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

I'm trying to understand a paper by Kawauchi and Matumoto, 'An estimate of infinite cyclic coverings and knot theory.' In one of their proofs they have a manifold $E$ which is the complement of a codimension-2 $n$-knot, with its infinite cyclic cover $\widetilde{E}$. They show that $\widetilde{H}_{*}(\widetilde{E};\mathbb{Z})=0$. Then they claim that because of this we must have that $E$ is homotopy equivalent to $S^{1}$.

It is clear that this shows $E$ is homologically equivalent to $S^1$, but why do we also get the result that it is homotopy equivalent?

share|improve this question
    
I don't know the paper, but are you sure that there is not an additional assumption that the $n$-knot is simple? –  Scott Carter Jun 30 '12 at 8:09
4  
The statement as written is clearly false -- the complement of the trefoil is not homotopy equivalent to S^1, as its fundamental group is not Z. If one knows (in addition to what you've written) that $\pi_1(E) = \mathbb Z$, we can conclude $E \simeq S^1$; otherwise it seems there's something missing. –  Hiro Lee Tanaka Jun 30 '12 at 12:32
1  
For what it is worth, I'm peeking at the paper mentioned and the result mentioned is: Let n $\geq$ 3. A ribbon n-knot $K$ is unknotted if $\pi_1(S^{n+2}-K) \cong \mathbb{Z}$ –  Aru Ray Jun 30 '12 at 14:06
    
I was very careless and missed the crucial assumption that Aru Ray has filled in... I guess staring at the paper for hours had me taking it for granted. –  Blake Jun 30 '12 at 16:26
add comment

1 Answer

up vote 2 down vote accepted

The result being proved is:

Let $n ≥ 3$. A ribbon $n$-knot $K$ is unknotted if $\pi_1(\mathbb{S}^{n+2}−K)\cong > \mathbb{Z}$

Let $E$ denote $\mathbb{S}^{n+2}−K$, then $\pi_1(E) \cong \mathbb{Z}$ by hypothesis. Let $\tilde{E}$ denote the infinite cyclic cover of $E$. $\pi_2(E)\cong \pi_2(\tilde{E})\cong H_2(\tilde{E})=0$, since $\tilde{E}$ covers $E$ and by the Hurewicz theorem, since (it is proved in the paper as mentioned by the OP) $\tilde{H}(\tilde{E};\mathbb{Z})\cong 0$ and $\pi_1(\tilde{E})\cong 0$ (as the infinite cyclic cover of a space with $\pi_1 = \mathbb{Z}$). Keep doing this to get that $\pi_1(E)\cong \mathbb{Z}$ and $\pi_i(E)\cong 0$, for all $i>1$. This means that $E$ is a $K(\mathbb{Z},1)$. All of those are homotopy equivalent to each other, so $E$ is homotopy equivalent to $\mathbb{S}^1$

share|improve this answer
    
Ok - the point I am missing is: why do we get that $\pi_{2}(\widetilde{E})=0$ just from the triviality of $\pi_1$ and $\widetilde{H}(\widetilde{E};\mathbb{Z})$? Does this come from a long exact sequence of some sort? Because otherwise the second homology could be zero, without the second homotopy group vanishing. Then likewise once we know the first $n$ homotopy groups vanish, how does the triviality of the homology groups imply the next homology group vanishes? –  Blake Jun 30 '12 at 21:50
    
Check out en.wikipedia.org/wiki/Hurewicz_theorem . In short, the relevant bit is that for $n \geq 2$, if a space is $n-1$ connected, i.e. the first $n-1$ homotopy groups vanish, then there is an isomorphism between $\pi_n$ and $H_n$. Above we have that $\pi_1(\tilde{E})=0$, and so $\pi_2(\tilde{E})\cong H_2(\tilde{E})$ which is known to be zero. –  Aru Ray Jun 30 '12 at 23:58
    
Oh, of course! I forgot the Hurewicz theorem applies to higher homotopy groups, and was just thinking about it in its application to the abelianization of the first homotopy group. Thanks! –  Blake Jul 1 '12 at 1:46
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.