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I wholeheartedly agree with both of Chris Gerig's suggestions. The small McDuff-Salamon book on holomorphic curves ("J-holomorphic curves and quantum cohomology", available on McDuff's webpage) has a …

According to Proposition 3.9 of Ozsvath and Szabo's original paper "Holomorphic disks and topological invariants for closed three-manifolds" there are indeed topological conditions one can put on the …

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 …

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 …

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 …

I noticed this question has been bumped up to the front page, and the most recent answer is about 8 years old: the subject has moved on since then, and more has been written. Here is my understanding …

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 …