This has got me stumped

In what sense are these dual? Recall that reversal of arrows in a “categorical” definition of an immetry did not produce a submetry. Let’s try another approach. Let our metric spaces be pointed: they all contain the point which is fixed by all maps under consideration. The Lipschitz dual of a metric space consists of all Lipschitz maps . This is naturally a vector space. Moreover, it is a Banach space with the norm . A Lipschitz map induces a bounded linear operator . If is 1-Lipschitz, then so is (i.e., its operator norm is at most 1).

It would be nice to have the following:

• is an immetry iff is a submetry
• is a submetry iff is an immetry

but that’s too much to hope for. For example, the inclusion of into is not a submetry, but it induces the identity map . Let’s go through these one by one.

Suppose is an immetry. Then we think of as a subset of , and is simply the restriction operator. To prove that it’s a submetry, we should verify the 2-point lifting property: given any and any that extends , we must find that extends and satisfies . This is easy: extend in a norm-preserving way (by McShane-Whitney) and add .

I also wrote down an (easy) proof that being a submetry implies that is an immetry, but WP ate it. Specifically, having pressed “Save Draft”, I was asked to re-login (the cookie expired). Having done so, I was presented with a 10 min old draft.

We already know that being an immetry does not imply that is a submetry. Whether being an submetry implies that is an immetry is left as an exercise for the reader.

Conversely, suppose is not compact. Pick an orthonormal basis and let be as above. The non-compactness of implies that there exists such that for all (because has finite rank). Let .

Pick a unit vector such that . Since is a basis, for any vector we have as . Thus, there exists such that . Let be the vector normalized and note that .

Continuing in this way, we form a sequence of unit vectors such that and lies in the range of projection (this is a projection because all commute). These ranges are mutually orthogonal, because for any we have , which implies and therefore for all . Thus, is the desired sequence.

In some sense (formal or informal) the following classes of maps are dual to each other.

• Injective : Surjective
• Immersion : Submersion
• Monomorphism : Epimorphism

An isometric embedding of metric space into a metric space is a map such that for all . This concept belongs to the left column of the table above. What should be its counterpart?

Candidate 1. Observe that a 1-Lipschitz map is an isometric embedding iff it does not factor through any 1-Lipschitz surjection (for any space ) unless is an isometric isomorphism. Reversing the order of arrows, we arrive at the following concept:

A 1-Lipschitz map is a metric quotient map if it does not factor through any 1-Lipschitz injection (for any space ) unless is an isometric isomorphism.

This can be reformulated as follows: is the greatest pseudo-metric on subject to

1. 
2. if

This seems reasonable: does as little damage as possible, given the structure of its fibers. There is also a natural way to construct for any reasonable fibering of : begin by defining for points in different fibers and otherwise. Then force the triangle inequality by letting subject to and . As long as the fibers stay at positive distance from one another, this will be a metric. The corresponding metric quotient map sends each point of onto its fiber.

Here is a simple example, in which a part of an interval is mapped to a point.

These aren't the quotients I'm looking for

However, the above example made me unhappy. The only nontrivial fiber is the interval . Both points and belong to trivial fibers, but the distance between them decreases from 3 to 2. This looks like a wrong kind of quotient to me.

Candidate 2 already appeared in my post on Lipschitz quotient, but wasn’t recognized at the time. It could be called -Lipschitz quotient, but a better name is available. A map is a submetry if where the balls are closed (using open balls yields something almost equivalent, but generally weaker). Such need not be an isometry: consider orthogonal projections in a Euclidean space. It does have to be 1-Lipschitz. The additional property that distinguishes it from general 1-Lipschitz maps is the 2-point lifting property: for every and every there is such that . Incidentally, this shows that is an isometric embedding of into the hyperspace which I covered earlier (“This ain’t like dusting crops, boy“).

The concept and the name were introduced by В.Н. Берестовский (V.N. Berestovskii) in his paper “Submetries of space-forms of nonnegative curvature” published in 1987 in Siberian Math. J. Among other things, he proved that a submetry between spheres (of any dimension) is an isometry. Of course, there are many submetries of onto other spaces: take the quotient by a subgroup of , which can be either discrete or continuous. Are there any submetries that are not quotients by isometries?

Yes, there are. I’ll describe a (modified) example given by Berestovskii and Guijarro (2000). Let be the hyperbolic plane realized as the upper half-plane with the metric . Define by

Don’t panic; this is just the signed distance function (in the metric of ) to the fat green curve below. I also drew two other level sets of , to the best of my Friday night line-drawing ability.

Weird submetry

To convince yourself that is a submetry, first consider for which the submetry property is clear (it’s the quotient by horizontal translation), and then note that the inversion in the unit circle exchanges horizontal lines (horocycles at infinity) with horocycles at 0. An interesting feature of this submetry that it is not very smooth: but not .

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:

1. They do not want to die.
2. They want to maximize their own profit.
3. They like to kill people, including crewmates.
4. They prefer to be on good terms with the higher ranked crewmates.

The question is: what division do you, as the captain, suggest?

It is convenient to number the pirates in the reverse order of importance, so that the captain is th. This way, if the captain gets killed, the problem neatly reduces to the case . For Rule 4 does not come into effect, so the answers are identical to Colin’s, then they begin to diverge, but for large the answers are again the same. I summarized the solution in a Google spreadsheet, where the numbers give the distribution of booty, green background indicates who votes for the decision, and black spots are self-explanatory.

The captain survives if and only if one of the following holds:

• , trivially.
• . He gets to keep PoE, except in the cases and when he can do better, keeping and , respectively.
• for . Except for , the captain gets nothing, and is only able to survive because of the support of those who are sure to walk the plank after him. Indeed, he can bribe pirates and still needs votes from the rest of the crew. These votes come from those who see a black spot hanging over them.

The distribution of looty for looks a little messy, but eventually it becomes distributed between the pirates that are not directly threatened by a black spot.

1. v is removed and each oriented path of length two through v is contracted into an edge. That is, the stopover at v is eliminated.
2. Step 1 may create some 2-edge oriented cycles, which must be deleted. That is, we cancel the pairs of arrows going in opposite directions.
3. The replacement vertex v’ is inserted, connected to the rest of the graph in the same way that v was, but with opposite orientation. In practice, one simply reuses v for this purpose.

Some quivers have a finite set of mutation equivalent ones; others an infinite one. Perhaps the simplest nontrivial case is the oriented 3-cycle with edges of multiplicities . The finiteness of its equivalents has to do with the Markov constant (not this time), which is invariant under mutation. This is investigated in the paper “Cluster-Cyclic Quivers with Three Vertices and the Markov Equation” by Beineke, Brűstle and Hille. The appendix by Kerner relates the Markov constant to Hochschild cohomology, which I take as a clue for me to finish this post.

So I’ll leave you to play with the mutation applet linked in the embedded tweet below.

Tired of Sudoku? Try quiver mutation puzzles math.jussieu.fr/~keller/quiver… and maybe win €20. Mutation rules & more puzzles at mutatingquivers.com—
Syracuse Mathematics (@SUmathematics) February 14, 2012

This equation is symmetric with respect to unknown terms , , ; therefore, knowing one of its solutions

it is easy to find the following five:

Although these six solutions may be different, we will consider them as one, denoted by

“

The above is a quote from A.A.Markov’s paper “Sur les formes quadratiques binaires indéfinies” (part 2). Back in 1880 people were patient enough to write out all permutations of three symbols… If we fix and , the equation for becomes

which admits the second integer root . We can also write which does not immediately tell that is an integer, but it does tell us that is positive. For example, from the obvious solution we get . Now it would not do us any good to mutate in the third variable again, for it would bring us back to . But we can mutate in the second variable, replacing it with . Having understood so well the symmetry of the equation, we write this new solution as , in the increasing order. Now we can mutate either 1 or 2, and so it goes…

Markov number tree, from http://en.wikipedia.org/wiki/File:MarkoffNumberTree.png

All triples, except for the two at the beginning, consist of distinct numbers (thus, they do generate six distinct solutions if order matters). The tree contains all solutions of the Markov equation. The Wikipedia article also points out the occurrence of Fibonacci numbers along the top branch, as well as a curious identity discovered by Don Zagier: let ; then (for triples written in increasing order)

Looks like a fun problem on simplification of inverse hyperbolic trigonometric functions.

Also, it’s still unknown whether two distinct Markov triples can have the same maximum . Looks like a fun problem for amateur number theorists.

To wrap this up, I will describe how Markov (the one of Markov chains fame, not his identically-named son of 4-manifold undecidability fame) came across the equation. Let be a quadratic form with real coefficients normalized by . What is the best upper bound on

In 1873 Korkin and Zolotarev published a paper showing that the best bound is , attained by the form . Looks like the case is closed. But what if is not (precisely, not equivalent to under the action of )? Then the best bound improves to , attained by (this is also due to KZ). Well, what if is not equivalent to either or ? Then the bound improves to (found by Markov), and we could go on…

But rather than continue in this fashion, Markov looked for the threshold at which the number of inequivalent forms with minimum becomes infinite. And he found it: (for comparison, ). Specifically, there are only finitely many forms with minimum above , for every . But there are infinitely many forms with minimum exactly , such as . It was the iterative process of getting more and more of these forms that led Markov to the Diophantine equation .

The number and its square also appear in Zagier’s identity with … But enough numerology for today.

Building block

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.

Laakso graph (click to magnify)

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:

Slit pseudo-carpet: beginning

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:

Slit pseudo-carpet (click to magnify)

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). Warning: copying and pasting from WP can make the single quotes ‘ unusable in scilab. Same goes for line breaks.

function laakso(angle,steps)
s=1/(2+2*cos(angle));
xoffset = [0,1-s,s,s,1/2,1/2];
yoffset = [0,0,0,0,s*sin(angle),-s*sin(angle)];
rotation =[0,0,angle,-angle,-angle,angle];
sc=s*cos(rotation);
ss=s*sin(rotation);
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))];
end
plot(point(:,1),point(:,2),’linestyle’,'none’,'markstyle’,’.’,'marksize’,1);
endfunction

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.

• Matthew Badger
• Laurent Bartholdi
• Heinz H. Bauschke
• Jonathan Borwein
• Marius Buliga
• Graziano Crasta
• David Damanik
• Danny Calegari
• Djalil Chafaï
• David Drasin
• David Eppstein
• Tamas Erdelyi
• Alexandre Eremenko
• Nassif Ghoussoub
• Martin Hairer
• Hakån Hedelman
• Piotr Indyk
• Robert Jerrard
• Palle Jørgensen
• Michael Jury
• David Kalaj
• Peter Kuchment
• Russell Lyons
• Juan Manfredi
• Donald Marshall
• Gaven Martin
• John McCarthy
• Manor Mendel
• Giuseppe Rosario Mingione
• Enzo Mitidieri
• Massimiliano Morini
• Frederic Mynard
• Piotr Nowak
• Jani Onninen
• Dmitry Panchenko
• Yuval Peres
• Pietro Poggi-Corradini
• Andrew Putman
• Luis Silvestre
• Ilya Spitkowski
• Jan Vybiral
• Xianfu Wang
• JD Ward
• Ruhan Zhao
• Joel Zinn
• Maciej Zworski

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.

Imagine that

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.