Let $M$ be a Ricci-flat Riemannian manifold and $N \subset M$ a totally geodesic submanifold. Is $N$ also Ricci-flat?

A partial result in that direction is that the Ricci curvature of $N$ is given by $$\operatorname{Ric}^N(Y, Z) = \operatorname{tr}(TN \ni X \mapsto R(X, Y)Z \in TN),$$ where $R$ is the Riemmanian curvature tensor of $M$. So $\operatorname{Ric}^N(Y, Z)$ is the trace of the restriction of $X \mapsto R(X, Y)Z$ to $TN$. But the restriction of a trace-free map is not necessarily trace-free.


1 Answer 1


You can find an explicit counterexample in the Riemannian Schwarzschild solution.

Let $(z,r,\omega) \in \mathbb{R} \times (1,\infty) \times \mathbb{S}^2$, denote by $h$ the standard sphere metric. Consider the following Riemannian metric

$$ ds^2 = (1- r^{-1})~dz^2 + (1-r^{-1})^{-1} ~dr^2 + r^2 h $$

One can explicitly compute that this metric is Ricci-flat using:

  • The metric is a warped product of $\mathbb{R}\times (1,\infty)$ against the sphere,
  • Exercise 5.5 from O'Neill's Semi-Riemannian Geometry
  • Standard formulae for the Gauss curvature of a diagonal metric in two dimensions.

As a warped product, fixing any $\omega\in \mathbb{S}^2$, the submanifold $\mathbb{R}\times (1,\infty) \times \{\omega\}$ is totally geodesic, but it has non-vanishing Gauss curvature $K = \frac{1}{r^3}$ and so is not Ricci flat.

(I think you also have a co-dimension one example if you look at a $z$-level set, which is totally geodesic due to the reflection symmetry in $z$. I am pretty sure that this slice also has non-vanishing Ricci curvature.)

Because any $z$-level set is asymptotically flat, if such a set were not Ricci flat then Bishop-Gromov comparison would imply that the level set is isometric to Euclidean space. This is clearly false, as can be seen by, say, calculating the Riemann curvature, or by computing the ADM mass of the level set.


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