Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|cite|improve this question

1 Answer 1

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:

share|cite|improve this answer
Thank you for the relevant article. –  Dheeraj Kulkarni Aug 26 '11 at 8:46

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.