## Posts Tagged ‘David Gabai’

### A train track on twice punctured torus

April 22, 2012

This is a non-technical post about how I started off trying to prove a lemma and ended up painting this:

One of my favorite books of all time is Thurston‘s ‘Geometry and Topology of 3-manifolds‘ (and I just can’t resist to add here, Thurston, who happen to be my academic grandfather, is in my taste simply the coolest mathematician on earth!) Anyways, for those of you who aren’t topologists, the book is online and I have also blogged about bits and parts of it in some old posts such as this one.

I still vividly remember the time I got my hands on that book for the first time (in fact I had the rare privilege of reading it from an original physical copy of this never-actually-published book, it was a copy on Amie‘s bookshelf, which she ‘robbed’ from Benson Farb, who got it from being a student of Thurston’s here at Princeton years ago). Anyways, the book was darn exciting and inspiring; not only in its wonderful rich mathematical content but also in its humorous, unserious attitude — the book is, in my opinion, not an general-audience expository book, but yet it reads as if one is playing around just to find out how things work, much like what kids do.

To give a taste of what I’m talking about, one of the tiny details which totally caught my heart is this page (I can’t help smiling each time when flipping through the book and seeing the page, and oh it still haunts me >.<):

This was from the chapter about Kleinian groups, when the term ‘train-track’ was first defined, he drew this image of a train(!) on moving on the train tracks, even have smoke steaming out of the engine:

To me such things are simply hilarious (in the most delightful way).

Many years passed and I actually got a bit more into this lamination and train track business. When Dave asked me to ‘draw your favorite maximal train track and test your tube lemma for non-uniquely ergodic laminations’ last week, I ended up drawing:

Here it is, a picture of my favorite maximal train track, on the twice punctured torus~! (Click for larger image)

Indeed, the train is coming with steam~

Since we are at it, let me say a few words about what train tracks are and what they are good for:

A train track (on a surface) is, just as one might expect, a bunch of branches (line segments) with ‘switches’, i.e. whenever multiple branches meet, they must all be tangent at the intersecting point, with at least one branch in each of the two directions. By slightly moving the switches along the track it’s easy to see that generic train track has only switches with one branch on one side and two branches on the other.

On a hyperbolic surface $S_{g,p}$, a train track is maximal if its completementry region is a disjoint union of triangles and once punctured monogons. i.e. if we try to add more branches to a maximal track, the new branch will be redundant in the sense that it’s merely a translate of some existing branch.

As briefly mentioned in this post, train tracks give natural coordinate system for laminations just like counting how many times a closed geodesic intersect a pair of pants decomposition. To be slightly more precise, any lamination can be pushed into some maximal train track (although not unique), once it’s in the track, any laminations that’s Hausdorff close to it can be pushed into the same track. Hence given a maximal train track, the set of all measured laminations carried by the train track form an open set in the lamination space, (with some work) we can see that as measured lamination they are uniquely determined by the transversal measure at each branch of the track. Hence giving a coordinate system on $\mathcal{ML})(S)$.

Different maximal tracks are of course them pasted together along non-maximal tracks which parametrize a subspace of $\mathcal{ML}(S)$ of lower dimension.

To know more about train tracks and laminations, I highly recommend going through the second part of Chapter 8 of Thurston’s book. I also mentioned them for giving coordinate system on the measured lamination space in the last post.

In any case I shall stop getting into the topology now, otherwise it may seem like the post is here to give exposition to the subject while it’s actually here to remind myself of never losing the Thurston type childlike wonder and imagination (which I found strikingly larking in contemporary practice of mathematics).

### Filling and unfilling measured laminations

April 10, 2012

(images are gradually being inserted ~)

I’m temporarily back into mathematics to (try) finish up some stuff about laminations. While I’m on this, I figured maybe sorting out some very basic (and cool) things in a little post here would be a good idea. Browsing through the blog I also realized that as a student of Dave’s I have been writing surprisingly few posts related to what we do. (Don’t worry, like all other posts in this blog, I’ll only put in stuff anyone can read and hopefully won’t be bored reading :-P)

Here we go. As mentioned in this previous post, my wonderful advisor has proved that the ending lamination space is connected and locally connected (see Gabai’08).

Definition: Let $S_{g,p}$ be a hyperbolic surface of genus $g$ and $p$ punctures. A (geodesic) lamination $L \subseteq S$ is a closed set that can be written as a disjoint union of geodesics. i.e. $L = \sqcup_{\alpha \in I} \gamma_\alpha$ where each $\gamma_\alpha$ is a (not necessary closed) geodesic, $\gamma$ is called a leaf of $L$.

Let’s try to think of some examples:

i) One simple closed geodesic

ii) A set of disjoint simple closed geodesics

iii) A non-closed geodesic spirals onto two closed ones

iV) Closure of a single simple geodesic where transversal cross-sections are Cantor-sets

An ending lamination is a lamination where
a) the completement $S \backslash L$ is a disjoint union of discs and once punctured discs (filling)
b) all leaves are dense in $L$. (minimal)

Exercise: example i) satisfies b) and example iv) as shown satisfies both a) and b) hence is the only ending lamination.

It’s often more natural to look at measured laminations, for example as we have seen in the older post, measured laminations are natural generalizations of multi-curves and the space $\mathcal{ML}(S)$ is homeomorphic to $\mathbb{R}^{6g-6+2p}$ (Thurston) with very natural coordinate charts (given by train-tracks).

Obviously not all measured laminations are supported on ending laminations (e.g. example i) and ii) with atomic measure on the closed curves.) It is well known that if a lamination fully supports an invariant measure, then as long as the base lamination satisfies a), it automatically satisfies b) and hence is an ending lamination. This essentially follows from the fact that having a fully supported invariant measure and being not minimal implies the lamination is not connected and hence won’t be filling.

Exercise:Example iii) does not fully support invariant measures.

Scaling of the same measure won’t effect the base lamination, hence we may eliminate a dimension by quotient that out and consider the space of projective measured laminations $\mathcal{PML}(S) \approx \mathbb{S}^{6g-7+2p}$. Hence we may decompose measured laminations into filling and unfilling ones. i.e.

$\mathcal{PML}(S) = \mathcal{FPML}(S) \sqcup \mathcal{UPML}(S)$

where $\mathcal{FPML}(S)$ projects to the ending laminations via the forgetting measure map $\pi$.

This decomposition of the standard sphere $\mathbb{S}^{6g-7+2p}$ is mysterious and very curious in my opinion. To get a sense of this, let’s take a look at the following facts:

Fact 1: $\mathcal{UPML}$ is a union of countably many disjoint hyper-discs (i.e. discs of co-dimension $1$).

Well, if a measured lamination is unfilling, it must contain some simple closed geodesic as a leaf (or miss some simple closed geodesic). For each such geodesic $C$, there are two possible cases:

Case 1: $C$ is non-separating. The set of measured laminations that missed $C$ is precisely the set of projective measured laminations supported on $S_{g-1, p+2}$, hence homeomorphic to $\mathbb{S}^{6g-13+2p+4} = \mathbb{S}^{(6g-7+2p)-2}$ we may take any such measured lamination, disjoint union with $C$, we may assign any ratio of wrights to $C$ and the lamination. This corresponds to taking the cone of $\mathbb{S}^{(6g-7+2p)-2}$ with vertex being the atomic measure on $C$. Yields a disc of dimension $(6g-7+2p)-1$.

Case 2: $C$ is separating. Similarly, the set of measured laminations missing $C$ is supported on two connected surfaces with total genus $g$ and total punctures $p+2$.

To describe the set of projective measured laminations missing $C$, we first determine the ratio of measure between two connected components and then compute the set of laminations supported in each component. i.e. it’s homeomorphic to $[0,1] \times \mathbb{S}^{d_1} \times \mathbb{S}^{d_2}/\sim$ where $d_1+d_2 = 6g-2*7+2(p+2) = 6g-10+2p$ and $(0, x_1, y) \sim (0, x_2, y)$ and $(1, x, y_1) \sim (1, x, y_2)$.

Exercise: check this is a sphere. hint: if $d_1 =d_2 = 1$, we have:

Again we cone w.r.t. the atomic measure corresponding to $C$, get a hyper disc.

At this point you may think ‘AH! $\mathcal{UPML}$ is only a countable union of hyper-discs! How complicated can it be?!’ Turns out it could be, and (unfortunately?) is, quite messy:

Fact 2: $\mathcal{UPML}$ is dense in $\mathcal{PML}$.

This is easy to see since any filling lamination is minimal, hence all leaves are dense, we may simply take a long segment of some leaf where the beginning and end point are close together on some transversal, close up the segment by adding a small arc on the transversal, we get a simple closed geodesic that’s arbitrarily close to the filling lamination in $\mathcal{PML}$. Hence the set of simple closed curves with atomic measure are dense, obviously implying $\mathcal{UPML}$ dense.

So how exactly does this decomposition look like? I found it very mysterious indeed. One way to look at this decomposition is: we know two $\mathcal{UPML}$ discs can intersect if and only if their corresponding curved are disjoint. Hence in some sense the configuration captures the structure of the curve complex. Since we know the curve complex is connected, we may start from any disc, take all discs which intersect it, then take all discs intersecting one of the discs already in the set, etc.

We shall also note that all discs intersecting a given disc must pass through the point corresponding to the curve at the center. Hence the result will be some kind of fractal-ish intersecting discs:

(image)

Yet somehow it manages to ‘fill’ the whole sphere!

Hopefully I have convinced you via the above that countably many discs in a sphere can be complicated, not only in pathological examples but they appear in ‘real’ life! Anyways, with Dave’s wonderful guidance I’ve been looking into proving some stuff about this (in particular, topology of $\mathcal{FPML}$). Hopefully the mysteries would become a little clearer over time~!

### Haken manifolds and virtual Haken conjecture

November 21, 2011

Hi people~ My weekends have been unfortunately filled up with grading undergrad assignments for the last couple of weeks >.< I'll try to catch up on blogging by finding some other time slot during the week.

As a grand-student of Thurston's I feel obligated to end my ignorance regarding Haken manifolds. I guess it's a good idea to start by writing my usual kids-friendly exposition here.

In the rest of the post, $M$ is a compact (so perhaps with boundary), orientable, irreducible (meaning each embedded 2-sphere bounds a ball) 3-manifold.

Definition: A properly embedded oriented surface $S \subseteq M$ is incompressible if $S$ is not the 2-sphere and any simple closed curve on $S$ which bounds an embedded disc in $M \backslash S$ also bounds one in $S$.

Figure 1

In other words, together with Dehn’s lemma this says the map $\varphi: \pi_1(S) \rightarrow \pi_1 (M)$ induced by the inclusion map is injective.

Note that the surface $S$ could have boundary, for example:

Figure 2

Definition: $M$ is Haken if it contains an incompressible surface.

Okay, at this point you should be asking, what’s good about Haken manifolds? The beauty about it is that, roughly speaking, once you find one incompressible surface in the manifold, you can just keep finding them until the manifold is completely chopped up into balls by incompressible surfaces.

Theorem: (Haken) Any Haken 3-manifold $M$ contains a hierarchy $S_0 \subseteq S_1 \subseteq \cdots \subseteq S_n$ where

1.$S_0$ is an incompressible surface in $M$
2.$S_i = S_{i-1} \cup S$ where $S$ is an incompressible surface for the closure of some connected component $K$ of $latex $M \backslash S_{i-1}$ 3.$M \backslash S_n$ is a union of 3-balls Sketch of proof: This is much simpler than it might appear to be. The point is (at least in my opinion), except for trivial cases as long as a manifold has boundary it must be Haken. Lemma: If $\partial M$ has a component that’s not $\mathbb{S}^2$ then $M$ is Haken. The proof of the lemma is merely that any such $M$ will have infinite $H_2$ hence by the sphere theorem it will contain an embedded surface with non-trivial homology, if such surface is compressible then we just cut along the boundary of the compressing disc and glue two copies of it. This does not change the homology. Hence at the end we will arrive at a non-trivial incompressible surface. Figure 3 Now back to proving of the theorem, so we start by setting $S_0$ to be an incompressible surface given by $M$ being Haken. Now since $M$ is irreducible, we cut along $S_0$, i.e. take the closure of each component (may have either one or two components) of $M\backslash S_0$. Those will have a non-spherical boundary component, hence by lemma containing homologically non-trivial incompressible surface. This process continuous as long as some pieces has non-spherical boundary components. But since $M$ is irreducible, any sphere bounds a 3-ball in $M$, hence all components with sphere boundary are 3-balls. (In particular, the case where a component have multiple sphere boundary components cannot occur since the first boundary component bounds a 3-ball hence it can’t have any non-trivial incompressible surfaces on both sides.) Now the only remaining piece is to show that this process terminates. We apply a standard ‘normal surface argument’ for this. Essentially if we fix a triangulation of $M$, A normal surface in $M$ is one that intersects each 3-simplex in a disjoint union of following two shapes: Figure 4 There can’t be infinitely many non-parallel disjoint normal surfaces in $M$ (in fact there can be no more than 6 times the number of 3-simplexes since each complementry component need to contain at least one non-I-bundle part from one 3-simplex). Figure 5 However, if the above process do not terminate, we would obtain a sequence of non-parallel non-spherical boundary components: Figure 6 They represent different homology classes hence can be represented by disjoint normal which results in a contradiction. In general, this gives a way to prove theorems about Haken manifolds by using inductionL i.e. one may hope to just show the property trivially holds for 3-balls and is invariant under gluing two pieced along an incompressible surface. Note that the gluing surface being incompressible is in fact quite strong hence making the induction step possible in many cases. For example, by applying an incredible amount of brilliant techniques, Thurston was able to prove his revolutionary result: Hyperbolization theorem for Haken manifolds: Any Haken manifold $M$ with tori boundary components that does not contain incompressible tori admits a complete hyperbolic structure of finite volume in its interior. In other words, this is saying that given a Haken manifold, we cut along any incompressible tori, the resulting manifold with tori boundary must have a complete hyperbolic structure with cusps near each boundary component, This is the best we could hope for since manifolds with incompressible tori would have their fundamental group split over $Z^2$ which of course imply they can’t be hyperbolic. Now the more manifolds being Haken means the better this theorem is. Many evidences show that in fact a lot of manifolds are indeed Haken, in perticular we have: Virtual Haken Conjecture: $M$ is finitely covered by a Haken manifold as long as $\pi_1(M)$ is infinite. We can see that together with Thurston’s hyperbolization theorem, this would give full solution to the geometrization conjecture for general 3-manifolds. However, although now Perelman has proved the geometrization conjecture, the virtual Haken conjecture remains open. But in light of Perelman’s result now we are able to try to ‘back-solve’ the puzzle and only prove the virtual Haken conjecture for hyperbolic manifolds. (to be continued) ### A report from the Workshop in Geometric Topology @ Utah (part 1) May 29, 2011 I went to Park City this passed week for the Workshop in Geometric Topology. It was a quite cool place filled with ski-equipment stores, Christmas souvenir shops, galleries and little wooden houses for family winter vacations. Well, as you may have guessed, the place would look very interesting in summer. :-P As the ‘principal speaker’, Professor Gabai gave three consecutive lectures on his ending lamination space paper (this paper was also mentioned in my last post). I would like to sketch some little pieces of ideas presented in perhaps couple of posts. Classification of simple closed curves on surfaces Let $S_{g,p}$ denote the (hyperbolic) surface of genus $g$ and $p$ punchers. There is a unique geodesic loop in each homotopy class. However, given a geodesic loop drew on the surface, how would you describe it to a friend over telephone? Here we wish to find a canonical way to describe homotopy classes of curves on surfaces. This classical result was originally due to Dehn (unpublished), but discovered independently by Thurston in 1976. For simplicity let’s assume for now that $S$ is a closed surface of genus $g$. Fix pants decomposition $\mathcal{T}$ of $S$, $\mathcal{T} = \{ \tau_1, \tau_2, \cdots, \tau_{3g-3} \}$ is a disjoint union of $3g-3$ ‘cuffs’. As we can see, any simple closed curve will have an (homology) intersection number with each of the cuffs. Those numbers are non-negative integers: Around each cuff we may assign an integer twist number, for a cuff with intersection number $n$ and twist number $z$, we ‘twist’ the curve inside a little neighborhood of the cuff so that all transversal segments to the cuff will have $z$ intersections with the curve. Negative twists merely corresponds to twisting in the other direction: Theorem: Every simple closed curve is uniquely defined by its intersection number and twisting number w.r.t each of the cuffs. Conversely, if we consider multi-curves (disjoint union of finitely many simple closed curves) then any element in $\mathbb{Z}^{3g-3} \times \mathbb{Z}_{\geq 0}^{3g-3}$ describes a unique multi-curve. To see this we first assume that the pants decomposition comes with a canonical ‘untwisted’ curve connecting each pairs of cuffs in each pants. (i.e. there is no god given ‘0’ twist curves, hence we have to fix which ones to start with.) In the example above our curve was homotopic to the curve $((1,2), (2,1), (1,-4))$. In other words, pants decompositions (together with the associated 0-twist arcs) give a natural coordinate chart to the set of homotopy class of (multi) curves on a surface. i.e. they are perimetrized by $\mathbb{Z}^{3g-3} \times \mathbb{Z}_{\geq 0}^{3g-3}$. For the converse, we see that any triple of integers can be realized by filling the pants with a unique set of untwisted arcs: In fact, this kind of parametrization can be generalized from integers to real numbers, in which case we have measured laminations instead of multi-curves and maximal train trucks on each pants instead of canonical untwisted arcs. i.e. Theorem: (Thurston) The space of measured laminations $\mathcal{ML}(S)$ on a surface $S$ of genus $g$ is parametrized by $\mathbb{R}^{3g-3} \times \mathbb{R}_{\geq 0}^{3g-3}$. Furthermore, the correspondence is a homeomorphism. Here the intersection numbers with the cuffs are wrights of the branches of the train track, hence it can be any non-negative real number. The twisting number is now defined on a continuous family of arcs, hence can be any real number, as shown below: As we can see, just as in the case of multi-curves, any triple of real numbers assigned to the cuffs can be realized as the weights of branches of a train track on the pants. ### Gromov boundary of hyperbolic groups May 16, 2011 As we have seen in pervious posts, the Cayley graphs of groups equipped with the word metric is a very special class of geodesic metric space – they are graphs that have tons of symmetries. Because of that symmetry, we can’t construct groups with any kind of Gromov boundary we want. In fact, there are only few possibilities and they look funny. In this post I want to introduce a result of Misha Kapovich and Bruce Kleiner that says: Let $G$ be a hyperbolic group that’s not a semidirect product $H \ltimes N$ where $N$ is finite or virtually cyclic. (In those cases the boundary of $G$ can be obtained from the boundary of $H$$).

Theorem: When $G$ has $1$-dimensional boundary, then the boundary is homeomorphic to either a Sierpinski carpet, a Menger curve or $S^1$.

OK. So what are those spaces? (don’t worry, I had no clue about what a ‘Menger curve’ is before reading this paper).

The Sierpinski carpet

(I believe most people have seen this one)

Start with the unit square, divide it into nine equal smaller squares, delete the middle one.

Repeat the process to the eight remaining squares.

and repeat…

Of course we then take the infinite intersection to get a space with no interior.

Proposition: The Sierpinski carpet is (covering) 1-dimensional, connected, locally connected, has no local cut point (meaning we cannot make any open subset of it disconnected by removing a point).

Theorem: Any compact metrizable planar space satisfying the above property is a Sierpinski carpet.

The Menger curve

Now we go to $\mathbb{R}^3$, the Menger curve is the intersection of the Sierpinski carpet times the unit interval, one in each of the $x, y, z$ direction.

Equivalently, we may take the unit cube $[0,1]^3$, subtract the following seven smaller cubes in the middle:

In the next stage, we delete the middle ‘cross’ from each of the remaining 20 cubes:

Proceed, take intersection.

Proposition: The Menger curve is 1-dimensional, connected, locally connected, has no local cut point.

Note this is one dimensional because we can decompose the ‘curve’ to pieces of arbitrary small diameter by cutting along thin rectangular tubes, meaning if we take those pieces and slightly thicken them there is no triple intersections.

Theorem: Any compact metrizable nowhere planar (meaning no open set of it can be embedded in the plane) space satisfying the above property is a Menger curve.

Now we look at our theorem, infact the proof is merely a translation from the conditions on the group to topological properties of the boundary and then seeing the boundary as a topological space satisfies our universal properties.

A group being Gromov hyperbolic implies the boundary is compact metrizable.

No splitting over finite or virtually cyclic group implies the boundary is connected, locally connected and if it’s not $S^1$, then it has no local cut point.

Now what remains is to show, for groups, if the boundary is not planar then it’s nowhere planar. This is an easy argument using the fact that the group acts minimally on the boundary.

Please refer to first part of their paper for details and full proof of the theorem.

Remark: When study classical Polish-school topology, I never understood how on earth would one need all those universal properties (i.e. any xxx space is a xxx, usually comes with a long condition include ten or so items >.<). Now I see in fact such thing can be powerful. i.e. sometimes this allows us to actually get a grib on what does some completely unimaginable spaces actually look like!

Another wonderful example of this is the recent work of S. Hensel and P. Przytycki and the even more recent work of David Gabai which shows ending lamination spaces are Nobeling curves.

### Stabilization of Heegaard splittings

May 9, 2011

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 $ and area $ 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.

### A report of my Princeton generals exam

April 26, 2011

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 $50$ $0$s 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…

February 14, 2011

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

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.

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.

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

### 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.