[proofplan] At each point of the hypersurface, the tangent space $T_x\Sigma$ is a hyperplane inside the symplectic [vector space](/page/Vector%20Space) $T_xM$. The symplectic dimension formula shows that its symplectic complement is one-dimensional, and nondegeneracy forces that complement to lie inside the hyperplane. This identifies the kernel of the restricted two-form with a one-dimensional subspace. Smoothness is checked locally by choosing a defining function for $\Sigma$ and observing that its Hamiltonian vector field spans the kernel; ordinary integral curves of this nonvanishing local vector field give the characteristic foliation. [/proofplan] [step:Identify the pointwise kernel with the symplectic complement of the tangent hyperplane] Fix $x\in\Sigma$, and set $V:=T_xM$ and $W:=T_x\Sigma$. Then $(V,\omega_x)$ is a symplectic vector space of dimension $2n$, and $W\subset V$ is a hyperplane, so $\dim W=2n-1$. By the symplectic complement dimension formula [citetheorem:10039], $\dim W+\dim W^{\omega_x}=\dim V$. Therefore $\dim W^{\omega_x}=1$. We next prove that $W^{\omega_x}\subset W$. Let $v\in W^{\omega_x}$. If $v\notin W$, then $V=W\oplus \mathbb R v$ because $W$ is a hyperplane. For every $w\in W$, $\omega_x(v,w)=0$ by the definition of $W^{\omega_x}$, and also $\omega_x(v,v)=0$ because $\omega_x$ is alternating. Hence $\omega_x(v,u)=0$ for every $u\in V$, contradicting the nondegeneracy of $\omega_x$ unless $v=0$. Since this argument applies to every nonzero $v\in W^{\omega_x}$, we have $W^{\omega_x}\subset W$. Now $\ker\left(\omega_x|_{T_x\Sigma}\right)=\{v\in W:\omega_x(v,w)=0\text{ for every }w\in W\}=W\cap W^{\omega_x}$. Since $W^{\omega_x}\subset W$, this gives $\ker\left(\omega_x|_{T_x\Sigma}\right)=W^{\omega_x}$. Thus $\mathcal L_{\Sigma,x}$ is one-dimensional for every $x\in\Sigma$. [guided] Fix a point $x\in\Sigma$. The relevant linear algebra takes place in the symplectic vector space $V:=T_xM$ with symplectic form $\omega_x$. Define the hyperplane $W:=T_x\Sigma\subset T_xM$. Because $\Sigma$ is a hypersurface in the $2n$-dimensional manifold $M$, we have $\dim W=2n-1$. We apply the symplectic complement dimension formula [citetheorem:10039] to the subspace $W\subset V$. Its hypotheses are satisfied because $(V,\omega_x)$ is a finite-dimensional symplectic vector space and $W$ is a linear subspace. The formula gives $\dim W+\dim W^{\omega_x}=\dim V$. Substituting $\dim W=2n-1$ and $\dim V=2n$ gives $\dim W^{\omega_x}=1$. The next point is special to hypersurfaces. We must show that this one-dimensional symplectic complement actually lies inside the hyperplane $W$. Let $v\in W^{\omega_x}$. Suppose, toward a contradiction, that $v\notin W$. Since $W$ has codimension one in $V$, the vector $v$ together with $W$ spans all of $V$: \begin{align*} V=W\oplus \mathbb R v. \end{align*} For every $w\in W$, the definition of $W^{\omega_x}$ gives \begin{align*} \omega_x(v,w)=0. \end{align*} Also $\omega_x(v,v)=0$ because $\omega_x$ is alternating. Therefore $v$ pairs to zero with every vector in $W$ and with $v$ itself, hence with every vector in $V=W\oplus \mathbb R v$. This contradicts the nondegeneracy of the symplectic form $\omega_x$ unless $v=0$. Thus every nonzero vector in $W^{\omega_x}$ lies in $W$, and so \begin{align*} W^{\omega_x}\subset W. \end{align*} Finally, the kernel of the restricted two-form is exactly the set of tangent vectors in $W$ that pair to zero with every vector in $W$: \begin{align*} \ker\left(\omega_x|_{T_x\Sigma}\right) = \{v\in W:\omega_x(v,w)=0\text{ for every }w\in W\} = W\cap W^{\omega_x}. \end{align*} Since $W^{\omega_x}\subset W$, this intersection is just $W^{\omega_x}$. Hence \begin{align*} \ker\left(\omega_x|_{T_x\Sigma}\right)=W^{\omega_x}, \end{align*} and this space is one-dimensional. [/guided] [/step] [step:Use a local defining function to show the line field varies smoothly] Let $x_0\in\Sigma$. Since $\Sigma$ is an embedded hypersurface, there exist an open neighbourhood $U\subset M$ of $x_0$ and a smooth function $f:U\to\mathbb R$ such that \begin{align*} \Sigma\cap U=f^{-1}(0) \end{align*} and $df_x\ne 0$ for every $x\in\Sigma\cap U$. Let $\mathfrak X(U)$ denote the space of smooth vector fields on the [smooth manifold](/page/Smooth%20Manifold) $U$. Define the Hamiltonian vector field $X_f\in\mathfrak X(U)$ by \begin{align*} \iota_{X_f}\omega=df, \end{align*} that is, \begin{align*} \omega_x(X_f(x),v)=df_x(v) \end{align*} for every $x\in U$ and every $v\in T_xM$. This vector field is smooth because $\omega$ is a smooth nondegenerate two-form and $df$ is a smooth one-form. For $x\in\Sigma\cap U$ and $v\in T_x\Sigma$, the defining equation $f|_{\Sigma\cap U}=0$ gives \begin{align*} df_x(v)=0. \end{align*} Hence \begin{align*} \omega_x(X_f(x),v)=0 \end{align*} for every $v\in T_x\Sigma$, so $X_f(x)\in (T_x\Sigma)^{\omega_x}=\mathcal L_{\Sigma,x}$. Also $X_f(x)\ne 0$, because $X_f(x)=0$ would imply $df_x=0$ by nondegeneracy of $\omega_x$, contradicting the defining-function condition. Since $\mathcal L_{\Sigma,x}$ is one-dimensional, we obtain \begin{align*} \mathcal L_{\Sigma,x}=\operatorname{span}\{X_f(x)\} \end{align*} for every $x\in\Sigma\cap U$. Thus $\mathcal L_\Sigma$ is locally spanned by a smooth nonvanishing vector field, so it is a smooth rank-one distribution on $\Sigma$. [/step] [step:Integrate the smooth line distribution by local vector-field flows] Let $x_0\in\Sigma$, and choose $U\subset M$, $f:U\to\mathbb R$, and $X_f\in\mathfrak X(U)$ as above. Let $\mathfrak X(\Sigma\cap U)$ denote the space of smooth vector fields on the smooth manifold $\Sigma\cap U$. Restrict $X_f$ to the submanifold $\Sigma\cap U$; since $X_f(x)\in T_x\Sigma$ for every $x\in\Sigma\cap U$, this gives a smooth nonvanishing vector field $Y\in\mathfrak X(\Sigma\cap U)$ defined by $Y(x):=X_f(x)$. By the flow-box theorem for nonvanishing smooth vector fields, for every $x_0\in\Sigma\cap U$ there are an open neighbourhood $V\subset\Sigma\cap U$ of $x_0$, an [open set](/page/Open%20Set) $A\subset\mathbb R^{2n-2}$, a number $\varepsilon>0$, and a smooth coordinate chart $\Phi:A\times(-\varepsilon,\varepsilon)\to V$ such that the curves $t\mapsto\Phi(a,t)$ are integral curves of $Y$. Each such curve is an immersion because its velocity is $Y$, which is nowhere zero. Its tangent line at $\Phi(a,t)$ is $\operatorname{span}\{Y(\Phi(a,t))\}=\mathcal L_{\Sigma,\Phi(a,t)}$. Therefore the smooth line distribution $\mathcal L_\Sigma$ defines a local foliation of $\Sigma$ by immersed curves tangent to $\ker(\omega|_{T\Sigma})$. [/step]