# Ricci curvature of totally geodesic submanifold

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

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.)