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The broad, generic and badly posed question may be formulated in this way:

Let $X$ be a compact Kälher manifold (or even a projective one). If one considers the Hodge decomposition $H^k(X, \mathbb{C}) = \bigoplus_{p+q=k} H^{p,q}(X)$ and interprets the left hand side as Betti cohomology $\text{Hom}(\pi_1(X), \mathbb{C})$, which homomorphisms fit into which direct summand?

I'm sure this is well studied, but couldn't find any reference. Without a precise question there is probably not much to say, so let us look at a few motivation and give more precise statements below.

Motivations

I was trying to study the easy cases of Simpson's non-abelian cohomology (cfr. for example his paper "Moduli of representations of the fundamental group of a smooth projective variety"), and the first possible idea is to reduce oneself to the abelian case. So some questions which are of interest in the non-abelian case should become easy in the abelian context, and the above provides the basics for one of them: Simpson defines Betti and Dolbeaut non-abelian cohomology, then proves that the constructed spaces are in fact naturally homeomorphic, and studies interactions between objects defined in these "different worlds". In the abelian context this boils down to interpreting the cohomology as singular cohomology on one side and as the direct sum of the Dolbeaut cohomology on the other.

More specific questions

One can in particular consider any given subgroup $G \subset \mathbb{C}$, and seek only those homomorphisms which take values in this subgroup (if $G$ is a subfield this means considering $H^k(X, G)$, otherwise there may be some torsion in this cohomology group which is ignored if we just take the subgroup of $H^k(X, \mathbb{C})$ as above). Then, can one say whether this subgroup, call it $H_G \subset H^k(X, \mathbb{C})$, is contained in some pure part of the Hodge decomposition $H^{p,q}(X)$? I would have thought that this depended entirely on the nature of $X$, but on second thought it seems that something can always be said: for example, if $k = 1$, then $H_{\mathbb{R}}$ is never contained in any pure part of some weight (since they are interchanged by conjugation), while some work I did would lead one to think that if $G = \lbrace k \pi i \rbrace_{k \in \mathbb{Z}}$, then $H_G \subset H^{0,1}(X)$, at least for $X$ projective such that $H_1(X, \mathbb{Z})$ has no torsion (I'm absolutely not sure of the rightness of this proof, and if this fact was true there should really be a more straight idea then the one I thought of). Furthermore, for $k = 2$, then always $H_{\mathbb{R}} \cap H^{1,1}(X) \neq \emptyset $ since there is the Kähler form, while the question if $H_{\mathbb{Z}}$ is a subset of $H^{1,1}(X)$ is by Kodaira embedding theorem equivalent to $X$ being projective.

Hence a slightly more focused question could be:

For which classes of compact Kähler manifolds and which (in case cyclic) subgroups $G \subset \mathbb{C}$ one knows the Dolbeaut type of the elements of $ H_G $ inside some $H^k(X, \mathbb{C})$?

The question may become more interesting (hence, probably won't have any sensitive answer) if one allows (in case small) variations of the Hodge structure, by considering a family $X \to S$. I do not know much about variations of Hodge structures, so this could be well known. In this case, what is surely known is that the question for $k = 2$ and $G = \mathbb{Z}$ is false (small deformations of projective varieties need not be projective) and for $G = \mathbb{R}$ is true (small deformation of Kähler manifolds are Kähler). In this case one may want to drop the smoothness hypothesis, or, on the other hand, require the morphism $X \to S$ to be both projective and smooth.

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2 Answers

up vote 2 down vote accepted

This is really a comment on Donu Arapura's answer, but it seemed large enough to deserve it's own post. Working again in the case of $GL_1$, Simpson considers three spaces:

$M_{betti}$: The space of $\mathbb{C}^*$-local systems on $X$. If you like, you can think of this as smooth complex line bundles with an integrable connection.

$M_{DR}$: The space of holomorphic line bundles $L$ on $X$, equipped with an integrable holomorphic connection. (Being compatible with a holomorphic connection forces $c_1(L)$ to be $0$ in $H^2(X)$.)

$M_{Dol}$: The space of holomorphic line bundles $L$ on $X$, with $c_1(L)= 0$ in $H^2(X)$, and equipped with an $\mathrm{End}(L)$-valued $1$-form. $\mathrm{End}(L)$ is naturally isomorphic to $\mathcal{O}$, so this is just a $1$-form, but it is the $\mathrm{End}(L)$ version which generalizes to higher $GL_n$.

The first space is $\mathrm{Hom}(\pi_1(X), \mathbb{C}^*) = H^1(X, \mathbb{Z}) \otimes \mathbb{C}^*$. In this latter form, it has a natural algebraic structure, as a multiplicative algebraic group.

The second space is an affine bundle over $\mathrm{Pic}^0(X)$. Each fiber is a torsor for $H^0(X, \Omega^1)$, so we can describe this space by giving a class in $H^1(\mathrm{Pic}^0(X), \mathcal{O}) \otimes H^0(X, \Omega^1)$. By GAGA, this cohomology group on $\mathrm{Pic}$ is the same algebraically or analytically; viewing it algebraically, we get an algebraic structure on $M_{DR}$.

The third space is simply $\mathrm{Pic}^0(X) \times H^0(X, \Omega^1)$ (for larger $n$, this vector bundle can be nontrivial). For obvious reasons, this has an algebraic structure.

The relations between these spaces are the following: All three are diffeomorphic. $M_{betti}$ and $M_{DR}$ are isomorphic as complex analytic varieties, but have different algebraic structure. $M_{Dol}$ and $M_{DR}$ are not isomorphic as complex analytic varieties, rather, $M_{Dol}$ is the vector bundle for which the affine bundle $M_{DR}$ is a torsor.

You might enjoy writing this all down in coordinates for $X$ an elliptic curve. As smooth manifolds, all three spaces should be $(\mathbb{C}^*)^2$.

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Okay, so it seems that this is turning to a discussion on my thesis... I will stop before the discussion on MO overtake my work on the subject, otherwise I risk to become out of work ;-) I take from the absence of an answer that there is no hope of having a satisfying one? Indeed, I doubt of it. But it still puzzles me if it is really true that for any projective manifold $X$ (if needed, with $H_1(X)$ torsion-free) it is true that the homomorphisms in $H^1(X, \mathbb{C})$ with values in $\lbrace k \pi i \rbrace_{k \in \mathbb{Z}}$ correspond to form of type $(0,1)$, that seems an odd result! –  Marco Spinaci Feb 2 '11 at 11:10
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No, certainly not. Look at an elliptic curve: $\mathbb{C}/\langle 1, \tau \rangle$. Write $\alpha$ and $\beta$ for the cycles which, in the universal cover, run from $0$ to $1$ and $\tau$ respectively. Let $f$ be a map $\mathbb{Z}^2 \to \mathbb{C}$. Then $f$ comes from a $(0,1)$-form if and only if there is a complex number $a$ such that $f(\alpha)=a$ and $f(\beta) = a \overline{\tau}$. Or did I misunderstand you? –  David Speyer Feb 2 '11 at 12:15
    
Obviously you're right, and just as obviously the fact was much easier than anything I was thinking of... And suddenly I realize that most of the questions written above are meaningless! Well, let's hope that they will be useful to other confused grad students as me! –  Marco Spinaci Feb 2 '11 at 13:05
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These answers are fine, but somehow do not seem to address directly the original question, which is actually asking about unipotent representations of $\pi_1$ of index two. I haven't looked carefully at it, but I think a pretty good answer is contained in the paper arxiv.org/PS_cache/math/pdf/0305/0305258v1.pdf by Silke Lekaus (which complements earlier work of Hain). I believe the short answer is that the two parts of the Hodge decomposition correspond to unipotent bundles with trivial underlying bundle and those with trivial Higgs field. –  Minhyong Kim Feb 2 '11 at 19:46
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Marco,

Since I don't fully understand what you are asking, let me instead discuss what you say is motivating you. If it isn't relevant, you can say so. When $X$ is a compact Kaehler manifold then what Simpson calls the Betti moduli space $M_B(X)$ is the set of semisimple representations in $Hom(\pi_1(X),GL_n(\mathbb{C}))/conjugacy$. This can be made into a variety (actually a scheme). When $n=1$, this is simply $$Hom(\pi_1(X),\mathbb{C}^*)=H^1(X,\mathbb{C}^*)$$ On the other side, the "Dolbeault moduli space" $M_{Dol}(X)$ is the moduli space of Higgs bundles (satisfying appropriate conditions). Again, in the rank one case, this simply the cotangent bundle of the Picard torus $T^*Pic^0(X)= Pic^0(X)\times H^0(X,\Omega_X^1)$. The correspondence in this case can be deduced from standard Hodge theory; it is sketched on page 21 of Simpson's Higgs bundles and local systems, and it is worthwhile to fill in the details yourself.

One thing you will notice is that this story doesn't really have much to do with cohomology beyond $H^1$. Actually, this isn't quite true. Even though the FAQ discourages discussion of open problems, let me mention one that may have some relevance to your question:

Is the image of the natural map $H^*(\pi_1(X),\mathbb{C})\to H^*(X,\mathbb{C})$ a sub Hodge structure?

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Thanks for the clarification. In fact, I was only interested in $H^1(X, \mathbb{C})$, but it seemed natural to make a parallel with higher cohomology. The motivation is just what led me to the question, my original problem was of understanding the intersection between a subset of $Hom(\pi_1(X), \mathbb{C}^*)$ and one of $T^* Pic^0(X)$ and I tried to translate it into an "abelian" Hodge theory question which sounded as above. This didn't help, and I solved the original problem otherwise, but the curiosity for what is known remained - furthermore I didn't mean to ask help on my thesis work ;-) –  Marco Spinaci Feb 1 '11 at 17:27
    
Just added some comments about higher cohomology. –  Donu Arapura Feb 1 '11 at 17:31
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