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Yesterday and today Marty Scharlemann gave two talks on the Schoenflies Conjecture, one of the great open problems in low dimensional topology. These talks were very clear and inspiring, and I thought it would be useful to summarize what Marty said in a blog post, just for my own benefit.
The story starts with the following classical theorem, usually called the Jordan curve theorem, or Jordan-Schoenflies theorem:
Theorem (Jordan-Schoenflies): Let P be a simple closed curve in the plane. Then its complement has a unique bounded component, whose closure is homeomorphic to the disk in such a way that P becomes the boundary of the disk.
In order to make the relationship between the two complementary components more symmetric, one could express this theorem by saying that a simple closed curve P in the 2-sphere separates the 2-sphere into two components X and Y, each of which has closure homeomorphic to a disk with P as the boundary.
Based on this simple but powerful fact in dimension 2, Schoenflies asked: is it true for every n that every n-sphere P in the (n+1)-sphere splits the (n+1)-sphere into two standard (n+1)-balls?
There is an old puzzle which starts by asking: what is the next number in the sequence 1,2,4,? We are supposed to recognize the start of the sequence and answer that the next number is surely 8, because the first three numbers are consecutive powers of 2, and so the next number should be the cube of 2 which is 8. The puzzler then explains (contrary to expectations) that the successive terms in the sequence are actually the number of regions into which the plane is divided by a collection of lines in general position (so that any two lines intersect, and no three lines intersect in a single point). Thus:
So the “correct” answer to the puzzle is 7 (and the sequence continues 11, 26, ). This is somehow meant to illustrate some profound point; I don’t quite see it myself. Anyway, I would like to suggest that there is a natural sense in which the “real” answer should actually be 8 after all, and it’s the point of this short blog post to describe some connections between this puzzle, the theory of cube complexes (which is at the heart of Agol’s recent proof of the Virtual Haken Conjecture), and the location of the missing 8th region.
Ian gave his second and third talks this afternoon, completing his (quite detailed) sketch of the proof of the Virtual Haken Theorem. Recall that after work of Kahn-Markovic, Wise, Haglund-Wise and Bergeron-Wise, the proof reduces to showing the following:
Theorem (Agol): Let G be a hyperbolic group acting properly discontinuously and cocompactly on a CAT(0) cube complex X. Then there is a finite index subgroup G’ so that X/G’ is special; in other words, G is virtually special.
I am in Paris attending a workshop at the IHP where Ian Agol has just given the first of three talks outlining his proof of the Virtual Haken Conjecture and Virtual Fibration Conjecture in 3-manifold topology (hat tip to Henry Wilton at the Low Dimensional Topology blog from whom I first learned about Ian’s announcement last week). I think it is no
under overstatement to say that this marks the end of an era in 3-manifold topology, since the proof ties up just about every loose end left over on the list of problems in 3-manifold topology from Thurston’s famous Bulletin article (with the exception of problem 23 — to show that volumes of closed hyperbolic 3-manifolds are not rationally related — which is very close to some famous open problems in number theory). The purpose of this blog post is to say what the Virtual Haken Conjecture is, and some of the background that goes into Ian’s argument. I hope to follow this up with more details in another post (after Agol gives talks 2 and 3 this coming Wednesday). Needless to say this post has been written in a bit of a hurry, and I have probably messed up some crucial details; but if that caveat is not enough to dissuade you, then read on.
Patrick Foulon and Boris Hasselblatt recently posted a preprint entitled “Nonalgebraic contact Anosov flows on 3-manifolds”. These are flows which are at the same time Anosov (i.e. the tangent bundle splits in a flow-invariant way into stable, unstable and flow directions) and contact (i.e. they preserve a contact form — that is, a 1-form for which is a volume form). Their preprint gives some very interesting new constructions of such flows, obtained by surgery along a Legendrian knot (one tangent to the kernel of the contact form) which is transverse to the stable/unstable foliations of the Anosov flow.
My student Steven Frankel has just posted his paper Quasigeodesic flows and Mobius-like groups on the arXiv. This
heartbreaking work of staggering genius interesting paper makes a deep connection between dynamics, hyperbolic geometry, and group theory, and represents the first significant progress that I know of on a conjectural program I formulated a few years ago.
One of the main results of the paper is to show that every quasigeodesic flow on a closed hyperbolic 3-manifold either has a closed orbit, or the fundamental group of the manifold admits an action on a circle with some very peculiar properties, namely that it is Mobius-like but not Mobius. The problem of giving necessary and sufficient conditions on a vector field on a 3-manifold to guarantee the existence of a closed orbit is a long and interesting one, and the introduction to the paper gives a brief sketch of this history as follows:
I’m in Melbourne right now, where I recently attended the Hyamfest and the preceding workshop. There were many excellent talks at both the workshop and the conference (more on that in another post), but one thing that I found very interesting is that both Michel Boileau and Cameron Gordon gave talks on the relationships between taut foliations, left-orderable groups, and L-spaces. I haven’t thought seriously about taut foliations in almost ten years, but the subject has been revitalized by its relationship to the theory of Heegaard Floer homology. The relationship tends to be one-way: the existence of a taut foliation on a manifold implies that the Heegard Floer homology of is nontrivial. It would be very interesting if Heegaard Floer homology could be used to decide whether a given manifold admits a taut foliation or not, but for the moment this seems to be out of reach.
Anyway, both Michel and Cameron made use of the (by now 20 year old) classification of taut foliations on Seifert fibered 3-manifolds. The last step of this classification concerns the case when the base orbifold is a sphere; the precise answer was formulated in terms of a conjecture by Jankins and Neumann, proved by Naimi, about rotation numbers. I am ashamed to say that I never actually read Naimi’s argument, although it is not long. The point of this post is to give a new, short, combinatorial proof of the conjecture which I think is “conceptual” enough to digest easily.
1. Mostow Rigidity
For hyperbolic surfaces, Moduli space is quite large and complicated. However, in three dimensions Moduli space is trivial:
Theorem 1 If is a homotopy equivalence of closed hyperbolic manifolds with , then is homotopic to an isometry.
In other words, Moduli space is a single point.
This post will go through the proof of Mostow rigidity. Unfortunately, the proof just doesn’t work as well on paper as it does in person, especially in the later sections.
1.1. Part 1
First we need a definition familiar to geometric group theorists: a map between metric spaces (not necessarily Riemannian manifolds) is a quasi-isometry if for all , we have
Without the term, would be called bilipschitz.
First, we observe that if is a homotopy equivalence, then lifts to a map in the sense that is equivariant with respect to (thought of as the desk groups of and , so for all , we have .
Now suppose that and are hyperbolic. Then we can lift the Riemannian metric to the covers, so and are specific discrete subgroups in , and maps equivariantly with respect to and .
Lemma 2 is a quasi-isometry.
Proof: Since is a homotopy equivalence, there is a such that . Perturbing slightly, we may assume that and are smooth, and as and are compact, there exists a constant such that and . In other words, paths in and are stretched by a factor of at most : for any path , . The same is true for going in the other direction, and because we can lift the metric, the same is true for the universal covers: for any path , , and similarly for .
Thus, for any in the universal cover ,
We see, then, that is Lipschitz in one direction. We only need the for the other side.
Since , we lift it to get an equivariant lift For any point , the homotopy between gives a path between and . Since this is a lift of the homotopy downstairs, this path must have bounded length, which we will call . Thus,
Putting these facts together, for any in ,
By the triangle inequality,
This is the left half of the quasi-isometry definition, so we have shown that is a quasi-isometry.
Notice that the above proof didn’t use anything hyperbolic—all we needed was that and are Lipschitz.
Our next step is to prove that a quasi-isometry of hyperbolic space extends to a continuous map on the boundary. The boundary of hyperbolic space is best thought of as the boundary of the disk in the Poincare model.
Lemma 3 A quasi-isometry extends to a continuous map on the boundary .
The basic idea is that given a geodesic, it maps under to a path that is uniformly close to a geodesic, so we map the endpoints of the first geodesic to the endpoints of the second. We first need a sublemma:
Lemma 4 Take a geodesic and two points and a distance apart on it. Draw two perpendicular geodesic segments of length from and . Draw a line between the endpoints of these segments such that has constant distance from the geodesic. Then the length of is linear in and exponential in .
Proof: Here is a representative picture:
So we see that . By Gauss-Bonnet,
Where the on the left is the sum of the turning angles, and is the geodesic curvature of the segment . What is this geodesic curvature ? If we imagine increasing , then the derivative of the length with respect to is the geodesic curvature times the length , i.e.
So . Therefore, by the Gauss-Bonnet equality,
so . Therefore, , which proves the lemma
With this lemma in hand, we move on the next sublemma:
Lemma 5 If is a quasi-isometry, there is a constant depending only on and such that for all on the geodesic from to in , is distance less than from any geodesic from to .
Proof: Fix some , and suppose the image of the geodesic from to goes outside a neighborhood of the geodesic from to . That is, there is some segment on between the points and such that maps completely outside the neighborhood.
Let’s look at the nearest point projection from to . By the above lemma, . Thus means that
On the other hand, because is a quasi-isometry,
So we have
Which implies that
That is, the length of the offending path is uniformly bounded. Thus, increase by times this length plus , and every offending path will now be inside the new neighborhood of .
The last lemma says that the image under of a geodesic segment is uniformly close to an actual geodesic. Now suppose that we have an infinite geodesic in . Take geodesic segments with endpoints going off to infinity. There is a subsequence of the endpoints converging to a pair on the boundary. This is because the visual distance between successive pairs of endspoints goes to zero. That is, we have extended to a map , where is the diagonal . This map is actually continuous, since by the same argument geodesics with endpoints visually close map (uniformly close) to geodesics with visually close endpoints.
1.2. Part 2
Now we know that a quasi-isometry extends continuously to the boundary of hyperbolic space. We will end up showing that is conformal, which will give us the theorem.
We now introduce the Gromov norm. if is a topological space, then singular chain complex is a real vector space with basis the continuous maps . We define a norm on as the norm:
This defines a pseudonorm (the Gromov norm) on by:
This (pseudo) norm has some nice properties:
Lemma 6 If is continuous, and , then .
Proof: If represents , then represents .
Thus, we see that if is a homotopy equivalence, then .
If is a closed orientable manifold, then we define the Gromov norm of to be the Gromov norm .
Here is an example: if admits a self map of degree , then . This is because we can let represent , so , so represents . Thus . Notice that we can repeat the composition with to get that is as small as we’d like, so it must be zero.
Theorem 7 (Gromov) Let be a closed oriented hyperbolic -manifold. Then . Where is a constant depending only on .
We now go through the proof of this theorem. First, we need to know how to straighten chains:
Lemma 8 There is a map (the second complex is totally geodesic simplices) which is -equivariant and – equivariantly homotopic to .
Proof: In the hyperboloid model, we imagine a simplex mapping in to . In , we can connect its vertices with straight lines, faces, etc. These project to being totally geodesics in the hyperboloid. We can move the original simplex to this straightened one via linear homotopy in ; now project this homotopy to .
Now, if represents , then we can straighten the simplices, so represents , and , so when finding the Gromov norm it suffices to consider geodesic simplices. Notice that every point has finitely many preimages, and total degree is 1, so for any point , .
Next, we observe:
Lemma 9 If given a chain , there is a collection such that and is a cycle homologous to .
Proof: We are looking at a real vector space of coefficients, and the equations defining what it means to be a cycle are rational. Rational points are therefore dense in it.
By the lemma, there is an integral cycle , where is some constant. We create a simplicial complex by gluing these simplices together, and this complex comes together with a map to . Make it smooth. Now by the fact above, , so . Then
on the one hand, and on the other hand,
The volume on the right is at most , the volume of an ideal simplex, so we have that
This gives the lower bound in the theorem. To get an upper bound, we need to exhibit a chain representing with all the simplices mapping with degree 1, such that the volume of each image simplex is at least .
We now go through the construction of this chain. Set , and fix a fundamental domain for , so is tiled by translates of . Let be the set of all simplices with side lengths with vertices in a particular -tuple of fundamental domains . Pick to be a geodesic simplex with vertices , and let be the image of under the projection. This only depends on up to the deck group of .
Now define the chain:
With the to make it orientation-preserving, and where is an -invariant measure on the space of regular simplices of side length . If the diameter of is every simplex with has edge length in , so:
- The volume of each simplex is if is large enough.
- is finite — fix a fundamental domain; then there are only finitely many other fundamental domains in .
Therefore, we just need to know that is a cycle representing : to see this, observe that every for every face of every simplex, there is an equal weight assigned to a collection of simplices on the front and back of the face, so the boundary is zero.
By the equality above, then,
Taking to zero, we get the theorem.
1.3. Part 3 (Finishing the proof of Mostow Rigidity
We know that for all , there is a cycle representing such that every simplex is geodesic with side lengths in , and the simplices are almost equi-distributed. Now, if , and represents , then represents , as is a homotopy equivalence.
We know that extends to a map . Suppose that there is an tuple in which is the vertices of an ideal regular simplex. The map takes (almost) regular simplices arbitrarily close to this regular ideal simplex to other almost regular simplices close to an ideal regular simplex. That is, takes regular ideal simplices to regular ideal simplices. Visualizing in the upper half space model for dimension 3, pick a regular ideal simplex with one vertex at infinity. Its vertices form an equilateral triangle in the plane, and takes this triangle to another equilateral triangle. We can translate this simplex around by the set of reflections in its faces, and this gives us a dense set of equilateral triangles being sent to equilateral triangles. This implies that is conformal on the boundary. This argument works as long as the boundary sphere is at least 2 dimensional, so this works as long as is 3-dimensional.
Now, as is conformal on the boundary, it is a conformal map on the disk, and thus it is an isometry. Translating, this means that the map conjugating the deck group to is an isometry of , so is actually an isometry, as desired. The proof is now complete.