Conformal-symplectic geometry ? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T22:27:17Z http://mathoverflow.net/feeds/question/53655 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/53655/conformal-symplectic-geometry Conformal-symplectic geometry ? Qfwfq 2011-01-28T22:00:30Z 2011-01-28T22:31:05Z <p>I think in priciple it's possible to consider a theory of "conformal-symplectic manifolds", in an analogous fashion as the usual conformal geometry. </p> <p>To spell out the spontaneous definitions: say that two symplectic forms $\omega_1$, $\omega_2$ on a smooth manifold $M$ are <em>conformal</em> 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 <em>conformal-symplectic</em> manifold. A smooth map $\varphi : M \to N$ between conformal-symplectic manifolds $(M,[\omega_1])$ and $(N,[\omega_2])$ is <em>conformal-symplectic</em> if $\varphi^*(\omega_2)\in [\omega_1 ]$.</p> <p>Just out of curiosity, I would like to ask:</p> <blockquote> <p>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)? </p> </blockquote> http://mathoverflow.net/questions/53655/conformal-symplectic-geometry/53656#53656 Answer by Igor Rivin for Conformal-symplectic geometry ? Igor Rivin 2011-01-28T22:05:58Z 2011-01-28T22:05:58Z <p>Yes, this has been considered (hasn't everything). See the following antique reference:</p> <p><a href="http://archive.numdam.org/ARCHIVE/CTGDC/CTGDC_1978__19_3/CTGDC_1978__19_3_223_0/CTGDC_1978__19_3_223_0.pdf" rel="nofollow">http://archive.numdam.org/ARCHIVE/CTGDC/CTGDC_1978__19_3/CTGDC_1978__19_3_223_0/CTGDC_1978__19_3_223_0.pdf</a></p> http://mathoverflow.net/questions/53655/conformal-symplectic-geometry/53661#53661 Answer by Eric O. Korman for Conformal-symplectic geometry ? Eric O. Korman 2011-01-28T22:31:05Z 2011-01-28T22:31:05Z <p>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 $df = 0$.</p>