The Delta Epsilon, Issue 1

CONTENTS 1

Contents

Letter From The Editors 2

Interview with Professor Niky Kamran 3

Group Theory and Quantum Chromodynamics 6

Predicting the Lifespan of AIDS Patients with Survival Analysis 10

Benford’s Law 14

Generators of SL(2, Z) 16

Mathematical digest 19The Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . 19Linear Differential Equations and the Minimal Polynomial . . . . . . . . . . . . . . . . . 19The first isomorphism theorems for rings and groups . . . . . . . . . . . . . . . . . . . . 20Vanilla Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20The Change of Variables Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

Interview with Professor Henri Darmon 24

A Field of Six Elements? 29

Euler’s Brick 30

Ten Proofs of the Infinitude of Primes 34

Living without Math 36

Interview with Professor Nilima Nigam 37

The Birth of Quaternions 39

Getting Acquainted with Φ 42

Book review: George Polya’s How to Solve it? 44

The Adventures of A and B 45

Crossnumbers 47

Credits 48

Acknowledgements 48

The δelta-ǫpsilon McGill Mathematics Magazine

2 Letter From The Editors

Letter From The Editors

Dear mathematics students of McGill,

One fine morning last April, as the snows were melting, the birds were singing and the ODEexam was nearing, a few math undergraduates turned envious eyes towards the Faculty of Artsand asked “Why do they have journals and not us? Aren’t we part of that faculty too?”. Half ayear later, you’re holding the pilot edition of The Delta-Epsilon. Ain’t time remarkable.

This magazine is intended as a place to publish summer research by undergraduates, to learnwhat your professors or fellow students are doing, and to share ideas about course material. Thereare also articles on the history and culture of mathematics, the required jokes and book reviews,and some interesting puzzles that may keep you busy in (or out) of class. Some of the articles aremore involved than others, and some are aimed at people who’ve taken a specific course; however,our main objective was to produce a magazine which you could keep on your bookshelf and consultfrom time to time as the content becomes relevant (or understandable) to you.

Also, we would like to encourage all undergraduates to think about writing an article for nextyear’s edition, scheduled for release in September, 2007.

So enjoy the articles and let us know what you think.

The Delta-Epsilon Editing Board

Letter from SUMS

On behalf of the Society of Undergraduate Math Students (SUMS) I would like to thank thestaff of The Delta-Epsilon for their hard work and commitment to putting together this excitingundergraduate mathematics journal for the McGill mathematics community. It’s hard work fromstudents such as these that enriches the McGill mathematics experience.

SUMS is an organization aimed at improving the undergraduate experience for mathematicsstudents. We offer many services including the organization of lectures, social events, tutorials, atutor list, a website and, starting this year, a notes scanning project.

I would like to encourage everyone with even a passing interest in mathematics to come joinus at 1B20 in Burnside Hall in our lounge area anytime you want to connect. If you need helpwith your work, chances are you’ll find someone there who has been through the same trials andtribulations, and will be willing and able to help you out. On the other hand, if you’ve finishedyour work for the day and just want to unwind and play games or discuss current topics, we’rethere for that too.

Please check out our website at http://sums.math.mcgill.ca

Marc-Andre RousseauSUMS President



The beauty of math:

An Interview with Professor Niky KamranAlexandra Ortan

“What makes math interesting is what’s unexpected and harmonious at the sametime. It is the same in music. The interesting parts in a piece of music are thesingular points. It’s when things modulate, when something happens that you didn’texpect. In math, if you just stay in familiar territory where you can guess the answerto every question, it’s rather boring.” And Professor Niky Kamran, whose passionnext to math is playing the violin, is well-placed to know how interesting math canget when you step out of that familiar territory.

What got prof. Kamran interested in math-ematics is precisely this æsthetic component.“I thought it was beautiful” he says about thewhole new body of mathematics he encounteredin high school. “The thing that really amazedme when I was being taught the rudiments ofcalculus, was the realization that you could com-pute exactly the velocity of an object that’smoving and twisting in space, by knowing its po-sition as a function of time. I thought that wasamazing. The point is that there is the physicalintuition that you have, and then there comesboth a set of hypotheses that you formulate anda framework. Within that framework, you’reable to discover and confirm facts. The processwhereby you do this is fun, because it’s like play-ing a game. You’re given the rules, you havean objective (or maybe you don’t have an ob-jective, but you surmise an objective) and thenyou use the rules of the game to get there. Thenthere are the facts that you’ve been taught andthat you’ve in some cases discovered, which arereally beautiful, and that’s the æsthetic compo-nent to it. It’s the combination of the two whichappealed to me.”

Thus, it is not surprising to find prof. Kam-ran, many years later, still trying to figure outhow objects move in space, though we are nolonger talking about the rigid bodies and New-tonian space of our high school years, but ratherabout wave-particles and curved space-time –a notch or two higher in complexity. In fact,when asked what particular branch of mathe-matics he is working on, prof. Kamran repliesthat his interests lie at the confluence of ge-ometry, analysis and a bit of algebra. “What Ilike to do, for example, is to look at differentialequations (DEs) through the prism of geome-try, to see what geometry can tell me about theproperties and the attributes of the DE. I’m also

TRIVIAErdos number: 3Favourite millennium problem: “TheRiemann hypothesis. It really fascinates mebecause the statement is so easy to under-stand; every math student has to be fasci-nated by that. But the millennium problemwhich has greater relevance to what I do isthe problem of mass gap for Yang-Mills the-ories, because I’ve been thinking about themass gap in a much simpler context.”

interested in looking at DEs that originate inphysics, particularly in relativity, and seeinghow the concept of the curvature of space-timewill affect the behaviour of the solutions to a


4 Interview with Professor Niky Kamran

DE governing, for example, the propagation ofwaves. So there are problems that have theirroots in physics, but that I look at from theperspective of geometry, analysis and algebra.”

“Here’s a simple question on which I workedand for which my collaborators and I got a pre-cise answer. Suppose you’re in ordinary flatMinkowski space (a 4 dimensional metric spacewith complex time-dimension) and you look atthe propagation of a wave. What you can show,and you learn this in a PDEs course, is thatthe amplitude of the wave at any location ofspace will decay proportionally to t−3/2. Nowif you’re in Kerr geometry (see framed text),you wonder ‘is the space-time curvature goingto affect the rate at which your wave is goingto decay?’ We expected that the rate would beaffected by the curvature, but we didn’t knowhow. So the bets were open. For a long time, wethought that the amplitude should decay fasterthan in Minkowski space, but the answer is thatit decays slower. In fact the rate of decay ist−5/6. You can justify it by some non-rigorousarguments, but to give a real solid mathematicalproof takes some effort, and this is what we didin some of our papers: we quantified preciselythe effects of space-time curvature on the prop-agation of certain types of waves in the Kerr-geometry.”

“Now this result works for the simplest kindof waves, scalar waves. A really interestingproblem is to know what happens if you lookat a wave that corresponds to a gravitationalperturbation. So you’ve got the Kerr blackhole and then you have a gravitational fieldwhich is felt by the space-time geometry of theKerr black hole through incoming gravitationalwaves. Things are going to interact, the event-horizon is going to go hay-wire, all sorts of thingsare going to go wrong. The question is to knowwhether eventually this thing will settle downor not. This is called the black hole stabilityproblem, and it’s one of the problems I’m work-ing on. It’s a very fascinating problem, and formathematicians it’s a real challenge.”

So now that you’ve found a fascinating prob-lem, you might wonder how to go about solv-ing something which, at least to the neophyte,seems out of reach. “Well, typically when you doresearch in mathematics, you don’t take a hugeleap in one shot” says prof. Kamran. “You buildon a very large body of known results. I don’twake up one morning and while shaving decide

KERR GEOMETRY

When a sufficiently massive star exhaustsits nuclear fuel and collapses gravitationally,you sometimes end up with a black hole. TheKerr geometry is the exact solution of theEinstein equations of gravitation, which de-scribes the space-time geometry outside therotating black hole in equilibrium. So as aspace-time geometry, it’s axi-symmetric, be-cause there is rotation. It’s stationary be-cause it’s an equilibrium configuration, andit describes a black hole because it has anevent horizon, which hides a singularity. Butwhat’s amazing is that it’s an exact solution,i.e. a solution that you can write in closedform. What this solution means is that whenthe star collapses and you end up with a blackhole, all the complicated degrees of freedom ofthe star have been radiated away and you’releft with mass, angular momentum and charge- if it is charged. That’s an absolutely amaz-ing result. It’s the uniqueness theorem forthe Einstein equations. The history of itsdiscovery is very interesting. The Kerr ge-ometry was discovered by a mathematicianwho was interested in Lorentzian geometry infour dimensions, Roy Kerr. He discovered itin 1963, from purely mathematical premises.What happened afterwards is that in 1968and 1972, Werner Israel and Brandon Carterproved that the parameterized family of Kerrsolutions is the unique solution of the bound-ary value problem for the Einstein equationsthat corresponds to black hole equilibriumstate. So when you’re observing a black holein equilibrium, it’s characterized uniquely bythese parameters (mass, angular momentumand charge if applicable).

that I’m going to solve the stability problem forthe Kerr black hole. You spend a lot of timestudying the literature and then you let thingssit for a while. Then typically what you needfirst is a strategy. That’s in some sense themost important part. So you try to map outwhat would be the main steps that you’d haveto go through. Once that is done and you’refairly confident that your strategy is the rightone, you pick up your shovel and you start dig-ging. You try to go from step one to step twoin a process which will take you through onehundred steps. And sometimes you take step 34for granted to see what happens in step 71 and



if it looks like everything else is going to workthen you move forward and you come back totry to fix things.” As to the part played by in-tuition in the whole process, prof. Kamran de-clares it to be crucial. “You work with intuition,and intuition is based on the wealth of experi-ence, it doesn’t come directly from heaven. Itcomes from a collection of facts which are partof your memory, things that you’ve seen, that’swhat intuition is built upon. You can’t do anymath without intuition. Otherwise, you couldjust feed the statement of a theorem into a com-puter and the computer would go on and proveit for you. There is a branch of theoretical com-puter science called automated theorem provingwhere this type of process is looked at, but forresults of the type that I’m interested in, it’suseless.”

As most of our readers are likely to be youngaspiring mathematicians, undertaking yet an-other (or first) year on the long road of mathe-matical training, and are perhaps still not surewhat it is they’ve gotten themselves into, weasked prof. Kamran what it’s really like to bea mathematician, and whether the urban leg-ends speaking of a reclusive, solitary being haveany foundation. “Not necessarily” he answers.“More and more people do mathematics in col-laboration, but I think it’s fair to say that whenyou think hard about a problem, you’re on yourown. The fun part however comes when you’re

able to collaborate with someone who’s exper-tise is complementary to your own, then you re-alize that what you’ve been thinking about res-onates into a broader context. You can thenbring two points of view together and you canprove good theorems and solve interesting prob-lems. So there is a reclusive, solitary componentto it, but it’s not the whole story by a long shot.Although mathematicians can be a bit strangesometimes, there is a whole social component tothe activity.”

However prof. Kamran warns the aspir-ing mathematician that “First and foremost youhave to love math. If you don’t love it, you won’tsurvive it. It’s serious work, it’s competitive.You have to sit down and spend a lot of time inapprenticeship. Secondly, you have to be able tobe realistic and see if this is really for you or not.That can only come if you’re a good apprenticeand you learn to do mathematics. If it comes toyou with some effort, but the pleasure exceedsthe pain and you’re getting something valuableout of it, then it’s well worth it. It’s a nobleactivity. It’s the same thing for becoming a mu-sician in some sense. I’ve seen many people whowere good enough to become professional mu-sicians but who’ve lost the sparkle and love forthe activity, and that’s one path to depressionand unhappiness. It’s the same thing for math.You need to keep loving doing mathematics.”


6 Group Theory and Quantum Chromodynamics

Les Confessions, ou de la theorie des groupes en chromo-

dynamique quantiqueMichael McBreen

This article divides into two parts: a popularized introduction to gauge field theo-ries for the mathematics student, followed by a technical exposition of an aspect ofthe authors NSERC-sponsored (USRA) work on gauge theory. None of the resultspresented here are original.

An Introduction to Gauge Theory

For pedagogical reasons, I will avoid mentioningquark fields until after the introduction of gaugesymmetry. Any resulting inaccuracies shouldhopefully be rectified at that point.

The atom is composed of electrons and a nu-cleus, and this nucleus is made of protons andneutrons, collectively known as nucleons. Ac-cording to the Standard Model1, the nucleonsthemselves are made of particles called quarks.There are many species of quarks, each with itsown mass, charge and so forth, but from nowon you can assume I’m always talking about thesame species.

Any given quark has a property called“colour”, which corresponds to a direction in anabstract three-dimensional “colour-space” iso-morphic to C3. I will sometimes say that a quark“points” in some direction, meaning its colouris given by that direction, or write q to indi-cate the corresponding unit vector. All coloursor orientations are equivalent in that only therelative orientation (i.e. the inner product (q2,q2)) of two quarks will affect their interaction.In other words, there’s no absolute orientation,just as there’s no absolute position, orientationor velocity in ordinary space-time. This meansthat if we magically“rotate”2 every quark by thesame angle, no observables will change: noth-ing will get hotter or colder, go faster or slower,get bigger or smaller. We call the freedom tovary a parameter of our model without chang-ing the physical situation a symmetry. It es-sentially means that the parameter is superflu-ous: it’s an artifact of our formalism. We callthe corresponding group of transformations thesymmetry group (SU(3) in this case). For in-stance, consider the interaction between two iso-

lated stars: only the relative distance betweenthe stars matters, not their “absolute” position.The latter is therefore a superfluous parameter,and the symmetry group is the group of trans-lations.

In fact, quarks possess a stronger form ofsymmetry called local gauge invariance. Notonly can we rotate all quarks by the same angle,we can rotate quarks at different positions bydifferent angles (i.e. apply a gauge transforma-tion). But wait - didn’t I claim that the relativeorientation of two quarks is important? In fact,there’s an object called the gauge field U , whichpermeates all space, and which“remembers”therelative orientation of the quarks when we ro-tate them. More precisely, when we rotate thequarks we must also transform this field - I’ll ex-plain how in a second - and this transformationsomehow cancels out the rotation of the quarks.

You might ask why we bother postulatinggauge invariance, if we also need to postulatea field that cancels out any gauge transforma-tions we might make. The justification is thatthis gauge field can be observed in nature. Infact, if we require U to evolve over time accord-ing to a certain equation of motion, we find thatit reproduces the behaviour of the strong forcethat binds quarks together. Better still, it turnsout that all fundamental forces seem to originatefrom gauge fields.

But back to U . What sort of a field do weneed to cancel out gauge transformations? Saywe have two quarks pointing in different direc-tions, and we rotate them so the directions co-incide. We want to recover the initial relativeorientation, so we’d like the field to somehowencode a rotation that brings the vectors backto their initial (relative) positions. The field Udoes this as follows. U = U(p) is a function

1The Standard model is the currently accepted description of particles and forces, though rumour has it thatcracks are showing in the theory, and a replacement will urgently be needed.

2The analogue of the group of rotations relevant here is the group SU(3) of unitary operators on C3 withdeterminant 1.



that associates a gauge group element g (a rota-tion) to each path p in spacetime. If p2 beginswhere p1 ends, then U(p1)U(p2) = U(p1 · p2),where p1 ·p2 is the path obtained by going alongp1 first and p2 second. U(ptrivial) = 1 whereptrivial is the trivial path from x to x, and fi-nally U(p) varies continuously as p is length-ened. Basically, U lets us compare two quarksat different points x and y by taking the quark atx, rotating it by U(px→y) where px→y is somepath3 from x to y, and then taking the rela-tive orientations. When we rotate the quarks atx and y by R(x) and R(y), U(px→y) becomesR(y)U(px→y)R(x)−1, so that we’ll get the sameanswer if we compare our quarks after the rota-tion:

(U(px→y)vx, vy) = (R(y)U(px→y)R(x)−1

R(x)vx, R(y)vy).

(0.1)

Now, I have a confession to make: the pictureI’ve been painting isn’t quite right. In fact,rather than having many point-like quarks fly-ing about and interacting, we really have one bigobject called the quark field ~φ, which (roughly)associates a quark density with each point inspace.4 What we would call a quark is a bump inthe field (in the density) that travels along, col-lides with other bumps, bounces off them, etc.,like ripples on a lake. The time evolution of ~φis governed by an equation linking its spacetime“slope”and its absolute value at each point. Forinstance, an isolated bump will flatten out intoexpanding ripples, while a long straight wavewill travel along at a constant speed. Why is ~φ avector field? Remember that each quark pointsin some direction. We therefore have three sep-arate fields, φ1, φ2, φ3, giving the quark densityalong each of the colour axes, and we choose togroup them into the vector field ~φ = (φ1, φ2, φ3).Local gauge invariance really means that youcan rotate this vector by different angles at dif-ferent points without affecting the physical stateof the system.5

As before, the gauge field U(p) allows localgauge invariance. The description of U(p) aboveis still valid, but we need only consider infinites-imal paths pr→r+dr since the motion of the fieldis governed by local laws.6 We avoid worry-ing about the path independence of U(pr→r+dr)

by always choosing the infinitesimal “straight”path. Of course, the gauge field itself must begauge invariant (i.e. all measurable properties ofthe field should be invariant under local gaugetransformations).

One more confession: what I’ve described upto now is a classical field, but we’re really afterthe quantum field. The quantum field, muchlike the quantum particle, has a wavefunctionalΦ(U(p)) that associates a probability to eachclassical parameter (we’ll simply call it a wave-function). In this case, the function U(p) is theclassical parameter. The classical parametersof a system are grouped into subsets of com-patible parameters, meaning they can simulta-neously have definite measurable values, whilenon-compatible parameters satisfy a generalizedversion of the Heisenberg uncertainty principle.

That’s gauge field theory for you – nowcomes my own summer research. We studiedvarious gauge fields in a “vacuum”, i.e. with noquarks present. The goal was to take a fieldwith state U0 at time t = 0, and determine UT

at t = T . We have an equation for this, but wecan’t solve it, so we make three major simplify-ing approximations and modifications:

1. Space is discrete. We replace the contin-uum of space with a lattice of discrete points(x, y, z) with spacing a. The field U(p) is de-fined on the set of edges (a edge is a straightpath p joining two adjacent points). This way,each edge is associated with a group element g ofSU(3). The wavefunction Φ(U(p)) associates aprobability to each such configuration of the lat-tice. If we were including quarks in our model,we would define the field ~φ on the vertices.

2. We neglect the interaction of the fieldwith itself. In other words, any excitations ofthe field (waves or bumps) will travel straightthrough each other rather than bounce or devi-ate. To make this simplification, we remove allreferences to U(p) as such (the potential terms)from the equation of motion, leaving only thederivatives of U(p) (the kinetic terms).

3. We use Euclidian time. The equations ofmotion for the wavefunction involve a time vari-able t. We analytically continue the functionsof this variable to the complex plane, so thatt → z, and evaluate the functions at z = it

3Don’t worry about path independence - I’ll get to that later.4I’ll explain the vector notation in a minute.5We do require that the rotations change continuously as we go from point to point.6In other words, dφ

dt|x depends only on an infinitesimal spacetime region around x.


8 Group Theory and Quantum Chromodynamics

rather than t. We lose some information thisway, but we can still obtain many interestingproperties such as the energy of the field.

Within this simplified model, the equations ofmotion can be solved completely for importantgauge fields such as the SU(3) field discussedabove, and the electromagnetic and weak fields,whose gauge groups are respectively U(1) andSU(2).

LexiconSymmetry group: Consider a model of aphysical system in which every physical stateis given by a (possibly infinite) number of pa-rameters, such as position x, speed v, temper-ature T , etc. A symmetry of the system is atransformation of the parameters that leavesthe physical state invariant. The set of allsymmetries of the system form a group undercomposition, called the symmetry group.Gauge invariance and gauge group:Consider a model which associates a set ofparameters to each point in space - or space-time, if you prefer. If we can transform theparameters at each point independently (withcertain restrictions) while leaving the physicalstate invariant, we say the system is gauge in-variant. The associated group of symmetriesis the gauge group.U(1): The multiplicative group of all com-plex numbers with norm 1. It is isomorphic tothe group of rotations of the real plane aboutthe origin.SU(2): The group of unitary linear trans-formations with determinant 1 acting on C2.This can be thought of as a complex versionof the group of rotations in 3D.SU(3): The group of unitary linear transfor-mations with determinant 1 acting on C3.Group representation: A representation isan action of the group on a vector space V , i.e.a homomorphism ρ : G → L(V ). The homo-morphism property notably implies ρ(g1g2) =ρ(g1)ρ(g2). If V is an n-dimensional complexspace, ρ : G → GLn(C) associates a ma-trix ρ(g) with each group element g. Givena vector in Cn, we can let the group act onit through some n-dimensional representationρ, in which case we say the vector belongs toρ.There are a number of distinguished represen-tations. For instance, the trivial represen-tation is the homomorphism to GL1(C) thatsends G to the identity. The left regular

Lexicon - continuedrepresentation is the action of G on thespace of functions on G defined by ρreg(h) :f(g) → f(h−1g). It’s an infinite dimensionalreducible representation.Irreducible representation: A representa-tion that does not preserve any non-trivialproper subspaces of V . The name irreduciblecomes from the fact that non-irreducible (re-ducible) representations can often be ex-pressed as “sums” of irreducible representa-tions.Representative functions of a represen-tation: The functions ρ(g)ij : G → C thatgive the matrix elements ρ(g)ij of the ma-trix ρ(g) ∈ GLn(C). The Peter-Weyl The-orem tells us the representative functions ofthe irreducible representations are an orthog-onal set of functions spanning the full spaceof “nice” functions on G. It applies to all com-pact Lie groups, including the gauge groupswe’re working with.

Harmonic Analysis on a Group

Manifold

We take space to be a cubic lattice with sidelength N+1 vertices vi (or N edges lij). Ourfield U(p) is defined on the edges.

The wave function Φ : G × G · · · × G → C

associates a complex number with each classi-cal configuration of the lattice (there’s a copy ofG in the function domain for each edge in thelattice). We denote a generic classical configu-ration, which associates some group element gij

to each edge lij on the lattice, by−−→U(p). We write

Φ(−−→U(p)) for the wavefunction, and the probabil-

ity density of a configuration is |Φ(−−→U(p))|2. The

set V all such wavefunctions is a vector space



over C, with inner product defined as

(Φ1, Φ2) =

∫

G

dg12

∫

G

dg23

∫

G

dg34 · · ·∫

G

dgnm

Φ∗1(g12 × g23 × · · · gnm)Φ2(g12 × g23 × · · · gnm)

where the star denotes complex conjugation.The integration measure dg is called the Haarmeasure, and is invariant under left and rightgroup composition: d(hg) = d(gh) = dg whereh is a fixed group element.

A local gauge transformation Λ3 at vertexv3 sends g23 to g23Λ3, g43 to g43Λ3, etc. – thegauge group element at each edge connected tovertex i is transformed, as we can see from (0.1).Λ3 accordingly sends Φ(· · · × g23 × g43 · · · ) toΦ(· · · × g23Λ3 × g43Λ3 · · · ). If Φ is gauge invari-ant, then it should be sent into itself. V containsboth invariant and non-invariant wavefunctions,but the non-invariant ones are artifacts of ourformalism and are not found in nature. Whatwe are really interested in is the subspace Vinvar

of gauge invariant functions. I wont explain howto find this subspace here, but the hardy readeris referred to [4].

The time evolution of a wavefunction (itsequation of motion) is given by

Φ(t = T ) = e−kTP

ij∆ij Φ(t = 0) (0.2)

where k is a constant and ∆ is a generalizedversion of the Laplacian called the Laplace-Beltrami operator on the group manifold. Itcan be thought of as measuring the convexityof a function of the group. Each ∆ij acts on asingle edge lij .

We want to find a basis of “harmonic”wave-function with simple time evolution properties.This would allow us to find the evolution of anarbitrary wavefunction by decomposing it intoharmonics. Luckily, the time evolution operatoris self-adjoint, so we can find a basis of eigen-functions for it spanning V .

To construct this basis, we need a cen-tral result of group representation theory: thePeter-Weyl Theorem. The PWT generalizes thetheory of Fourier transformations to arbitrarygroup manifolds. Fourier analysis tells us thatany sufficiently nice function of the real inter-val [−2π, 2π] can be expressed as an (infinite)sum of orthogonal harmonics or sinusoidal func-tions. That is, the sinusoidal functions are insome sense an orthogonal basis for the space of

functions with that domain. The PWT tells usthat any nice function of the group manifold canbe expressed as an (infinite) sum of representa-tive functions of irreducible representations ofthe group, and that these representative func-tions are orthogonal.7 For the gauge groupsU(1), SU(2) and SU(3), we have a simple clas-sification of all non-isomorphic irreducible rep-resentations.

Now, the Laplace-Beltrami happens to be aCasimir operator for the group G, meaning thatit commutes with all ρ(g), no matter the repre-sentation. By a simple theorem of group repre-sentation theory called Schur’s Lemma, any op-erator that commutes with an irreducible repre-sentation of a group is a multiple of the identityon that representation. Using the fact that rep-resentations preserve the group action, we caneasily show that the representative functions ofany irreducible representation ρ transform un-der the action of G as the representation ρ itself.According to Schur’s Lemma, these are eigen-vectors of all Casimir operators, including theLaplace-Beltrami operator.

We therefore have a basis of eigenvectors of∆ij spanning the functions on G. Φ is a functionof the direct product G×G× · · ·×G, which forphysical reasons factors into a product of func-tions of G. We can hence decompose any Φ intoa sum of products of representative functionsusing the PWT, and each product is an eigen-vector of the sum of Laplace-Beltrami operatorsappearing in (0.2). In other words, we have afull basis of eigenvectors labeled by the repre-sentations and corresponding matrix indices ap-pearing in the product. This was the objective.

References

[1] Montvay, I, Munster, G Quantum Fields ona Lattice , Cambridge University Press, Cam-bridge (1997)

[2] Creutz, M Quarks, Gluons, and Lattices,Cambridge University Press, (1986)

[3] Hall, B Lie groups, Lie algebras, and repre-sentations. An elementary introduction., Grad-uate Texts in Mathematics, 222. Springer,(2003)

[4] Baez, J Spin Network States in Gauge The-ory, arXiv:gr-qc/9411007 v1 2 Nov 94

7See the lexicon if you’re not familiar with group representation theory.


10 Predicting the Lifespan of AIDS Patients with Survival Analysis

Predicting the Lifespan of AIDS Patients with Survival

AnalysisMireille Schnitzer

To infer expected survival time after the onset of a disease such as acquired im-mune deficiency syndrome (AIDS), a researcher will recruit a number of patientsand eventually analyze their sample failure times. This paper will briefly describethe difficulties that arise when attempting a statistical analysis of such a study, andwill explain one method of dealing with right-censorship. The product limit esti-mate (developed in 1958 by Kaplan and Meier) will be used to demonstrate a way ofapproximating expected failure time.

Survival is a common concern in many sci-entific, social and industrial disciplines. Forinstance, a biologist might consider the aver-age lifetime of wild deer, or an economist couldstudy how long the average person remains onwelfare. The field of survival analysis is a branchof statistics that evaluates the results of suchstudies and the data loss that might occur. For-mally, survival analysis is the study of measur-ing a time-to-event, or the time lapse betweentwo events.

One of the most important applications ofsurvival analysis is in the medical field, where itplays a fundamental role in determining the esti-mated survival time after the onset of a disease.Such studies are designed to work in conjunc-tion with hospitals or clinics where patients areasked to participate.

Let’s consider the example of measuring theexpected survival time after a patient developsAIDS (acquired immune deficiency syndrome).We will work with a hospital and acquire a sam-ple of patients, recording their times of deathas the study progresses. The notation generallyused is as follows: we will observe n individualswhose failure times (random survival times afterdeveloping the disease) are the random variablesT1, ..., Tn. These will be assumed to be indepen-dent and identically distributed with commondensity f(t) for time t. Note that we relabel thetime of onset as t0 = 0 so that the time of deathcorresponds to the failure time.

To simplify our analysis, we will assume thatthere is no bias in the way the patients are se-lected. This is usually a bad assumption to makein medical studies (as it is generally more likelyto recruit patients with longer life spans), butit is necessary to simplify our analysis at thislevel. We will also assume that every patienthas a known time of onset, such that we know

the time at which they developed the disease.The form of data loss that we will be con-

cerned with is called right-censoring and it oc-curs when we never learn the failure time of anindividual. The only information we have is thetime at which we lose the individual, who isnow referred to as being censored. This mightarise if a patient chooses to end his visits to anAIDS clinic so that their previous doctor can nolonger follow them. We will never know theirfailure time, but we do know a time at whichthey were still alive. Censoring could also occurif the study ends before all of our patients havedied. In any case, a censoring at time a meansthat the only information we have is that Ti > afor the ith individual.

A New Function

In statistics, we’ve been taught to find the cu-mulative density function, P (X ≤ x). But insurvival analysis, it is more meaningful (and,eventually, neater) to study the survivor func-tion:

F (t) = P (T > t)

which is the probability of an individual surviv-ing past some time t. In the discrete case, wherewe have failure times a1, a2, ...ar, this functionbecomes

F (t) = P (T > t) =∑

i|ai>t

P (T = ai).

The characteristics of the survivor function in-clude:

1. The function is non-increasing (it cannot bemore likely to survive past a later time)

2. F (0) = 1 (everyone is surviving initially)

3. The limit approaches zero (at time infinity,there are no survivors)



Clearly,

F (t) = 1 − P (T ≤ t)

so that

dF (t)

dt= −f(t)

where f(t) is the distribution function for T .

Since we are concerned with estimating thisfunction using the data from a study, we mustuse empirical functions. Before we develop moresophisticated machinery, consider the situationwhere there is absolutely no data loss. To es-timate the survivor function, we calculate theproportion of patients surviving after time t. Todo so, consider a discrete model where the fail-ure times t1, t2, ...tr are specifically the observedtimes where we have recorded a patient’s death.Then, at any given time ti, we can compute thenumber of deaths, di, and the number of sur-vivors immediately after that time, ni.

So, the empirical survivor function is

Fn(t) =no. ti > t

n

(which, you might notice, is also 1 − Fn(t), theempirical distribution function).

The empirical survivor function is a stepfunction that begins at 1 and decreases by di

nat every ti. When the population has expired,the curve is at zero.

To illustrate, suppose we have observed 15individuals (Table 1) and witnessed each oftheir failure times in months (there is no dataloss). Please note that these numbers have beenarbitrarily selected and do not represent realmeasures of survival.

We can measure the survivorship at anygiven time simply by counting. Hence, we getthe empirical survivor function in the next fig-ure.

Notice that all members of the populationsurvive at time zero, and that by 401 months,all have died. The average (mean) failure timeis 163.87 months.

0 100 200 300 400

0.0

0.2

0.4

0.6

0.8

Empirical Survivor Function for AIDS patients

Time (T) in months

Pro

po

rtio

n s

urv

ivin

gFrom the empirical survivor function, we can

derive the empirical failure distribution by sim-ply noting that P (T = ti) = F (t−i ) − F (ti) =di/n where F (a−

i ) = limt→a−

i. That is, the

probability of failure at any given time is liter-ally the height of the step in the survivor func-tion at that time (it’s how many patients diedafter that many months).

Incorporating Right-Censorship

Assuming that we can collect every failure timeis clearly incorrect, and we will now incorpo-rate the possibility of right-censoring into ourscenario. We have a pair of times for each indi-vidual; a failure time (Ti) and a censored time(Ci), and we assume that these are independentof each other. The independence here meansthat, for instance, patients do not change hospi-tals if their condition worsens (and their failuredraws near). We only actually observe one ofthese times.

The notation is then as follows: we watch in-dividuals X1, ...Xn assumed to be independentand identically distributed with common prob-ability distribution function f(t). For each ran-dom variable, we observe the data (δi, Xi). Instudies with censoring, δi refers to the indicatorthat equals one if the individual’s failure is ob-served, and zero if the individual is censored. Xi

is then either the failure time Ti or the censoredtime Ci (the last time the individual was knownto be alive), depending on which information wehave, so that Xi = min(Ti, Ci).

Using this notation, we can use the likeli-hood to develop a maximum likelihood estima-


12 Predicting the Lifespan of AIDS Patients with Survival Analysis

tor for the survivor function for a parametricmodel. In order to build this estimator, we makethe assumption that the censoring mechanismcarries no information about the parameter. Inother words, any function of the distribution ofcensoring is a constant with respect to θ, theparameter.

We observe the pair (Xi, δi) for every indi-vidual, and we consider the likelihood of suchan observation. If a failure is not censored, itscontribution to the likelihood is

Pθ[Xi = t, δi = 1]

where θ is the parameter. An uncensored obser-vation means that Ci is larger than Ti (which isactually observed), so the corresponding contri-bution to the likelihood is,

Pθ[Ti = t, Ci > t].

Now, using the assumption that the censoringscheme is independent, this is equal to,

f(t, θ)Gi(t)

where f is the p.d.f. for the failure time T andGi is the survivor function of the censored times.Recall that we’ve assumed that Gi holds no in-formation about θ.

Now, suppose that we are considering a fail-ure that is censored. The resulting probabilityof such an occurrence is

Pθ[Xi = t, δi = 0]

which, through the same reasoning, correspondsto,

Pθ[Ci = t, Ti > t].

As before, this gives us

gi(t)F (t, θ)

where gi is the distribution function of the cen-soring random variable for the ith individual.

So now we can construct the likelihood func-tion:

L =

n∏

i=1

P [Xi = ti, δ = δi]

=

n∏

i=1

(P [Ci = ti, δi = 0])1−δi ×

×(P [Ti = ti, δi = 1])δi

∝n∏

i=1

f(ti, θ)δiF (ti, θ)

1−δi .

The last line is proportional due to the assump-tion that the censoring is uninformative with re-spect to θ. It then does not influence the max-imization of the function (since gi and Gi areconstant with respect to θ).

This demonstrates that the contribution ofthe censoring to the likelihood function is of theform P (T > t). The contribution of a normalobserved (uncensored) failure is P (T = t).

Now, consider the results of a real study. Wehave t1 < t2 < ... < tk which are observed (un-censored) failure times. Then, at each of theseobserved failure times, we might have more thanone failure, so we refer to the number of deathsat time tj as dj . In between these observed fail-ures, there are a number of censorings that mayoccur. We will refer to the number of censoringsin the interval [tj , tj+1] as mj . These censoredtimes are tj1, ..., tjmj

for j = 0, ..., k, t0 = 0 andtk+1 = ∞. The number of individuals at riskjust prior to time tj is nj (nj is equal to all fu-ture deaths and censorings).

Recall that

P (T = ti) = F (t−i ) − F (ti)

so that the proportional likelihood can be writ-ten as

L ∝k∏

j=0

{(F (t−j ) − F (tj))dj

mj∏

l=1

F (tjl)}.

For each observed failure, we must run throughand multiply each censored contribution in theinterval between failures.

To maximize this function for the parame-ter, we first find the non-parametric MLE forthe survivor function that maximizes L. Thisfunction is then substituted into the likelihood.This simplifies the expression so that it can thenbe maximized in terms of its parameter. In theend, our MLE is the empirical survivor functionwritten in terms of ni and di. So, the productlimit estimate or Kaplan-Meier estimate is

F (t) =∏

j|tj≤t

1 − dj

nj.

(Note that the fractiondj

njis specifically the

proportion of failures at every observed failuretime.)

This function looks very much like the em-pirical survivor function and one can use it in



the same way. One difference is that the Kaplan-Meier estimator may not go to zero if the last ob-servation is a censoring. There are several sug-gested ways of dealing with this, but the mostcommon is to take the function as undefined fortime after the last censoring (Kalbfleisch andPrentice, 2002).

Now, as an illustration with right-censoring,consider the data set in Table 2.

Four of our 15 data points have been cen-sored. Since censoring occurs more often as timepasses, ignoring the censored individuals wouldbe a bad idea as it would result in an underesti-mation. Furthermore, if we treated the censoredtimes just like failure times, we would again beunderestimating. The following graph comparesthe approximated survivor function found by us-ing the KM estimator (solid line) to the empir-ical survivor function (dotted line) that treatsthe censored data as simple failure times.

0 100 200 300 400

0.0

0.2

0.4

0.6

0.8

KM Estimator − AIDS Patients Survival

Time (T) in months

Pro

port

ion

If we do treat the censored times as failuretimes, we would expect the average survival time

of patients (E(T )) to be just 142.47 months.But, using the KM estimator, we find that theexpected failure time is 177.41 months (I usedthe fact that the jumps in the graph correspondto the p.d.f. of T, as mentioned earlier). Bymistreating the data, we arrive at a value that ismuch smaller than the method that takes right-censoring into account.

The KM estimator is asymptotically unbi-ased, and its bias decreases exponentially as nincreases (Zhou, 1988). Therefore, we can getvery good results for high sample sizes. In areal medical study, the number of participantsis far greater so that this method can be usedto accurately predict an expected survival timeafter the onset of disease.

References

[1] A. G. Babiker, J. H. Darbyshire, T. E. A.Peto and A. Sarah, 1998, Issues in the De-sign and Analysis of Therapeutic Trials in Hu-man Immunodeficiency Virus Infection: WalkerJournal of the Royal Statistical Society. SeriesA (Statistics in Society), v. 161(2), pp. 239-249.

[2] P. Hougaard, 1999, Fundamentals of SurvivalData: Biometrics, v. 55, pp. 13-22.

[3] J. D. Kalbfleisch and R. L. Prentice, 2002,The Statistical Analysis of Failure Time Data,second ed., John Wiley and Sons, Inc., New Jer-sey.

[4] E. L. Kaplan and P. Meier, 1958, Nonpara-metric Estimation from Incomplete Observa-tions: Journal of the American Statistical As-sociation, v. 53(282), pp. 457-481.

[5] M. Zhou, 1988, Two-Sided Bias Bound ofthe Kaplan-Meier Estimator: Probability The-ory and Related Fields, v. 79, pp. 165-173.

Table 1:n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

failure time 5 26 79 81 83 90 102 102 124 134 259 263 349 360 401

Table 2:n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

failure time 5 26 15 81 83 90 102 102 124 57 259 263 289 240 401δi 1 1 0 1 1 1 1 1 1 0 1 1 0 0 1


14 Benford’s Law

Benford’s LawJoel Perras

Benford’s law states that for large sets of data, the distribution of the First SignificantDigits (FSD) within this data follows a logarithmic relationship. The FSD frequencyis determined by P (FSD = d) = log10

(

1 + 1d

)

, where d = 1, 2, 3, . . . , 8, 9. Moreover,Benford’s Law may be generalized to find the probability for the nth significant digitor combinations of significant digits.

Discovery

Simon Newcomb, the original discoverer ofBenford’s Law, was an astronomer and a physi-cist who lived from 1835-1909. At that time,most calculations of the movements and pre-cessions of celestial bodies involved long andcomplex derivations, which frequently requiredthe use of Napier tables, now more commonlyreferred to as log tables. After extensive useof these tables, Newcomb came to the real-ization that, for some reason or another, theactual pages in his copy of logarithmic tableswere not being used equally; instead, some ofthe pages were more stained and folded, show-ing an increase in usage. Upon further analy-sis, Newcomb postulated that the distributionof significant digits of his collected data wasnon-uniform; rather, the resulting logarithms ofthose digits was. In fact, Newcomb was able topublish an empirically-derived equation whichdescribed for this behaviour. However, withouta proper proof or even possibility of application,no one paid attention to this very interesting re-sult.

Frank Benford, who independently discov-ered the phenomenon in 1938, in contrast toNewcomb, did not simply publish his empiri-cally derived formula; rather, he “compiled alist of nearly 20 000 different observations, cov-ering everything from river drainage areas andthe addresses of notable scientists to numbersappearing in an issue of Reader’s Digest”[1]. Us-ing this data, he was able to empirically verifythis empirically determined phenomena; whilethis was not a ground-breaking proof, it was astep in the right direction. The actual proof of

Benford’s Law was found nearly sixty years af-ter Benford’s rediscovery of the phenomena, andwas published by T.P. Hill [3], who had alreadywritten several other papers on the remarkableproperties of this phenomenological law.

What is Benford’s Law?

Surprisingly, the actual formulation of Benford’sLaw is quite simple. Moreover, Benford’s Lawitself is merely a generalization of a distributionpattern that arise from mixtures of uniform dis-tributions [2], and is most often the null hypoth-esis in testing for human influence on data.

Benford’s Equation

Originally, Benford’s empirical formula was putforth to determine the distribution pattern offirst significant digits (FSDs) from a collectionof data sources, whose probability distributionswere non-uniform and varied in nature:

P (FSD = d) = log10

(

1 +1

d

)

where d = 1, 2, 3, . . . , 8, 9.

(0.3)

While this in and of itself is quite an achieve-ment, the formula lacked any capability for de-termining the frequency of second, third or nth

significant digits, as well as combinations of twoor more digits (such as the frequency of the twoconsecutive digits of 99 appearing). As can beobserved from eq. (0.3), for any probabilisticprediction of an nth significant digit other thanthe first, a more generalized equation must beput forth. This was precisely what Hill derived

8A σ-algebra U is defined as follows: Let S be a set, and U be a non-empty collection of subsets of S such thatthe following are true:

1. Closed under complements

2. Closed under countable unions

3. Should contain the total set


Benford’s Law 15

in 1996. Moreover, he also formulated the cor-rect σ-algebra 8 set that described the probabil-ity domain of Benford’s Law, thus establishinga proper mathematical proof for the empiricallyderived law.

Hill was able to show that the set A is thesmallest σ-algebra generated by the countableunions of the ith significant digits [3]. From this,he derived his General Significant Digit Law :

P

[

k⋂

i=1

{Db = di}]

=

= logb

1 +

(

k∑

i=1

bk−i · di

)−1

where k ǫ N and di ǫ {1, 2, 3, . . . , 9}.(0.4)

This equation, when compared with (0.3), hasthe remarkable property of being able to predictthe frequency of appearance of any significantdigit, or combination of digits. Furthermore, italso illustrates the fact that the nth significantdigit is dependent on the n − 1 previous signifi-cant digits.

To calculate unconditional probabilities forthe nth significant digit, the sum of the proba-

bilities for the digits before the nth significantdigit must be calculated:

P (nthSD = d) =

9∑

k=1

log10

(

1 +1

10k + d

)

Where d = 0, 1, . . . , 9.

(0.5)

Using this formula, we find that the (uncon-ditional) maximum probability for the secondsignificant digit occurs when d = 0, with a prob-ability of 0.1197. As can be discerned from (0.5),the significant digit probabilities move exponen-tially towards a uniform distribution as i → ∞[3].

References

[1] R. Matthews, “Benford Bend”, May/June2000: EBSCO Research Database.

[2] R. J. Rodriguez, Ricardo, “First SignificantDigit Patterns From Mixtures of Uniform Dis-tributions”, The American Statistician, 58.1February 2004. 64-71.

[3] T.P. Hill, “The Significant Digit Phe-nomenon”, The American MathematicalMonthly, 102.4 April 2004, 322-327.

Jokes

Q: How can you tell a sailor used to be a mathematician?A: Instead of saying “aye, aye, captain” he says “negative one, captain!” 2

A math professor organized the seminar in hydrodynamics in his University. Among the regularattendees there were two men in uniform, obviously military engineers. They never discussed theproblems they were working on. But one day they asked the professor to help them with a mathproblem. They explained that the solution of a certain equation oscillated and asked how theyshould change the coefficients to make it monotonic. The professor looked at the equation andsaid, “Make the wings longer!” 2

A chemist, a physicist and a mathematician are stranded on an island when a can of food rollsashore. The chemist and the physicist comes up with many ingenious ways to open the can. Thensuddenly the mathematician gets a bright idea: “Assume we have a can opener...” 2

A tragedy of mathematics is a beautiful conjecture ruined by an ugly fact. 2

The dean, to the physics department: “Why do I always have to give you guys so much money forlaboratories and expensive equipment and stuff? Why couldn’t you be like the math department– all they need is money for pencils, paper and waste-paper baskets; or even better, like thephilosophy department. All they need is pencils and paper!” 2


16 Generators of SL(2, Z)

Generators of SL(2, Z)Agnes F. Beaudry

We first prove that SL(2, Z) is generated by two elements. Then we motivate thestudy of this group by describing its action on the upper-half plane.

Why SL(2, Z)?

If you studied algebra, you have probably en-countered the group of matrices called the gen-eral linear group of degree 2, GL(2, F ), the set ofinvertible 2×2 matrices with entries in a field F .One important subgroup of GL(2, F ) is the spe-cial linear group, SL(2, F ), the set of matriceswith determinant equal to one. Now, if F = R,the set of matrices in SL(2, R) with integer en-tries, denoted SL(2, Z), is also a subgroup andplays an important role in the geometry of theupper-half plane

H = {z = x + yi ∈ C | y > 0}.

I will start by discussing the structure of thegroup SL(2, Z), showing that it is generated

by the matrices T =

(

1 10 1

)

and S =(

0 −11 0

)

. Then I will quickly try to convince

you that this group is interesting.

The Generators

SL(2, Z) is closed under matrix multiplication,since the determinant of a product is the prod-uct of the determinants. With identity I =(

1 00 1

)

, all elements

(

a bc d

)

have an in-

verse,

(

d −b−c a

)

in SL(2, Z). This gives

SL(2, Z) the desired group structure.

Theorem. Let T =

(

1 10 1

)

and S =(

0 −11 0

)

. Any matrix in SL(2, Z) can be

written as a product of positive powers of T andS.

This is stronger than asking that T and Sgenerate SL(2, Z), so it implies it. The proofwill use the fact that all elements of SL(2, Z)are invertible, so they can be brought to theidentity by a series of row operations. Then wewill use the fact that Z is Euclidean, i.e. that

there is a Euclidean division algorithm on Z. Asa reminder, this means that for any pair of inte-gers α and β, there exists a list [q1, ..., qn, qn+1]and [r1, ...rn] such that

α = q1β + r1

β = q2r1 + r2

...

rn−2 = qnrn−1 + rn

rn−1 = qn+1rn,

with 0 ≤ r1 < β, 0 ≤ ri < ri−1 and rn =gcd (α, β).

Before we start, it is useful notice that mul-tiplication on the right by S interchanges thecolumns of the matrix (changing signs to pre-serve the determinant), while multiplication onthe left interchanges the rows.

Proof. If any B−1 ∈ SL(2, Z) can be written asa product of powers of S and T , then since eachA ∈ SL(2, Z) is the inverse of A−1 ∈ SL(2, Z),all A can be expressed as such a product. Inthe following, we denote the desired matrix byB and let its inverse be A. Writing A−1 as aproduct of S and T is just writing B as such.What we will do is equivalent to transform-ing A into the identity using a product of el-ementary matrices. The whole set of elemen-tary operations, i.e. interchanging rows, mul-tiplying by a non-zero scalar or multiplying arow and adding it to another, cannot be usedbecause in the two first cases, the relevant ele-mentary matrices are not in SL(2, Z) (unless thescalar is one). So we must only allow the thirdkind of operation, together with multiplicationby powers of S. For this we need the matri-

ces T z =

(

1 z0 1

)

and T zt =

(

1 0z 1

)

for

z ∈ Z, M t denoting the transpose of the matrixM . The following identities together with the

fact that T n =

(

1 n0 1

)

(and similarly for the

other matrices below), show that these elemen-tary matrices are products of powers of S andT .


Generators of SL(2, Z) 17

• S−1 = S3

• T−1 =

(

1 −10 1

)

= S2 · STSTS

• T−1t=

(

1 0−1 1

)

= S2 · STS

• T t =

(

1 01 1

)

= TST.

Let A =

(

a bc d

)

. Because det(A) = ad −bc = 1, we know that gcd (a, b) = gcd (a, c) =gcd (b, c) = gcd (c, d) = 1 (if x|a and x|b, thenx|(ad − bc) = 1, thus x = ±1. The same ar-gument holds for the other pairs.) First sup-

pose that c = 0, then A =

(

±1 b0 ±1

)

.

Then

(

1 ∓b0 1

)

×(

±1 b0 ±1

)

= ±I. Since

S2 = −I we are done.Now suppose c 6= 0. Because Z is euclidean

and gcd (a, c) = 1, we have a list of integers[q1, . . . , qn, qn+1] which gives us the above equa-tions with last remainder rn = 1. The followingprocedure is just the division algorithm carriedout on a column of A.

T−q1A =

(

r1 b1

c d1

)

where r1 = a− q1c is the first remainder andb1 and d1 are the appropriate new entries (notethat d1 = d, but we re-label for simplicity ofnotation). The rows remain relatively prime be-cause the determinant of the two matrices is 1.Then

T−q2tA =

(

r1 b2

r2 d2

)

.

As above, r2 = c − q2r1. Because gcd (a, c) = 1,this process must terminate with (1, 0)t or (0, 1)t

in the first column of A. In the latter case,multiplication on the left by S3 finally brings

A to A′ =

(

±1 b′n0 ±1

)

(because det(A′) = 1).

From here it is obvious that multiplication bypossibly S2, and then T−bn gives the iden-tity. 2

The Upper-half plane

One of the interesting things about SL(2, Z) isits group action on the upper-half plane H. A

group action of a group G on a set X is a func-tion G × X ⇒ X , denoted g · x = y for g ∈G, /x, y ∈ X , satisfying both g ·(h·x) = (g◦h)·xand i · x = x, where g, h are in G, i is the iden-tity element of G and ◦ is the composition of thegroup. The action of SL(2, Z) on H is

(

a bc d

)

· (z) =az + b

cz + d, z ∈ H.

We need to mod out by ±I since −IS ·z = S ·z.We denote the “quotient” by PSL(2, Z), oftencalled the modular group.

Now, a fundamental region R for the ac-tion of PSL(2, Z) in H is a region of H suchthat, for every z ∈ H, there exists a uniqueA ∈ PSL(2, Z) and z′ ∈ R such that A · z′ = z.If we want to fix R, the region

R = {z = x + yi|x2 + y2 ≥ 1 and |x| ≤ 1

2},

with some restrictions on the boundaries, is theusual choice.

The interesting point, which you should lookup, is that by applying elements of PSL(2, Z) to

both the vertices of R in H (eπi3 and e

2πi3 ), and

to ∞, which we view as the third vertex of the“triangle”R, (defining A ·∞ = a

c ), we get a tes-sellation of the upper-half plane, i.e. the upper-half plane is divided into hyperbolic triangles.Of course, since S and T generate the group,it is sufficient to do the exercise with powers ofthese two elements (try it with T to see how R isjust translated throughout H). You will noticethat this extends H to include Q. Accordingly,we call H∪∞∪Q the extended upper-half plane.It needs to be shown that this is indeed a tessel-lation, i.e. that it is space-filling and that thereare no overlaps. I hope this is enough to sparkyour interest.


Mathematical digest 19

Mathematical digestMichael McBreen

The mathematical digest is a collection of articles on undergraduate course material.The articles aim to give an intuitive understanding of the material, not to proveanything rigorously. Enjoy.

The Fundamental Theorem of Cal-

culus

The FTC states that∫ b

a

f(x)dx = F (b) − F (a)

where dF (x)dx = f(x). We can also write this as

d

db

∫ b

a

f(x)dx = f(b)

d

da

∫ b

a

f(x)dx = −f(a)

We’re not going to prove the FTC here.We’re going to show why it makes sense. Weonly consider the positive f(x) case for sim-plicity. Consider the area under a curve f(x)bounded by lines at x = a and x = b,

or∫ b

af(x)dx. If we keep a fixed, the area

is obviously a function of b, which we’ll callIntegral(b). As we increase b, the right linemoves along the axis and the area increases.

The FTC tells us what the derivative ofIntegral(b) is. In other words, how fast doesIntegral(b) increase when we increase b? Moreprecisely, how many times faster (or slower)than b does it increase?

Well, the bigger f(b) is, the faster it in-creases. In fact, if you increase b by an infinites-imal amount δb, you’re increasing the area byadding a rectangle9 with height f(b) and widthδb, as shown in the following figure:

Symbolically,

Integral(b + δb) = Integral(b) + f(b)δb

Of course, this means that Integral(b) in-creases f(b) times faster than b, or in symboliclanguage,

d

db

∫ b

a

f(x)dx = f(b)

And that’s the FTC. Well, that’s half of it.Hopefully, you can explain the other half, con-cerning F (a), to yourself - exactly the same rea-soning applies.

Linear Differential Equations and

the Minimal Polynomial

We’re taught in ODE how to solve equations ofthe form

P (D)y = 0

where P (D) is a non-trivial polynomial in thelinear differential operator D = d

dx . Our aimis to find the solution space V = ker(P (D)) ofall functions annihilated by P (D). V happensto be a vector space, because if y and z are so-lutions, then so are y − z and λy for λ ∈ C.To find V , we simply factor P (D) into mutu-ally prime components pi(D) such as (a + D)m,find the solution space ker(pi(D)) of each in-dividual factor and then take V to be the di-rect sum of all the individual solution spaces:V = ker(p1(A))⊕ker(p2(A))⊕. . .⊕ker(pn(A)).

Why can we do this? Recall the PrimaryDecomposition Theorem from Linear Algebra:Let P (A) be the minimal polynomial of a lin-ear operator A acting on a finite dimensionalvector space W . Let P (A) be the product of mu-tually prime factors p1(A)p2(A) · · · pn(A). ThenW = ker(p1(A))⊕ker(p2(A))⊕. . .⊕ker(pn(A))

Clearly, if we can show that V is finite di-mensional and that P (D) = λM(D) whereM(D) is the minimal polynomial of D over V ,

9It’s really a trapezium, but from the picture you can see that the difference in area is negligible.


20 Mathematical digest

then the two situations will be completely equiv-alent.

To show dim(V ) < ∞, recall that if P (D) isof degree n, then we can reduce the equation

P (D)y = 0

to a system of n first order equations which wewrite concisely as

D~y = ~F (~y, x)

where ~y is the n-component vector of unknownfunctions of x. Given an initial vector ~y0, thisequation obviously has a single solution. Inother words, n initial conditions (the n com-ponents of ~y0) suffice to determine a solutioncompletely. If dim(V ) = ∞, then n initial con-ditions would be insufficient to pick out a singlefunction, so dim(V ) must be finite.

To show P (D) = λM(D), note that P (D)annihilates everything in V , so M(D) must di-vide P (D). We just have to show that if P (D) =Q(D)M(D) with degQ(D) > 0, then P (D)annihilates some functions that M(D) leavesstanding, which contradicts M(D) being theminimal polynomial of D for V = ker(P (D)).This can easily be proven using the fundamen-tal existence and uniqueness theorem; I won’tdo it here.

And that’s why we can do what we do inODE.

The first isomorphism theorems for

rings and groups

If you look at the FIT the right way, it seemsalmost trivial. First, the claim for rings: Theimage of a surjective homomorphism ρ : R → Kis isomorphic to R/ ker(ρ).Now, the explanation. Consider a finite ring.Any finite ring is completely described by itsaddition and multiplication tables. Since an iso-morphism preserves these tables, the only thingit can do to a ring is change the names of the el-ements (i.e. 1 → a ,2 → b ,3 → c ,4 → d ,etc).Since an isomorphism is reversible, it must givedistinct names to distinct elements. It’s sim-ply a name switch. Now, consider a surjectivehomomorphism. What can this homomorphismdo that an isomorphism can’t? It still can’t ac-tually modify the + and × tables, but it canforget that certain elements are distinct: it cangive the same name to distinct elements. Thereare certain obvious consistency requirements: if

you forget that a and b are distinct, then a - bmust be sent to 0 (i.e. you must also forget thata - b and 0 are distinct). Likewise, if you send a- b to 0, then a and b must be merged together.Finally, if you send a to 0, you have to send abto 0 as well. These consistency requirementsimply that what your homomorphism “forgets”is entirely determined by what it sends to 0, i.e.by its kernel, i.e. your whole homomorphism isdetermined by its kernel. Now, where else havewe taken a ring, sent some of its elements to 0and then studied the result? You will kindly re-call that we do the exact same thing when wetake the quotient of a ring by an ideal. Notethat the conditions that define an ideal corre-spond precisely to the consistency requirementson our homomorphism. For instance,

a ∈ I → ba ∈ I

simply means that if you send a to 0, thenyou must also send ba to 0. Once we’ve speci-fied a kernel for our homomorphism, i.e. whichideal we’ll be modding out with, we have onlyone freedom left: we can name the resulting el-ements as we wish (provided we give distinctnames to distinct elements in the target ring-we’ve already chosen which elements we wantto merge). That’s why we say that the targetring is identical to the quotient ring “up to anisomorphism”(a name switch). And that, essen-tially, is the first isomorphism theorem for rings.In the group case, there’s only one major differ-ence: the “kernel” is the set mapped to 1 andnot 0. This is why the consistency requirementson a normal subgroup are different from thoseon an ideal. For instance,

g ∈ N → hgh−1 ∈ N

means (roughly) that if you send g to 1, youmust also send hgh−1 to 1 , which makes per-fect sense.

Vanilla Taylor Series

This article is about the meaning of Taylor se-ries, strange as it may seem. For a more conven-tional proof of the formula, see the box at theend of the article. We start with some context.

Picture yourself gazing wistfully throughyour window at a narrow stretch of road below,with your chronometer on (Why? That’s notfor me to say). A car passes by at time t = 100seconds sharp, and then disappears from view.



Now, the question is: how far from the housewill the car be at t = 105 sec? How far was itat 95 sec? In other words, what is f(t) = d, thefunction that gives car distance d with respectto time t? Using the magic of Taylor series, wecan approximate f(t) step by step by replacingit with simpler functions Pn(t) – polynomials int, more precisely.

1. 0th order10 approximation in t : We cansay that 5 sec later, the car will still be roughlyin front of the house (at d = 0 m). In otherwords,

P0(t) = f(100) = 0.

Not so great.

2. 1st order approximation in t : Knowingthe car’s speed f ′(t) as it passed (at t = 100sec), we can assume it keeps that speed and let

P1(t) = f(100) + [f ′(100)](t − 100)

This is better. Notice that we must subtract 100sec from t to get the actual time the car spentdriving away from us.

3. 2nd order approximation in t : What ifthe car is accelerating? In other words, what iff ′′(100) 6= 0 ? In that case, we can let

P2(t) = f(100) + [f ′(100)](t − 100)

+1

2[f ′′(100)](t − 100)2

What do the factor of 12 and the square mean?11

The square means that at large times (eg. 1000sec), your rate of acceleration will have a lotmore impact than your initial speed. A carmay zoom past your window while a bus crawlsslowly by, but if the car is constantly slowingdown while the bus is constantly accelerating,the bus will eventually leave the car far behind.As for the 1

2 factor, it (very, very roughly) meansthat in the short term, speed is more importantthan acceleration.

4. 3rd order approximation in t : Here wedepart from high-school physics. If the car is ac-celerating faster and faster, we can take it into

account by setting

P3(t) = f(100) + [f ′(100)](t − 100)

+1

2[f ′′(100)](t − 100)2

+1

6[f ′′′(100)](t − 100)3

The meaning of the cube and 16 factor are much

the same as above.

5. nth order approximation in t : We cancontinue this process forever, taking into ac-count ever higher order rates of change. Eachsuccessive Pn(t) is a polynomial in t, with coef-ficients determined by successive derivatives off(t) at 100 sec. Higher derivatives correspond tomore removed forms of “acceleration”and henceare multiplied by smaller and smaller fractions.If you don’t see why, think of it like this: imag-ine three cars at the starting line of a race. Onestarts at speed 200km/h, the second at accelera-tion 200km/h2 and speed 0, and the third startswith 0 speed, acceleration 0 and rate of increaseof acceleration 200km/h3. In the short term,the first car will lead. After a while the secondcar will have gained enough speed to overtakethe first, and will eventually leave it far behind.But later still, the third car will lead, becauseits acceleration is always increasing. The higherthe order of the rate of change, the later theeffect.

The formula for the nth order approxima-tion, should you really want it, is:

nthorder Pn(t) = f(100) + [f ′(100)](t

−100) +1

2[f ′′(100)](t − 100)2

+ . . . +1

N ![f (N)(100)](t − 100)N

(0.6)

If we’re lucky12, as we let n go to infinity, Pn(t)should accurately give us the car’s position atany time. We call this P∞(t) the Taylor seriesof f(t) around t = 100 sec. Something could gowrong, though. Let’s see what.

Possible Failure and Analytic Functions

First off, what if the driver does somethingunexpected? We only have information about

10The order, in this case, is simply the degree n of the polynomial Pn(t).11You can explain the square by noting that acceleration has dimensions of m/s2 (if you believe the physicists),

but it’s not a particularly intuitive reason.12I’ll explain what this means in a moment.


22 Mathematical digest

what he’s doing as the car passes our window.If he chooses to stop at Roarin’ Willy’s Road-side Pub 15 seconds later, our predictions fail:no amount of Taylor series can save him. In fact,if our Taylor series really does predict what thedriver will do in the far future, we have a verystrange driver on our hands. The f(t) of thispredictable driver is called an analytic function:its value at any point is fully determined by itsbehaviour in some small region.

Textbook Taylor SeriesHere’s the usual proof that if f(x) has

a power series expansion around x = c (i.e.can be expressed as an “infinite polynomial”in (x− c)), then that power series is given bythe Taylor series

f(x) = f(c) + f ′(c)(x − c) +f ′′

2(c)(x − c)2

+ . . . +f (n)

n!(c)(x − c)n + . . .

Let f(x) be expressed by the following powerseries:

f(x) = a0 + a1(x − c) + a2(x − c)2 + . . .

+a3(x − c)n + . . .

(0.7)

Set x = c to get f(c) = a0.Then, differentiate (0.7) once and set x =

c to get f ′(c) = a1 In general, differentiate(0.7) n times and set x = c to get f (n)(c) =n!an or

an =f (n)(c)

n!

which completes the proof. To prove thatsome function f(x) does have a power seriesexpansion is more involved, and we won’t gointo it here.

However, even if our driver does act pre-dictably, our predictions could yield infinitequantities (gibberish) past a given time (eg.200 sec). The causes of that failure, non-convergence, are beyond the reach of this article.

The Change of Variables Formula

Before we begin: infinitesimals. I’m goingto throw a lot of infinitesimals around: dx, dy, dθand so on. If you like, you can think of themas extremely small numbers. They’re meant tohelp your intuition along, not to be elements ofa rigorous proof, so don’t worry too much about

the fine print. If the change of variables formulamakes sense to you by the end of this article,then all is well.

We use the change of variables formula(CoVF) when we’re integrating some function of3-space, and wish to go from Cartesian coordi-nates (x, y, z) to spherical coordinates (ρ, θ, φ),for instance. We’re going to stick to that ex-ample for the rest of the article, but it shouldbe obvious that nothing I say will be specific tothese two coordinate systems or to 3D space.

Now, the CoVF tells us that when we changecoordinates, we have to multiply our integrandby |detJC→S(ρ, θ, φ)| where JC→S(ρ, θ, φ) is theJacobian matrix of our coordinate change (Iwrite C → S for “Cartesian to Spherical”):

∫

V

f(x, y, z) dx dy dz =

=

∫

V ′

f(r, θ, φ) |detJC→S(ρ, θ, φ)| drdθdφ

where

JC→S(r, θ, φ) =

∂x∂r

∂x∂θ

∂x∂φ

∂y∂r

∂y∂θ

∂y∂φ

∂z∂r

∂z∂θ

∂z∂φ

Why does this Jacobian factor arise? We’re inte-grating over a volume, so the integrand f(x, y, z)can be thought of as the “density” of some sub-stance (call it gob). When we integrate, we di-vide our integration region into tiny boxes withsides dx, dy and dz. You build such a boxby choosing a point (x, y, z), varying x by dx,y by dy and z by dz and taking the volumeyou’ve just “swept out” to be your box. Thisbox contains an amount f(x, y, z) dx dy dz =f(x, y, z) dV of gob, and the integral sums thecontributions from all boxes to give the total gobin the region.

Now, let’s change variables to ρ, θ and φ. Weget our new boxes by varying ρ by dρ, θ by dθand φ by dφ. The problem is that these boxeswill no longer all have the same volume. If youmake the same small variations at a point withsmall ρ and at a point with large ρ, your secondbox will be bigger, and chances are that neitherwill be the size of the Cartesian boxes they’re re-placing. Clearly, we need a function that givesthe change in volume of the new boxes with re-spect to the old boxes. We’re going to show thatthis function is the Jacobian JC→S(ρ, θ, φ).



The first thing to note is that JC→S(ρ, θ, φ)is a coordinate change matrix: it takes an in-finitesimal vector with spherical coordinates andre-expresses it in Cartesian coordinates. To seethis, stare hard at the following equation:

JC→S(ρ, θ, φ)

dρdθdφ

=

=

∂x∂ρ

∂x∂θ

∂x∂φ

∂y∂ρ

∂y∂θ

∂y∂φ

∂z∂ρ

∂z∂θ

∂z∂φ

dρdθdφ

=

∂x∂ρdρ + ∂x

∂θ dθ + ∂x∂φdφ

∂y∂ρdρ + ∂y

∂θ dθ + ∂y∂φdφ

∂z∂ρdρ + ∂z

∂θ dθ + ∂z∂φdφ

=

dxdydz

The last equality follows from the chain rule(or simply from the fact that a smooth function -such as x(ρ, θ, φ) - varies linearly when we makeinfinitesimal changes in its arguments). Weneed to show that when this coordinate changematrix J acts on a box with sides dx, dy, dz, thebox’s volume increases by a factor |detJ |. Wecan use the following theorem

Theorem. Polar Decomposition Theorem Anyreal or complex matrix S can be expressed as aproduct of the self-adjoint operator

√S∗S and

some isometry13 U :

S = U√

S∗S

I will only sketch the proof and leave the de-tails to you. First note that

‖√

S∗Sv‖2 = 〈√

S∗Sv,√

S∗Sv〉 = 〈S∗Sv, v〉= 〈Sv, Sv〉 = ‖Sv‖2

(0.8)

Define U ′ : Im(√

S∗S) → Im(S) byU ′(

√S∗Sv) = Sv. Using (0.8), you can check

that U ′ is in fact an isometry. We can easilyextend U ′ to an isometry U of the full vectorspace. This proves the theorem.

We can therefore write J = U√

J∗J forsome U . Since U is an isometry (analogousto a rotation), it obviously preserves volumes.Hence, Volume(J dV ) = Volume(U

√J∗J dV ) =

Volume(√

J∗J dV ). But since√

J∗J is self-adjoint (as you should see), the real spectraltheorem tells us that it has a basis of orthonor-mal eigenvectors! Such an operator expandsor contracts volumes by a factor equal to theproduct of its eigenvalues λ1λ2 . . . λn, or moreprecisely |λ1λ2 . . . λn| (since a negative factorsimply means certain directions have been re-versed). Look at the following figure to see why.

Now, the determinant of a diagonal matrix issimply the product of its eigenvalues, so we have

|λ1λ2 . . . λn| = | det√

J∗J | =√

det(J∗J)

=√

det(J∗) det(J)

=√

(detJ)∗ det(J)

= | det J |(0.9)

Equation (0.9) gives precisely the change of vari-ables formula.

13An isometry is a linear operator that preserves the length of vectors or more generally the inner product of twovectors. Rotations and reflections are isometries acting on R3.


24 Interview with Professor Henri Darmon

The Birch and Swinnerton-Dyer Conjecture:

An Interview with Professor Henri DarmonAgnes F. Beaudry

If you made a poll of number theorists and asked them, “What’s your favorite problemin number theory?”, you would probably have it equally divided between the RiemannHypothesis and the Birch and Swinnerton-Dyer conjecture, which are two of the Mil-lennium Prize problems in number theory. My favorite problem is the Birch andSwinnerton-Dyer conjecture.

-Prof. Henri Darmon

The Birch and Swinnerton-Dyer conjecture(BSD) was stated in the sixties by PeterSwinnerton-Dyer and Bryan Birch who gatheredconsiderable evidence, based on numerical datafrom the EDSAC computer at Cambridge, sug-gesting a relation between the rational solutionsof special Diophantine equations called ellipticcurves and their solutions in Zp for differentprime values. In the early eighties, the work ofVictor Kolyvagin, Benedict H. Gross and DonZagier, combined with that of Andrew Wiles fi-nally created tools to approach the previouslyobscure problem, eventually bringing it to theforefront, the Millennium prize stamping it asone of the most important problems of the cen-tury. prof. Henri Darmon has been working onthis problem. We met with him to try to under-stand what the BSD really means and perhapsget a glimpse of his work on the problem.

Elliptic curves and projective mod-

els

In the official description by Andrew Wiles, theBSD is described as a relation between the L-function of an elliptic curve, terms I will clar-ify below, and its rank, a number that, to someextent, measures the size of the set of rationalsolutions (solutions in Q) of that elliptic curve.Pr. Darmon explains:

“It started without involving L-functions atall, these are just part of the baggage that youneed to make this conjecture very precise. Butthe idea is simple. You start with an an ellipticcurve

E = y2 = x3 + ax + b, a, b ∈ Q.

The set of rational solutions of these equations,E(Q), has a very nice structure, an abelian

group law. To obtain this group, one needs tolook at the projective model of the equation.”

The projective model of an equation is ob-tained by adding an extra variable, z, in orderto transform it into a homogenous equation ofdegree three:

y2z = x3 + axz2 + bz3.

This equation has a trivial solution, (x, y, z) =(0, 0, 0), which we ignore. Also, if (x, y, z) is asolution to this equation, then so is (λx, λy, λz).We let two solutions be equivalent if they differby a non-zero scalar.

There are two possibilities, either z 6= 0 orz = 0. If z 6= 0 for a solution P = (x, y, z), thenP is equivalent to a solution (x′, y′, 1), becausewe can multiply P by z−1. This solves the origi-nal equation, thus we get a bijection between the



solutions of the projective model with z 6= 0,and the solutions original elliptic curve calledthe affine model. On the other hand, if z = 0,things get interesting: “Here we get new solu-tions which are of the form Q = (x, y, 0). If wesubstitute z = 0 into the equation, it becomesx3 = 0, and x also has to be zero. ThereforeQ = (0, y, 0), where y 6= 0 (since we are notallowing the zero solution), which is equivalentto (0, 1, 0). We therefore have this new solu-tion called the point at infinity of the projectivemodel, which did not appear in the affine model.

THE GROUP LAW

To find the group law on the set of solu-tions, we first look at the given elliptic curve

E = y2 = x3 + ax + b, a, b ∈ Q

over C. Given two points P and Q on Eand the line determined by these two points,L = y = mx + n, L must intersect E at an-other point R since C is algebraically closed(and the intersection (mx+n)2 = x3 + ax+ bis of degree three.) Note that these threepoints need not be distinct, i.e. we countmultiplicities. Now we draw the line L′ con-necting R and ∞ (which lies on E since we arelooking at the projective model.) This lineintersects E at a third point, P ⊕ Q, and theoperation ⊕ thus defined is the compositionlaw of the group with the identity element∞ = (0, 1, 0). It satisfies all the axioms of anabelian group. Strangely enough, the hardthing to verify is the associativity property.

One must then show that if P and Q arerational, then P ⊕ Q ∈ Q, i.e. the set ofrational solutions E(Q) is a subgroup ofE(C). To verify this, you can write theequations of the E, L and L′ and verify thatthe intersections must be in Q. For moredetails, see [1].

That extra point is a distinguished point,which plays the role of the identity element forthe addition law of the group of rational solu-tions. That’s why, for the elliptic curve, we al-ways consider the projective model. The hardthing to show is that this group is finitely gen-erated, it’s that finiteness result.”

The rank and its role in the BSD

That E(Q) is finitely generated was proved byLouis Mordell in 1922. It implies that

E(Q) ≃ Zr ⊕ T,

where T is the torsion part of the group, i.e. theset of elements of E(Q) with finite order, and ris the smallest number of elements needed togenerate the non-torsion part of E(Q). We callr the rank. It’s similar to the dimension of avector space: it measures the size of the spaceof solutions. “What you want to be able to com-pute is this rank, how many solutions you needto generate all the other ones by repeated appli-cations of these group laws.” The BSD proposesan answer to this question.

Now, Diophantine equations in Q are not asmalleable as solutions in certain other fields, soone hopes to find tools in these nicer fields toget to the rational solutions:“You have an equation, you try to understand itby studying its complex solutions, its real solu-tions maybe, and its solutions over a finite field.These are very easy solutions, much less subtlethan those over Q. Over C you get this nicesurface with topological invariants, and some-how you just care about the shape. Over thefinite fields you have the finite sets, the cardi-nality. Then you’d like to understand the solu-tions over Q, and the principles that allow youto derive this information are very deep. This isthe main philosophical question in the study ofDiophantine equation, this passage from under-standing the behaviour of the solutions in finitefields to their behaviour over Q.

Birch and Swinnerton-Dyer’s first insightwas thinking that maybe you could get at therank by counting the number of solutions overZp. So they defined Np to be the number of(x, y) where p|f(x, y), in other words, wheref(x, y) ≡ 0 (mod p). A priori, that has nothingto do with it, because here you are looking atthe solutions over Q and there you are lookingat the corresponding congruence classes. Their



insight was that if r is big, then you get a sys-tematic contribution to E(Zp), or the solutionsover Q reduced modulo p.

Take the solutions over Q. If r = 0, then youhave essentially no solutions, i.e. finitely many,but if r = 1, 2, then you have many, many ratio-nal solutions. You take these rational solutions(x, y), (supposing that p does not divide the de-nominator of x and y) and reduce them mod p.If r is large, then the number of solutions overthe different Zp should have a tendency to belarge. What Birch and Swinnerton-Dyer did isthey tried to make that precise and quantitative:they looked at the product of Np over p.

Roughly, how big is Np? You can let x rangefrom 0 to p − 1. For every value of x, you geta number modulo p. You’re asking whether thisequation, y2 = x3 + ax + b has a solution in Zp.You’re asking whether this number is a squareor not (a quadratic residue or non-residue). Halfof the elements in Zp are squares, and the otherhalf are not. Thus when you run over all thevalues of x, roughly one out of two times youare going to win, you’ll get two solutions, theone and its negative, i.e. if (x, y) is a solutionto Y 2 = X3 + aX + b then so is (x,−y). Theother half of the time you’ll loose, you’ll get nosolutions. On average you expect to get roughlyp solutions, but there’s an “error term” definedto be

ap = 1 + p − Np.

It turns out that ap < p since in the worst ofcases, for every trial you’ll get two points. Itcan actually be shown that |ap| is at most 2

√p.

Birch and Swinnerton-Dyer looked at thisproduct over the primes less than a given χ.They observed that this quantity grows roughlyas a constant that depends on E, times a powerof log χ, and that the exponent seems to be therank.

∏

p<χ

Np

p∼ CE(log χ)r

In particular, they found that the productseems to be bounded when r = 0, in other wordswhen E(Q) is finite. When r = 1, it appears togrow as log χ, when r = 2 it appears to grow likelog χ2 and so on. This means that if you knoweverything there is to know about this equa-tion over the congruence equations, then youwill know something about the equation over Q.This was found experimentally, and that’s theBSD, that’s really what it’s about. It’s aboutrelating the solutions over Q to the solutions of

the corresponding congruence equations. If youlook at finitely many of these Np, you’re not go-ing to be able to tell what the rank is, but hereyou’re looking at the asymptotic of these finiteproducts, and in the asymptotic the Np do knowabout the rank.”

L-functions

To attack the problem, mathematicians intro-duced analytical tools to study the behaviour ofthe solutions over the Zp. That’s where the L-functions come in. There’s no real definition of ageneral L-funtion. Roughly, L-functions are in-finite products indexed by the primes, but theycan’t be any such product, they have to be some-how “natural”. The best way to describe themis to give examples. Historically, the first L-function was the Riemann zeta-function.

ζ(s) =

∞∑

n=1

1

ns(0.1)

“The main result about the Riemann zeta-function is that it factors into a product overall the primes.”

ζ(s) =∏

p

(1 − p−s)−1 (0.2)

This equality, together with the fact that (0.1)converges for s > 1 is a restatement of theunique factorization theorem.

“This was proven by Euler, the first to havegiven a factorization formula for an L-function.These factors, which are indexed by the primes,are therefore called Euler factors. For Diophan-tine equations it’s the same idea as for the Rie-mann zeta-function, you look at solutions overvarious finite fields and you package togetherthis information by making a generating series.

There is a very concrete recipe which takesNp and converts it into some polynomial. Theinteresting case is when we have an elliptic curveequation. Here the polynomial is defined to bePp(x) = 1 − apx + px2, with the L-function de-fined as

L(E, s) =∏

p

Pp(p−s),

(convergent for s > 3/2). This definition of theL-function is the object of interest in the BSD.The very deep fact, which was proved by Wiles,is that it actually extends to an analytic func-tion of the complex variable s over the entirecomplex plane.”



THE ζ-FUNCTION AND UPF

ζ(s) =

∞∑

n=1

1

ns=∏

p

(1 − p−s)−1 (0.3)

if and only if every n ∈ N has a uniqueprime factorization.

Proof. First assume unique prime factoriza-tion (UPF). We can expand each factor in theleft of (0.3) into a geometric series

1

1 − p−s= 1 + p−s + p−2s + . . . + p−ns + . . .

Let pi be the primes. If we expand theproduct in (0.2), we get exactly one term(pm1

1 pm2

2 pm3

3 · · · pmk

k . . .)−s

for each list of pos-itive integers mj with a finite number of non-zero entries. Unique prime factorization tellsus that every n ∈ N can be expressed as one ofthese combinations in one and only one way.Also, every such combination corresponds toan integer. Therefore there is exactly on term1

ns for each n in N on the left of (0.3), and theequality follows.

Now suppose that (0.3) holds. Then ex-panding the product as above, we get that∏

p(1 − p−s)−1 =∑∞

n=1fn

ns where fn de-notes the number of distinct prime factoriza-tions for n. Therefore, when the sum con-verges, i.e. when s > 0, we can write (0.3)as T (s) =

∑∞n=1

an

ns ≡ 0 with an = 1 − fn.Letting s → ∞, T (s) = a1 + a2

2s + . . . = a1.Thus a1 = 0 and T (s) = a2

2s + a3

3s + . . . = 0. Asimple inductive argument shows that an = 0for all n, and therefore fn = 1. This provesthe theorem. 2

The Conjecture and the Corollary

With this fact, the BSD can be formulated inits current form:

Conjecture. For an elliptic curve E, the Tay-lor expansion of the L-function L(E, s), s ∈ C,around s = 1 is

L(E, s) = c(s − 1)r + higher order terms

where r is the rank of E(Q) and c 6= 0.

From this we get the immediate corollarythat

L(E, 1) = 0 ⇔ E(Q) is infinite,

because if E(Q) is finite (i.e. r = 0), and theBSD implies that L(E, 1) = c. Conversely, ifE(Q) is infinite, then r > 0 and all terms of theTaylor expansion vanish.

“This is one of the most striking parts of theBSD, because it’s actually easy to test numeri-cally whether L(E, 1) is 0 or not. Also, there isa theorem (independent of the conjecture) say-ing that if L(E, s) 6= 0, the group of solutionsE(Q) is finite. The converse is really, really ex-citing, since with a computer, it’s easy to com-pute L(E, 1) and see whether it’s zero or not. Ifit’s zero, this implies that there is a solution. Soeven if it’s really hard to compute it, you knowit’s there, and that would already be an amaz-ing result. Just that implication would alreadybe amazing.”

It is because of the proved implication thatprof. Darmon began working on the BSD:“My advisor at Harvard, Benidict H. Gross,proved part of the first implication. Thenthere was a Russian mathematician called Vic-tor Kolyvagin who brought another piece of thepuzzle. The combination of these two results,Gross and Kolyvagin, proved that implication,called the “easy” implication. That was aroundmy second year of graduate school and was veryexciting. It brought the BSD at the forefront.Suddenly, there were tools and techniques to ex-plore it. Then, when I was a post doc at Prince-ton, Wiles proved Fermat’s Last Theorem byshowing that the L-function had the analyticcontinuation. That was another big piece of thepuzzle for the BSD. When you’re a graduate youalways try to gravitate towards the areas wherethere’s a lot of activity, because there’s a lot ofaction.”

His Work on the BSD

When asked about his contribution to the BSDproblem, prof. Darmon laughed and explainedhis work:

“It’s a bit technical. There are all kinds ofvariants and generalizations for either cases ofthe BSD. The case I proved has to do with ellip-tic curves which are defined over number fields.Instead of looking at Q you can look at exten-sions of the rational numbers. The number fieldsI was considering were quadratic fields, i.e. theextension of the rationals by

√d for some d ∈ Z.

You can have an elliptic curve over that field andlook at the Np’s. They’re indexed by the primesof that field rather than the primes of Q. You



can do exactly the same thing and make thesame conjecture. For quadratic number fields,it’s even more mysterious. Even that “easy”implication is not completely understood, butthere were certain cases where we were able toprove it.”

I asked prof. Darmon what draws him to theBSD:

“I think it’s always important to work onproblems that are very central and have a lot ofmystery. The important thing about the BSDis that even the major ideas have been found inthe last twenty years and there’s still a lot ofuncertainty in the conjecture, it’s by no meansa done deal. When you’re doing research, youalways try to gravitate towards problems thathave this mystery. Even if you can’t prove theRiemann Hypothesis, even if you can’t provethe BSD, you keep being led to rich and use-ful results. It’s a very motivating idea, becauseyou feel like you’re really digging into somethingthat we absolutely don’t understand.

However, you want to make a compromisebetween that and not working on a problemwhich hasn’t been understood at all and whereabsolutely no progress has been made. Thereare a lot of questions in number theory of thatsort, we can come up with all sorts of questionsthat no one has any idea how to solve, where nostructure has ever been approached that wouldsay anything about the problem. So one tries toavoid those also. The BSD is a great problem:on the one hand, it’s a fundamental mystery and

at the same time there are all kinds of incrediblyrich structures that have been developed to saysomething about it. All the fundamental con-cepts of number theory have been used here atsome time or another.

Of course, how do you determine what’s animportant problem? It comes from experience.When a problem seems to be very difficult, andat the same time seems to create a lot of interest-ing mathematics, then you get this feeling thatit’s a fundamental problem. There’s a feelingabout the BSD that if we understood it, if wewere able to prove it or even prove some specialcase, then that would lead to a lot of progress,we would understand a lot of other things aswell. It’s like the Riemann hypothesis, to whichit is related by the L-functions. The BSD tries tolink two objects that seem to be part of differentworlds. One of them is an analytically definedobject, which is the L-function, the other is analgebraically defined object which is the ellipticcurve and the solutions over Q. We don’t reallyknow how to make a bridge between these twoand if we understood that mechanism, I thinkthe insight that would emerge would say some-thing about the Riemann Hypothesis. Workingon the BSD or the Riemann Hypothesis, one isbound to find something very rich and intricate.”

References

[1] Silverman JH. The Arithmetic of EllipticCurves, Springer, 1985


A Field of Six Elements? 29

A Field of Six Elements?Agnes F. Beaudry

Have you ever tried to find a field of six ele-ments? Well now is a good time to stop search-ing, because I will show here that there is nosuch field.

First, I will prove that any field contains asubfield which is isomorphic to Q or Zp for aprime p. Then, recall from your algebra classthat any field K containing another field F canbe viewed as an F -vector space, i.e. we can seeK as vector space over F . We will use this factto prove that no field can have six elements.

Theorem. Any field K contains a subfield Fisomorphic to either Q or Zp.

Proof. I’ll sketch the proof and leave the de-tails to you. You can map N0 = N ∪ {0} intothe field K through the function φ : N0 → K,φ(n) = n · 1 where n · 1 means addition of themultiplicative identity n times with itself in K.Two things can happen, either φ(n) = 0 if andonly if n = 0, or there exists an n in N such thatφ(n) = 0. In the first case, we can see that thereis a subring of K isomorphic to Z. The field gen-erated by this subring must also be in K and isisomorphic to Q. I’ll let you think about thisand move on.

If the second case holds, then take the small-est such n. If n was not prime, i.e. n = qs for

neither q, s 6= 1, then the elements q · 1 andc · 1 would be zero divisors. You can figure thisout easily. Since K is a field, this can not hap-pen, so n must be prime. Thus there is a sub-field of K isomorphic to Zp for some prime p,namely {0, 1, 2 · 1, ..., (p − 1) · 1}, and our claimis proven. 2

This theorem tells us that our field K con-tains the field Zp for some prime p (if you’renot convinced, think really hard how a field ofsix elements could contain a copy of Q. If youfind an answer, please write to us at [email protected].) Thus we can view Kas a vector space over Zp. Since K is finite,this vector space is finite dimensional, thus ithas a basis, α1, . . . , αk. Every element of K isdetermined uniquely by a linear combination,p1α1 + . . . + pkαk with coefficients pi ∈ Zp.Also, every such linear combination determinesa unique element of K. There are pk choices forthese, therefore the cardinality of K is pk. NowI leave it to the reader to prove that there doesnot exist a prime p such that pk = 6.

Remark. Nothing special about the number 6was used in this proof. This means that for anyfinite field K, the cardinality of K is a primepower. I think this is a very interesting result!

Jokes

Q: Why did the chicken cross the Mobius strip?A: To get to the other... uh.... 2

A logician sees a sign on his way to fish that reads, “All the worms you want for 1 dollar.” Hestops his car and orders 2 dollars’ worth. 2

A physicist has been conducting experiments and has worked out a set of equations which seemto explain his data. He asks a mathematician to check them. A week later, the mathematiciancalls: “I’m sorry, but your equations are complete nonsense.” “But these equations accuratelypredict results of experiments. Are you sure they are completely wrong?” asks the physicist. Themathematician replies, “To be precise, they are not always a complete nonsense. But the only casein which they are true is the trivial one where the field is Archimedean...” 2

A team of engineers is trying to measure the height of a flag pole, but they can’t keep the measuringtape on the pole, since it kept falling off. A mathematician passes by, asks them what the problemwas, then proceeds to remove the pole and lay it on the ground. After he leaves, an engineer saysto another, “Just like a mathematician! We need the height, and he gives us the length!” 2


30 Euler’s Brick

Euler’s brick

Alexandra Ortan and Vincent Quenneville-Belair

An Euler brick is a parallelepiped with integer sides whose face diagonals are alsointegers. Euler is the first to have thoroughly investigated the problem and has thusbequeathed it his name. If such a parallelepiped’s body diagonal also happens to bean integer, it is called an Euler integer brick, or perfect cuboid. In spite of numerousefforts, no perfect cuboid has yet been discovered.

Finding instances of an Euler brick (EB) isequivalent to finding solutions to the first threeof the following system of Diophantine equa-tions. If the last one is also satisfied, one hasa perfect cuboid (PC).

a2 + b2 = d2

a2 + c2 = e2

b2 + c2 = f2

a2 + b2 + c2 = g2.

a

g

f

e

d

c

b

If the edges a, b and c are not relativelyprime, i.e. there exists a t such that t|a, b, c,then the equations above can all be divided byt2 on both sides and thus the perfect cuboid isscaled down to a smaller one whose edges are rel-atively prime. Such cuboids are called primitivecuboids, and are the most interesting ones, sinceany other can be obtained by taking a primitivecuboid and scaling it up.

Pythagorean triples

Before attempting to solve the whole Diophan-tine system for the perfect cuboid, it is instruc-tive to take a closer look at the individual equa-tions. One observes that the first three equa-tions describe Pythagorean triples (PT), i.e.triples of positive integers that form a right-angled triangle. Here’s a way to generate thesetriples.

Consider the integers x, y, z such that x2 +y2 = z2. Once again, one is only really in-terested in primitive triples, so the set of so-lutions can be restricted to triples such thatgcd(x, y, z) = 1. This implies the following:

Theorem 1. gcd(x, y) = gcd(y, z) =gcd(x, z) = 1.

Proof. Suppose t|x, y; then x2 + y2 = z2 ⇒t2(x2

t2 + y2

t2 ) = z2 ⇒ t|z. 2

Theorem 2. z is odd and x, y have oppositeparity.

Proof. Assume x = y = z = 1(mod2); this im-plies that x2 = y2 = z2 = 1 (mod2) which leadsto 1 + 1 = 1 (mod2), a contradiction. The sameargument holds for one side odd and two even.Thus, only one side can be even (if they were alleven, they would not be a primitive triple). As-sume z = 0 (mod2) and x = y = 1 (mod2); thenz2 = 0 (mod4) and x2 = y2 = 1 (mod4), whichis a contradiction because 1 + 1 6= 0 (mod4).Hence, the hypotenuse of a primitive PT is oddand one of the legs is even. 2

Theorem 3. All primitive PTs can be gener-ated by two integers p, q of opposite parity whosegcd is 1.

Proof. Finding PT is equivalent to finding in-teger points on the circle x2 + y2 = z2, or ra-tional points on the circle X2 + Y 2 = 1 (whereX = x

z and Y = yz ). Any rational point (a

b , cd )

can be joined to the point (0, 1) by a line withrational slope - just let p = cb and q = ad + bd.Then Y = p

q (X +1) is the equation of that line,

where gcd(p, q) = 1. This line intersects the

circle precisely where X2 + p2

q2 (X + 1)2 = 1 or

X = ±q2−p2

p2+q2 ; since only positive solutions are ofinterest, we will keep the + sign. Together withY = 2pq

p2+q2 , this forms a rational point on the

circle X2 + Y 2 = 1. To recover integer points(x, y), one need only choose z = p2 + q2.


Euler’s Brick 31

(-1,0)

(X,Y)

y

x

slopepq=

Since gcd(p, q) = 1, p and q cannot both beeven. Assume p = q = 1 (mod2); then p2 =q2 = 1 (mod2), but z = p2 + q2 = 0 (mod2)which implies that the PT is not primitve andcontradicts statement 2. Hence, p and q haveopposite parity. 2

To sum up, here are a few useful properties ofprimitive Pythagorean triples.

Property 1. One leg is odd, the other is evenand the hypotenuse is odd.

Property 2. One leg is divisible by 3.

Proof. Assume neither x nor y are divisibleby 3. Then x2 = y2 = 1 (mod3), and z2 =2 (mod3). However, this equation has no solu-tions in Z3. 2

Property 3. One leg is divisible by 4.

Proof. The PT’s formula states that y = 2pq,where p and q have opposite parity. Thus,4|y. 2

Property 4. One member of the triple is divis-ible by 5.

Proof. Assume any two are not, sox 6= 0 (mod5) and y 6= 0 (mod5)where x and y can be any two edges.Thus the third member is z2 = ±x2 ± y2.We have x2, y2 ∈ {0, 1, 4} (mod5) ⇒z2 ∈ {±1 ± 1,±1 ± 4,±4 ± 4} (mod5), so z2

can only take on the values 0,2,3. However,among those, only 0 is a square in Z5, whichimplies that z must be divisible by 5. There-fore, at least one edge is divisible by 5. On theother hand, if 5 divides two or more edges, thetriple is no longer primitive. 2

Euler bricks

Having thus gleaned some information about thedivisibility of PTs, one can now wonder how thisbrings one any closer to finding an Euler brick.In fact, it is now possible to derive a series ofproperties of the edges of Euler bricks. Hereare a few of these properties, along with theirproofs.

Theorem 4. There is exactly one odd edge.

Proof. Every pair of edges is also a pair of legsof a PT. Since at least one edge must be evenin any PT, exactly two edges must be even in aprimitive EB. 2

Theorem 5. One edge must be divisible by 4and another by 16.

Proof. Exactly two edges of the EB are divisi-ble by 4, by the same argument as above. Con-sider the PT formed by these two edges and di-vide it by the gcd of its sides - one obtains aprimitive PT, which must still have one leg di-visible by 4. Therefore, one of the edges of theEB must be divisible by 16. 2

Theorem 6. One edge must be divisible by 3and another by 9.

Proof. One can apply the same method asabove. 2

Theorem 7. One edge must be divisible by 5.

Proof. Suppose none are. Property 10 impliesthat all face diagonals of this EB must be di-visible by 5. An edge can be congruent to ei-ther ±1 or ±2 (mod5). Two edges cannot bothbe congruent to ±1 (mod5) since their corre-sponding face diagonal would not be an integer,d2 = a2 + b2 = 2 (mod5) not having any solu-tions. The same argument stands for two edgesboth congruent to ±2 (mod5). Some edge musttherefore be congruent to 0 (mod5). 2

Theorem 8. One edge must be divisible by 11.

Proof. In Z11, the only perfect squares are{0, 1, 3, 4, 5, 9}, so the square of the hypotenuseof a PT can only take on those values. Abit of calculation reveals that the only pairsof legs whose squares add up to one of thosenumbers are {(0,±1), (0,±2), (0,±3), (0,±4),(0,±5), (±1,±2), (±1,±5), (±2,±4), (±3,±4),(±3,±5)}. Supposing no edge is divisibleby 11 reduces those possibilities to {(±1,±2),


32 Euler’s Brick

(±1,±5), (±2,±4), (±3,±4), (±3,±5)}. How-ever, one can verify that if an edge takes onany of those values, the two other edges can-not themselves also form a pair of legs of a PT.Hence no such EB exists. 2

Perfect cuboids

In addition to being an Euler brick, a perfectcuboid’s body diagonal must also be an inte-ger. This diagonal would have to be the hy-potenuse of three Pythagorean triples (not nec-essarily primitive), each of which must have a legwhich is in turn the hypotenuse of another PT.

a

g

c

b

The first condition is in principle easily compliedwith: if an integer has n distinct prime factors ofthe form 4k+1, then it can be the hypotenuse of2n−1 distinct primitive PTs [9]. The second con-dition, however, is more elusive: three of thosePTs must have an edge of the cuboid as one oftheir legs, such that the sum of the squares ofthose legs is precisely a2 + b2 + c2 = g2.

The problem can be viewed the other wayaround: find a number whose square is thesum of three squares which in turn, paired twoby two, generate three PTs. In this case, onewould have to first look at integers g such thatg2 = a2+b2+c2. A derivation similar to that forPTs, but involving rational points on a spherethis time, shows that for integers p, q, r withgcd(p, q, r) = 1, a = r2−p2−q2, b = 2qr, c = 2prand g = r2 + p2 + q2 satisfy g2 = a2 + b2 + c2.It is still unclear precisely when the integersa, b, c thus obtained are the legs of three differ-ent right-angled triangles.

Results to date

The smallest Euler brick is (240,117,44) andwas discovered by Paul Halcke in 1719. Later,Saunderson found a parametric solution thatalways generates Euler bricks, but not all ofthem. Starting with a PT (x, y, z) he showed

that (a, b, c) = (x(4y2 − z2), y(4x2 − z2), 4xyz)is an Euler brick with face diagonals dab = z3,dac = x(4y2 + z2), dbc = y(4x2 + z2). Lagrange,however, proved that none of those, nor any de-rived cuboids are perfect cuboids [6]. Some twocenturies later, Korec used a computer to showthat the smallest edge of a perfect cuboid canbe no smaller than 106, proving along the waya few numTheorems to speed up the algorithm.More recently, Rathbun has increased the lowerbound to 232.

Theorem 9. The following is equivalent to find-ing the PC:

1. Let a = p2+q2

2pq and b = p2−q2

2pq . The questionis to know whether it is possible to get ab and a

bin the form of b.

2. Find non-trivial integer solutions to (a2c2−b2d2)(a2d2 − b2c2) = (a2b2 − c2d2)2

Theorem 10. If there exists a PC, the follow-ing also exist:

1. A classical rational cuboid with edges x1,x2, x3 and a square z2 such that z2 − x2

i are allsquares.

2. A set of four non-zero squares whose differ-ences are all squares.

3. Two ratios of the form p2−q2

2pq , whose prod-uct and ratios are of this form.

4. A set of four squares whose sums in pairsare also square.

5. An arbitrarily long or infinite sequence ofsquares, where the sum of any two (or three)adjacent members is also a square.

6. A cycle of integer solutions toα2

1−β2

1

2α1β1

α2

3−β2

3

2α3β3

=α2

2+β2

2

2α2β2

with α1

β1

= p2−q2

2pq andα2

β2

= r2+s2

2rs .

7. The validity of Conjecture C in [3] (whichis far beyond the scope of this article)...

Solutions to the PC with relaxation (oneedge or one face diagonal irrational) have alsobeen studied.

The problem of finding a PC with one edgeirrational is equivalent to finding an integer so-lution to x2

1 + x22 = y2

3 with t + x21, t + x2

2 andt+y2

3 where t is an integer – the square of the ir-rational edge. One solution (x1, x2, y3) is (124,957, 13852800). It can be extended to a one-parameter family of solutions.


Euler’s Brick 33

Finding a PC with one face diagonal irra-tional corresponds to solving the following sys-tem of equations: x2

1+x22 = y2

3 , x21+x2

3 = y22 and

x21 + x2

2 + x23 = z2. There are no conditions on

x22 + x2

3. One can see that the equations implythat 2(z2 + x2

1), 2(z2 − x21) and 2

∣

∣x22 − x2

3

∣

∣ havetheir sums and differences square. The sums inpairs are 2z, 2x1, 2y2, 2y3, 2x2 and 2x3. Notethat the differences of z, y2, x3 and z, y3, x2 areall squares.

Conclusion

If the problem of finding a perfect cuboid hasremained the same for many centuries, the tech-niques employed to tackle it have evolved muchsince Euler first gave it some serious thought.More recently it has been viewed through thelens of algebraic geometry, where solutions tothe perfect cuboid translate into rational pointson curves, a question much beyond the scope ofthis article.

If such pursuits as this appear appealingto the reader, the following problems have asimilar flavor. With money helping motivation,there is a million dollars price for the solutionof the Birch and Swinnerton-Dyer Conjecture.[http://www.claymath.org/millennium/Birchand Swinnerton-Dyer Conjecture]. For only$100,000, the Beal Conjecture is a good deal.It goes as follows: let A, B, C, x, y and zbe positive integers where x, y and z are allgreater than 2 such that Ax + By = Cz ,then A, B and C have a common divisor.[http://www.bealconjecture.com/]. There isalso a small hundred dollars from Martin Gard-ner for finding a 3x3 magic square with 9 distinctsquare entries. Moreover, Sierpinski asks if there

are non-trivial solutions to (x + y + z)3 = xyz.Finally you can try to find the solutions for1n + 2n + ... + kn = (k + 1)n with n > 1 anda characterization of the integers d for whichx2 − dy2 = −1 has an integer solution.

References

[1] N. Saunderson, The Elements of Algebra, Vol2, Cambridge, 1740.

[2] J. Lagrange, Sur le derive du cuboıde Eu-lerien, Canadian Mathematical Bulletin, Vol 22,1979.

[3] J. Leech, The Rational Cuboid Revisited,The American Mathematical Monthly, Vol 84,No 7, p. 518-533, 1977.

[4] P. Halcke, Deliciae Mathematicae; oder,Mathematisches sinnen-confect. Hamburg, Ger-many, N. Sauer, 1719.

[5] M. Kraitchik, On certain Rational Cuboids,Scripta Math. 11, 1945.

[6] I. Korec, Nonexistence of small Perfect Ra-tional Cuboid I/II, Acta Mathematica Universi-tatis Comenianae, 1983/1984.

[7] Jay R. Goldman, The queen of mathemat-ics : an historical motivated guide to numbertheory, Wellesley, 1998.

[8] R. Luijk, On Perfect Cuboids, doctoral the-sis, University of Ultrecht, 2000.

[9] A. H. Beiler, The Eternal Triangle, Ch. 14in Recreations in the Theory of Numbers: TheQueen of Mathematics Entertains, New York:Dover, 1966.

Jokes

“Have solved the Riemann Hypothesis” – G. H. Hardy 2

A physicist and an engineer are in a hot-air balloon. Soon, they find themselves lost in a canyonsomewhere. They yell out for help: “Helllloooooo! Where are we?” 15 minutes later, they hear anechoing voice: “Helllloooooo! You’re in a hot-air balloon!” The physicist says, “That must havebeen a mathematician.” The engineer asks, “Why do you say that?” The physicist replies, “Theanswer was absolutely correct, and it was utterly useless.” 2


34 Ten Proofs of the Infinitude of Primes

Ten Proofs of the Infinitude of PrimesNan Yang

A prime number is a counting number, greaterthan 1, that cannot be divided except by 1 anditself. Prime numbers have fascinated amateurand professional mathematicians for thousandsof years. While I am far from being qualifiedeven as an amateur mathematician, I would liketo share my fascination with these figures bypresenting a collection of proofs of the followingtheorem, which has been known since ancienttimes:

Theorem. There exists an infinite number ofprimes.

We will begin with a modernized version ofEuclid’s original proof which appeared as propo-sition 20 in book 9 of The Elements over 2000years ago. Although Euclid used slightly differ-ent notations (he proved the case where n = 3),the idea of the proof has not changed since thattime:

Proof. 1 Suppose there are only n primesp1, ..., pn. Then p1...pn + 1 is either prime ornot. If it is prime, then we have found anotherprime, hence the hypothesis must be false. If itis not prime, then it must be divisible by someprime pi where 1 ≤ i ≤ n. But since pi alsodivides p1...pn, it must divide the difference, 1,which is impossible, and hence the hypothesismust be false. Therefore the number of primescannot be finite. 2

By using ideas similar to those used in thefirst proof, it is possible to produce a muchshorter proof:

Proof. 2 Any prime divisor of n! + 1 is greaterthan n. 2

Hidden in the statement of the previousproof the fact that a divisor of any number isobviously less than or equal to the number it-self. Hence there is at least one prime betweenn and n! + 1, and another between n! + 1 and(n! + 1)! + 1, etc.

It is possible to produce a proof based on Eu-clid’s method, but without adding a unit. Stielt-jes first published such a proof, of which a vari-ant is shown here:

Proof. 3 If there are only n primes p1, ..., pn,then p1...pi + pi+1...pn is not divisible by any ofthe n primes, since each prime divides exactlyone of the summands. 2

Metrod gave a proof similar to that of Stieltjes:

Proof. 4 Suppose there are only n primesp1, ..., pn. Let N =

∏

i≤n pi and let Qi = N/pi.Thus each of the n primes does not divide ex-actly one of the summands of S =

∑

i≤n Qi;therefore, S is not divisible by any of the primesp1, ..., pn. 2

Hardy and Wright’s Introduction to the The-ory of Numbers contains a very intricate proofwhich uses ideas that are different from those ofEuclid:

Proof. 5 Suppose that 2, 3, ..., pj are the firstj primes and let N(x) be the number of n ≤ xsuch that n is not divisible by any prime p ≥ pj .Any such n can be expressed in the form n = a2bwhere b is squarefree, i.e. b = 2e13e2 ...p

ej

j where

ei = 0 or 1. Hence, there are 2j possible val-ues of b. Since a ≤ √

n ≤ √x, there are not

more than√

x different values of a. ThereforeN(x) ≤ 2j

√x.

Now suppose there are only j primes, thenN(x) = x for all x. This implies that x ≤ 22j ,which is false for x ≥ 22j + 1. 2

One way of proving that the primes are in-finite is by constructing an infinite sequence ofnumbers that are pairwise coprime, that is, theprime factors of an element of that sequence isunique to that element. A very well known suchsequence is the Fermat numbers Fn, whichare defined as Fn = 22n

+ 1.

Proof. 6 Suppose Fn and Fn+k have a commonfactor m. Let x = 22n

. We then have

Fn+k − 2

Fn=

x2k − 1

x + 1= x2k−1 − x2k−2 + ... − 1.

Hence m divides Fn+k − 2, which means m = 2,but this is impossible since all Fermat numbersare odd. 2

A number is a Fermat prime if it is a Fer-mat number and is prime. An open questionabout Fermat primes is that whether they are


Ten Proofs of the Infinitude of Primes 35

infinite. Although Fermat himself conjecturedthat all numbers of the form 22n

+ 1 are prime,this is now known to be false. In fact, so far onlythe first five are known to be prime, the largestbeing F4 = 65537. Many mathematicians be-lieve that all Fermat numbers greater than F4

are composite; the largest known such compos-ite is F23471.

It is also possible to prove the infinitudeof primes by constructing arbitrarily long se-quences whose elements are pairwise coprime.Schorn produced one such sequence as follows:

Proof. 7 If 1 ≤ i < j ≤ n, then any divisor ofn!i + 1 and n!j + 1 must divide the difference,which is n!(j − i). However, by proof 2, any di-visor of n! + 1 can not divide n! and must begreater than n. Therefore

gcd(n!i + 1, n!j + 1) = 1,

and the n integers n!i+1(i = 1, 2, ..., n) are pair-wise coprime. 2

Recall the following lemma: gcd(a, b) = 1 ⇔there exists s, t such that sa+ tb = 1. With thisin mind:

Proof. 8 Let q1 = 3, qn+1 = q1...qn − 1. With-out loss of generality let i < j. Then

j∏

k=1

qk − qj = 1.

Let A =∏j

k=1 qk/qi. We thus have Aqi−qj = 1.Therefore, any two elements qi, qj are coprimeby the converse of the previous lemma. Since(qn) is increasing for n > 2, there exists in-finitely many primes. 2

A more exotic proof came from Furstenberg,based on topological ideas, in 1955. Here is avariant:

Proof. 9 Define a topology on the integers Z

by taking the set of arithmetic progressions from−∞ to ∞ as a basis. One can check that thisgives an actual topological space. Note that thecomplement of an arithmetic progression is theunion of the other arithmetic progressions withthe same step size, so that arithmetic progres-sions are both closed and open. Now, considerA =

⋃

p Ap where p runs through all the primes≥ 2, and Ap is the (closed) set of all multiples of

p. The complement of A is {−1, 1}, because allother numbers are primes or multiples of primes.Since {−1, 1} is obviously not open, its com-plement A cannot be closed. However, a finiteunion of closed sets is closed, so there must beinfinitely many Ap, i.e. there are infinitely manyprimes. 2

Recall the geometric series identity

∞∑

k=0

1

pk=

1

1 − (1/p).

If p, q are two primes, then

1+1

p+

1

q+

1

p2+

1

pq+

1

q2+ ... =

1

1 − 1p

× 1

1 − 1q

.

By the unique factorization theorem and bymultiplying the sums of the reciprocals of ev-ery possible prime to every possible power, weobtain14

∑

n∈N

1

n=∏

p∈P

1

1 − (1/p).

Euler was the first person to discover this iden-tity. The left-hand side of Euler’s identity isa sum taken over the counting numbers whilethe right-hand side is a product taken over theprimes. We will thus conclude with an analyticproof of the infinitude of primes:

Proof. 10 By comparing the harmonic series toan integral, we obtain:

∫ x

1

dn

n + 1<

x∑

n=1

1

n

for all x. Since the integral diverges as x → ∞,so does the sum. By Euler’s identity, the num-ber of primes cannot be finite. 2

References

[1] Hardy, G. H. and Wright, E. M. “An Intro-duction to the Theory of Numbers.” Oxford atthe Clarendon Press, 1960.

[2] Ribenboim, Paulo. “The New Book of PrimeNumber Records.” Springer, 1996.

[3] Ore, Oystein. “Number Theory and Its His-tory.” Dover Publications, 1988.

14See A. Beaudry, Interview with prof. Henri Darmon for a proof.


36 Living without Math

Living without MathVincent Quenneville-Belair

Living without math is possible – as surprising as it seems. The 200 members of thePiraha tribe, gathered in groups of ten to twenty near the Amazon, live without theconcepts of numbers and counting.

Daniel Everett, an American linguistic an-thropologist, and Peter Gordon, a psycholin-guist from New York’s Columbia University,studied the Piraha. The former lived with thistribe for 27 years; the latter did some experi-ments with them over a three-year period.

No Math

Gordon’s study was conducted on a group ofmen only, because cultural taboos excludedwomen and children. Everett tried to teachthem how to count for eight months, but “inthe end, not a single person could count to ten.”One of the experiments consisted in duplicat-ing a line of batteries. Beyond two or three,the men started making mistakes. The prob-lem seemed to come from their lack of wordsfor counting. The word they use for “one” iscloser to “a small amount” and the word for“two” is like “a relatively bigger amount”. Itis impossible to know if a Piraha is designat-ing one fish, a small fish or two fishes. Theyalso have trouble drawing: “Producing simplestraight lines was accomplished only with greateffort and concentration, accompanied by heavysighs and groans.” (Gordon)

Other characteristics

The Pirahas do not have a written language;they communicate by singing, whistling andhumming. Their pronouns seem to originatefrom another language, and some of them com-bine singular and plural: “he” and “they” are

the same; “more”, “several” and “all” are inexis-tent. Furthermore, their collective memory doesnot extend more than two generations back. Noequivalent to our art and fiction seems to ex-ist. Also, they simply say, when urged to ex-plain their history, that“everything is the same”.Everett explains that the Piraha culture limits“communication to non-abstract subjects in im-mediate experience for the interlocutors”.

Notable Facts

The particularities of the Pirahas are not ex-plained by social or genetic isolation, becausethey have been trading goods and women withBrazilians for 200 years. Their motivation tolearn came from such exchanges with outsiders:they wanted to know if they were being cheatedbecause they did not understand non-barter eco-nomic relations. Their enthusiasm made themattend daily classes given by Everett and hisfamily with great interest. However, adding 3to 1 remained impossible for them, and the Pi-raha concluded that they could not learn thematerial.

References

[1] D. Everett. “Cultural constraints on gram-mar and cognition in Piraha: Another look atthe design features of human language”, CurrentAnthropology 46 (4): 621-46, 2005.

[2] J. Crow, 2006. [www.jcrows.com/ without-numbers.html]


Interview with Professor Nilima Nigam 37

Interview with professor Nilima Nigam

Nilima Nigam is a professor of mathematics at McGill, where she teaches differentialequations and numerical analysis to terrified (but increasingly enthusiastic) under-graduates. She was kind enough to grant the Delta-Epsilon an interview about herwork and – yes – life as a mathematician.

The δelta-ǫpsilon: What area of research areyou working on now? Can you tell us aboutsome of your projects? What led you to thatspecific area and project?

Nilima Nigam: This year, I’m focusing on fourprojects, three of which are in the area of nu-merical analysis (the study of algorithms andtheir convergence), and one of which is a col-laboration with Prof. Komarova in the Facultyof Dentistry. The latter is an investigation intothe growth dynamics of bone cells. It’s a funproject- my collaborator has actual cultures ofthese cells, and lots of experimental data. Myjob is to build mathematical models to describewhat we see, and help predict specific thingsabout the system. The models lead to new bi-ological questions, which in turn suggest newexperiments. The results of these experimentsmay confirm our model, or require us to refineit. This summer, the SUMS treasurer, Tayeb,was deeply involved with this project.

In the long term, I’m deeply interested in themathematics of wave interactions with boundedobjects. The waves could be sound, electromag-netic, elastic, gravitational or pressure waves.This field is quite old, and has motivated thedevelopment of many branches of mathematics.I’ve been thinking about problems in this areafor well over a decade, and every time I thinksomething is settled, another interesting ques-tion arises.

While performing numerical simulations ofwave-obstacle interactions, one is constrained todescribe the physical problem on a bounded re-gion. Think of sound scattering off a hedgehog.In theory, the scattered wave can propagateforever. Computationally, though, we need toput a box around the hedgehog, and try to cap-ture the behaviour of the scattered wave insidethe box. This process introduces errors; it’snot at all obvious what boundary conditions toprescribe on the box, and finally creating algo-rithms which are provably convergent and accu-rate is hard. Two of my projects concern tech-niques for this problem; I’m collaborating with

a graduate student, Simon Gemmrich, on oneof these. There are many interesting analyticalquestions at the PDE level, and then a wholehost of other questions concerning the numericalanalysis of the methods. This work is strictlyof a theorem-proof nature, though of course Idesign experiments to test the methods.Associated with scattering problems, but alsomore generally applicable, are a class of nu-merical methods for PDE known as finite ele-ment methods. These work by approximatingthe solution in terms of compactly supported(polynomial) basis functions. Recently, muchwork has been done to develop a finite elementexterior calculus using the tools of differen-tial geometry and homological algebra. In thisframework, discretizations of PDE are designedto be compatible with the geometric, algebraicand topological properties of the actual solu-tions of the PDE. Existing work requires thepolynomial basis functions to be defined onsimplicial objects–tetrahedra or boxes in 3D. Inwork with Joel Phillips, another graduate stu-dent, we’re trying to extend this framework tonon-simplicial objects such as pyramids.


38 Interview with Professor Nilima Nigam

Fig.1 A plane wave incident on a hedgehogand the associated scattered wave. The dashedline indicates the artificial boundary we’d putaround Hedgie while computing the scatteredwave.

δ − ǫ: How does your research relate to whatis taught in undergraduate courses here?NN: I admire the undergraduate programbroadly for instilling a sense of mathematicalfearlessness. On the many occasions when I’mstuck on a research problem, I think to my-self - ‘if this were a homework problem, and Iwere a McGill undergrad, when would I quit?’A lot of my work requires functional analysisand PDE (not just separation of variables!).You see some of the analytical tools involvedin courses like Math 564/565. At an advancedlevel, the study of PDE merges with analysis(Math 575/580/581). In addition, courses in nu-merical analysis or matrix computation (Math317/387, 327/397, 578/579) contain many ofthe key concepts–stability, accuracy and conver-gence of numerical algorithms- which I use. In-creasingly, algorithms for PDE incorporate toolsfrom differential geometry, which is another fieldyou may see during your studies here.

I’m not sure if students in Mathematics takecourses in the Physics department. In an idealworld, budding mathematicians interested inthe mathematics of scattering theory would seeclassical mechanics, electrodynamics and quan-tum mechanics.

δ − ǫ: Why did you choose to become a math-ematician? What kind of a life is it?

NN: Becoming a mathematician didn’t reallyinvolve a choice. I’m very fortunate to be ableto do what I love, and would be miserable do-ing anything else. I started off as a student ofPhysics, realized I loved the mathematical as-pects of my training most, and ended up pur-suing a career in mathematics. Physics has big

problems, which filter into our collective con-sciousness - this decade, quantitatively inclineddreamers want to work in String Theory. A mil-lennium ago, High Energy Physics and Cosmol-ogy captured my imagination. By the time I gotthe necessary educational background to beginto understand the science behind these areas, Irealized mathematics was my deeper passion.

It’s a great life. I’m fortunate enough toenjoy the various aspects of my chosen career-doing research, teaching, and interacting withother mathematicians and scientists. Some mayfavour one part of this life, and regard the otherbits as distractions. I’m energized by mathe-matics, and by the belief that fun mathemat-ical questions can be found everywhere. Thismakes teaching and interacting with people partof the larger search for interesting mathematicalquestions and their resolution - any given lec-ture, a student could ask me something thought-provoking.δ − ǫ: Are there any particular open problemsyou’d like to see the solutions to?NN: There are several open and rich ques-tions in mathematics, and several technicalquestions in my field of interest I’d like tosee resolved. However, a particularly chal-lenging mathematical question concerns the ex-istence and regularity properties of solutionsof the Navier-Stokes equations in R3. Thissystem of partial differential equations gov-erns the motion of fluids; the study of theirsolutions will require huge advances in theanalysis of PDE. The question evades stan-dard methods of attack, and a successful ap-proach will likely be both surprising, and in-timately connected to many other branches ofmathematics. The problem was recently clas-sified as one of the Millennium problems bythe Clay Institute; the precise problem state-ment due to Charles Fefferman is available athttp://www.claymath.org/millennium/Navier-Stokes-Equations/navierstokes.pdf.

δ−ǫ: What major mathematical event (be-yond your own work) do you remember mostvividly? Alternatively, what such event had thebiggest impact on your life as a mathematician?NN: This one’s tricky... I cannot think of a sin-gle formative experience. Rather, many chanceencounters with mathematicians I admire haveimpacted my career. Reading about the lives offamous mathematicians and their work habitshas always inspired me.



The Birth of QuaternionsMichael McBreen

Where do vector spaces come from? Did Gauss wake from troubled slumber onenight, spring from his bed and shout, “Let there be a set V and a field F , and let+ : V × V → V be an associative, commutative operation such that ...”? No, no hedid not.

Modern vectors arguably evolved from Hamil-ton’s quaternions15, the set of numbers of theform

a + bi + cj + dk

with a, b, c, d ∈ R and

i2 = j2 = k2 = ijk = −1.

As we will see at the end of this article (keepreading), not only vectors but also the innerproduct and the cross product of vectors arehiding inside quaternions. Better still, they gaveus the concept of non-commutative number sys-tems. I write this to convince you, the reader,that it is worth your time to learn how quater-nions were discovered. The thing is, we areblessed with an unusually detailed account ofthe birth of quaternions16 from the correspon-dence of Hamilton himself, so we can follow histhought process step by step.

But first, a few words about the ancestors ofthe quaternions themselves, the complex num-bers. The square root of -1 was introducedby Cardano to solve cubic equations. Complexnumbers share many properties with the reals,such as

z + w = w + z

wz = zw

w(z + x) = wz + wx

∀z ∃z′ s.t. zz′ = z′z = 1

|wz| = |w||z|.

The last property was known to Hamilton as thelaw of the moduli, and will play a crucial role inwhat follows.

Mathematicians were very queasy aboutcomplex numbers at first – back then, even neg-ative numbers were suspicious – but eventuallya whole host of people developed a geometricinterpretation of complex numbers as lines in aplane (the Argand plane, for those of you whowent to high school). Each number a + bi cor-responds to a line stretching from the origin to(a, b), and multiplication corresponds to addi-tion of angles and multiplication of lengths inthe plane.

Well, that’s complex numbers for you. Nowfast forward to the 1800s, and enter WilliamRowan Hamilton. Given the interpretation ofcomplex numbers as directed lines in a 2D plane,it’s quite natural to desire a similar system for3D space. Hamilton suspected that physicalconcepts like velocity and force could find natu-ral expressions in this hypothetical system; theexisting notation was extremely cumbersome.

His goal was to forge a system of “triplets”a + bi + cj that shared most of the propertieslisted above, among them the law of the moduli,the existence of inverses and distributivity. Thecatch is that there’s no such system: Hurwitzwould prove half a century later that any finitedimensional normed division algebra17 over thereals has dimension either 1, 2, 4 or 8. Real andcomplex numbers are dimension 1 and 2 normeddivision algebras – Hamilton was going for 3 di-mensions, but his efforts would yield the 4 di-mensional quaternions instead.

Let’s see how he went about looking for histriplets. In order to define triplets completely,he only needed to choose (or find, if you will)the values of the products ij, ji, i2 and j2 – ev-

15For brevity’s sake we will neglect the other great ancestor of vectors, Grassman’s exterior calculus.16Much of this information is summarized by a nice article [1] from B.L. van der Waerden, which I encourage you

to seek out on the internet.17An algebra over the reals is roughly a vector space V over the reals equipped with a bilinear operation

∗ : V × V → V which you can think of as a multiplication of vectors. The complex numbers can be viewed as avector space over the reals with 1 and i as basis vectors. When we define 1 ∗ 1, 1 ∗ i, i ∗ 1 and i2 = i ∗ i and usedistributivity to extend the product to the whole vector space, it becomes an algebra. A division algebra is analgebra where very element except 0 has a multiplicative inverse. A normed algebra satisfies the law of the moduli.

18The “purely real” triplets a + 0i + 0j were presumed to commute with all other triplets.


40 The Birth of Quaternions

erything else would follow by distributivity andthe properties of real numbers.18 The challengewas to choose values that would satisfy the basicproperties listed above.

First he went after i2 and j2. With the normsquared defined as the sum of the squares of thecoefficients (this comes to us from Pythagoras’theorem), the law of the moduli for Hamilton’shypothetical triplets reads

|(a+bi+cj)(α+βi+γj)| = |a+bi+cj||α+βi+γj|

with

|a + bi + cj| =√

a2 + b2 + c2

Considering the subset of triplets of the forma + bi or a + bj, Hamilton used distributivity toget

(a + bi)(α + βi) = aα + (aβ + bα)i + bβi2.

Setting i2 = A + Bi, with A and B as yet un-determined, he then used the law of the modulito get

|(a + bi)(α + βi)| = (aα + bβA)2 + (aβ+

bα + bβB)2

= (a2 + b2)(α2 + β2).

To make this equality hold, Hamilton hadto set i2 = −1. The exact same procedure alsogave j2 = −1. Now, he had both i2 and j2, so hewas only missing ij and ji. If we assume com-mutativity, then (ij)2 must be 1, so ij shouldbe either 1 or −1. Alas, both of these optionsviolate the law of moduli (check for yourself).

So he forgot commutativity for a second andconsidered the square of the triplet a+bi+cj. Henoticed that the very demanding law of moduli

|(a + bi + cj)2| = |a + bi + cj|2

would be satisfied with ij = 0, and that further-more it gave the product a geometric interpre-tation, just like the complex number product.Just as the square of a complex number z lies attwice the angle with respect to the real axis as zitself, the square of a triplet (with ij = 0) lies attwice the angle with respect to the “real axis”,i.e. the axis of (a, 0, 0) triplets. But happinessis fleeting, as this quote reveals:

Behold me therefore tempted fora moment to fancy that ij = 0.

But this seemed odd and uncomfort-able, and I perceived that the samesuppression of the term which wasde trop might be attained by assum-ing what seemed to me less harsh,namely that ji = −ij. I made there-fore ij = k, ji = −k, reserving tomyself to inquire whether k was 0 ornot.

The reader will note that Hamilton’s ambi-tions first leaned toward poetry, but his friendWordsworth wisely advised him to study math-ematics instead. Anyway, Hamilton next con-sidered the product

(a + bi + cj)(x + bi + cj) = ax − b2 − c2+

b(a + x)i + c(a + x)j + (bc − bc)k

He wrote:

The coefficient of k still vanishes;and ax − b2 − c2, (a + x)b, (a +x)c are easily found to be the cor-rect coordinates of the product-pointin the sense that the rotation fromthe unit line to the radius vector ofa, b, c being added in its own planeto the rotation from the same unit-line to the radius vector of the otherfactor-point x, b, c conducts to theradius vector of the lately mentionedproduct-point; and that this latter ra-dius vector is in length the productof the two former. Confirmation ofij = −ji; but no information yet ofthe value of k.

Mysterious k. Well, the next thing Hamiltondid was to bravely consider the general productof two triplets:

(a + bi + cj)(x + yi + zj) = ax − by − cz+

(ay + bx)i + (az + cx)j + (bz − cy)k

He then checked whether the law of the mod-uli was happy with k being null, and foundthat with k = 0, the left-hand side’s norm wassmaller than the right-hand side’s by (bz− cy)2.Curious – exactly the square of the coefficientof k. As Hamilton and his wife strolled alongDublin’s Royal Canal on the 16th of October1843, inspiration struck: if he set k on equalfooting with i and j, and hence added a fourthdimension, then the norms would match. Thiswas the key result, the crucial step. With kgiven the respect it deserved, everything flowedsmoothly:



I saw that we had probably ik =−j, because ik = iij, and i2 = −1;and that in like manner we might ex-pect to find kj = ijj = −i;

Hamilton writes “probably” because he wasnot sure whether or not his quaternions – as theymust now be called – were associative. He wasa wise, cautious man and you would do well toemulate him.19 The same reasoning that led toi2 = j2 = −1 now led him to k2 = −1, and hehappily carved the defining equations

i2 = j2 = k2 = ijk = −1

in the stone flank of Broom Bridge.And what about vectors? Well, Hamilton

called the real component of his quaternions the“scalar part” whereas the i, j, k component wasthe “vector part”. Now, consider the product oftwo “pure vector” quaternions:

(bi + cj + dk)(βi + γj + δk) = −bβ − cγ − dδ+

(cδ − dγ)i + (dβ − bδ)j + (bγ − βc)k

Note that the scalar part of the productis simply minus the scalar product of the twoquaternions viewed as vectors with components(b, c, d) and (β, γ, δ), while the vector part istheir vector product (cross product). The ge-ometric interpretation of these operations was

known, and when the modern vector was born,they were carried over. Of course, this is an-other story, given with much scholarly detail byM. J. Crowe in [2]. But let us end on a high notewith this poem, by Hamilton himself, about hiscreations:

Or high Mathesis, with her charm severe,

Of line and number, was our theme; and we

Sought to behold her unborn progeny,

And thrones reserved in Truth’s celestial sphere:

While views, before attained, became more clear;

And how the One of Time, of Space the Three,

Might, in the Chain of Symbol, girdled be:

And when my eager and reverted ear

Caught some faint echoes of an ancient strain,

Some shadowy outlines of old thoughts sublime,

Gently he smiled to see, revived again,

In later age, and occidental clime,

A dimly traced Pythagorean lore,

A westward floating, mystic dream of FOUR.

References

[1] B. L. van der Waerden, Hamilton’s discov-ery of quaternions, Mathematics Magazine, 49(1976), pp. 227–234.

[2] M. J. Crowe, A History of Vector Analysis,Dover Publications, Reprint edition (1994)

Jokes

This is |BS|! 2

ex and x2 are walking down the road when they suddenly see a differential operator. x2 says to ex,“Run, or we’ll be differentiated!” ex calmly replies that he cannot be differentiated. So ex walksup to the differential operator and says, “Hi, I’m ex.” To which the differential operator replies,“Hi, I’m d

dy .” 2

Mathematics is made of 50 percent formulas, 50 percent proofs, and 50 percent imagination. 2

An engineer, a physicist and a mathematician find themselves in an anecdote, indeed an anecdotequite similar to many that you have no doubt already heard. After some observations and roughcalculations the engineer realizes the situation and starts laughing. A few minutes later thephysicist understands too and chuckles to himself happily as he now has enough experimentalevidence to publish a paper. This leaves the mathematician somewhat perplexed, as he hadobserved right away that he was the subject of an anecdote, and deduced quite rapidly the presenceof humor from similar anecdotes, but considers this anecdote to be too trivial a corollary to besignificant, let alone funny. 2

19It turns out that they really are associative.


42 Getting Acquainted with Φ

Getting Acquainted with ΦJuan Manuel Martinez

One can say so many cheesy things about the number Φ = 1+√

52 , that there’s no

way of picking which one to start with. Read on to find out what makes it so special.

At the beginning of the sixteenth century, theItalian mathematician Luca Pacioli publishedDivina Proportione in which he called Φ the“Di-vine Proportion”. Since the nineteenth century,Φ has been also called the “Golden Section, Ra-tio or Number”.

How to derive Φ

It was Euclid of Alexandria who firstgave a proper definition of the number Φin his famous book The Elements. Hederived it using the following geometricconstruction. Consider a straight line.

A BC

φ 1

If you divide the line so that the ratio of theentire line AB to that of the larger line segmentAC is the same as the ratio of the larger linesegment AC to that of the smaller line segmentCB, you will get Φ. The lines are then said tohave been cut in extreme and mean ratio.

Φ can also be derived from the Fibonacci se-quence 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, . . . Each term(except the first two) is the sum of the two pre-ceding ones. We’re going to prove here that theratio of successive terms in the Fibonacci se-quence converges to Φ. Consider the set V ofreal-valued sequences, {an}, n ≥ 0 satisfyingan+1 = an + an−1 for all n ≥ 1. It can eas-ily be shown that this set, equipped with theusual addition and scalar multiplication of se-quences, is a vector space of dimension 2. Ageometric progression is a sequence satisfyingan = αn. We will now build a basis for V con-sisting of two geometric progressions. By defini-tion, αn+1 = αn + αn−1. Dividing by αn−1 andsolving for α, we get α = Φ and its algebraic con-jugate, denoted Φ. To show that the sequencesgenerated by Φn and Φ

nform a basis for V ,

we need only show that they are linearly inde-pendent (since the space is 2-dimensional), i.e.show that if a(1+

√5)/2)n +b((1−

√5)/2)n = 0

then a = 0 and b = 0. We can show this easilyby successively setting n=0 and n=1. Since the

Fibonacci sequence is an element of V, we canexpress it as a linear combination of the two ba-sis vectors. Hence there exist u, v ∈ R such thatan = u(Φn) + v(Φ

n). Setting n = 0 and n = 1,

we have u + v = 0 and (Φ)u + (Φ)v = 1. Solv-ing for u and v, we get a closed expression forthe n-th term of the Fibonacci sequence givenby an = (Φn − Φ

n)/√

5.

We can write the ratio of consecutive termsof the Fibonacci sequence as (Φ/(1−(Φ/Φ)n))−Φ/((Φ/Φ)n − 1). Since ‖Φ‖ < Φ, the secondterm vanishes as n goes to infinity and the firstterm yields Φ. This proves the initial claim.

Φ can also be expressed in terms of vari-ous limits. Consider the following expression:√

1 +√

1 +√

1 + ... To find the value of this

expression, let x =

√

1 +√

1 +√

1 + .... Then

x2 = 1+√

1 +√

1 + ... and hence x2 = x+1 or

Φ =

√

1 +√

1 +√

1 + ....

A similar expression for Φ is the sequence1+ (1/(1+ (1/(1+ ...). Letting x be this expres-sion, we have that x = 1 + (1/x), or x2 = x + 1.Hence, the Golden ratio can also be expressedas a continued fraction.

The golden rectangle

A rectangle whose sides are in the ratio of thegolden number is known as a golden rectangle.Here is an unusual procedure. Consider a goldenrectangle.

Cut a square out of this rectangle and you’llget a golden rectangle whose dimensions aresmaller by a factor of Φ. Moreover, by join-ing the successive points of division we obtain alogarithmic spiral.


Getting Acquainted with Φ 43

Relationships between Φ and

trigonometric functions

Here are some exact trigonometric formulas in-volving Φ.

Φ = 2 cos(π/5)

= (1/2) sec(2π/5)

= (1/2) csc(π/10)

Here are some surprisingly simple and interest-ing expressions involving Φ and complex num-bers. Recall that sin(x) = (eix − e−ix)/(2i) andnote that (1/Φ) = Φ − 1.

sin(i logΦ) = (e− log Φ − elog(Φ)/(2i))

= ((1/Φ) − Φ)/(2i)

= (−1)/(2i) = i/2

An even more interesting expression obtained byusing the Euler formula eix = cosx + i sin x:

sin(π

2− i log(Φ)

)

=1

2i(eπi/2elog(Φ)

−e−πi/2elog(Φ−1))

Using the Euler formula again, we obtain

(iΦ + i/Φ)/(2i) = (Φ + (1/Φ))/2

= (2Φ − 1)/2

= (1 +√

(5) − 1)/2

=√

(5)/2

Occurrences of the Golden Ratio

The Golden Number makes multiple appear-ances in art, historical monuments and life ingeneral. Totally unrelated phenomena, such asthe petal arrangement in a red rose and thebreeding of rabbits, have this proportion in com-mon. The Great Pyramid at Giza in Egypt

has dimensions based on the Golden ratio. Infact, the area of a triangular lateral side is equalto the square of the height, as stated by theGreek historian Herodotus. With simple geo-metric manipulations, one can show that theheight of a lateral side divided by half the sideof the base gives Φ. This ratio is accurate toless than 0.1 percent. This suggests that theEgyptians knew about the Golden Section. Thespiral growth of sea shells is also based on theGolden Number.

It is widely believed that the Golden ratioappears in Leonardo Da Vinci’s“Mona Lisa”andin Salvador Dali’s “Sacrament of the Last Sup-per”. There are other cases where the Goldenratio’s role is still uncertain. A very good exam-ple is the Parthenon in Greece. Some sources,such as the Random House Encyclopedia, statethat the Golden Section appears in the buildingof the Parthenon. George Markowsky’s ”Mis-conceptions about the Golden Ratio”states thatthis is not true.

References

[1] Livio, Mario. The Golden Ratio: The Storyof Phi, The World’s Most Astonishing Number.New York: Broadway Books, 2002.

[2] Borowski, E.J. and Borwein, J.M. Collinsinternet-linked dictionary of Mathematics.Collins, 2nd ed. HarperCollins Publishers:Glasgow, 2002. 239-240.

[3] Weisstein, Eric W. ”Golden Ratio.”From Mathworld–A Wolfram Web Resource.[mathworld.wolfram.com/GoldenRatio.html]

[4] Weisstein, Eric W. ”Golden Rectangle.”From Mathworld–A Wolfram Web Resource.[mathworld.wolfram.com/GoldenRectangle.html]

Jokes

A mathematician is someone who thinks “A”, writes “B”, says “C” when it should be “D”. 2

A mathematician is asked to design a table. He first designs a table with no legs. Then he designsa table with infinitely many legs. He then spends the rest of his life generalizing results for thetable with N legs (where N is not necessarily a natural number). 2


44 Book review: George Polya’s How to Solve it?

Review of George Polya’s How to Solve it?

Nan Yang

Heuristics is not often taught systematically inmathematics, or indeed in general today, at leastnot directly; while we are mostly taught howto solve a problem or even a series of problems,‘how to solve problems’ is a question that is usu-ally left to the students, as if it is an instinct thatis best left to develop on its own. Evidently,George polya disagrees, and How to Solve It ishis answer to this unspoken question.

How to Solve It is literally a book of dia-logues; it contains dialogues between a fictionalteacher and a fictional student through whichPolya illustrates the process of how the studentis gradually nudged into the right direction bythe teacher; that is, his ideal process of learn-

ing. The dialogues need not be strictly betweenthe fictional characters; when the teacher says,‘What is the unknown? Can you think of a sim-ilar problem?’ it is obviously directed at thereader.

A recurring theme throughout the book isthat if you can not solve a problem, then youshould find an easier but similar one. ‘Do youknow a related problem?’ polya would ask. Forexample, suppose the student has just learnedthe Pythagorean theorem, and is now asked tofind the length of the spatial diagonal of a par-allelepiped. A small amount of ingenuity is re-quired to make the jump from the plane to thespace, and the student is naturally stumped.‘Do you know a problem with a similar un-known?’ the teacher asks. The student getsa brilliant insight – a previously solved prob-lem. ‘Good! Here’s a problem related to yoursand solved before. Can you use it?’ the teacherpresses on. Eventually the solution is found, andall is well. Of course, the point of that passage isnot to introduce to the readers the Pythagoreantheorem in higher dimensions, but to show thereaders the process with which one can use tofind an ‘auxiliary problem’ that has been solvedand use it to solve the harder problem. polyahimself is often accredited the quote, ‘For everyproblem you can’t solve there exists an easierproblem that you can: find it.’

Unfortunately, the effectiveness of the bookis debatable. As Feynman said, ‘You can’t learnto solve problems by reading about it.’ There isno other way of gaining problem-solving experi-ence save for actually solving problems. Hence,ironically, it is hard to appreciate the book un-less you no longer need it.

References

[1] G. Polya, How to Solve it, 2nd ed., PrincetonUniversity Press, 1957.


The Adventures of A and B 45

The Adventures of A and BJoel Perras and Nan Yang

authors’ note: A and B are fictitious characters. Any resemblance to any real-world persons is purely coincidental.

On a bright and sunny afternoon in early May,two students sat in the corner of a tiny restau-rant, far back where the deadly rays of sun couldnot touch them – deadly because years of iso-lation in the bleak dungeons of Burnside base-ment had left them extremely vulnerable to nat-ural light. Over a meal consisting of poutineand hot-dogs, they celebrated the end of the fi-nal examination period. With that dreaded or-deal behind them, they were now able to con-centrate on the independent projects they hadplanned for the summer. Somehow, the con-versation drifted onto lottery numbers. Bothhaving studied probability and statistics, theylamented over the superstitions that the generalpublic usually attached to the choice of num-bers.

‘What possible difference would it makewhat numbers you choose?’ A said. ‘For in-stance, I make my point to buy the numbersone through six. Uniform distribution... youknow...’ He trailed off as he dug up anotherfork-load of poutine.

‘I agree,’ said B. ‘But I always choose nineeight seven six five. That way, if I win, I willhave a better chance of getting the whole prize.’

A nodded, but after a second hesitated andlooked at B quizzically, ‘Better chance? Whatdo you mean? The distribution is uniform.Surely you’re not one of those superstitioustypes...’ He looked at B as if he were Goldsteinhimself and was considering whether or not todenounce him to Big Brother.

‘You don’t know about Benford’s law?’ Basked with traces of genuine surprise in his voice.A shaked his head, still suspicious of the truemotives of B.

‘Why, Benford’s law states that:

theorem: (Benford’s law) In listings, tablesof statistics, etc., the digit 1 tends to occur with≈ 30% probability , much greater than the ex-pected 11.1% (i.e., one digit out of 9).

Surprising, is it not? Hence, since most peo-ple see numbers that begin with a 1 more oftenthan any other single number, they are more

likely to choose a number that begins with 1when given the choice.’

‘Impossible!’

‘It’s true. Why, I might even publish an ar-ticle in that upcoming delta-epsilon magazineabout it.’ B said as he shifted his attention backto the poutine.

A period of silence lingered between themwhile an idea formed in A’s head. Suddenlyhe looked up and said, ‘Gee, I wonder of thislaw applies to the prime numbers. Let’s see...2, 3, 5, 7, 11, 13, 17, 19, 23. Why, of the first tenprimes, four of them begin with 1!’

‘Interesting... but we better not jump to con-clusions here. We need more data.’

‘Yes. We’re going to need quite a list, maybefifty million or so,’ A replied.

On the very next day, they went to Rosen-thall library and asked the librarian whetherthey kept a list of at least fifty million primes.

‘Fifty million primes?’ the librarian said,shocked. ‘I don’t think anyone keeps anythinglike that on hand. It’s much faster just to gen-erate them yourselves unless your a dispropor-tionally fast connection.’

‘Oh.’

So they went to the computer lab and beganworking on a prime list generator. Fortunately,B had some programming experience; they be-gan working and half an hour later they were al-ready generating primes. Eight hours later, thefirst ten million primes were ready. An analy-sis of the first ten million primes showed thatover 40 percent of them began with a 1. A andB were stunned, but were too caught up in themoment of excitement to think properly. Allthey knew was that they needed a bigger sam-ple size. But their program would take at leastanother 80 hours to cough up the next forty mil-lion primes. Surely a better algorithm exists,they thought. A bit of googling revealed sucha program which, after only 5 minutes, was ableto generate fifty million primes.

‘Well, that would have saved us some time,’B said.


46 The Adventures of A and B

After analyzing the new set of data, theywere shocked to discover that almost fifty per-cent of the first fifty million primes began with 1.A and B had some past experience with num-ber theory, and never had they heard of thisstartling result. Surely something this trivialcould not have gone unnoticed since the time ofEuclid!

Before they had time to contemplate further,they realized that they had to attend a Mon-treal Symphony Orchestra concert. A profes-sional mathematician would never have given upnew math for a concert but, as it were, they putdown the problem temporarily and went. Theirspeculations and visions of grandeur had to wait;they could not forfeit their monthly allowance ofculture and appreciation of the arts.

Nonetheless, while the bodies of A and Bwere in the most expensive section they can af-ford (the balcony), their minds were decidedlyelsewhere. After an hour of almost-listening andnear-appreciation, A gripped the arms of hisseat, his eyes bulging wide with apprehension.

‘B, when we generated that list of primes,did you specify that we wanted the first fiftymillion primes, or all the primes up to fifty mil-lion?’

‘The first fifty million primes. Why do you– we made a sampling mistake, didn’t we?’ Bcaught on.

‘I think so. After this concert, we need to goverify this,’ he answered.

After the curtain fell and the applause stillringing in their ears, both rushed to the mostimportant place they needed to be on a Satur-day night: a computer lab at McGill. B ran theprime generator program once again, but this

time instructed it to halt after producing all theprimes up to one billion. The results showedwhat they both feared: a near uniform distribu-tion of the first digits.

‘There goes our theory,’ said A.‘Oh well, it was fun and interesting while it

lasted,’ commented B. ‘Let’s go have some morepoutine.’

Several weeks later A mentioned their nullresult to a mathematician, who readily replied,‘The logarithmic density of the primes actuallydoes follow Benford’s law. It’s an old result.You can check it up in the February 1972 issueof American Mathematical Monthly, page 150.I believe it’s in a paper by R. E. Whitney.’

Unbelieving, A browsed the shelves ofRosenthall Library, quite sure that the math-ematician he had spoken to had been mistaken.He scanned the volumes of journals until hefound the one with the article in question andopened it to the correct page. It read:

‘It is well known that the logarithmic den-sity of D in the sequence of positive integers islog10(1 + 1/a). The purpose of this note is toshow that the relative logarithmic density of Din P is also log10(1+1/a). This is an unusual re-sult because of the irregular distribution of theprimes. As a consequence of this result, onemight say that 1 is the preferred initial digit forthe sequence of primes.’

A closed the journal and swore in frustration,‘ǫ < 0!’

The article, more than thirty years old, dis-proved in two pages the fallacy that had tiedthem in ropes for days. In the end, A and Blearned a valuable lesson:

corollary: Never abuse statistics.2

Jokes

Several scientists were all posed the following question: “What is 2*2?” The engineer whips outhis slide ruler and shuffles it back and forth, and finally announces, “4.01.” The physicist consultshis technical references, sets up the problem on his computer, and announces, “it lies between 3.98and 4.02.” The mathematician cogitates for a while, then says, “I don’t know what the answeris, but I can tell you an answer exists!” The philosopher smiles and replies, “But what do youmean by 2*2?” The logician replies, “Please define 2*2 more precisely.” The sociologist says, “Idon’t know, but it was nice talking about it.” The medical student replies, “4!” All the others areamazed. “How did you know?” they asked. The medical student replies, “I memorized it.” 2


Crossnumbers 47


48 Acknowledgements

Credits

The Delta-Epsilon Editing Team

In alphabetical order

Agnes F. Beaudry

Juan Manual Martinez

Michael McBreen

Alexandra Ortan

Joel Perras

Vincent Quenneville-Belair

Nan Yang

Cover Art & Design

Michael McBreen

Mathieu Menard

Acknowledgements

It’s a little exaggerated to claim that some accomplishment would not have been possible withoutthe help of person x and y. If an“event”isn’t“independent”of the participation of some individual,it might remain possible, but become less probable. Therefore, I would like to thank the followingpeople for their help in increasing the probability of the publication of this journal, but aboveall, for surely increasing its quality. Thank you prof. Henri Darmon, prof. Niky Kamran andprof. Nilima Nigam for taking the time to speak with us as well as for smiling for the picture.Thank you SUMS and especially Marc-Andre Rousseau for supporting the project from beginningto end. We are grateful to the Science Undergraduate Society (SUS), the Faculty of Science andthe McGill Mathematics Department for providing the funding. Graduate students of McGill,in particular David Campbell, Daniel Simeone, Dennis The and Michael Lennox Wong, manymistakes have been avoided thanks to your keen eyes (and those that remain are left as exercises.)Samuel Altarac-Hofmann, your assistance for printing the magazine is much appreciated. Finally,reader, thank Adele Flannery pour la beaute de la rhetorique, or at least for polishing our broken,mathematical English: she tried to wean us from our ways and sometimes she succeeded, so bethankful to her that you can actually read this!


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