Distinguishing Riemannian Manifolds by Curvature Relationships

[tasguitar7]

Consider two pseudo-Riemmannian manifolds, $M$ and $N$. Suppose that in coordinates $y^\mu$ on $M$ and $x^\mu$ on $N$, the Riemann curvatures $R^M$ and $R^N$ of $M$ and $N$ are related by a coordinate transformation $y = y(x)$:
\begin{equation*}
R^N_{\rho\mu\sigma\nu} = R^M_{\alpha\beta\gamma\lambda}\frac{\partial y^\alpha}{\partial x^\rho}\frac{\partial y^\beta}{\partial x^\mu}\frac{\partial y^\gamma}{\partial x^\sigma}\frac{\partial y^\lambda}{\partial x^\nu}.
\end{equation*}
This is intended to mean that the curvatures are related everywhere by coordinate transformation, although some care may need to be taken with respect to this condition when changing charts in the atlas.

Anyway, if two manifolds have such a relationship everywhere between their curvatures, does that imply that their metrics are related by coordinate transformation:
\begin{equation*}
g^N_{\mu\nu} = g^M_{\alpha\beta}\frac{\partial y^\alpha}{\partial x^\mu}\frac{\partial y^\beta}{\partial x^\nu}?
\end{equation*}
If so or if not, how can you show it? The question is essentially a generalized version of that an everywhere vanishing Riemann curvature implies flatness.

 

[andrewkirk]

Consider two pseudo-Riemmannian manifolds, $M$ and $N$. Suppose that in coordinates $y^\mu$ on $M$ and $x^\mu$ on $N$, the Riemann curvatures $R^M$ and $R^N$ of $M$ and $N$ are related by a coordinate transformation $y = y(x)$:
\begin{equation*}
R^N_{\rho\mu\sigma\nu} = R^M_{\alpha\beta\gamma\lambda}\frac{\partial y^\alpha}{\partial x^\rho}\frac{\partial y^\beta}{\partial x^\mu}\frac{\partial y^\gamma}{\partial x^\sigma}\frac{\partial y^\lambda}{\partial x^\nu}.
\end{equation*}
This is intended to mean that the curvatures are related everywhere by coordinate transformation, although some care may need to be taken with respect to this condition when changing charts in the atlas.

Anyway, if two manifolds have such a relationship everywhere between their curvatures, does that imply that their metrics are related by coordinate transformation:
\begin{equation*}
g^N_{\mu\nu} = g^M_{\alpha\beta}\frac{\partial y^\alpha}{\partial x^\mu}\frac{\partial y^\beta}{\partial x^\nu}?
\end{equation*}
If so or if not, how can you show it? The question is essentially a generalized version of that an everywhere vanishing Riemann curvature implies flatness.

I don't think that can be the case. If it were, it would imply that the curvature tensor field fully determines the geometry of the manifold, and hence the metric. This is hinted at by the fact that there is a nice formula that expresses the curvature tensor in terms of second derivatives of the metric, but no formula that expresses the metric in terms of the curvature tensor.

Responses to this stack overflow discussion indicate that the curvature tensor field does not always uniquely determine the metric.

I imagine there are some constraints you could put on your manifolds and their curvature fields so that each would determine a unique metric, but I don't know what those constraints are.

 

[sweet springs]

If so or if not, how can you show it?

So. Why do not you check out the relation
$$ds^2=g_{\alpha\beta}^Ndx^\alpha dx^\beta=g_{\alpha\beta}^M dy^\alpha dy^\beta$$
?