# Exact Deformation of Lagrangian Submanifolds

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!

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 $$j_t = \psi^{H}_t \circ j_0$$.