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

  1. First we compute the generators of \(CF^\bullet(L_{(a,b),\theta_1},L_{(c,d),\theta_2})\), i.e. the intersection points \(L_{(a,b),\theta_1}\cap L_{(c,d),\theta_2}\). By translational invariance we may assume that \(\theta_1=0\). Let now \(A\in SL(2,\ZZ)=Symp(\RR^2)\) be such that \(A\begin{pmatrix} a\\ b \end{pmatrix}= \begin{pmatrix} 1\\ 0 \end{pmatrix}\), which exists because \(gcd(a,b)=1\). Then \[ A \begin{pmatrix} c\\ d \end{pmatrix} = \begin{pmatrix} a & *\\ b & * \end{pmatrix}^{-1} \begin{pmatrix} c\\ d \end{pmatrix} =\begin{pmatrix} *\\ ad-bc \end{pmatrix} \] and therefore \[L_{(a,b),\theta_1}\cap L_{(c,d),\theta_2}\cong L_{(a,b),0}\cap L_{(c,d),\theta_2-\theta_1}\cong L_{(1,0),0}\cap L_{(*,ad-bc),*}\] has precisely \(ad-bc\) points. To conclude we note that the differential of the Floer cochain complex is trivial, as all the generators live in the same degree (\(1\) or \(0\), depending on whether \(ad-cb\gtrless0\)). Therefore \[ \HF(L_{(a,b),\theta_1},L_{(c,d),\theta_2})=\ZZ^{ad-cb} \]
  2. Since Lagrangian surgery preserves homology, we have \([L_0\# L_1]=[L_0]+[L_1]=(1,1)\in H_1(T^2;\ZZ)\), so \((a,b)=(1,1)\). One sees that the curve \(L_{(1,1),\frac{1}{2}}\) is Lagrangian isotopic to \(L_0\#L_1\) through a flux zero isotopy, giving the result.
  3. Note that discs lift to the universal cover \(\RR^2\) of \(T^2\). Here it is easy to visualize the potential triangles that would contribute to the given operations. An orientation argument shows these cannot be pseudo-holomorphic, so there are no triangles contributing to \(m^2\).
  4. The result no longer needs to hold true. For instance, for \(L_2'=L_{(1,1),0}\) there is an infinite sequence of holomorphic discs with contributing to \[m^2(x_{01},x'_{12}),x'_{02}\], all of them with positive energy, thus \(m^2(x_{01},x'_{12})\neq 0\).