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Oct 30, 2017 at 11:50 vote accept Nick L
Oct 30, 2017 at 11:23 answer added Tobias Diez timeline score: 0
Oct 29, 2017 at 22:13 comment added Nick L The lowest dimension is 4, I can give argument if you want.
Oct 29, 2017 at 21:53 comment added YCor @RobertBryant it gets a little to do with symplectic geometry if you wonder about the minimal dimension of a symplectic manifold with such an action, for example.
Oct 29, 2017 at 21:51 history edited YCor
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Oct 29, 2017 at 14:38 comment added Nick L Nice! Thanks for pointing this out.
Oct 29, 2017 at 14:30 comment added Robert Bryant The question really has nothing to do with symplectic geometry: Given any connected $n$-manifold $N^n$ with a smooth $S^1$-action with infinitely many stabilizer types, then the canonical extension of this action to $T^*N$ will be an example of a Hamiltonian $S^1$-action with infinitely many stabilizer types.
Oct 29, 2017 at 13:53 comment added Nick L Actually now I think about it, the problem can be solved in dimension 4, by gluing together infinitely many Hamiltonian $4$-manifolds via symplectomorphic reduces spaces (with level sets with the same Seifert structure). However, I will leave the question up to see if there is a more elegant solution.
Oct 29, 2017 at 13:39 comment added Nick L Thank you, Yes it does seem there is a Delzant construction for open manifolds there, but after 10 minutes of reading I couldn't extract the statement about the stabilisers.
Oct 29, 2017 at 13:13 comment added Thomas Rot Hmm, googling made me believe that this is a bit more complex than I thought. See this paper: arxiv.org/pdf/0907.2891.pdf . However, I was thinking of a toric manifold with moment map similar to figure 1 in this paper.
Oct 29, 2017 at 13:04 comment added Thomas Rot I guess such an example arises in (real) dimension four. I don't know much about this stuff, but I think there is a delzant type theorem for non-compact toric manifolds. Then you can find a fan in the upper right quadrant in $\mathbb{R}^2$, with vertices something like $(0,0) , (1,1), (3,2), (6,3),\ldots$. Then taking the induced $S^1$ action corresponding to the vector $(0,1)$ should satisfy this condition. But I am someone more knowledgable will chime in.
Oct 29, 2017 at 12:58 history edited Nick L CC BY-SA 3.0
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Oct 29, 2017 at 12:58 comment added Thomas Rot You want $M$ to be connected as well I presume. Otherwise you can take a disjoint copy of $\mathbb{CP^1}$ for every $n\in \mathbb{N}$. The action on this sphere will be the $n$-fold rotation over this sphere.
Oct 29, 2017 at 12:54 comment added YCor I guess it's standard as I found it on Google but it didn't come with a definition (as it's natural to guess it). Here in an abelian group it can be restated as "infinitely many distincts point stabilizers". For non-abelian group actions, probably it means infinitely many up to conjugation.
Oct 29, 2017 at 12:48 history edited Nick L CC BY-SA 3.0
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Oct 29, 2017 at 12:47 comment added Nick L Thanks for the comment. I will clarify in the question, maybe there is a more standard terminology for this?
Oct 29, 2017 at 12:42 comment added YCor Could you define "stabilizer type"?
Oct 29, 2017 at 12:34 history asked Nick L CC BY-SA 3.0