Take a complex nilpotent or solvable group $G$ with the right action by a co-compact lattice $\Gamma$ and conisder the quotient $G/\Gamma$. On this quotient right-invariant $1$-forms give a subspace of $H^{1,0}$. The group $G$ is acting on $G/\Gamma$ on the left and if it would presrve all the $1$-forms, $G$ would be abelian.
Torsten Ekedahl expained that what is following IS NOT CORRECT (the article of Hasegawa tells something different)
In fact, the simplest example of this kind is given by primary Kodaira surfaces (http://en.wikipedia.org/wiki/Kodaira_surface), they have two holomorphic $1$-forms.
These surfaces are described as quotinets of sovlable groups, for example, in an article of Keizo Hasegawa http://arxiv.org/PS_cache/math/pdf/0401/0401413v1.pdf
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 $(\mathbb{C}^*)^2$.
Best Answer
I think I've got it. This is only a sketch, and I am not an expert on singular complex analytic things, but it seems about right.
Let $f:X\to Y$ be a proper holomorphic map between complex manifolds of the same dimension $n$, and assume that some fiber has positive dimension. I want to find a rationally nontrivial element of the kernel of $f_\ast :H_{2k}(X)\to H_{2k}(Y)$ for some $k$.
Let $S\subset X$ be the union of all fibers having positive dimension. I presume that $S$ is a closed analytic subset of $X$. Let $Z$ be an irreducible component of $S$ of maximal dimension. Let $m$ be the dimension of $f(S)$. Let $k<n-m$ be the codimension of $Z$ in $X$. Choose a holomorphic map $j:D\to X$, where $D$ is the closed unit disk in $\mathbb C^k$, such that $j(0)\in Z$ and $j(D\backslash 0)\cap S=\emptyset$.
$f\circ j:D\to Y$ gives a relative $2k$-cycle for the pair $(Y,Y\backslash f(S))$, which I'll call $\Delta$. The $(2k-1)$-cycle $\partial \Delta$, given by the restriction of $f\circ j$ to the sphere $\partial D$, represents zero in $H_{2k-1}(Y\backslash f(S))$, because $H_{2k}(Y,Y\backslash f(S))=0$, because $2n-2m$, the (real) codimension of $f(S)$ in $Y$, is greater than $2k$. So there is a $2k$-chain $c$ in $Y\backslash f(S)$ with $\partial c=\partial\Delta $. In fact, the whole cycle $\Delta-c$ can be taken to be in a little ball, so that it is trivial in $H_{2k}(Y)$.
Because $f$ is finite to one over the complement of $f(S)$, we can arrange for $c$ to have a sort of branched cover, a chain $\tilde c$ in $X\backslash S$ whose boundary is a branched cover of $\partial\Delta$.
Now I want a chain $\tilde \Delta$ in $X$, an analytic chain that is a branched cover of $\Delta$ and has the same boundary as $\tilde c$. For this, make the fiber product of $f:X\to Y$ and $f\circ j:D\to Y$. In this fiber product let $\tilde D$ be the closure of the preimage of $D\backslash 0$. The resulting map $\tilde D\to X$ gives the desired $\tilde \Delta$.
The cycle $\tilde \Delta-\tilde c$ represents an element of $H_{2k}(X)$ which must have infinite order because its intersection number with the fundamental class of $Z$ is positive, because the only part of $\tilde \Delta-\tilde c$ that intersects $Z$ is the analytic part $\tilde \Delta$ (and it does intersect at least once, at $j(0)\in Z$).