## Posts Tagged ‘foliation’

### Fibering the figure-8 knot complement over the circle

October 11, 2010

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.

### On plaque expansiveness

May 5, 2010

This note is mostly based on parts of (RH)^2U (2006) and conversations with R. Ures while he was visiting Northwestern.

Let $\mathcal{F}$ be a foliation of the manifold $M$, for $p \in M$, a plaque in of $\mathcal{F}$ through $p$ is a small open neighborhood of $p$ in the leaf $\mathcal{F}_p$ that’s pre-image of a disc via a local foliation chart. (i.e. plaques stuck nicely to make open neighborhoods where the foliation chart is defined.) For $\varepsilon$ small enough, whenever the leaves of $\mathcal{F}$ are $C^1$, the path component of $B(p, \varepsilon)$ containing $p$ is automatically a plaque, we denote this by $\mathcal{F}_\varepsilon(p)$.

Given a partially hyperbolic diffeomorphism $f: M \rightarrow M$, suppose the center integrates to foliation $\mathcal{F}^c$.

Definition: An $\varepsilon$-pseudo orbit w.r.t. $\mathcal{F}^c$ is a sequence $(p_n)$ where for any $n \in \mathbb{Z}$, $f(x_n) \in \mathcal{F}^c_\varepsilon(x_{n+1})$.

i.e. $p_{n+1}$ is the $f$-image of $p_n$ except we are allowed to move along the center plaque for a distance less than $\varepsilon$.

Definition: $f$ is plaque expansive at $\mathcal{F}^c$ if there exists $\varepsilon>0$ s.t. for all $\varepsilon$-pseudo orbits $(p_n), (q_n)$ w.r.t. $\mathcal{F}^c$, $d(p_i, q_i)<\varepsilon$ for all $i \in \mathbb{Z}$ then $p_0 \in \mathcal{F}^c_\varepsilon(q_0)$.

i.e. any two pseudo-orbits in different plagues will eventually (under forward or backward iterates) be separated by a distance $\varepsilon$.

In the book Invariant Manifolds (Hirsch-Pugh-Shub), it’s proven that

Theorem: If a partially hyperbolic system has plaque expansive center foliation, then the center being integrable and plaque expansiveness are stable under perturbation (in the space of diffeos). Furthermore, the center foliation of the perturbed system $g$ is conjugate to the center foliation of the origional system $f$ in the sense that there exists homeomorphism $h: M \rightarrow M$ where

1) $h$ sends leaves of $\mathcal{F}^c_f$ to leaves of $\mathcal{F}^c_g$ i.e. for all $p \in M$,

$h(\mathcal{F}^c_f(p)) = \mathcal{F}^c_g(p)$

2) $h$ conjugates the action of $f$ and $g$ on the set of center leaves i.e. for all $p \in M$,

$h \circ f \ (\mathcal{F}^c_f(p)) = g \circ h \ ( \mathcal{F}^c_f(p))$

(both sides produce a $\mathcal{F}^c_g$ leaf)

Morally this means plaque expansiveness implies structurally stable in terms of permuting the center leaves.

It’s open whether or not any partially hyperbolic diffeomorphism with integrable center is plaque expansive w.r.t. its center foliation.

Another problem, stated in HPS about plaque expansiveness is:

Question: If $f$ is partially hyperbolic and plaque expansive w.r.t. center foliation $\mathcal{F}_c$, then is $\mathcal{F}_c$ the
unique $f$−invariant foliation tangent to $E^c$?

(RH)^2U has recently gave a series of super cool examples where the 1-dimensional center bundles of a $C^1$ partially hyperbolic diffeomorphism 1) does not integrate OR 2) integrates to a foliation but leaves through a given point is not unique (there is other curves through the point that’s everywhere tangent to the bundle). I will say a few words about the examples without spoil the paper (which is still under construction).

Start with the cat map on the $2$-torus (matrix with entries $( 2, 1, 1, 1)$, take the direct product with the source-sink map on the circle, we obtain a diffeo on the $3$ torus. For the purpose of our map, we make the expansion in the source-sink map weaker than that of the cat map and the contraction stronger.

Then we perturb the map by adding appropriate small rotations to the system, the perturbation vanish on the $\mathbb{t}^2$ fibers corresponding to the two fixed points in the source-sink map. This will make our system partially hyperbolic, with center bundles as shown below:

To construct a non-integrable center, we make a perturbation that gives center boundle (inside the unstable direction of the cat map times the circle):

For intergrable but have non-unique center leaves, we simply rotate the upper and bottom half in opposite directions and obtain:

Note that in this case, all center leaves are merely copies of $S^1$. The example is plaque expansive due to to fact that all centers leaves are compact (and of uniformly bounded length). However, although the curve through any given point tangent to the bundle is non-unique, there is only one possible foliation of the center. Hence this does not give a counter example to the above mentioned question in HPS.

I think there are hopes to modify the example and make one that has similar compact leafs but non-unique center foliation, perhaps by making the unique integrability fail not only on a single line.

### Hausdorff dimension of projections

March 30, 2010

A few days ago, professor Wilkinson asked me the following question on google talk (while I was in Toronto):

Say that a set in $\mathbb{R}^n$ is a $k$-zero set for some integer $k if for every $k$-dimensional subspace $P$, saturating the set $X$ by planes parallel to $P$ yields a set of $n$-dimensional Lebesgue measure zero. How big can a $k$-zero set be?”

On the spot my guess was that the Hausdorff dimension of the set is at most $n-k$. In deed this is the case:

First let’s note that $n$-dimensional Lebesgue measure of the $P$-saturated set is $0$ iff the $n-k$ dimensional Lebesgue measure of the projection of our set to the $n-k$ subspace orthogonal to $P$ is $0$.

Hence the question can be reformulated as: If a set $E \subseteq \mathbb{R}^n$ has all $n-k$ dimensional projection being $n-k$ zero sets, how big can the set be?

Looking this up in the book ‘The Geometry of Fractal Sets’ by Falconer, indeed it’s a theorem:

Theorem: Let $E \subseteq \mathbb{R}^n$ compact, $\dim(E) = s$ (Hausdorff dimension), let $G_{n,k}$ be the Garssmann manifold consisting of all $k$-dimensional subspaces of $\mathbb{R}^n$, then
a) If $s \leq k$, $\dim(\mbox{Proj}_\Pi E) = s$ for almost all $\Pi \in G_{n,k}$

b) If $s > k$, $\mbox{Proj}_\Pi E$ has positive $k$-dimensional Lebesgue measure for almost all $\Pi \in G_{n,k}$.

In our case, we have some set with all $n-k$-dimensional projection having measure $0$, hence the set definitely does not satisfy b), i.e. it has dimensional at most $n-k$. Furthermore, a) also gives that if we have a uniform bound on the dimension of almost all projections, this is also a bound on the dimension of our original set.

This is strict as we can easily find sets that’s $n-k$ dimensional and have all such projections measure $0$. For example, take an $n-k$ subspace and take a full-dimension measure $0$ Cantor set on the subspace, the set will have all projections having measure $0$.

Also, since the Hausdorff dimension of any projection can’t exceed the Hausdorff dimension of the original set, a set with one projection having positive $n-k$ measure implies the dimension of the original set is $\geq n-k$.

Question 2: If one saturate a $k$-zero set by any smooth foliations with $k$-dimensional leaves, do we still get a set of Lebesgue measure $0$?

We answer the question in the affective.

Given foliation $\mathcal{F}$ of $\mathbb{R}^n$ and $k$-zero set $E$. For any point $p \in E$, there exists a small neighborhood in which the foliation is diffeomorphic to the subspace foliation of the Euclidean space. i.e. there exists $f$ from a neighborhood $U$ of $p$ to $(-\epsilon, \epsilon)^n$ where the leaves of $\mathcal{F}$ are sent to $\{\bar{q}\} \times (-\epsilon, \epsilon)^k$, $\bar{q} \in (-\epsilon, \epsilon)^{n-k}$.

By restricting $f$ to a small neighborhood (for example, by taking $\epsilon$ to be half of the origional $\epsilon$), we may assume that $f$ is bi-Lipschitz. Hence the measure of the $\mathcal{F}$-saturated set inside $U$ of $U \cap E$ is the same as $f(U \cap E)$ saturated by parallel $k$-subspaces inside $(-\epsilon, \epsilon)^n$. Dimension of $f(E)$ is the same as dimension of $E$ which is $\leq n-k$, if the inequality is strict, then all projections of $f(E)$ onto $n-k$ dimensional subspaces has measure $0$ i.e. the saturated set by $k$-planes has $n$ dimensional measure $0$.

…to be continued