# Question about the dimension of a Contact (Symplectic) manifold

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.

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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 $\alpha \wedge (d\alpha)^k$ to be nowhere vanishing for the largest possible $k$ allowed by dimension, or even rank, with the same condition on $(d\alpha)^k$. In the former case you have a contact structure and in the latter an exact symplectic structure. Symplectic forms are nondegenerate by definition, so this can only happen if the dimension is even. (I'm assuming finite-dimensionality throughout.) –  José Figueroa-O'Farrill Feb 21 '11 at 2:16
@Jose: This should probably be an answer, not a comment. Since that the answer is a comment, I'm tempted to vote to close as "no longer relevant". –  Theo Johnson-Freyd Feb 21 '11 at 2:44
I agree that José should post his comment as an answer. –  Deane Yang Feb 21 '11 at 3:06
@Jose: That is great. Thank you. –  Phi Le Feb 21 '11 at 4:50

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:

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

2. 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.

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By the way, something I should have mentioned, if only in passing, is that one statement of the second law of thermodynamics is that a certain nowhere-vanishing one-form $\alpha$ (the so-called heat form) should fail (maximally) to be a contact structure; that is, $\alpha \wedge d\alpha = 0$. –  José Figueroa-O'Farrill Feb 22 '11 at 1:19