Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Every manifold that I ever met in a differential geometry class was a homogeneous space: spheres, tori, Grassmannians, flag manifolds, Stiefel manifolds, etc. What is an example of a connected smooth manifold which is not a homogeneous space of any Lie group?

The only candidates for examples I can come up with are two-dimensional compact surfaces of genus at least two. They don't seem to be obviously homogeneous, but I don't know how to prove that they are not. And if there are two-dimensional examples then there should be tons of higher-dimensional ones.

The question can be trivially rephrased by asking for a manifold which does not carry a transitive action of a Lie group. Of course, the diffeomorphism group of a connected manifold acts transitively, but this is an infinite-dimensional group and so doesn't count as a Lie group for my purposes.

Orientable examples would be nice, but nonorientable would be ok too.

share|improve this question
3  
Well, a compact surface of genus at least two is the quotient of $\mathbb{H}$ by a group action, and $\mathbb{H}$ is a homogeneous space for $\text{SL}_2(\mathbb{R})$, so in some sense these examples still come from Lie groups. –  Qiaochu Yuan Feb 24 '12 at 0:48
10  
Well a compact orientable surface of genus two or more has no transitive action of a compact Lie group because such an action would necessarily preserve a Riemann metric and therefore a conformal structure. The group of conformal automorphisms of such a surface is finite. –  Tom Goodwillie Feb 24 '12 at 0:52
8  
I must say I am surprised that this question has garnered two votes to close as "not a real question." Would anybody care to explain? The answers below seem to imply (to me, at least) that the question isn't so trivial. –  MTS Feb 24 '12 at 5:04
4  
@MTS: Sometimes people underestimate a question, oversimplifying it. I suspect in this case it's just a mistake on the part of the people who cast votes to close. But if they don't say anything it's hard to tell. I don't think you have to worry about this thread being closed. –  Ryan Budney Feb 24 '12 at 5:11
1  
As you mention, locally symmetric spaces give a huge class. The example you describe with the genus at least two is actually $M = \Gamma \backslash PSL_2( \mathbb{R}) /PSO(2)$, where $\Gamma \cong \pi_1(M)$. –  Marc Palm Feb 24 '12 at 13:03

7 Answers 7

up vote 46 down vote accepted

$\pi_2$ of a Lie group is trivial, so $\pi_2(G/H)$ is isomorphic to a subgroup of $\pi_1(H)$, which is finitely generated (isomorphic to $\pi_1$ of a maximal compact subgroup of the identity component of $H$). But $\pi_2$ of a closed manifold is often not finitely generated. For example, the connected sum of two copies of $S^1\times S^2$ has as a retract a punctured $S^1\times S^2$, which is homotopy equivalent to $S^1\vee S^2$ and so has universal cover homotopy equivalent to an infinite wedge of copies of $S^2$.

EDIT This ad hoc answer can be extended as follows: All I really used was that $\pi_2(G)$ and $\pi_1(G)$ are finitely generated. But $\pi_n(G)$ is finitely generated for all $n\ge 1$ (reduce to simply connected case and use homology), so $\pi_n(G/H)$ is finitely generated for $n\ge 2$. That leads to a lot more higher-dimensional non simply connected examples.

share|improve this answer
2  
Something puzzles me. You seem to be using that a subgroup of a finitely generated group is finitely generated. But this is false (e.g. commutator subgroup of $F_2$). What am I missing? (I upvoted your answer back in the day, so hopefully I understood it then...) –  Mark Grant Jul 18 '13 at 15:46
2  
@MarkGrant, $\pi_1(H)$ is abelian. –  Mariano Suárez-Alvarez Jul 18 '13 at 21:14
    
@Mariano: Ah yes, thanks. –  Mark Grant Jul 19 '13 at 6:17

Apart from already mentioned non simply connected examples most simply connected manifolds are also not homogeneous. One easy criterion is that simply connected homogeneous spaces are rationally elliptic, i.e. they have finite dimensional total rational homotopy. That is because any connected Lie group is rationally homotopy equivalent to a product of odd dimensional spheres. so a homogeneous space is elliptic by a long exact homotopy sequence.

Most simply connected manifolds are not rationally elliptic. For example the connected sum of more than two $CP^2$'s or $S^2\times S^2$'s. This is easily seen by looking at their minimal models. But even without computing minimal models it's known that an elliptic manifold $M^n$ has nonnegative Euler characteristic and has the total sum of its Betti numbers $\le 2^n$. So anything that violates either of these conditions such as the connected sum of several $S^3\times S^3$'s is definitely not rationally elliptic and hence can not be a homogeneous space or even a biquotient.

As for higher genus surfaces it should not be hard to show that they can not be homogeneous spaces $G/H$ even if you don't assume that $G$ acts by isometries. If $G/H=S^2_g$ and the $G$ action is effective then for any proper normal $K\unlhd G$ which does not act transitively on $S^2_g$ we must have $K/(K\cap H)=S^1$. But then $G/K$ is also 1-dimensional and hence also a circle which is obviously impossible. This reduces the situation to the case of $G$ being simple which can also be easily ruled out for topological reasons.

share|improve this answer
    
In the last paragraph why does $K/(K\cap H)$ have to be compact? –  Tom Goodwillie Feb 24 '12 at 13:01
    
sorry, I meant to say that that we should look at a closed proper normal subgroup $K$. –  Vitali Kapovitch Feb 24 '12 at 13:21
1  
Even so, how do you rule out the possibility that $K/(K\cap H)$ is a noncompact $1$-manifold? –  Tom Goodwillie Feb 24 '12 at 13:38
1  
hmm, I thought this was obvious but it does require some argument. how about this then. in such situation all orbits of $K$ must be 1-dimensional submanifolds of $S^2_g$ which gives a foliation of $S^2_g$ which is impossible since its Euler characteristic is not zero. –  Vitali Kapovitch Feb 24 '12 at 14:23

Atiyah and Hirzebruch gave a rather dramatic answer to your question in their paper "Spin Manifolds and Group Actions": if $M$ is a compact smooth spin manifold of dimension $4k$ whose $\hat{A}$-genus is nonzero then no compact Lie group can act on $M$ nontrivially, let alone transitively! The proof uses Atiyah and Bott's Lefschetz fixed point theorem in a clever way.

Unfortunately I don't have a simple example of such a manifold lying around, though I know that there are plenty of examples among 4-manifolds. It's possible that some 4 dimensional lens spaces would do the job.

share|improve this answer
    
I guess my answer doesn't rule out the possibility that $M$ is a homogeneous space for a non-compact Lie group, but perhaps it's still interesting. –  Paul Siegel Feb 24 '12 at 3:03
5  
there are indeed many such 4-manifolds. connected sum of any number of $K3$-surfaces is the most standard example. so if anything like this is a homogenous space $G/H$ then the maximal compact subgroup of $G$ must be zero-dimensional which means that $G$ must be contractible. This is of course impossible which means that none of such manifolds can be homogeneous. –  Vitali Kapovitch Feb 24 '12 at 3:13
3  
@Paul : You should perhaps say "no compact connected Lie group", as there might be finite order symmetries. This is equivalent to "no non trivial $S^1$ action". –  BS. Feb 24 '12 at 11:10

It is a result of Mostov that any compact homogeneous manifold must have nonnegative Euler characteristic:

http://www.ams.org/mathscinet-getitem?mr=2174096

That should provide plenty of counterexamples. :)

share|improve this answer
    
Now I see this is also implied by Vitali's post... I wonder if Mostov knew this? –  Dylan Wilson Feb 24 '12 at 6:59
    
This is also proven in R. Hermann "Compactification of homogeneous spaces. I." J. Math. Mech. 14 1965 655–678. Apparently Mostow did not know that paper. –  Johannes Ebert Feb 24 '12 at 17:08
    
Are you sure they aren't assuming the group is compact? Mostow doesn't assume this. He also has some classification results. –  Dylan Wilson Feb 24 '12 at 19:20
2  
@Dylan: I am pretty sure Hermann, as well as Mostow, does not assume compactness of $G$, only of $G/H$. If $G$ is compact, then in fact the result is much older and, nowadays, fairly easy. If $G$ is connected and compact and the rank of $H$ equals the rank of $G$, then there is a fibre bundle $H/T \to G/T \to G/H$, where $T$ is the maximal torus. By Hopf-Samelson, the Euler numbers of $G/T$ and $H/T$ are given by the order of the Weyl groups, hence both positive, so $\chi(G/H)>0$. –  Johannes Ebert Feb 24 '12 at 20:54
2  
If $H$ has smaller rank, $S \subset H$ is a maximal torus of $H$, $S \subset T \subset G$ a maximal torus of $G$. The fibre bundle $T/S G/S \to G/T$ shows that $\chi(G/S)=0$. The fibre bundle $H/S \to G/S \to G/H$ shows $0=\chi(G/S) = \chi(H/S) \chi(G/H) = # W_H \chi(G/H)$ and so $\chi(G/H)\geq 0$. –  Johannes Ebert Feb 24 '12 at 20:57

I would think that many examples from 3-manifold theory would work. Take any compact, oriented, irreducible 3-manifold $M$ whose torus decomposition is nontrivial and has at least one hyperbolic piece. Such examples at least have no locally homogeneous Riemannian metric, as a consequence of Thurston's analysis of the 8 geometries of 3-manifold theory. A specific example of this sort can be obtained from a hyperbolic knot complement in $S^3$ by deleting an open solid torus neighborhood of the knot and doubling across the resulting 2-torus boundary; the doubling torus produces a characteristic $Z^2$ subgroup of $\pi_1(M)$. These examples have universal cover homeomorphic to $R^3$, and so they have trivial $\pi_2$. By the homotopy exact sequence there would be a quotient group $\pi_1(G) / \pi_1(H)$ identified with a subgroup of $\pi_1(G/H)$ whose quotient set is $\pi_0(H)$. Perhaps, in order to get a proof, one can analyze this situation by considering the intersection of the $Z^2$ subgroup with the $\pi_1(G) / \pi_1(H)$ subgroup.

share|improve this answer
    
This is a bit over my head, but a nice answer nonetheless. Thank you. –  MTS Feb 24 '12 at 22:01
    
np. I'll throw in that this idea of analyzing the $Z^2$ subgroup intersected with $\pi_1(G) / \pi_1(H)$ is kind of what is going on in the proof of Thurston's 8 geometries theorem. It seems to me that your question, in the 3-manifold context, is a kind of generalization of the 8 geometries theorem. –  Lee Mosher Feb 24 '12 at 23:28

Here is a proof that any closed surface $S$ of genus at least 2 cannot support a homogeneous Riemannian metric. In fact, let $g$ be any such metric. Being homogeneous, $g$ has constant curvature, and e.g. by Gauss-Bonnet such a curvature must be negative. Therefore, $(S,g)$ is homotetic to a hyperbolic surface, and it is well-known that the isometry group of any closed hyperbolic surface is finite (in fact, if $h$ is the genus of $S$, then $(S,g)$ admits at most $84(h-1)$ isometries).

share|improve this answer
    
This is a nice argument, but I am not asking for a homogeneous Riemannian metric. This doesn't rule out the possibility of having a transitive action of a Lie group that is not an action by isometries for any Riemannian metric. –  MTS Feb 24 '12 at 1:05
    
@MTS: if $M$ is a homogeneous space for a Lie group $G$ and $x \in M$ is a point with compact stabilizer, then you can choose a $G$-invariant inner product on $T_x(M)$ and homogeneity gives you a $G$-invariant Riemannian metric on $M$, doesn't it? I guess it is possible that there are no points with compact stabilizers, but in any case this shows at least that any compact surface of genus at least two is not a homogeneous space of a compact Lie group. –  Qiaochu Yuan Feb 24 '12 at 1:36
    
Qiaochu, I think what you are saying is the following: if $M = G/H$, then the tangent bundle of $M$ is the homogeneous vector bundle induced by the adjoint representation of $H$ on $\mathfrak{g}/\mathfrak{h}$. If $H$ is compact then we can integrate with respect to the Haar measure of $H$ to get an invariant inner product on $\mathfrak{g}/\mathfrak{h}$. Then translating around with $G$ gives the invariant metric on $M$. Is that right? I don't think you can choose a $G$-invariant inner product on $T_x(M)$ since $G$ doesn't preserve the point $x$ in general. –  MTS Feb 24 '12 at 4:45
    
@MTS: sorry, I meant a $\text{Stab}(x)$-invariant inner product. –  Qiaochu Yuan Feb 24 '12 at 17:42

In the homotopy exact sequence of the fiber bundle $G\to G/H$ the group $\pi_i(G/H)$ sits between $\pi_i(G)$ and $\pi_{i-1}(H)$. For example, if $i=1$, then $\pi_1(G)$ is abelian, and $\pi_0(H)$ is finite (as $H$ is compact). Thus $\pi_1(G/H)$ has abelian subgroup of finite index. Surely there are lots of manifolds that do not have this property, e.g. any closed negatively curved manifold does not. Connected sum of several lens spaces is another example.

By the way, it is my opinion that the class of compact homogeneous spaces is very rich, and their finer topological properties are still poorly understood. For example, classifying homogeneous spaces up to diffeomorphism in a given homotopy type is quite challenging, and it is not easy to cook up a homogeneous spaces with prescribed topological features.

share|improve this answer
1  
the OP is interested in a more general situation than homogeneous Riemannian manifolds (see his comments above). thus one can not assume that $H$ is compact. So things like quotients by uniform lattices in semisimple Lie groups or nilmanifolds are fair game. –  Vitali Kapovitch Feb 24 '12 at 19:30
    
@Vitali: I was answering to OP's request "in finding an example of a compact manifold which is not a homogeneous space of any compact Lie group". If a compact Lie group acts transitively on a manifold, then the isotropy subgroup is closed, hence compact. –  Igor Belegradek Feb 24 '12 at 22:55
    
@Igor sorry, this was not clear from your answer. the OP states the question 3 times and only mentions compactness once. He should modify the question IMO. It's also clear from this comments that he is interested in the general case and it creates confusion because some answers assume that G is compact and others do not. –  Vitali Kapovitch Feb 25 '12 at 0:21

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.