The homology of the universal covering space, why so difficult to compute

Question

Let suppose that we are given a connected CW-complex $X$, such that we know

1. All its homology groups.
2. All its homotopy groups, in particular we know $\pi_{1}(X)$.

As far as I know there is no spectral sequence converging to the homology of the universal covering $\tilde{X}$ of $X$. Why it is so difficult (I guess there is no a clear method in general) to compute the homology groups of $\tilde{X}$ ?

Edit: I would like to thank all the authors for their answers, I did learn a lot. I had to choose one answer. I am aware that my question was vague enough, but at the and it seems that the answer I was looking for corresponds more to the one given by M. Rivera.

Answer 1

This is not a complete answer to your question, which I am not sure how to answer without having a precise meaning for what you mean by "computing" and what you mean by "knowing" $\pi_1(X)$, as indicated in the comments above, and given that you have not included the data of any action of $\pi_1(X)$.

However, I think you may be interested in taking a look at my paper with Mahmoud Zeinalian "Singular chains and the fundamental group".

In this article, we prove that for any path connected pointed space $(X,b)$, the $E_{\infty}$-coalgebra structure (in fact, the $E_{2}$ part of it) of (a pointed version of) the singular chains $C_*(X,b)$ extending the Alexander-Whitney coproduct determines $\pi_1(X,b)$ functorially and in complete generality. The correct notion of weak equivalence under which this information is preserved is defined via the cobar functor.

In particular, the natural algebraic structure on the singular chains $C_*(X,b)$ determines a chain complex (by purely algebraic means) which calculates the homology of the universal cover of $X$ at $b$. The construction is the following: consider the differential graded connected coassociative coalgebra $C=(C_*(X,b), \partial, \Delta)$ and take its cobar construction $\Omega C$. This is a dg associative algebra such that $H_0(\Omega C)\cong \mathbb{Z}[\pi_1(X,b)]$, the fundamental group ring. There is a natural "twisting cochain" (in the sense of Brown) $\tau: C \to H_0(\Omega C)$ through which we may construct a twisted tensor product $(C \otimes_{\tau} H_0(\Omega C), \partial_{\tau})$. The homology of this chain complex is the homology of the universal cover. Note that the differential $\partial_{\tau}$ uses both the dg coassociative coalgebra structure of $C$ and the algebra structure of $H_0(\Omega C)$ (which in turn, was obtained from the dg coassociative coalgebra structure of $C$). Also, this construction is invariant in the following "Koszul duality" sense: if $f: C\to C'$ is a map of dg coalgebras such that $\Omega f: \Omega C \to \Omega C'$ is a quasi-isomorphism of dg algebras then the induced map $$(C \otimes_{\tau} H_0(\Omega C), \partial_{\tau}) \to (C' \otimes_{\tau} H_0(\Omega C'), \partial_{\tau})$$ is a quasi-isomorphism. This notion is strictly stronger that ordinary quasi-isomorphisms of dg coalgebras. In some sense, this suggests that you need more information than just the homology to determine the action of $\pi_1$ on the universal cover.

This fact can be used to show an extension of classical theorem of Whitehead, namely we can now prove that a continuous map of pointed path connected spaces $f: (X,b) \to (Y,c)$ is a weak homotopy equivalence if and only if the induced map at the level of dg coassociative coalgebras of pointed singular chains $f: C_*(X,b) \to C_*(Y,c)$ becomes a quasi-isomorphism after applying the cobar functor.

Note that I didn't use the homotopy groups of $X$ directly but I used more information than the homology of $X$, namely, the chain level natural algebraic structure of the singular chains. Again, the question of "how computable this is" is a different one, which I am not sure how to formulate precisely (and it may be tricky given that group theory is not decidable).

Answer 2

Perhaps an example could illuminate.

Let $k \in H^4(K(\mathbb{Q},2);\mathbb{Q}) \cong \mathbb{Q}$. Let also $a,b \in \mathbb{Q}^\times$ satisfy $k (a^2 - b) = 0$, and let $G := \mathbb{Z}$ act on the pointed spaces $K(\mathbb{Q},2)$ and $K(\mathbb{Q},4)$ in a way that the generator acts by multiplication by $a$ and $b$ on the respective non-trivial homotopy group. There is then no obstruction to choosing a $G$-equivariant map $$f: EG \times K(\mathbb{Q},2) \to K(\mathbb{Q},4)$$ representing the cohomology class $k$, since the relevant cohomology group agrees with the $G$-equivariant cohomology group, under our assumption. The homotopy fiber $\widetilde{X} := \mathrm{hofib}(f)$ then inherits a $G$-action, and we may define $X = (\widetilde{X})_{hG}$ as the homotopy orbit space (a mapping torus, in this case). The universal cover of $X$ is then homotopy equivalent to $\widetilde{X}$.

If $a^2 = b \neq 0$, some fiddling with e.g. the Serre spectral sequence shows that the canonical map $X \to S^1$ induces an isomorphism in homology with any constant coefficients, no matter what $k$ is. Choosing $k=0$ and choosing $k\neq 0$ leads to $X$ and $X'$ isomorphic homotopy groups and isomorphic cohomology rings, but their universal covering spaces will have non-isomorphic homology groups. In fact, $H^*(\widetilde{X}) \cong \mathbb{Q}[x,y]$ is polynomial on two generators of degree 2 and 4 when $k = 0$ while $H_*(\widetilde{X}') \cong H_*(S^2;\mathbb{Q})$ when $k \neq 0$.

Therefore there cannot be a functor outputting the homology of the universal covering space in terms of the input data you allow. The lesson of the example is that when the action of $\pi_1(X)$ on the homology of the universal cover is "sufficiently non-trivial", the homology of $X$ does not contain enough information.

Answer 3

Denote $G=\pi _1(X)$. Then we have a fibration $\tilde{X}\to X \to BG$, which leads to the Eilenberg-Moore spectral sequence $$ Tor ^{H^*(BG)}(H^*(X),H^*(pt))\Rightarrow H^*(\tilde{X})$$ provided that we have the Kunneth isomorphism for $H^*(X^n)$, for example, if we take $Z/p$ coefficients or if $H^*(X,Z)$ is free.

Answer 4

One way to think about why this is hard is to consider the relationship between $C_*(\tilde X)$ and $C_*(X)$.

The group $G := \pi_1 X$ acts on $C_*(\tilde X)$ and $C_*(X)$ is quasi-isomorphic to the space of coinvariants, i.e. $H_i(X) = H_i(C_*(X)/G)$. Thus an analogous question is: given a group $G$ and and complex $D \in Ch(Ab^G)$, can we recover $D$ from $D_G$?

The answer to this question is clearly no. For instance, if $G$ is finite, and $D$ a non-trivial irreducible $G$ representation, we have $D_G = 0$. In general, the most you can hope to recover are the "unipotent" $G$ representations-- i.e. those which are a finite extension of direct sums of trivial representations. On the subcategory of chain complexes whose homology is unipotent, coinvariant functor is conservative in a derived sense So you can hope to recover $D$ from $D_G$, together with some extra algebraic structure.

This is why Eilenberg-Moore spectral sequence will only converge to what you want when $\pi_1 X$ acts unipotently on $H_*(\tilde X)$. In general, it will converge to something different--which we might call the homology of the unipotent completion of $C_*(\tilde X)$.