The product raised to the twenty-fourth power

Every partition, hidden in a product read the partition numbers off Euler’s product: the reciprocal of $(1 - q)(1 - q^2)(1 - q^3)\cdots$ is $\sum p(n) q^n$, because expanding each factor $1/(1 - q^k) = 1 + q^k + q^{2k} + \cdots$ chooses how many parts of size $k$ to use. The terms that cancel almost everything then expanded the product itself and found it almost entirely empty — only the pentagonal exponents survive, each with coefficient $\pm 1$.

Between those two extremes lie the powers of the product. Its cube is still sparse, by Jacobi’s identity. Its twenty-fourth power, multiplied by $q$, is not sparse at all:

$$q \prod_{n \ge 1} (1 - q^n)^{24} = \sum_{n \ge 1} \tau(n)\, q^n = q - 24 q^2 + 252 q^3 - 1472 q^4 + 4830 q^5 - 6048 q^6 - \cdots$$

Srinivasa Ramanujan computed the coefficients up to $n = 30$ in 1916, called them $\tau(n)$, and noticed three things in his table that nobody could prove: that they multiply, that they are never larger than a precise bound, and that they agree with a quite different sum modulo 691. All three are now theorems, the second one of the deepest in twentieth-century mathematics, and a fourth question he did not ask — whether $\tau(n)$ is ever zero — is still open.

Computing a coefficient nobody can write down #

There is no formula for $\tau(n)$ in terms of $n$, and expanding a product of twenty-four copies of an infinite product is not a sensible way to get one. The logarithmic derivative is. Taking $q \frac{d}{dq} \log$ of both sides turns the product into a sum,

$$q \frac{d}{dq} \log\Big(q \prod (1 - q^n)^{24}\Big) = 1 - 24 \sum_{n \ge 1} \sigma(n) q^n,$$

where $\sigma(n)$ is the sum of the divisors of $n$ — the same move that gives the partition numbers, whose size has no formula, a recurrence through the divisor sums. Multiplying back and comparing coefficients gives

$$(n - 1)\, \tau(n) = -24 \sum_{k=1}^{n-1} \sigma(k)\, \tau(n - k),$$

which computes each value from all the earlier ones in whole numbers. The division by $n - 1$ always comes out exact, which the computation checks at every step. To $n = 10{,}000$ it takes under two seconds; $\tau(10{,}000)$ has twenty-two digits.

The figure divides by $n^{11/2}$ because that is how fast the values grow. At the primes the scaled values lie between $-2$ and $2$ — every one of the 1,229 primes below ten thousand — and at other numbers they stay within the number of divisors of $n$; the largest scaled value below ten thousand is 3.28, at $n = 4{,}841 = 47 \times 103$, which has four divisors. Those are Ramanujan’s conjecture in its two forms, and the reason for the exponent $11/2$ is that the series is a modular form of weight 12, half of which is 6, less a half.

The growth is worth pausing on, because it is the opposite of the partition numbers’. The reciprocal of Euler’s product has coefficients $p(n)$ that grow like $e^{\pi\sqrt{2n/3}}$, faster than any power of $n$; the product itself has coefficients that are only $0$ and $\pm 1$. The twenty-fourth power sits between, and its coefficients grow like a fixed power of $n$ — but with a sign that changes erratically, so that the sum $\tau(1) + \cdots + \tau(n)$ grows far more slowly than $n^{13/2}$. Raising the product to a power multiplies out a great deal of cancellation, and what survives the cancellation is exactly the size $n^{11/2}$ that the figure divides by.

The coefficients multiply #

The first surprise is in the table’s first row. $\tau(2) = -24$ and $\tau(3) = 252$, and $\tau(6) = -6048 = -24 \times 252$. That is not a coincidence of small numbers: whenever $m$ and $n$ have no common factor, $\tau(mn) = \tau(m)\,\tau(n)$. The computation checks it on all 21,935 coprime pairs whose product is at most ten thousand, numbers of up to twenty-two digits agreeing to the last digit. Where $m$ and $n$ share a factor the equality fails, in all 345 such pairs up to thirty, and what replaces it is a rule for prime powers:

$$\tau(p^{k+1}) = \tau(p)\, \tau(p^k) - p^{11}\, \tau(p^{k-1}),$$

checked on all 51 prime powers in range. So $\tau$ at every number is determined by $\tau$ at the primes, the way a number is determined by its one factorisation.

Nothing about the definition suggests this. The product $\prod (1 - q^n)^{24}$ is an additive object — its coefficients count signed ways of writing numbers as sums — and multiplicativity is a property of products of numbers. Louis Mordell proved it in 1917, the year after Ramanujan’s paper, and the proof is the beginning of the theory of Hecke operators: the series is an eigenvector of a family of linear maps indexed by the primes, and its coefficients are the eigenvalues. A sum of divisor powers behaves the same way for a simpler reason, and the divisor sums $\sigma_{11}$ will return below.

Why the multiplication is forced #

The reason $\tau$ multiplies can be said without the full theory of modular forms, and it is worth saying because it is an argument from counting dimensions.

The series $\Delta = \sum \tau(n) q^n$, read as a function of a complex variable through $q = e^{2\pi i z}$, changes in a prescribed way when $z$ is replaced by $-1/z$: it is multiplied by $z^{12}$. Functions with that property and a few others form a vector space, the modular forms of weight 12, and the ones that vanish at $q = 0$ — the cusp forms — form a subspace of dimension exactly one. Weight 12 is the first weight at which that subspace is not empty, and the twenty-fourth power is what reaches it: the product $\prod (1 - q^n)$ by itself picks up a twenty-fourth root of unity under the same substitution, and only its twenty-fourth power is free of it.

Hecke found, in the 1930s, a family of linear maps $T_p$, one for each prime, that carry modular forms of weight 12 to modular forms of weight 12 and cusp forms to cusp forms. A linear map from a one-dimensional space to itself can only multiply by a number. So $T_p \Delta = \lambda_p \Delta$ for some $\lambda_p$, and reading off coefficients shows $\lambda_p = \tau(p)$ and gives both of Mordell’s identities at once. The coefficients multiply because there is nowhere else for the series to go. At weight 24 the space of cusp forms has dimension two, and its natural basis does not multiply; the miracle at weight 12 is a miracle of small dimension.

The angles at the primes #

Deligne’s bound says that at every prime $\tau(p) = 2p^{11/2} \cos\theta_p$ for an angle $\theta_p$ between nought and $\pi$. Where do the angles fall?

Not evenly. They avoid the ends — angles near nought or $\pi$, where $\tau(p)$ would be near its bound — and crowd the middle, following the density $\frac{2}{\pi}\sin^2\theta$, with a chi-squared distance of 8.8 over eighteen bins. That is the Sato–Tate law, the same semicircle that a curve counted modulo every prime obeys, and the same distribution as the angle of a unit quaternion chosen uniformly at random — half the angle of a rotation of space chosen at random, a group that seems to have nothing to do with partitions. Richard Taylor and his collaborators proved it for $\tau$ in 2011, building on the methods that proved Fermat’s Last Theorem.

The bound itself was harder. Pierre Deligne proved it in 1974 by showing that $\tau(p)$ is, up to sign, the trace of a Frobenius on a piece of the cohomology of a variety over a finite field, and then proving the Weil conjectures — the generalisation of the Riemann hypothesis to such varieties — of which that bound is a special case. A question about a power series that a schoolchild could expand turned out to need the whole of modern algebraic geometry.

Modulo 691 #

Every point lies on the diagonal: $\tau(n)$ and $\sigma_{11}(n) = \sum_{d \mid n} d^{11}$ leave the same remainder on division by 691, for every one of the ten thousand values computed. Modulo neighbouring primes the two agree only occasionally — eighteen times in two thousand modulo 101, twice modulo 683 — so 691 is genuinely special.

The reason is the number $691$ itself: it is the numerator of the Bernoulli number $B_{12} = -691/2730$. The divisor sums $\sigma_{11}$ are the coefficients of another modular form of weight 12, an Eisenstein series, whose constant term has $B_{12}$ in its denominator. Multiply it by 691 and subtract the right multiple of $\Delta = \sum \tau(n) q^n$, and what is left is divisible by 691 coefficient by coefficient. The congruence is the shadow of two objects of the same weight that are equal modulo one prime and different everywhere else, and it is the same kind of statement as Ramanujan’s congruences for the partition numbers, which come from the partition product at weights where the theory forces a prime to divide.

The other primes that bind #

691 is not the only prime modulo which $\tau$ has a closed form. The figure’s last two lines record two more, checked on every value: $\tau(n) \equiv n\,\sigma_3(n) \pmod 7$ and $\tau(n) \equiv n\,\sigma_9(n) \pmod 5$. Similar congruences hold modulo powers of 2 and of 3. Jean-Pierre Serre and Peter Swinnerton-Dyer showed in the early 1970s that the primes admitting such a congruence — modulo which $\tau(p)$ is determined by a simple power of $p$ — are exactly 2, 3, 5, 7, 23 and 691, and that for every other prime $\ell$ the values $\tau(p)$ modulo $\ell$ are as unconstrained as they could possibly be. Five of the six give congruences with divisor sums. The sixth, 23, gives something stranger.

Modulo 23, a cubic equation #

The strangest of the congruences is modulo 23. Take the cubic $x^3 - x - 1$, whose discriminant is $-23$, and ask for each prime $p$ how many roots it has modulo $p$. When it has three, $\tau(p) \equiv 2$; when it has none, $\tau(p) \equiv -1$; when it has one, $\tau(p) \equiv 0 \pmod{23}$. The computation finds 193, 408 and 627 primes of the three kinds below ten thousand, and the rule holds for every one.

The shares approach a sixth, a third and a half, which are the proportions of the six symmetries of the cubic’s three roots that fix all three, none or one of them — the counting rule of how a polynomial breaks modulo the primes, Chebotarev’s theorem, for the group of all permutations of three things. The congruence says that $\tau(p)$ modulo 23 is a function of which symmetry Frobenius picks: the trace, in a two-dimensional representation of that group of six symmetries, reduced modulo 23. It is the simplest case of the correspondence between modular forms and Galois representations that the Langlands programme generalises, and it was found by J. R. Wilton in 1930 by comparing tables.

Does τ ever vanish? #

No $\tau(n)$ below ten thousand is zero. Derrick Lehmer asked in 1947 whether any is, and showed that the first zero, if there is one, must be at a prime: since $\tau$ is multiplicative, a zero at a composite number forces a zero at one of its prime powers, and the prime-power rule then forces one at the prime. The closest approach below ten thousand is at $p = 7{,}993$, where $|\tau(p)|$ is about three ten-thousandths of its bound — an angle within a fiftieth of a degree of a right angle.

Computers have checked that $\tau(p) \ne 0$ for every $p$ below about $8 \times 10^{23}$, using the congruences above and others like them to rule out whole classes of primes at once. The Sato–Tate law gives no reason for a zero and no reason against one: angles near a right angle are as common as any, and a zero needs one to be exactly a right angle, which for an integer of thirty digits is a coincidence of probability about one in $p^{11/2}$. The sum of those probabilities over all primes converges, so a heuristic count predicts that there are probably no zeros at all — but heuristics have no force against a single exception, and none is proved impossible.

What τ counts #

$\tau$ was not invented for its own sake. Ramanujan met it in counting the ways of writing a number as a sum of twenty-four squares. For four or eight squares the count is a divisor sum exactly — Jacobi’s formula and its relatives — because the corresponding generating function is an Eisenstein series and nothing else. For twenty-four squares the generating function has weight 12, the space it lives in has room for $\Delta$, and the count is a divisor sum built from $\sigma_{11}$ plus a correction that is a combination of $\tau(n)$ and $\tau(n/2)$. So the number of representations by twenty-four squares has no formula in divisor sums alone, and the part that resists is exactly $\tau$ — which is why Ramanujan cared about its size: a bound on $\tau(n)$ is an error term for counting points of a lattice in twenty-four dimensions on a sphere.

That count is the beginning of a long road. The lattice in twenty-four dimensions with no vectors of the shortest lengths, John Leech’s, has a theta function whose coefficients are built from $\sigma_{11}$ and $\tau$, and the twenty-four in the exponent of Euler’s product turns up again as the dimension of that lattice, one of several appearances of the number that the theory of modular forms ties together.

What the computations show and assume #

Every value of $\tau(n)$ in the figures is exact, computed in integers of any size, and every identity is checked on every case in range, not sampled. The checks prove nothing beyond ten thousand; they are the evidence Ramanujan had, extended. The theorems are what make the patterns true everywhere: Mordell’s for the multiplication, Deligne’s for the bound, Barnet-Lamb, Geraghty, Harris and Taylor’s for the angles, and for the congruences an explicit identity of modular forms in each case.

The figure of angles is the least conclusive. A histogram of 1,229 angles fits the Sato–Tate density to sampling error and would fit a few nearby densities too; the law is a theorem about the limit, and the figure shows only that the first twelve hundred primes do nothing to contradict it. The same is true of the closest approaches to zero: the smallest scaled value in range says nothing about whether a smaller one exists at a larger prime.

Still open: Lehmer’s question #

Is $\tau(n) \ne 0$ for every $n$? The question has resisted every approach for seventy-five years. It would follow from showing that $\tau(p)$ avoids nought modulo enough primes to cover every case, and the congruences found so far rule out only a set of primes of density less than one. It would also follow from information about how often a Galois representation’s trace vanishes, and the general results there — Jean-Pierre Serre showed in 1981 that for forms like this the primes with a zero trace have density nought — stop short of proving the set empty. Serre’s bound is the strongest statement available: if $\tau$ vanishes at all, it does so at a set of primes too sparse to count. Whether that set has any members is open.

The bound that Deligne proved has its own open cousins. The same inequality, stated for the coefficients of every cusp form of every weight, is a theorem for the holomorphic forms like $\Delta$ and an open problem for the non-holomorphic ones, the Maass forms, where the best known bounds still fall short of the conjectured exponent by a small but stubborn fraction. The generalisation to all automorphic forms, the generalised Ramanujan conjecture, is one of the central open problems of the Langlands programme, and Ramanujan’s observation about a column of thirty numbers is its first instance.

A product that knows about primes #

The twenty-fourth power of Euler’s product has coefficients that behave as if they had been built from primes rather than from sums. They multiply on coprime numbers and obey a three-term rule on prime powers, so the primes determine everything; at each prime they are bounded by $2p^{11/2}$ and their angles follow the semicircle law of random rotations; modulo 691 they equal a divisor sum, and modulo 23 they are decided by how $x^3 - x - 1$ factors. All of it was in Ramanujan’s table of thirty numbers, and all of it except whether the coefficients ever vanish is now proved.