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Let $j_{t}:L \rightarrow P$ be a family of Lagrangian submanifolds. I'm trying to show that the form $j_{t}^{*}(i(X_{t})\omega)$, $X_{t}(j_{t}(x)):=\frac{dj_{t}(x)}{dt}$ is exact for all $t$ if and only if there is a family of Hamiltonian diffeomorphisms $\psi_{t}^{H}:P \rightarrow P$ such that $j_{t}(L)=\psi_{t}^{H}(L)$.

What I have so far for one implication is that if the form is exact, it is of the form $j_{t}^{*}(i(X_{t})\omega)=dh_{t}$. The idea I have is to extend $h_{t}$ to the family of Hamiltonians $H_{t}$ I need using the canonical almost complex structure on P, but I'm struggling with the technical details. Thank you for your help in advance!

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Let $\psi^{H}_t : P \to P$ be a Hamiltonian isotopy, i.e. there exists a smooth time-dependent function $H_t : P \to \mathbb{R}$ satisfying $i(X_t)\omega = dH_t$ where $X_t$ is defined via $X_t(\psi^{H}_t(x)) = d(\psi^{H}_t(x))/dt$ for $x \in P$, such that $j_t = \psi^{H}_t \circ j_0 : L \to P$. Then $j_t^*(i(X_t)\omega) = j_t^*(dH_t) = d(j_t^*H_t)$ is exact on $L$.

Conversely, let $j_t : L \to P$ be a smooth family of Lagrangian submanifolds satisfying $j_t^*(i(X_t)\omega) = dh_t$ where $X_t \in C^{\infty}(j_t^* TP)$ is defined via $X_t(j_t(x)) = d(j_t(x))/dt$ for $x \in L$ and $h_t : L \to \mathbb{R}$ is a smooth time-dependent function. (Here, $j_t^*TP$ is the pullback tangent bundle, namely the vector bundle over $L$ whose fiber over $x \in L$ is the tangent space $T_{j_t(x)}P$.) Note that it makes sense to let the 1-form $A_t := i(X_t)\omega$ act on any element in $j_t^*TP$, and the equation $j_t^*(i(X_t)\omega) = dh_t$ is then a constraint on $A_t$: it means that the restriction of $A_t$ to the subbundle $j_t^* TL_t = j_t^* (j_t)_* TL$ is exact. Because of this constraint, the circle of ideas known as "Whitney extension theorem" shows that we can extend $h_t$ (thought of as a function on the submanifold $L_t$) to a function $H_t$ on $P$ such that $dH_t$ is given along $L_t$ by $A_t$. We can therefore define the vector field $X_t$ on the whole of $P$ via $i(X_t) \omega = dH_t$ and define $\psi^{H}_t$ to be the isotopy generated by $X_t$. Since this $X_t$ clearly extends the vector field $X_t$ which was already defined along each $L_t$, it follows that $j_t = \psi^{H}_t \circ j_0$.

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