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Suppose I have the symplectic manifold $(M, \omega)$. Now consider a function $C: M \rightarrow \mathbb{R}$ whose differential is non-zero. Then restricting to the submanifold of $M$ given by $C=0$ is not a symplectic manifold right? Since the resulting submanifold is not even dimensional, right?

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Indeed, an odd-dimensional manifold cannot be symplectic. – user31373 Jun 26 '12 at 20:50
up vote 2 down vote accepted

Yes, you are correct. In fact, given a smooth function $C$ on any smooth manifold $M$, the set given by $N=\{x\in M|C(x)=0\}$ may not be a smooth manifold. You can think of some examples. It's a good exercise.

Back to your question. Even if $N$ is a submanifold of $M$, it will not be symplectic. Because as you have said, it is not even dimensional. In fact, we have $\dim M=\dim N-1$. You can consider the following example: $\mathbb{R}^{2n}$ is symplectic manifold with the standard symplectic form $\omega=dx_1\wedge dy_1+\cdots dx_n\wedge dy_n$, where $(x_1,\cdots,x_n,y_1,\cdots, y_n)$ is coordinates of $\mathbb{R}^{2n}$. Consider the function $C(x_1,\cdots,x_n,y_1,\cdots, y_n)=x_1$ which is smooth. And $N=\{C=0\}$ is a smooth manifold which is nothing but $\mathbb{R}^{2n-1}$, which cannot be symplectic.

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Thanks, the example is really helpful. With the case were there is no smooth manifold $N$ you mean that the differential is 0 zero right? – Novo Jun 27 '12 at 18:23

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