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 $B$.



Reference request for cohomology of coverings
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.




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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.


