Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Can anyone explain to me what the Hodge decomposition form of a symplectic form in a special symplectic manifold looks like?

share|improve this question
What exactly are you asking? What does a special symplectic manifold mean? Do you mean a Kaehler manifold? The standard Hodge decomposition does not make sense unless you have a Riemannian metric. Do you mean some other, less common, Hodge decomposition? If you mean the usual Hodge decomposition on a Kaehler manifold, then the symplectic (Kaehler) form is already harmonic, so it does not decompose further. –  Spiro Karigiannis Dec 6 '11 at 12:24
Mirjana, I guess you are referring to the notion of "special symplectic manifold" introduced by Alekseevsky et al as a variant of the more popular "special Kaehler" geometry. It would be good to edit your question to include a full definition and references, explaining how you know that such a complex manifold actually has a Hodge decomposition. The need for this level of detail is illustrated by the fact that Spiro, an expert on manifolds with special holonomy, does not understand what you are asking. –  Tim Perutz Dec 6 '11 at 15:23
@Tim, "how you know that such a complex manifold actually has a Hodge decomposition" I thought that every symplectic form can be written as a direct sum of a closed, coclosed and harmonic form? –  Mirjana Dec 7 '11 at 5:52
Mirjana: Ah, so I misunderstood too. But what metric do you want to use for the Hodge theory? Note that Prop. 4 of the paper of Alekseevsky et al is about the type decomposition w.r.t. $J$, not the Hodge decomposition. [You've been leaving comments, but please edit the question to make it clearer.] –  Tim Perutz Dec 7 '11 at 14:00
@Tim In Proposition 4 it is explicitly written that it is the Hodge decomposition... –  Mirjana Dec 7 '11 at 15:48
show 1 more comment

2 Answers

up vote 1 down vote accepted

Using the additional information that the OP provided in the comments to Yael Fregier's answer, I can elaborate as follows:

I still don't know what "special complex manifold" means, but in any case, I will assume the following. If $(M, J, \nabla)$ is a complex manifold with a connection $\nabla$ coming from a metric $g$, (that is, $\nabla$ is the Levi-Civita connection of $g$), then we get an associated symplectic form $\omega(X,Y) = g(JX, Y)$, and $\omega$ is parallel with respect to $\nabla$ if and only if $J$ is parallel with respect to $\nabla$, if and only if $J$ is integrable and $\omega$ is closed. That is, $(M, g, J, \omega)$ is Kaehler. In this case, $\omega$ is harmonic, so its Hodge decomposition is $\omega = \omega \in \Delta_2$, where $\Delta_2$ is the space of harmonic $2$-forms on $M$, using the notation of the OP.

If $\nabla$ does not come from a metric, you still need some metric to define the co-derivative $d^* = \delta$ of $d$, and to define the Laplacian $\Delta$. One can indeed do this with a different connection $\nabla$, as long as you have a metric. But in this case it is not clear to me what the symplectic form $\omega$ is, and how it is related to $J$ and $\nabla$.

Added later: I think I just realized that the OP is not asking about the Hodge decomposition of the form $\omega$ in particular, just the "Hodge decomposition" for a "special symplectic manifold." There is a version of "symplectic Hodge theory." See, for example, these notes by Victor Guillemin: http://www-math.mit.edu/~vwg/shlomo-notes.pdf --- I don't know if this is the same thing mentioned in Yael Fregier's answer. Otherwise, I remain confused by the question.

share|improve this answer
Thanks for the reference to Victor Guillemin's notes. I have taken a look, what I talked about corresponds to sections 4,5 and 8 of these notes (modulo taking duals). But he does more, in particular what I described was the linear version, whereas he also treats the global aspects in section 7. –  yael fregier Dec 6 '11 at 22:55
I found the definition and the Hodge decomposition in this paper: Special complex manifolds, D.V Alekseevsky, V. Cortes, C. Devchand. And I get confused when I start to read the proof of Proposition 4. –  Mirjana Dec 7 '11 at 0:31
add comment

I agree with Spiro Karigiannis that the question would need some clarifications. I slightly modify the question by replacing the first "form" by "for". In this case I also agree with Spiro about the fact that a metric would be needed to speak of Hodge decomposition... Unless you refer to the Hodge-Lepage decomposition which has a meaning for a symplectic manifold without metric :

Given a symplectic vector space $(V,\omega)$, one can associate to it two operators $\omega^+$ and $\omega^-$ which act on the exterior algebra on $V^*$ by respectively left multiplying a given form $\alpha$ by $\omega$ (i.e. $\omega^+(\alpha)=\omega\wedge\alpha$) or by contracting $\alpha$ by the Poisson bivector $\pi$ associated to $\omega$ ($\omega^-(\alpha)=i_\pi(\alpha)$). These two operators satisfy the relations of the Lie algebra $sl(2)$ and they cut out the space of differential forms into irreducible $sl(2)$-modules which are also modules over the Lie algebra of symplectomorphisms since the operators $\omega^+$ and $\omega^-$ are invariant under the action of this Lie algebra. This decomposition is called the Hodge-Lepage decomposition, and the highest weight vectors are called effective forms. One can find all the details, and some explicit formulas in Darboux coordinates in chapter 5 of the book "Contact Geometry and Nonlinear Differential Equations" by Kushner, Lychagin and Roubtsov (encyclopedia of Mathematics and its Applications (No. 101)) at Cambridge University Press.

share|improve this answer
You can't edit comments, so it's best to write it in a separate TeX editor, check that it looks good, and then paste it in. You can then delete your previous comments. –  Spiro Karigiannis Dec 6 '11 at 14:21
Ok, here is the definition of special symplectic manifold: A special symplectic manifold (M,J,∇,ω)is a special complex manifold (M,J,\nabla) together with a \nabla−parallel symplectic structure \omega. And there is a theorem(Hodge decomposition)which asserts that \Omega^{k}=Im d_{k-1} + Im \delta_{k+1} + \Delta_{k},where \Omega^{k}is a set of all k−forms,d_{k}maps \Omega^{k} to Ω^{k+1}, δ^{k}=∗ d ∗ maps Ω^{k} to Ω^{k−1}, where ∗ is Hodge star operator. \Delta_{k} is the space of k-harmonic forms. –  Mirjana Dec 6 '11 at 14:53
add comment

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.