A slightly modified version of the riddle discussed in detail by Colin Carroll. I recap the key parts of his description below. Rule 4 was added by me in order to eliminate any probabilistic issues from the problem.
You are the captain of a pirate ship with a crew of N people ordered by rank. Your crew just managed to plunder 100 Pieces of Eight. Now you are to propose a division of the 100 PoE, and the crew will vote on the division. The captain doesn’t vote except to break a tie. If the proposal fails, the captain walks the plank, and the first mate becomes captain, the third in command becomes first mate, and so on. Each pirate votes according to the following ordered priorities:
They do not want to die.
They want to maximize their own profit.
They like to kill people, including crewmates.
They prefer to be on good terms with the higher ranked crewmates.
The question is: what division do you, as the captain, suggest?
In early 2000s Tomi Laakso published two papers which demonstrated that metric spaces could behave in very Euclidean ways without looking Euclidean at all. One of his examples became known as the Laakso graph space, since it is the limit of a sequence of graphs. In fact, the best known version of the construction was introduced by Lang and Plaut, who modified Laakso’s original example to make it more symmetric. The building block is this graph with 6 edges:
Each edge is assigned length 1/4, so that the distance between the leftmost and rightmost points is 1. Next, replace each edge with a copy of the building block. This increases the number of edges by the factor of 6; their length goes down by the factor of 4 (so that the leftmost-rightmost distance is still 1). Repeat.
The resulting metric space is doubling: every ball can be covered by a fixed number (namely, 6) balls of half the size. This is the typical behavior of subsets of Euclidean spaces. Yet, the Laakso space does not admit a bi-Lipschitz embedding into any Euclidean space (in fact, even into any uniformly convex Banach space). It remains the simplest known example of a doubling space without such an embedding. Looking back at the building block, one recognizes the cycle as the source of non-embeddability (a single cycle forces a certain amount of distortion; adding cycles withing cycles ad infinitum forces infinite distortion). The extra edges on the left and right are necessary for the doubling condition.
In some sense, the Laakso space is the antipode of the Cantor set: instead of deleting the middle part repeatedly, we duplicate it. But it’s also possible to construct it with a ‘removal’ process very similar to Cantor’s. Begin with the square and slit it horizontally in the center; let the length of the slit be 1/3 of the sidelength. Then repeat as with the Cantor set, except in addition to cutting left and right of the previous cut, we also do it up and down. Like this:
Our metric space if the square minus the slits, equipped with the path metric: the distance between two points is the infimum of the length of curves connecting them within the space. Thus, the slits seriously affect the metric. This is how the set will look after a few more iterations:
I called this a slit pseudo-carpet because it has nonempty interior, unlike true Sierpinski carpets. To better see the similarity with the Laakso space, multiply the vertical coordinate by and let (equivalently, redefine the length of curves as instead of ). This collapses all vertical segments remaining in the set, leaving us with a version of the Laakso graph space.
Finally, some Scilab code. The Laakso graph was plotted using the Chaos game, calling the function below with parameters laakso(0.7,50000).
xoffset = [0,1-s,s,s,1/2,1/2];
yoffset = [0,0,0,0,s*sin(angle),-s*sin(angle)];
point = zeros(steps,2);
vert = grand(1,steps,'uin',1,6);
for j = 2:steps
point(j,:) = point(j-1,:)*[sc(vert(j)),ss(vert(j));-ss(vert(j)),sc(vert(j))] + [xoffset(vert(j)), yoffset(vert(j))];
I have 111 MathSciNet reviews posted, and there are three more articles on my desk that I should be reviewing instead of blogging. Even though I think of canceling my AMS membership, I don’t mind helping the society pay their bills (MathSciNet brings about 37% of the AMS revenue, according to their 2010-11 report.)
Sure, reviews need to be edited, especially when written by non-native English speakers like myself. Still, I’m unhappy with the edited version of my recent review:
This was the approach taken in the foundational paper by J. Heinonen et al. [J. Anal. Math. 85 (2001), 87-139]
The paper was written by J. Heinonen, P. Koskela, N. Shanmugalingam, and J. T. Tyson. Yes, it’s four names. Yes, the 14-letter name is not easy to pronounce without practice. But does the saving of 45 bytes justify omitting the names of people who spent many months, if not years, working on the paper? Absolutely not. The tradition of using “et al” for papers with more than 3 authors belongs to the age of typewriters.
P.S. I don’t think MathSciNet editors read my blog, so I emailed them.
P.P.S. The names are now restored. In the future I’ll be sure to add in “comments to the editor” that names should not be replaced by et al.
The hyperspace is a set of sets equipped with a metric or at least with a topology. Given a metric space , let be the set of all nonempty closed subsets of with the Hausdorff metric: if no matter where you are in one set, you can jump into the other by traveling less than . So, the distance between letters S and U is the length of the longer green arrow.
The requirement of closedness ensures for . If is unbounded, then will be infinite for some pairs of sets, which is natural: the hyperspace contains infinitely many parallel universes which do not interact, being at infinite distance from one another.
Every continuous surjection has an inverse defined in the obvious way: . Yay ambiguous notation! The subset of that consists of the singletons is naturally identified with , so for bijective maps we recover the usual inverse.
Exercise: what conditions on guarantee that is (a) continuous; (b) Lipschitz? After the previous post it should not be surprising that
Even if is open and continuous, may be discontinuous.
If is a Lipschitz quotient, then is Lipschitz.
Proofs are not like dusting crops—they are easier.
A map is continuous if the preimage of every open set is open. If the topology is defined by a metric, we can reformulate this as: the inverse image of an open ball contains an open ball . Like this:
But bringing these radii and into the picture will not serve any purpose unless we use them to quantify continuity. For example, if we insist that for a fixed constant , we arrive at the definition of a Lipschitz map.
But why do we look at the inverse image; what happens if we take the direct image instead? Then we get the definition of an open map: the image of every open set is open. Recasting this in metric terms: the image of an open ball contains an open ball . Like this:
If we quantify openness by requiring for a fixed , we arrive at the definition of a co-Lipschitz map. [Caution: some people use “co-Lipschitz” to mean , which is a different condition. They coincide if is bijective.]
I don’t know if openness without continuity is good for anything other than torturing students with exercises such as: “Construct an open discontinuous map from to .” We probably want both. At first one can hope that open continuous maps will have reasonable fibers : something -dimensional when going from dimensions to , with . The hope is futile: an open continuous map can squeeze a line segment to a point (construction left as an exercise).
A map that is both Lipschitz and co-Lipschitz is called a Lipschitz quotient; this is a quantitative analog of “open continuous”. It turns out that for any Lipschitz quotient the preimage of every point is a finite set. Moreover, factors as where is a complex polynomial and is a homeomorphism.
This is encouraging… but going even one dimension higher, it remains unknown whether a Lipschitz quotient must have discrete fibers. For an overview of the subject, see Bill Johnson’s slides.
Takagi (高木) curves are fractals that are somehow less known than Cantor sets and Sierpinski carpets, yet they can also be useful as (counter-)examples. The general 高木 curve is the graph of a function that is built from triangular waves. The th generation wave has equation where means the distance to the nearest integer. Six of these waves are pictured below.
Summation over creates the standard 高木 curve , also known as the blancmange curve:
Note the prominent cusps at dyadic rationals: more on this later.
General 高木 curves are obtained by attaching coefficients to the terms of the above series. The simplest of these, and the one of most interest to me, is the alternating 高木 curve :
The alternation of signs destroys the cusps that are so prominent in . Quantitatively speaking, the diameter of any subarc of is bounded by the distance between its endpoints times a fixed constant. The curves with this property are called quasiarcs, and they are precisely the quasiconformal images of line segments.
Both and have infinite length. More precisely, the length of the th generation of either curve is between and . Indeed, the derivative of is just the Rademacher function . Therefore, the total variation of the sum is the norm of . With the sharp form of the Хинчин inequality from the previous post yields
For the upper bound I added 1 to account for the horizontal direction. Of course, the bound of real interest is the lower one, which proves unrectifiability. So far, a construction involving these curves shed a tiny bit of light on the following questions:
Which sets have the property that any quasiconformal image of contains a rectifiable curve?
I won’t go (yet) into the reasons why this question arose. Any set with nonempty interior has the above property, since quasiconformal maps are homeomorphisms. A countable union of lines in the plane does not; this is what 高木 curves helped to show. The wide gap between these results remains to be filled.