In an earlier thread I had asked whether or not one can find a smooth 4-dimensional submanifold of $S^4$ whose fundamental group has an unsolvable word problem. The answer is yes, and the reference is in that thread. 


see: [Word problem for fundamental group of submanifolds of the 4-sphere](https://mathoverflow.net/questions/32787/word-problem-for-fundamental-group-of-submanifolds-of-the-4-sphere)


I'm curious if one can go a step further. Can one find a smooth $4$-dimensional submanifold $M$ of $S^4$ such that both $\pi\_1 M$ and $\pi\_1 (S^4 \setminus M)$ have unsolvable word problems? 


My motivation for this question has to do with the decidability of a classical 3-manifold theory problem. Given a smooth $3$-manifold, determine whether or not it admits a smooth embedding into $S^4$. 


It's possible to algorithmically generate all smooth embeddings of $3$-manifolds in $S^4$ using normal surface theory for triangulations of the $4$-sphere ((a) start with a given triangulation of $S^4$, (b) enumerate vertex-normal $3$-manifolds, (c) barycentrically subdivide, goto (b)). This is an unfortunately horribly memory-intensive algorithm, but it's exponential run-time so it could be a lot worse.


Whether or not the embedding problem is algorithmically decidable boils down to whether or not there is a computable function $\beta : \mathbb N \to \mathbb N$ such that if a $3$-manifold $M$ has a triangulation with $n$ tetrahedra, you want to know that you only need to barycentrically subdivide a triangulation of $S^4$ at most $\beta(n)$ times to find $M$ as a vertex-normal 3-manifold in the triangulated $4$-sphere. 


Another way to look at this is that perhaps some $3$-manifolds have to embed in $S^4$ in extremely "twisted" ways. If $M$ embeds in $S^4$ such that both $4$-manifolds that it separates $S^4$ into have unsolvable word problem, that qualifies for "really twisted" in my opinion. 


This also gets to the issue of invariants. If you want to say that a $3$-manifold does not embed in $S^4$, perhaps invariants sensitive to the fundamental group should be important. But any such invariant would have to be relatable to $\pi\_1$ of the two $4$-manifolds that $M$ separates $S^4$ into. So issues of computability also come up here. At present I know of no obstruction to a $3$-manifold embedding in $S^4$ that sees really "deeply" into $\pi\_1 M$ but it seems like such obstructions are a realistic possibility. Alexander modules are known to play a role as an obstruction. Twisted Alexander modules should play a role as well, for example, although the literature on this is currently contained in the (closely related) knot-concordance world. 


Presumably this is an open problem (my apologies) but I'm curious if there are any insights out there. 


An offshoot of this question would be whether or not every $3$-manifold that admits an embedding in $S^4$ has *some* embedding in $S^4$ where the two $4$-manifolds it splits $S^4$ into have solvable fundamental group word problems.



4-manifolds in the 4-sphere such that it, *and* its complement have unsolvable word problem
The answer to the question in the beginning should be YES and it follows from the answer to your previous question. We just need to use the fact that if $G$ has unsolvable word problem then $G\*F$ too, where $F$ is a group and $G\*F$ is the free product.


To construct the example take the solution to the previous question, namely a $4$-manifold $M^4 $embeddable in $S^4$ with unsolvable word problem. Now take an open ball $B^4$ in $M^4$ and cut from it a small copy of $M^4$, that we call $N^4$. Finally dig a wormhole that connects $N^4$ with $S^4\setminus M^4$. This divides the sphere into two connected 4-manifolds each of which is homotopic to a connected sum of two manifolds, one of which has same fundamental group as $M^4$. 


