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.


2 Answers 2


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.

  • $\begingroup$ I knew I was missing something simple. Thanks. $\endgroup$ Jul 29, 2011 at 4:09

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.


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