\(
\def\CC{{\mathbb C}}
\def\RR{{\mathbb R}}
\def\NN{{\mathbb N}}
\def\ZZ{{\mathbb Z}}
\def\TT{{\mathbb T}}
\def\CF{{\operatorname{CF}^\bullet}}
\def\HF{{\operatorname{HF}^\bullet}}
\def\SH{{\operatorname{SH}^\bullet}}
\def\ot{{\leftarrow}}
\def\st{\;:\;}
\def\Fuk{{\operatorname{Fuk}}}
\def\emprod{m}
\def\cone{\operatorname{Cone}}
\def\Flux{\operatorname{Flux}}
\def\li{i}
\def\ev{\operatorname{ev}}
\def\id{\operatorname{id}}
\def\grad{\operatorname{grad}}
\def\ind{\operatorname{ind}}
\def\weight{\operatorname{wt}}
\def\Sym{\operatorname{Sym}}
\def\HeF{\widehat{CHF}^\bullet}
\def\HHeF{\widehat{HHF}^\bullet}
\def\Spinc{\operatorname{Spin}^c}
\def\min{\operatorname{min}}
\def\div{\operatorname{div}}
\def\SH{{\operatorname{SH}^\bullet}}
\def\CF{{\operatorname{CF}^\bullet}}
\def\Tw{{\operatorname{Tw}}}
\def\Log{{\operatorname{Log}}}
\def\TropB{{\operatorname{TropB}}}
\def\wt{{\operatorname{wt}}}
\def\Span{{\operatorname{span}}}
\def\Crit{\operatorname{Crit}}
\def\into{\hookrightarrow}
\def\tensor{\otimes}
\def\CP{\mathbb{CP}}
\def\eps{\varepsilon}
\)
SympSnip:
We show one direction here, which is that \(\omega\left(\frac{d}{dt}\li_t,v\right)=dH_t(v)\) implies \(\Flux_{\li_t}\) vanishes in cohomology.
Let \(c:S^1\to L\) be any 1-cycle in \(L\).
Parameterize a 2-chain \(\li_t\circ c: S^1\times I \to X\) with coordinates \((\theta, t)\).
The flux class applied to \(c\) can be explicitly computed:
\begin{align*}
\Flux_{\li_t}(c)=&\int_{\li_t\circ c}\omega
=\int_{I \times S^1} (\li_t\circ c )^* \omega\\
=&\int_I \int_{S^1} c^*\circ (\li_t)^* \omega
=\int_I \int_{S^1} (c^* \iota_{\frac{d}{dt}\li_t}\omega ) dt\\
=&\int_I \left(\int_{S^1} (c^* dH_t) \right)dt
=\int_I \left(\int_{S^1} d(c^*H_t)\right) dt
\end{align*}
By Stoke's theorem, the integral of an exact form over the circle is zero.
For the reverse direction, fix a base point \(x_0\in L\).
For every point \(x\in L\), pick a path \(\gamma_x: [0,1]\to L\) with \(\gamma_x(0)=x_0\) and \(\gamma_x(1)=x\).
Define the function \(H_t: L\to \RR\) by
\[dH_t(x_1):=\int_{\li_t\circ \gamma} \omega.\]
Because the flux of the isotopy is zero, this integral does not depend on the choices of paths \(\gamma_x\) and gives a well defined function on \(L\).
We now show that this function generates the Lagrangian isotopy.
The vector field \(\frac{d}{dt}\li_t\) is determined by the form \(\iota_{\frac{d}{dt}\li_t}\omega\).
Since \(\iota_{\frac{d}{dt}\li_t}\) is closed,
\begin{align*}
\int_{\gamma_x} \iota_{\frac{d}{dt}\li_t}\omega =&
\end{align*}
\todo{We now check that this these two things match up by computing the vector field.}