I have the following definitions for a contact structures and hypersurface of contact type in my lecture:
1)A contact structure on a manifold $W^{2n+1}$ is a hyperplane field $\xi \subset TW$ which is maximally non-integrable, meaning that the local defining forms satisfy \begin{align*} \alpha \wedge d \alpha ^n \end{align*} is a local volume form.
An equivalent condition says that for any local defining form $\alpha$, the form $d \alpha$ is symplectic on the fibers of $\xi$.
My first question: Why is this equivalent? $d \alpha$ symplectic form on fibers of $\xi$ means that $(d \alpha)^n$ is a volume form on fibers of $\xi$.
2) $(M, \omega)$ symplectic. A hypersurface $W \subset M$ is called of contact type, if there exists a vector field $Y$ defined on a neighborhood of $W$ in $M$, s.t.
- $Y \pitchfork X$
- $L_Y \omega = \omega$
As a comment, it says that then $W$ is a contact manifold with the contact structure $ker ( \iota_Y \omega)$.
My second question: For this to give a contact structure on $W$, is the first condition necessary?
The first question is just a point-wise issue. If you take a basis $\{v_i\}$ of a fiber of $\xi$, then you can extend it by any tangent vector $v_0$ outside of $\xi$ to a basis of the tangent space in question. Now inserting these vectors into $\alpha\wedge (d\alpha)^n$ the only contribution comes from the term where $v_0$ is inserted into $\alpha$ (since all $v_i$ insert trivially into $\alpha$). But by construction $\alpha(v_0)\neq 0$, so since $(d\alpha)^n$ is a volume form on the fiber of $\xi$, we get a non-zero result.
In the situation of question 2, you first of all need $Y$ to be nowhere vanishing, since otherwise $i_Y\omega$ has zeros. Moreover, you get $d(i_Y\omega)=L_Y\omega=\omega$. If $i_Y\omega$ is a contact form on $W$, then $(i_Y\omega)\wedge\omega^{n-1}$ has to be a volume form on $W$. But $Y$ clearly inserts trivially into that form, so it can't be tangent to $W$ anywhere.