Archive for December 13th, 2010

Cutting the Knot

December 13, 2010

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 ydirection, 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.