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In the comments on the MathSE question "Is Seifert-Weber space homogeneous for a Lie group?", it is claimed that if $ M $ is a manifold which admits a finite volume hyperbolic metric (constant negative sectional curvature) then there can exist no Lie group $ G $ acting (smoothly) transitively on $ M $. Why is that?

Thoughts on the case of compact surfaces: A compact hyperbolic surface must have negative Euler characteristic. Mostow, G. D., A structure theorem for homogeneous spaces, Geom. Dedicata 114, 87-102 (2005). ZBL1086.57024 proves that a compact manifold admitting a transitive action by a Lie group must have nonnegative Euler characteristic $ \chi \geq 0 $. Thus compact hyperbolic surfaces are never homogeneous.

The same Mostow paper above is used by Moishe Kohan to prove this result for 3 manifolds here https://mathoverflow.net/a/409329/387190

Update: More generally, a Lie group cannot act transitively on any complete finite volume negatively curved manifold

Ian Gershon Teixeira
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  • @AndyPutman I suppose that's obvious because you could just take a compact hyperbolic surface and remove a point to create a finite volume hyperbolic surface which is not compact. I've edited the question to reflect this. – Ian Gershon Teixeira Sep 10 '23 at 23:14
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    Simply removing a point from a compact hyperbolic surface gives a metric that isn't complete. But there are plenty of complete hyperbolic metrics on punctured surfaces too. – Andy Putman Sep 10 '23 at 23:22
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    One can assume $G$ to be simply connected. Then the action lifts to the universal cover $X$ of $M$. Then $G$ must normalise the group of deck transformations $\Gamma$. As $\Gamma$ is discrete and $G$ connected, it must centralise $\Gamma$. In other words, $\Gamma$ and $G$ commute on $X$. This sounds weird. –  Sep 11 '23 at 17:02
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    Here’s a sketch when $M$ is compact (as Seifert—Weber space is). Let $f$ be an isometry homotopic to the identity, and let $F$ be its natural lift to the universal cover. By compactness, $F$ moves every point a bounded distance. Now note that the identity is the only isometry of hyperbolic space that moves every point a bounded distance. – HJRW Sep 11 '23 at 20:47
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    One more approach. A theorem of Kobsyashi states that a homogeneous Riemannian manifold with negative definite Ricci tensor and nonpositive sectional curvature is simply connected.

    Apply this to $M$ hyperbolic to conclude $M = \Bbb H^n$ which is not finite volume. Kobayashi's proof echoes the argument of the comment two above mine.

     – mme Sep 12 '23 at 00:06
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    @mme but what if $ G $ is not acting by isometries? Does this argument still work to show that a hyperbolic manifold with a transitive $ G $ action must be simply connected? If so then I certainly agree that a simply connected hyperbolic manifold must be $ \mathbb{H}^n $ and thus is not finite volume. – Ian Gershon Teixeira Dec 20 '23 at 22:59
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    Good point. ${}$ – mme Dec 20 '23 at 23:43
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    There's a theorem of Borel (1969) that would allow you to conclude the Lie group would have to be contractible. His theorem states that if a circle acts effectively on a closed aspherical manifold then the inclusion of an orbit (of the action) into the manifold induces an injection on the fundamental group, with image in the centre of the fundamental group of the manifold. So that rules out all manifolds with negative curvature. – Ryan Budney Dec 21 '23 at 07:35
  • @RyanBudney That sounds very interesting! (and I love aspherical manifolds!) Do you know what the name of the Borel reference is? – Ian Gershon Teixeira Dec 22 '23 at 21:07
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    It's called "On periodic maps of certain $K(\pi,1)$". It's from Borel Oeuvres Vol III, i.e. it was unpublished. You can probably get it from your library but if you have any trouble send me an e-mail. – Ryan Budney Dec 22 '23 at 22:40
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    In many cases one should be able to use Yanos result that any manifold with a non trivial $S^1$ action has vanishing simplicial volume. – ThorbenK Jan 02 '24 at 18:59

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