Incompressible surfaces in an open subset of R^3

Question

Let $U$ be a connected open subset of $\mathbb R^3$. Furthermore, we have:

1. $\mathbb R^3\setminus U$ has exactly two connected components (thus by Alexander duality, $H_2(U;\mathbb Z)=\mathbb Z$).
2. $U$ may be "very complicated": we make no assumptions on the regularity of $\partial U$.

I would like to understand the incompressible surfaces in $U$ representing a generator of $H_2(U;\mathbb Z)$.

Question: If I have a collection of incompressible surfaces $F_1,\ldots,F_p$ (all representing the same generator of $H_2(U)$), is there a canonical "lower envelope" incompressible surface $F$? What I want morally is that $F=\partial\left(\bigcap_{i=1}^p\operatorname{int}(F_i)\right)$. Of course, $\partial\left(\bigcap_{i=1}^p\operatorname{int}(F_i)\right)$ depends on how the $F_i$'s are embedded, and it may not be incompressible, so it's not the right choice. Perhaps though as long as there aren't any intersections $F_i\cap F_j$ which are inessential in either $F_i$ or $F_j$, then this guarantees (the isotopy class of) $\partial\left(\bigcap_{i=1}^p\operatorname{int}(F_i)\right)$ is unchanged by isotoping $F_i$.

Motivation: Secretly there is a group action on $U$, and I can easily construct incompressible surfaces $F_1,\ldots,F_p$ which are cyclically permuted by the group, but what I really need is a single incompressible surface which is fixed by the action. Taking some sort of "lower envelope" of $F_1\ldots,F_p$ is my first try at constructing such a surface.

Answer 1

Yes, I think if I understand your question, what you're asking for is true. Take minimal area representatives for your surfaces $F_1,\ldots, F_p$, and take the boundary of the component of the complement of $F_1\cup\cdots \cup F_p$ which contains infinity (I think this is the same as your surface $F$). If this surface is compressible to the "inside", then you may choose a maximal compression body to the inside. The inner boundary will be unique up to isotopy, and will be incompressible inwardly. One can show that each compression may be made disjointly from $F_i$ for all $i$ so this surface may be made disjoint from isotopic copies of the $F_i$ simultaneously. Repeat this compressing outwardly and inwardly, until you obtain an incompressible surface $F'$, and isotopic surfaces $F_i$ which are disjoint. Now it's a fact that if two incompressible surfaces are disjoint, then their minimal area representatives will also be disjoint. So a minimal area representative of $F'$ will be disjoint from the original $F_i$s. So I think this will give your the desired "lower envelope", modulo checking certain details.

Addendum: One would like to see that the surface $F$ is unique up to isotopy, in particular independent of the Riemannian metric. I think this is true, but it requires some proof. I'll sketch a possible argument. There is a partial ordering on isotopy classes of incompressible surfaces in the homology class, where $[E]\leq [F]$ if there are disjoint representatives $E'\sim E, F'\sim F$ such that $E' \cap F'=\emptyset$, and $E'$ is ``inside" $F'$. Now, one needs to show that if $[E]\leq [F], [F]\leq [E]$, the $[E]=[F], i.e. E\sim F$ are isotopic. This follows from standard facts.

The key property here is that this partial order is a lattice: given isotopy classes of incompressible surfaces $[F_1],\ldots, [F_k]$, there is a unique surface $[F]$ such that $[F]\geq [F_i], i=1,\ldots,k$ (and similarly there is a unique infimum). The argument I gave before shows the existence of some $[F]\geq [F_i]$. If we have two such $[F], [F']\geq [F_i]$, then a similar argument shows that there exists $[F'']$ such that $[F]\geq [F''], [F']\geq [F'']$, and $[F'']\geq [F_i]$. Repeating this argument, and using a compactness argument, I think one can show that there exists a unique minimal such surface with is $\geq [F_i]$. This is independent of the Riemannian metric, since a 3-manifold has a unique smooth structure, and the partial ordering is independent of the differentiable structure.