TITLE: Embedding the product of three circles in the 4-sphere. 
QUESTION [10 upvotes]: Let $M$ be a smooth submanifold of the 4-sphere $S^4$.  I'm going to demand that $M$ be diffeomorphic to $S^1 \times S^1 \times S^1$.   By Jordan-Brouwer separation, $M$ separates the 4-sphere into two compact 4-manifolds $V_1$ and $V_2$, i.e. $V_1 \cup V_2 = S^4$, $V_1 \cap V_2 = M$, $\partial V_1 = \partial V_2 = M$. 
The question is, is it possible for the rank of $H_1(V_1, \mathbb Z)$ to be zero?  
A little Mayer-Vietoris sequence argument will convince you that $H_i M \simeq H_i V_1 \oplus H_i V_2$ for $i \in \{1,2\}$, the map given by inclusion.  
I believe all known embeddings of $(S^1)^3$ in $S^4$ have $rank(H_1(V_i, \mathbb Z)) \geq 1$ for both $i$ -- so one will have rank $1$, the other rank $2$.   
Off the top of my head I don't see a reason why that should always be true. 
This is a question that came up in a discussion with Jonathan Hillman.

REPLY [2 votes]: No embedding of a product $M=T_g\times{S^1}$ in $S^4$ can have one complementary component
$X$ with $H_1(X)=0$.
For otherwise, the other component $Y$ would have $H_2(Y)=0$.
But then the inclusions of $M$ and of a wedge of $2g+1$ circles into $Y$ would induce
isomorphisms on the lower central series quotients of the fundamental groups, 
by an old theorem of Stallings.
This cannot be so, as $\pi_1(M)$ has a central factor,
whereas the free group $F(2g+1)$ does not.