# Higher homology groups of infinite cyclic cover

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.

• 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. Commented May 12, 2015 at 22:38
• 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 $H_2\bar X=0$. Commented May 12, 2015 at 22:43
• 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 $[S] = 0 \in H_2\bar X$. Commented May 12, 2015 at 23:12
• @Dan: Very cool arguments. If you want to repost those comments as an answer, I'll accept it.
– Kyle
Commented May 23, 2015 at 13:36

## 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.
• 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! Commented May 13, 2015 at 10:08