[proofplan] We prove that the pulled-back two-form $\varphi_t^*\omega$ has zero derivative with respect to the flow time. The derivative of pullbacks along a flow is governed by the Lie derivative, and Cartan's homotopy formula gives \begin{align*} \mathcal{L}_{X_H}\omega = d(\iota_{X_H}\omega)+\iota_{X_H}(d\omega). \end{align*} The Hamiltonian identity $\iota_{X_H}\omega=dH$ and the closedness condition $d\omega=0$ force this Lie derivative to vanish, so the pulled-back form is constant in time and equals $\omega$ at time $0$. [/proofplan] [step:Localize the flow on a fixed time interval] Let $D \subset \mathbb{R}\times M$ denote the flow domain, and for $r \in \mathbb{R}$ define $D_r := \{q \in M : (r,q) \in D\}$. Fix $t \in \mathbb{R}$ and $p \in D_t$. Let $J \subset \mathbb{R}$ be the compact interval with endpoints $0$ and $t$. Since the maximal flow curve through $p$ is defined at time $t$, the flow property implies $(s,p) \in D$ for every $s \in J$. Thus $J\times\{p\}$ is a compact subset of the [open set](/page/Open%20Set) $D$. For each $s_0 \in J$, choose an open interval $I_{s_0} \subset \mathbb{R}$ containing $s_0$ and an open neighbourhood $U_{s_0} \subset M$ of $p$ such that $I_{s_0}\times U_{s_0}\subset D$. Choose finitely many points $s_1,\dots,s_N \in J$ such that $J\subset \bigcup_{k=1}^N I_{s_k}$, and set \begin{align*} I := \bigcup_{k=1}^N I_{s_k}, \qquad U := \bigcap_{k=1}^N U_{s_k}. \end{align*} Then $I$ is an open interval containing $J$, $U$ is an open neighbourhood of $p$, and $I\times U\subset D$. Hence $\varphi_s$ is defined on $U$ for every $s \in I$. [guided] Let $D \subset \mathbb{R}\times M$ be the domain of the local flow, so $(s,q)\in D$ means that the time-$s$ value of the integral curve starting at $q$ exists. For each $r\in\mathbb{R}$, define the time slice $D_r:=\{q\in M:(r,q)\in D\}$. We fix $t\in\mathbb{R}$ and $p\in D_t$, and we let $J$ be the compact interval with endpoints $0$ and $t$. The point of this localization step is to replace one point $p$ by a whole neighbourhood of points whose flow exists for all times near the interval from $0$ to $t$. Since the maximal flow curve through $p$ is defined at time $t$, the flow property gives existence at every intermediate time: $(s,p)\in D$ for each $s\in J$. Therefore $J\times\{p\}$ is a compact subset of the open set $D$. Now use openness of $D$ around this compact set. For each $s_0\in J$, choose an open interval $I_{s_0}\subset\mathbb{R}$ containing $s_0$ and an open neighbourhood $U_{s_0}\subset M$ of $p$ such that $I_{s_0}\times U_{s_0}\subset D$. The intervals $I_{s_0}$ cover the compact interval $J$, so finitely many of them, say $I_{s_1},\dots,I_{s_N}$, still cover $J$. Define \begin{align*} I := \bigcup_{k=1}^N I_{s_k}, \qquad U := \bigcap_{k=1}^N U_{s_k}. \end{align*} Then $I$ is an open interval containing both $0$ and $t$, and $U$ is an open neighbourhood of $p$. If $s\in I$ and $q\in U$, then $s\in I_{s_k}$ for some $k$ and $q\in U_{s_k}$, hence $(s,q)\in I_{s_k}\times U_{s_k}\subset D$. Thus $\varphi_s$ is defined on all of $U$ for every $s\in I$. [/guided] For each $s \in I$, define the smooth two-form \begin{align*} \alpha_s \in \Omega^2(U) \end{align*} by \begin{align*} \alpha_s := \varphi_s^*(\omega|_{\varphi_s(U)}). \end{align*} It is enough to prove $\alpha_t=\alpha_0$, because $\alpha_0=\varphi_0^*\omega=\operatorname{id}_U^*\omega=\omega|_U$, and then evaluation at the arbitrary point $p$ gives the asserted identity on $D_t$. [/step] [step:Differentiate the pulled-back symplectic form along the Hamiltonian flow] We use the standard flow-pullback differentiation identity for differential forms. Applied to the time-independent two-form $\omega$ and the flow of $X_H$, it gives, for every $s \in I$, \begin{align*} \frac{d}{ds}\alpha_s = \varphi_s^*(\mathcal{L}_{X_H}\omega). \end{align*} Here $\mathcal{L}_{X_H}\omega \in \Omega^2(M)$ denotes the Lie derivative of $\omega$ along the vector field $X_H$. [guided] We want to show that $\alpha_s=\varphi_s^*\omega$ does not change with $s$. The correct derivative formula is not the ordinary derivative of coefficients in one coordinate chart; it is the intrinsic identity saying that differentiation of a pulled-back form along the flow is the pullback of the Lie derivative. More precisely, for the smooth vector field $X_H \in \mathfrak{X}(M)$ with flow $\varphi_s$, and for the fixed two-form $\omega \in \Omega^2(M)$, the standard flow-pullback identity gives \begin{align*} \frac{d}{ds}\varphi_s^*(\omega|_{\varphi_s(U)}) = \varphi_s^*(\mathcal{L}_{X_H}\omega). \end{align*} This identity applies because $\omega$ is a smooth differential form and $\varphi_s:U\to \varphi_s(U)$ is a smooth local diffeomorphism for each $s \in I$. With the notation $\alpha_s:=\varphi_s^*(\omega|_{\varphi_s(U)})$, this becomes \begin{align*} \frac{d}{ds}\alpha_s = \varphi_s^*(\mathcal{L}_{X_H}\omega). \end{align*} Thus the problem has been reduced to proving that the Lie derivative $\mathcal{L}_{X_H}\omega$ vanishes. [/guided] [/step] [step:Use Cartan's formula and the Hamiltonian identity to make the Lie derivative vanish] We use Cartan's homotopy formula for differential forms: \begin{align*} \mathcal{L}_{X_H}\omega = d(\iota_{X_H}\omega)+\iota_{X_H}(d\omega). \end{align*} Since $X_H$ is the Hamiltonian vector field associated to $H$, we have $\iota_{X_H}\omega=dH$. Since $\omega$ is symplectic, it is closed, so $d\omega=0$. Therefore \begin{align*} \mathcal{L}_{X_H}\omega = d(dH)+\iota_{X_H}(0). \end{align*} The [exterior derivative](/theorems/1525) satisfies $d^2=0$, hence $d(dH)=0$, and contraction with the zero three-form is the zero two-form. Thus \begin{align*} \mathcal{L}_{X_H}\omega=0. \end{align*} [guided] The Lie derivative is now computed using Cartan's homotopy formula. For the vector field $X_H \in \mathfrak{X}(M)$ and the two-form $\omega \in \Omega^2(M)$, the formula states \begin{align*} \mathcal{L}_{X_H}\omega = d(\iota_{X_H}\omega)+\iota_{X_H}(d\omega). \end{align*} Each term has a specific meaning: $\iota_{X_H}\omega \in \Omega^1(M)$ is contraction of the two-form $\omega$ with the vector field $X_H$, and $d\omega \in \Omega^3(M)$ is the exterior derivative of $\omega$. Now we use the two defining pieces of Hamiltonian symplectic geometry. First, by definition of the Hamiltonian vector field in this theorem, \begin{align*} \iota_{X_H}\omega=dH. \end{align*} Second, because $(M,\omega)$ is symplectic, $\omega$ is closed: \begin{align*} d\omega=0. \end{align*} Substituting these two identities into Cartan's formula gives \begin{align*} \mathcal{L}_{X_H}\omega = d(dH)+\iota_{X_H}(0). \end{align*} The exterior derivative squares to zero, so $d(dH)=0$. Also, contraction is linear, so contraction of the zero three-form is the zero two-form. Hence \begin{align*} \mathcal{L}_{X_H}\omega=0. \end{align*} This is the essential cancellation: Hamiltonian vector fields preserve $\omega$ because their contraction with $\omega$ is exact, and exact one-forms have zero exterior derivative after applying $d$ again. [/guided] [/step] [step:Integrate the zero derivative and recover the initial value] Combining the previous two steps, for every $s \in I$, \begin{align*} \frac{d}{ds}\alpha_s = \varphi_s^*(0)=0. \end{align*} Therefore the smooth family $s \mapsto \alpha_s$ is constant on $I$. Since $\varphi_0=\operatorname{id}_U$, we have \begin{align*} \alpha_0=\varphi_0^*\omega=\omega|_U. \end{align*} Consequently \begin{align*} \varphi_t^*(\omega|_{\varphi_t(U)})=\alpha_t=\alpha_0=\omega|_U. \end{align*} Evaluating at the arbitrary point $p \in D_t$ gives \begin{align*} \varphi_t^*(\omega|_{D_{-t}})_p=\omega_p. \end{align*} Because $p \in D_t$ was arbitrary, this proves \begin{align*} \varphi_t^*(\omega|_{D_{-t}})=\omega|_{D_t}. \end{align*} If $X_H$ is complete, then $D_t=M$ for every $t \in \mathbb{R}$, and the identity becomes $\varphi_t^*\omega=\omega$ on $M$. [guided] From the differentiation identity and the vanishing of the Lie derivative, for every $s\in I$ we have \begin{align*} \frac{d}{ds}\alpha_s = \varphi_s^*(0)=0. \end{align*} This says that every coefficient of the smooth two-form family $s\mapsto\alpha_s$ is constant in $s$ after evaluation on smooth vector fields on $U$. Hence $\alpha_s$ is constant on the interval $I$. At time $0$, the flow is the identity map on $U$, so \begin{align*} \alpha_0=\varphi_0^*\omega=\operatorname{id}_U^*\omega=\omega|_U. \end{align*} Since $t\in I$, constancy gives \begin{align*} \varphi_t^*(\omega|_{\varphi_t(U)})=\alpha_t=\alpha_0=\omega|_U. \end{align*} Evaluating this identity at the originally fixed point $p\in D_t$ gives \begin{align*} \varphi_t^*(\omega|_{D_{-t}})_p=\omega_p. \end{align*} The point $p\in D_t$ was arbitrary, so the equality holds on the whole time-$t$ domain: \begin{align*} \varphi_t^*(\omega|_{D_{-t}})=\omega|_{D_t}. \end{align*} If the Hamiltonian vector field $X_H$ is complete, then $D_t=M$ for every $t\in\mathbb{R}$, and this restricted identity is exactly \begin{align*} \varphi_t^*\omega=\omega. \end{align*} [/guided] [/step]