Cutting the Knot

December 13, 2010December 14, 2010 Conan10 Comments

Recently I came across a paper by John Pardon – a senior undergrad here at Princeton; in which he answered a question by Gromov regarding “knot distortion”. I found the paper being pretty cool, hence I wish to highlight the ideas here and perhaps give a more pictorial exposition.

This version is a bit different from one in the paper and is the improved version he had after taking some suggestions from professor Gabai. (and the bound was improved to a linear one)

Definition: Given a rectifiable Jordan curve $\gamma: S^1 \rightarrow \mathbb{R}^3$ , the distortion of $\gamma$ is defined as

$\displaystyle \mbox{dist}(\gamma) = \sup_{t,s \in S^1} \frac{d_{S^1}(s,t)}{d_{\mathbb{R}^3}(\gamma(s), \gamma(t))}$ .

i.e. the maximum ratio between distance on the curve and the distance after embedding. Indeed one should think of this as measuring how much the embedding ‘distort’ the metric.

Given knot $\kappa$ , define the distortion of $\kappa$ to be the infimum of distortion over all possible embedding of $\gamma$ :

$\mbox{dist}(\kappa) = \inf\{ \mbox{dist}(\gamma) \ | \ \gamma \ \mbox{is an embedding of} \ \kappa \ \mbox{in} \ \mathbb{R}^3 \}$

It was (somewhat surprisingly) an open problem whether there exists knots with arbitrarily large distortion.

Question: (Gromov ’83) Does there exist a sequence of knots $(\kappa_n)$ where $\lim_{n \rightarrow \infty} \mbox{dist}(\kappa_n) = \infty$ ?

Now comes the main result in the paper: (In fact he proved a more general version with knots on genus $g$ surfaces, for simplicity of notation I would focus only on torus knots)

Theorem: (Pardon) For the torus knot $T_{p,q}$ , we have

$\mbox{dist}(T_{p,q}) \geq \frac{1}{100} \min \{p,q \}$

To prove this, let’s make a few observations first:

First, fix a standard embedding of $\mathbb{T}^2$ in $\mathbb{R}^3$ (say the surface obtained by rotating the unit circle centered at $(2, 0, 0)$ around the $z$ -axis:

and we shall consider the knot that evenly warps around the standard torus the ‘standard $T_{p,q}$ knot’ (here’s what the ‘standard $T_{5,3}$ knot looks like:

By definition, an ’embedding of the knot’, is a homeomorphism $\varphi:\mathbb{R}^3 \rightarrow \mathbb{R}^3$ that carries the standard $T_{p,q}$ to the ‘distorted knot’. Hence the knot will lie on the image of the torus (perhaps badly distorted):

For the rest of the post, we denote $\varphi(T_{p,q})$ by $\kappa$ and $\varphi(\mathbb{T}^2)$ by $T^2$ , w.l.o.g. we also suppose $p<q$ .

Definition: A set $S \in T^2$ is inessential if it contains no homotopically non-trivial loop on $T^2$ .

Some important facts:

Fact 1: Any homotopically non-trivial loop on $\mathbb{T}^2$ that bounds a disc disjoint from $T^2$ intersects $T_{p,q}$ at least $p$ times. (hence the same holds for the embedded copy $(T^2, \kappa)$ ).

As an example, here’s what happens to the two generators of $\pi_1(\mathbb{T}^2)$ (they have at least $p$ and $q$ intersections with $T_{p,q}$ respectively:

From there we should expect all loops to have at least that many intersections.

Fact 2: For any curve $\gamma$ and any cylinder set $C = U \times [z_1, z_2]$ where $U$ is in the $(x,y)$ -plane, let $U_z = U \times \{z\}$ we have:

$\ell(\gamma \cap C) \geq \int_{z_1}^{z_2} | \gamma \cap U_z | dz$

i.e. The length of a curve in the cylinder set is at least the integral over $z$ -axis of the intersection number with the level-discs.

This is merely saying the curve is longer than its ‘vertical variation’:

Similarly, by considering variation in the radial direction, we also have

$\ell(\gamma \cap B(\bar{0}, R) \geq \int_0^{R} | \gamma \cap \partial B(\bar{0}, r) | dr$

Proof of the theorem

Now suppose $\mbox{dist}(T_{p,q})<\frac{1}{100}p$ , we find an embedding $(T^2, \kappa)$ with $\mbox{dist}(\kappa)<\frac{1}{100}p$ .

For any point $x \in \mathbb{R}^3$ , let

$\rho(x) = \inf \{ r \ | \ T^2 \cap (B(x, r))^c$ is inessential $\}$

i.e. one should consider $\rho(x)$ as the smallest radius around $x$ so that the whole ‘genus’ of $T^2$ lies in $B(x,\rho(x))$ .

It’s easy to see that $\rho$ is a positive Lipschitz function on $\mathbb{R}^3$ that blows up at infinity. Hence the minimum value is achieved. Pick $x_0 \in \mathbb{R}^3$ where $\rho$ is minimized.

Rescale the whole $(T^2, \kappa)$ so that $x_0$ is at the origin and $\rho(x_0) = 1$ .

Since $\mbox{dist}(\kappa) < \frac{1}{100}p$ (and note distortion is invariant under scaling), we have

$\ell(\kappa \cap B(\bar{0}, 1) < \frac{1}{100}p \times 2 = \frac{1}{50}p$

Hence by fact 2, $\int_1^{\frac{11}{10}} | \kappa \cap \partial B( \bar{0}, r)| dr \leq \ell(\kappa \cap B(\bar{0}, 1)) < \frac{1}{40}p$

i.e. There exists $R \in [1, \frac{11}{10}]$ where the intersection number is less or equal to the average. i.e. $| \kappa \cap \partial B(\bar{0}, R) | \leq \frac{1}{4}p$

We will drive a contradiction by showing there exists $x$ with $\rho(x) < 1$ .

Let $C_z = B(\bar{0},R) \cap \{z \in [-\frac{1}{10}, \frac{1}{10}] \}$ , since

$\int_{-\frac{1}{10}}^{\frac{1}{10}} | U_t \cap \kappa | dt \leq \ell(\kappa \cap B(\bar{0},1) ) < \frac{1}{50}p$

By fact 2, there exists $z_0 \in [-\frac{1}{10}, \frac{1}{10}]$ s.t. $| \kappa \cap B(\bar{0},1) \times \{z_0\} | < \frac{1}{10}p$ . As in the pervious post, we call $B(\bar{0},1) \times \{z_0\}$ a ‘neck’ and the solid upper and lower ‘hemispheres’ separated by the neck are $U_N, U_S$ .

Claim: One of $U_N^c \cap T^2, \ U_S^c \cap T^2$ is inessential.

Proof: We now construct a ‘cutting homotopy’ $h_t$ of the sphere $S^2 = \partial B(\bar{0}, R)$ :

i.e. for each $t \in [0,1), \ h_t(S^2)$ is a sphere; at $t=1$ it splits to two spheres. (the space between the upper and lower halves is only there for easier visualization)

Note that during the whole process the intersection number $h_t(S^2) \cap \kappa$ is monotonically increasing. Since $| \kappa \cap B(\bar{0},R) \times \{z_0\} | < \frac{1}{10}p$ , it increases no more than $\frac{1}{5}p$ .

Observe that under such ‘cutting homotopy’, $\mbox{ext}(S^2) \cap T^2$ is inessential then $\mbox{ext}(h_1(S^2)) \cap T^2$ is also inessential. (to ‘cut through the genus’ requires at least $p$ many intersections at some stage of the cutting process, but we have less than $\frac{p}{4}+\frac{p}{5} < \frac{p}{2}$ many interesections)

Since $h_1(S^2)$ is disconnected, the ‘genus’ can only lie in one of the spheres, we have one of $U_N^c \cap T^2, \ U_S^c \cap T^2$ is inessential. Establishes the claim.

We now apply the process again to the ‘essential’ hemisphere to find a neck in the $y$ direction, i.e.cutting the hemisphere in half in $(x,z)$ direction, then the $(y,z)$ -direction:

The last cutting homotopy has at most $\frac{p}{5} + 3 \times \frac{p}{4} < p$ many intersections, hence has inessential complement.

Hence at the end we have an approximate $\frac{1}{8}$ ball with each side having length at most $\frac{6}{5}$ , this shape certainly lies inside some ball of radius $\frac{9}{10}$ .

Let the center of the $\frac{9}{10}$ -ball be $x$ . Since the complement of the $\frac{1}{8}$ ball intersects $T^2$ in an inessential set, we have $B(x, \frac{9}{10})^c \cap T^2$ is inessential. i.e.

$\rho(x) \leq \frac{9}{10} <1$

Contradiction.

Extremal length and conformal geometry

December 6, 2010December 6, 2010 Conan2 Comments

There has been a couple of interesting talks recently here at Princeton. Somehow the term ‘extremal length’ came up in all of them. Due to my vast ignorance, I knew nothing about this before, but it sounded cool (and even somewhat systolic); hence I looked a little bit into that and would like to say a few words about it here.

One can find a rigorous exposition on extremal length in the book Quasiconformal mappings in the plane.

Let $\Omega$ be a simply connected Jordan domain in $\mathbb{C}$ . $f: \Omega \rightarrow \mathbb{R}^+$ is a conformal factor on $\Omega$ . Recall from my last post, $f$ is a Lebesgue measurable function inducing a metric on $\Omega$ where

$\mbox{Vol}_f(U) = \int_U f^2 dx dy$

and for any $\gamma: I \rightarrow \Omega$ ( $I \subseteq \mathbb{R}$ is an interval) with $||\gamma'(t)|| = \bar{1}$ , we have the length of $\gamma$ :

$l_f(\gamma) = \int_I f dt$ .

Call this metric $g_f$ on $\Omega$ and denote metric space $(\Omega, g_f)$ .

Given any set $\Gamma$ of rectifiable curves in $U$ (possibly with endpoints on $\partial U$ ), each comes with a unit speed parametrization. Consider the “ $f$ -width” of the set $\Gamma$ :

$\displaystyle w_f(\Gamma) = \inf_{\gamma \in \Gamma} l_f(\gamma)$ .

Let $\mathcal{F}$ be the set of conformal factors $f$ with $L^2$ norm $1$ (i.e. having the total volume of $\Omega$ normalized to $1$ ).

Definition: The extremal length of $\Gamma$ is given by

$\mbox{EL}(\Gamma) = \displaystyle \sup_{f \in \mathcal{F}} w_f(\Gamma)^2$

Remark: In fact I think it would be more natural to just use $w_f(\Gamma)$ instead of $w_f(\Gamma)^2$ since it’s called a “length”…but since the standard notion is to sup over all $f$ , not necessarily normalized, and having the $f$ -width squared divide by the volume of $\Omega$ , I can’t use conflicting notation. One should note that in our case it’s just the square of sup of width.

Definition:The metric $(\Omega, g_f)$ where this extremal is achieved is called an extremal metric for the family $\Gamma$ .

The most important fact about extremal length (also what makes it an interesting quantity to study) is that it’s a conformal invariant:

Theorem: Given $h: \Omega' \rightarrow \Omega$ bi-holomorphic, then for any set of normalized curves $\Gamma$ in $\Omega$ , we can define $\Gamma' = \{ h^{-1}\circ \gamma \ | \ \gamma \in \Gamma \}$ after renormalizing curves in $\Gamma'$ we have:

$\mbox{EL}(\Gamma) = \mbox{EL}(\Gamma')$

Sketch of a proof: (For simplicity we assume all curves in $\Gamma'$ are rectifiable, which is not always the case i.e. for bad maps $h$ the length might blow up when the curve approach $\partial \Omega'$ this case should be treated with more care)

This is indeed not hard to see, first we note that for any $f: \Omega \rightarrow \mathbb{R}^+$ we can define $f' : \Omega' \rightarrow \mathbb{R}^+$ by having

$f^\ast (z) = |h'(z)| (f \circ h) (z)$

It’s easy to see that $\mbox{Vol}_{f^\ast}(\Omega') = \mbox{Vol}_{f}(\Omega)$ (merely change of variables).

In the same way, $l_{f^\ast}(h^{-1}\circ \gamma) = l_f(\gamma)$ for any rectifiable curve.

Hence we have

$w_{f^\ast}(\Gamma') = w_f(\Gamma)$ .

On the other hand, we know that $\varphi: f \mapsto f^\ast$ is a bijection from $\mathcal{F}_\Omega$ to $\mathcal{F}_{\Omega'}$ , deducing

$\mbox{EL}(\Gamma) = \displaystyle ( \sup_{f \in \mathcal{F}} w_f(\Gamma))^2 = \displaystyle ( \sup_{f' \in \mathcal{F}'} w_{f'}(\Gamma'))^2 = \mbox{EL}(\Gamma')$

Establishes the claim.

One might wonder how on earth should this be applied, i.e. what kind of $\Gamma$ are useful to consider. Here we emphasis on the simple case where $\Omega$ is a rectangle (Of course I would first look at this case because of the unresolved issues from the last post :-P ):

Theorem: Let $R = (0,w) \times (0, 1/w)$ , $\Gamma$ be the set of all curves starting at a point in the left edge $\{0\} \times [0, 1/w]$ , ending on $\{1\} \times [0, 1/w]$ with finite length. Then $\mbox{EL}(\Gamma) = w^2$ and the Euclidean metric $f = \bar{1}$ is an extremal metric.

Sketch of the proof: It suffice to show that any metric $g_f$ with $\mbox{Vol}_f(R) = 1$ has at least one horizontal line segment $\gamma_y = [0,w] \times \{y\}$ with $l_f(\gamma_y) \leq w$ . (Because if so, $w_f(\Gamma) \leq w$ and we know $w_{\bar{1}}(\Gamma) = w$ for the Euclidean length)

The average length of $\gamma_y$ over $y$ is

$w \int_0^{1/w} l_f(\gamma_y) dy$

$= w \int_0^{1/w} (\int_0^w f(t, y) dt) dy = w \int_R f$

By Cauchy-Schwartz this is less than $w (\int_R f^2)^{1/2} |R|^{1/2} = w$

Since the shortest curve cannot be longer than the average curve, we have $w_f(\Gamma) \leq w$ .

Hence $\mbox{EL}(\Gamma) = \displaystyle \sup_{f \in \mathcal{F}}w_f(\Gamma)^2 = w^2$

Note it’s almost the same argument as in the proof of systolic inequality on the 2-torus.

Corollary: Rectangles with different eccentricity are not conformally equivalent (i.e. one cannot find a bi-homomorphic map between them sending each edge to an edge).

Remark: I was not aware of this a few days ago and somehow had the silly thought that there are conformal maps between any pair of rectangles while discussing with Guangbo >.< then tried to see what would those maps look like and was of course not able to do so. (there are obviously Riemann maps between the rectangles, but they don't send conners to conners, i.e. can't be extended to a conformal map on the closed rectangle).

An add-on: While I came across a paper of Odes Schramm, applying the techniques of extremal length, the following theorem seemed really cool.

Let $G = (V, E)$ be a finite planar graph with vertex set $V$ and edges $E \subseteq V^2$ . For each vertex $v$ we assign a simply connected domain $D_v$ .

Theorem: We can scale and translate each $D_v$ to $D'_v$ so that $\{ D_v \ | \ v \in V \}$ form a packing (i.e. are disjoint) and the contact graph of $D'_v$ is $G$ . (i.e. $\overline{D'_{v_1}} \cap \overline{D'_{v_2}} \neq \phi$ iff $(v_1, v_2) \in E$ .