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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?

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up vote 6 down vote accepted

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 $\partial M_1 \times M_2$.

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.

Here's a relevant article by Alex Oancea:

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Thank you for the relevant article. – Dheeraj Kulkarni Aug 26 '11 at 8:46

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