"Rounding the corners" to get contact boundary - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T22:29:15Z http://mathoverflow.net/feeds/question/73673 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/73673/rounding-the-corners-to-get-contact-boundary "Rounding the corners" to get contact boundary Dheeraj Kulkarni 2011-08-25T17:14:49Z 2011-08-25T17:39:05Z <p>Suppose we have symplectic manifolds $(M_1, \omega_1)$ and $(M_2, \omega_2)$ with non-empty boundary of contact . Often we need to deal with the product $M_1 \times M_2$ with the product symplectic structure. Can we round the corners to get a contact manifold as boundary? </p> http://mathoverflow.net/questions/73673/rounding-the-corners-to-get-contact-boundary/73678#73678 Answer by Tim Perutz for "Rounding the corners" to get contact boundary Tim Perutz 2011-08-25T17:39:05Z 2011-08-25T17:39:05Z <p>In that generality, the answer is no: a symplectic form $\omega$ on $X$ which has contact-type boundary is exact on $\partial X$. Yet $\omega_1 \oplus \omega_2$ need not be exact on $M_1\times \partial M_2$, nor on $\partial M_1 \times M_2$.</p> <p>It is possible, however, if $M_1$ and $M_2$ are Liouville domains, i.e., if the symplectic form $\omega_i$ is given as $d\theta_i$ for 1-forms $\theta_i$ whose dual vector field $\lambda_i$ points strictly outwards along the boundary. In fact, if you round corners sensibly, $\theta_1 \oplus \theta_2$ will have those same properties on the product.</p> <p>Here's a relevant article by Alex Oancea: <a href="http://arxiv.org/abs/math/0403376" rel="nofollow">http://arxiv.org/abs/math/0403376</a></p>