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Question

The question asked is:

On a manifold $M$ equipped with a Riemann metric $g$ and a symplectic structure $\omega$, with $\ast$ the Hodge star and $\ast_s$ the symplectic star, does $\ast=\ast_s$ iff $(M,g,\omega)$ is Kähler?

Answer: no

For the reason posted below.

Definitions

Here the star operators $\ast$ and $\ast_s$ are defined in the usual way (see for example, equation 2.9 of Tseng and Yau Cohomology and Hodge Theory on Symplectic Manifolds: II):

$$ A\wedge \ast B := \langle A,B\rangle_{g^{-1}} dV_g$$ $$A\wedge \ast_{s} B := \langle A,B\rangle_{\omega^{-1}} dV_\omega $$

where $A$ and $B$ are $k$-forms, $\langle\cdot,\cdot\rangle_{g^{-1}}$ is the inner product associated to $g$, $dV_g$ is the volume form associated to $g$, and $\langle\cdot,\cdot\rangle_{\omega^{-1}}$ and $dV_\omega$ are defined analogously with respect to $\omega$.

Motivation

In seeking a mathematically natural link between Hilbert-space expositions of quantum mechanics and product space expositions, both the Hodge star operator and the symplectic star operator enter naturally (for example, in the context of Onsager theory). Moreover, in cases of practical quantum systems engineering interest it is commonly observed that:

  • the two star operations are identical, and
  • the dynamical state-manifold is Kählerian.

Does each of these observations mathematically imply the other? An engineer-friendly reference for this fact (if it is a fact) would be very welcome---it's not easy to find discussion of the practical relevance of the symplectic star operation.

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Just as a remark, I am beginning to appreciate that the postulate is (rather obviously) not true. For supposing that M is Kahler, then the volume forms $dV_g$ and $dV_\omega$ are identical. In which case for $B=A$ the definition yields $A\wedge\ast A \ne A\wedge\ast_s A$. I will verify this reasoning against a few concrete cases (like $S^2$), and post it as an answer. –  John Sidles May 17 '12 at 16:46
    
Hi, Hodge star and symplectic star can be defined pointwise, but Kahler identity requires integrability which is not a pointwise condition –  Guangbo Xu Jun 1 '12 at 13:17

2 Answers 2

up vote 3 down vote accepted

Dear John,

The symplectic Hodge star and Riemannian Hodge star (associated to a Kahler metric) are never identical to each other.

In fact, it is proved in Brylinski's paper on symplectic Hodge theory that on a Kahler manifold, for a $(p,q)$ form $\alpha$ we have that

$$* \alpha =(-1)^{p-q}*_s \alpha.$$ One can also find a detailed proof of the above identity in Guillemin's lecture notes on Hodge theory, which is available at

http://math.mit.edu/~vwg/shlomo-notes.pdf

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Congratulations on your first accepted answer! :) –  John Sidles May 29 '12 at 17:24
1  
I should probably add an remark that the above-mentioned Brylinski identify holds at the linear algebra level. As observed by Guangbo Xu here, the definition of Hodge star does not involve the integrability of the underlying complex structure. –  Yi Lin Jun 1 '12 at 14:55

To answer the question myself, the answer is (trivially) "no". A fast way to see this is via the following identities, valid for $k$-forms on a manifold of dimension $n$:

$$\ast\ast = (-1)^{k(n-k)}\ \mathrm{Id}$$

$$\ast_s\ast_s = \mathrm{Id}$$

Thus for odd-$k$ forms on even-$n$ manifolds, we have $\ast \ne \ast_s$.

Perhaps some amended version of the postulate could be proved … if anyone posts such amended postulate (or even provides some nice references), then I will rate that answer as "accepted".

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