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 scalar curvature.)


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