- 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 \(m^2(x_{01},x'_{12})\neq 0\).
Symplectic snippets