I’m reading Anahita Eslami Rad’s book Symplectic and Contact Geometry: A Concise Introduction. In Definition 4.1 (p. 116), she gives a definition of contact structure:
Let $M$ be a $(2n-1)$-dimensional manifold. A contact structure $\xi \subset TM$ on $M$ is a hyperplane field that is maximally non-integrable.
As she states, the maximally non-integrable condition in the definition means that (at least locally) there is no $n$-dimensional integral submanifold, and at most we have up to $(n-1)$-dimensional integral submanifolds. She also says that , as a corollary of the Frobenius theorem, an $n$-dimensional integral submanifold exists if and only if $\alpha \wedge (d\alpha)^{n-1}=0$, where $\xi = \mathrm{ker}\alpha$. I know other books and papers also use the condition that $\alpha \wedge (d\alpha)^{n-1}\neq 0$ to define a contact manifold. Here are my questions
- An $n$-dimensional integral submanifold exists if and only if $\alpha \wedge (d\alpha)^{n-1}=0$;
- There always exists $(n-1)$-dimensional integral submanifold even if $\alpha \wedge (d\alpha)^{n-1}\neq 0$.
Sadly, I have no idea about the second question. But I have some ideas about the first question. For the necessary direction, suppose that there exists an $n$-dimensional integral submanifold $N$ with respect to $\xi$. Then there exist $(n-2)$ $1$-forms $\alpha_1,…, \alpha_{n-2}$, such that $\{\alpha, \alpha_1,…, \alpha_{n-2}\}$ are linearly independent and they make up a closed differential ideal. So we have $$d\alpha= \eta\wedge \alpha+ \eta_1\wedge \alpha_1+ \cdots\eta_{n-2}\wedge \alpha_{n-2},$$ for some $1$-forms $\eta,\eta_1,…,\eta_{n-2}$. Then it follows immediately that $\alpha \wedge (d\alpha)^{n-1}=0$, since the $(n-1)$-power of $d\alpha$ means that some $\eta_j\wedge \alpha_j$ will be taken more that one time.
For the sufficient direction, suppose that we have $\alpha \wedge (d\alpha)^{n-1}=0$, which is equivalent to $(d\alpha)^{n-1}|_{\xi} =0$, i.e. $d\alpha$ is degenerate on $\xi$. We know that the anti-symmetric bilinear map $d\alpha$ has even rank on $\xi$. So the degenerate $2$-form $d\alpha$ may have rank $2n-2-2k$, with $k$ ranging from $1$ to $n-1$. Let’s consider the case where $d\alpha$ has rank $2n-4$. In this case the null subspace of $d\alpha$ is of dimension $2$, while the nondegenerate subspace $P$ of $d\alpha$ is of dimension $2n-4$. Note that in order to find an $n$-dimensional integral submanifold with respect to $\xi= \mathrm{ker}\alpha$, we only need to find other $(n-2)$ $1$-forms on $\xi$, such that $\alpha$ together with these forms makes up a closed differential ideal on $\xi$. We take a symplectic basis $u_1,v_1,…,u_{n-2}, v_{n-2}$ of $d\alpha$ on $P$, and expand it to a basis $u_1,v_1,…,u_{n-1}, v_{n-1}$ of $\xi$. Let $x_1,y_1,…,x_{n-1}, y_{n-1}$ be the dual basis. Then we may write $d\alpha$ on $\xi$ as $$ d\alpha= x_1\wedge y_1 + x_{n-2}\wedge y_{n-2}. $$ I’v wondered whether these $(n-2)$ $1$-forms $\{x_1,…,x_{n-2}\}$ together with $\alpha$ would make up a closed differential ideal. Unfortunately, they didn’t. And $\{x_1+ y_1,…,x_{n-2}+ y_{n-2} \}$ didn’t, too. Would somebody please complete my analysis of the first question or help me with the second question? Thank you very much!
