Author: Kowalski

Today I was invited to give a short lecture on the occasion of the award ceremony for the Swiss Mathematical Olympiads (the national olympiads from which the Swiss participants to the International Mathematical Olympiad are selected). So I talked (of course…) about prime numbers. I’ve put the presentation on the web (it is in French, but of course there is only a minimal amount of text so it should be fairly understandable; if there’s any interest, I’ll prepare an English translation).

Suppose you are asked to describe an infinite sequence of signs + or -. How likely is it that you will end up with the same sequence as the one I am now thinking about? Well, according to all natural probabilistic models, the probability should be zero. Read on, then…

Quite a while ago, I wrote a short note to try to understand the distribution of the root number (also known as the sign of the functional equation) for the Jacobian of the modular curve X₀(p), where p is prime. My guess was that this should be +1 or -1 more or less equally often, and it should have been easy to confirm it by computing explicitly this sign using the Eichler-Shimura relation for the L-function of this jacobian and the trace formula for the Fricke involution, except that the latter turns out to involve class numbers, and after watching the dust settle, one sees that the approximate equidistribution is equivalent to asking some questions about the equidistribution modulo 4 or 8 of some class numbers of imaginary quadratic fields for which one knows the residue modulo 2 or 4, respectively.

And in particular, one needs to know the distribution modulo 4 of h(-p) for p congruent to 3 modulo 4, in which case genus theory says that the class number is odd.

And now for the coincidence: having reached this question, it seemed fairly natural to drop by the office (two doors up on the other side of the corridor) of one of the foremost expert on the distribution of class numbers, Henri Cohen, and ask him if the answer was already known? As it turned out, Henri was just in the midst of using Pari/GP to compute the values

$latex \Gamma_p(1/2)$

of the p-adic Gamma function at 1/2. These, although they should (in theory) be elements of p-adic fields, satisfy

$latex \Gamma_p(1/2)^2=1,$

or, in other words, these values form a sequence of signs. And the two sequences are the same!: we have

$latex \Gamma_p(1/2)\equiv h(-p)\text{ mod } 4\text{ for } p\equiv 3\text{ mod } 4.$

After this, I can not be impressed when hearing of people who just happen to think of their great-aunt who has been lost in the jungle of New Guinea for twenty years just a few minutes before receiving a phone call from her.

At the FIM Tea, this afternoon, the observation of the almost-never-erased blackboards in front of the seminar room led to the following historical question of clear importance: when was the blackboard, as a tool for lectures, invented? For instance, did Newton (who was a professor at Cambridge around 1670) ever lecture on a blackboard?

“Every sum is a trace” is a well-known folklore saying in certain automorphic circles (echoed by a no less convincing “Every sum is an expectation” around probabilists); in this spirit, let’s have a look at one of the most famous sums

$latex \zeta(2)=\sum_{n\geq 1}{\frac{1}{n^2}},$

which Euler was the first to evaluate.

It is possible to see it, and then compute it, as a trace, and I’m sure this has been done many times; here is (a variant of) the way I presented things for an exercise for my Spectral Theory class.

Consider the Hilbert space

$latex H=L^2([0,1])$

(for the Lebesgue measure), and the Volterra operator T:

$latex Tf(x)=\int_0^x{f(y)dy}$

which is a linear operator acting on H, in fact a Hilbert-Schmidt operator with kernel

$latex k(x,y)=\mathbf{1}_{\{y\leq x\}}$

which is bounded, and therefore certainly belongs to the space

$latex L^2([0,1]^2,dxdy).$

The operator T is therefore compact, and the operator S=T^*T is also compact, and in fact positive, so that the trace is well-defined as a non-negative real number, or infinity. The trace is well-known to be expressible in different ways: (i) as a sum of the series formed with the eigenvalues of S (with multiplicity); (ii) as the sum of the series

$latex \sum_{n}{\langle S(f_n),f_n\rangle}=\sum_{n}{||Tf_n||^2},$

for an arbitrary choice of orthonormal basis (f_n)_n of H; (iii) as the integral

$latex \int_0^1\int_0^1{|k(x,y)|^2dxdy}.$

This last integral is of course completely elementary: it is the area of the lower-half triangle in the square [0,1]² below the diagonal; in other words, it is 1/2.

For an alternate expression (hence an identity), we look at the series above for the easiest orthonormal basis available:

$latex f_n(t)=e^{2i\pi nt}\quad\quad\text{ for } n\in\mathbf{Z}.$

For the special case n=0, we have

$latex Tf_0(x)=x,$

hence

$latex \langle Sf_0,f_0\rangle =||Tf_0||^2=\int_0^1{x^2dx}=\frac{1}{3}.$

For non-zero n, we have

$latex Tf_n(x)=\frac{e^{2i\pi nx}-1}{2i\pi n},$

and therefore (Parseval, if you wish, or direct computation):

$latex \langle Sf_n,f_n\rangle = ||Tf_n||^2=\frac{1}{4\pi^2 n^2}\times (1+1)=\frac{1}{2\pi^2n^2}.$

Summing over all n and identifying the two expression for the trace, we get

$latex \frac{1}{2}=\frac{1}{3}+\sum_{n\not=0}{\frac{1}{2\pi^2n^2}}=\frac{1}{3}+\frac{1}{\pi^2}\zeta(2),$

and hence — unsurprisingly, I presume — we get

$latex \zeta(2)=\frac{\pi^2}{6}.$

(I said unsurprisingly, but I first managed to get confused enough about the computation — for a slightly different operator — that, for a while, I almost convinced myself that ζ(2)=π²/12).

As a proof (which, I repeat, is certainly not new, though it is not found in this collection), this is fairly close in flavor to the Fourier-expansion proofs, where one expands (typically) the function x-1/2 on [0,1] into Fourier series before applying the Parseval identity. (In fact, it seems this is “dual” in some simple way which could be made precise).

Like the Fourier-expansion argument, it has the nice feature of showing almost immediately that it will be possible to generalize the argument to

$latex \zeta(2k)=\sum_{n\geq 1}{\frac{1}{n^{2k}}}$

for k> 0 integer, using T^k instead of T; it also hints quite strongly that the result will be of the type

$latex \zeta(2k)=\alpha_k\pi^{2k},$

for some rational number α_k. But it is equally obvious that this will not work at all like this for zeta evaluated at odd positive integers, as it should…