Projective embedding of symplectic manifolds - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T00:26:46Z http://mathoverflow.net/feeds/question/59733 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/59733/projective-embedding-of-symplectic-manifolds Projective embedding of symplectic manifolds YCho 2011-03-27T15:52:08Z 2011-10-26T20:22:12Z <p>Let $(M^{2n},\omega)$ be a symplectic manifold with an integral symplectic form $\omega$. Due to the work of M.Gromov and D.Tischler (M.Gromov "A topological technique for the construction of solutions of differential equations and inequalities", D.Tischler "Closed 2-forms and an embedding theorem for symplectic manifolds"), there exists a symplectic embedding $$(M,\omega) \rightarrow (\mathbb{C}P^{2n+1},\omega_{FS}),$$ where $\omega_{FS}$ denote by the Fubini-Study form on the projective space. For example, Kodaira-Thurston manifold is a symplectic submanifold of $\mathbb{C}P^5$. </p> <p>My questions are as follows : </p> <ol> <li><p>Is there an example of non-Kaehler symplectic manifold $(M,\omega)$ which can be embedded into $\mathbb{C}P^n$ for some $n \leq 4$? (There is no restriction of the dimension of $M$.) </p></li> <li><p>Is there an example of non-Kaehler symplectic manifold $(M,\omega)$ of dimension $2n$ which can be embedded into $\mathbb{C}P^{n+1}$? (I mean, $M$ is a submanifold of codimension 2)</p></li> </ol> <p>I really appriciate for your any comments. </p> http://mathoverflow.net/questions/59733/projective-embedding-of-symplectic-manifolds/61532#61532 Answer by Eigenbunny for Projective embedding of symplectic manifolds Eigenbunny 2011-04-13T10:44:41Z 2011-04-13T10:44:41Z <p>You can embed T^2 into CP^2, and I think by a suitable fibering construction, you should be able to embed the Kodaira-Thurston symplectic non-Kaehler manifold symplectically into a CP^2-bundle over CP^2. I think the bundle can be made trivial, yielding CP^2 \times CP^2 (have not checked the details). Obviously, it's hard to say what the cohomology class of the resulting symplectic form would be.</p>