Is there a Riemannian metric $g$ on $\mathbb{R}^{2}$ with corresponding volume form $\omega= \sqrt{det(g_{ij})} dx \wedge dy$ and the corresponding Laplace operator $\Delta$ such that the space of $\Delta$ harmonic functions would be closed under the Poisson bracket arising from the symplectic form $\omega$?
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5$\begingroup$ No, this does not exist, as you can tell by using the fact that the $\Delta$harmonic functions depend only on the conformal structure, which you can assume to be the standard one. I.e., you can assume that the metric is of the form $g = \lambda(x,y)(\mathrm{d}x^2+\mathrm{d}y^2)$. Then the $\Delta$harmonic functions $u(x,y)$ are just the harmonic functions in the usual sense, i.e., $u_{xx}+u_{yy}=0$, and you can now check that there is no nonzero $\lambda(x,y)$ that yields the property you want. $\endgroup$ – Robert Bryant Jul 25 '16 at 20:26

$\begingroup$ @RobertBryant Thank you very much for your comment. What about if we replace the plane by an arbitrary orientable 2 manifold?Moreover, in order to generalize to $2n$ dimension, assume that $\Omega$ is the volume form associated with the metric $g$. Is there a symplectic two form $\omega$ on $\mathbb{R}^{2n}$ such that $\omega^{n}=\Omega$? $\endgroup$ – Ali Taghavi Jul 25 '16 at 21:16