MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 11 down vote accepted

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

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.