Tag: David Gabai

Stabilization of Heegaard splittings

ConanLeave a comment

In the last lecture of a course on Heegaard splittings, professor Gabai sketched an example due to Hass-Thompson-Thurston of two genus g Heegaard splittings of a 3-manifold that requires at least g stabilization to make them equivalent. The argument is, in my opinion, very metric-geometric. The connection is so striking (to me) so that I feel necessary to give a brief sketch of it here.

(Side note: This has been a wonderful class! Although I constantly ask stupid questions and appear to be confused from time to time. But in fact it has been very interesting! I should have talked more about it on this blog…Oh well~)

The following note is mostly based on professor Gabai’s lecture, I looked up some details in the original paper ( Hass-Thompson-Thurston ’09 ).

Recall: (well, I understand that I have not talked about Heegaard splittings and stabilizations here before, hence I’ll *try to* give a one minute definition)

A Heegaard splitting of a 3-manifold M is a decomposition of the manifold as a union of two handlebodies intersecting at the boundary surface. The genus of the Heegaard splitting is the genus of the boundary surface.

All smooth closed 3-manifolds has Heegaard splitting due the mere existence of a triangulation ( by thicken the 1-skeleton of the triangulation one gets a handlebody perhaps of huge genus, it’s easy exercise to see its complement is also a handlebody). However it is of interest to find what’s the minimal genus of a Heegaard splitting of a given manifold.

Two Heegaard splittings are said to be equivlent if there is an isotopy of the manifold sending one splitting to the other (with boundary gluing map commuting, of course).

A stabilization of a Heegaard splitting (H_1, H_2, S) is a surgery on S that adds genus (i.e. cut out two discs in S and glue in a handle). Stabilization will increase the genus of the splitting by 1)

Let M be any closed hyperbolic 3-manifold that fibres over the circle. (i.e. M is F_g \times [0,1] with the two ends identified by some diffeomorphism f: F_g \rightarrow F_g, g\geq 2):

Let M'_k be the k fold cover of M along S^1 (i.e. glue together $k$ copies of F_g \times I all via the map f:

Let M_k be the manifold obtained by cut open M'_k along $F_g$ and glue in two handlebodies H_1, H_2 at the ends:

Since M is hyperbolic, M'_k is hyperbolic. In fact, for any \varepsilon > 0 we can choose a large enough k so that M_k can be equipped with a metric having curvature bounded between 1-\varepsilon and 1+\varepsilon everywhere.

( I’m obviously no in this, however, intuitively it’s believable because once the hyperbolic part M'_k is super large, one should be able to make the metric in M'_k slightly less hyperbolic to make room for fitting in an almost hyperbolic metric at the ends H_1, H_2). For details on this please refer to the original paper. :-P

Now there comes our Heegaard splittings of M_k!

Let k = 2n, let H_L be the union of handlebody H_1 together with the first n copies of M, H_R be H_2 with the last n copies of M. H_L, H_R are genus g handlebodies shearing a common surface S in the ‘middle’ of M_k:

Claim: The Heegaard splitting \mathcal{H}_1 = H_L \cup H_R and \mathcal{H}_2 = H_L \cup H_R cannot be made equivalent by less than g stabilizations.

In other words, first of all one can not isotope this splitting upside down. Furthermore, adding handles make it easier to turn the new higher genus splitting upside down, but in this particular case we cannot get away with adding anything less than g many handles.

Okay, not comes the punchline: How would one possible prove such thing? Well, as one might have imagined, why did we want to make this manifold close to hyperbolic? Yes, minimal surfaces!

Let’s see…Suppose we have a common stabilization of genus 2g-1. That would mean that we can sweep through the manifold by a surface of genus (at most) 2g-1, with 1-skeletons at time 0, 1.

Now comes what professor Gabai calls the ‘harmonic magic’: there is a theorem similar to that of Pitts-Rubinstein

Ingredient #1: (roughly thm 6.1 from the paper) For manifolds with curvature close to -1 everywhere, for any given genus g Heegaard splitting \mathcal{H}, one can isotope the sweep-out so that each surface in the sweep-out having area < 5 \pi (g-1).

I do not know exactly how is this proved. The idea is perhaps try to shrink each surface to a ‘minimal surface’, perhaps creating some singularities harmless in the process.

The ides of the whole arguement is that if we can isotope the Heegaard splittings, we can isotope the whole sweep-out while making the time-t sweep-out harmonic for each t. In particular, at each time there is (at least) a surface in the sweep-out family that divides the volume of M'_n in half. Furthermore, the time 1 half-volume-surface is roughly same as the time 0 surface with two sides switched.

We shall see that the surfaces does not have enough genus or volume to do that. (As we can see, there is a family of genus 2g surface, all having volume less than some constant independent of n that does this; Also if we have ni restriction on area, then even a genus g surface can be turned.)

Ingredient #2: For any constant K, there is n large enough so no surfaces of genus <g and area <K inside the middle fibred manifold with boundary M'_n can divide the volume of M'_n in half.

The prove of this is partially based on our all-time favorite: the isoperimetric inequality:

Each Riemannian metric \lambda on a closed surface has a linear isoperimetric inequality for 1-chains bounding 2-chains, i.e. any homologically trivial 1-chain c bounds a 2 chain z where

\mbox{Vol}_2(z) \leq K_\lambda \mbox{Vol}_1(c).

Fitting things together:

Suppose there is (as described above) a family of genus 2g-1 surfaces, each dividing the volume of M_{2n} in half and flips the two sides of the surface as time goes from 0 to 1.

By ingredient #1, since the family is by construction surfaces from (different) sweep-outs by ‘minimal surfaces’, we have \mbox{Vol}_2(S_t) < 5 \pi (2g-2) for all t.

Now if we take the two component separated by S_t and intersect them with the left-most n copies of M in M'_{2n} (call it M'_L), at some t, S_t must also divide the volume of M'_L in half.

Since S_t divides both M'_2n and M'_L in half, it must do so also in M'_R.

But S_t is of genus 2g-1! So one of S_t \cap M'_L and S_t \cap M'_R has genus < g! (say it's M'_L)

Apply ingredient #2, take K = 5 \pi (2g-2), there is n large enough so that S_t \cap M'_L, which has area less than K and genus less than g, cannot possibly divide M'_L in half.

Contradiction.

A report of my Princeton generals exam

ConanLeave a comment

Well, some people might be wondering why I haven’t updated my blog since two weeks ago…Here’s the answer: I have been preparing for this generals exam — perhaps the last exam in my life.

For those who don’t know the game rules: The exam consists of 3 professors (proposed by the kid taking the exam, approved by the department), 5 topics (real, complex, algebra + 2 specialized topics chosen by the student). One of the committee member acts as the chair of the exam.

The exam consists of the three committee members sitting in the chair’s office, the student stands in front of the board. The professors ask questions ranging in those 5 topics for however long they want, the kid is in charge of explaining them on the board.

I was tortured for 4.5 hours (I guess it sets a new record?)
I have perhaps missed some questions in my recollection (it’s hard to remember all 4.5 hours of questions).

Conan Wu’s generals

Commitee: David Gabai (Chair), Larry Guth, John Mather

Topics: Metric Geometry, Dynamical Systems

Real analysis:

Mather: Construct a first category but full measure set.

(I gave the intersection of decreasing balls around the rationals)

Guth: F:S^1 \rightarrow \mathbb{R} 1-Lipschitz, what can one say about its Fourier coefficients.

(Decreasing faster than c*1/n via integration by parts)

Mather: Does integration by parts work for Lipschitz functions?

(Lip imply absolutely continuous then Lebesgue’s differentiation theorem)

Mather: If one only has bounded variation, what can we say?

(f(x) \geq f(0) + \int_0^x f'(t) dt)

Mather: If f:S^1 \rightarrow \mathbb{R} is smooth, what can you say about it’s Fourier coefficients?

(Prove it’s rapidly decreasing)

Mather: Given a smooth g: S^1 \rightarrow \mathbb{R}, given a \alpha \in S^1, when can you find a f: S^1 \rightarrow \mathbb{R} such that
g(\theta) = f(\theta+\alpha)-f(\theta) ?

(A necessary condition is the integral of g needs to vanish,
I had to expand everything in Fourier coefficients, show that if \hat{g}(n) is rapidly decreasing, compute the Diophantine set \alpha should be in to guarantee \hat{f}(n) being rapidly decreasing.

Gabai: Write down a smooth function from f:\mathbb{R}^2 \rightarrow \mathbb{R} with no critical points.

(I wrote f(x,y) = x+y) Draw its level curves (straight lines parallel to x=-y)

Gabai: Can you find a such function with the level curves form a different foliation from this one?

(I think he meant that two foliations are different if there is no homeo on \mathbb{R}^2 carrying one to the other,
After playing around with it for a while, I came up with an example where the level sets form a Reeb foliation, and that’s not same as the lines!)

We moved on to complex.

Complex analysis:

Guth: Given a holomorphic f:\mathbb{D} \rightarrow \mathbb{D}, if f has 50 0s inside the ball B_{1/3}(\bar{0}), what can you say about f(0)?

(with a bunch of hints/suggestions, I finally got f(0) \leq (1/2)^{50} — construct polynomial vanishing at those roots, quotient and maximal modulus)

Guth: State maximal modulus principal.

Gabai: Define the Mobius group and how does it act on \mathbb{H}.

Gabai: What do the Mobius group preserve?

(Poincare metric)

Mather: Write down the Poincare metric, what’s the distance from \bar{0} to 1? (infinity)

(I don’t remember the exact distance form, so I tried to guess the denominator being \sqrt{1-|z|}, but then integrating from 0 to 1 does not “barely diverge”. Turns out it should be (1-|z|^2)^2.)

Gabai: Suppose I have a finite subgroup with the group of Mobius transformations acting on \mathbb{D}, show it has a global fixed point.

(I sketched an argument based on each element having finite order must have a unique fixed point in the interior of \mathbb{D}, if two element has different fixed points, then one can construct a sequence of elements where the fixed point tends to the boundary, so the group can’t be finite.)

I think that’s pretty much all for the complex.

Algebra:

Gabai: State Eisenstein’s criteria

(I stated it with rings and prime ideals, which leaded to a small discussion about for which rings it work)

Gabai: State Sylow’s theorem

(It’s strange that after stating Sylow, he didn’t ask me do anything such as classify finite groups of order xx)

Gabai: What’s a Galois extension? State the fundamental theorem of Galois theory.

(Again, no computing Galois group…)

Gabai: Given a finite abelian group, if it has at most n elements of order divisible by n, prove it’s cyclic.

(classification of abelian groups, induction, each Sylow is cyclic)

Gabai: Prove multiplicative group of a finite field is cyclic.

(It’s embarrassing that I was actually stuck on this for a little bit before being reminded of using the previous question)

Gabai: What’s SL_2(\mathbb{Z})? What are all possible orders of elements of it?

(I said linear automorphisms on the torus. I thought it can only be 1,2,4,\infty, but turns out there is elements of order 6. Then I had to draw the torus as a hexagon and so on…)

Gabai: What’s \pi_3(S^2)?

(\mathbb{Z}, via Hopf fibration)

Gabai: For any closed orientable n-manifold M, why is H_{n-1}(M) torsion free?

(Poincare duality + universal coefficient)

We then moved on to special topics:

Metric Geometry:

Guth: What’s the systolic inequality?

(the term ‘aspherical’ comes up)

Gabai: What’s aspherical? What if the manifold is unbounded?

(I guessed it still works if the manifold is unbounded, Guth ‘seem to’ agree)

Guth: Sketch a proof of the systolic inequality for the n-torus.

(I sketched Gromov’s proof via filling radius)

Guth: Give an isoperimetric inequality for filling loops in the 3-manifold S^2 \times \mathbb{R} where S^2 has the round unit sphere metric.

(My guess is for any 2-chain we should have

\mbox{vol}_1(\partial c) \geq C \mbox{vol}_2(c)

then I tried to prove that using some kind of random cone and grid-pushing argument, but soon realized the method only prove

\mbox{vol}_1(\partial c) \geq C \sqrt{\mbox{vol}_2(c)}.)

Guth: Given two loops of length L_1, L_2, the distance between the closest points on two loops is \geq 1, what’s the maximum linking number?

(it can be as large as c L_1 L_2)

Dynamical Systems:

Mather: Define Anosov diffeomorphisms.

Mather: Prove the definition is independent of the metric.

(Then he asked what properties does Anosov have, I should have said stable/unstable manifolds, and ergodic if it’s more than C^{1+\varepsilon}…or anything I’m familiar with, for some reason the first word I pulled out was structurally stable…well then it leaded to and immediate question)

Mather: Prove structural stability of Anosov diffeomorphisms.

(This is quite long, so I proposed to prove Anosov that’s Lipschitz close to the linear one in \mathbb{R}^n is structurally stable. i.e. the Hartman-Grobman Theorem, using Moser’s method, some details still missing)

Mather: Define Anosov flow, what can you say about geodesic flow for negatively curved manifold?

(They are Anosov, I tried to draw a picture to showing the stable and unstable and finished with some help)

Mather: Define rotation number, what can you say if rotation numbers are irrational?

(They are semi-conjugate to a rotation with a map that perhaps collapse some intervals to points.)

Mather: When are they actually conjugate to the irrational rotation?

(I said when f is C^2, C^1 is not enough. Actually C^1 with derivative having bounded variation suffice)

I do not know why, but at this point he wanted me to talk about the fixed point problem of non-separating plane continua (which I once mentioned in his class).

After that they decided to set me free~ So I wandered in the hallway for a few minutes and the three of them came out to shake my hand.

When k looks and smells like the unknot…

Conan5 Comments

Valentine’s day special issue~ ^_^

Professor Gabai decided to ‘do some classical topology before getting into the fancy stuff’ in his course on Heegaard structures on 3-manifolds. So we covered the ‘loop theorem’ by Papakyriakopoulos last week. I find it pretty cool~ (So I started applying it to everything regardless of whether a much simpler argument exists >.<)

Let M be a three dimensional manifold with (non-empty) boundary. In what follows everything is assumed to be in the smooth category.

Theorem: (Papakyriakopoulos, ’58)
If f: \mathbb{D}^2 \rightarrow M extends continuously to \partial \mathbb{D} and the image f(\partial \mathbb{D}) \subseteq \partial M is homotopically non-trivial in \partial M. Then in any neighborhood N(f(\mathbb{D})) we can find embedded disc D \subseteq M such that \partial D is still homotopically non-trivial in \partial M.

i.e. this means that if we have a loop on \partial M that is non-trivial in \partial M but trivial in M, then in any neighborhood of it we can find a simple loop that’s still non-trivial in \partial M and bounds an embedded disc in M.

We apply this to the following:

Corollary: If a knot k \subseteq \mathbb{S}^3 has \pi_1(\mathbb{S}^3 \backslash k) = \mathbb{Z} then k is the unknot.

Proof: Take tubular neighborhood N_\varepsilon(k), consider M=\mathbb{S}^3 \backslash \overline{N_\varepsilon(k)}, boundary of M is a torus.

By assumption we have \pi_1(M) = \pi_1(\mathbb{S}^3 \backslash k) = \mathbb{Z}.

Let k' \subseteq \partial M be a loop homotopic to k in N_\varepsilon(k).

Since \pi_1(M) = \mathbb{Z} and any loop in M is homotopic to a loop in \partial M = \mathbb{T}^2. Hence the inclusion map i: \pi_1(\mathbb{T}^2) \rightarrow \pi_1(M) is surjective.

Let l \subseteq \partial M be the little loop winding around k.

It’s easy to see that i(l) generates \pi_1(M). Hence there exists n s.t. k'-n \cdot l = 0 in \pi_1(M). In other words, after n Dehn twists around l, k' is homotopically trivial in M i.e. bounds a disk in M. Denote the resulting curve k''.

Since k'' is simple, there is small neighborhood of k'' s.t. any homotopically non-trivial simple curve in the neighborhood is homotopic to k''. The loop theorem now implies k'' bounds an embedded disc in M.

By taking a union with the embedded collar from k to k'' in N_\varepsilon(k):

We conclude that k bounds an embedded disc in \mathbb{S}^3 \backslash k hence k is the unknot.

Establishes the claim.

Happy Valentine’s Day, Everyone! ^_^

Cutting the Knot

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 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.

Extremal length and conformal geometry

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.