# How can I tell whether a manifold is homogeneous?

I have been influenced by this question with many beautiful answers.

Are there any useful practical criteria to say positively that a real connected paracompact smooth manifold $X$ is homogeneous?

I can think of something silly like existence of a finite dimensional Lie algebra of vector fields spanning $T_aX$ at each $a\in X$ but I don't see any practical way of constructing such subalgebra besides giving a homogeneous structure:-)

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{\em Homogeneous} means that there exists a smooth connected Lie group $G$ with a smooth transitive action on $X$. – Bugs Bunny Jul 18 '12 at 9:13
I deleted my answer, since you seem to be interested more interested to decide without knowing the Lie group explicitly in advance. Is this correct? I know only the approach: 1. Compute the sutomorphism group. 2. Show that it acts transitively. – Marc Palm Jul 18 '12 at 10:09
Correct, I am starting with a space. – Bugs Bunny Jul 18 '12 at 10:16
You may be interested in this question: mathoverflow.net/questions/89345/… – MTS Jul 18 '12 at 13:17
@YC: The notion automorphism depends on the category! There are situation, where the diffeomorphism group is quite small and not possibly transitive, e.g. $SL_2(\mathbb{Z})\backslash \mathbb{H}$ is an example. @MTS: That was already included in the question:) – Marc Palm Jul 18 '12 at 13:28

Is Eberlein's theorem relevant to your question? If $M$ is a compact Riemannian manifold of nonpositive section curvature, Eberlein's theorem MR0674166 characterizes when $M$ is a Riemannian symmetric space of noncompact type.

And if that is of interest, there is a more recent and more powerful version due to Farb and Weinberger, MR2456886.

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Good one! Albeit not very useful for what I had in mind... I am voting it up but wait for more answers... – Bugs Bunny Jul 19 '12 at 8:23
@Bugs Bunny: that's the point to put the question CW: you don't have to choose a correct answer; indeed answers can typically sufficient conditions of independent interest. Anyway they will still be ranked by votes. – YCor Jul 19 '12 at 9:04
There's a local characterization of the curvature tensor of a symmetric space - its covariant derivative vanishes: mathoverflow.net/questions/22087/constant-curvature-manifolds/… if the space is simply connected, then it's also globally a symmetric space. I guess Bugs might be interested in a similar characterization of the curvature tensor of a general homogeneous space? – Ian Agol Jul 19 '12 at 17:05

as mentioned in various comments there are many geometric conditions of various type that imply homogeneity (or local homogeneity). Constant curvature, parallel curvature tensor, almost flat metric, quarter-pinched curvature, nonnegative bisectional curvature and so on. But in purely topological terms the best you can hope for is necessary conditions of the kind mentioned in the question you linked. Necessary and sufficient conditions are pretty much impossible because topological recognition problems are hard.

For example, if you look at compact 2-connected homogeneous spaces (or even biquotients) then there are only finitely many of them in every dimension since nonabelian simple Lie groups have only finitely many irreducible representations in every dimension. so in principle the recognition problem for such manifolds ought to be straightforward: just check that that $\pi_1(M)=\pi_2(M)=0$ and compare $M$ to a finite list. However, that last step is actually quite tricky. Surgery theory sort of tells us how to do it but the method is hardly practical.

To give a specific example from my own experience. When classifying biquotients with singly generated rational cohomology rings I and Wolfgang Ziller had to deal with one specific biquotient $M^{11}=G_2//SU(3)$ given by a representation $\rho:SU(3)\to G_2\times G_2$ where the representation on the left has index 2 and on the right has index 3. We were able to show that $M^{11}$ is almost diffeomorphic to the unit tangent bundle $T^1S^6$ which is the homogeneous space $SO(7)/SO(5)$. this means that they differ at most by a connected sum with an exotic sphere. but we couldn't decide if they are actually diffeomorphic. to do that one needs to compute the Eells-Kuiper invariant of $M$ which would couldn't do as that requires writing $M$ as a boundary.

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Not really an answer, but too long to be a comment.

I have thought about this question before, without much progress. It seems difficult. Even just to recognize if a manifold has the structure of a Lie group seems to be a difficult problem. And a manifold can be a Lie group in different ways.

For instance, the exponential map from the Lie algebra $\mathfrak{n}$ of strictly upper triangular $n \times n$ matrices is a diffeomorphism with the group $N$ of unipotent upper triangular matrices, so $N$ has its natural Lie group structure as well as the abelian Lie group structure obtained by transporting the addition from $\mathfrak{n}$ through the exponential map.

Thinking about it algebraically, asking for a manifold to be a Lie group is the same as asking for its function algebra to be a Hopf algebra (with some topology, perhaps), and the responses to this question seem to indicate that it is not very easy to recognize a Hopf algebra from the algebra structure alone.

If you want a homogeneous space rather than a Lie group, you are then asking for the function algebra to be a (right or left) coideal subalgebra of a Hopf algebra. That seems even more difficult than looking for a Hopf algebra structure. And again, there can be no canonical construction because the same manifold can be a homogeneous space of different Lie groups: for example, $S^2$ is a homogeneous space of $SU(2)$, $SL(2,\mathbb{C})$, $O(3)$, and $SO(3)$.

So I am pessimistic about there being a nice characterization, or even about the existence of simply stated sufficient conditions. But I'll be watching this thread and hoping that somebody comes up with something!

Edit: Oh yeah! I forgot to mention something. Thinking further about the problem of identifying a Lie group, given only the manifold structure: the multiplication map on a Lie group $G$ turns the cohomology ring of $G$ into a graded Hopf algebra. I don't know the details, but my understanding is that (perhaps with some assumption on the cohomology groups being finitely generated free modules over the coefficient ring?) that if the cohomology ring of a manifold has the structure of a graded Hopf algebra, then the manifold is an $H$-space, i.e. it has a multiplication and a unit map, but it is not necessarily associative. For instance $S^7$ is an $H$-space, but not a Lie group. So cohomology gets you a certain distance, but not all the way.

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A connected manifold carries a Lie group structure iff it is diffeomorphic to $\mathbf{R}^n\times K$ with $K$ compact Lie group. This boils down this specific question to the compact case for which there's a complete classification. For homogeneous spaces, it's more complicated, even sticking to the compact case. In small dimension, we can sort of enumerate the possibilities. Various sufficient conditions of independent interest could be stated so the question might be made CW. – YCor Jul 19 '12 at 5:48
OK, interesting. Where can I find more details about that? And (this may be a naive question) is it easy to recognize when a manifold splits as a product of $\mathbb{R^n}$ with something compact? – MTS Jul 19 '12 at 6:14
Hopf algebraic thinking is productive for proving abstract properties but may not be useful here. I suspect that Yves' statement should be a relatively straightforward exercise on Lie algebras (Malcev decomposition, followed by Iwasawa decomposition)... which brings me back to my question – Bugs Bunny Jul 19 '12 at 8:12