Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers

up vote 11 down vote accepted

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.

share|improve this answer
    
I knew I was missing something simple. Thanks. –  Ryan Budney Jul 29 '11 at 4:09
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.