Proving the tameness conjecture

ConanLeave a comment

I have recently went through professor Gabai’s wonderful paper that gives a proof of the tameness conjecture. (This one is a simplified version of the argument given in Gabai and Calegari, where everything is done in the smooth category instead of PL). It’s been a quite exciting reading with many amazing ideas, hence I decided to write a summary from my childish viewpoint (as someone who knew nothing about the subject beforehand).

We say a manifold is tame if it an be embedded in a compact manifold s.t. the closure of the embedding is the whole compact manifold.

To motivate the concept, let’s look at surfaces: Any compact surface is, of course, tame. However, if we “shoot out” a few points of the surface to infinity, as the figure below, it become non-compact but still tame, as we can embed the infinite tube to a disk without a point.

Of course, we can also make a surface non-compact by shooting any closed subset to infinity (e.g. a Cantor set), but such construction will always result in a tame surface. (This can be realized using similar embeddings as above, we may embed the resulting surface into the original surface with image being the original surface subtract the closed set. If the closed set has interior, we further contract each interior components.)

On the other hand, any surface with infinite genus would be non-tame since if there is an embedding into a compact set, the image of ‘genesis’ would have limit points, which will force the compact space fail to be a manifold at that point.

Hence in spirit, being tame means that although the manifold may not be compact itself, but all topology happens in bounded regions (we can think of a complete embedding of the manifold into some \mathbb{R}^N so bounded make sense)

As usual, life gets more complicated for three-manifolds.

Tameness conjecture: Every complete hyperbolic 3-manifold with finitely generated fundamental group is tame.

A bubble chart for capturing the structure of the proof:

A few highlights of the proof: The key idea here is shrinkwrapping, very roughly speaking, to prove an geometrically infinite end is tame one needs to find a sequence of simplicial hyperbolic surfaces exiting at the end. Bonahon’s theorem gives us a sequence of closed geodesics exiting the end. By various pervious results, one is able to produce (topological) surfaces that are ‘in between’ those geodesics. Shrinkwrapping takes the given surface and shrinks it until it’s ‘tightly wrapped’ around the given sequence of geodesics. The fact that each of the curve the surface is wrapping around is a geodesic guarantees the resulting surface simplicial hyperbolic. (think of this as folding a piece of paper along a curve would effect its curvature, but alone a straight line would not; geodesics are like straight lines).

Once we have that, the remaining part would be showing the position of the surfaces are under control so that they would exit the end. Since simplicial hyperbolic surfaces has curvature \leq -1, by Gauss-Bonnet they have uniformly bounded area (given our surfaces also has bounded genus). By passing to a subsequence, we may choose the sequence of geodesics to be separated by some uniform constant, which will guarantee the wrapped surfaces are not too thin in the thick parts of the manifold, hence we have control over the diameter of the surface, from which we can conclude that the surfaces must exit the manifold.

Remark: Note that in general, unlike in two dimensions, a three manifold with finitely generated fundamental group does not need to be tame as the Whitehead manifold is homotopic to \mathbb{R}^3 (hence trivial fundamental group) but is not tame. On the other hand, if we have infinitely generated fundamental group, then the manifold can never be tame. The theorem says all examples of non-tame manifolds with finitely generated fundamental group does not admit hyperbolic structure.

Fibering the figure-8 knot complement over the circle

Conan5 Comments

As I was making some false statements about how I think geometrically finite ends of a hyperbolic three manifold would look like, professor Gabai pointed out this super cool fact (proved by Cannon and Thurston, 2007) that the figure-eight knot complement admits a hyperbolic structure and fibers over the circle, but if we lift any fiber (which would be a surface) into the hyperbolic 3-space, the resulting surface would be an embedded topological disc with limit set being the whole limit 2-sphere (!) i.e. if we see \mathbb{H}^3 as a Euclidean open ball, then the boundary of such a disc is a Peano curve that covers the whole 2-sphere bounding \mathbb{H}^3.

I have read about the hyperbolic structure on the figure-8 knot complement in Thurston’s notes (4.3) (A similar construction can be found in my pervious post about hyperbolic structure on the Whitehead link complement), but I didn’t know the fibering over circle part, so I decided to figure out what this fibration would look like.

After playing with chicken wire and playdo for a few days, I am finally able to visualize the fibration. Here I want to point out a few simple points discovered in the process.

Start with the classical position of the figure-8 knot (two ends extends to infinity and meet at the point infinity in \mathbb{S}^3):

To find a fibration over the circle, we need to give a surface that spans the knot (such surface is called a Seifert surface) and a homotopy of the surface \varphi: S \times [0,1] \rightarrow \mathbb{S}^3 \backslash K which restricts to a bijection from S \times [0,1) to \mathbb{S}^3 \backslash K and \varphi(S\times\{1\}) = \varphi(S\times\{0\}).

For quite some time, I tried with the following surface:

Since it’s perfectly symmetric (via a rotation by \pi), we only need to produce a homotopy that sends S to the symmetric Seifert surface in the upper half plane. I was not able to find one. (I’m still curious if there is such homotopy, if so, then there are more than one way the knot complement can fiber)

It turns out that there are in fact non-homeomorphic (hence of course non-homotopic) Seifert surfaces spanning the knot, the one I end up using for the fibration is the following surface:

Or equivlently, we may connect the two ends at a finite point.

To see the boundary is indeed the figure-8 knot:

Note that this surface is not homeomorphic to the pervious one because this one is orientable and the pervious is not.

Now I’ll leave it as a brain exercise to see the homotopy. (well…this is largely because it takes forever to draw enough pictures for expressing that) A hint on how the it goes: think of the homotopy as a continuous family of disjoint Seifert surfaces that ‘swipes through’ the whole \mathbb{S}^3 \backslash K and returns to the initial one. As in the picture above, our surface is like a disc with two intertwined stripe handles on it, each handle is two twists in it. The major step is to see that one can ‘pass’ the disc through a double-twisted handle by making the interior of the old disc to become the interior of the new handle. i.e. we can homotope the bowl from under the strap to above the strap with a family of disjoint surfaces with same boundary.

In in figure-8 knot case, the disc would need to pass through both straps and return to itself.

“Exact dimension” of measures

ConanLeave a comment

Mike throw this question to me today which I have a huge lack of background knowledge to understand >.< … Since think there are some cool stuff there which I’m unable to find much reference, to prevent me from forgetting anything, I guess it’s a good idea to record and organize the discussion here:

Definition: Given measure \mu on X, we say \mu has exact dimension \alpha (denoted \mbox{Edim} (\mu) = \alpha if for \mu-almost every x,

\displaystyle \lim_{r \rightarrow 0} \frac{\log(\mu(B(x,r)))}{\log(r)} = \alpha

–as radius approach to 0, the volume of the ball behaves more and more like r^\alpha, but no further restriction on the rate it approaches r^\alpha.

To illustrate the concept, we look at the following:

Given a measure-preserving system (X, \mathcal{B}, \mu, T) and a partition \mathcal{P} = \{P_1, P_2, \cdots, P_n \},

Notation: For any x \in X, \mathcal{P}(x) denotes the set P_i \in \mathcal{P} containing x.

T^{-i}(\mathcal{P}) = \{ T^{-i}(P_j) \ | \ 1 \leq j \leq n \} i.e. pulling back the partition by T^{-i}, note that T^{-i}(\mathcal{P}) is itself a partition of X, with elements having same measures as \mathcal{P}.

Given two petitions \mathcal{P}, \ \mathcal{P}', let \mathcal{P} \vee \mathcal{P}' denote the smallest common refinement of \mathcal{P} and \mathcal{P}'.

The entropy of a partition \mathcal{P} is defined as

H(\mathcal{P}) = - \displaystyle \sum_{i=1}^n \mu(P_i) \log(\mu(P_i))

–Since Log approaches -\infty as \mu(P_i) gets small, as we would expect, the entropy measures how ‘fine’ the partition is. The smaller the parts gets, the larger the entropy; To get a sense of the rate of growth as partition gets finer: if one bisects each set in the partition to form a new partition, the entropy would increase by \log(2).

The measure-theoretic entropy of the system w.r.t. \mathcal{P} is defined as:

h(T, \mathcal{P}) = \displaystyle \lim_{n \rightarrow \infty} H( \bigvee_{i=1}^n T^{-i} (\mathcal{P}))

Observe that for partitions \mathcal{P}, \ \mathcal{P}' if each element of \mathcal{P}' is divided by elements of \mathcal{P} into equal pieces, then H(\mathcal{P} \vee \mathcal{P}') = H(\mathcal{P}) + H(\mathcal{P}') and this is as large as it can get.

Hence H( \bigvee_{i=1}^n T^{-i} (\mathcal{P})) \leq n \times H(\mathcal{P}) –as remarked above, H(T^{-1}(\mathcal{P}) = H(\mathcal{P}). i.e. we have h(T, \mathcal{P}) is at most H(\mathcal{P}).

*If the sets in \mathcal{P} intersects as much as they can and divides all parts uniformly smaller, then the entropy will equal to the entropy of the partition, in other cases, the more they ‘coincide’ under the action by T^{-1}, the smaller the entropy gets. When T just permutes the P_is, entropy is, of course, 0.

Shannon-McMillan-Breiman theorem:

Let T be ergodic, for \mu almost all x \in X,

\displaystyle \lim_{n \rightarrow \infty} - \log (\mu( \vee_{i=1}^n T^{-i}\mathcal{P}(x)))/n = h(T, \mathcal{P})

I would not try to dig into the proof of this here…

(Note that ergodicity is clearly necessary since we can have T fixing P_1 and ‘evenly mixes’ P_2, \cdots, P_n, for points in P_1, the limit converges to 0, for points not in P_1, the limit converges to \displaystyle \sum_{i=2}^n \mu(P_i) \log(\mu(P_i))…I’m not sure if it suffices to require T to be ergodic)

The measure-theoretic entropy of a system is defined as the sup over all partitions of the entropy w.r.t. the partition. i.e.

\displaystyle h(T) = \sup_{\mathcal{P}} h(T, \mathcal{P})

–Although H(\mathcal{P}) blows up to infinity as \mathcal{P} gets finer, but there is a limited amount of growth, i.e. when the partition is fine, it’s harder for T^{-1}(P_i) to intersect each P_j. Hence one should believe that H(T) is (at least normally) finite.

Consider the angle doubling map f(x) = 2x ( \ \mbox{mod} \ 1). Let \mu be an invariant measure, then \mbox{Edim}(\mu) = h(\mu, T) / \log(2). i.e. exact dimension exists for each such measure.

This fact can be seen by considering the space of binary sequences, our measure assigns a weight to each initial sequence, take sequence of refining partitions to be just the all initial sequences of length n to obtain the entropy.

Question:1. Find an example of two measures with exact dimension where a projection of the product does not have exact dimension.

2. What about the projection of the product of the same measure?

3. For measure \mu as defined above (invariant measure of angle doubling map), is it true that any projection of \mu \times \mu has exact dimension?

A few interesting items from the ICM

ConanLeave a comment

So, as many people know, as part of my India vacation, I went to the ICM in Hyderabad.

On this last day of conference, I decided to write a small note of a few cool items I picked up in some talks: (although there are in fact many, many other cool facts I might write more once I got back :-P)

1. Renormalization (on Artur Avila‘s talk) So we look at one-dimensional systems, one may zoom in at a part of the interval that maps into itself, which yields a similar (or not) system, the idea is called ‘renormalization’. i.e. we have the ‘renormalization operator’ acting on a certain class of systems (‘renormalizable systems’) and this gives a map on the function space. Now we study the dynamics there(!) At the first glance, it doesn’t look like solving a one-dimensional problem in infinite dimensional space would help in any useful way, but it does(!). As an example, we look at space of circle diffeomorphisms, they have a rotation number, it’s not hard to see, in this case, the linear rotations form a circular attractor for the renormalization operator, further more, the dynamics is perfectly understood on the circle (Gauss map), the operator permutes (infinite dimensional) fibers with equal rotational number. It turns out we know enough about the dynamics on the function space to get useful information to the original problem! Super cool~

2. Differentiating Lipschitz functions and decomposing Kakeya sets (on Marianna Csornyei‘s talk, for details please refer to their ICM paper) They had a through study of exactly which sets in \mathbb{R}^n can be contained in the discontinuity set of a Lipschitz function f: \mathbb{R}^n \rightarrow \mathbb{R}^m. I found the following unbelievable at the first glance: Given a cone C in \mathbb{R}^n (a set of rays from \bar{0}), the C-width of set E \subseteq \mathbb{R}^n is, roughly speaking, the \sup of lengths of E \cap \gamma where \gamma is a Lipschitz curve going only in directions in C. (A more precise definition requires a generalized notion of ‘tangent’ for Lipschitz curves and can be found in the paper). They proved that:

Theorem: Any Lebesgue 0 set in \mathbb{R}^2 can be decomposed into two sets A and B that A has C-width 0 for C= [0, \pi/2] and B has C'-width 0 for C' = [\pi/2, \pi].

Why does this surprise me? Well, of course the first thing I donsider is: what would happen for the Kakeya set? Our null set contains a line segment in each direction, hence even if we just requiring the decomposed sets A and B to intersect all straight lines in directions of C, C' in length 0 sets would give pretty much only one possible decomposition: we have to take A to be the union of all segments in direction of C', each missing a linear 0 set, and same for B and C(! not much freedom, right?). It’s already hard to believe such A and B can be made satisfying the property, not to mention that in fact they can be made intersecting all Lipschitz curves in null sets (!) (At first I thought it was an obvious counterexample to the theorem, but after discussing with her after the talk, this is indeed what the theorem does) Amazing…

List to be filled in:

3. Boundary rigidity via filling volume (On Sergei Ivanov‘s talk. For details please refer to his paper on the ArXiv)

4. Constant main curvature surfaces

…to be continued…

On minimal surfaces

Conan1 Comment

Let D \subseteq \mathbb{R}^2 be a domain. f: D \rightarrow \mathbb{R}, f \in C^2(D).

Recall: from last talk, Zhenghe described the Lagrange’s Equation, in this case the equation is written: (we denote \frac{\partial f}{\partial x} as f_x)

(1+ f_y^2)f_{xx}-2f-xf-yf_{xy}+(1+f_x^2)f_{yy}=0

Theorem: The graph of f is area-minimizing then f satisfies Lagrange’s equation.

Proof: Since f is area-minimizing,

A(f) = \int_D(1+f_x^2+f_y^2)^{\frac{1}{2}}dx dy

is minimized by f for given boundary values. Hence the variation \delta A of A due to an infinitesimal \delta f of f where \delta f |_{\partial D} = 0. i.e.

\delta A = \int_D \frac{1}{2}(1+f_x^2+f_y^2)^{-\frac{1}{2}}(2 f_x \delta f_x + 2 f_y \delta f_y) dxdy

= \int_D \{ [ (1 + f_x^2 + f_y^2)^{-\frac{1}{2}} f_x ] \delta f_x

+ [ (1 + f_x^2 + f_y^2)^{-\frac{1}{2}} f_y ] \delta f_y \}dxdy =0

Let G(f) = - \frac{\partial}{\partial x}[ (1 + f_x^2 + f_y^2)^{-\frac{1}{2}} f_x ] - \frac{\partial}{\partial y}[ (1 + f_x^2 + f_y^2)^{-\frac{1}{2}} f_y ]

= -(1 + f_x^2 + f_y^2)^{-\frac{3}{2}} [(1+f_y^2)f_{xx}

- 2f_x f_y f_{xy}+(1+f_x^2)f_{yy}]

Apply integration by parts, since \delta f vanishes on \partial D, the constant term vanishes, we have:

\int_D G(f) \delta f dxdy= 0 for all \delta f with \delta f |_{\partial D} = \bar{0}, hence G(f) = \bar{0}. i.e.

(1+ f_y^2)f_{xx}-2f-xf-yf_{xy}+(1+f_x^2)f_{yy}=0

which is the Lagrange’s equation.

We should note that the converse of the theorem is, in general, not true.

Example: two rectangles, star-shaped 4-gon.

Theorem: For D convex, any f satisfying Lagrange’s equation has area-minimizing graph.

Let \varphi be 2-form in \mathbb{R}^3 s.t. d \varphi = 0 and \sup(\varphi) = 1. i.e. \varphi acts on the unit Grassmannian space of oriented planes in \mathbb{R}^3.

Definition: An immersed surface S is calibrated by \varphi if \varphi(P) = 1 for all P in the unit tangent bundle of S.

*All calibrated surfaces are automatically area-minimizing.

Let \varphi: D \times \mathbb{R} \rightarrow (\Lambda^2 \mathbb{R}^3)^\ast be the two-form

\varphi(x,y,z) =\frac{ -f_x dydz - f_y dzdx + dxdy}{f_x^2+f_y^2+1}

By construction, for all (x,y,z) \in D \times \mathbb{R}, v, w \in S^2, we have \varphi(x,y,z)(v,w) \leq 1, \varphi(x,y,z)(v,w) = 1 when the plane spanned by v, w is tangent to the graph of f at (x,y,f(x,y)).

d \varphi = - \frac{\partial}{\partial x}f_x(f_x^2+f_y^2+1)^{-\frac{1}{2}}- \frac{\partial}{\partial y}f_y(f_x^2+f_y^2+1)^{-\frac{1}{2}}

= -(1 + f_x^2 + f_y^2)^{-\frac{3}{2}} [(1+f_y^2)f_{xx}

- 2f_x f_y f_{xy}+(1+f_x^2)f_{yy}]

which is 0 by Lagrange’s equation. Hence \varphi is closed.

Let S be the graph of f, since \varphi(p)(v,w) = 1 whenever the plane spanned by v, w is tangent to S at p, we have

A(S) = \int_S \varphi

Suppose S is not area-minimizing, there exists 2-chain T with \partial T = \partial S with smaller area than that of S.

Since D is convex, any T not contained in D \times \mathbb{R} cannot be area-minimizing (by projecting to the cylinder). Hence we may assume T \subseteq D \times \mathbb{R} (So that \varphi is well-defined on T)

Since S-T bounds a 3-chain, \varphi is closed, hence \int_S \varphi = \int_T \varphi

Beacuse \varphi(p)(v,w) \leq 1 hence \int_T \varphi \leq A(T).

Therefore we have A(S) \leq A(T). i.e. S is area-minimizing.

Definition: A minimal surface in \mathbb{R}^3 is a smoothly immersed surface which is locally the graph of a solution to the Lagrange’s equation.

Note that small pieces of minimal surfaces are area-minimizing but lager pieces may not be.

Example: Enneper’s surface

Theorem: Let C be a rectifiable Jordan curve in \mathbb{R}^3, there is a area-minimizing 2-chain S \subseteq \mathbb{R}^3 with \partial S = C

Sketch of proof:

There exists rectifiable 2-chain with boundary being C. -Take a point in \mathbb{R}^3 and take the cone of the curve.

Define flat norm on the space of 2-chains in \mathbb{R}^3 by F(T) = \inf\{A(C_2)+V(C_3) \ | \ T = C_2+\partial C_3 \} i.e. if two chains are close together, they would almost bound a 3-chain with small volume, hence the difference has small norm.

Fact: \mathcal{T} = \{ T \subseteq B^3(\bar{0}, R) \ | \ A(T) < K and l(\partial T)0, for any chain T, we may find a chain T' inside the grid of mesh \delta where F(T-T') < \varepsilon (hence the area of T' is also bounded). Since there are only finitely many such chains, we have:

\mathcal{T} is totally bounded under the flat norm F.

Hence \mathcal{T} is compact.

Now we choose sequence (T_n) of rectifiable chains with boundary C and area decreasing to \inf \{ A(T) \ | \ \partial T = C \}

Choose R large enough s.t. C \subseteq B^3(\bar{0}, R). Project \mathbb{R}^3 \backslash B^3(\bar{0}, R) radially onto S^2(\bar{0}, R) the projection does not increase area.

Hence A(\pi(T_i)) \leq A(T_i) for all i. i.e. (A(\pi(T_i))) \rightarrow \inf \{ A(T) \ | \ \partial T = C \} and \partial(\pi(T_i)) = \partial (T_i) = C.

Since \mathcal{T} = \{ T \subseteq B^3(\bar{0}, R) \ | \ A(T) < K and l(\partial T)< K \} is compact, there exists subsequence (T_{n_i}) converging to a rectifiable chain S \in \mathcal{T}.

We can prove that: \partial S = C (continuity of \partial under the flat norm).

A(S) = \inf \{ A(T) \ | \ \partial T = C \} (lower-semicontinuity of area under the flat norm).

Therefore S is an area-minimizing surface with \partial S = C.