# Homology of the universal cover

$$k$$ is a field. Let $$X$$ be a connected pointed $$CW$$-complex such that the homology $$H_{n}(X;k)$$ is a finite dimensional $$k$$-vector space for any $$n\in \mathbb{N}$$. Suppose that we have continuous mapping $$r: X\rightarrow K(\pi_{1}(X),1)$$ from the space $$X$$ to the Eilenberg-MacLane space $$K(\pi_{1}(X),1)$$ inducing an isomorphism on fundamental group $$\pi_{1}$$. Suppose also that there is a mapping $$i: K(\pi_{1}(X),1) \rightarrow X$$ such that:

1. $$r\circ i= id$$
2. the homotopy cofiber of $$i$$ is homotopy equivalent to a finite $$CW$$-complex.
3. and $$H_{n}(r;k): H_{n}(X;k)\rightarrow H_{n}(K(\pi_{1}(X),1);k)$$ is an isomorphism for any $$n\in \mathbb{N}$$.

My question is the following: Is the homology of $$\tilde{X}$$ (the universal covering of $$X$$) finite dimensional? i.e. $$H_{n}(\tilde{X};k)$$ is a finite dimensional $$k$$-vector space for any $$n\in \mathbb{N}$$?

• What about $S^2 \vee S^1 \to S^1$? Am I missing something? – Najib Idrissi Jun 11 '19 at 20:26
• @NajibIdrissi Sorry, I forgot the most important condition (3). – GSM Jun 11 '19 at 20:30

If we replace the field $$k$$ with the ring of integers $$\Bbb Z$$, then no.

There are non-trivial high dimensional knots $$K: S^n \to S^{n+2}$$, whose complements $$X = S^{n+2}-K(S^n)$$ have $$\pi_1(X) \cong \Bbb Z$$ ($$n > 2$$). The map $$X \to S^1$$ defining the generator of $$H^1(X) \cong \Bbb Z$$ has a section, and the mapping cone of this map has the homotopy type of a finite complex.

The homotopy fiber of this map is $$\tilde X$$, the universal abelian cover. It has the homotopy type of a finite complex iff $$K$$ is a fibered knot (meaning that there is a representative of the map which is a fiber bundle having compact manifold fibers which are Seifert surfaces of $$K$$; this is a result of Browder and Levine).

Since $$\tilde X$$ is $$1$$-connected, it has the homotopy type of a finite complex iff its homology is finitely generated over $$\Bbb Z$$ in each degree (this is an easy case of a result due to Wall). Since there are non-fibered knots with $$\pi_1(X) \cong \Bbb Z$$, any such knot will have the property that the homology of $$\tilde X$$ in some degree will fail to be finitely generated over $$\Bbb Z$$.

Remarks:

(1) It could very well be that these examples work over any field $$k$$, but I do cannot seem to deduce that statement.

(2) Observe what we really constructed is an example satisfying your criteria such that $$\tilde X$$ fails to have the homotopy type of a finite complex.

Addendum (June 12, 2019): Danny Ruberman points out that Milnor settled Remark (1) in the negative. In other words, there are no examples satisfying the original question having $$\pi_1(X) \cong \Bbb Z$$.

• In this case $X\rightarrow S^1$ is a homology isomorphism ? – GSM Jun 11 '19 at 22:36
• For any knot the map $X\to S^1$ is a homology isomorphism by Alexander duality. – John Klein Jun 11 '19 at 23:51
• Milnor (Infinite Cyclic Coverings, Assertion 5) shows that with field coefficients, the homology of the infinite cyclic covering of any knot complement is finitely generated. The proof is a nice application of the Milnor exact sequence in homology derived from the SES $0 \to C_{*}(\tilde{X}) \overset{(t_*-1)}{\to} C_{*}(\tilde{X}) \overset{p_*}{\to} C_{*}(X) \to 0$. – Danny Ruberman Jun 12 '19 at 0:59
• @DannyRuberman Yes, I had forgotten that. So these examples are not finitely generated over $\Bbb Z$ in some degree but are so over any field, implying that there is torsion at in infinite number of primes in that degree. – John Klein Jun 12 '19 at 2:06
• @DannyRuberman I misphrased my previous comment. The statement about primes was rubbish. I meant to write: 1) The universal cover, which is 1-connected, is finite dimensional. 2) The total homology with Z coefficients is not finitely generated. 3) The total homology with any field coefficients is finitely generated. so we could get an example wit, say, a copy of $\Bbb Q$ in its homology and 2 and 3 will hold. – John Klein Jun 14 '19 at 17:49

Take $$Y=S^1\times S^2$$. Take a map from a sphere $$S^2\to Y$$ given by one of the fibers $$*\times S^2$$ and glue on a disk: $$X=S^1\times S^2\cup_{S^2} D^3$$. Then $$H_•(X)=H_•(S^1)$$. There is a section from $$S^1$$ inducing a homology isomorphism, so the cofiber is contractible. For an alternate description, think of gluing as $$X=S^1\times S^2\cup_{S^1\vee S^2}S^1\vee D^3$$. From this point of view, the universal cover is seen to be a pushout of universal covers: $$\tilde X =\widetilde{S^1\times S^2}\bigcup_{\widetilde{S^1\vee S^2}}\widetilde{S^1\times D^3} =S^2\times\mathbb R^1\bigcup_{\widetilde{S^1\vee S^2}}\mathbb R^1\times D^3$$ The three constituents are homotopy equivalent to $$S^2$$, $$\bigvee_\infty S^2$$, and a point, so we can compute the homology with Mayer-Vietoris and it is infinite rank, all in degree $$3$$.

• The cofiber of the map $\ast \times S^2 \to S^1 \times S^2$ is $S^1 \wedge (S^2_+) \simeq S^1 \vee S^3$ (which is still a finite complex). So $H_\bullet(X) \neq H_\bullet(S^1)$. Moreover it's clear from this that $\tilde X$ is homotopy equivalent to countable wedge of copies of $S^3$. But your example fails to satisfy condition (3). – John Klein Jun 13 '19 at 2:22
• Oh, yeah. But it would work if the fundamental group were acyclic... – Ben Wieland Jun 13 '19 at 22:11
• @BenWieland How? could you say more, please ? – GSM Jun 14 '19 at 10:15