The sums that obey a smaller equation

The thirteenth roots of unity other than $1$ are the twelve points $\zeta, \zeta^2, \dots, \zeta^{12}$ on the unit circle, where $\zeta = e^{2\pi i/13}$. The polygon an equation forces showed that they are exactly the roots of

$$x^{12} + x^{11} + \dots + x + 1 = 0,$$

and that this equation cannot be factored with rational coefficients: no smaller equation with whole-number coefficients is satisfied by any one of the twelve. Each of them is, in that sense, an algebraic number of full complexity twelve.

Every third coefficient added roots of unity over a structured set to pick out part of a polynomial; here the structured sets are chosen so that the sums themselves become simple numbers.

Now add some of them. Not any four — the four whose exponents are $1, 5, 8, 12$. The sum is a real number, about $0.274$. Add the four with exponents $2, 3, 10, 11$ and the sum is about $-2.651$; add the remaining four, $4, 6, 7, 9$, and it is about $1.377$. And these three sums are the three roots of

$$x^3 + x^2 - 4x + 1 = 0,$$

an equation of degree three, with whole-number coefficients. Four points of full complexity twelve, added in the right groups, produce a number of complexity three.

Gauss called these sums periods, and in the seventh section of the Disquisitiones Arithmeticae of 1801 he used them to do something nobody had done in two thousand years: write down a construction of the regular seventeen-gon. This essay is about why the right groups exist, why their sums obey small equations, and why that is exactly what a ruler and compass need.

The right groups are the cosets of a subgroup #

The exponents $1, 5, 8, 12$ are not arbitrary. Modulo $13$ they are exactly the cubes, and equally exactly the numbers $x$ with $x^4 \equiv 1$ — for instance $5^2 = 25 \equiv 12$ and so $5^4 \equiv 144 \equiv 1$. The cleanest description is multiplicative. The nonzero remainders modulo $13$ form a group under multiplication, and that group is cyclic: there is a remainder, $2$, whose powers

$$2, 4, 8, 3, 6, 12, 11, 9, 5, 10, 7, 1$$

run through all twelve. Every subgroup of a cyclic group of order twelve is cyclic too, and there is exactly one of each order dividing twelve. The subgroup of order four is generated by $2^3 = 8$: its elements are $8, 64 \equiv 12, 96 \equiv 5, 40 \equiv 1$ — that is, $\{1, 5, 8, 12\}$.

A subgroup of order four in a group of order twelve has three cosets: the subgroup itself, and its images under multiplication by $2$ and by $4$. Multiplying $\{1, 5, 8, 12\}$ by $2$ gives $\{2, 10, 3, 11\}$; multiplying by $4$ gives $\{4, 7, 6, 9\}$. Those are the three groups in the figure.

So the recipe is general. For a prime $p$, pick a divisor $d$ of $p - 1$. The nonzero remainders modulo $p$ have exactly one subgroup $H$ of index $d$ — of size $f = (p-1)/d$ — and it has $d$ cosets. The $d$ periods of length $f$ are the sums

$$\eta_j = \sum_{h \in H} \zeta^{g^j h}, \qquad j = 0, 1, \dots, d - 1,$$

where $g$ is a primitive root. Every non-trivial $p$-th root of unity appears in exactly one period, so the $d$ periods always add up to the sum of all $p - 1$ non-trivial roots, which is $-1$.

The heptagon’s cubic, from three pairs #

The smallest example with a real payoff is the heptagon. Modulo $7$, the subgroup of order two is $\{1, 6\}$, that is, $\{1, -1\}$. Its three cosets pair each root with its complex conjugate, and each pair adds to twice a cosine.

The three periods are $2\cos(2\pi/7)$, $2\cos(4\pi/7)$ and $2\cos(6\pi/7)$, and they are the roots of

$$x^3 + x^2 - 2x - 1 = 0.$$

That cubic is the reason the regular heptagon cannot be drawn with ruler and compass. The number $2\cos(2\pi/7)$ satisfies a cubic with no rational root, so it has degree three over the rationals, and a tower whose degrees multiply showed that anything a ruler and compass reach has degree a power of two. Which polygons can be drawn states this as a condition on $p - 1$; the periods are the mechanism behind the condition. A cubic period equation is exactly what $p - 1 = 6$ produces when it is split by its factor $3$.

The group drawn as a map drew a cyclic group as a single loop of arrows; the nonzero remainders modulo a prime form exactly such a loop, generated by a primitive root, and choosing a subgroup is choosing to step round the loop $d$ places at a time.

The cosine of a regular polygon’s angle is always a period — the period of the subgroup $\{1, -1\}$ — and the equation it satisfies has degree $(p-1)/2$. The heptagon’s is a cubic; the thirteen-gon’s is a sextic; the seventeen-gon’s has degree eight.

Multiplying two periods by hand #

The cubic can be checked without any numerics, and doing it once shows where the whole-number coefficients come from. Write the heptagon’s three periods as

$$\eta_0 = \zeta + \zeta^6, \qquad \eta_1 = \zeta^3 + \zeta^4, \qquad \eta_2 = \zeta^2 + \zeta^5,$$

grouped by the cosets of $\{1, 6\}$: the coset of $3$ is $\{3, 18 \equiv 4\}$ and the coset of $2$ is $\{2, 12 \equiv 5\}$. Multiply two of them, using $\zeta^7 = 1$ to reduce every exponent:

$$\eta_0 \eta_1 = \zeta^4 + \zeta^5 + \zeta^9 + \zeta^{10} = \zeta^4 + \zeta^5 + \zeta^2 + \zeta^3 = \eta_1 + \eta_2.$$

The four roots in the product fall into whole periods again. That is not luck: multiplying every exponent in the product by a generator of the subgroup — here by $6$, that is, by $-1$ — permutes the terms of the product among themselves, so the product is a union of whole cosets. Every product of periods is therefore a whole-number combination of periods, plus possibly a whole number of copies of $\zeta^0 = 1$.

With that, the coefficients are bookkeeping. The sum $\eta_0 + \eta_1 + \eta_2 = -1$. The three pairwise products are $\eta_1 + \eta_2$, $\eta_2 + \eta_0$ and $\eta_0 + \eta_1$ by the same computation shifted, and they add to $2(\eta_0 + \eta_1 + \eta_2) = -2$. The product of all three is $\eta_2(\eta_1 + \eta_2) = \eta_1\eta_2 + \eta_2^2$, and $\eta_2^2 = \zeta^4 + 2 + \zeta^{10} = \eta_1 + 2$, while $\eta_1\eta_2 = \eta_2 + \eta_0$ by the shifted computation, so the triple product comes to $\eta_0 + \eta_1 + \eta_2 + 2 = -1 + 2 = 1$. So the periods are the roots of $x^3 + x^2 - 2x - 1$, as the figure found numerically.

The periods behave like a small number system of their own. Multiplying is turning made the roots of unity a group under multiplication; the periods are closed under multiplication too, but only as combinations, because the product of two cosets is a union of several.

When the subgroup misses −1, the periods leave the line #

The periods are real exactly when every coset is closed under taking the complex conjugate, and that happens exactly when $-1$ is in the subgroup. When it is not, the periods come in complex pairs.

Modulo $7$ the subgroup of index two is the set of squares, $\{1, 2, 4\}$, and $-1 \equiv 6$ is not a square. So the two periods

$$\eta_0 = \zeta + \zeta^2 + \zeta^4, \qquad \eta_1 = \zeta^3 + \zeta^5 + \zeta^6$$

are complex conjugates of each other, and they satisfy $x^2 + x + 2 = 0$: they are $(-1 \pm \sqrt{-7})/2$. Their difference is $\sqrt{-7}$. For $p = 13$, where $-1$ is a square, the same construction gives $x^2 + x - 3 = 0$ and a difference of $\sqrt{13}$.

That difference — the sum of the roots at the squares minus the sum at the non-squares — is the quadratic Gauss sum that one sum, squared two ways follows to the reciprocity law. There it was a single sum whose square is $\pm p$; here it is the gap between the two periods of index two. The Gauss sum is the square root that the quadratic period equation needs, and whether it is real or imaginary is whether $-1$ is a square modulo $p$.

Every divisor has its equation #

For $p = 13$ the divisors of $12$ other than $1$ and $12$ are $2, 3, 4$ and $6$, and each gives a period equation.

The pattern in the leading terms is forced — every equation starts $x^d + x^{d-1}$ because the periods add to $-1$ — but the rest is not predictable from $d$ alone. The quartic’s constant term is $3$; the cubic’s is $1$; the sextic’s is $-1$. What is predictable is that they are whole numbers, and the figure checks this by multiplying out the linear factors numerically and requiring every coefficient to land within a ten-millionth of an integer.

Why whole numbers? Consider the map that sends each root $\zeta^a$ to $\zeta^{ga}$ — raising every root to the power $g$. It permutes the $p - 1$ roots, and because it respects addition and multiplication of sums of roots, it is a symmetry of the whole number system built from $\zeta$. Applied to a period, it sends $\eta_j$ to $\eta_{j+1}$, shifting every coset to the next. So it permutes the periods cyclically, and every coefficient of $\prod (x - \eta_j)$ — being a symmetric expression in the $\eta_j$ — is left unchanged.

A number built from $\zeta$ and left unchanged by every such symmetry is rational: that is the Galois correspondence of the lattice that runs the other way, read at its bottom level. And a coefficient is a sum of products of roots of unity, hence an algebraic integer; a rational algebraic integer is an integer. The period equation has whole-number coefficients because the periods are shuffled among themselves by every symmetry and by nothing else.

One field for each divisor, and every abelian group #

The periods do more than satisfy small equations. The field they generate — all the numbers that can be written with the periods of index $d$, fractions and sums allowed — has degree exactly $d$ over the rationals, and it is the only field of degree $d$ inside the field generated by $\zeta$.

This is the Galois correspondence made concrete. The symmetries of $\mathbb{Q}(\zeta)$ form a cyclic group of order $p - 1$, isomorphic to the nonzero remainders under multiplication. A cyclic group has exactly one subgroup of each order, so the field has exactly one subfield of each degree $d$ dividing $p - 1$, and a subfield of degree $d$ sits inside one of degree $e$ exactly when $d$ divides $e$. The periods of index $d$ are a basis for that subfield: its elements are exactly the combinations $c_0 \eta_0 + \dots + c_{d-1}\eta_{d-1}$ with rational $c_j$.

Each of these subfields has a cyclic group of symmetries — the shift $\eta_j \to \eta_{j+1}$ generates it — so each period equation is an example of a polynomial whose Galois group is cyclic of order $d$. And since there are primes $p$ with $p - 1$ divisible by any given $d$ — infinitely many, by Dirichlet’s theorem on primes in progressions — every cyclic group is the Galois group of some period equation. Taking composites of fields from different primes gives products of cyclic groups, and so every finite abelian group occurs as a Galois group over the rationals.

The converse is a deep theorem. Leopold Kronecker stated in 1853, Heinrich Weber proved in 1886 and David Hilbert repaired in 1896 that every field with an abelian symmetry group lies inside some field of roots of unity. So the periods and their composites are not merely examples: they are all the abelian extensions of the rationals.

The seventeen-gon, as four quadratics #

For $p = 17$, the group of nonzero remainders has order $16 = 2^4$, and its subgroups form a single chain of index $2, 4, 8, 16$. Each subgroup has index two in the one above it, so each level of periods splits the level above in pairs.

This is Gauss’s construction, and it is worth following once in full. The single period at the top is $-1$, the sum of all sixteen roots.

First split. The two periods of length eight add to $-1$ and multiply to $-4$, so they solve $x^2 + x - 4 = 0$: they are $(-1 \pm \sqrt{17})/2$, about $1.562$ and $-2.562$.

Second split. Each of those splits into two periods of length four. The two children of each parent add to that parent, and — the figure solves for this and checks it — they multiply to $-1$. So they solve $x^2 - \eta x - 1 = 0$, where $\eta$ is the parent, already known as a square root expression.

Third split. Each period of length four splits into two of length two, which are the numbers $2\cos(2\pi k/17)$. Their sum is the parent; their product is not a constant this time but another period of length four — one of the siblings of the parent. Since all four are known by now, the quadratic’s coefficients are known.

Four levels, four square roots nested inside one another, and at the bottom $2\cos(2\pi/17)$ — from which a compass marks off the side of the polygon. Every split is a quadratic because every step down the chain of subgroups halves the size, and that is why $17 = 2^4 + 1$ works while $13$, whose chain has a step of three, does not.

Why the construction needs no guessing #

It is worth saying what the method replaces. Before Gauss, a construction of a regular polygon was a geometric discovery: someone found, by insight, a sequence of circles and lines that produced the right angle. The periods turn it into a computation that is guaranteed to succeed whenever it can.

For any prime $p$, the chain of subgroups of the cyclic group of order $p - 1$ can be refined into steps whose sizes are the prime factors of $p - 1$. Each step splits every period into that many, and the children are the roots of an equation of that degree whose coefficients lie in the field of the level above. When every prime factor is $2$ — when $p - 1$ is a power of two, so $p$ is a Fermat prime — every step is a quadratic and the polygon is constructible. When a factor of $3$ appears, a cubic step appears, and with it a cube root; that is the heptagon and the thirteen-gon, which three ways to pair four roots would recognise as the cubic resolvent’s kind of obstacle.

The method also says exactly how each step is to be solved. The product of each pair in the seventeen-gon’s tower is determined by multiplying out two sums of roots of unity and regrouping — mechanical work, which Gauss did by hand and which the figures do by solving a small linear system and checking that its solution consists of whole numbers.

What the figures can and cannot show #

The period equations are computed, not quoted. Each is obtained by multiplying out $\prod(x - \eta_j)$ with the periods evaluated numerically, and each coefficient is required to be within $10^{-7}$ of a whole number and the imaginary parts to vanish to the same accuracy. That is strong evidence in each case, and the argument above — cyclic permutation of the periods, rationality of symmetric expressions, integrality of algebraic integers — is the proof.

The fields are not drawn. A period is drawn as a point, and the field it generates is not a thing a figure can show. The statement that the periods of index $d$ span the unique subfield of degree $d$ is proved, not illustrated.

Only primes. For a composite $n$, the group of remainders prime to $n$ need not be cyclic — modulo $8$ it is not — and the periods are then defined over subgroups of a non-cyclic group, with a lattice of subfields that is no longer a single chain of divisors.

Still open: which groups appear at all #

Every abelian group is the symmetry group of some field built from roots of unity, and every group a polynomial can have is a subgroup of the symmetric group on its roots. Whether every finite group is the Galois group of some polynomial with rational coefficients is the inverse Galois problem, and it is open.

Much is known. Igor Shafarevich proved in 1954 that every solvable group occurs. Every symmetric and alternating group occurs, by Hilbert’s irreducibility theorem, and so do most of the sporadic simple groups, some of them — the Monster among them — by explicit rigidity arguments. But for the Mathieu group $M_{23}$, a group of about ten million elements, no polynomial with rational coefficients is known whose Galois group it is, and no proof that one exists. The periods settle the abelian case completely and in one stroke; nothing comparable is known for any general class beyond the solvable groups.

Sums that remember their subgroup #

The habit worth keeping is to look for the combination that a symmetry permutes.

Twelve roots of unity, each of degree twelve, are too complicated to handle one at a time. Add them in the groups a subgroup cuts out, and the sums are shuffled among themselves by every symmetry of the whole — so they satisfy an equation as small as the number of groups, with whole-number coefficients. The subgroup decides the equation, the chain of subgroups decides the tower of equations, and whether that tower uses only square roots decides whether a compass can draw the polygon.