Posts Tagged ‘extremal length’

A convergence theorem for Riemann maps

January 17, 2011

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.

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Extremal length and conformal geometry

December 6, 2010

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.