Concept

Quadratic form

A homogeneous quadratic expression in several variables, written as a symmetric matrix sandwiched between a vector and itself. Its level sets are the conics and quadrics, and the signs of its eigenvalues decide which - a classification unchanged by any invertible change of coordinates.

Named by 18 essays across 5 fields — each of them below, with the objects they name alongside it.

The level curve, and the axes the matrix chooses. The curve xᵀAx = 1 for the matrix [2, 0.8, 0.8, 1.4], drawn by solving for the radius at each angle, with the two eigen-directions marked; they cross at a right angle and are the axes of the curve.

Symmetry forces a right angle

A matrix equal to its own reflection across the diagonal always has real stretches and always has perpendicular directions to stretch along. Neither is true of matrices in general, and both follow from one line of algebra.

algebra · Eigenvectors
One sign decides which curve it is. 3 conics drawn from the general quadratic, each labelled with its discriminant B² − 4AC and the curve that sign names, checked against how many times the curve meets a large circle.

One sign decides which curve

The general quadratic in two variables has six coefficients and draws a conic. Which of the four it draws is settled by a single combination of three of them, and the other three cannot change the answer however they are chosen.

geometry · Conic sections
The zeros of x² + y² + z² over GF(5), and of x² + y² over GF(7). Grids of every point over a small prime field with the solutions of a quadratic equation filled in: the three-variable equation drawn as one slice per value of z, beside a two-variable equation with far fewer solutions.

Solutions that come in multiples of p

Count the solutions of x² + y² + z² = 0 in the field with five elements and there are 25; with seven, there are 49. Whenever a system of equations has more unknowns than its total degree, its number of solutions is a multiple of the characteristic — which forces a solution besides zero, and the reason is a sum over the field that vanishes because its non-zero elements form one cycle.

computation · Finite fields
The states the continued fraction of √61 can be in. A grid of whole-number pairs with the band of reduced states shaded, the pairs that qualify marked, and the cycle of states visited by the expansion of the square root numbered in order.

Why the expansion has to repeat

The continued fraction of √61 runs 7; 1, 4, 3, 1, 2, 2, 1, 3, 4, 1, 14 and then starts again. It must: each step's state is a pair of whole numbers trapped in a small band, and only 14 pairs fit. The expansion of √61 visits 11 of them in a cycle, the other 3 form a cycle of their own, and the period reads the same backwards before its last term, which is twice the first.

number · Pell
100 as three triangular numbers. 100 drawn as three triangles of dots with 36, 36 and 28 dots. There are 6 such decompositions.

Three triangular numbers, and no fewer

On 10 July 1796 Gauss wrote in his diary: ΕΥΡΗΚΑ — num = Δ + Δ + Δ. Every whole number is a sum of three triangular numbers. Two are not enough, and not by a little: the numbers that are sums of two thin out to a share of nought. Both facts are statements about squares in disguise, and one picture translates them.

geometry · Figurate numbers
The largest value on each plane, over every direction of space. A longitude–latitude map of directions u in three dimensions, each shaded by the largest value of xᵀAx on the plane perpendicular to u. The smallest such value is the middle eigenvalue 1.829, reached at ±v₁; the largest is 3.622, reached along a great circle.

The highest point on the sphere is an eigenvalue

For a symmetric matrix, walk a unit arrow over every direction and record the value of xᵀAx. The highest value reached is the largest eigenvalue, the lowest is the smallest, and every eigenvalue in between is a saddle height, a minimum of maxima. From that one description comes a theorem no formula for the roots could give: delete a row and its column, and every eigenvalue of what is left sits between two of the original's.

algebra · Eigenvectors
Signature against unknotting, for every knot to seven crossings. A table of the fourteen prime knots with up to seven crossings, listing determinant, signature, the lower bound on crossing changes it gives, the number of changes found by search, and whether the two agree.

How many changes undo a knot

Cut the string at a crossing, pass it through the other strand and join it up again, and any knot can be undone by doing that often enough. The fewest changes needed is the unknotting number, and proving that fewer will not do needs a number that each change can move only a little. The signature moves by at most two per change — enough to settle thirteen of the fourteen knots up to seven crossings, and not the fourteenth.

topology · Knots
From a rational point on x² + y² = 17 to a whole one. The circle of radius √17 on the integer lattice, a rational point on it, and 4 reflections through nearest lattice points, the denominators 3757, 205, 25, 5, 1, ending at (1, 4).

A fraction on the circle forces a whole point

If a number is a sum of two squares of fractions, it is a sum of two squares of whole numbers. Draw the circle, mark the rational point, join it to the nearest lattice point and follow the line to where it meets the circle again: the new point is rational too, with a smaller denominator. Repeat, and the denominators fall until they reach one. The argument needs no primes at all — only the fact that every point of the plane is within distance one of the lattice.

number · Sums of two squares
The 4 reduced forms of discriminant −84, each in its place. The reduced binary quadratic forms of discriminant −84 (x² + 21y²; 2x² + 2xy + 11y²; 3x² + 7y²; 5x² + 4xy + 5y²) plotted at their roots in the upper half-plane, all inside the modular fundamental region.

Counting the classes that break factorisation

The class number measures how badly unique factorisation fails in a field, and defined through ideals it looks impossible to compute. Gauss computed it by hand, for every field he wanted, by counting quadratic forms — and every form can be squeezed, by changes of variable that keep its values, into exactly one small standard shape.

number · Unique factorisation
The plane of order 3 as a table, and the table times its transpose. The 13 × 13 incidence table of the projective plane of order 3 and its product with its transpose, which has 4 on the diagonal and 1 in every other cell.

The orders a plane cannot have

Every counting condition allows a projective plane of order six, and there is none. The proof that rules it out looks at one matrix identity — each point on seven lines, each two points on one — and turns it, by way of Lagrange's four squares, into the statement that six would have to be a sum of two squares. Run on the planes that do exist, the same argument hands back their orders as sums of two squares; run on six, it asks for something no arithmetic can supply.

computation · Finite geometry
The fewest pentagonal numbers adding to each number up to 120. Fewest pentagonal numbers summing to 1..120; the most ever needed up to 20000 is 5, by 9, 21, 31, 43, 55, 89.

Six numbers that need five pentagons

In 1638 Fermat wrote that every whole number is a sum of three triangular numbers, four squares, five pentagonal numbers, six hexagonal numbers, and so on for every polygon — and that he had a proof he would not write down. The claim is true; Cauchy proved it in 1813. What the claim hides is how unequal the cases are. Triangles and squares need their full count infinitely often. For pentagons, only six numbers ever need all five — 9, 21, 31, 43, 55 and 89 — and from hexagons on, two apiece.

geometry · Figurate numbers
The fifty-four sums of four weighted squares that reach every number. The 54 universal diagonal quaternary forms, as coefficient lists: 1,1,1,1; 1,1,1,2; 1,1,1,3; 1,1,1,4; 1,1,1,5; 1,1,1,6; 1,1,1,7; 1,1,2,2; 1,1,2,3; 1,1,2,4; 1,1,2,5; 1,1,2,6; 1,1,2,7; 1,1,2,8; 1,1,2,9; 1,1,2,10; 1,1,2,11; 1,1,2,12; 1,1,2,13; 1,1,2,14; 1,1,3,3; 1,1,3,4; 1,1,3,5; 1,1,3,6; 1,2,2,2; 1,2,2,3; 1,2,2,4; 1,2,2,5; 1,2,2,6; 1,2,2,7; 1,2,3,3; 1,2,3,4; 1,2,3,5; 1,2,3,6; 1,2,3,7; 1,2,3,8; 1,2,3,9; 1,2,3,10; 1,2,4,4; 1,2,4,5; 1,2,4,6; 1,2,4,7; 1,2,4,8; 1,2,4,9; 1,2,4,10; 1,2,4,11; 1,2,4,12; 1,2,4,13; 1,2,4,14; 1,2,5,6; 1,2,5,7; 1,2,5,8; 1,2,5,9; 1,2,5,10.

Fifteen numbers decide every number

Lagrange proved that x² + y² + z² + w² takes every whole value. Ramanujan asked which other sums of four weighted squares do, and in 1917 listed fifty-five. One of them is wrong: x² + 2y² + 5z² + 5w² misses 15. The mistake points at a theorem found eighty years later — to know whether such a form reaches every number, it is enough to check that it reaches 1, 2, 3, 5, 6, 7, 10, 14 and 15. Nine checks decide infinitely many cases, and the nine numbers come out of an escalation any one can run.

geometry · Figurate numbers
Class numbers of x² + ny², with Euler's idoneal numbers marked. A scatter of class number against n up to 2000, on a logarithmic scale, with the 65 idoneal numbers marked on the power-of-two levels.

Euler's sixty-five convenient numbers

For some n, whether a prime can be written as x² + ny² is settled by its remainder on division by 4n alone, the way a prime's remainder on division by four settles whether it is a sum of two squares. Euler found sixty-five such n, from 1 to 1,848, called them convenient, and used the largest to prove that 18,518,809 is prime. Every one of them is a number whose class group has no element of order more than two — and whether the list is complete is still not known.

number · Unique factorisation
How often a prime divides the class number. p 3: 39.1% against 1/p 33.3% and predicted 44.0%; p 5: 22.6% against 1/p 20.0% and predicted 24.0%; p 7: 15.3% against 1/p 14.3% and predicted 16.3%; p 11: 9.1% against 1/p 9.1% and predicted 9.9%.

Class groups drawn at random

The class number of an imaginary quadratic field measures how badly unique factorisation fails there, and it looks arbitrary: 1, 2, 4, 3, 5, 12, 1. Count how often 3 divides it and the answer is not one time in three but noticeably more — heading, by a conjecture of Cohen and Lenstra, for 44 per cent. The reason is that class groups behave like finite groups chosen at random with each group weighted by one over its number of symmetries, so groups with few symmetries, like the cyclic group of order 3, turn up more often than the naive count suggests.

number · Unique factorisation
One form completed three ways, and the same straddle each time. The plane coloured by the sign of 2x² + 6xy + 3y². Three pairs of lines, one for completing with x first, one for y first and one along the eigenvectors; their coefficients are 2 and −3/2; 3 and −1; 5.541 and −0.541, and every pair has one line in a positive wedge and one in a negative wedge.

The signs no completion can change

A quadratic in several variables can be completed square by square in many orders, and each order hands back different coefficients. What no order changes is how many come out positive and how many negative — and that count, read off a completion of A − tI, says how many eigenvalues lie below t without finding a single one.

algebra · Completing the square
Every slice of a tilted bell is a bell, centred off the axis. Contour ellipses of a two-variable bell with correlation 0.6, five vertical slices drawn as bells, their centres on the line y = 0.6x, and the ellipses' long axis on the diagonal y = x.

Regression is a square completed halfway

A bell in two variables, sliced at a fixed x, is a bell in y — and completing the square in y alone says where it is centred: at ρx, on a line flatter than the ellipse's own axis. That gap is regression to the mean, and the same completion says it runs backwards in time just as well as forwards.

algebra · Completing the square
Three cycles of forms for the prime 229. Reduced forms of discriminant 229 in 3 cycles of lengths 6, 6, 2; checked: h(5) = h(13) = 1, h(229) = h(257) = 3, h(401) = 5, h(577) = 7, h(1129) = 9.

Real fields that factorise uniquely

Among the fields Q(√−d) only nine have unique factorisation, and Gauss guessed as much. Among the real fields Q(√p) he guessed the opposite — infinitely many — and the guess is still unproved. Counting cycles of reduced quadratic forms for the 16,900 primes p ≡ 1 mod 4 below 400,000 finds 79 per cent with class number one, drifting slowly down towards the 75.45 per cent that Cohen and Lenstra's heuristic predicts — and finds the reason the real fields behave so differently: their units are enormous.

number · Pell
Right triangles of area five, six and seven. area 5: 3/2, 20/3, 41/6; area 6: 3, 4, 5; area 7: 24/5, 35/12, 337/60.

The areas a right triangle can have

A right triangle with sides 3, 4 and 5 has area 6; one with sides 3/2, 20/3 and 41/6 has area 5. No right triangle with rational sides has area 1, 2 or 3. Which whole numbers are areas is a thousand-year-old question, and its best answer — two counts of solutions to quadratic equations — rests on one of the great unproved conjectures.

geometry · Pythagoras

Named alongside it

The objects these essays reach for when they reach for this one.

EigenvalueExhaustive searchSums of two squaresClass groupClass numberConjectureDiscriminantCompleting the squareContinued fractionsEigenvectorExistence proofInvariant

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