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

Question

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.

Answer 1

A formula for the second derivatives of the metric in terms of the Riemann tensor cannot exist.

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.

Answer 2

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.