# Hodge decomposition of a symplectic form.

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

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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

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.

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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

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.

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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