# Can the Ricci tensor determine a manifold up to homeomorphism?

I don't know how to ask this question. Maybe there are many improper statements.

Let $(M, g)$ and $(N, h)$ be two Riemannian manifolds and $f:M \to N$ a bijection such that $\operatorname{Ric}(x)$ is equal to $\operatorname{Ric}(f(x))$ up to a coordinate transformation. Are $M$ and $N$ homeomorphic?

In fact, I feel "up to a coordinate transformation" is not suitable, because, locally, I feel (not sure) that I can choose suitable coordinates making $\operatorname{Ric}(x) = \operatorname{Ric}(f(x))$. But I also don't know how to ask it. I just want to know whether the Ricci tensor can determine a manifold up to homeomorphism.

• There are several theorems in Riemannian geometry (going back to Riemann) of the form "curvature determines the metric", but they only hold locally because you can always pull-back Riemannian metrics via local diffeomorphisms. See math.stackexchange.com/questions/162175/… and links therein. – Moishe Kohan Nov 15 '16 at 15:18

This is not true. For example, not all Ricci-flat manifolds are homeomorphic. More explicitly, for any bijection $f$ between $\mathbb{R}^n$ and $(S^1)^n$, $\operatorname{Ric}(x) = \operatorname{Ric}(f(x)) = 0$, but $\mathbb{R}^n$ and $(S^1)^n$ are not homeomorphic.
• I remeber that the surface is determined by its curvature in $\mathbb R^3$ up to a translation. Is right ? – lanse7pty Nov 15 '16 at 7:42
• @lanse7pty: Definitely not. A cylinder (or cone) and a plane are both flat (curvature $0$). – Ted Shifrin Feb 5 '18 at 2:32