2
$\begingroup$

Let $M$ be a projective nonsingular complex variety. Let $I$ and $J$ be two complex structures on M. We then have the corresponding Kälher classes $\omega_I$ and $\omega_J$ in $H^2(M, \mathbb{R})$, and cup product with these define the Lefschetz operators $L_I$ and $L_J$, which are endomorphisms of the real cohomology of $M$ of degree $2$. By the Hard Lefschetz theorem there exist adjoint operators $\Lambda_I$ and $\Lambda_J$ of degree $-2$. It is clear from definition that $[L_I, L_J]=0$. I am asking for an example in which $[\Lambda_I, \Lambda_J]$ is not zero.

Motivation for this question: I am currently reading the article https://link.springer.com/article/10.1007/s002220050166. After the first general sections, they show that for abelian varieties and hyperkähler manifolds the $\Lambda$ operators relative to any two complex structures commute. This seems to be a fact peculiar to these varieties; for abelian varieties, it follows after identifying the $\Lambda$ operators on $A$ with Lefschetz operators on the dual abelian variety, while for hyperkälher, the proof is reduced to the case when $I, J$ are part of a hyperkälher triple and then the assertion can be checked by an explicit computation.

However, following Huybrechts' book "Complex Geometry", we have $\Lambda_I = \star^{-1} L_I \star$, where $\star$ is a Hodge operator. Such a Hodge operator is determined by the Riemannian metric on the underlying real manifold $M$ (indipendent of the complex structure) and by the choice of an orientation, i.e., a generator of the top cohomology of $M$ normalized so to have integral 1 on $M$. Any complex structure determines an orientation via the class $\frac{\omega_I^n}{n!}$. This argument implies that, as long as the two complex structures determine the same orientation on $ M$, the corresponding $\Lambda$ operators commute because the Lefschetz operators do so. But I have the feeling that something must be wrong...

Could somebody help me clarifyg these things and give me some suggestions to build an explicit example in which these $\Lambda$ operators do not commute?

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ I thought that the reason $\Lambda_I$ and $\Lambda_J$ do not commute is that they are scaled by a factor depending on which part of the Lefschetz decomposition you're in. Thus, in general they should not commute on classes that are primitive for $\omega_1$ but not for $\omega_2$. $\endgroup$
    – R. van Dobben de Bruyn
    Commented May 6, 2018 at 15:59
  • 2
    $\begingroup$ Huybrechts explains in section 2 of this paper that different $\Lambda_I$ and $\Lambda_J$ commute on the level of forms, but this does not descend to cohomology because the Laplacian for $\omega_I$ does not commute with $L_J$ and $\Lambda_J$. But I would still love see an actual example worked out... $\endgroup$
    – R. van Dobben de Bruyn
    Commented May 6, 2018 at 20:38
  • $\begingroup$ Thank you very much! This clearifies my ideas. I will think more about examples, and try to post the outcome. $\endgroup$
    – S. S.
    Commented May 6, 2018 at 23:51

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .