Can a flat Heegaard splitting surface be totally geodesic?

Question

I'm trying to understand (Meeks, Simon, Yau, 1982), page 652, Corollary of Theorem 5:

Suppose $N$ is a compact 3-dimensional orientable Riemannian manifold with non-negative Ricci curvature whose boundary, possibly empty, has non-negative mean curvature with respect to the outward normal. If $\Sigma$ is a compact embedded orientable minimal surface in $N$, then one of the following holds:

(1) $\Sigma$ is a Heegaard surface.

(2) $N$ is flat and $\Sigma$ is totally geodesic.

(3) $N$ is isometric to $S^2\times S^1$ or $S^2\times I$ with a product metric and $\Sigma$ is one of the sphere factors.

(4) $N$ is diffeomorphic to $P^3$ minus a ball or is diffeomorphic to $P^3\#P^3$ where $P^3$ is a 3D projective space. In this case, $\Sigma$ is a totally geodesic sphere in $N$ such that each component of $N\sim \Sigma$ is isometric to the nontrivial interval bundle over $P^2$ induced as a $Z_2$-quotient of $S^2 \times [0,1]$ with a product metric by an isometry $(x, t)\mapsto (-x, -t)$.

The phrasing makes it sound like the 4 cases are mutually exclusive, and I can understand why, except for cases (1) and (2). I see no reason why a Heegaard surface may not be totally geodesic, or why Heegaard surfaces may not exist in flat space, or why a Heegaard surface can't be flat and totally geodesic at the same time. Still, I'm not sure. So there are two ways I can see case (2) should be interpreted. Either

$N$ is flat, and $\Sigma$ is totally geodesic and not a Heegaard surface, although it may have been.

or

$N$ is flat and $\Sigma$ is totally geodesic, so it can't possibly be a Heegaard surface

Which is correct and why?

Answer 1

First of all, the authors do not claim that these cases are mutually exclusive. However, if $N$ is a (closed, oriented, connected) flat 3-d manifold, it cannot have a genus 1 Heegaard decomposition. In particular, it cannot have a flat (e.g. totally geodesic) Heegaard surface. The reason is quite simple: It follows from Van Kampen's theorem that each genus 1 3-d manifold has cyclic fundamental group (finite or infinite). On the other hand, a compact flat manifold of dimension $>1$ cannot have cyclic fundamental group. This is a nice exercise to work out: if $\gamma$ is an isometry of $E^3$ generating a group $\Gamma$ acting freely, then $\gamma$ preserves a line in $E^3$, hence, the quotient space $E^3/\Gamma$ is a plane bundle over $S^1$, hence, cannot be compact.

At the same time:

1. A Heegaard surface may be totally geodesic (just not in a flat manifold).

2. Heegaard surfaces do exist in every closed connected 3-manifold, even in a flat one. (Just such a surface cannot be flat.)

3. A Heegaard surface can be flat and totally geodesic at the same time: Just not in a flat manifold. For instance, one can find such a surface in $S^3$ equipped with a suitably chosen Riemannian metric.