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Let $X$ be a compact Kahler manifold and $A\subset H_1(X,\mathbb{Z})$ be a subgroup. Corresponding to $A$ there is an abelian covering $X_A \to X$ with $Deck(X_A)=H_1(X,\mathbb{Z})/A$. For example if $A=0$, $X_0$ is the maximal abelian cover of $X$ obtained as the quotient of the universal cover with respect to the commutator subgroup of $\pi_1(X)$.

What I am trying to figure out is when $H^*(X_A,\mathbb{C})$ is finitely generated (in terms of $A$ and $X$ and how $A$ is sited in $H_1(X,\mathbb{Z})$).

Some relevant background:

If the covering is free of rank $k$, Dwyer and Fried [1] looked at $H^*(X_A,\mathbb{C})$ as a module over $\mathbb{C}[x_1^{\pm1},\ldots, x_n^{\pm 1}]$ where $n=rank ~H_1(X,\mathbb{Q})$ and obtained a criterion in terms of how the support of this module intersects the image of $\mathbb{C}[x_1^{\pm1},\ldots, x_r^{\pm 1}]$ (which corresponds to embedding of $A$).

Suciu [2] expands [1] into more details; he introduces characteristic varieties of different kind which tell you about betti numbers of $H^*(X_A,\mathbb{C})$. These varieties can be very nasty in general. But thats too much details compared to the finitely generated question and I expect this particular question to have somewhat simpler characterization.

Finally, Green and Lazarsfeld [3] show that that in the Kahler case, these characteristic varieties are just translated tories; but they don't tell you about what the dimensions or the number of their components are.

If you too believe that there is no simple characterization of finitely generation property, can you give examples where this support variety is very nasty?

[1] HOMOLOGY OF FREE ABELIAN COVERS, I

[2] Characteristic Varieties and Betti Numbers of Free Abelian Covers

[3] Higher obstructions to deforming cohomology groups of line bundles

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  • $\begingroup$ Are you asking for finite-generation of the cohomology groups, over the group-ring of covering transformations? $\endgroup$ Commented Oct 31, 2014 at 16:12
  • $\begingroup$ No, over $\mathbb{Q}$ or $\mathbb{C}$; that one is f.g for sure. $\endgroup$ Commented Oct 31, 2014 at 16:16

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