Let me try to sum up the comments and understand at least the first half of the proof, i.e. how we get from a Liouville field to a contact form. First of all, we need two things.
Firstly, $\iota(X)$ is not a map applied to $X$, but rather the interior product, or interior derivative, of $\omega$ and $X$. Given a field $X$, the interior derivative of a form $\omega$ with respect to $X$, or rather the interior product or antiderivative, is the form:
$$\iota_X\omega(X_2,\dotsc,X_k)=\omega(X,X_2,\dotsc,X_k),$$
where $\omega$ is a $k$-form and $\iota_X\omega$ is, therefore, a $k-1$-form. So the field takes the place of the first vector in the $k$-tuple of vectors the form is applied to. This means that for each point $p$ we have $\iota_X(p)(X_2(p),\dotsc,X_k(p))=\omega(p)(X(p),X_2(p),\dotsc,X_k(p))$ if $X_2,\dotsc,X_k$ are fields, and of course with no evaluation of $X_i$ when $X_i$ is a vector. Naturally, if $\omega$ is a 1-form, the result is a function, which is often indicated by $\langle X,\omega\rangle$ or $\langle\omega,X\rangle$.
Secondly, $\mathcal{L}_X\omega$ is the Lie derivative of $\omega$ with respect to $X$, that is:
$$\mathcal{L}_X\omega=\frac{d}{dt}\Big|_{t=0}\phi^\ast_t\omega,$$
where $\phi_t$ is the flow of the field $X$. This gives a form of the same "rank" as $\omega$, so it makes sense to compare it to $\omega$. In particular, if $\omega$ is a symplectic form (i.e. a closed nondegenerate 2-form), a field for which $\mathcal{L}_X\omega=\omega$ is called a Liouville vector field. The Lie derivative relates to the exterior derivative and interior antiderivative by Cartan's (magic) formula:
$$\mathcal{L}_X\omega=d(\iota_X\omega)+\iota_Xd\omega.$$
This is a theorem, and can be proved as follows, by the arguments I took from this pdf. Firstly, we need a lemma.
Lemma
For any forms $\omega,\eta$ and field $X$, one has:
\begin{align*}
d\mathcal{L}_X\omega={}&\mathcal{L}_Xd\omega. \tag{1} \\
\mathcal{L}_X(\omega\wedge\eta)={}&\mathcal{L}_X\omega\wedge\eta+\omega\wedge\mathcal{L}_X\eta. \tag{2}
\end{align*}
Proof.
We know that $\phi^\ast d\omega=d\phi^\ast\omega$ by Lemma 9.14 on Lee's introduction to manifolds. Let us write out number (1):
$$d\mathcal{L}_X\omega=d\left(\frac{d}{dt}\Big|_{t=0}\phi^\ast_t\omega\right)=\frac{d}{dt}\Big|_{t=0}d(\phi^\ast_t\omega),$$
since $d$ and $\frac{d}{dt}\Big|_{t=0}$ are essentially derivatives and so can be exchanged, but due to Lee's lemma $d$ swaps with the pull-back, and the result of that swapping is precisely $\mathcal{L}_Xd\omega$, proving (1). As for (2), Lemma 9.9 from Lee gives us $\phi^\ast_t(\omega\wedge\eta)=\phi^\ast_t\omega\wedge\phi^\ast_t\eta$. With this:
$$\mathcal{L}_X(\omega\wedge\eta)=\frac{d}{dt}\Big|_{t=0}\phi^\ast_t(\omega\wedge\eta)=\frac{d}{dt}\Big|_{t=0}\phi^\ast_t\omega\wedge\phi^\ast_t\eta,$$
and I bet that at this point all that is left to do is to write things in a coordinate chart and compute the derivatives, and the equality will reduce to Leibniz's rule, the well-known rule for derivatives of products. $\square$
Now we can finally proceed to prove Cartan's formula. This is done by induction. According to here:
The interior product of a vector and a 0-form is understood to be 0, since there are no $(-1)$-forms.
This makes perfect sense. THis also makes the case of 0-forms in Cartan's formula easy, since the formula reduces to simply $\mathcal{L}_Xf=\iota_Xdf$, but:
$$\iota_Xdf=df(X)=Xf=\mathcal{L}_Xf,$$
the last equality being proved here as Theorem 2.3. Getting back to Cartan's formula, now we have it is true for $k$-forms with $k=0$. Suppose it works for $(k-1)$-forms. If $\omega$ is a $k$-form, it is a sum of terms of the form $fdx^{i_1}\wedge\dotso\wedge dx^{i_k}$, so by linearity of both sides of the formula it suffices to prove the formula for $\omega=fdx^1\wedge\dotso\wedge dx^k=dx^1\wedge\omega_1$, where $\omega_1=fdx^2\wedge\dotso\wedge dx^k$. By the above lemma:
$$\mathcal{L}_X\omega=\mathcal{L}_Xdx^1\wedge\omega_1+dx^1\wedge\mathcal{L}_X\omega_1.$$
Let us take the other side of the formula. We have:
\begin{align*}
d\iota_X(dx^1\wedge\omega_1)+\iota_Xd(dx^1\wedge\omega_1)\stackrel{(1)}{=}{}&d(\iota_Xdx^1\wedge\omega_1-dx^1\wedge\iota_X\omega_1)-\iota_X(dx^1\wedge d\omega_1)={} \\
{}\stackrel{(2)}{=}{}&d(\iota_Xdx^1)\wedge\omega_1+\underline{\iota_Xdx^1\wedge d\omega_1}+dx^1\wedge d\iota_X\omega_1-{} \\
&{}+\underline{\iota_Xdx^1\wedge\omega_1}+dx^1\wedge\iota_Xd\omega_1={} \\
{}\stackrel{(3)}{=}{}&d\mathcal{L}_Xx^1\wedge\omega_1+dx^1\wedge\mathcal{L}_X\omega_1={} \\
{}\stackrel{(4)}{=}{}&\mathcal{L}_Xdx^1\wedge\omega_1+dx^1\wedge\mathcal{L}_X\omega_1
\end{align*}
Let's follow the passages one by one:
- It is pretty easy to show that if $\beta$ is a $k$-form and $\gamma$ is a $h$-form then $\iota_X(\beta\wedge\gamma)=\iota_X\beta\wedge\gamma+(-1)^k\beta\wedge\iota_X\gamma$, and in our case we have $\beta=dx^1$ which is a 1-form so $(-1)^k=(-1)^1=-1$. Same line of thought for the second term, plus the fact that $d(dx^1)=0$, which follows from the general property that $d^2=0$.
- Indeed, $d$ and $\iota_X$ "distribute" in the same way over a wedge product. Apply this to each term and you get to the right of (2). Remember $\iota_Xdx^1$ is a 0-form so you get a $(-1)^0=1$.
- Remember $\iota_Xdx^1=Xx^1=\mathcal{L}_Xx^1$, which deals with the first term. The underlined terms cancel each other out. The wedge product is bilinear so you can collect a $dx^1$ and the other side of the wedge is $\iota_Xd\omega_1+d\iota_X\omega_1$, which by induction is $\mathcal{L}_X\omega_1$
- Finally remember how, by the lemma above, $\mathcal{L}_X$ and $d$ commute, and we finish these passages, which lead to the same result as we got for $\mathcal{L}_X\omega$, proving Cartan's formula for $\omega$, which proves its validity for $k$-forms by the above arguments.
We can finally proceed to proving the first implication of the proposition:
If $X$ is a Liouville vector field defined on a neighborhood of $Q$, which is a hypersurface in $M$, and $X$ is transverse to $Q$, then there exists a contact form $\alpha$ on $Q$ such that $d\alpha=\omega|_Q$.
Proof. We set $\alpha=\iota_X\omega|_Q$. Then:
$$d\alpha=d\iota_X\omega|_Q\stackrel\ast=d\iota_X\omega+\iota_Xd\omega=\mathcal{L}_X\omega=\omega,$$
$\ast$ being because $d\omega=0$ ($\omega$ is symplectic and therefore closed, which is exactly $d\omega=0$), and the last equality being since $X$ is Liouville. So $d\alpha=\omega$. Well, $\omega|_Q$ and $\omega$ are being identified. Oh wait. $\alpha$ is on $M$, not on $Q$. Now we have to show $\alpha|_Q$ is a contact form. What does that mean? Unfortunately, my copy of McDuff is unsearchable, so I have to ask Wikipedia. This is what I understand by reading there:
A contact form is one that gives a contact structure on $Q$.
A contact structure is a distribution of contact elements which is integrable at no point.
A contact element is simply a hyperplane in the tangent space to a point $p\in Q$, which is called its contact point.
A distribution means assigning to each point a subspace of the tangent space (in this case, a contact element) in such a way that for any point $p$ in the manifold one can find a neighborhood $U$ of $p$ and $k$ (in our case, 1 less than the dimension of $Q$) linearly independent vector fields $X_1,\dotsc,X_k$ on $U$ such that for $q\in U$ we have $X_1(q),\dotsc,X_k(q)$ spanning $\Delta_q$, where $\Delta_q$ is the subspace of $T_qQ$ which our distribution assigns.
A distribution is integrable at a point $p$ if one can find a manifold $N$, which is then called an integral manifold of the distribution an $p$, and $F:N\to M$ a one-to-one immersion which sends $T_qN$ inside $\Delta_{F(q)}$ for any $q\in N$.
Now, pulling back $\omega$ to $Q$ gives a closed form, so $d\omega|_Q=0$. $Q$ is a hypersurface in a symplectic manifold, so according to this comment it cannot be symplectic, and being closed and a 2-form $\omega|_Q$ therefore can't be nondegenerate, which implies the existence of a vector $Y$ such that $\omega|_Q(Y,T_qQ)=0$ for all $q$. Let us rename the vector to $Y_q$. Assuming "transversal to $Q$" means $X(q)\not\in T_qQ$ for all $q\in Q$, then since $\omega$ is nondegenerate it is evident that $\omega(X(q),Y)$ can't be 0, otherwise that would contradict the nondegenerateness, since for $v=Y$ we would have $\omega(w,v)=0$ for all $w\in T_qM$.
If you are lost, remember we are assuming $X$ is a Liouville field in a neighborhood of the hypersurface $Q$ and transverse to its tangent space at any point, and after showing $d\alpha=\omega$, where $\alpha=\iota_X\omega$, we are attempting to show $\alpha|_Q$ gives a contact form, for the definition of which see blockquotes above.
We set:
$$\xi_q=\{v\in T_qQ:\omega(X(q),v)=0\}.$$
This is by definition contained in $T_qQ$. Now, $\omega$ with $X(q)$ fixed is a 1-form, which goes from $T_qM$ to $\mathbb{R}$. It is either 0 or surjective. Since $\omega(X(q),Y)\neq0$, it is surjective, and therefore has a hyperplane as its kernel at each point. The kernel is precisely $\xi_q$. Leaving the "symplectic complement" thing alone, this $\xi_q$ is a distribution on $Q$. Why it is not integrable at any point is probably because non-integrability is equivalent to $\alpha\wedge(d\alpha)^k\neq0$, where $(d\alpha)^k$ is the wedge product of $d\alpha$ with itself $k$ times. How this is shown still escapes me. Let us get the symplectic complement thing done. The symplectic complement of a vector space is simply, if I guess correctly, the set of vectors which give 0 if given to the symplectic form as an argument together with any vector of the subspace, i.e. the vectors $v$ such that for all $w$ in the subspace one has $\omega(w,v)=0$. We know $\omega(X(q),Y)\neq0$, and the rest is $\xi_q$, so $\xi_q$ is surely the symplectic complement of $\mathrm{span}Y$. Since $\omega(X(q),X(q))=0$ by alternatingness, adding $X(q)$ to $\mathrm{Span}Y$ and getting $\mathrm{span}(Y,X(q))$ doesn't change the fact we have a symplectic complement.
So I am left with two questions:
- How can one prove symplectic (i.e. closed non-degenerate) 2-forms are only possible on even-dimensional manifolds?
- How does integrability equate to that wedge product condition, or otherwise, how does one show the distribution of $\xi_q$s is non-integrable at any point of $Q$?
Update:
Question 1 is solved by this.
Update 2:
Rereading and editing this answer, I realized I have another question, so I am going to rewrite all questions.
- How does non-integrability equate to that wedge product condition, or otherwise, how does one show the distribution of $\xi_q$s is non-integrable at any point of $Q$? Here is (hopefully) the answer.
- How does one prove that form, $\alpha\wedge(d\alpha)^k$, is nonzero?
Update 3:
After being answered in the link above, I have an answer for question 2 as well: $d\alpha=\omega$, so $d\alpha$ is nondegenerate on $\xi_q$, which equates to $\alpha\wedge(d\alpha)^k\neq0$, and to $(\Sigma,\alpha)$ being a contact manifold, and the proof is finally complete. Now let me see how much I understand of the converse.
