordinates at the point in question), all the Christoffel symbols $\Gamma^a_{bc}$ (but not their derivatives) vanish at A. The presence of the $g_{de}$ is to lower the index and get $R_{abcd}$ from $R^d_{abc}$.

Answer 2

Multiplying $R_{abc}^e=(\partial_a\Gamma_{bc}^e+\Gamma_{af}^e\Gamma_{bc}^f)-a\leftrightarrow b$ by $g_{de}$ with contraction over $e$ gives $R_{abcd}=(g_{de}\partial_a\Gamma_{bc}^e+g_{de}\Gamma_{af}^e\Gamma_{bc}^f)-a\leftrightarrow b$, so you're asking why $\Gamma_{af}^e\Gamma_{bc}^f$ is $a\leftrightarrow b$-symmetric at $A$. As Ted Shifrin notes, this expression is locally $0$ in the Riemann normal coordinates at $A$.