Question about the dimension of a Contact (Symplectic) manifold

2011-02-21T01:52:29Z

I am reading about contact geometry and I have a question: Why do we only consider contact structure of an odd-dimension manifold? and the same question for definition of symplectic geometry?

I think for the contact geometry case, a reason is that we want $\alpha \wedge (d\alpha)^n$ to be a volume form. Am I right? I am not sure about that.

Thank you so much for your help.

P.S there is no tag for Contact-geometry.

Answer by José Figueroa-O'Farrill for Question about the dimension of a Contact (Symplectic) manifold

2011-02-21T09:52:21Z

And by popular request, here's my comment as an answer :)

Your guess is correct. Contact structures are structures associated to a one-form $\alpha$ with maximal rank. There are two cases:

for odd rank, you want $\alpha∧(d\alpha)^k$ to be nowhere vanishing for the largest possible $k$ allowed by dimension, and
for even rank, you want the same condition on $(d\alpha)^k$.

In the former case you have a contact structure and in the latter an exact symplectic structure.

More generally, symplectic forms are nondegenerate by definition. You can understand nondegeneracy of a 2-form $\omega$ pointwise, where it turns into the statement that an antisymmetric matrix has nonzero determinant. This can only happen if the dimension is even.

I'm assuming finite-dimensionality throughout. There is a reasonably well-developed theory of infinite-dimensional symplectic manifolds and presumably also of contact manifolds.

Question about the dimension of a Contact (Symplectic) manifold - MathOverflow

Question about the dimension of a Contact (Symplectic) manifold

Answer by José Figueroa-O'Farrill for Question about the dimension of a Contact (Symplectic) manifold