Can the Ricci tensor determine a manifold up to homeomorphism?

Question

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 http://math.stackexchange.com/questions/162175/does-the-curvature-determine-the-metric and links therein. — Moishe Kohan, Nov 15 '16 at 15:18

Michael Albanese · Accepted Answer · 2018-02-04T21:01:36.217

5

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.

edited Feb 04 '18 at 21:01

answered Nov 15 '16 at 03:19

Michael Albanese

99,526

I remeber that the surface is determined by its curvature in $\mathbb R^3$ up to a translation. Is right ? – Enhao Lan Nov 15 '16 at 07:42
@lanse7pty: Definitely not. A cylinder (or cone) and a plane are both flat (curvature $0$). – Ted Shifrin Feb 05 '18 at 02:32

Can the Ricci tensor determine a manifold up to homeomorphism?

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