Trouble understanding proof of this proposition on contact type hypersurfaces

Question

I have finally started working on my thesis. I am reading the first reference book, which is Dusa McDuff, D. Salamon, Introduction to Symplectic Topology. It's a tough struggle, given my not-too-great experience with differential forms. I will recall a few concepts. A manifold $M$ is called symplectic if it is equipped with a closed non-degenerate differential 2-form $\omega$, which is also called symplectic form. A submanifold of $M$ is a subset $S\subseteq M$ such that the inclusion is an embedding, i.e. is a one-to-one immersion which is also a homeomorphism of $S$ onto its image. Alternately, it can be defined as a subset for which each point $p\in S$ has a neighborhood $U\subseteq M$ and a chart $\phi:U\to\mathbb{R}^n$ which is "adapted to $S$", that is for which $S\cap U$ maps exactly to the locus of points in $\phi(U)$ having the last $k$ coordinates all equal to 0. $\dim S$ is then $\dim M-k$. A differential 2-form is a field of bilinear and alternating forms, i.e. a map associating to each $p\in M$ a bilinear and alternating form on $T_pM$ such that locally, i.e. in a coordinate neighborhood, this filed is represented by $\sum_{i<j}a_{ij}(x^1,\dotsc,x^n)dx_i\wedge dx_j$ and $a_{ij}$ are differentiable. The exterior derivative is defined well enough on the linked Wikipedia page. That is, I'd say, just about all we need for the proof. Here are screenshots of the book containing the statement and proof of the troublesome proposition.

$\textbf{Proposition 3.58}$ Let $(M,\omega)$ be a symplectic manifold and $Q\subset M$ be a compact hypersurface. Then the following are equivalent.

$\bf (i)$ There exists a contact form $\alpha$ on $Q$ such that $d\alpha=\omega|_Q$.

$\bf (ii)$ There exists a Liouville vector field $X:U\to TM$, defined in a neighbourhood $U$ of $Q$, which is transverse to $Q$.

If these conditions are satisfied then $Q$ is said to be of $\textbf{contact type}$.

$\bf Proof:$ First assume that $\rm (ii)$ is satisfied and define $\alpha=\iota(X)\omega$. Then $$d\alpha=d\iota(X)\omega+\iota(X)d\omega={\cal L}_X\omega=\omega.$$ Since $T_qQ$ is odd-dimensional there exists a nonzero vector $Y\in T_qQ$ such that $\omega_q(Y,v)=0$ for every $v\in T_qQ$. Since $\omega$ is nondegenerate we have $\alpha_q(Y)=\omega_q(X(q),Y)\neq 0$. Hence $$\xi_q=\{v\in T_qQ|\omega_q(X(q),v)=0\}$$ is a hyperplane field on $Q$ and $Y$ is transversal to $\xi_q$. In fact $\xi_q$ is the symplectic complement of $\operatorname{span}\{X(q),Y\}$. This implies that $\omega=d\alpha$ is nondegenerate on $\xi_q$ and hence $\alpha$ restricts to a contact form on $Q$.

$\quad$ Conversly, suppose that $\alpha\in\Omega^1(Q)$ is a contact form such that $d\alpha=\omega|_Q$. Let $Y\in{\cal X}(Q)$ be the Reeb field of $\alpha$: $$\iota(Y)d\alpha=0,\qquad\alpha(Y)=1.$$ Choose a vector field $X\in{\cal X}(M)$ such that $$\omega(X,Y)=1,\qquad \omega(X,\xi)=0$$ on $Q$. (First choose any vector field $X_0$ such that $\omega(X_0,Y)=1$ on $Q$. Then for every $q\in Q$ there exists a unique vector $X_1(q)\in\xi_q$ such that $\omega(X_0+X_1,v)=0$ for every $v\in\xi_q$. Define $X=X_0+X_1$ on $Q$ and extend to a vector field on $M$.) Define $\phi:Q\times\Bbb R\to M$ by $$\phi(q,\theta)=\exp_q(\theta X(q)).$$ Then a simple calculation shows that $$\phi^{ \ast }\omega|_{Q\times\{0\}}=d\alpha-\alpha\wedge d\theta.$$ By Moser's argument there exists a local diffeomorphism $\psi:Q\times(-\varepsilon,\varepsilon)\to M$ such that $$\psi(q,0)=q,\qquad \psi^\ast\omega=e^\theta(d\alpha-\alpha\wedge d\theta).$$ The required Liouville vector field is $\psi_\ast(\partial/\partial\theta).$ $\tag*{$\square$}$

Oh well, a hypersurface in a manifold $M$ of dimension $n$ is simply an $(n-1)$-dimensional submanifold $H\subseteq M$. I am assuming $\iota$ here is the inclusion of $Q$ into $M$, i.e. $\iota:Q\hookrightarrow M$. My questions are numerous:

1. How do I interpret $d\iota(X)\omega$ and $\iota(X)d\omega$? $\iota:Q\to M$, so $d\iota:T_pQ\to T_pM$, $X$ is a vector field, so $d\iota(X)$ would be a vector field. However, $X$ would have to be on $Q$ to do that, and instead it is on $U$, a neighborhood of $Q$ in $M$. And anyway, even if $d\iota(X)$ were a field on $M$, how do I multiply it with $\omega$? What does the juxtaposition of those two things mean? $\iota(X)$ may be interpreted as $\iota^\ast(X)$, which means I composte $X$ and $\iota$. And again, what is $\iota(X)$ "multiplied by" (juxtaposed to) $d\omega$?
2. What is that $\mathcal{L}_X\omega$?
3. «Since $T_qQ$ is odd-dimensional, there exists» etc. Why?
4. The following «Since» is also obscure.
5. What does it mean for $Y$ to be «transversal to $\xi_q$»?
6. How does one prove the existence of the Reeb field for a 1-form?
7. The rest of the proof is like Elvish to me. In other words, almost evertything in this proof is incomprehensible to me. Any help is appreciated.

Answer 1

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:

1. 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 $d^2=0$.
2. 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 $(-1)^0=1$.
3. 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 $\mathcal{L}_X\omega_1$
4. 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:

1. How can one prove symplectic (i.e. closed non-degenerate) 2-forms are only possible on even-dimensional manifolds?
2. 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.

1. 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.
2. 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.