[proofplan] The Liouville identity restricts to $\Sigma$ and gives $d\alpha=i^*\omega$, so the kernel of $d\alpha$ is exactly the kernel of the restricted symplectic form on $T\Sigma$. At each point, the restriction of a symplectic form to a hypersurface has a one-dimensional kernel; transversality of $Y$ forces $\alpha$ to be nonzero on that line. A symplectic complement to the characteristic line then shows that $\alpha\wedge(d\alpha)^{n-1}$ is nowhere zero. Finally, the Reeb equations say precisely that $R_\alpha$ is the unique vector in this characteristic line normalized by $\alpha(R_\alpha)=1$. [/proofplan]

[step:Restrict the Liouville identity to the hypersurface]Let $i:\Sigma\to M$ be the inclusion. Since $\alpha=i^*(\iota_Y\omega)$ and exterior differentiation commutes with pullback, \begin{align*} d\alpha=d\,i^*(\iota_Y\omega)=i^*d(\iota_Y\omega)=i^*\omega. \end{align*} Thus, for every $x\in\Sigma$ and every $u,w\in T_x\Sigma$, \begin{align*} (d\alpha)_x(u,w)=\omega_x(u,w). \end{align*} Consequently, \begin{align*} \ker((d\alpha)_x)=\{u\in T_x\Sigma:(d\alpha)_x(u,w)=0\text{ for all }w\in T_x\Sigma\} \end{align*} equals \begin{align*} \ker(\omega_x|_{T_x\Sigma})=\{u\in T_x\Sigma:\omega_x(u,w)=0\text{ for all }w\in T_x\Sigma\}. \end{align*} Since this holds for every $x\in\Sigma$, the bundle kernels satisfy \begin{align*} \ker(d\alpha)=\ker(\omega|_{T\Sigma}). \end{align*}[/step]

[guided]We first translate the Liouville condition into an identity on the hypersurface. The one-form $\alpha\in\Omega^1(\Sigma)$ is defined by pulling back the one-form $\iota_Y\omega\in\Omega^1(U)$ along the inclusion map $i:\Sigma\to M$: \begin{align*} \alpha=i^*(\iota_Y\omega). \end{align*} The [exterior derivative](/theorems/1525) is natural with respect to pullback, so \begin{align*} d\alpha=d\,i^*(\iota_Y\omega)=i^*d(\iota_Y\omega). \end{align*} Because $Y$ is assumed to be Liouville on $U$, we have \begin{align*} d(\iota_Y\omega)=\omega \end{align*} on $U$. Therefore \begin{align*} d\alpha=i^*\omega. \end{align*} This identity means the following pointwise statement. If $x\in\Sigma$ and $u,w\in T_x\Sigma$, then \begin{align*} (d\alpha)_x(u,w)=\omega_x(u,w). \end{align*} Hence a vector $u\in T_x\Sigma$ lies in $\ker((d\alpha)_x)$ exactly when it pairs to zero with every tangent vector $w\in T_x\Sigma$ under $\omega_x$. That condition is precisely the definition of the kernel of the restricted two-form $\omega_x|_{T_x\Sigma}$: \begin{align*} \ker((d\alpha)_x)=\ker(\omega_x|_{T_x\Sigma}). \end{align*} Since this equality is pointwise for every $x\in\Sigma$, it gives the claimed equality of line subbundles inside $T\Sigma$: \begin{align*} \ker(d\alpha)=\ker(\omega|_{T\Sigma}). \end{align*}[/guided]

[step:Identify the restricted kernel as a characteristic line] Fix $x\in\Sigma$, and set \begin{align*} V:=T_xM, \qquad H:=T_x\Sigma. \end{align*} Then $(V,\omega_x)$ is a finite-dimensional symplectic [vector space](/page/Vector%20Space) and $H\subset V$ is a hyperplane. Define the symplectic complement of $H$ in $V$ by \begin{align*} H^{\omega_x}:=\{v\in V: \omega_x(v,h)=0 \text{ for all } h\in H\}. \end{align*} By the [dimension formula for symplectic complements](/theorems/10039) from [citetheorem:10039], \begin{align*} \dim H+\dim H^{\omega_x}=\dim V. \end{align*} Since $\dim H=2n-1$ and $\dim V=2n$, we get \begin{align*} \dim H^{\omega_x}=1. \end{align*} Moreover $H^{\omega_x}\subset H$. Indeed, if a nonzero $u\in H^{\omega_x}$ were not in $H$, then $V=H\oplus \mathbb R u$. Since $\omega_x(u,H)=0$ by definition of $H^{\omega_x}$ and $\omega_x(u,u)=0$ by skew-symmetry, we would have $\omega_x(u,V)=0$, contradicting nondegeneracy of $\omega_x$. Therefore \begin{align*} \ker(\omega_x|_H)=H\cap H^{\omega_x}=H^{\omega_x} \end{align*} is one-dimensional. Denote this line by \begin{align*} L_x:=\ker(\omega_x|_{T_x\Sigma}). \end{align*} [/step]

[step:Show that the restricted Liouville form is nonzero on the characteristic line] Let $v\in L_x$ with $v\ne 0$. Since $Y$ is transverse to $\Sigma$, the vector $Y_x\in T_xM$ does not lie in $T_x\Sigma$, and hence \begin{align*} T_xM=T_x\Sigma\oplus \mathbb R Y_x. \end{align*} If $\omega_x(Y_x,v)=0$, then for every vector $z\in T_xM$ we can write $z=w+cY_x$ with $w\in T_x\Sigma$ and $c\in\mathbb R$, and obtain \begin{align*} \omega_x(v,z)=\omega_x(v,w)+c\,\omega_x(v,Y_x)=0. \end{align*} Here $\omega_x(v,w)=0$ because $v\in L_x$, and $\omega_x(v,Y_x)=-\omega_x(Y_x,v)=0$ by skew-symmetry. This contradicts the nondegeneracy of $\omega_x$, since $v\ne 0$. Thus \begin{align*} \omega_x(Y_x,v)\ne 0. \end{align*} By the definition of $\alpha$, \begin{align*} \alpha_x(v)=\omega_x(Y_x,v), \end{align*} so $\alpha_x$ is nonzero on $L_x$. [/step]

[step:Evaluate the contact volume form using a symplectic complement] Choose a linear complement $E_x\subset H$ to $L_x$, so that \begin{align*} H=L_x\oplus E_x. \end{align*} The restriction $\omega_x|_{E_x}$ is nondegenerate: if $e\in E_x$ satisfies $\omega_x(e,e')=0$ for every $e'\in E_x$, then also $\omega_x(e,\ell)=0$ for every $\ell\in L_x$ because $L_x=\ker(\omega_x|_H)$. Hence $\omega_x(e,h)=0$ for every $h\in H$, so $e\in L_x$. Since $E_x\cap L_x=\{0\}$, this gives $e=0$. Thus $(E_x,\omega_x|_{E_x})$ is a symplectic vector space of dimension $2n-2$. Choose a basis \begin{align*} e_1,\dots,e_{2n-2} \end{align*} of $E_x$ for which \begin{align*} (\omega_x|_{E_x})^{n-1}(e_1,\dots,e_{2n-2})\ne 0. \end{align*} Choose $v\in L_x$ with $v\ne 0$. Since $d\alpha=i^*\omega$ and $\alpha_x(v)\ne 0$, the alternating product satisfies \begin{align*} (\alpha\wedge(d\alpha)^{n-1})_x(v,e_1,\dots,e_{2n-2}) = \alpha_x(v)\,(\omega_x|_{E_x})^{n-1}(e_1,\dots,e_{2n-2}). \end{align*} The right-hand side is nonzero. Therefore \begin{align*} \alpha\wedge(d\alpha)^{n-1} \end{align*} is nonzero at $x$. Since $x\in\Sigma$ was arbitrary, $\alpha$ is a contact form on $\Sigma$. [/step]

[step:Normalize the characteristic line to obtain the Reeb vector field] Because $\alpha$ is a contact form, its Reeb vector field $R_\alpha\in\mathfrak X(\Sigma)$ is the unique vector field satisfying \begin{align*} \alpha(R_\alpha)=1 \end{align*} and \begin{align*} \iota_{R_\alpha}d\alpha=0. \end{align*} The second condition says exactly that \begin{align*} R_\alpha(x)\in\ker((d\alpha)_x) \end{align*} for every $x\in\Sigma$. By the kernel equality proved above, \begin{align*} R_\alpha(x)\in\ker(\omega_x|_{T_x\Sigma})=L_x. \end{align*} Conversely, $\alpha_x$ is nonzero on the one-dimensional line $L_x$, so there is a unique vector in $L_x$ satisfying $\alpha_x(v)=1$. This vector is $R_\alpha(x)$. Hence $R_\alpha$ spans the characteristic line field \begin{align*} \ker(\omega|_{T\Sigma})\subset T\Sigma. \end{align*} This completes the proof. [/step]