\( \def\CC{{\mathbb C}} \def\RR{{\mathbb R}} \def\NN{{\mathbb N}} \def\ZZ{{\mathbb Z}} \def\TT{{\mathbb T}} \def\CF{{\operatorname{CF}^\bullet}} \def\HF{{\operatorname{HF}^\bullet}} \def\SH{{\operatorname{SH}^\bullet}} \def\ot{{\leftarrow}} \def\st{\;:\;} \def\Fuk{{\operatorname{Fuk}}} \def\emprod{m} \def\cone{\operatorname{Cone}} \def\Flux{\operatorname{Flux}} \def\li{i} \def\ev{\operatorname{ev}} \def\id{\operatorname{id}} \def\grad{\operatorname{grad}} \def\ind{\operatorname{ind}} \def\weight{\operatorname{wt}} \def\Sym{\operatorname{Sym}} \def\HeF{\widehat{CHF}^\bullet} \def\HHeF{\widehat{HHF}^\bullet} \def\Spinc{\operatorname{Spin}^c} \def\min{\operatorname{min}} \def\div{\operatorname{div}} \def\SH{{\operatorname{SH}^\bullet}} \def\CF{{\operatorname{CF}^\bullet}} \def\Tw{{\operatorname{Tw}}} \def\Log{{\operatorname{Log}}} \def\TropB{{\operatorname{TropB}}} \def\wt{{\operatorname{wt}}} \def\Span{{\operatorname{span}}} \def\Crit{\operatorname{Crit}} \def\into{\hookrightarrow} \def\tensor{\otimes} \def\CP{\mathbb{CP}} \def\eps{\varepsilon} \) 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.}