4
$\begingroup$

Prove that all homology groups of the infinite cyclic cover of a knot complement are trivial except $H_1$.

I've posted an answer below using Mayer-Vietoris. If you know of other arguments, please share! I've believe there is a proof that uses a long exact sequence associated with an action on $H_*(\tilde X)$ that comes from covering transformations $\tilde X \to \tilde X$. Rolfsen also suggests an argument in Knots and Links. Is it possible to use a direct limit? Please share any ideas!

Why? I had typed up a long question about one step in proving this with Mayer-Vietoris, then realized I was making a silly mistake. I've shared my argument so that my $\LaTeX$ efforts are not in vain.

Background: Construction of the cover

Let $K \subset S^3$ be a knot with complement $X = S^3 \setminus K$. Given a Seifert surface $\Sigma$ for $K$, we can construct an infinite cyclic cover $\tilde X \to X$: Let $N\subset S^3$ be a collar neighborhood of the interior $\mathring \Sigma= \Sigma \setminus K$ homeomorphic to $\mathring \Sigma \times (-1,1)$ and define \begin{align*} Y &= S^3 \setminus \Sigma \\ N^+ &= \mathring \Sigma \times (0,1) \subset N\\ N^- &= \mathring \Sigma \times (-1,0) \subset N. \end{align*} Now take countably many copies of the triples $(N,N^+,N^-)$ and $(Y,N^+,N^-)$, obtaining $$\tilde N = \cup_{i \in \mathbb{Z}} N_i \qquad \text{and} \qquad \tilde Y = \cup_{i \in \mathbb{Z}} Y_i.$$ Finally construct $\tilde X$ by identifying $N_i^+ \subset Y_i$ with $N^+_i \subset N_i$ via the identity and likewise $N_i^- \subset Y_i$ with $N^-_{i+1} \subset N_{i+1}$. See the figure below.

enter image description here

$\endgroup$
4
  • 1
    $\begingroup$ 1: If the knot is fibered, the infinite cyclic covering has the homotopy type of a surface with boundary, namely a genus minimizing Seifert surface. $\endgroup$ Commented May 12, 2015 at 22:38
  • 1
    $\begingroup$ 2: it is an easy exercise that the complement has homotopy type of a CW cplx with $1$ 0-cell, $n$ 1-cells and 0 3-cells. Write down the boundary maps of the lifted cellular complex and observe that $H_2\bar X=0$. $\endgroup$ Commented May 12, 2015 at 22:43
  • 1
    $\begingroup$ 3: Assume $i: S \to \bar X$ represents a homology class $[S] \in H_2X$, then by compactness of $S$ the image is in some connected finite union $\cup X_i$ where $X_i$ is the closure of a fundamental domain containing $Y_i$. Add up the dimensions in the long exact sequence of the pair $(\cup X_i,\partial)$ using duality and you get triviality of $H_2(\cup X_i) $ and by functoriality $[S] = 0 \in H_2\bar X$. $\endgroup$ Commented May 12, 2015 at 23:12
  • $\begingroup$ @Dan: Very cool arguments. If you want to repost those comments as an answer, I'll accept it. $\endgroup$
    – Kyle
    Commented May 23, 2015 at 13:36

1 Answer 1

2
$\begingroup$

Computing the homology of $\, \boldsymbol{\tilde X}$ via Mayer-Vietoris:

The cover $\tilde X$ is a non-compact 3-manifold, so $H_*(\tilde X)$ vanishes for $\, * \geq 3$. We'll compute $H_2(\tilde X)$ using part of the Mayer-Vietoris sequence associated to the decomposition $\tilde X = \tilde Y \cup \tilde N$: $$H_2(\tilde Y) \oplus H_2(\tilde N) \overset{f}{\longrightarrow} H_2(\tilde X) \overset{\partial}{\longrightarrow} H_1(\tilde Y \cap \tilde N) \overset{g}{\longrightarrow} H_1(\tilde Y) \oplus H_1(\tilde N).$$ We'll first show $H_2(\tilde Y)$ and $H_2(\tilde N)$ are zero, implying that the connecting homomorphism $\partial$ is injective. Then we'll show that $g$ is injective, which will give us $$H_2(\tilde X) \cong \operatorname{im} \partial \cong \ker g =0.$$ Since $N_i$ deform retracts onto the non-compact surface $\Sigma_i$, we have $$H_2(\tilde N) \cong \oplus_{i \in \mathbb{Z}} H_2(N_i) \cong \oplus_{i \in \mathbb{Z}} H_2(\Sigma_i) = 0.$$ A basic argument using duality, excision, and the cohomology LES for $(S^3,\Sigma)$ gives us $$H_*(S^3 \setminus \Sigma) \cong H^{3-*}(S^3,\Sigma) \cong \begin{cases} \mathbb{Z} & *=0 \\ H_1(\Sigma) & *=1 \\ 0 & *\geq 2.\end{cases} \tag{1}$$ Therefore we have $H_2(\tilde Y) \cong \oplus_{j \in \mathbb{Z}} H_2(Y_j) \cong 0$. This shows that $\partial$ is injective.

Before analyzing $\ker g$, let's set up some useful notation. The intersection $\tilde Y \cap \tilde N$ is a countable disjoint union of subspaces $\tilde Y \cap N_i$, where $\tilde Y \cap N_i$ deform retracts onto $\mathring \Sigma_i^- \cup \mathring \Sigma_i^+$, a pair of pushoffs of $\mathring \Sigma_i$. Given $a_i \in H_1(\mathring \Sigma_i)$, let $a_i^\pm$ denote the image of $a_i$ in $H_1(\mathring \Sigma_i^\pm)$. It follows that we can represent an element of $H_1(\tilde Y \cap \tilde N)$ as a sum $\sum_{k \in \mathbb{Z}} (a_k^-,b_k^+)$, where $a_k,b_k \in H_1(\mathring \Sigma_k)$ are zero for all but finitely many $k \in \mathbb{Z}$.

Now suppose $\sum_k (a_k^-,b_k^+)$ lies in $\ker g$. Then, for each $i \in \mathbb{Z}$, it must also lie in the kernel of the projection $$H_1(\tilde Y \cap \tilde N) \to H_1(\tilde Y) \oplus H_1(\tilde N) \to H_1(\tilde N) \to H_1(N_i)$$ given by $\sum_k(a_k^-,b_k^+)\mapsto a_i+b_i$. It follows that $b_i=-a_i$ for all $i$, i.e. $\sum_k(a_k^-,b_k^+)=\sum_k (a_k^-,-a_k^+)$. We also have a projection $$H_1(\tilde Y \cap \tilde N) \to H_1(\tilde Y) \oplus H_1(\tilde N) \to H_1(\tilde Y) \to H_1(Y_j)$$ given by $\sum_k (a_k^-,b_k^+)\mapsto [a_{j+1}]_{Y_j}+ [b_j]_{Y_j}$ where $[\cdot]_{Y_j}$ denotes inclusion into $H_1(Y_j)$, which is isomorphic to $H_1(\Sigma)$ by (1). If $\sum_k(a_k^-,-a_k^+)$ is in the kernel of this projection, then $$[a_{j+1}]_{Y_j}+[-a_j]_{Y_j}=0 \quad \Leftrightarrow \quad [a_{j+1}]_{Y_j}=[a_j]_{Y_j}.$$ We finish with an inductive argument: Because all but finitely many $a_k$ are zero, there is some $N \gg0$ such that $a_k=0$ for all $k \geq N$. Then $$[a_{N-1}]_{Y_{N-1}}=[a_N]_{Y_{N-1}}=[0]_{Y_{N-1}}=0,$$ which implies $a_{N-1}=0$ because the inclusion $H_1(\Sigma_{N-1}) \to H_1(Y_{N-1})$ is an isomorphism. A simple induction argument shows that $a_k=0$ for all $k$. This implies that $\ker g=0$, so we conclude that $H_2(\tilde X)$ is zero. $\blacksquare$

Remark. If we extended the Mayer-Vietoris sequence on the right, we would see that $H_1(\tilde X)$ is isomorphic to the quotient of $H_1(\tilde Y)\oplus H_1(\tilde N)$ by $\operatorname{im} g$. This is essentially how Rolfsen does his "Calculation Using Seifert Surfaces" for $H_1(\tilde X)$ in $\S$6.B of Knots and Links.

$\endgroup$
1
  • $\begingroup$ Since you mention direct limits (=colimits) and actions, and ask to share ideas, you could look at the book advertised at pages.bangor.ac.uk/~mas010/nonab-a-t.html, particularly Chapter 8, e.g. the example of $\pi_n$ of the universal cover of $S^n\vee S^1\vee S^1$ in the Introduction to that chapter. I'd like to see these ideas applied to knots! $\endgroup$ Commented May 13, 2015 at 10:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .