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

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 Möbius 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.

share|improve this question
    
I hope you don't mind that I changed M\"obius to Möbius, throughout. –  José Figueroa-O'Farrill May 5 '10 at 12:57
    
Jose -- thanks! Have to learn how to use html tags some day. –  algori May 5 '10 at 13:23
2  
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 '10 at 13:35
    
(Forgot link: jstor.org/stable/1999172) –  macbeth May 5 '10 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 '10 at 14:12

5 Answers 5

up vote 11 down vote accepted

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.

share|improve this answer
    
Thanks, Agol! I fixed the connected sum character. –  algori May 5 '10 at 20:36
    
I re-fixed the connected sum character to adapt to the new iteration of MO :-) –  Mariano Suárez-Alvarez Aug 10 '13 at 16:41

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 $\mathbb{RP}^3$.

share|improve this answer
    
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 '10 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 '10 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 $\mathbb{RP}^3\#\mathbb{RP}^3$ has no $\mathbb{RP}^3$ structure. The article gives further references, but I couldn't find a reference for the Cooper result.

share|improve this answer
    
Thanks, Sam! It would be interesting to see a reference for this. –  algori May 5 '10 at 17:59

The questions about manifolds admitting flat conformal or projective structures have been answered, so I'll just address your final question, which is whether $\mathrm{PGL}(n{+}1,\mathbb{R})$ and $\mathrm{O}(n{+}1,1)/\{\pm I_{n+2}\}$ are 'maximal' finite dimensional groups acting on $n$-manifolds.

The answer depends on what you mean by 'maximal'. They are maximal in the sense that any finite dimensional Lie subgroup of $\mathrm{Diff}(\mathbb{RP}^n)$ that contains $\mathrm{PGL}(n{+}1,\mathbb{R})$ must be equal to $\mathrm{PGL}(n{+}1,\mathbb{R})$ and any finite dimensional Lie subgroup of $\mathrm{Diff}(S^n)$ that contains $\mathrm{O}(n{+}1,1)/\{\pm I_{n+2}\}$ must be equal to $\mathrm{O}(n{+}1,1)/\{\pm I_{n+2}\}$. This is essentially due to Élie Cartan, who proved the corresponding statement that the corresponding Lie algebras of vector fields are maximal finite dimensional Lie algebras in $\mathrm{Vect}(M)$ in his series of papers on so-called 'infinite groups' of diffeomorphisms. For a more modern treatment, see Singer and Sternberg's "The infinite groups of Lie and Cartan, Part I (the transitive case)".

However, if my maximal, you mean 'highest (finite) dimensional Lie subgroup of $\mathrm{Diff}(M^n)$', then this is false, as there is no upper bound for trivial reasons: Take $k$ disjoint open sets $U_j\subset M$ each with compact closure, and let $X_j$ be a nonzero vector field on $M$ that is supported in $U_j$. Then the flows of the $X_j$ commute and generate an effective action of $\mathbb{R}^k$ on $M^n$. Since $k$ can be arbitrary, it follows that there is no upper bound on the dimension of Lie subgroups of $\mathrm{Diff}(M)$.

Of course, this example is not satisfying, and you well might ask whether there is an upper bound for transitive actions. Even then, the answer is 'no'. For example, for a fixed integer $m$, consider the transformations of $\mathbb{R}^2$ that are of the form $$ \Phi(x,y) = (\ ax{+}b,\ cy + d_m x^m + d_{m-1} x^{m-1} + \cdots + d_1 x + d_0\ ) $$ where $a,b,c, d_0,\ldots, d_m$ are constants with $ac\not=0$. These constitute a Lie group of dimension $m{+}4$ acting effectively and transitively on $\mathbb{R}^2$. A similar example can obviously be constructed on $\mathbb{R}^n$ for any $n>1$. One can construct transitive examples for any $n>1$ on compact manifolds as well.

The key missing condition necessary for classification is to ask that the group act both transitively and primitively, i.e., that the group action not preserve any nontrivial foliation of $M$. Then there is a classification, due in large part to Élie Cartan, but finally completed to modern standards of rigor by Ochiai in the 70s. The conformal and projective transformation groups are indeed maximal among primitive, transitive actions, but there are many others. For example, the action of $\mathrm{PGL}(n,\mathbb{R})$ on the Grassmannian of $k$-planes in $\mathrm{R}^n$ is maximal, primitive, transitive for every pair $(k,n)$. For a given dimension $n$ of the underlying manifold, the projective group, at dimension $n^2{+}2n$, is the largest primitive transitive Lie subgroup in overall dimension.

share|improve this answer
    
Informative answer as always! Concerning your last sentence do I interpret it correctly: for any smooth $n$-manifold there is a transitive, primitive faithful action of $PGL(n^2+2n,\mathbb{R})$ on it, and there is no group of equal or higher dimension which has the same property? –  Michael Bächtold Aug 11 '13 at 11:34
    
My suggested interpretation is probably unrealistically strong. –  Michael Bächtold Aug 11 '13 at 12:13
    
I guess the wording in my last sentence is not clear. I meant to say that there is no primitive transitive Lie transformation group of dimension greater than $n^2{+}2n$ acting in dimension $n$, and the only such action that attains that maximal dimension is $\mathrm{PGL}(n{+}1,\mathbb{R})$ acting on $\mathbb{RP}^n$ (and, of course, the lifted action of $\mathrm{GL}(n{+}1,\mathbb{R})$ on the nontrivial double cover $\mathbb{S}^n$, i.e., the oriented lines). –  Robert Bryant Aug 11 '13 at 12:55

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).

share|improve this answer
    
I meant projective and conformal actions, not actions by diffeomorphisms, which is a different story. –  Igor Belegradek May 5 '10 at 22:01

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.