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http://dx.doi.org/10.4169/amer.math.monthly.119.01.003

January 2012] A LETTER FROM THE EDITOR 3

Invariant Histograms Daniel Brinkman and Peter J Olver

Abstract. We introduce and study a Euclidean-invariant distance histogram function for curves For a sufficiently regular plane curve, we prove that the cumulative distance histograms based on discretizing the curve by either uniformly spaced or randomly chosen sample points converge to our histogram function We argue that the histogram function serves as a simple, noise-resistant shape classifier for regular curves under the Euclidean group of rigid motions Extensions of the underlying ideas to higher-dimensional submanifolds, as well as to area his- togram functions invariant under the group of planar area-preserving affine transformations, are discussed.

1 INTRODUCTION. Given a finite set of points contained in Rn, equipped withthe usual Euclidean metric, consider the histogram formed by the mutual distancesbetween all distinct pairs of points An interesting question, first studied in depth by

Boutin and Kemper [4, 5], is to what extent the distance histogram uniquely determines

the point set Clearly, if the point set is subjected to a rigid motion—a combination oftranslations, rotations, and reflections—the interpoint distances will not change, and

so two rigidly equivalent finite point sets have identical distance histograms However,there do exist sets that have identical histograms but are not rigidly equivalent (Thereader new to the subject may enjoy trying to find an example before proceeding fur-ther.) Nevertheless, Boutin and Kemper proved that, in a wide range of situations, theset of such counterexamples is “small”—more precisely, it forms an algebraic sub-variety of lower dimension in the space of all point configurations Thus, one cansay that, generally, the distance histogram uniquely determines a finite point set up torigid equivalence This motivates the use of the distance histogram as a simple, robust,noise-resistant signature that can be used to distinguish most rigidly inequivalent fi-nite point sets, particularly those that arise as landmark points on an object in a digitalimage

The goal of this paper is to develop a comparable distance histogram function forcontinua—specifically curves, surfaces, and higher-dimensional submanifolds of Eu-clidean spaces Most of the paper, including all proofs, will concentrate on the simplestscenario: a “regular” bounded plane curve Regularity, as defined below, does allowcorners, and so, in particular, includes polygons We will approach this problem usingthe following strategy We first sample the curve using a finite number of points, andthen compute the distance histogram of the sampled point set Stated loosely, our mainresult is that, as the curve becomes more and more densely sampled, the appropriatelyscaled cumulative distance histograms converge to an explicit function that we name

the global curve distance histogram function Alternatively, computing the histogram

of distances from a fixed point on the curve to the sample points leads, in the limit, to

a local curve distance histogram function, from which the global version can be

ob-tained by averaging over the curve Convergence of both local and global histograms

is rigorously established, first for uniformly sampled points separated by a commonarc length distance, and then for points randomly sampled with respect to the uniformarc length distribution

http://dx.doi.org/10.4169/amer.math.monthly.119.01.004

MSC: Primary 53A04, Secondary 68U10

The global curve distance histogram function can be computed directly through anexplicit arc length integral By construction, it is invariant under rigid motions Hence,

a basic question arises: does the histogram function uniquely determine the curve up torigid motion? While there is ample evidence that, under suitably mild hypotheses, such

a result is true, we have been unable to establish a complete proof, and so must state it

as an open conjecture A proof would imply that the global curve histogram function,

as approximated by its sampled point histograms, can be unambiguously employed as

an elementary, readily computed classifier for distinguishing shapes in digital images,and thus serve as a much simpler alternative to the joint invariant signatures proposed

in [15] Extensions of these ideas to subsets of higher-dimensional Euclidean spaces,

or even general metric spaces, are immediate Moreover, convergence in sufficientlyregular situations can be established along the same lines as the planar curve casetreated here

Following Boutin and Kemper [4], we also consider area histograms formed by

triangles whose corners lie in a finite point set In two dimensions, area histogramsare invariant under the group of equi-affine (meaning area-preserving affine) transfor-mations We exhibit a limiting area histogram function for plane curves that is alsoequi-affine invariant, and propose a similar conjecture Generalizations to other trans-formation groups, e.g., similarity, projective, conformal, etc., of interest in image pro-

cessing and elsewhere [9, 16], are worth developing The corresponding discrete

his-tograms will be based on suitable joint invariants—for example, area and volume crossratios in the projective case—which can be systematically classified by the equivariant

method of moving frames [15] Analysis of the corresponding limiting histograms will

be pursued elsewhere

Our study of invariant histogram functions has been motivated in large part by thepotential applications to object recognition, shape classification, and geometric mod-eling Discrete histograms appear in a broad range of powerful image processing al-

gorithms: shape representation and classification [1, 23], image enhancement [21, 23], the scale-invariant feature transform (SIFT) [10, 18], object-based query methods [22], and as integral invariants [11, 19] They provide lower bounds for and hence estab-

lish stability of Gromov–Hausdorff and Gromov–Wasserstein distances, underlying

an emerging new approach to shape theory [12, 13] Local distance histograms derly the method of shape contexts [2] The method of shape distributions [17] for

un-distinguishing three-dimensional objects relies on a variety of invariant histograms,including local and global distance histograms, based on the fact that objects withdifferent Euclidean-invariant histograms cannot be rigidly equivalent; the converse,however, was not addressed Indeed, there are strong indications that the distance his-togram alone is insufficient to distinguish surfaces, although we do not know explicitexamples of rigidly inequivalent surfaces that have identical distance histograms

2 DISTANCE HISTOGRAMS. Let us first review the results of Boutin and

Kem-per [4, 5] on distance histograms defined by finite point sets For this purpose, our

ini-tial setting is a general metric space V , equipped with a distance function d(z, w) ≥ 0, for z, w ∈ V , satisfying the usual axioms.

Definition 1. The distance histogram of a finite set of points P = {z1, , z n} ⊂ Visthe functionη = ηP: R+→ N defined by

η(r) = #{(i, j) | 1 ≤ i < j ≤ n, d(z i , z j ) = r}. (2.1)

In this paper, we will restrict our attention to the simplest situation, when V =

Rm is endowed with the usual Euclidean metric, so d(z, w) = kz − wk We say that

January 2012] INVARIANT HISTOGRAMS 5

two subsets P, Q ⊂ V are rigidly equivalent, written P ' Q, if we can obtain Q by applying an isometry to P In Euclidean geometry, isometries are rigid motions: the

translations, rotations, and reflections generating the Euclidean group [25] Clearly,

any two rigidly equivalent finite subsets have identical distance histograms Boutinand Kemper’s main result is that the converse is, in general, false, but is true for abroad range of generic point configurations

Theorem 2 LetP(n)=P(n)(Rm ) denote the space of finite (unordered) subsets P ⊂

Rm of cardinality #P = n If n ≤ 3 or n ≥ m + 2, then there is a Zariski dense open

subsetR(n)⊂P(n) with the following property: if P ∈R(n) , then Q ∈P(n) has the same distance histograms,ηP =ηQ , if and only if the two point configurations are rigidly equivalent: P ' Q.

In other words, for the indicated ranges of n, unless the points are constrained by

a certain algebraic equation, and so are “nongeneric,” the distance histogram uniquelydetermines the point configuration up to a rigid motion Interestingly, the simplestcounterexample is not provided by the corners of a regular polygon For example,the corners of a unit square have 4 side distances of 1 and 2 diagonal distances of

√

2, and so its distance histogram has valuesη(1) = 4, η(√2) = 2, while η(r) = 0 for r 6= 1,√2 Moreover, this is the only possible way to arrange four points withthe given distance histogram A simple nongeneric configuration is provided by thecorners of the kite and trapezoid quadrilaterals shown in Figure 1 Although clearlynot rigidly equivalent, both point configurations have the same distance histogram,with nonzero valuesη(√2) = 2, η(2) = 1, η(√10) = 2, η(4) = 1 A striking one-

dimensional counterexample, discovered in [3], is provided by the two sets of

inte-gers P = {0 , 1, 4, 10, 12, 17}, Q = {0, 1, 8, 11, 13, 17} ⊂ R, which, as the reader can

check, have identical distance histograms, but are clearly not rigidly equivalent

√ 2

Figure 1. Kite and trapezoid.

To proceed, it will be more convenient to introduce the (renormalized) cumulative

3P (r) − 3 P (r − δ) for sufficiently smallδ  1 (2.3)

We further introduce a local distance histogram that counts the fraction of points in P that are within a specified distance r of a given point z ∈ R m:

λP (r, z) = 1

n # j | d(z, z j ) ≤ r = 1

n#(P ∩ Br (z)), (2.4)

B r (z) = v ∈ V | d(v, z) ≤ r (2.5)

denotes the ball (in the plane, the disk) of radius r centered at the point z Observe that

we recover the cumulative histogram (2.2) by averaging its localization:

In this paper, we are primarily interested in the case when the points lie on a curve

Until the final section, we restrict our attention to plane curves: C ⊂ V = R2 A finite

subset P ⊂ C will be called a set of sample points on the curve We will assume throughout that the curve C is bounded, rectifiable, and closed (Extending our results

to non-closed curves is straightforward, but we will concentrate on the closed case inorder to simplify the exposition.) Further mild regularity conditions will be introduced

below We use z(s) to denote the arc length parametrization of C, measured from some base point z(0) ∈ C Let

l (C) =I

C

denote the curve’s length, which we always assume to be finite

Our aim is to study the limiting behavior of the cumulative histograms constructedfrom more and more densely chosen sample points It turns out that, under reason-able assumptions, the discrete histograms converge, and the limiting function can beexplicitly characterized as follows

Definition 3. Given a curve C ⊂ V , the local curve distance histogram function based

at a point z ∈ V is

h C (r, z) = l (C ∩ B r (z))

i.e., the fraction of the total length of the curve that is contributed by those parts

con-tained within the disk of radius r centered at z The global curve distance histogram

function of C is obtained by averaging the local version over the curve:

Modulo the definition of “regular,” to be presented in the following section, anddetails on how “randomly chosen points” are selected, provided in Section 4, our mainconvergence result can be stated as follows

January 2012] INVARIANT HISTOGRAMS 7

Theorem 4 Let C be a regular plane curve Then, for both uniformly spaced and

randomly chosen sample points P ⊂ C , the cumulative local and global histograms converge to their continuous counterparts:

λP (r, z) −→ h C (r, z), 3P (r) −→ H C (r), (2.10)

as the number of sample points goes to infinity.

3 UNIFORMLY SPACED POINTS. Our proof of Theorem 4 begins by ing convergence of the local histograms In this section, we work under the assumptionthat the sample points are uniformly spaced with respect to arc length along the curve.Let us recall some basic terminology concerning plane curves, mostly taken from

establish-Guggenheimer’s book [8] We will assume throughout that C ⊂ R2has a piecewise C2

arc length parametrization z(s), where s belongs to a bounded closed interval [0, L], with L = l(C) < ∞ being its overall length The curve C is always assumed to be

simple , meaning that there are no self-intersections, and closed, so z(0) = z(L), and thus a Jordan curve We use t (s) = z0(s) to denote the unit tangent, and1κ(s) = z0(s) ∧

z00(s) the signed curvature at the point z(s) Under our assumptions, both t(s) and κ(s)

have left- and right-hand limiting values at their finitely many discontinuities A point

z (s) ∈ C where either the tangent or curvature is not continuous will be referred to

as a corner We will often split C up into a finite number of nonoverlapping curve

segments, with distinct endpoints

A closed curve is called convex if it bounds a convex region in the plane A curve segment is convex if the region bounded by it and the straight line segment connecting its endpoints is a convex region A curve segment is called a spiral arc if the curvature

functionκ(s) is continuous, strictly monotone,2and of one sign, i.e., eitherκ(s) ≥ 0

orκ(s) ≤ 0 Keep in mind that, by strict monotonicity, κ(s) is only allowed to vanish

at one of the endpoints of the spiral arc

Definition 5. A plane curve is called regular if it is piecewise C2and the union of afinite number of convex spiral arcs, circular arcs, and straight lines

Thus, any regular curve has only finitely many corners, finitely many inflection

points , where the curvature has an isolated zero, and finitely many vertices, meaning

points where the curvature has a local maximum or minimum, but is not locally stant In particular, polygons are regular, as are piecewise circular curves, also known

con-as biarcs [14] (But keep in mind that our terminological convention is that polygons

and biarcs have corners, not vertices!) Examples of irregular curves include the graph

of the infinitely oscillating function y = x5sin 1/x near x = 0, and the nonconvex

spiral arc r = e− θfor 0 ≤θ < ∞, expressed in polar coordinates

Theorem 6 If C is a regular plane curve, then there is a positive integer m C such that the curve’s intersection with any disk having center z ∈ C and radius r > 0, namely

C ∩ B r (z), consists of at most m C connected segments The minimal value of m C will

be called the circular index of C

1 The symbol ∧ denotes the two-dimensional cross product, which is the scalar v ∧ w = v 1 w 2 − v 2 w 1 for

v = (v 1 , v 2 ), w = (w 1 , w 2 ).

2Guggenheimer [8] only requires monotonicity, allowing spiral arcs to contain circular subarcs, which we

exclude Our subsequent definition of regularity includes curves containing finitely many circular arcs and straight line segments.

Proof. This is an immediate consequence of a theorem of Vogt ([24], but see also [8,

Exercise 3-3.11]) that states that a convex spiral arc and a circle intersect in at most 3

points Thus, m C ≤3 j + 2 k, where j is the number of convex spiral arcs, while k is the number of circular arcs and straight line segments needed to form C.

Example 7. Let C be a rectangle A disk B r (z) centered at a point z ∈ C will intersect

the rectangle in either one or two connected segments; see Figure 2 Thus, the circular

index of a rectangle is m C =2

Figure 2.Intersections of a rectangle and a disk.

For each positive integer n, let P n = {z1, , z n} ⊂C denote a collection of n

uniformly spaced sample points, separated by a common arc length spacing1l =

L /n.

Proposition 8 Let C be a regular curve Then, for any z ∈ C and r > 0, the

cor-responding cumulative local histograms based on uniformly spaced sample points

where m C is the circular index of C.

By assumption, since z ∈ C, the intersection C ∩ B r (z) = S1∪ · · · ∪S k consists

of k connected segments whose endpoints lie on the bounding circle S r (z), where

1 ≤ k ≤ m C Since the sample points are uniformly spaced by1l = L/n, the number

of sample points n i contained in an individual segment S ican be bounded by

Lλn (r, z) − k 1l ≤ L h C (r, z) < L λ n (r, z) + k 1l,

from which (3.2) follows

Example 9. Let C be a circle of radius 1 A set of n evenly spaced sample points P n⊂

C forms a regular n-gon Using the identification R2

' C, the cumulative histogram

which, by symmetry, is independent of the point z ∈ C.

In Figure 3, we plot the discrete cumulative histogramλn (r, z) for n = 20, along with the bounds h C (r, z) ± 1l/(2π) coming from (3.2) and the fact that a circle has circular index m C =1 In the first plot, the center z coincides with a data point, while the second takes z to be a distance.01 away, as measured along the circle Observethat the discrete histogram stays within the indicated bounds at all radii, in accordancewith our result

0.4 0.6 0.8 1.0

Figure 3. Local histogram functions for a circle.

We now turn our attention to the convergence of the global histograms Again, wework under the preceding regularity assumptions, and continue to focus our attention

on the case of uniformly spaced sample points P n ⊂C

First, we observe that the local histogram function h r (s) = h C (r, z(s)) is piecewise continuous as a function of s Indeed, h r (s) is continuous unless the circle of radius

r centered at z(s) contains one or more circular arcs that belong to C, in which case

h r (s) has a jump discontinuity whose magnitude is the sum of the lengths of such arcs.

By our assumption of regularity, C contains only finitely many circular arcs, and so

h r (s) can have only finitely many jump discontinuities On the other hand, regularity

implies that the global histogram function is everywhere continuous

Therefore, the global histogram integral (2.9) can be approximated by a Riemannsum based on the evenly spaced data points:

Since C has finite length, 1l = L/n → 0 as n → ∞, and so the Riemann sums

con-verge On the other hand, (3.1) implies that the local histogram function can be imated by the (rescaled) cumulative point histogramλn (r, z), and hence we should be

approx-able to approximate the Riemann sum in turn by

Example 10. Let C be a unit square, so that L = l (C) = 4 Measuring the arc length

s along the square starting at a corner, the local histogram function h r (s) = h C (r, z(s))

can be explicitly constructed using elementary geometry, distinguishing several

differ-ent configurations For 0 ≤ s ≤ 1

(3.7)

It is interesting that, while the local histogram function has six intervals with differentanalytical formulas, the global function has only three

Figure 4 plots the global cumulative histograms of a square based on n = 20 evenly

spaced points, along with the bounds 1

41l and 1

21l Observe that the discrete

his-togram stays within 1

41l of the curve histogram, a tighter bound than we are able

to derive analytically Interestingly, a similarly tight bound appears to hold in all theexamples we have looked at so far

4 RANDOM POINT DISTRIBUTIONS. We have thus far proved, under suitableregularity hypotheses, convergence of both the local and global cumulative histogramsconstructed from uniformly spaced sample points along the curve However, in prac-tice, it may be difficult to ensure precise uniform spacing of the sample points For ex-

ample, if C is an ellipse, then this would require evaluating n elliptic integrals Hence,

January 2012] INVARIANT HISTOGRAMS 11

0.5 1.0 1.5 2.0 0.2

0.4

0.6

0.8

1.0

Figure 4. Global histogram bounds for a square.

for practical shape analysis, we need to examine more general methods of histogram

creation In this section, we analyze the case of sample points P n = {z1, , z n} ⊂C

that are randomly chosen with respect to the uniform arc length distribution

In this case, we view the cumulative local histogramλn (r, z) as a random variable representing the fraction of the points z i that lie within a circle of radius r centered at the point z Indeed, we can write

Similarly, to construct a statistical variable whose expectation approximates the

global histogram function H C (r), consider

whereσi , j (r) is a random variable that is 1 if d(z i , z j ) ≤ r and 0 otherwise As above,

its expected value is

i ,i0, j, j0 all distinct

As above, the terms in the first summation are all 0, whereas those in the second are

bounded As there are O n3

of the latter, we conclude thatVar[3n (r)] = O n−1

Thus,3n (r) converges to H C (r) in the sense that, for any given value of r, the

proba-bility of3n (r) lying in any interval around H C (r) approaches 1 as n → ∞.

Example 11. Let C be a 2 × 3 rectangle In Figure 5, we graph its global curve togram function H C (r) in black and the approximate histograms 3 n (r), based on

his-n =20 sample points, in gray The first plot is for evenly distributed points, in whichthe approximation remains within1l of the continuous histogram function, while the

second plot is for randomly generated points, in which the approximation stays within

21l Thus, both methods work as advertised.

January 2012] INVARIANT HISTOGRAMS 13

Figure 5.Comparison of approximate histograms of a rectangle.

5 HISTOGRAM–BASED SHAPE COMPARISON. In this section, we discussthe question of whether distance histograms can be used, both practically and the-oretically, as a means of distinguishing shapes up to rigid motion We begin with thepractical aspects As we know, if two curves have different global histogram functions,they cannot be rigidly equivalent For curves arising from digital images, we will ap-proximate the global histogram function by its discrete counterpart based on a reason-ably dense sampling of the curve Since the error in the approximations is proportional

to1l = L/n, we will calculate the average difference between two histogram plots,

normalized with respect to1l Our working hypothesis is that differences less than 1

represent histogram approximations that cannot be distinguished

Tables 1 and 2 show these values for a few elementary shapes We use randompoint distributions3to illustrate that identical parameterizations do not necessarily giveidentical sample histograms This is also evident from the fact that the matrix is notsymmetric—different random sample points were chosen for each trial However, sym-metrically placed entries generally correlate highly, indicating that the comparison isworking as intended

Table 1 is based on discretizing using only n = 20 points As we see, this is too

small a sample set to be able to unambiguously distinguish the shapes Indeed, the

2 × 3 rectangle and the star appear more similar to each other than they are to a secondrandomized version of themselves On the other hand, for the star and the circle, thevalue of 5.39 is reasonably strong evidence that they are not rigidly equivalent

Table 1. 20-point comparison matrix.

As we increase the number of sample points, the computation time increases (in

proportion to n2 for calculating the histograms and n for comparing them), but our

3More precisely, we first select n uniformly distributed random numbers s i∈ [ 0, L], i = 1, , n, and

then take the corresponding n random points z(s i ) ∈ C based on a given arc length parameterization In our

experiment, the shapes are sufficiently simple that the explicit arc length parameterization is known.

ability to differentiate shapes increases as well In Table 2, based on n = 500 sample

points, it is now clear that none of the shapes are rigidly equivalent to any of the others.The value of 4 for comparing the 1 × 3 rectangle to itself is slightly high, but it is stillsignificantly less than any of the values for comparing two different shapes

Table 2. 500-point comparison matrix.

are identical However, the curve histograms H C (r) based on their outer polygons can

easily be distinguished In Figure 6, we plot the approximate global histograms3n (r) based on n = 20 uniformly spaced sample points; the kite is dotted and the trapezoid

Figure 6. Curve histograms for the kite and trapezoid.

While we have as yet been unable to establish a complete proof, there is a variety

of credible evidence in favor of the following:

Conjecture Two regular plane curves C and e C have identical global histogram tions, so H C (r) = H Ce(r) for all r ≥ 0, if and only if they are rigidly equivalent:

func-C ' e C

One evident proof strategy would be to approximate the histograms by samplingand then apply the convergence result of Theorem 4 If one could prove that the sampleJanuary 2012] INVARIANT HISTOGRAMS 15

points do not, at least when taken sufficiently densely along the curve, lie in the tional set of Theorem 2, then our conjecture would follow.

excep-A second strategy is based on our observation that, even when the corners of a gon lie in the exceptional set, the associated curve histogram still appears to uniquelycharacterize it Indeed, if one can prove that the global distance histogram of a simpleclosed polygon (as opposed to the discrete histogram based on its corners) uniquelycharacterizes it up to rigid motion, then our conjecture for general curves would follow

poly-by suitably approximating them poly-by their interpolating polygons

To this end, let K be a simple closed polygon of length L = l(K ) all of whose angles are obtuse, as would be the case with a sufficiently densely sample polygon of

a smooth curve Let l?be the minimum side length, and d?be the minimum distance

between any two nonadjacent sides Set m?=min{l?, d?} Then any disk B r (z) tered at a point z ∈ K of radius r with 0 < r <1

cen-2m?intersects K in either one or two sides, the latter possibility only occurring when z is within a distance r of the nearest corner Let z1, , z n be the corners of K , and letθj > 1

2π denote the interior angle at

z j—see Figure 7

y j θj

r

Figure 7. Intersection of a polygon and a disk.

Then, for r > 0 sufficiently small, and all z ∈ K ,

Thus, for small r, the global histogram function (2.9) for such an “obtuse polygon”

takes the form

9(θj , r) =Z r

0

x + y j (x) dx, (5.4)

with y j = y j (x) for x = x j implicitly defined by (5.2) (There is, in fact, an explicit,but not very enlightening, formula for this integral in terms of elementary functions.)

Observe that (5.3) is a symmetric function of the polygonal anglesθ1, , θn, i.e.,

it is not affected by permutations thereof Moreover, for distinct angles, the integrals9(θj , r) can be shown to be linearly independent functions of r This implies that

one can recover the set of polygonal angles {θ1, , θn}from knowledge of the global

histogram function H K (r) for small r In other words, the polygon’s global histogram

function does determine its angles up to a permutation

The strategy for continuing a possible proof would be to gradually increase the size

of r Since, for small r, the histogram function has prescribed the angles, its form is fixed for all r ≤ 1

2m? For r > 1

2m?, the functional form will change, and this will

serve to characterize m?, the minimal side length or distance between nonadjacent

sides Proceeding in this fashion, as r gradually increases, more and more sides of the

polygon can be covered by a disk of that radius, providing more and more geometricinformation about the polygon from the resulting histogram This points the way to

a proof of our polygonal histogram conjecture, and hence the full curve conjecture.However, the details in such a proof strategy appear to be quite intricate

Barring a resolution of the histogram conjecture, let us discuss what properties of

the curve C can be gleaned from its histogram First of all, the curve’s diameter is equal

to the minimal value of r for which H C (r) = 1 Secondly, values where the derivative

of the histogram function is very large usually have geometric significance In the

square histogram in Figure 4, this occurs at r = 1 In polygons, such values often

correspond to distances between parallel sides, because, at such a distance, the diskcentered on one of the parallel sides suddenly begins to contain points on the oppositeside For shapes with multiple pairs of parallel sides, we can see this effect at several

values of r — such as when r = 2 and r = 3 in the case of a 2 × 3 rectangle shown

in Figure 5 The magnitude of the effect depends on the overall length of the parallel

sides; for instance, the slope at r = 3 is larger than that at r = 2 However, not every

value where the derivative is large is the result of such parallel sides The histogramfunction of the Boutin–Kemper kite shown in Figure 6 has two visible corners, but thekite has no parallel sides

In a more theoretical direction, let us compute the Taylor expansion of the global

histogram function H C (r) at r = 0, assuming that C is sufficiently smooth The

coef-ficients in the expansion will provide Euclidean-invariant quantities associated with asmooth curve We begin by constructing the Taylor series of the local histogram func-

tion h C (r, z) based at a point z ∈ C To expedite the analysis, we apply a suitable rigid motion to move the curve into a “normal form” so that z is at the origin, and the tan- gent at z is horizontal Thus, in a neighborhood of z =(0, 0), the curve is the graph

of a function y = y(x) with y(0) = 0 and y0(0) = 0 As a consequence of the moving

frame recurrence formulae developed in [7]—or working by direct analysis—we can

write down the following Taylor expansion

Lemma 12 Under the above assumptions,

We use this formula to find a Taylor expansion for the local histogram function

h C (r, z) at r = 0 Assume that r is small The curve (5.5) will intersect the circle of

January 2012] INVARIANT HISTOGRAMS 17

radius r centered at the origin at two points z± =(x±, y±) = (x±, y(x±)), which arethe solutions to the equation

We now substitute (5.6) to produce

written at any point z ∈ C.

To obtain the Taylor expansion of the global histogram function, we substitute (5.7)back into (2.9), resulting in

where we can use integration by parts and the fact that C is a closed curve to simplify

the expansion coefficients Each integral appearing in the Taylor expansion (5.8) is

invariant under rigid motion, and uniquely determined by the histogram function Aninteresting question is whether the resulting collection of invariant integral moments,depending on curvature and its arc length derivatives, uniquely prescribes the curve up

to rigid motion If so, this would establish the validity of our conjecture for smoothcurves

6 EXTENSIONS. There are a number of interesting directions in which this search program can be extended The most obvious is to apply it to more substantialpractical problems in order to gauge whether histogram-based methods can competewith other algorithms for object recognition and classification, particularly in noisy

re-images In this direction, the method of shape distributions [17], touted for its

in-variance, simplicity, and robustness, employs a variety of discrete local and invariantglobal histograms for distinguishing three-dimensional objects, including distances be-tween points, areas of triangles, volumes of tetrahedra, and angles between segments

An unanswered question is to what extent the corresponding limiting histograms canactually distinguish inequivalent objects, under the appropriate transformation group:Euclidean, equi-affine, conformal, etc

6.1 Higher Dimensions. Extending our analysis to objects in three or more sions requires minimal change to the methodology For instance, local and global his-

dimen-togram functions of space curves C ⊂ R3are defined by simply replacing the disk of

radius r by the solid ball of that radius in the formulas (2.8) and (2.9) For example,

consider the saddle-like curve parametrized by

z (t) = cos t, sin t, cos 2 t
, 0 ≤ t ≤ 2π (6.1)

In Figure 8, we plot the discrete approximations3n (r) to its global histogram function, based on n = 10, 20, and 30 sample points, respectively, indicating convergence as

0.4 0.6 0.8 1.0

0.5 1.0 1.5 2.0 2.5 3.0 0.2

0.4 0.6 0.8 1.0

Figure 8.Approximate distance histograms for the three-dimensional saddle curve.

January 2012] INVARIANT HISTOGRAMS 19

We can also apply our histogram analysis to two-dimensional surfaces in

three-dimensional space We consider the case of piecewise smooth surfaces S ⊂ R3 with

finite surface area Let P n ⊂S be a set of n sample points that are (approximately)

uniformly distributed with respect to surface area We retain the meaning ofλn (r, z)

as the proportion of points within a distance r of the point z, (2.4), and3n (r) as its

average, (2.6) By adapting our proof of Theorem 4 and assuming sufficient regularity

of the surface, one can demonstrate that the discrete cumulative histogramsλn (r, z) and

3n (r) converge, as n → ∞, to the corresponding local and global surface distance

The discrete approximations3n (r) for the unit sphere S2= {kzk =1} ⊂ R3, based

on n = 10, 30, and 100 sample points, are plotted in Figure 9 The global histograms are evidently converging as n → ∞, albeit at a slower rate than was the case with

0.4 0.6 0.8 1.0

0.5 1.0 1.5 2.0 2.5 3.0 0.2

0.4 0.6 0.8 1.0

Figure 9. Approximate distance histograms of a sphere.

Future work includes rigorously establishing a convergence theorem for surfacesand higher-dimensional submanifolds of Euclidean space along the lines of Theorem 4.Invariance under rigid motions immediately implies that surfaces with distinct distancehistograms cannot be rigidly equivalent However, it seems unlikely that distance his-tograms alone suffice to distinguish inequivalent surfaces, and extensions to distancehistograms involving more than two points, e.g., that are formed from the side lengths

of sampled triangles, are under active investigation An interesting question is whetherdistance histograms can be used to distinguish subsets of differing dimensions Or, tostate this another way, can one determine the dimension of a subset from some innateproperty of its distance histogram?

6.2 Area Histograms. In image processing applications, the invariance of objects

under the equi-affine group, consisting of all volume-preserving affine transformations

of Rn , namely, z 7→ A z + b for det A = 1, is of great importance [6, 9, 16] Planar

equi-affine (area-preserving) transformations can be viewed as approximations to jective transformations, valid for moderately tilted objects For example, a round plateviewed at an angle has an elliptical outline, which can be obtained from a circle by

pro-an equi-affine trpro-ansformation The basic plpro-anar equi-affine joint invaripro-ant is the area

of a triangle, and hence the histogram formed by the areas of triangles formed by alltriples in a finite point configuration is invariant under the equi-affine group Similar

to Theorem 2, Boutin and Kemper [4] also proved that, in most situations, generic

pla-nar point configurations are uniquely determined, up to equi-affine transformations, bytheir area histograms, but there is a lower-dimensional algebraic subvariety of excep-tional configurations

For us, the key question is convergence of the cumulative area histogram based ondensely sampled points on a plane curve To define an area histogram function, wefirst note that the global curve distance histogram function (2.9) can be expressed inthe alternative form

where ˆs, ˆs0, ˆs00now refer to the equi-affine arc length of the curve [8], while L = H C d ˆs

is its total equi-affine arc length (In local coordinates, if the curve is the graph of a

function y(x) then the equi-affine arc length element is given by d ˆs =√3

a circle, based on n = 10, 20, and 30 sample points, respectively.

Let us end by illustrating the equi-affine invariance of the curve area histogramfunction Since rectangles of the same area are equivalent under an equi-affine trans-formation, they have identical area histograms In Figure 11, we plot discrete areahistograms for, respectively, a 2 × 2 square, a 1 × 4 rectangle, and a.5 × 8 rectangle,

using n = 30 sample points in each case As expected, the graphs are quite close.

ACKNOWLEDGMENTS.The authors would like to thank Facundo Memoli, Igor Pak, Ellen Rice, Guillermo Sapiro, Allen Tannenbaum, and Ofer Zeitouni, as well as the anonymous referees, for helpful

January 2012] INVARIANT HISTOGRAMS 21

0.4 0.6 0.8 1.0

0.5 1.0 1.5 2.0 0.2

0.4 0.6 0.8 1.0

Figure 10. Area histogram of a circle.

comments and advice The paper is based on the first author’s undergraduate research project (REU) funded

by the second author’s NSF Grant DMS 08–07317.

2 Belongie, S.; Malik, J.; Puzicha, J., Shape matching and object recognition using shape contexts, IEEE

Trans Pattern Anal Mach Intell.24(2002), 509–522, available at http://dx.doi.org/10.1109/ 34.993558.

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6 Calabi, E.; Olver, P J.; Shakiban, C.; Tannenbaum, A.; Haker, S., Differential and numerically invariant

signature curves applied to object recognition, Int J Computer Vision 26 (1998), 107–135, available at

http://dx.doi.org/10.1023/A:1007992709392.

7 Fels, M.; Olver, P J., Moving coframes II Regularization and theoretical foundations, Acta Appl Math.

55(1999), 127–208, available at http://dx.doi.org/10.1023/A:1006195823000.

8 Guggenheimer, H W., Differential Geometry McGraw–Hill, New York, 1963.

9 Kanatani, K., Group–theoretical methods in image understanding Springer–Verlag, New York, 1990.

10 Lowe, D G., Object recognition from local scale-invariant features, in: Integration of Speech and Image

Understanding, IEEE Computer Society, Los Alamitos, CA, 1999, 1150–1157.

11 Manay, S.; Cremers, D.; Hong, B.-W.; Yezzi, A.; Soatto, S., Integral invariants and shape matching, in:

Statistics and Analysis of Shapes, H Krim and A Yezzi, eds., Birkh¨auser, Boston, 2006, 137–166.

0.4 0.6 0.8 1.0

0.0 0.5 1.0 1.5 2.0 2.5 3.0 0.2

0.4 0.6 0.8 1.0

Figure 11. Area histograms of affine-equivalent rectangles.

12 M´emoli, F., On the use of Gromov–Hausdorff distances for shape comparison, in: Symposium on Point

Based Graphics, M Botsch, R Pajarola, B Chen, and M Zwicker, eds., Eurographics Association,

Prague, Czech Republic, 2007, 81–90.

13 M´emoli, F., Gromov-Wasserstein distances and the metric approach to object matching, Found Comput.

Math.11(2011), 417–487, available at http://dx.doi.org/10.1007/s10208-011-9093-5.

14 Nutbourne, A W.; Martin, R R., Differential geometry applied to curve and surface design, Vol 1:

Foundations Ellis Horwood, Chichester, UK, 1988.

15 Olver, P J., Joint invariant signatures, Found Comput Math 1 (2001), 3–67, available at http://dx.

doi.org/10.1007/s10208001001.

16 Olver, P J.; Sapiro, G.; Tannenbaum, A., Differential invariant signatures and flows in computer vision: A symmetry group approach, in: Geometry–Driven Diffusion in Computer Vision, B M Ter Haar Romeny,

ed., Kluwer, Dordrecht, Netherlands, 1994, 255–306.

17 Osada, R.; Funkhouser, T.; Chazelle, B.; Dobkin, D., Shape distributions, ACM Trans Graphics 21

(2002), 807–832, available at http://dx.doi.org/10.1145/571647.571648.

18 Pele, O.; Werman, M., A linear time histogram for improved SIFT matching, in: Computer Vision— ECCV 2008, part III, D Forsyth, P Torr, A Zisserman, eds., Lecture Notes in Computer Science, vol.

5304, Springer–Verlag, Berlin, 2008, 495–508.

19 Pottmann, H.; Wallner, J.; Huang, Q.; Yang, Y.-L., Integral invariants for robust geometry processing,

Comput Aided Geom Design26(2009), 37–60, available at http://dx.doi.org/10.1016/j.cagd 2008.01.002.

20 Rustamov, R M., Laplace–Beltrami eigenfunctions for deformation invariant shape representation, in:

SGP ’07: Proceedings of the Fifth Eurographics Symposium on Geometry Processing, Eurographics

Association, Aire-la-Ville, Switzerland, 2007, 225–233.

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Cambridge, 2001.

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query-by-shape-and-color in image and video databases, Image Vision Comput 23 (2005), 1170–1180, available at http:

25 Yale, P B., Geometry and symmetry Holden–Day, San Francisco, 1968.

January 2012] INVARIANT HISTOGRAMS 23

DAN BRINKMANreceived B.S degrees in Mathematics and Physics from the University of Minnesota in

2009, and is currently studying for his Ph.D in Applied Mathematics and Theoretical Physics from the versity of Cambridge in England His mathematical interests include partial differential equations, especially applications to physical problems His current research involves reaction–diffusion modeling of organic pho- tovoltaic devices In his free time he enjoys spending time in Europe, traveling for both academic and leisure activities.

Uni-Department of Applied Mathematics and Theoretical Physics, University of Cambridge, Cambridge, England D.Brinkman@damtp.cam.ac.uk

PETER J OLVERreceived his Sc.B from Brown University in 1973 and his Ph.D from Harvard University

in 1976, and is currently Professor and Head of Department at the University of Minnesota He is the author

of over 100 research papers in a wide range of subjects, mostly concerned with applications of symmetry, as

well as five books, including two undergraduate texts: Applied Linear Algebra, coauthored with his wife Cheri Shakiban, and the forthcoming Introduction to Partial Differential Equations On those rare occasions when

he is not doing mathematics or administrating, he relaxes by playing the piano, gardening, and boating on the Mississippi.

School of Mathematics, University of Minnesota, Minneapolis, MN 55455

I scream, “i is complex!”

This excuse is the root of my error

—Submitted by Dan McQuillan, Norwich University, Northfield, Vermont

Zariski Decomposition:

A New (Old) Chapter of Linear Algebra Thomas Bauer, Mirel Caib˘ar, and Gary Kennedy

Abstract In a 1962 paper, Zariski introduced the decomposition theory that now bears his

name Although it arose in the context of algebraic geometry and deals with the configuration

of curves on an algebraic surface, we have recently observed that the essential concept is purely within the realm of linear algebra In this paper, we formulate Zariski decomposition as

a theorem in linear algebra and present a linear algebraic proof We also sketch the geometric context in which Zariski first introduced his decomposition.

1 INTRODUCTION Oscar Zariski (1899–1986) was a central figure in 20th tury mathematics His life, ably recounted in [10], took him from a small city in White

cen-Russia, through his advanced training under the masters of the “Italian school” of gebraic geometry, and to a distinguished career in the United States, the precursor

al-of a tide al-of emigrant talent fleeing political upheaval in Europe As a pral-ofessor atJohns Hopkins and Harvard University, he supervised the Ph.D.s of some of the mostoutstanding mathematicians of the era, including two Fields Medalists, and his math-

ematical tribe (traced through advisors in [7]) now numbers more than 800 Zariski

thoroughly absorbed and built upon the synthetic arguments of the Italian school, and

in [13] he gave a definitive account of the classical theory of algebraic surfaces In the

course of writing this volume, however, despite his admiration for their deep geometricinsight he became increasingly disgruntled with the lack of rigor in certain arguments

He was thus led to search for more adequate foundations for algebraic geometry, taking(along with Andre Weil) many of the first steps in an eventual revolutionary recasting

of these foundations by Alexander Grothendieck and others

In a 1962 paper [12], Zariski introduced the decomposition theory that now bears

his name Although it arose in the context of algebraic geometry and deals with theconfiguration of curves on an algebraic surface, we have recently observed that theessential concept is purely within the realm of linear algebra (A similar observation

has been made independently by Moriwaki in Section 1 of [9].) In this paper, we

formulate Zariski decomposition as a theorem in linear algebra and present a linearalgebraic proof To motivate the construction, however, we begin in Section 2 with abreezy account of the the original geometric situation, and eventually return to thissituation in Section 7 to round off the discussion and present one substantive example

We give only a sketchy description which lacks even proper definitions; one needs aserious course in algebraic geometry to treat these matters in a rigorous way But, asalready indicated, the thrust of the paper is in a far different direction, namely towarddisentangling the relatively elementary linear algebra from these more advanced ideas.Beginning in Section 3, our treatment is both elementary and explicit; a basic course

in linear algebra, which includes the idea of a negative definite matrix, should be asufficient background After laying out the definitions and the main idea, we present a

simple new construction (which first appeared in [2]) and show that it satisfies the

re-quirements for a Zariski decomposition We look at a few elaborations, and we presentZariski’s original algorithm (shorn of its original geometric context)

http://dx.doi.org/10.4169/amer.math.monthly.119.01.025

MSC: Primary 15A63, Secondary 14J99; 14C20

January 2012] ZARISKI DECOMPOSITION 25

2 THE ORIGINAL CONTEXT The study of algebraic curves, with its ties to the

theory of Riemann surfaces and many other central ideas of mathematics, has ancientroots, but our understanding of algebraic surfaces has developed more recently One

of Zariski’s main concerns was how to extend well-known fundamental theories fromcurves to surfaces In trying to understand such a surface, one is naturally led to studythe algebraic curves which live on it, asking what sorts of curves there are, how theymeet each other, and how their configurations influence the geometry of the surface.For example, in the plane1(the simplest example of an algebraic surface) an algebraic

curve is the solution set of a polynomial equation f (x, y) = 0 One can calculate that the vector space of all polynomials in two variables of degree not exceeding d is a

vector space of dimension d+2

2

Since two such polynomials define the same curve

if and only if one is a multiple of the other, we say that the set of all such curves

forms a linear system of dimension d+2

2
−1 (In general, the dimension of a linearsystem is one less than the dimension of the corresponding vector space of functions.)

More generally, for each curve D on an algebraic surface one can naturally define

an associated linear system of curves which are equivalent in a certain sense to D, denoting it by |D| This linear system depends not just on the curve as a set of points but also on the equation which defines it: the equation f (x, y) n =0 defines a larger

linear system than does f (x, y) = 0, and we denote this larger linear system by |nD| (For a curve of degree d in the plane, |n D| consists of all curves of degree nd.)

His student David Mumford (in an appendix to [10]) says that “Zariski’s papers

on the general topic of linear systems form a rather coherent whole in which one canobserve at least two major themes which he developed repeatedly One is the Riemann-Roch problem: to compute the dimension of a general linear system and especially

to consider the behavior of dim |n D| as n grows The other is to apply the theory of

linear systems in the 2-dimensional case to obtain results on the birational geometry

of surfaces and on the classification of surfaces In relation to his previous work, thisresearch was, I believe, something like a dessert He had worked long setting up manynew algebraic techniques and laying rigorous foundations for doing geometry—andlinear systems, which are the heart of Italian geometry, could now be attacked.”

Zariski’s paper [12] is concerned with the following question: for a specified curve

D on an algebraic surface, what is the order of growth of dim |n D| as a function of

n ? His answer involved a decomposition: he showed that D, considered as an element

of a certain vector space, could be written as a sum P + N of a “positive part” and a

“negative part,” so that the answer to his question was determined by P alone ically, he showed that the order of growth was the “self-intersection number” of P In

Specif-the heart of this paper, we will give an account of Zariski’s decomposition, assumingthat we already have been given the relevant “intersection theory” on the surface Inthe last section of the paper we will resume this account of the original context Inparticular we will say something about how this intersection theory arises, and give aprecise statement of Zariski’s formula on the order of growth

3 THE DECOMPOSITION We now forget about the original context, and lay out

an elementary theory within linear algebra In Section 7 we will resume our account

of the geometry which motivates the following definitions

Suppose that V is a vector space over Q (the rational numbers) equipped with a

symmetric bilinear form; we denote the product of v and w by v · w Suppose

fur-thermore that there is a basis E with respect to which the bilinear form is an

inter-1 We mean the complex projective plane Our equation is given in affine coordinates, but we intend for the curve to include appropriate points at infinity The reader who hasn’t encountered these notions will need to take our assertions in this section on faith.

Figure 1 Oscar Zariski in 1960 (frontispiece photo of [10], credited to Yole Zariski).

section product, meaning that the product of any two distinct basis elements is negative Most of our examples will be finite-dimensional, but we are also interested

non-in the non-infnon-inite-dimensional case If V is fnon-inite-dimensional then we will assume that

E has an ordering, and using this ordered basis we will identify V with Q n (where

nis its dimension); a vector v will be identified with its coordinate vector, written as

a column We can then specify the bilinear form by writing its associated symmetric

matrix M with respect to the basis, calling it the intersection matrix Thus the

prod-uct of v and w is vT Mw, where T denotes the transpose With this interpretation, the

form is an intersection product if and only if all off-diagonal entries of M are

non-negative

In any case, whether V is finite- or infinite-dimensional, each element v ∈ V can

be written in a unique way as a linear combination of a finite subset of the basis,

with all coefficients nonzero We will call this finite subset the support of v, and the finite-dimensional subspace of V which it spans is called the support space of v If all coefficients are positive, then v is said to be effective.2In particular each basis element

is effective, and the zero vector is also considered to be effective (since we may sumover the empty set)

A vector w is called nef with respect to V if w · v ≥ 0 for every effective

vec-tor v Note that to check whether a vecvec-tor satisfies this condition it suffices to check

whether its product with each basis element is nonnegative In the finite-dimensional

case (using the identification of V with Q n, as described above) the definition can be

formulated in terms of the intersection matrix: since the entries of Mw are the

prod-ucts of the basis elements with w, we observe that a vector w is nef with respect to V

precisely when Mw is effective In particular if M is nonsingular, then w is nef with

respect to V if and only if there is an effective vector v ∈ V for which M−1v = w.

Now suppose that W is a subspace of V spanned by some subset of the basis and

containing the support space of a vector w (for example, W could be the support space

2 In the motivating application, the basis vectors will be certain curves on the algebraic surface, and hence

an arbitrary vector v ∈ V will be a linear combination of such curves The combinations that use nonnegative

coefficients may be interpreted geometrically, while the others are just “virtual curves.”

January 2012] ZARISKI DECOMPOSITION 27

itself) If w is nef with respect to V then it is nef with respect to W , but the opposite

implication may not be correct

Example 3.1 Suppose that the intersection matrix is

M = −2 11 1

,

and let W be the one-dimensional subspace spanned by the first basis element e1 Then

−e1is nef with respect to W , but it is not nef with respect to V

We do, however, have a partial converse

Lemma 3.2 If w ∈ W is effective and nef with respect to the subspace W , then it is

nef with respect to the entire space V

Proof By hypothesis, the product of w and a basis element for W is nonnegative Since w is effective, its intersection product with any other basis element of V is like-

wise nonnegative

In view of this lemma, we may simply call such a vector effective and nef.3

Here is our main theorem

Theorem 3.3 For each effective element v ∈ V , there is a unique way to write it as

a sum

v = p + n

of elements satisfying the following conditions:

1 p is nef with respect to V ;

2 n is effective;

3 p · e = 0 for each basis element e in the support of n;

4 the restriction of the intersection product to the support space of n is negative

definite.

Furthermore p is effective.

This is called the Zariski decomposition of v; the elements p and n are called its

positive and negative parts We note that both extremes are possible: for example, if v itself is nef with respect to V , then p = v and the support space of n is trivial.

Example 3.4 Again suppose that

M = −2 11 1

,

and let v = 2e1+e2 Since v · v = −3, the vector v is not nef But since e2·e2 is

positive, e2 cannot be in the support of n Thus n = xe1 and p =(2 − x)e1+e2for

3 Some say that the neologism “nef” is short for “numerically effective,” but this gives a misleading sion of its meaning (since an effective vector is not necessarily nef) Others insist that it should be thought of

impres-as an acronym for “numerically eventually free.”

some number x By the third condition p · e1 = −2(2 − x) + 1 = 0 Thus

p = 1

2e1+e2 and n =

3

2e1

It’s instructive to look at all elements xe1+ye2, where x ≤ 2 and y ≤ 1 (Since the

coordinates of n must be nonnegative, these are the only possibilities for p.) If the

corresponding points (x, y) are plotted in the plane, then the nef elements form a

triangle, and the element p corresponds to the upper right vertex See Figure 2.

n

Figure 2 An example of Zariski decomposition The picture on the left shows the candidates for the positive

part of v The middle picture shows the nef vectors The shaded triangle in the right picture is their overlap.

4 PROOF OF THE MAIN THEOREM Recall that if V is finite-dimensional then

we will identify it with Qn In particular the basis element ej is identified with the

column vector having 1 in position j and 0 elsewhere We begin the proof with a pair

of lemmas

Lemma 4.1 If M is a negative definite matrix whose off-diagonal entries are

non-negative, then all entries ofM− 1are nonpositive.

Proof. (adapted from the Appendix of [3]) Write M−1ej as a difference of effective

vectors q − r with no common support vector Then qTMr ≥ 0 Hence (since M is negative definite) for q 6= 0 we have

qTMq − qTMr< 0

But this is qTej , the jth entry of q, which is nonnegative Thus q = 0, which says that all the entries of column j of M− 1are nonpositive

Lemma 4.2 Suppose M is a symmetric matrix whose off-diagonal entries are

non-negative Suppose that M is not negative definite Then there is a nonzero vector q for

which q and Mq are both effective.

Proof. If the top left entry of M is nonnegative then we can take q = e1 Otherwise let

M0be the largest upper left square submatrix which is negative definite, and write

Denote the dimension of M0by m0 Since M0is nonsingular, there is a vector

January 2012] ZARISKI DECOMPOSITION 29

in the kernel of the map defined by [ M0 A ], where q0 has length m0 Letting A1

denote the first column of A, we see that M0q0= −A1, and thus q0 = −M0−1A1 By

Lemma 4.1 we see that all entries of q0are nonnegative Thus the same is true of q.

Turning to Mq, we know that it begins with m0 zeros Thus the product qTMq

computes entry m0+1 of Mq Now note that by the choice of M0there is a vector

bilinear form is negative definite Thus by bilinearity

qTMq = (q − w)TMq − (q − w)TM (q − w) + wTMw + qTM (q − w) ≥ 0.

As for the remaining entries of Mq, each one is a sum of products of nonnegative

numbers; thus these entries are all nonnegative

Corollary 4.3 Suppose that the restriction of an intersection product to a

finite-dimensional subspace is not negative definite Then there is a nonzero effective and nef element in this subspace.

We now present a procedure for constructing the Zariski decomposition of an

ef-fective element v = Pn i=1c iei We will momentarily allow arbitrary real numbers ascoefficients, but we will soon show that rational coefficients suffice Consider a “can-didate” for the positive part: Pn i=1x iei, where

maximized Let p be the corresponding element of V , and let n = v − p We claim

that this is a Zariski decomposition

By construction, the first two conditions in Theorem 3.3 are satisfied Regarding the

third condition, note (since p maximizes P x i) that if ej is in the support of n then, for

 > 0 and sufficiently small, the element p + ejis not nef But(p + ej) · ei ≥0 for

all i 6= j Thus(p + ej) · ej < 0 for all sufficiently small positive , and this implies

that p · ej ≤0 Since p is nef we have p · ej =0

To prove that the restriction of the intersection product to the support space of n is

negative definite, we argue by contradiction Supposing that the restriction of the form

is not negative definite, Corollary 4.3 tells us that there is a nonzero effective and nef

element q in the support space of n Then for small  > 0 the element p + q is nef and n − q is effective But this contradicts the maximality of p.

To prove the remaining claims of Theorem 3.3 (and the implicit claim that all

coefficients of p and n are rational numbers), we need the following idea Define

the maximum of two elements v = P n i=1x iei and v0= Pn

Lemma 4.4 If p and p0are both nef, then so ismax(p, p0).

Proof The jth inequality in (4.2) involves at most one negative coefficient, namely

ej·ej Suppose that p = Pn i=1x ieiand p0=Pn

Zariski decompositions of v Let max(p, p0) = p + P x iei, where the sum is over the

support of n and the coefficients are nonnegative Since max(p, p0) is nef, we know

that for each element ej of the support of n we have

X x iei·ej =max(p, p0

) · ej ≥0

Thus

X x iei ·X x jej =X X x i x jei·ej ≥0

Since the intersection product is negative definite on the support space of n, all

x i =0 Thus p = max(p, p0) Similar reasoning shows that p0=max(p, p0), and thus

x i(ei·ej) = 0 for each basis element ejin the support of n,

x j =c j for each basis element ej not in the support of n

In matrix form (and with the basis suitably reordered), we have the following equation:

,January 2012] ZARISKI DECOMPOSITION 31

where N is negative definite, 0 is a zero matrix, and I is an identity matrix This is

a nonsingular system in which all entries are rational numbers, and we know that itsunique solution gives the positive part of the Zariski decomposition

5 ZARISKI’S ORIGINAL ALGORITHM Our construction gives the Zariski

de-composition of an effective vector in one fell swoop In Zariski’s original paper, bycontrast, he built up the negative part in stages.4Our exposition of his algorithm relies

on the last chapter of [1] Let us call a finite subspace of V a special subspace if it

is spanned by a subset of the basis We say that a subspace is negative definite if the

restriction of the intersection product to the subspace is negative definite The basicidea is to work toward the correct support space for the negative part of the specifiedvector, through an increasing sequence of negative definite special subspaces

Example 5.1 Suppose that V is finite-dimensional with intersection matrix

Figure 3 The lattice of negative definite subspaces in Example 5.1 The subspace spanned by basis vectors

e1and e3 , for example, is indicated by 13.

The algorithm relies on three lemmas

Lemma 5.2 (cf Lemma 14.9 of [1]) Let N be a negative definite special subspace,

and suppose that n ∈ N is a vector for which n · e ≤ 0 for every basis element e ∈ N

Then n is effective.

Proof. As in the proof of Lemma 4.1, write n = q − r, where q and r are effective but have no common support vector Then q · r ≥ 0 Hence

r · r ≥ r · r − q · r = −n · r ≥ 0

Since the subspace is negative definite this implies that r = 0, i.e., that n is effective.

4 This comparison is somewhat unfair, since our construction simply says to maximize a certain linear function on a polytope To actually discover the location of the maximum one would have to invoke a step-by- step algorithm such as the simplex method.

Lemma 5.3 Suppose that v ∈ V is an effective vector Suppose that N is a negative

definite special subspace of the support space of v Suppose that p is a vector satisfying

these two conditions:

1 p · e = 0 for each basis element e ∈ N ;

2 v − p is an element of N

Then p is effective.

Proof. Work in the support space of v, which is finite-dimensional Rearrange the basis

so that the intersection matrix is



M A

AT B

,

where M is the negative definite intersection matrix for the subspace N Write p as a

column matrix with respect to this basis:

p =



X Y



Then, since p · e = 0 for each basis element e ∈ N ,



=



0 Z

,

and thus X = −M− 1AY We know that all entries of A and Y are nonnegative (Note that the column vector representing v would likewise include Y.) By Lemma 4.1, all entries of M− 1are nonpositive Thus all entries of X are nonnegative.

The following more technical lemma is akin to Lemma 14.12 of [1], but we give a

more elementary proof

Lemma 5.4 Suppose that N ⊂ W are two special subspaces, with N being negative

definite Suppose there is an effective vector v ∈ V with the following properties:

1 v · e ≤ 0 for each basis element e ∈ N ;

2 v · e< 0 for each basis element in e ∈ W \ N.

Then W is also a negative definite subspace.

Proof We give a proof by contradiction Suppose that W is not negative definite.

Then by Corollary 4.3 there is a nonzero effective and nef element q in W Since N is

a negative definite subspace, q/∈ N Thus v · q < 0, but this contradicts the fact that q

is nef

Here is Zariski’s algorithm for the decomposition of a specified effective vector v.

If v is nef, then the decomposition is given by p = v and n = 0 Otherwise let N1 be

the subspace spanned by all basis vectors e for which v · e< 0 Since v is effective, N1

is a subspace of its support space and hence has finite dimension By Lemma 5.4 (with

W = N1 and N trivial), it is a negative definite subspace Since the restriction of the intersection product to N1is nonsingular, there is a unique vector n1 ∈N1 satisfyingthis system of equations:

n1·e = v · e for each basis vector e ∈ N1.January 2012] ZARISKI DECOMPOSITION 33

By Lemma 5.2, n1is effective Let v1 =v − n1, which by Lemma 5.3 is an effective

vector If v1 is nef with respect to V , then we have found the Zariski decomposition:

p = v1and n = n1

Otherwise proceed inductively as follows By an inductive hypothesis, vk−1 is an

effective vector satisfying vk−1·e = 0 for each basis vector e ∈ N k−1 Let N k be the

subspace spanned by N k−1and by all basis vectors e for which vk−1·e< 0 Again N k

is finite-dimensional By Lemma 5.4 (with N = N k−1and W = N k ), the subspace N k

is negative definite Hence there is a unique vector nk ∈ N k satisfying this system ofequations:

nk·e = vk−1·e for each basis vector e ∈ N k

By Lemma 5.2, nk is effective Let vk = vk−1−nk, which is effective by Lemma

5.3 If vk is nef with respect to V , then the Zariski decomposition is p = v k and n =

n1+ · · · +nk Otherwise vk·e = 0 for each basis vector e ∈ N k, which is the required

inductive hypothesis Since the sequence of subspaces N1 ⊂ N2 ⊂ · · · is strictly

in-creasing and contained in the support space of v, this process eventually terminates Example 5.5 Using the same intersection matrix as in Example 5.1, we apply

Zariski’s algorithm to the vector







Thus the algorithm works in just one step

For the vector

6 NUMERICAL EQUIVALENCE We continue to suppose that V is a vector

space over Q equipped with an intersection product with respect to a fixed basis We say that two elements v and v0of V are numerically equivalent in V if v · w = v0

sat-decomposition, we must have n = n0

Example 6.2 Suppose that V is a 5-dimensional vector space with intersection matrix

Let v = 3e1+e2+e3+e4, and let v0be the numerically equivalent vector 2e1+e5.Then the Zariski decompositions are as follows:

Using the notion of numerical equivalence, we can extend Zariski decomposition to

a potentially larger set of vectors We say that a vector w ∈ V is quasi-effective in V if

w · v ≥ 0 for every element v ∈ V which is nef with respect to V 5In particular eacheffective element is quasi-effective; more generally, any vector numerically equivalent

to an effective vector is quasi-effective

Proposition 6.3 Suppose that M is an intersection matrix for a finite-dimensional

vector space V Then w is quasi-effective in V if and only if w T Mv ≥ 0 whenever Mv

is effective In particular if M is nonsingular, then w is quasi-effective in V if and only

if it is effective.

Proof. The first sentence uses the definitions, together with the previous observation

that a vector v is nef with respect to V if and only if Mv is effective If the matrix is

nonsingular then each effective element can be written as Mv for some nef element v.

Thus in this case w is quasi-effective in V if and only if w Tv ≥ 0 for each effective element v An element w satisfying the latter condition must be effective.

In general, however, there may be quasi-effective vectors which are not effective

In Example 6.2, for instance, the vector w = 7

2e1+ 32e2+ 32e3+ 32e4− 12e5 is

quasi-effective, since it is numerically equivalent to the effective vector 2e1+e5

Here is another example, which shows that the notion of quasi-effectiveness is

“volatile” as one passes to subspaces

Example 6.4 Start with the following(2k) × (2k) matrix:

Use row and column operations to construct an intersection matrix M2k as follows:

beginning at i = 2, replace column 2i by itself plus column 1 minus column 2, and

do the corresponding operation on row 2i; continue this up until i = k Here is an illustration when k = 3:

Let Mj and Pj denote the upper left j × j submatrices (noting that this is

consis-tent with our previous usage when j is even) Note that det M j =det Pj for all j In

particular the matrix Mj is singular if and only if j is even.

Now let V j denote the subspace spanned by the first j basis vectors, and consider

the vector w = e1−e2 If j > 1 is odd, then w is not quasi-effective in V j, since

5 We have heard “quef” as a short form The terminology “pseudo-effective” is also in use.

the matrix is nonsingular and w is not effective If j > 2 is even, however, then w is

numerically equivalent in V j to the effective vector ej; hence w is quasi-effective in

V j (It’s also quasi-effective in V2, being numerically equivalent to e1.)

Proposition 6.5 If a vector w is numerically equivalent to an effective vector, then it

has a unique Zariski decomposition, i.e., there is unique way to write it as a sum of a nef vector p and an effective vector n satisfying conditions (1) through (4) of Theorem

3.3.

Note, however, that the positive part does not have to be effective In particular if w

is nef but not effective, then its positive part is itself

Proof. Suppose that w = v + t, where v is effective and t is numerically trivial, and let v = q + n be the Zariski decomposition of v Putting p = q + t, we see that p and n satisfy the four conditions Conversely, if w = p + n is a Zariski decomposi- tion then v = (p − t) + n must be the unique decomposition of v Thus the Zariski decomposition of w is unique.

For a detailed treatment of Zariski decomposition for quasi-effective vectors (in the

original context, where these vectors represent curves on surfaces), see [5].

7 THE ORIGINAL CONTEXT (CONTINUED) We now resume our informal

account of the original context in which Zariski developed his theory of tion Figure 4 shows two plane curves of degree three The polynomial

decomposi-f (x, y) = (y − x2)(3y − x − 3) (7.1)defining the curve on the right can be factored, with the visible result that the curve is

the union of a line and a conic (a curve of degree 2); we say that these are the

compo-nents of the curve The other curve has a single component: we call it irreducible.

Figure 4 Two plane curves of degree three The curve on the left is irreducible, while the curve on the right

has two components In the next figure, we show what happens if this curve is blown up at the indicated point.

Suppose that C is defined by

f (x, y) = ( f1(x, y)) n1( f2(x, y)) n2· · ·( f k (x, y)) n k,

where each f i is an irreducible polynomial and thus defines an irreducible curve C i,

one of the components of C We associate to f the formal linear combination

calling it the divisor of f Note that all coefficients are nonnegative; thus this is an

effective divisor For example, the divisor associated to the polynomial in (7.1) is C1+

C2, where C1 is the conic and C2 is the line A similar recipe works for any other

surface For an effective divisor in the plane we define its degree to be the degree of the defining polynomial; thus the degree of P n i C i is P n i deg C i

Given two distinct irreducible curves C and D on an algebraic surface, they have

an intersection number C · D Intuitively, this is the number of points in which the

curves intersect, and indeed in many cases that is its precise meaning, but to definethis number carefully one needs to consider exotic possibilities, so that for example

a tangency between the curves gets counted as “two points” (or even more) Thus to

an algebraic curve we can associate a matrix recording the intersection numbers of itscomponents In the plane6the intersection number between curves of degrees c and d

is cd, a fundamental result of ´Etienne B´ezout dating to 1776 Hence for the curves in

Figure 4 these matrices are

For distinct irreducible curves, the intersection number is always a nonnegative teger Thus the off-diagonal entries in these matrices are nonnegative, and they are

in-intersection matrices as defined in Section 3 The diagonal entries are self-in-intersection

numbers In our example we have calculated them using B´ezout’s formula, but on otheralgebraic surfaces one has the startling fact: a self-intersection number may be nega-

tive! The simplest example of this comes from a process called blowing up, in which a given surface is modified by replacing one of its points p by an entire curve E having self-intersection number −1, called an exceptional curve (This process is the basic

operation of the “birational geometry” to which Mumford alludes in the quotation in

Section 2.) Each irreducible curve C on the original surface which contains p can be

“lifted” to a curve on the new surface meeting E We will abuse notation by referring

to the lifted curve with the same letter C, but a remarkable thing happens to C · C: it

is reduced in value (typically by 1) For example, if one blows up the plane at one ofthe two intersection points shown in Figure 4, then the intersection matrix for the two

original components and the new curve E is as follows:

The definition of intersection number between a pair of irreducible curves extends

by linearity to any pair of divisors (effective or not) If one has the intersection trix, then the calculation is simply a matrix multiplication For example, the self-

ma-intersection of the divisor C1+C2associated to the polynomial in (7.1) is



=9

Note that the result is the square of its degree

6 As in Section 2, we mean the complex projective plane.