\(
\def\CC{{\mathbb C}}
\def\RR{{\mathbb R}}
\def\NN{{\mathbb N}}
\def\ZZ{{\mathbb Z}}
\def\TT{{\mathbb T}}
\def\CF{{\operatorname{CF}^\bullet}}
\def\HF{{\operatorname{HF}^\bullet}}
\def\SH{{\operatorname{SH}^\bullet}}
\def\ot{{\leftarrow}}
\def\st{\;:\;}
\def\Fuk{{\operatorname{Fuk}}}
\def\emprod{m}
\def\cone{\operatorname{Cone}}
\def\Flux{\operatorname{Flux}}
\def\li{i}
\def\ev{\operatorname{ev}}
\def\id{\operatorname{id}}
\def\grad{\operatorname{grad}}
\def\ind{\operatorname{ind}}
\def\weight{\operatorname{wt}}
\def\Sym{\operatorname{Sym}}
\def\HeF{\widehat{CHF}^\bullet}
\def\HHeF{\widehat{HHF}^\bullet}
\def\Spinc{\operatorname{Spin}^c}
\def\min{\operatorname{min}}
\def\div{\operatorname{div}}
\def\SH{{\operatorname{SH}^\bullet}}
\def\CF{{\operatorname{CF}^\bullet}}
\def\Tw{{\operatorname{Tw}}}
\def\Log{{\operatorname{Log}}}
\def\TropB{{\operatorname{TropB}}}
\def\wt{{\operatorname{wt}}}
\def\Span{{\operatorname{span}}}
\def\Crit{\operatorname{Crit}}
\def\into{\hookrightarrow}
\def\tensor{\otimes}
\def\CP{\mathbb{CP}}
\def\eps{\varepsilon}
\)
SympSnip:
- 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
\[
\HF(L_{(a,b),\theta_1},L_{(c,d),\theta_2})=\ZZ^{ad-cb}
\]
- 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.
- 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 \(m^2\).
- 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 \(m^2(x_{01},x'_{12})\neq 0\).