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.
This has been a long Monday, toward the end of which I confused knot width with its bridge number. This post is meant to remind me which is which. Here is a trefoil knot:
The crossing number is the smallest number of self-intersections that the knot can have when thrown on a plane. For the trefoil cr=3.
Here is the same knot on which points of locally maximal height are marked with red, and points of locally minimal height are in blue. There are two extrema of each kind.
The bridge number of a knot is the smallest number of maxima in any of its diagrams. The trefoil has b=2. A knot is in the bridge position if its diagram achieves b, and in addition all the maxima are above all the minima. This allows one to neatly cut the knot by its bridge sphere (in orange), leaving b untangled arcs hanging from either side.
The third invariant also has to do with the extrema of the height function. Now we pick a regular value in each interval between each pair of consecutive critical values, and count the multiplicity of said value. The sum of all these multiplicities is the width of the knot. For the trefoil w=8.
The knot is in a thin position when it attains w, as on the diagram above. Despite their apparent similarity, two different positions of the knot may be required to attain b and w. Also, the invariants behave in different ways under connected sums. Here is a connected sum of two knots, with trefoil on the left:
It’s obvious that the crossing number is subadditive: . It remains an open problem whether it’s in fact additive, i.e., whether the equality always holds.
The bridge number is the best behaved of the three: it’s known that (one-sided inequality is easy: connected sum can kill the absolute maximum on the lower-placed knot). Horst Schubert proved this equality in 1954. Much more recently (in 2004) Jennifer Schultens used modern machinery to give a 6-page proof of Schubert’s theorem.
The width turns out to misbehave. By putting one of the summands way below the other, one sees at once that . In many examples this is in fact an equality. However, very recently (2011) Blair and Tomova proved that strict inequality may hold; moreover, the Scharlemann-Schultens lower bound is attained for infinitely many pairs of knots.
If you ask a random passerby to give you an orthonormal basis for , they will probably respond with , . There is a lot to like about this exponential basis: most importantly, it diagonalizes the operator: . This property makes the exponential basis indispensable in the studies of differential equations. However, I prefer to describe the Walsh basis, which has several advantages:
the basis functions take just two values , which simplifies the computation of coefficients
the proof of the basis property is easier than for the exponential basis
there is a strong connection to probability: the Walsh expansion can be seen as conditional expectation, and the partial sums form a Doob martingale
partial sums converge a.e. for any function, which is not the case for the exponential basis.
First, introduce the Rademacher functions , (The enumeration is slightly different from what I used in class.) These are :
Alternatively, one can define as the function which takes the values alternatively on the dyadic intervals .
To define the th Walsh function , express the index as the sum of powers of 2, i.e., and let . For example, because . Since the binary representation is unique, the definition makes sense. We also have because the product of an empty set of numbers is 1.
In class I checked that the set is orthonormal. Also, for any integer the linear span of is the space of all functions that are constant on the dyadic intervals of length . This follows by observing that and that the dimension of is .
To prove that the Walsh basis is indeed a basis, suppose that is orthogonal to all . Since for all , the integral of over any dyadic interval is zero (note that the characteristic function of any dyadic interval belongs to some ). But any subinterval can be written as a disjoint countable union of dyadic intervals: just take all dyadic intervals that are contained in . (You don't necessarily get the right type of endpoints, but as long as we work with integrals, the difference between open and closed intervals is immaterial.) Thus, the integral of over any subinterval of is zero. By the Lebesgue differentiation theorem, for a.e. we have . Thus as required.
The proof is even simpler if we use the non-centered form of the Lebesgue differentiation theorem: for a.e. the average approaches as in such a way that . Armed with this theorem, we can consider the sequence of dyadic intervals containing , and immediately obtain a.e.
Having proved that is a basis, let’s expand something in it. For example, this moderately ugly function :
I used Maple to compute the coefficients and plotted the partial sums for :
Such partials sums (those that use basis functions) are particularly nice: they are obtained simply by averaging over each dyadic interval of length . In probability theory this is known as conditional expectation. The conditional expectation is a contraction in any space, including which gives so much trouble to the exponential basis. The highly oscillatory parts of are killed by the dyadic averaging; in contrast, when integrated against the exponentials, they may cleverly accumulate and destroy the convergence of partial sums.
In my previous post the 14 possible triangulations of hexagons were arranged by Whitehead moves. But if asked, most people would probably group them in this way:
Well, some may decide to put the S- and Z-shapes together since they are mirror images of each other. But I won’t. It’s easy to rotate a piece of paper with a triangulation on it; it’s not nearly as easy to fetch a mirror and study triangulations through it. So let it be recorded that the cyclic group acts on triangulations by rotating the hexagon. The orbits have lengths 6,3,3,2.
In the previous post I also mentioned that the number of triangulations of an (n+2)-gon is the Catalan number . With n=4 we get
Amazingly, the formula (*) does not just count triangulations: it also knows how they behave under rotations of the hexagons. But it will only tell us if we don’t rush to its evaluation. Instead, replace each number on the right of (*) by its q-analog
The result is
The fact that we get a polynomial can be explained: since the Catalan number is an integer, every prime divisor of the denominator in (*) also appears in the numerator (to equal or higher power), and this can be translated into the order of vanishing of polynomials in (**) at the roots of unity. It’s not as easy to explain why the coefficients end up nonnegative.
Now, both the cyclic group and the roots of unity were mentioned. Let’s bring them together by introducing which generates by multiplication. The cyclic sieving phenomenon (in this special case) is the following: is the number of triangulations fixed by the action of . Indeed, evaluating at we get . So, no triangulations are invariant under rotation by , two are invariant under rotation by , etc. It’s easy to read the entire orbit structure from this:
there are no fixed points
2 elements of order 2 form a single orbit of size 2
6 elements of order 3 form two orbits of size 3
there are no elements of order 4, since their number is
there are elements of order 6, which form a single orbit of size 6
And it’s not just triangulations. Consider noncrossing matchings of 2n points — these can be observed experimentally by sitting 2n people at a round table and asking them all to shake hands with someone with no arms crossing.
The number of such matchings is also the Catalan number (hence 14 matchings for 8 people). But the orbit structure is completely different! How can the same polynomial work here? It can because we now act by and therefore take . Evaluation of at the powers of yields . As above, we see the orbit structure in these numbers:
2 elements of order 2 form a single orbit
4 elements of order 4 form a single orbit
8 elements of order 8 form a single orbit
And it’s not just the Catalan numbers. Suppose that among those 8 guests at the round table are 2 men and 6 women. The number of seating arrangements is . Don’t evaluate to 28 just yet. Replace each integer with its q-analog and you’ll get the polynomial , which will tell you that under rotation, 4 arrangements form an orbit of size 4, while the other 24 form 3 orbits of size 8.
And there is much more. The cyclic sieving phenomenon was established recently (in 2004) by three mathematicians from Minnesota (Reiner, Stanton, and White), and there are still paths waiting to be explored.