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

This is a basically an adjusted version of my earlier question about how to define a convolution algebra on a general Riemannian manifold. The motivation for asking such a question of course comes from the observation that if G is a group and X is a manifold and the action of G on X is transitive, then the pullback from each point in X to its orbit is faithful. This then cuts out an ideal in the convolution algebra on G which would (hopefully) correspond to some type of general convolution on X (and that would be pretty handy to have for many obvious reasons).

My intuition is that for 2D surfaces (which is the case I am most interested in right now), the group is going to be something like $PSL(2) / \pi(X)$ with the action obtained by the pushforward of the action of $PSL(2)$ on $RP^2$ by the universal covering of $RP^2 / \pi(X)$. Of course, trying to work all of this out via quotient relations is an enormous pain in the neck, so it would be nice to maybe avoid some headaches and instead try to find maybe some standard references for this sort of thing (if they exist at all).

share|improve this question
add comment

4 Answers 4

up vote 13 down vote accepted

At least in the compact case, there's a topological obstruction. In a 2005 paper, Mostow proved that a compact manifold that admits a transitive Lie group action must have nonnegative Euler characteristic. Here's the reference:

MR2174096 (2007e:22015) Mostow, G. D. A structure theorem for homogeneous spaces. Geom. Dedicata 114 (2005), 87--102.

Of course, even if X admits a transitive Lie group action, most Riemannian metrics on X will not be homogeneous. (You didn't say whether you wanted actions by isometries, but I assume that's what you're interested in, because otherwise the Riemannian structure on X is irrelevant.) In the 2D case, the only compact, connected, homogeneous Riemann surfaces are the sphere, $RP^2$, the torus (and maybe the Klein bottle?)*, all with constant-curvature metrics. In general, the group has to be compact, because the isometry group of a compact Riemannian manifold is itself compact.

*EDIT 3: An earlier paper by Mostow constructed a transitive group action on the Klein bottle, but I doubt that this action preserves a Riemannian metric. It's a complicated construction, so I haven't had a chance to work through it in detail, but here's the reference:

Mostow, G.D., The Extensibility of Local Lie Groups of Transformations and Groups on Surfaces. Ann. Math., Second Series, (52) No. 3 (1950), 606-636.

I don't know what's known in the noncompact case.

share|improve this answer
    
Thanks! Though I guess this kind of blows the idea out of the water. It would be nice if the theorem went in the opposite direction so that surfaces embeddable in R3 could be equipped with a group action, but I suppose that it is the best one can do. Anyway, I will read through the paper and maybe try to understand a bit better what is going on. –  Mikola Sep 4 '10 at 15:16
    
Jack -- how can a Lie group act transitively on the Klein bottle? –  algori Sep 4 '10 at 16:08
1  
David -- I don't know what the full isometry group is in that case, but here's a proof that the isometry group of a compact Riemannian manifold M must be compact. Let G be the isometry group and H the isotropy group of a point, so G/H is diffeomorphic to M. Then the action of an element of H is determined by what it does to tangent vectors at the base point, so H injects into the group of orthogonal transformations of the tangent space at the base point, and thus it's compact. Since H and G/H are both compact, it follows that G is compact. –  Jack Lee Sep 5 '10 at 0:23
1  
David -- Good question -- my proof certainly doesn't make it obvious that the image of $H$ is closed. Here's a sketch of an argument that it is. Suppose $\phi$ is a limit point in the image of $H$ in the group of orthogonal transformations of the tangent space at the base point $p$, and let $\Phi_i$ be elements of $H$ such that $d\Phi_i \to \phi$. Define a map $X\to X$ by sending each geodesic $\gamma$ starting at $p$ to the geodesic with initial tangent vector $\phi(\gamma'(0))$. This map is a uniform limit of isometries, and thus is an isometry itself. –  Jack Lee Sep 5 '10 at 4:53
1  
David -- yes, the limit map is well-defined. It's uniquely defined away from the cut locus by the recipe I gave; in there it's a uniform limit of isometries and thus an isometry. Then use the fact that an isometry defined on a dense open subset extends by continuity to an isometry on the whole space. To prove uniformity, use comparison theory to show that if $\gamma$ and $\tilde\gamma$ are geodesics starting at the same point, the distance between $\gamma(t)$ and $\tilde\gamma(t)$ can be estimated in terms of $t$, $|\gamma'(0)-\tilde\gamma'(0)|$, and bounds on the curvature. –  Jack Lee Sep 5 '10 at 15:21
show 12 more comments

I think that one of the nicest results in this direction may be the one in Ambrose and Singer's paper "On homogeneous Riemannian manifolds". They give a local NASC for a complete Riemannian manifold to be homogeneous. There is a follow-up book by Tricerri and van Hecke called Homogeneous structures on Riemannian manifolds.

share|improve this answer
add comment

If a connected complete (edit) Riemannian manifold $M$ has an isometric transitive effective action of a Lie group $G$, then the manifold is diffeomorphic to $G/G_x$, where $G_x$ is the stabilizer of a point $x\in M$. Since $G_x$ preserves the metric at $T_x M$, then $G_x$ must be compact (the action $G_x:T_xM\to T_xM$ is faithful by considering the exponential map). Thus, one obtains all such manifolds by quotienting a Lie group by a compact subgroup. Another viewpoint is that the Lie algebra of the identity component of $G$ gives rise to a subspace of the Killing vector fields on $M$. When you restrict the vector space of Killing fields to a point $x$, then the image should map onto the tangent space $T_x M$. I think this is necessary and sufficient.

An interesting example is the Poincare dodecahedral space. In fact, one can decide which elliptic 3-manifolds are homogeneous, as the ones of the form $SU(2)/\Gamma$, where $\Gamma \leq SU(2)$ is finite.

share|improve this answer
    
You said: "...the image should map onto the tangent space $T_xM$. I think this is necessary and sufficient." Necessary, yes, but not sufficient unless $M$ is complete and simply connected. To see that completeness is necessary, just think of an open interval. As for simply connected, the Killing fields will lift to Killing fields on the universal cover, and by a monodromy argument, the local diffeos that they generate will fit together to define global isometries of the universal cover. But, unless these isometries "preserve" the deck translations, these isometries need not descend to $M$. –  Dick Palais Sep 5 '10 at 18:12
    
@ Palais: You're right about the completeness - I suppose I was implicitly assuming this. As for the simple connectivity, I don't believe this is necessary (at least for compact manifolds). A smooth vector field on a compact manifold is complete, and therefore generates a 1-parameter family of diffeomorphisms. If the vector field is a Killing field with respect to a Riemannian metric, then the diffeomorphims are Riemannian isometries. I'll have to think about the non-compact case (i.e. whether Killing fields are complete) - maybe this is what you're getting at? –  Ian Agol Sep 5 '10 at 21:04
    
Thanks, Agol. Yes, Killing vector fields on a complete manifold generate one parameter groups of isometries. So what I said was incorrect. What I MEANT to say is that M is LOCALLY homogeneous iff the LOCAL Killing fields near each point x span $TM_x$. Since these lift to the universal cover, the same is true there, where the condition implies (by a monodromy argument) that the isometry group G is transitive. Whether M itself is homogeneous depends on whether the subgroup of G that commutes with deck transformations is transitive. (For an examples, think of the Klein Bottle and Lens spaces.) –  Dick Palais Sep 6 '10 at 5:59
add comment

RETRACTED: This is nonsense. Note to self: don't answer MO questions before 8am.

If you want your action to be smooth then, by the Poincare--Hopf Theorem, the manifold has $\chi = 0$. This rules out most surfaces.

share|improve this answer
1  
What about $S^2$ ? –  BS. Sep 4 '10 at 14:21
1  
I don't follow. Why couldn't you have a Lie group acting, but every the vector field of every infinitesimal action has fixed points? –  David Speyer Sep 4 '10 at 14:21
    
Oh, right. This is nonsense. Apologies. –  HJRW Sep 4 '10 at 14:29
2  
@Henry: It happens to many of us. –  Ryan Budney Sep 4 '10 at 18:47
add comment

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.