# Reference request for cohomology of coverings

Let $B$ be a compact manifold, and $\hat{B}\to B$ be the maximal abelian covering of $B$; i.e. $\hat{B}$ is the quotient of the universal cover with respect to the commutator subgroup of $\pi_1(B)$. Given $H\subset H_1(B)$, we can further quotient $\hat{B}$ with respect to $H$ to get a covering $X\to B$ with group of deck transformation $G=H_1(B)/H$.

Is it true\false that every $G$-invaraint $\mathbb{Q}$-cohomology class on $X$ is a pull-back from $B$.

• For a finite group $G$ and coefficients prime to the order of $G$, this is explained in the answers to MO-question mathoverflow.net/questions/57071 – Matthias Wendt Nov 15 '14 at 16:59
• In general, there is the Cartan-Leray spectral sequence $H^p(G,H^q(X,M))\Rightarrow H^{p+q}(X/G,M)$, see Theorem 7.9, Section VII of Brown's book on cohomology groups. $H^0(G,H^q(X,M))$ appears in the $E^2$-term, but there may be differentials interfering... – Matthias Wendt Nov 15 '14 at 17:10
• Could you please formulate your situation precisely? I do not really understand the formulation with the quotients of maximal abelian covering... Generally, the problem will be that if $X$ and $G$ have sufficiently complicated rational cohomology, it will be difficult to see why the differentials $H^0(G,H^p(X,M))\to H^2(G,H^{p-1}(X,M))$ (and then the higher ones...) should be trivial. – Matthias Wendt Nov 15 '14 at 19:24
• For the special case of an Z covering, there is an easy approach due to Milnor (in his beautiful paper called Infinite Cyclic Coverings). For $X'\to X$ a Z-covering with covering group generated by t, there is a short exact sequence of chain groups $0 \to C_*(X') \to C_*(X') \to C_*(X) \to 0$ where the first map is $t_* -1$ and the second is $p_*$. A quick look at the associated long exact sequence gives the requested result. Perhaps this can be iterated (to get $Z^n$ coverings) and combined with easy results about finite coverings to get the general case. – Danny Ruberman Nov 17 '14 at 1:12
• @MatthiasWendt: Yes, of course you're right that this can be done via a spectral sequence. My point was that there's a more elementary approach where you see what's going on at the chain level. Plus, I really like that Milnor paper! – Danny Ruberman Nov 17 '14 at 16:01

This is false. It is false for first cohomology and the maximal abelian cover of the classifying space of the integral Heisenberg group. The integral Heisenberg group $H(\mathbb Z)$ of $3\times 3$ upper triangular matrices with integer entries and 1s on the diagonal. The abelianization is $\mathbb Z^2$ given by taking the two entries just above the diagonal. The commutator subgroup is $\mathbb Z$, that last entry. The infinite cyclic group has nontrivial homomorphisms to $\mathbb Q$, that is first cohomology classes. Since it is the commutator subgroup, these homomorphisms do not extend to the whole group, which is to say that the cohomology classes do not restrict from the whole group. The cohomology group is 1-dimensional, so there is not much room for the abelianization to act. Indeed, that the extension is central means that the action is trivial.

In geometric terms, the integral Heisenberg group is a subgroup of the real Heisenberg group $H(\mathbb R)$, the same thing, but with real entries. The real Heisenberg group is diffeomorphic to $\mathbb R^3$, parameterized by the three entries above the diagonal, and thus contractible. Thus the coset space $B=H(\mathbb R)/H(\mathbb Z)$ is a classifying space for the integral group. It maps to the coset space for the abelianization, with fiber the coset space for the the center. That is, it is a circle bundle over a torus; hence compact. The maximal abelian cover $\hat B$ is a circle bundle over $\mathbb R^2$, and thus has $H^1(\hat B;\mathbb Q)=\mathbb Q$. The action of the abelianization of the fundamental group is trivial. But this invariant class is not pulled back from $H^1(B;\mathbb Q)$, for it detects a commutator.

The other answers show that this is true when the cover is infinite cyclic, but induction fails. The Heisenberg group demonstrates how it fails. I will switch from invariants to coinvariants and cohomology to homology (abelianization). We want to analyze the Heisenberg group by peeling off the abelianization one step at a time. If we express it as an extension of a 1-dimensional group, it is a semidirect product $H(\mathbb Z)=\mathbb Z^2\rtimes \mathbb Z$, where the action is by shearing (ie, not diagonalizable). It shears the non-central elements into the center. Thus the coinvariants are 1-dimensional, the other part of the abelianization. The center is killed by the shearing. But when we take the second step to just the center, the element that sheared into it and killed it is gone, so center now contributes to the coinvariants.

• Thank you, Ben. I think I see where my thinking with the induction went wrong. $H^*(\hat{B})^{\mathbb Z}$ comes from $H^*(\hat{B}/{\mathbb Z})$, but only $H^*(\hat{B}/{\mathbb Z})^{\mathbb Z}$ comes from $H^*(\hat{B}/{\mathbb Z}^2)$. More relevantly, one needs to show that $H^*(\hat{B})^{\mathbb{Z}^2}$ comes from $H^*(\hat{B}/{\mathbb{Z}})^{\mathbb{Z}}$, and there was nothing in the inductive step about some auxiliary group. – Aleksey Nov 18 '14 at 0:43
• Is there a statement in the spirit of the original one which is true? It is true if the rank of $G$ is at most one. Is it true if the abelianization of the commutator subgroup of $\pi_1(B)$ is finite? or perhaps even the commutator subgroup itself is finite? – Aleksey Nov 18 '14 at 0:48
• @Aleksey There are examples with higher cohomology when the fundamental group is just $\mathbb Z^n$. – Ben Wieland Nov 18 '14 at 18:08
• @BenWieland. It sounds like there is no reasonable statement along the lines of the original question then. This is very enlightening. Many thanks, Ben, as well as Matthias and Dan. – Aleksey Nov 19 '14 at 3:21

I am not too comfortable with spectral sequences and in particular I do not see the obstruction statement in Matthias's answer. It seems there is a direct construction of an invariant cocycle for the total space of the fibration Matthias suggested. I would guess it is a translation of Matthias's answer into less technical terms.

Suppose $B=X/{\mathbb Z}$ is a CW-complex and ${\mathbb Z}$ acts freely on $X$ with generator $g: X\to X$. Let $\widetilde{B}\!=\!({\mathbb R}\!\times\!X)/{\mathbb Z}$ be the corresponding Borel construction with $(t,x)\sim(t\!-\!1,g\cdot x)$. Since ${\mathbb Z}$ acts freely on $X$, $\widetilde{B}$ is homotopy equivalent to $B$ by the projection to the second factor and the projection $X\!\to\!B$ corresponds to the inclusion $X\!\to\!\widetilde{B}$ of a fiber for the fibration $\widetilde{B}\!\to\!S^1$.

Let $\eta$ be a $k$-cocycle on $X$ representing a ${\mathbb Z}$-invariant cohomology class. Thus, $g^*\eta\!-\!\eta\!=\!\delta\mu$ for some $(k\!-\!1)$-cochain $\mu$. The chain groups $C_k(X,{\mathbb Q})$ and $C_{k-1}(X,{\mathbb Q})$ are freely generated by the simplices $g^s\!\circ\!\sigma_i$ and $g^s\!\circ\!\tau_j$ for some $k$-cells $\sigma_i$ and $(k\!-\!1)$-cells $\tau_j$. Define $$\widetilde\eta\in C^k(\mathbb{R}\times X)$$ by $$\widetilde\eta(\{r\}\!\times\!g^s\!\circ\!\sigma_i)=\eta(g^{r+s}\!\circ\!\sigma_i),\quad \widetilde\eta([r,r\!+\!1]\!\times\!g^s\!\circ\!\tau_i)=\mu(g^{r+s}\!\circ\!\tau_i).$$ This cochain is ${\mathbb Z}$-invariant and is a cocycle; the latter needs to be checked only on the $(k\!+\!1)$ cells of the form $[r,r\!+\!1]\!\times\!g^s\!\circ\!\sigma_i$. Thus, $\widetilde\eta$ descends to a cocycle on $\widetilde{B}$. The latter restricts to $\eta$ along the fiber of $\widetilde{B}\!\to\!S^1$ over $0\!\in\!S^1$.

Is this correct? Is this really not written in somewhere published?

• This answer was meant to give an alternative description of Matthias's answer for cyclic coverings. As Ben points out, this does not suffice for the inductive step and there is no hope of extending this construction even to ${\mathbb{Z}}^2$-coverings without additional assumptions. – Aleksey Nov 18 '14 at 0:51