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In a certain sense, symplectic geometry (or safer to say symplectic topology) as we know it now was not existing before Arnold formulated these conjectures. So many would say that Arnold conjectures g …

There are counter-examples, hope they answer your question completely, just take any non-simply connected $G$ and consider its action on $T^*G$. The simplest case is: Let $M$ be the cylinder $S^1\tim …

This is not a complete answer to the question, and I don't know if a complete answer is written down anywhere in the literature. In the first revison of the answer I tried to adress all the 10 commen …

Here are some remarks about your definition. 1) $H_1(A,\partial A)$ is just one-dimensional, it is generated by a path that joins two sides of $A$. 2) The definition that you gave works for the ann …

I think, that in order to answer this question it is worth to conisder the complex algebraic analog of this question. Namely, suppose we have a $\mathbb C^*$ action on a projective manifold $V^n$. Can …

Yes, it is always possible. Of course, if $M$ has dimension $2$ we need to assume that $\gamma$ is embedded in $M$ in which case the statement is easy (but if $\gamma$ have self-intersections, the sta …

This answer is rewritten and include more details First of all I highly recommend you the article of Paul Biran From Symplectic Packing to Algebraic Geometry and Back available on the page http://ww …

Edited. For the definitions that you mention "Simplectic Fano" can be non-montone. For example, you can take a $4$-dimensional Kahler non-agebraic torus that does not have complex curves at all. Such …

I am pretty sure that the answer to this question is unknown. And I would guess it should be hard (via impossible) to construct such a manifold. Here is some argumentation: In dimension $6$ the class …

This can always be done. Let's first treat the case when $\Sigma$ is not a torus. Then take any symplectic $4$-manifold $(M,\omega)$ where $\Sigma$ can be embedded as a Lagrangian surface. Now, take a …

This doesn't need to hold. For example, if one takes a $(T^4,\omega)$ with a constant symplectic structure $\omega$, in order for it to have a fibration by Lagrangian tori one should be able to find a …

I'll give a positive answer for two generic totally real planes in $\mathbb C^2$. I believe this generalises to larger $n$, though I don't prove it - just give a possible plan of a proof with one step …

These two symplectic manifolds are canonically symplectomorphic. Notice first, that the map $\mu$ vanishes on the sub-bundle of $T^* M$ of 1-forms vanishing on the fibers of the fibration $M\to X$. …

The answer to this question is YES. I assume you want $A$ and $B$ to be connected. Already in the case of four manifolds two symplectic surfaces can have negative intersection. To construct an examp …

There are some very good reasons why the majority of calculations are done for algebraic manifolds. Maybe the most naive reason is as follows: it is harder to solve PDEs than to draw lines through tw …