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. 



Embedding the product of three circles in the 4-sphere. 
No. Suppose that the rank of $H^1(V\_1)$ is zero, so that the rank of $H^1(V\_2)$ is three and (by looking at the Mayer-Vietoris sequence again) the rank of $H^2(V\_2)$ is zero. Take two independent elements of $H^1(V\_2)$. Their product in $H^2(V\_2)=0$ is trivial, but its image in $H^2(M)$ is nontrivial, being the product of two independent elements of $H^1(M)$. Contradiction.


