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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 …

I had a look at the paper of Givental, https://math.berkeley.edu/~giventh/papers/tor.pdf and don't see this statement... If this statement were true, Conjecture 6.1 of Eliashberg from 2015 would be …

I hope that the following answers some parts of the question. 1) a) It is not true that a generic contactomorphim doesn't have fixed points. For example, let $M$ be the three-dimensional torus that …

Nick Lindsay and have just proved that such a manifold indeed exists. And surprise, surprise, this is Tolman's manifold. See Theorem 1.3 and Corollary 1.4 of our paper: https://arxiv.org/abs/1912.027 …

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 …

A positive answer to this question should follow from Berger's holonomy classification and the following statement, which I believe is correct: Statement. For any dimension $n$ there exists a positi …

One way to prove this is as follows. First, from the assumption $B(1)\subset \mathbb C^2$ and the centre of $B(1)$ is $(0,0)$. Now we need the following two claims. Claim 1. Any straight geodesic u …

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 …

Concerning 2) one can, of course, take the product of two curves of higher genus to get a counter-example. In general, to have a statement as you want, one should look for rigid complex surfaces of ge …

Let us consider the case of toric varieties of real dimension $4$ and prove they can not be represented as such a quotient unless they have second Betti number $1$ or $2$. Proof. Let us introduce …

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 …

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 …

I think the answer to your question should be positive and below is a sketch of what should work (I think). Any real analytic manifold can be realised as the real part of a complex projective manifol …

I have to apologize, in fact the answer to the second question is still unknown. Namely, up to now all known symplectic manifolds of dimension 4 that have negative Euler characteristic are blow ups of …

I tried to investigate the question for some time and sent emails to several experts in the field. For the moment the conclusion seems to be the following: The statement is well known to experts, was …