A convergence theorem for Riemann maps

January 17, 2011November 1, 2011 ConanLeave a comment

So~ After 2.5 weeks of wonderful math discussions with Amie and Charles, I finished my winter vacation and got back to Princeton! (and back to my normal blogging Sundays ^^)

One thing I would like to shear here is that we (me and Charles) finally got an answer to the following question that’s been haunting me for a while:

Question: Given Jordan curve $C \subseteq \mathbb{C}$ containing a neighborhood of $\bar{0}$ in its interior. Given parametrizations $\gamma_1:S^1 \rightarrow C$ .

Is it true that for all $\varepsilon >0$ , there exists $\delta >0$ s.t. any Jordan curve $C'$ with a parametrization $\gamma_2:S^1 \rightarrow C_2$ so that $||\gamma_1-\gamma_2||<\delta$ in the uniform norm implies the Riemann maps $R, R'$ from $\mathbb{D}$ to the interiors of $C, C'$ that fixes the origin and have positive real derivatives at $\bar{0}$ would be at most $\varepsilon$ apart?

i.e. Is the projection map from the space of parametrized Jordan curves (with the uniform metric) to the space of unparametrized Jordan curves (with metric given by taking uniform distance between the canonical Riemann maps) continuous?

First, I think the development and problem-solving process for this one is quite interesting and worth mentioning (skip this if you just want to see math):

—Begin story—

The problem was initially of interest because I stated a lemma on our Jordan curves paper which asserts the above projection map is continuous at smooth curves. To my surprise, I was unable to prove this seemingly-direct lemma. I turned to Charles for help, after a day or so of thinking he proved it for smooth curves (via a very clever usage of cross-cuts as in the proof of Carathedory’s theorem) and asked back whether the map is actually continuous at all points.

This seemed to be such a natural question but we couldn’t find it in the literature. For a day or so we were both feeling negative about this since the cross-cut method fails when the Jordan curve has positive measure, which “should” happen a lot. In any case, I posted a question on mathoverflow to see if there is a standard theorem out there implying this. Almost right after I posted the question, during a wonderful lunch-conversation with Charles, I got this wonderful idea of applying extremal length techniques not to the semi-circular crosscut but only to the ‘feet’ of it. Which later that day turned out to be a proof of the continuity.

The next morning, after confirming the steps of the proof and made sure it works, I was thrilled to find that Thurston responded to the post and explained his intuition that the answer is positive. Although having solved the problem already, I am still amazed by his insights ^^ (It’s the second question I asked there, he left an comment again! It just feels great to have your idol giving you ideas, isn’t it? :-P)

Later on, McMullen pointed out to us that in fact a book by Pommerenke contains the result. Nevertheless, it was great fun proving this, hence I decided to sketch the proof here ^^

—End story—

Ingredients of the proof: We quote the following well-known but weaker theorem (one can find this in, for example Goluzin’s classical book, p228)

Theorem: If the Jordan domains converge (in the sense that parametrizations of the boundaries converge uniformly) then the Riemann maps converge uniformly on compact sets.

We also use the following topological lemma:

Lemma: Given Jordan curve $C \subseteq \hat{\mathbb{C}}$ , $\gamma: S^1 \rightarrow C$ be a parametrization. For all $\varepsilon > 0$ , there exists $\mu >0$ s.t. for all $\gamma' : S^1 \rightarrow \hat{\mathbb{C}}$ with $|| \gamma - \gamma'|| < \mu$ ( denote C' = \gamma'(S^1)$) , for all $p, q \in C'$ , $d(p,q) < \mu \Rightarrow \mbox{diam}(A(p,q)) < \varepsilon$

where $A(p,q)$ is the short arc in $C'$ connecting $p, q$ .

The proof of the lemma is left as an exercise

Proof of the Theorem:

Given $C$ and $\varepsilon$ as in the theorem, apply the lemma to $(C, \varepsilon/6)$ , we obtain a $\mu < \varepsilon / 6$ so that all curves $\mu$ -close to $C$ has the property that the arc connecting any two points less than $\mu$ -apart has diameter no more than $\varepsilon/100$ .

By compactness of $\partial \mathbb{D}$ , we can choose finitely many crosscut neignbourhoods $\{ H_1, H_2, \cdots, H_N \}$ , $H_i \subseteq \bar{\mathbb{D}}$ are "semi-discs" around points in $\partial \mathbb{D}$ as shown:

By extremal length, we can choose the cross-cuts $C_i$ bounding $H_i$ with length $\ell(R(C_i)) < \mu/4$ where $R: \bar{\mathbb{D}} \rightarrow \hat{\mathbb{C}}$ is the canonical Riemann map corresponding to $C$ . Hence by lemma, we also get $\mbox{diam}(R(H_i) < \varepsilon/3$ .

Let $\{ f_1, f_2, \cdots, f_{2N} \}$ be endpoints of $\{C_1, \cdots, C_N \}$ .

Let $d = \min \{ d(f_i, f_j) \ | \ 1 \leq i < j \leq 2N \}$ .

Choose $\sigma < \mu d / 40$ and $\{ B( \bar{0}, 1-2\sigma), H_1, \cdots, H_N \}$ covers $\bar{\mathbb{D}}$ . Let $R = 1-\sigma$ :

By the above theorem in Goluzin, since $B_R = \bar{B(0, R)}$ is compact, there exists a $0 < \delta < \min \{\mu/4, d/10 \}$ s.t.

$|| \gamma' - \gamma || < \delta$ $\Rightarrow ||R|_{B_R} - R'|_{B_R}|| < \mu/4$ .

Fix a $(C', \gamma')$ with $|| \gamma - \gamma'|| < \delta$ . Let $R'$ be the canonical Riemann map corresponding to $C'$ .

Claim: $||R-R'|| < \varepsilon$ .

First note that assuming the theorem in Goluzin, it suffice to show $||R|_{\partial \mathbb{D}} - R'|_{\partial \mathbb{D}}|| < \varepsilon$ .

For any $1 \leq i \leq N$ , let $f_1, f_2$ be endpoints of $C_i$ . Apply the extremal length to the set of radial segments in the almost-rectangle $[f_1, f_1+d/10] \times [0,\sigma]$ .

We conclude there exists $e_1 \in [f_1, f_1+d/10]$ s.t. the segment $s_1 = \{e_1\} \times [0, \sigma]$ has length

$\ell(R'(s_1)) \leq 2 \sigma (d/10) m_2(R'([f_1, f_1+d/10] \times [0,\sigma]))$ .

Since $\sigma < \mu d / 40$ and $m_2(R'([f_1, f_1+d/10] \times [0,\sigma])) \leq 1$ , we have

$\ell(R'(s_1)) \leq \mu/4$ .

Similarly, find $e_2 \in [f_2 - d/10, f_2]$ where $\ell(R'(s_2)\leq \mu/4$ .

Connect $e'_1, e'_2$ by a semicircle contained in $H_i$ , denote the enclosed region by $V_i \subseteq H_i$ .

By construction, $\{ B_R, V_1, \cdots, V_N \}$ still covers $\bar{\mathbb{D}}$ .

Hence for all $p \in \partial \mathbb{D}$ , there exists $i$ where $p \in$ latex V_i$.

Since inside $V_i \cap B_R$ the two maps $R, R'$ are less than $d/10$ apart, we have $R(V_i) \cap R'(V_i) \neq \phi$ .

Hence $d(R(p), R'(p)) \leq \mbox{diam}(R(H_i)) + \mbox{diam}(R'(V_i))$ .

By construction, $\mbox{diam}(R(H_i)) < \varepsilon/2$ .

$\mbox{diam}(R'(V_i)) = \mbox{diam}(\partial V_i)$ , we will break $\partial V_i$ into three parts and estimate diameter of each part separately.

Since $||\gamma-\gamma'|| < \delta$ , $\tau = \gamma' \circ \gamma^{-1} \circ R|_{\partial \mathbb{D}}$ is another parametrization of $C'$ with $|| \tau - R|_{\partial \mathbb{D}}|| < \delta$ .

The arc connecting $e'_1$ to $e'_2$ is contained in $B_R \cap V_i$ , the arc in $C'$ connecting $\tau(e_1), \tau(e_2)$ is $\delta$ away from $R(H_i)$ hence the union of the two has diameter at most $\mbox{diam}(R(V_i)) + \delta < \varepsilon/6 + \varepsilon/6 = \varepsilon/3$

Length of the arcs $R'(s_1), R'(s_2)$ are less than $\mu/4 < \varepsilon/12$ .

Hence $d(\tau(e_1), R'(e_1)) < \ell(R'(s_1)) + \delta < \mu$ . By lemma, this implies the arc in $C'$ connecting $\tau(e_1), R'(e_1)$ has length at most $\varepsilon/12$ .

Hence altogether the we have $\mbox{diam}(R'(V_i)) \leq \varepsilon/3+\varepsilon/12+\varepsilon/12 = \epsilon/2$ .

We deduce $d(R(p), R'(p)) \leq \mbox{diam}(R(H_i)) + \mbox{diam}(R'(V_i)) < \varepsilon$ .

Q.E.D.

Stats monkey: this blog is doing awesome

January 2, 2011January 2, 2011 ConanLeave a comment

Hi everyone! As you may have noticed, I have been taking winter break from blogging…Sorry about that and I expect to resume regular blog-writing-Sundays once school starts.

Meanwhile, I would like to thank you all for your support this year! The following is a cute blog-health e-mail I received this morning from WordPress. I would like to share this with you and wish for a wonderful blogging year of 2011.

The stats helper monkeys at WordPress.com mulled over how this blog did in 2010, and here’s a high level summary of its overall blog health:

Healthy blog!

The Blog-Health-o-Meter™ reads This blog is doing awesome!.

Crunchy numbers

A Boeing 747-400 passenger jet can hold 416 passengers. This blog was viewed about 3,800 times in 2010. That’s about 9 full 747s.

In 2010, there were 30 new posts, growing the total archive of this blog to 37 posts. There were 79 pictures uploaded, taking up a total of 33mb. That’s about 2 pictures per week.

The busiest day of the year was April 21st with 71 views. The most popular post that day was About me.

Where did they come from?

The top referring sites in 2010 were math.princeton.edu, en.wordpress.com, google.com, zhenghezhang.wordpress.com, and bbs.gter.net.

Some visitors came searching, mostly for conan777, larry guth ias, conan wu 777, and whitehead manifold.

Attractions in 2010

These are the posts and pages that got the most views in 2010.

About me June 2009

Systolic inequality on the 2-torus March 2010
5 comments

Systoles and the generalized Geroch conjecture October 2010
3 comments

Whitney’s extension theorem revisited March 2010

On length and volume November 2010
2 comments

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 $(v_1, v_2) \in E$ .

Note: This is vastly stronger than producing a circle packing with prescribed structure.

On length and volume

November 29, 2010December 1, 2010 Conan2 Comments

About a year ago, I came up with an simple argument for the following simple theorem that appeared in a paper of professor Guth’s:

Theorem: If $U$ is an open set in the plane with area $1$ , then there is a continuous function $f$ from $U$ to the reals, so that each level set of $f$ has length at most $10$ .

Recently a question of somewhat similar spirit came up in a talk of his:

Question: Let $\langle \mathbb{T}^2, g \rangle$ be a Riemannian metric on the torus with total volume $1$ , does there always exist a function $f: \mathbb{T}^2 \rightarrow \mathbb{R}$ s.t. each level set of $f$ has length at most $10$ ?

I have some rough thoughts about how might a similar argument on the torus look like, hence I guess it would be a good idea to review and (somewhat carefully) write down the original argument. Since our final goal now is to see how things work on a torus (or other manifolds), here I would only present the less tedious version where $U$ is bounded and all boundary components of $U$ are smooth Jordan curves. Here it goes:

Proof: Note that if a projection of $U$ in any direction has length (one-dimensional measure) $\leq 10$ , then by taking $f$ to be the projection in the orthogonal direction, all level sets are straight with length $\leq 10$ (see image below).

Hence we can assume any $1$ -dimensional projection of $U$ has length $\geq 10$ . A typically bad set would ‘span’ a long range in all directions with small area, it can contain ‘holes’ and being not connected:

Project $U$ onto $x$ and $y$ -axis, by translating $U$ , we assume $\inf \pi_x(U) = \inf \pi_y(U) = 0$ . Look at the measure $1$ set $S$ in the middle of $\pi_y(U)$ (i.e. a measure 1 set $[a,b] \cap \pi_y(U)$ with the property $m_1(\pi_y(U) \cap [0,a]) = m_1(\pi_y(U) \cap [b, \infty]$ )

By Fubini, since the volume $\pi_y^{-1}(S)$ is at most $1$ , there must be a point $p\in S$ with $m_1(\pi_y^{-1}(p))\leq 1$ :

Since the boundary of $U$ is smooth, we may find a very small neighborhood $B_\delta(p) \subseteq \mathbb{R}$ where for each $q \in B_\delta(p), m_1(\pi_y^{-1}(q) \leq 1+\epsilon$ . (we will call this pink region a ‘neck’ of the set for it has small width and is roughly in the middle)

Now we define a $\varphi_1: U \rightarrow \mathbb{R}^2$ that straches the neck to fit in a long thin tube (note that in general $\pi_y(U)$ may not be connected, but everything is still well-defined and the argument does go through.) and then bend the neck to make the top chunk vertically disjoint from the bottom chunk.

We can take $\varphi$ so that $\varphi^{-1}$ sends the vertical foliation of $\varphi(U)$ to the following foliation in $U$ (note that here we drew the neck wider for easier viewing, in fact the horizontal lines are VERY dense in the neck).

If the $y$ -projection of the top or bottom chunk is larger than $2$ , we repeat the above process t the chunks. i.e. Finding a neck in the middle measure $1$ set in the chunk, starch the neck and shift the top chunk, this process is guaranteed to terminate in at most $m_1(\pi_y(U))$ steps. The final $\varphi$ sends $U$ to something like:

Where each chunk has $y$ -width $L$ between $1$ and $2$ .

Define $f = \pi_x \circ \phi$ .

Claim: For any $c \in \mathbb{R}, m_1(f^{-1}(c)) \leq 5$ .

The vertical line $x=c$ intersects $\varphi(U)$ in at most one chunk and two necks, taking $\varphi^{-1}$ of the intersection, this is a PL curve $C$ with one vertical segment and two horizontal segment in $U$ :

The total length of $f^{-1}(c) = C \cap U$ is less than $2+2\delta$ (length of $U$ on the vertical segment) $+ 2 \times (1+\epsilon)$ (length of $U$ on each horizontal segment). Pick $\epsilon, \delta$ both less than $1/4$ , we conclude $m_1(f^{-1}(c)) < 5$ .

Establishes the theorem.

Remark:More generally,any open set of volume $V$ has such function with fibers having length $\leq 5 \sqrt{V}$ . T he argument generalizes by looking at the middle set length $\sqrt{V}$ set of each chunk.

Moving to the torus

Now let’s look at the problem on $\langle \mathbb{T}^2, g \rangle$ , by the uniformization theorem we have a flat torus $T^2 = \mathbb{R}^2/\Gamma$ where $\Gamma$ is a lattice, $\mbox{vol}(T^2) = 1$ and a function $h: T^2 \rightarrow \mathbb{R}^{+}$ s.t. $\langle T^2, h g_0 \rangle$ is isometric to $\langle \mathbb{T}^2, g \rangle$ . $g_0$ is the flat metric. Hence we only need to find a map on $T^2$ with short fibers.

Note that

$\int_{T^2} h^2 d V_{g_0} = 1$

and the length of the curve $\gamma$ from $p$ to $q$ in $\langle T^2, h \dot{g_0} \rangle$ is

$\int_I h |\gamma'(t)| dt$ .

Consider $T^2$ as the parallelogram given by $\Gamma$ with sides identified. w.l.o.g. assume one side is parallel to the $x$ -axis. Let $L$ be a linear transformation preserving the horizontal foliation and sends the parallelogram to a rectangle.

Let $F$ be a piece-wise isometry that “folds” the rectangle:

(note that $F$ is four-to-one except for on the edges and the two medians)
Since all corresponding edges are identified, $lates F$ is continuous not only on the rectangle but on the rectangular torus.

Now we consider $F \circ L$ , pre-image of typical horizontal and vertical lines in the small rectangle are union of two parallel loops:

Note that vertical loops might be very long in the flat $T^2$ due to the shear while the horizontal is always the width.

(to be continued)

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Crunchy numbers

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