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.

Today’s technology should make it possible to use the native transcription of names like Хинчин without inventing numerous ugly transliterations. The inequality is extremely useful in both analysis and probability: it says that the norm of a linear combination of Rademacher functions (see my post on the Walsh basis) can be computed from its coefficients, up to a multiplicative error that depends only on . Best of all, this works even for the troublesome ; in fact for all . Formally stated, the inequality is

where the constants depend only on . The orthogonality of Rademacher functions tells us that , but what are the other constants? They were not found until almost 60 years after the inequality was proved. The precise values, established by Haagerup in 1982, behave in a somewhat unexpected way. Actually, only does. The upper bound is reasonably simple:

The lower bound takes an unexpected turn:

The value of is determined by the continuity of , and is not far from : precisely, . Looks like a bug in the design of the Universe.

For a concrete example, I took random coefficients and formed the linear combination shown above. Then computed its norm and the bounds in the Khintchine inequality. The norm is in red, the lower bound is green, the upper bound is yellow.

Take a finite set of vectors in some vector space. Call a set independent if the corresponding vectors are linearly independent. The collection of all independent sets is denoted ; what properties must it have? Well, it must be a matroid:

A matroid with ground set is a nonempty collection of subsets which is both hereditary and augmentable.

Hereditary: if , then any subset of is in

Augmentable: if and has fewer elements than , then can take an element from and still remain in .

The augmentation property brings to mind wealth redistribution (at least if you grew up in the USSR). Anyway, it is fairly obvious that any collection that arises from vectors is a matroid. Is the converse true? Not obvious at all.

Behold the Fano plane, the smallest projective plane in existence: 7 points, 7 lines, any two lines meet at one point, through any two points there is a unique line.

The Fano matroid has these 7 vertices as the ground set. A set of vertices is independent if it has at most three points and is not a line. The hereditary property is clear. The augmentation isn’t hard either: given a 2-point set , let be the line through it; since has 3 points and is not a line, it must contain a point outside of , and this is the point we add to .

Can the Fano matroid be realized by 7 vectors in a Euclidean space? Guess what… no.

Suppose we do have such a set of 7 vectors. Then they span a 3-dimensional space, since there is no linearly independent subset with 4 vectors. The three vertices of the triangle are independent, and we may assume they are realized by the standard basis vectors . Enumerate the other points/vectors so that is the midpoint opposite for . Observe that is a linear combination of , etc. Thus, are the columns of the matrix

But wait, there’s more! Denote and observe that the linear dependence of forces to be a scalar multiple of . Similarly for . So the matrix must be of the form

But wait, there’s even more! The vectors are also dependent (that inscribed circle is actually a line). The determinant of the 4,5,6 columns is . Can any of be zero? No, because that would make a linear combination of two of , which is not supposed to happen. Can any of be zero? No, because that would create a zero vector, and all of the 1- and 2-subsets are independent. So, the Fano matroid cannot be realized by vectors…

unless 2=0. Over there is no problem:

(P.S. All of this stuff is in Whitney’s 1935 paper that introduced matroids.)

Let’s admit it: it’s hard to keep track of signs when multiplying numbers. Being lazy people, mathematicians seek ways to avoid this chore. One popular way is to work in the enchanted world of , where . I’ll describe another way, which is to redefine multiplication by letting the factors reach a consensus on what the sign of their product should be.

If both and are positive, let their product be positive. And if they are both negative, the product should also be negative. Finally, if the factors can’t agree on which sign they like, they compromise at 0.

In a formula, this operation can be written as , but who wants to see that kind of formulas? Just try using it to check that the operation is associative (which it is).

But I hear someone complaining that is just an arbitrary operation that does not make any sense. So I’ll reformulate it. Represent real numbers by ordered pairs , for example becomes and becomes . Define multiplication component-wise. Better now? You don’t have to keep track of minus signs because there aren’t any.

This comes in handy when multiplying the adjancency matrices of quivers. The Wikipedia article on Quiver illustrates the concept with this picture:

But in mathematics, a quiver is a directed graph such as this one:

Recall that the adjancency matrix of a graph on vertices has if there is an edge between and , and otherwise. For a directed graph we modify this definition by letting if the arrow goes from to , and if it goes in the opposite direction. So, for the quiver shown above we get

For an undirected graph the square counts the number of ways to get from to in exactly 2 steps (and one can replace 2 by n). To make this work for the directed graph, we represent numbers as pairs and and carry on multiplying and adding:

For instance, there are two ways to get from 3 to 4 in two steps, but none in the opposite direction. This works for any powers, and also for multigraphs (with more than one edge between same vertices). Logically, this is the same as separating the adjacency matrix into its positive and negative parts, and multiplying them separately.

The last example is matrix mutation from the theory of cluster algebras. Given a (usually, integer) matrix and a positive integer , we can mutate in the direction by doing the following:

Dump nuclear waste on the th row and th column

To each non-radioactive element add , that is, the product of the radioactive elements to which is exposed.

Clean up by flipping the signs of all radioactive elements.

The properties of should make it clear that each mutation is an involution: mutating for the second time in the same direction recovers the original matrix. However, applying mutations in different directions, one can obtain a large, or even infinite, class of mutation-equivalent matrices.

This is an orthonormal basis for . Since the measure of is infinite, functions will have to decay at infinity in order to be in . The Hermite functions are

where is the nth Hermite polynomial, defined by
.
The goal is to prove that the functions can be obtained from via the Gram-Schmidt process. (They actually form a basis, but I won’t prove that.)

One can observe that the term would be unnecessary if we considered the weighted space with weight and the inner product . In this language, we orthogonalize the sequence of monomials and get the ON basis of polynomials with being a normalizing constant. But since weighted spaces were never introduced in class, I’ll proceed with the original formulation. First, an unnecessary graph of ; the order is red, green, yellow, blue, magenta.

Claim 1. is a polynomial of degree with the leading term . Proof by induction, starting with . Observe that

where the first term has degree and the second . So, their sum has degree exactly , and the leading coefficient is . Claim 1 is proved.

In particular, Claim 1 tells us that the span of the is the same as the span of .

Claim 2. for . We may assume . Must show . Since is a polynomial of degree , it suffices to prove

(*) for integers .

Rewrite (*) as and integrate by parts repeatedly, throwing the derivatives onto until the poor guy can't handle it anymore and dies. No boundary terms appear because decays superexponentially at infinity, easily beating any polynomial factors. Claim 2 is proved.

Combining Claim 1 and Claim 2, we see that belongs to the -dimensional space , and is orthogonal to the -dimensional subspace . Since the “Gram-Schmidtization'' of has the same properties, we conclude that agrees with this “Gram-Schmidtization'' up to a scalar factor.

It remains to prove that the scalar factor is unimodular ( since we are over reals).

Claim 3. for all . To this end we must show . Expand the first factor into monomials, use (*) to kill the degrees less than n, and recall Claim 1 to obtain
.
As in the proof of Claim 2, we integrate by parts throwing the derivatives onto . After n integrations the result is
, as claimed.

P.S. According to Wikipedia, these are the “physicists’ Hermite polynomials”. The “probabilists’ Hermite polynomials” are normalized to have the leading coefficient 1.