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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.