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## Möbius and projective 3-manifolds

A projective 3-manifold is a smooth manifold that admits an atlas with values in the real projective 3-space such that all transition maps are restrictions of projective transformations. A M\"obius 3-manifold is defined similarly, with the projective space replaced with the standard 3-sphere and projective transformations replaced with Möbius transformations. (Recall that a Möbius transformation of the sphere $S^n$ is a self-diffeomorphism of the sphere that preserves the angles of the standard metric; such transformations form a Lie group isomorphic to $SO_{n+1,1}(\mathbf{R}))/\pm I$).

Both projective and Möbius manifolds are particular cases of manifolds admitting an $(M,G)$-structure in the sense of W. Thurston.

Every closed (=compact, orientable and without boundary) 2-surface admits both a Möbius structure and a projective one. I vaguely remember having been to a talk some time ago where the speaker said that (conjecturally?) the situation in dimension 3 is similar. But I don't remember the details at all. So I would like to ask if anyone knows whether either of the statements (each closed 3-manifold admits a Möbius, resp. projective structure) is a theorem, a conjecture or becomes one or the other after eliminating some counter-examples.

A related question: if memory serves, in the same talk it was mentioned that the $PGL_{n+1}(\mathbf{R})$ and the Möbius group are (conjecturally?) the maximal groups that can act faithfully on an $n$-manifold. I was wondering if anyone knows a reference for this.

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I hope you don't mind that I changed M\"obius to Möbius, throughout. – José Figueroa-O'Farrill May 5 2010 at 12:57
Jose -- thanks! Have to learn how to use html tags some day. – algori May 5 2010 at 13:23
Rephrasings are useful, so I'll ask about one that occurs to me. Is it true that in dim>2 a "Möbius" (respectively, "projective") manifold is the same as a conformally flat (respectively, projectively flat) manifold? That is, a manifold equipped with a conformal equivalence class of metrics (respectively, a projective equivalence class of connections), including for each point a representative which is flat in a n/hd of that point. If so, here's an '83 paper producing a few counterexamples in the Möbius 3-case. – macbeth May 5 2010 at 13:35
(Forgot link: jstor.org/stable/1999172) – macbeth May 5 2010 at 13:36
macbeth -- thanks. Re the Mobius part of the question: I've taken a brief look at the paper by Goldman and it seems the answer is yes since according to the paper the representation corresponding to the developing map is by conformal diffeomorphisms. – algori May 5 2010 at 14:12
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The first examples of closed 3-manifolds not admitting a conformally flat (Mobius) structure were given by Goldman: 3-manifolds modeled on the Nil geometry (this link was given by Macbeth in the comments above). Sol manifolds also don't have a Mobius structure, whereas 3-manifolds modeled on the other six geometries do. Misha Kapovich has many results on conformally flat structures. In his thesis, he shows that certain classes of Haken 3-manifolds have finite-sheeted covers which are (uniformizable) conformally flat, and he shows there is a graph manifold which has no conformally flat structure, but which has a finite-sheeted cover which does. Kulkarni showed that connect sums of conformally flat manifolds are conformally flat. On the other hand, Kapovich's student Hwang showed that for any 3-manifold $M$, there is a 3-manifold $N$ such that $M {\#} N$ is conformally flat. I don't know of any recent activity on the topic.

As for your question on maximal Lie group actions on 3-manifolds, I've heard this before too, but I don't know a reference. I think you can prove it by analyzing the action of the isotropy group of a point on the jet space at that point.

Addendum: I had a discussion with Cooper about the last question (maximal Lie groups acting faithfully on an $n$-manifold), and we have an idea how to approach it (at least for smooth actions). If a Lie group $G$ acts smoothly and faithfully on a manifold $M$, then one obtains a homomorphism $\Phi: G \to Diff(M)$. Then we get a map $\phi:g \to Vect(M)$, where $g$ is the Lie algebra of $G$, and $Vect(M)$ is the Lie algebra of $Diff(M)$. So one wants to classify maximal finite-dimensional Lie algebras of $Vect(M)$. First, if $g$ is not semisimple, then it has a non-trivial center $c$, which is generated by a non-zero smooth vector field $V$. One should then be able to take a quotient $M/V$ of $M$ with action on $M/V$ by the Lie algebra $g/c$ and apply an inductive argument. Actually, one should try to do this only locally, since the quotient might not be nice. Then assume $g$ is semisimple. Its Cartan subalgebra gives $R^m$ acting on $R^n$. This gives $m$ commuting vector fields, and in particular gives $m$-dimensional coordinates at a point, so $m\leq n$. Now, one needs to appeal to the classification of semisimple Lie algebras to finish off the proof (the Cartan subalgebra together with Weyl group determines the Lie algebra), and then apply the inductive argument to deal with the radical of the Lie algebra. I haven't worked out how to do this, but it seems like a plausible approach. I suspect an argument like this may be well-known in the right circles.

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 Thanks, Agol! I fixed the connected sum character. – algori May 5 2010 at 20:36

If true, such a statement would immediately imply the Poincare conjecture. Indeed, suppose X were a closed, simply connected 3-manifold with a Mobius structure. Near any point of $X$, there would then be a Mobius map from $X$ to $S^3$. Analytically continuing this map (see Thurston's notes, chapter 3) would give a local homeomorphism $\phi: X \to S^3$. Since both spaces are compact, $\phi$ would be a covering map. Since $S^3$ is simply connected it would be a homeomorphism. A similar argument applies to projective structures, with $S^3$ replaced by $\mathbb{RP}^3$.

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 Thanks, Lucas! That's true, but the Poincare conjecture holds (or so they say..), so this does not provide a counterexample;) On a more serious note: yes, the Poincare conjecture for Mobius or projective manifolds is immediate, but what I hoping for is that at least in dimension 3 there is some topological condition which ensures that a manifold admits a Mobius or a projective structure, maybe something similar to the hyperbolization theorem. – algori May 5 2010 at 18:17 Yeah, sorry for the lack of a real answer. I was just pointing out that if there was a proof it wouldn't be easy. – Lucas Culler May 6 2010 at 3:43

William Goldman's article "What is a projective structure?" in the Notices of the AMS states that Daryl Cooper has shown that $RP^3 \# RP^3$ has no $RP^3$ structure. The article gives further references, but I couldn't find a reference for the Cooper result.

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 Thanks, Sam! It would be interesting to see a reference for this. – algori May 5 2010 at 17:59

The answer to your last question is in the paper by Vladimir Matveev, "Proof of the projective Lichnerowicz-Obata conjecture", [J. Differential Geom. 75 (2007), no. 3, 459--502]. The introduction contains an extensive historical survey, including references for the conformal case which was settled long ago. For example, for a closed Riemannian manifold of dimension $>2$ the group of conformal automorphisms is compact unless the manifold is conformal to a sphere (as proved by J. Lelong-Ferrand and M. Obata).

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 I meant projective and conformal actions, not actions by diffeomorphisms, which is a different story. – Igor Belegradek May 5 2010 at 22:01