# Metric tensor second derivative in terms of riemann curvature tensor NOT in a local coordinate system

I know that the Riemann curvature tensor, in a general coordinate system is: $$R_{abcd}=\frac{1}{2}(g_{ad,bc}+g_{bc,ad}-g_{ac,bd}-g_{bd,ac})+(...)$$ The dots mean lower derivatives of $g$.
I am interested in getting the second derivative of the metric tensor in terms of the Riemann curvature tensor. I don't care about the terms in place of the dots or how they combine in the final result, or if they appear or not.I'd like something like this: $$g_{ab,cd}=R_{abcd}....+(...)$$ Is this is even possible? Thank you in advance.

• I don't think that it is possible, the way you wrote the question. According to the first formula, $R_{abcd}$ is essentially "part" of the $g_{ab,cd}$ obtain by some symmetrization procedure. So, it seems to me that this is like asking, can one recover a square matrix from just its symmetrization, say, and the answer is no, because you can always add an arbitrary skew-symmetric part to a matrix, without affecting the symmetrization. That being said, I did not give it much thought, which is why I left it as a comment. – Malkoun Jan 7 '16 at 18:17

Consider polar coordinates on flat $\mathbb{R}^2$, where the metric takes the form $\mathrm{d}s^2 = \mathrm{d}r^2 + r^2\mathrm{d}\phi^2$. The second derivative $g_{\phi\phi,rr} = 2$ is non-zero, but since this metric is flat, the Riemann tensor vanishes identically, i.e. the r.h.s of your sought-after formula is zero, so no formula of $g_{ab,cd}$ in terms of $R_{abcd}$ can exist in general.
A polynomial $P(R)$ in the Riemann tensor (and related tensors such as Ricci or Weyl) is itself a tensor. On the other hand, $g_{ab,cd}$ is not a tensor. But the equality $g_{ab,cd}=P(R)$ would mean $g_{ab,cd}$ is a tensor. Contradiction.