The question.

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.

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.

  • $\begingroup$ Mayebe I m totally wrong, but does the Lefschetz fix point formula help: Integrate up your vector field u until some fix points occur. Hopefully the action on $H^{p,q}$ of $u$ integrate up gives one the same. $\endgroup$ – Sebastian Mar 25 '10 at 7:16
  • $\begingroup$ @Sebastian. Are you hoping to prove that the action is necessarily trivial? As far as I understand, the Lefschetz fixed point formula gives the number of fixed points of a (suitable) map (counted appropriately) in terms of the trace of its action on cohomology. There is certainly a Dolbeault version - due to Atiyah-Bott (ams.org/bull/1966-72-02/S0002-9904-1966-11483-0/…) - but I don't quite see how it helps, since it will only ever tell you about the trace of the action on Hp;q $\endgroup$ – Joel Fine Mar 25 '10 at 15:27

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

  • $\begingroup$ A comment to Dmitri's answer is given below but is too long to give as a comment. I write here as I believe that this will mean that he will be informed of my comments. $\endgroup$ – Torsten Ekedahl Apr 19 '10 at 18:19
  • $\begingroup$ Dear Torsten, thanks a lot for your comment! Indeed what you have said about surfaces shows that this can not work, I must have misinterpreted Hasegawa's article... $\endgroup$ – Dmitri Panov Apr 19 '10 at 19:02

These are comments on Dmitri's answer.

I don't think the surface example can work as all holomorphic forms on a compact surface are closed (a result due to Kodaira I believe). The Cartan formula $L_v = d\iota_v+\iota_vd$ shows that if all holomorphic forms are closed then vector fields act trivially.

If I understand the Hasegawa paper correctly he does not claim that the primary Kodaire surfaces are of the form $G/\Gamma$ for a complex Lie group $G$, only that the manifold underlying $G$ can be given a complex structure such that $\Gamma$ acts holomorphically.

Dmiri's method does work however. One example is given by the complex Heisenberg group (i.e., strict upper triangular $3\times3$-matrices) divided by the subgroup whose matrices have entries in the Gaussian integers. This then also gives an example of a compact complex manifold with non-closed holomorphic $1$-forms. In fact, for translation invariant $1$-forms the exterior differential, which is the map $d\colon \mathfrak g^\ast \to \Lambda^2\mathfrak g^\ast$ dual to the Lie bracket.

  • $\begingroup$ Thank you very much for clarifying Dmitri's answer. I hope you don't mind, but I felt I should accept his, since it was first and I guess made the main point, even if the specific example he gave wasn't right. If I could accept both answers, I would have done! $\endgroup$ – Joel Fine Apr 21 '10 at 9:36

Both the above examples are for compact $X$, which the original problem didn't specify. If you allow $X$ not to be compact, the question is much simpler. Take $X = \mathbb{C}$. Then $H^{0,0}(X)$ is the vector space of entire holomorphic functions, and $\partial/\partial z$ acts nontrivially.

  • 1
    $\begingroup$ Yes, sorry, I forgot to specify that I was really looking for compact examples. $\endgroup$ – Joel Fine Feb 8 '12 at 16:13

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