# A followup on non-homogeneous spaces.

This question asks for an example of a manifold which is not a homogeneous space of any Lie Group, and many examples are given in the answers. However: is there a an example known with a metric of positive sectional curvature (this, apparently, is a question asked by Marcel Berger many years ago, so was open then, but I have no idea of its current status...)?

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"is not a homogeneous question of any Lie Group" ??? $\;$ – Ricky Demer Feb 26 '12 at 2:17
You probably want to assume simply connected too, otherwise I think there are likely some non-homogeneous spherical space forms. – Ian Agol Feb 26 '12 at 4:29
@Agol, yes, simply connected is a natural assumption, though had I asked the question that way, we would have missed out on @Yves' nice example... – Igor Rivin Feb 26 '12 at 6:08

Yes, Eschenburg constructed an infinite family of simply connected 7-dimensional examples and proved that many of them are not homotopy equivalent to homogeneous spaces. His examples are biquotients however. In fact the only known examples of closed positively curved manifolds are biquotients or spaces of cohomogeneity one and above dimension 13 only homogeneous examples (and space forms) are known.

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Yes, some quotients of the 3-sphere are examples. Namely, view $S^3$ as the group of quaternions of unit one. Consider two finite subgroups $F_1,F_2$ of $S^3$, with $F_1$ not abelian, and $F_2$ cyclic of odd prime order $p$ not dividing the order of $F_1$. So $F_1\times F_2$ acts freely on the sphere by $(g,h)\cdot z=gzh^{-1}$. I claim that the quotient $X=F_1\backslash S^3/F_2$ is not homeomorphic to a homogeneous space of any connected Lie group $G$.

Assume the contrary. First, by a result of Montgomery (Proc AMS 1950), any maximal compact subgroup of $G$ acts transitively (it's a general fact, using only that $X$ has finite fundamental group). So we can assume that $G$ is compact. Then a result of Montgomery-Samelson (Annals 1943) tells us that some simple $S$ subgroup acts transitively. (The result concerns actions on spheres, and is applied by considering the action of a finite covering of $G$ on $S^3$.) The isometry group of a 3-dimensional Riemannian manifold is at most 6-dimensional, and the only simple compact Lie group of dimension at most 6 is $SO(3)$ (and its universal covering). Thus $X=S^3/F$ for some finite subgroup $F$ of $S^3$, isomorphic to $F_1\times F_2$. But in $S^3$, the centralizer of any nontrivial element of odd order is abelian. This is a contradiction.

These examples are not simply connected, of course. It seems that the other answers deal with the much more delicate simply connected case.

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At least for compact groups, I think your question was answered by Peterson and Wilhelm where they showed the Gromoll-Meyer sphere has a metric of positive sectional curvature.

That, when combined with Wu-Chung Hsiang's work from the 60's says that exotic spheres are not homogeneous spaces. I'm not sure which is the best reference but Wu-Chung has a few on similar topics, for example, his paper "On Compact Subgroups of Diffeomorphism Groups of Kervaire Spheres" Annals 1967 mentions that compact subgroups of Diff of an exotic sphere has dimension bounded above by $m^2/8 + 1$, where $m$ is the dimension of the sphere.

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I would be careful citing Petersen-Wilhelm example. It has not been published yet and I don't know anybody who has verified the proof. – Vitali Kapovitch Feb 26 '12 at 4:27
also note that compactness of $G$ can always be assumed if the manifold in question is closed and simply connected. If $M=G/H$ and $G$ is connected and simply connected then H must also be connected since $M$ is simply connected. Then let $\bar H$ be a maximal compact subgroup of $H$. It's contained in $\bar G$ - a maximal compact subgroup of G. Then $G/H$ is homotopy equivalent to $\bar G/\bar H$ which means that $\bar G$ acts on $M$ transitively. – Vitali Kapovitch Feb 26 '12 at 4:50
I haven't read the Petersen-Wilhelm paper, but I also didn't think it would be contentious. – Ryan Budney Feb 26 '12 at 4:56
I would defer to Vitali on the question of existence of "good housekeeping seal of approval"... – Igor Rivin Feb 26 '12 at 6:07