Does any compact boundary become a minimal hypersurface iff each of the boundary component is a minimal hypersurface?

Question

There are researchers who made the following assumptions in their paper on the Penrose inequality:

Let $M$ be a complete, connected Riemannian $3$-manifold and suppose that the boundary $\partial M$ of $M$ is compact and consists of minimal surfaces, and $M$ contains no other compact minimal surfaces.

I'm not sure I understand these assumptions correctly. Are they talking about the (connected) components of $\partial M$? Since $\partial M$ is assumed to be compact, there are finitely many boundary components. Each of these components is a compact surface in $M$. If we assume all of them are minimal surfaces, is $\partial M$ going to be a minimal surface? The converse seems to be true since if $\partial M$ has mean curvature that is identically zero, it is reasonable to see that each boundary component has mean curvature that is also identically zero. Can we make it an iff statement? Also, I don't know why the authors assume that $M$ contains no other compact minimal surfaces. What if we join together some of the boundary components (assumed to be minimal surfaces) to make another minimal surface? I really need some help. Thank you.

Answer 1

Yes, by "$\partial M$ consists of minimal surfaces" they just mean that each connected component of $\partial M$ is a minimal surface - they're just emphasizing that $\partial M$ need not be connected. (Some people may also require connectedness in their definition of "minimal surface" - I'm not sure.)

Also, I don't know why the authors assume that M contains no other compact minimal surfaces. What if we join together some of the boundary components ( assumed to be minimal surfaces) to make another minimal surface?

If you want to get precise, their "no other minimal surfaces" should probably be phrased as "no minimal surfaces that are not subsets of $\partial M$". As to why they make this assumption: it's been 10 years since I looked at this paper, so maybe take this with a grain of salt; but from memory it's essentially ensuring that $\partial M$ includes all event horizons, so that it makes sense to use the area of its components in the Penrose inequality. Without this assumption you could have some other horizon in the interior of $M$ not being counted.

Answer 2

Just to add to Anthony's answer, in particular about the second part of your question. The manifold M cannot contain any other closed minimal surfaces indeed is to exclude any other horizons in the interior of the manifold. But the reason you would want to exclude this is because if there were another closed minimal surface enclosing (a connected component of) the boundary, then the ADM mass could not "see" beyond that outer minimal surface. Basically the best you could do would be a Penrose inequality where the horizon area refers only to the outer one, not the boundary itself. (You can construct explicit counterexamples if the minimal surface is not outermost)

If you're looking at the Huisken-Ilmanen proof, then such a minimal surface containing your boundary would cause the weak flow to immediately jump to that surface at time zero.