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? 



"Rounding the corners" to get contact boundary 
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:
<http://arxiv.org/abs/math/0403376>


