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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Sep 15, 2014 at 7:08 history edited მამუკა ჯიბლაძე CC BY-SA 3.0
Added a variation in light of the comment by @YCor
Sep 13, 2014 at 23:02 comment added YCor You get a free orbit as follows: for every $g\in G$, the set of fixed points of $G$ is a $\le 1$-dimensional submanifold, and so is the union over the finite group $G$. Thus any point outside this union is in a free orbit.
Sep 13, 2014 at 21:25 comment added მამუკა ჯიბლაძე @YCor Thank you, spectacular. The reference seems to be L.Greenberg, Maximal groups and signatures, Ann. Math. Studies 79 (1974) 207–226, and in fact if I am not mistaken it also can be even realized as the group of all isometries! (Except I do not quite see whether there always is a free orbit.) I have to think whether this can be used for my purposes. Maybe I should add positive curvature condition or some homogeneity or something like that...
Sep 13, 2014 at 14:30 comment added YCor Every finite group embeds in the isometry group of a compact hyperbolic surface. Hence $m'_2(\Sigma)\le m_2(\Sigma)\le 2$ for every finite group $\Sigma$. But most likely this is useless for combinatorial optimization purposes.
Sep 13, 2014 at 11:27 history asked მამუკა ჯიბლაძე CC BY-SA 3.0