# Conformal-symplectic geometry ?

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)?

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 $df = 0$.

• Oh, so it was indeed a "trivial" subject, at least for symplectic manifolds (not almost symplectic)... – Qfwfq Jan 29 '11 at 1:25

There is a notion of conformal symplectic structure related to what you are asking. I refer to locally conformally symplectic manifolds. These are manifolds $M$ equipped with a non-degenerate two-form $\omega$ and a good open cover $\left\{ U_{a}\right\}_{a\in I}$ such that for every $U_{a}$ there exists a function $e^{f_{a}}\in C^{\infty}(U_{a})$ satisfying

$d\left( e^{f_{a}}\omega|_{U_{a}}\right)=0$

This is equivalent to the existence of a flat real line bundle $L\to M$ with connection $\nabla$ that descends to a well-defined closed one-form $\varphi$ in $M$ satisfying

$d\omega + \varphi\wedge\omega =0$

One can define the coboundary operator $d_{\varphi} = d +\varphi$ on the complex of forms $\Omega^{\bullet}(M)$, whose cohomology is the so-called Lichnerowicz cohomology, which in general is not equivalent to the standard de Rahm cohomology. The two-form $\omega$ satisfies $d_{\varphi}\omega = 0$ and it is thus a cocycle. For further information you can check Izu Vaisman's papers from the 70's and 80's on locally conformally symplectic and K\"ahler manifolds.

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

• It doesn't look like the same definition. The author there defines a "conformally symplectic manifold" as a manifold that has a 2-form $\omega$ such that there's a 1-form $\rho$ such that $\mathrm{d} \omega = \rho \wedge \omega$. – Qfwfq Jan 28 '11 at 22:16
• Ops, sorry, what I said above was what I just spotted on page 4... But it was not a definition but a theorem. – Qfwfq Jan 28 '11 at 22:21
• Notice that this paper is talking about almost symplectic manifolds, so my comment does not apply. – Eric O. Korman Jan 28 '11 at 22:34