Catalan’s Constant and periods

There is a 60th birthday conference in honour of Frits Beukers in Utrech in July; I’m hoping to swing by there on the way to Oberwolfach. Thinking about matters Beukers made me reconsider an question that I’ve had for while.

There is a fairly well known explanation of why \zeta(3) should be irrational (and linearly independent of \pi^2) in terms of Motives. There is also a fairly good proof that \zeta(3) \ne 0 in terms of the non-vanishinjg of Borel’s regulator map on K_5(\mathbf{Z}). (I guess there are also more elementary proofs of this fact.) A problem I would love to solve, however, is to show that, for all primes p, the Kubota-Leopoldt p-adic zeta function \zeta_p(3) is non-zero. Indeed, this is equivalent to the injectivity of Soule’s regulator map

K_5(\mathbf{Z}) \otimes \mathbf{Z}_p \rightarrow K_5(\mathbf{Z}_p).

(Both these groups have rank one, and the cokernel is (at least for p > 5) equal to \mathbf{Z}_p/\zeta_p(3) \mathbf{Z}_p by the main conjecture of Iwasawa theory.) It is somewhat of a scandal that we can’t prove that \zeta_p(3) is zero or not; it rather makes a mockery out of the idea that the “main conjecture” allows us to “compute” eigenspaces of class groups, since one can’t even determine if there exists an unramified non-split extension

0 \rightarrow \mathbf{Q}_p(3) \rightarrow V \rightarrow \mathbf{Q}_p \rightarrow 0

or not. Well, this post is about something related to this but a little different. Namely, it is about the vaguely formed following question:

What is the relationship between a real period and its p-adic analogue?

Since one number is (presumably) in \mathbf{R} \setminus \mathbf{Q} and the other in \mathbf{Q}_p \setminus \mathbf{Q}, it’s not entirely clear what is meant by this. So let me give an example of what I would like to understand. One could probably do this example with \zeta(3), but I would prefer to consider the “simpler” example of Catalan’s constant. Here

G = \displaystyle{\frac{1}{1} - \frac{1}{3^2} + \frac{1}{5^2} - \frac{1}{7^2} \ldots } = L(\chi_4,2) \in \mathbf{R},

is the real Catalan’s constant, and

G_2 = L_2(\chi_4,2) \in \mathbf{Q}_2

is the 2-adic analogue. (There actual definition of the Kubota-Leopoldt zeta function involves an unnatural twist so that one could conceivably say that L_2(\chi_4,2) = 0 and that the non-zero number is \zeta_2(2), but this is morally wrong, as the examples below will hopefully demonstrate. Morally, of course, they both relate to the motive \mathbf{Q}(2)(\chi_4).)

So what do I mean is the “relation” between G and G_2. Let me give two relations. The first is as follows. Consider the recurrence relation (think Apéry/Beukers):

n^2 u_n = (4 - 32 (n-1)^2) u_{n-1} - 256 (n-2)^2 u_{n-2}.

It has two linearly independent solutions with a_1 = 1 and a_2 = -3, and b_1 = -2 and b_2 = 14. One fact concerning these solutions is that b_n \in \mathbf{Z}, and a_n \cdot \mathrm{gcd}(1,2,3,\ldots,n)^2 \in \mathbf{Z}. Moreover one has that:

\displaystyle{ \lim_{n \rightarrow \infty} \frac{a_n}{b_n}} = G_2 \in \mathbf{Q}_2.

The convergence is very fast, indeed fast enough to show that G_2 \notin \mathbf{Q}. What about convergence in \mathbf{R}, does it converge to the real Catalan constant? Well, a numerical test is not very promising; for example, when n = 40000 one gets 0.625269 \ldots, which isn’t anything like G = 0.915966 \ldots; for contrast, for this value of n one has a_n/b_n - G_2 = O(2^{319965}), which is pretty small. There are, however, two linearly independent solutions over \mathbf{R} given analytically by

\displaystyle{\frac{(-16)^n}{n^{3/2}}  \left( 1 + \frac{5}{256} \frac{1}{n^2} - \frac{903}{262144} \frac{1}{n^4}  + \frac{136565}{67108864} \frac{1}{n^6} - \frac{665221271}{274877906944} \frac{1}{n^8} + \ldots \right)},

\begin{aligned}  \frac{(-16)^n \cdot \log n}{n^{3/2}}  \left( 1 + \frac{5}{256} \frac{1}{n^2} - \frac{32261}{7864320} \frac{1}{n^4}  + \frac{136565}{67108864} \frac{1}{n^6} - \frac{665221271}{274877906944} \frac{1}{n^8} + \ldots \right)\\  +\frac{(-16)^n}{n^{3/2}} \left( -\frac{1}{768} \frac{1}{n^2} + \frac{32261}{7864320} \frac{1}{n^4}  - \frac{30056525}{8455716864}  \frac{1}{n^6} + \frac{1778169492137}{346346162749440}  \frac{1}{n^8} + \ldots \right) \end{aligned},

from which one can see that a_n/b_n must converge very slowly, and indeed, one has (caveat: I have some idea on how to prove this but I’m not sure if it works or not):

\displaystyle{\frac{a_n}{b_n} = G -  \frac{1}{(0.2580122754655 \ldots) \cdot \log n + 0.7059470639 \ldots}}

So one has a naturally occurring sequence which converges to G in \mathbf{R} and G_2 in \mathbf{Q}_2. So that is some sort of “relationship” alluded to in the original question. Here’s another connection. Wadim Zudilin pointed out to me the following equality of Ramanujan:

\displaystyle{G = \frac{1}{2} \sum_{k=0}^{\infty} \frac{4^k}{(2k + 1)^2 \displaystyle{\binom{2k}{k}}}} \in \mathbf{R}.

This sum also converges 2-adically. So, one can naturally ask whether

\displaystyle{G_2 =^{?} \frac{1}{2} \sum_{k=0}^{\infty} \frac{4^k}{(2k + 1)^2 \displaystyle{\binom{2k}{k}}}} \in \mathbf{Q}_2.

(It seems to be so to very high precision.) These are not random sums at all. Indeed, they are equal to

\displaystyle{ \frac{1}{2} \cdot F \left( \begin{array}{c} 1,1,1/2 \\ 3/2,3/2 \end{array} ; z \right)}

at z = 1. Presumably, both of these connections between G and G_2 must be the same, and must be related to the Picard-Fuchs equation/Gauss-Manin connection for X_0(4). This reminds me of another result of Beukers in which one compares values of hypergeometric functions related to Gauss-Manin connections and elliptic curves, and finds that they converge in \mathbf{R} and \mathbf{Q}_p for various p to algebraic (although sometimes different!) values. Of course, things are a little different here, since the values are (presumably) both transcendental. Yet it would be nice to understand this better, and see to what extent there is a geometric interpretation of (say) the non-vanishing of L_p(\chi,2) for some odd quadratic character \chi. Of course, one always has to be careful not to accidentally prove Leopoldt’s conjecture in these circumstances.

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