Irrationality of Zeta(3) using Modular Forms

The proof is based on the following lemma.

Let ${f_{0}(t), f_{1}(t), \ldots, f_{k}(t)}$ be power series in ${t}$ Suppose that for any ${{n} \in \mathbb{N}, i=0,1, \ldots}$ , the ${n}$ -th coefficient in the Taylor series of ${f_{i}}$ is rational and has denominator dividing ${d^{n}[1, 2, \ldots, n]^r}$ where ${r}$ , ${d}$ are certain fixed positive integers and ${[1, 2, \ldots, {n}]}$ is the lowest common multiple of ${1, 2, \ldots, n .}$ Suppose there exist real numbers ${\theta_{1}, \theta_{2}, \ldots, \theta_{k}}$ such that ${f_{0}(t)+\theta_{1} f_{1}(t)+\theta_{2} f_{2}(t)+\ldots+\theta_{k} {f}_{k}(t)}$ has radius of convergence ${\rho}$ and infinitely many nonzero Taylor coefficients. If ${\rho> de^r}$ , then at least one of ${\theta_{1}, \ldots, \theta_{\mathrm{k}}}$ is irrational.

The proof is easy. If they were all rational we would get that

$\displaystyle Dd^n[1,2, \cdots , n]^r \left|a_0(n)+\theta_{1} a_1(n)+\theta_{2} a_{2}(n)+\ldots+\theta_{k} a_{k}(n) \right|$

nis a integer with is bounded by ${Dd^n[1,2, \cdots , n]^r (\rho -\epsilon)^{-n} \rightarrow 0}$

We now construct the power series using modular forms involving $\zeta(3)$ as one of the $\theta_i$ !

Consider the Eisenstein series

$\displaystyle E_{4}(\tau)=1+240 \sum_{1}^{\infty} \sigma_{3}(n) q^{n}, E_{2}(\tau)=1-24 \sum_{1}^{\infty} \sigma(n) q^{n}$

and the functions defined in terms of them

$\displaystyle 40F(\tau)=\left(E_{4}(\tau)-36 E_{4}(36 \tau)-7\left(4 E_{4}(2 \tau)-9 E_{4}(3 \tau)\right)\right).$

$\displaystyle 24E(\tau)=\left(-5\left(E_{2}(\tau)-6 E_{2}(6 \tau)\right)+2 E_{2}(2 \tau)-3 E_{2}(3 \tau)\right).$

These are modular forms of weight 4 and 2 respectively.

Also consider the function

$\displaystyle t(\tau)=q \prod_{n=1}^{\infty}\left(1-q^{6 n+1}\right)^{12}\left(1-q^{6 n+5}\right)^{-12}$

${t(\tau)}$ is a modular function on ${\Gamma_1(6)} =$ of

THe L-function corresponding to ${F}$ satisfies

$\displaystyle L(F, s)=\sum_{n=1}^{\infty}\left(\frac{6 \sigma_{3}(n)}{n^{s}}-36 \frac{6 \sigma_{3}(n)}{(6 n)^{s}}-28 \frac{6 \sigma_{3}(n)}{(2 n)^{s}}+63 \frac{6 \sigma_{3}(n)}{(3 n)^{s}}\right)$

$\displaystyle = 6\left(1-6^{2-s}-7 \cdot 2^{2-s}+7 \cdot 3^{2-s}\right) \zeta(s) \zeta(s-3).$

Define ${f(\tau)}$ by the equation

$\displaystyle \left(\frac{d}{d \tau}\right)^{3} f(\tau)=(2 \pi i)^{3} F(\tau) ; \quad f(i \infty)=0$

${E}$ can be written as multivalued function of ${t}$ as a integral power series

$\displaystyle E(t)=1+5 t+73 t^{2}+1445 t^{3}+\cdots$

Now the modular transformation gives

$\displaystyle E(-1 / 6 \tau)(f(-1 / 6 {\tau})-\zeta(3))={E}(\tau)(f(\tau)-\zeta(3))$

because $\displaystyle L(F, 3)=6 \cdot(-1 / 3) \zeta(3) \zeta(0)=\zeta(3)$ .

Hence we can write consider the function $E(t)f(t)$ , which looks like

$\displaystyle 6 t+(351 / 4) t^{2}+\ldots ..$

And we have that ${[1,2, 3,, \cdots n]^3E(t)f(t)}$ is a power series with integer coefficients.

The function

$\displaystyle E(t)(f(t)-\zeta(3))$

has radius of convergence ${\rho > e^3}$ . (One has to look carefully at the definitions and geometry of the functions $t$ , $\tau(t)$ , the mutlivaluedness, branch points to argue this)