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



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> 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?
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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:



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> 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 $H^k(X, \mathbb{C})$?
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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.



Hodge decomposition in Betti cohomology
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 $(\mathbb{C}^\*)^2$.


