Though not quite as exciting as a possible proof of the Riemann hypothesis, a paper on realizing the mapping class group of a surface as a subgroup of the group of self-homeomorphisms [1] caught my eye a couple of days ago. Recall that the set of all homeomorphisms from a surface *S* to itself form a group that I’ll call *Aut(S)* or the *automorphism group*. The *mapping class group* of S, which I’ll write *Mod(S)*, is the quotient of this group by isotopies of the surface. The quotient construction implies a homomorphism from *Aut(S)* onto *Mod(S)*. The paper above proves that for a surface of genus at least two, there is no reverse homomorphism from *Mod(S)* into *Aut(S)* such that composing the maps produces the identity on *Mod(S)*.

I had though this was already known, but apparently it was only known for genus at least 5 (or 3 if you replace *Aut(S)* with the group *Diff(S)* of diffeomorphisms.) The introduction to the paper lists the previously known results. You can ask a similar question for subgroups of *Mod(S)* as well, i.e. whether there’s a map from the subgroup into *Aut(S)* or *Diff(S)* that composes to the identity on that subgroup. For infinite cyclic groups, the answer is an almost immediate yes (just pick any representative for a generator). For finite subgroups, the answer is a much harder to prove yes; this is the Nielsen Realization Theorem (which was proved by Steve Kerckhoff, not by Nielsen). I wonder which infinite, non-cyclic subgroups of *Mod(S)* can be realized as groups of homeomorphisms?

The proof in the paper examines a certain relation that is discussed in Farb and Margalit’s primer on mapping class groups: Given a separating loop in a surface, you can write a Dehn twist around that loop as a composition of Dehn twists along loops in the interior of one of the complementary components or along loops in the other complementary component. Since both give you a Dehn twist along the same loop, these give you a relation. The authors then use the machinery defined in [2] (where the “no” answer for genus at least 5 is proved) to show that such a relation cannot exist in *Aut(S)*.

[2] V. Markovic, Realization of the mapping class group by homeomorphisms. Inventiones Mathematicae

168 , no. 3, 523–566 (2007)

Although the mapping class group cannot act on the surface, it does act on its unit tangent bundle, which is a Seifert fibered space. Take the circle at infinity of the universal cover of the surface.

Take triples of points on this circle, which may be identified with the space of ideal triangles in the hyperbolic plane, and therefore with the unit tangle bundle. Each mapping class element lifts to a homeomorphism of the universal cover of the surface, and therefore of the triple points of the circle at infinity. This lift is only well-defined up to covering translations, so when we quotient the space of triple points by the covering translations, we get a well-defined canonical action on the unit tangent bundle to the surface.

Can this action be realized by diffeomorphisms? I suspect the answer is no, given the nature of the action of each element on the circle at infinity.

For the Heegaard splitters, can this action preserve a Heegaard splitting?

Comment by Ian Agol — July 4, 2008 @ 1:15 pm |

Juan Souto has now shown that the action of the mapping class group on the unit tangent bundle of a surface can’t be “smoothed” to act by diffeomorphisms.

The only caveat is that the genus of the surface must be >11, and one must take the full mapping class group, including the orientation reversing maps.

http://www-personal.umich.edu/~jsouto/papers.html

Comment by Ian Agol — October 3, 2009 @ 11:49 pm |

How does an ideal triangle give you a point in the unit tangent bundle?

Comment by Jesse Johnson — July 7, 2008 @ 1:14 pm |

The construction goes as follows (I haven’t yet figured out how to use TeX in my comments…sorry). By the space of ideal triangles, we really mean the space of ideal triangles with a counterclockwise ordering on the vertices. This may be identified with the space T of triples (x,y,z) of distinct points on the unit circle so that x–>y–>z goes around the circle in the counterclockwise direction. The map f:T–>UH is the following one. Consider (x,y,z) in T. Let G be the oriented geodesic from y to z. There is then a unique geodesic K originating from x which intersects G at a right angle. Let p be the point of intersection of K and G and let v be the unit tangent vector based at p going in the direction of G. We then define f(x,y,z)=(p,v).

Comment by Andy P. — July 7, 2008 @ 8:15 pm |

“… and let v be the unit tangent vector based at p going in the direction of G.”

Wouldn’t it better if v is “in the direction of K” so that this construction generalizes to higher dimensions?

Comment by JL — July 8, 2008 @ 5:30 pm |

I’m not sure how useful such a parameterization would be in higher dimensions. Let’s just think about 3 dimensions for a minute. Your construction would map a unit tangent vector (x,v) to a pair (a,s), where a is a point on the sphere at infinity and s is a round circle on the sphere at infinity not containing a. The point a would be the endpoint of the geodesic originating at x in the direction of v and s would be the boundary of the geodesic hyperplane through x which is orthogonal to v. This is a bijection (and similarly in higher dimensions).

However, I’m not sure what applications it would have. If M^3 is a finite volume hyperbolic 3-manifold, then one can deduce from Mostow rigidity that the mapping class group of M^3 is exactly the isometry group of M^3. In particular, it is a finite group and it acts on M^3 (no need to pass to the unit tangent bundle). For an infinite volume hyperbolic 3-manifold, one can still lift any diffeomorphism to the universal cover and get an action on S^3. However, that action need not preserve round circles (if it always did, then one could extend Mostow rigidity to infinite volume 3-manifolds, which is wildly false). Thus we would not get an action on our parameterization of the unit tangent bundle.

Comment by Andy P. — July 9, 2008 @ 10:51 am |

You’re right, Andy. I just thought, say in dimension 3, the analogous map which assigns each quadruple points of S^2 in general position to a unit tangent vector to H^3. But, of course, the dimensions of the two spaces do not quite match and this map is not injective. So the only reasonable generalization has to be the one you suggested above.

As you remarked, such a generalization does not seem to be useful in the study of hyperbolic structures on manifolds but rather has something to do with Mobius structures on manifolds. See, for example, Kulkarni and Pinkall, Propositions (3.3.ii) and (4.6):

http://www.ams.org/mathscinet-getitem?mr=1273468

Comment by JL — July 9, 2008 @ 2:54 pm |

Andy’s comment is the sort of thing I had in mind, although there’s not really a canonical identification. As he points out, I should have said ideal triangles with the vertices marked. The orientation preserving isometries of the hyperbolic plane act freely on the set of such marked triangles. But one may also identify the orientation preserving isometries with the unit tangent bundle in two dimensions, since an isometry is determined by where it takes a unit vector. The corresponding identification in higher dimensions is to take an orbit of a (marked) ideal simplex, which may be identified with the isometry group, and therefore the space of orthonormal frames. But of course this identification is only defined up to conjugacy, which corresponds to where one chooses the base frame to be (Andy has given one possible choice of base frame). The only use that I know of this in higher dimensions is in Gromov’s proof of the Mostow rigidity theorem.

Comment by Ian Agol — July 9, 2008 @ 8:34 pm |

That is not quite an analogous identification in higher dimensions, because an orbit of a (marked) ideal simplex (e.g. a regular one) does not take up the space of ideal simplices; in three dimensions the former has dimension 6 and the latter 8. Is there any meaningful interpretation of the set of ordered quadruple points of S^2?

Comment by JL — July 9, 2008 @ 10:05 pm |

In higher dimensions this is used in dynamics to prove that the geodesic flows of two homeomorphic compact negatively curved manifolds are orbit equivalent.

Comment by Juan Souto — August 15, 2008 @ 6:33 pm |