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

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.

• @ConnorMalin I'm not sure this spectral sequence would be helpful, unless i'm wrong... – GSM Mar 17 at 19:11
• I suppose it doesn’t actually do anything besides tell you you can compute the homology of the universal cover via local coefficients – Connor Malin Mar 17 at 21:01
• Absent knowlede of $\pi_2(X)$, the problem of computing $H_2(\widetilde{X})\cong\pi_2(X)$ is known to be undecidable. Perhaps there's a similar undecidibility result for higher-dimensional homology groups, even with knowledge of the higher homotopy groups? – HJRW Mar 18 at 10:22
• @GSM The problem is as follows: "Is there an algorithm that takes as input a finite 2-complex $X$, guaranteed to be aspherical, and decides whether or not $\pi_1X$ is trivial?". The problem is related to the Andrews--Curtis conjecture, and thence to the smooth 4-dimensional Poincare conjecture. – HJRW Mar 26 at 9:14
• @RyanBudney In some sense I wanted to use the information in the chain complex $C_{\ast}(X)$ and deduce a chain model $C_{\ast}(\tilde{X})$ without allowing a purely topological manipulation as lifting cells. My idea was that the homology of the covering space is encoded in $C_{\ast}(X)$ + extra structure such as "cocommutative" structure of the chain. I know my justification is maybe a little bit artificial :) – GSM Mar 26 at 22:12

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).

• Thank you! I was wondering if $C(X)\otimes_{\tau}\mathbb{Z}[\pi_{1}(X)]$ is quasi-isomorphic to $C(\tilde{X})$ as $E_{\infty}$-coalgebra ? – GSM Mar 26 at 9:04
• I was reading your nice article (arXiv version) I found the answer to my precedent comment. Is the "question 4" still open ? – GSM Mar 26 at 10:18
• @GSM yes $C_*(X) \otimes_{\tau} \mathbb{Z}[\pi_1(X)]$ is quasi-isomorphic to $C_*(\tilde{X})$ as en $E_{\infty}$-coalgebra. We are writing the details down but it is not so surprising based on the work of Benoit Freese which constructs a natural $E_{infty}$-coagebra structure on the cobar construction $\Omega C$ of the underlying $A_{\infty}$-coalgebra of a $E_{\infty}$-coagebra $C$. – Manuel Rivera Mar 26 at 10:54
• Technically the answer is not in the literature, but we are writing down the details and this has already been discussed in talks and private discussions! – Manuel Rivera Mar 26 at 13:31
• Thanks for the elaboration Manuel. Morally, the failure of Koszul equivalence to agree with ordinary weak equivalence is the same as the failure of the Eilenberg--Moore spectral sequence to converge. For this reason, I doubt that you can use your construction to produce a spectral sequence converging to $H_*(\tilde X)$ whose $E_2$ page only depends $H_*(X)$ and $\pi_*(X)$. – Phil Tosteson Mar 30 at 18:24

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.

• This will only converge if the fundamental group is nilpotent. – Phil Tosteson Mar 17 at 20:41
• As Phil Tosteson noticed, this spectral sequence is useful only in a very particular case. My question is about a general fundamental group... In general it is not useful as far as I understand, it was a motivation for my initial question. – GSM Mar 17 at 21:55
• Maybe this points at the issue though. There are precious few methods in algebraic topology which can deal with $\pi_1$ issues in a general, uniform way without assumptions like nilpotence. So perhaps it's unreasonable to demand such a method in this case. @PhilTosteson is that "$\pi_1(X)$ is nilpotent" or rather "$\pi_1(X)$ acts nilpotently on the higher homotopy"? – Tim Campion Mar 18 at 5:16
• @GSM So now two important additional bits of context have been raised which were absent from the initial question: (1) There is a solution to the problem if one puts restrictions on $\pi_1(X)$ and (2) These sort of restrictions seem to be necessary in a broad range of homotopical questions. In this light, I think your question would benefit from some editing and re-focusing. For instance, (and maybe this is too radical a change) it might be more to-the-point in view of (1) and (2) to ask something like "When should I expect a homotopical problem to require nilpotence-like restrictions?" – Tim Campion Mar 18 at 15:56
• @GSM At any rate, for a question of the form "Why is problem X hard?", an answer of the form "Problem X fits into a broad class of problems Y, all of which are hard." is a perfectly acceptable answer. It's an even better answer if it comes along with "the standard simplifying assumption on problems of type Y is Z, and if you assume Z, then X is not hard." If you want more information, you need to ask a more precise or perhaps a more general question. – Tim Campion Mar 18 at 16:03

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.

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 $$C_*(\tilde X)$$.

• I see your point! The $C_{\ast}(Y)$ has more structure then just the structure of chain complex and the action of the fundamental group acts by preserving the structure of "$E_{\infty}$"-coring. Moreover, in my question I assume that we know more then just the homology, we know also the homotopy groups... – GSM Mar 20 at 17:09
• @GSM That's a fair point, but the $E_\infty$-structure itself is not very accessible (you can get the cup product structure, maybe the Dyer-Lashof action, but there's plenty of further structure there that is hard to read), so it hardly seems an improvement... – Denis Nardin Mar 26 at 6:42