[proofplan] The proof is a pointwise linear-algebra argument on the regular level hypersurface $\Sigma=H^{-1}(c)$. At a point $x\in\Sigma$, the tangent space is $\ker dH_x$, and the defining identity $\omega(X_H,\cdot)=dH$ shows that $X_H(x)$ annihilates $T_x\Sigma$ under $\omega_x$. The same identity and skew-symmetry show that $X_H(x)$ is tangent to $\Sigma$, while regularity and nondegeneracy show that it is nonzero. Since the regular level set theorem makes $\Sigma$ a smooth hypersurface in the symplectic manifold $(M^{2n},\omega)$, the hypotheses of [citetheorem:10060] apply; its rank-one conclusion then implies that this nonzero element spans the characteristic line. [/proofplan]

[step:Identify the tangent space of the regular level hypersurface]Fix $x\in\Sigma$. Since $c$ is a regular value of $H$, the differential \begin{align*} dH_x:T_xM\to T_c\mathbb R\cong\mathbb R \end{align*} is nonzero. By the regular level set theorem, applied to the smooth map $H:M\to\mathbb R$ at the regular value $c$, $\Sigma$ is an embedded hypersurface of $M$ and \begin{align*} T_x\Sigma=\ker dH_x. \end{align*} Equivalently, for every vector $w\in T_xM$, \begin{align*} w\in T_x\Sigma \iff dH_x(w)=0. \end{align*}[/step]

[guided]Fix a point $x\in\Sigma$. The first structural fact we need is the tangent space of the level set. Because $c$ is a regular value of the smooth map $H:M\to\mathbb R$, the differential \begin{align*} dH_x:T_xM\to T_c\mathbb R\cong\mathbb R \end{align*} is a nonzero linear functional. The regular level set theorem therefore applies to $H$ at the value $c$: it says that $\Sigma=H^{-1}(c)$ is an embedded hypersurface and that its tangent space at $x$ consists exactly of the tangent vectors killed by $dH_x$. Thus \begin{align*} T_x\Sigma=\ker dH_x. \end{align*} This identification is the bridge between the differential equation defining the Hamiltonian vector field and the geometry of the hypersurface. It means that proving a vector lies in $T_x\Sigma$ is the same as proving that $dH_x$ evaluates to zero on that vector. The Hamiltonian vector field is defined by the identity $\omega(X_H,\cdot)=dH$. Evaluating this identity at $x$ and on the vector $X_H(x)\in T_xM$ gives \begin{align*} dH_x(X_H(x))=\omega_x(X_H(x),X_H(x)). \end{align*} Because $\omega_x$ is an alternating [bilinear form](/page/Bilinear%20Form), $\omega_x(X_H(x),X_H(x))=0$. Hence $dH_x(X_H(x))=0$, and the tangent-space identification gives $X_H(x)\in T_x\Sigma$. Now let $w\in T_x\Sigma$. Since $T_x\Sigma=\ker dH_x$, we have $dH_x(w)=0$. Applying the same Hamiltonian identity to $w$ gives \begin{align*} \omega_x(X_H(x),w)=dH_x(w)=0. \end{align*} Since this holds for every $w\in T_x\Sigma$ and $X_H(x)\in T_x\Sigma$, the vector $X_H(x)$ lies in the characteristic kernel $\ker(\omega_x|_{T_x\Sigma})=\mathcal L_{\Sigma,x}$. It remains to know that this vector is not zero. If $X_H(x)=0$, then for every $v\in T_xM$, \begin{align*} dH_x(v)=\omega_x(X_H(x),v)=\omega_x(0,v)=0. \end{align*} Thus $dH_x=0$, contradicting that $c$ is a regular value of $H$. Therefore $X_H(x)\ne 0$. Finally, $\Sigma$ is a smooth hypersurface in the symplectic manifold $(M^{2n},\omega)$ by the regular level set theorem, so [citetheorem:10060] applies and says that $\ker(\omega_x|_{T_x\Sigma})$ is one-dimensional. A nonzero vector in a one-dimensional real [vector space](/page/Vector%20Space) spans it, so \begin{align*} \mathbb R X_H(x)=\ker\left(\omega_x|_{T_x\Sigma}\right)=\mathcal L_{\Sigma,x}. \end{align*}[/guided]

[step:Show that the Hamiltonian vector field is tangent to the level hypersurface] By the defining identity for the Hamiltonian vector field, \begin{align*} dH_x(v)=\omega_x(X_H(x),v) \end{align*} for every $v\in T_xM$. Taking $v=X_H(x)$ gives \begin{align*} dH_x(X_H(x))=\omega_x(X_H(x),X_H(x)). \end{align*} Since $\omega_x$ is alternating, $\omega_x(X_H(x),X_H(x))=0$. Hence \begin{align*} dH_x(X_H(x))=0. \end{align*} Using $T_x\Sigma=\ker dH_x$, we obtain \begin{align*} X_H(x)\in T_x\Sigma. \end{align*} [/step]

[step:Place $X_H(x)$ in the characteristic kernel] Let $w\in T_x\Sigma$. From $T_x\Sigma=\ker dH_x$, we have \begin{align*} dH_x(w)=0. \end{align*} Using again the defining identity $\omega(X_H,\cdot)=dH$, evaluated at $x$ and applied to $w$, we get \begin{align*} \omega_x(X_H(x),w)=dH_x(w)=0. \end{align*} Since $w\in T_x\Sigma$ was arbitrary and $X_H(x)\in T_x\Sigma$, this proves \begin{align*} X_H(x)\in\ker\left(\omega_x|_{T_x\Sigma}\right)=\mathcal L_{\Sigma,x}. \end{align*} [/step]

[step:Use regularity and symplectic nondegeneracy to prove nonvanishing] We claim that $X_H(x)\ne 0$. Suppose instead that $X_H(x)=0$. Then for every $v\in T_xM$, \begin{align*} dH_x(v)=\omega_x(X_H(x),v)=\omega_x(0,v)=0. \end{align*} Thus $dH_x=0$, contradicting that $c$ is a regular value of $H$. Therefore \begin{align*} X_H(x)\ne 0. \end{align*} [/step]

[step:Conclude that the nonzero Hamiltonian vector spans the characteristic line] Since $c$ is a regular value of $H$, the regular level set theorem gives that $\Sigma$ is a smooth hypersurface in the symplectic manifold $(M^{2n},\omega)$. Therefore the hypotheses of [citetheorem:10060] are satisfied, and the kernel \begin{align*} \ker\left(\omega_x|_{T_x\Sigma}\right) \end{align*} is a one-dimensional subspace of $T_x\Sigma$. We have proved that $X_H(x)$ is a nonzero element of this kernel. A nonzero vector in a one-dimensional real vector space spans that vector space, so \begin{align*} \mathbb R X_H(x)=\ker\left(\omega_x|_{T_x\Sigma}\right)=\mathcal L_{\Sigma,x}. \end{align*} Since $x\in\Sigma$ was arbitrary, the conclusion holds for every $x\in\Sigma$. [/step]