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?

  • $\begingroup$ I don't know the paper, but are you sure that there is not an additional assumption that the $n$-knot is simple? $\endgroup$ – Scott Carter Jun 30 '12 at 8:09
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    $\begingroup$ 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. $\endgroup$ – Hiro Lee Tanaka Jun 30 '12 at 12:32
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    $\begingroup$ 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 $\pi_1(S^{n+2}-K) \cong \mathbb{Z}$ $\endgroup$ – Aru Ray Jun 30 '12 at 14:06
  • $\begingroup$ 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. $\endgroup$ – Blake Jun 30 '12 at 16:26

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 $\mathbb{S}^1$

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  • $\begingroup$ 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? $\endgroup$ – Blake Jun 30 '12 at 21:50
  • $\begingroup$ 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. $\endgroup$ – Aru Ray Jun 30 '12 at 23:58
  • $\begingroup$ 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! $\endgroup$ – Blake Jul 1 '12 at 1:46

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