A Lagrangian Klein bottle you can't squeeze (Q6601742)

If \(L\) is a closed, orientable, Lagrangian surface in a symplectic 4-manifold \((X,\omega)\), then the genus of \(L\) is \(g(L)=1+\tfrac{1}{2}[L]^2\) and any family of symplectic forms \(\omega_t\), with \(\omega_0=\omega\), on \(X\) with respect to which \(L\) is smoothly isotopic to a Lagrangian must satisfy \(\int_L[\omega_t]=0\) for all \(t\). If, on the other hand, \(L\) is a nonorientable Lagrangian surface, then \(H^2(L;\mathbb{R})=0\) and the topology of \(L\) admits a less rigid relationship with \([\omega]\). This observation leads to the following two guiding questions for the present paper:\N\NQuestion 1.1. Given a symplectic 4-manifold \((X,\omega)\) and a \(\mathbb{Z}/2\)-homology class \(\beta\in H_2(X;\mathbb{Z}/2)\), what is the minimal nonorientable genus of a nonorientable Lagrangian surface \(L\subset X\) with \([L]=\beta\)?\N\NQuestion 1.2. Given a symplectic 4-manifold \((X,\omega)\) and a nonorientable Lagrangian surface \(L\subset X\), how far can you deform \(\omega\) in cohomology before there is no Lagrangian smoothly isotopic to \(L\)?\N\NThe 4-manifold for which the paper is addressing these questions is \(X=S^2\times S^2\), where all symplectic forms are diffeomorphic to one of the form\N\[\N\omega = \lambda\,p_1^*\sigma + p_2^*\sigma,\N\]\Nwhere \(p_1,p_2\colon X\to S^2\) are the obvious projections, \(\sigma\in\Omega^2(S^2)\) is an area form, and \(\lambda>0\) is a constant. Letting \(\beta\) be the \(\mathbb{Z}/2\)-homology class \([\{*\}\times S^2]\), a Lagrangian Klein bottle \(L\) representing \(\beta\) is constructed whenever \(\lambda<2\). Namely, \(L\) is constructed as a visible Lagrangian and also as a tropical Lagrangian in the almost toric 4-manifold \((X,\omega)\). Because this construction breaks down when \(\lambda\geq 2\), the author conjectures that \(\beta\) cannot be represented by a Lagrangian Klen bottle when \(\lambda\geq 2\), a conjecture whose proof would provide an answer to the minimal genus question above.\N\NFor the nonsqueezing question, the paper again considers a Lagrangian Klein bottle in a subset of \(S^2\times S^2\). For any connected, open interval \(I\subset\mathbb{R}\) of length \(|I|\), we let \(U_I = S^2\times C_I\), where \(C_I=I\times(\mathbb{R}/2\pi\mathbb{Z})\) is a cylinder with total area \(|I|\). If \(|I|>1\), then the construction mentioned above produces a Lagrangian Klein bottle in the unique nontrivial class of \(H_2(U_I;\mathbb{Z}/2)\). The following result shows that this Klein bottle cannot be squeezed into \(U_I\) when \(|I|\leq 1\):\N\NTheorem 3.1. Suppose that \(|I|\leq 1\). If \(\iota\colon K\to U_I\) is a Lagrangian embedding of the Klein bottle in the unique nontrivial class of \(H_2(U_I;\mathbb{Z}/2)\), then \(\iota_*\colon H_1(K;\mathbb{Q})\to H_1(U_I;\mathbb{Q})\) is the zero map.\N\NThe proof of Theorem 3.1 consists of an SFT limit analysis, using neck stretching with respect to a carefully chosen sequence of almost complex structures on \(X\).\N\NFor the entire collection see [Zbl 1515.53004].