The question.
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Let $(X, J)$ be a complex manifold and $u$ a holomorphic vector field, i.e. $L\_uJ = 0$. The holomorphicity of $u$ implies that the Lie derivative $L\_u$ on forms preserves the (p,q) decomposition and also that it commutes with $\bar{\partial}$. From this it follows that $u$ acts infinitesimally on the Dolbeault cohomology groups $H^{p,q}(X)$ of $X$. My question is, does anyone know of an example in which this action is non-trivial?


Some context.
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To give some context, first note that the analgous action for de Rham cohomology is always trivial: If $M$ is any smooth manifold and $v$ any vector field, then the formula $L\_v = d \circ i\_v + i\_v \circ d$ shows that the infinitesimal action of $v$ on de Rham cohmology is trivial. (This is an instance of the more general fact that homotopic maps induce the same homomorphisms on singular cohomology. The field $v$ generates diffeomorphisms which are by construction isotopic to the identity map.)


Returning to Dolbeault cohomology, suppose we know that each Dolbeault class is represented by a $d$-closed form. (For example, this is true if $X$ is a compact Kähler manifold, by Hodge theory.) Then the action is necessarily trivial. The proof is as follows. Let $\alpha$ be a $\bar{\partial}$-closed (p,q)-form which is also $d$-closed. Then we know that $L\_u \alpha = d(i\_u \alpha)$ is also of type (p,q). So, 
$$
L\_u\alpha = \bar{\partial}\left((i\_u\alpha)^{p, q-1}\right) + \partial\left((i\_u \alpha)^{p-1, q}\right)
$$
and the other contributions $\bar{\partial}((i\_u\alpha)^{p-1,q}$) and $\partial((i\_u\alpha)^{p,q-1})$ vanish. Now the fact that $\bar\partial((i\_u\alpha)^{p-1,q}) = 0$ and our hypothesis imply that there is a (p-1, q-1)-form $\beta$ such that 
$$
(i\_u\alpha)^{p-1,q}+ \bar\partial \beta
$$
is closed. Hence
$$
\partial \left((i\_u\alpha)^{p-1,q}\right) = \bar\partial \partial \beta
$$
and so
$$
L\_u\alpha = \bar \partial \left( (i\_u \alpha)^{p,q-1} + \partial \beta\right)
$$
which proves the action of $u$ on $H^{p,q}(X)$ is trivial.



Holomorphic vector fields acting on Dolbeault cohomology
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>


