In finite dimensional symplectic quotient problems, convexity properties of the moment map prove uniqueness of its zeros (modulo the action of the real group) in a complexified group orbit. Translating this into our terms is not quite possible, because there are hamiltonian deformations of $L$ which are not given by the flow of a fixed hamiltonian on $L$. By this we mean $L_{0},\ L_{1}$ are deformations given by a constant hamiltonian $h\in C^{\infty}(L_{0};\mathbb{R})$ if the flow

takes $L_{0}=f_{0}(L)$ to $L_{1}=f_{1}(L)$. We have to be precise about the vector $dh{\mbox{\,}\_\hskip-4.2679pt\shortmid\hskip 1.5pt}\omega^{-1}$ (which is only defined up to vectors tangent to $L$): we choose it to be perpendicular to $L$ with respect to the Riemannian metric on $W$; i.e. we take the vector $J\,\widetilde{\!dh\,}$. All small deformations of a Lagrangian are of this form; for more general deformations we have to use a different proof of uniqueness of a SLag representative of a hamiltonian deformation class (Proposition 4.3 below), but for these constant hamiltonian deformations we describe the moment map proof to show how the formalism works.

So for $\theta$ lying in $(-\pi/2,\pi/2)$ this is always strictly positive, and $\int_{L}h\,\mathrm{Im}\,\Omega$ is zero at $t=0,\,1$. Thus the two SLags must in fact coincide.

However, we must show that $\theta$ stays in this range if it starts in it, and deal with the case when it is not so bounded. The way to do this in fact proves the whole Lemma in one go anyway: pick a maximum $x\in L$ of $h$. Then by (2.4) $\dot{\theta}\arrowvert_{x}\leq 0\ \ \forall t$, but $\theta$ must flow to its original value in a flow from one SLag to another (necessarily of the same phase, as it can be computed cohomologically). So the maximum of $h$ must in fact be degenerate. A similar argument shows the critical point must also be stationary to $n$th order for any $n$, and in fact $h$ must be constant and the hamiltonian deformation trivial. $\square$

However, we can do better by mirroring the algebro-geometric argument that a non-zero map between stable bundles of the same slope is an isomorphism, using the grading on Floer cohomology (3.4). This will appear to be slightly magical; the crux of the argument is the hamiltonian isotopy invariance of Floer cohomology, provided by precisely the holomorphic discs in the theory about which we have had so little to say.

So there is no isolated transverse intersection point. In fact, working in a small neighbourhood of the intersection, we may choose coordinates such that $L_{1}$ is the graph in $T^{*}L_{2}$ of a closed one-form $\sigma$ on $L_{2}$ which is also coclosed in a certain metric on $T^{*}L_{2}$ in a first order infinitesimal neighbourhood of $L_{2}$. In this small open set, write $\sigma=df$, so that $d^{*}df=0$ and $f$ is harmonic; thus by the maximum principle, it has no local maxima or minima. The Floer index (3.4) of intersection points $df=0$ now reduces to the Morse index of $f$ at isolated critical points. We also have to deal with very degenerate critical points of $f$, though. Assuming for a contradiction that the critical set of $f$ is not all of $L_{2}$, we may perturb $f$ inside any connected component of a small neighbourhood of its critical set such that its value is unchanged on the boundary, where it attains its global maximum and minimum, and is Morse in the interior. (That we may take the extrema to be on the boundary is a consequence of the maximum principle.) We can then perturb $f$ further to arrange its index 1 critical points to be lower (with respect to $f$) than all higher index points (by general position arguments [Mi] Theorem 4.8) and then cancel any local minima with them ([Mi] Theorem 8.1). (There must be index 1 critical points if there are any interior minima, by connectivity of our neighbourhood.)

The upshot is a hamiltonian perturbation of $L_{1}$, using this function, with no Floer index zero intersection points with $L_{2}$. Thus $HF^{0}(L_{1},L_{2})=0$, a contradiction, so in fact $f$ is locally constant and $L_{1}=L_{2}$.

The final statement follows from the fact that the obstructions of [FO3] live in $H^{2}(L)$, and $S^{n}$ is spin. $\square$

As Donaldson pointed out, this proof is similar in flavour to proofs of the Arnold conjecture. If the local situation (of all hamiltonian deformations coming from a fixed function) held globally, the proof would be ‘trivial’, i.e. that of Lemma 4.2 above. Even more simply, if one SLag is a graph in the cotangent bundle of another, we reduce the problem to the uniqueness of harmonic functions of integral zero on $L$, i.e. to $H^{0}(L;\mathbb{C}\,)=\mathbb{C}\,$. To extend this argument globally we need to replace de Rham cohomology $H^{0}(L;\mathbb{C}\,)$ by Floer cohomology $HF^{0}(L;\mathbb{C}\,)$.