I have found this identity in a paper that I am reading:
On a Riemannian $n$-manifold $(M,g)$, consider a normal coordinate system $\{x^i\}_{i=1,\dots,n}$centered at $p\in M$. Then $$g_{a b, c d}(p) + g_{a d,b c}(p) + g_{a c, d b}(p) = 0$$ Where $g_{a b, c d} = \frac{\partial^2 g(\frac \partial {\partial x^a }, \frac \partial {\partial x^b })} {\partial {x^d}\partial {x^c}}. $ How can I prove this statement?
I recall that in a normal coordinate system centered at $p$ the following identities hold (notice that are true only at $p$): $$ g_{i j}(p) = \delta_{i j} \text{ (Kronecker delta)}$$ $$ \partial_{x^k} g_{i j}(p) = 0 $$ $$ \Gamma_{i j}^k (p) = 0$$ $$\partial_{x^k}\Gamma_{i j}^l(p) = \frac 1 2 (g_{i l,j k}(p) + g_{j l,i k}(p)- g_{i j, l k}(p))$$ Where $\Gamma_{i,j}^k$ are the Christoffel symbols of the Levi-Civita connection induced by $g$.
So we have that (at the point $p$) $$g_{a b,c d} + g_{a d, b c} + g_{a c,d b} = 2 \partial_{x^d} \Gamma_{b c}^a + g_{b c, a d} + g_{a c,d b} = 2 \partial_{x^d} \Gamma_{b c}^a+ 2 \partial_{x^d} \Gamma_{b a}^c + g_{a b, c d} $$
The fourth equation seems to be promising but still I don't see how to conclude.