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I'm not sure if this is quite what you're looking for, but I always found it a useful example to keep in mind. Take the genus 1 invariant for degree 1 maps into $P^1$. There are no degree 1 holomorphi …

Not a full answer, but a partial answer to your question 1a. Torus-invariant Kaehler metrics (in particular complex structures) were constructed by Guillemin, just starting with data on the moment pol …

Sometimes it is, sometimes it isn't. The spheres living over the edges generate the second homology, so you can read off the action on $H_2$ from that. For $S^2\times S^2$ (square) the action on homol …

Though not about solving the nontransversality problem, Fukaya's paper Unobstructed immersed Lagrangian correspondence and filtered A infinity functor is the state of the art in why nontransversality …

In this case, the J-holomorphic curves are all constant, so the evaluation pseudocycle is the tridiagonal $\{(x,x,x) : x\in M\}$. You take cycles $A_1,A_2,A_3$ Poincare dual to $a_1,a_2,a_3$ respectiv …

Mutation doesn't even change the integral affine base, which is why it doesn't change the symplectic manifold. All you're doing is changing the way the integral affine base is drawn. If you're given a …

If you have a lattice polytope then you get a projective toric variety. If your polytope is not a lattice polytope then you can still construct a symplectic toric manifold (e.g. by performing symplect …

First, I assume you want $w$ to be holomorphic (or else it isn't true). If the target is really $S^2$ with boundary on the equator $L$ then this should be easy to prove directly. Take a point $p$ not …

The relative homology long exact sequence puts this group in between $H_2(\mathbb{CP}^n)=\mathbb{Z}$ and $\mathbb{Z}/2$. It maps surjectively to $\mathbb{Z}/2$. Let D be a generator for relative homol …

This is not possible. There is something called the growth rate of symplectic cohomology which is subexponential for affine varieties and exponential for cotangent bundles of higher genus surfaces (am …

I assume you're asking about embedded or at least immersed discs, in order to make sense of the normal bundle. If so, let's fix an immersion $\iota\colon\Sigma\to M$ and observe that the pullback bund …

On the suggestion of DamienC, I'm converting my comments into an answer. I didn't do this before because I don't really know the answer to the deeper question of why the microlocal sheaf category is t …

Since you tagged this with "symplectic geometry", I'm going to give an answer from a symplectic geometry perspective, which may not be what you're looking for, but (as a symplectic person) I find it i …

One classic thing to do is to take a sequence of holomorphic curves with a tangency condition and assume that the tangency condition still holds in the limit: if the limit is a multiple cover with a b …

I only just noticed this question, so maybe it's too late, but here's an answer. Note that some symplectic manifolds (like $\mathbf{CP}^2$) contain no Lagrangian spheres, so this complex is then empt …