Is the cup product of holomorphic $n$-forms with a fixed class injective?

Let $X$ be a compact Kahler manifold of complex dimension $n$. Fix a nonzero class $u \in H^1(X,T_X)$. This gives a linear morphism $$\phi_u : H^0(X,\Omega^n) \to H^{1}(X,\Omega^{n-1}), \quad \sigma \mapsto u \cup \sigma.$$ Is $\phi_u$ injective?

It is so for manifolds with $\Omega^n_X = \mathcal O_X$; the proof I've got is not hard but uses Ricci-flat Kahler metrics and the hard Lefschetz theorem so it cannot generalize to other situations. In the examples I know (curves, hypersurfaces in $\mathbb P^n$) we have $h^{n,0} \leq h^{n-1,1}$, so I haven't stumbled upon an obvious counterexample yet.

-

The answer to this question is negative in dimensions $\ge 3$. For example, take a quintic in $\mathbb CP^4$ and consider its blow up $X$ in $10^{100}$ points (just to be safe). Then the space $H^1(X, T_X)$ will be huge, since it parametrises deformations of the blown up variety and you can move points as you wish. So there will be non-zero $u\in H^1(X, T_X)$ so that you map is trivial.

Note that when you blow up the quintic $H^{2,1}$ does not change.

Also, this trick with blow ups will not work for Kahler surfaces as is explained for example, in appendix 1 in http://arxiv.org/pdf/1301.0478.pdf

-
I'd have upvoted just for "$10^{100}$ points (just to be safe)" alone. :D – Gunnar Þór Magnússon Jan 7 '13 at 9:25
I see, that Jason gave a different explanation that works for surfaces :) even though surfaces satisfy $h^{2,0}<h^{1,1}$ – Dmitri Jan 7 '13 at 9:26

The map $\phi_u$ is not always injective. Let $X$ be the blowing up along a large number of points inside a quartic surface in $\mathbb{P}^3$, or any surface $S$ with nonzero $h^{n,0}$. The weight $2$ Hodge structure of $X$ is a direct sum of the weight $2$ Hodge structure of $S$, the image of the pullback morphism, and a finite number of copies of the Hodge structure $\mathbb{Z}(-1)$, the Gysin images of the weight $0$ Hodge structures of the exceptional divisors. So, as you vary the blown up points in $S$, the corresponding variation of Hodge structures is trivial. Thus the Griffiths transversality map $\phi_u$ is zero for every $u$ coming from a variation of the points.

-