### 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.

nottrue. 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