Conformal-symplectic geometry ? - MathOverflow

Conformal-symplectic geometry ?

2011-01-28T22:00:30Z

I think in priciple it's possible to consider a theory of "conformal-symplectic manifolds", in an analogous fashion as the usual conformal geometry.

To spell out the spontaneous definitions: say that two symplectic forms $\omega_1$, $\omega_2$ on a smooth manifold $M$ are conformal to each other if there is a smooth positive function $\lambda \in \mathcal{C}^{\infty}(M,\mathbb{R}^{+})$ such that $\omega_1=\lambda\cdot \omega_2$ on $M$. Call a pair $(M,[\omega])$, with $[\omega]$ a conformal class of symplectic structures, a conformal-symplectic manifold. A smooth map $\varphi : M \to N$ between conformal-symplectic manifolds $(M,[\omega_1])$ and $(N,[\omega_2])$ is conformal-symplectic if $\varphi^*(\omega_2)\in [\omega_1 ]$.

Just out of curiosity, I would like to ask:

Has such a theory been considered or studied? What can be said about these structures (provided it doesn't turn out to be somehow a "trivial" subject)?

Answer by Igor Rivin for Conformal-symplectic geometry ?

2011-01-28T22:05:58Z

Yes, this has been considered (hasn't everything). See the following antique reference:

http://archive.numdam.org/ARCHIVE/CTGDC/CTGDC_1978__19_3/CTGDC_1978__19_3_223_0/CTGDC_1978__19_3_223_0.pdf

Answer by Eric O. Korman for Conformal-symplectic geometry ?

2011-01-28T22:31:05Z

If the manifold has dimension bigger than 2, I think the conformal class of $\omega$ is just $k\omega$ for constants $k$. Locally, by Darboux we can write $\omega = \sum_i dq^i \wedge dp_i$. If the dimension is greater than 2, the only way for $0 = d(f\omega) = df \wedge \omega$ is if $df = 0$.