We know every contact manifold admits a symplectic submanifold, e.g., by Giroux's bijection : if $\omega$ is a contact form for a 3-manifold $M^3$ , then $d \omega$ is symplectic on the fibers of the open book associated to the contact structure $(M^3 , \omega)$. Now, if we are given a symplectic manifold $M^{2n}$: when does $M^{2n}$ admit a contact submanifold? I think it has to see with the existence of a Liouville vector field, but I am not sure. Thanks for any answers, refs., etc.
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1$\begingroup$ You can hope for a hypersurface and a chosen primitive for $\omega$ in a neighborhood of it. This is in the spirit of the "canonical" example $(M,\lambda)\subset (\mathbb{R}\times M,d(e^s\lambda))$. $\endgroup$– Chris GerigDec 11, 2014 at 5:35
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$\begingroup$ Thanks, can this be extended to higher dimensions, i.e. 4-or-higher? $\endgroup$– ContactycDec 11, 2014 at 5:42
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As Chris Gerig indicates in his comment, the correct notion is that of a contact-type hypersurface. These always exist, even locally. You could just take the boundary of a Darboux ball, which always exists. This is a contact-type hypersurface. Or, if you can find a Lagrangian submanifold (which you always can, e.g. a small torus in a Darboux ball) then the boundary of a Weinstein neighbourhood is a contact-type hypersurface. Or, if you are in a projective variety and can find an ample normal crossing divisor Poincaré dual to the symplectic form then the boundary of a neighbourhood of the divisor is a contact-type hypersurface (see Seidel's "A biased view of symplectic cohomology").
answered Dec 11, 2014 at 8:36
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$\begingroup$ The title should be "A biased view of symplectic cohomology". $\endgroup$– YHBKJDec 16, 2014 at 3:26