Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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)?

share|improve this question
add comment

2 Answers 2

up vote 7 down vote accepted

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 $df = 0$.

share|improve this answer
    
Oh, so it was indeed a "trivial" subject, at least for symplectic manifolds (not almost symplectic)... –  Qfwfq Jan 29 '11 at 1:25
add comment

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

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.