Delta function, Multiplicativity of Ramanujan Tau, Congruences

Consider

$\displaystyle \Delta(z)=q \prod_{n=1}^{\infty}\left(1-q^{n}\right)^{24}$

where ${q=e^{2 \pi i z}}$

Write

$\displaystyle \Delta(z)=\sum_{n=1}^{\infty} \tau(n) q^{n}$

$\displaystyle L(s, \Delta)=\sum_{n=1}^{\infty} \tau(n) n^{-s}$

Ramanujan conjectured:

$\displaystyle L(s, \Delta)=\prod_{p: \text { prime }}\left(1-\tau(p) p^{-s}+p^{11-2 s}\right)^{-1}$

$\displaystyle |\tau(p)|<2 p^{\frac{11}{2}}$

$\displaystyle \tau(p) \equiv 1+p^{11} \bmod 691$

$\displaystyle \tau(n) \equiv \sigma_{11}(n) \bmod 691$

Mordell’s Proof of the multiplicativity (1917):

Consider the operator:

$\displaystyle (T(p) f)(z)=\frac{1}{p} \sum_{l=0}^{p-1} f\left(\frac{z+l}{p}\right)+p^{11} f(p z)$

We prove that

$\displaystyle T(p) \Delta=\tau(p) \Delta$

Proof:

$\displaystyle F(z)=\frac{(T(p) \Delta)(z)}{\Delta(z)}$

For

$\displaystyle \left(\begin{array}{ll} a & b \\ c & d \end{array}\right) \in S L_{2}(\mathbb{Z})$

we have the modularity relations:

$\displaystyle \Delta\left(\frac{a z+b}{c z+d}\right)=(c z+d)^{12} \Delta(z)$

$\displaystyle \left(T(p) \Delta\right)\left(\frac{a z+b}{c z+d}\right)=(c z+d)^{12} \left(T(p) \Delta\right)(z)$

One way to prove the modularity of ${\Delta}$ is to use Jacobi triple product type relations to relate products to theta functions and the modularity of theta functions comes from Poisson summation. The second fact about the fact that ${T(p)}$ preserves modularity comes follows from the defintion of ${T(p)}$ – the definition of ${T(p)}$ encodes averaging over some cosets.

From these we deduce the modular invariance of ${F,}$

$\displaystyle F\left(\frac{a z+b}{c z+d}\right)=F(z)$

Observing that

$\displaystyle F(z) \rightarrow \tau(p) ~~~\text{as}~~ q \rightarrow 0$

we get that ${F(z)}$ is a constant equal to ${\tau(p)}$ .

Therefore

$\displaystyle T(p) \Delta=\tau(p) \Delta.$

With this information, we can prove the multiplicativity by the following computations:

$\displaystyle \begin{aligned} (T(p) \Delta)(z) &=\frac{1}{p} \sum_{l=0}^{p-1} \sum_{n=1}^{\infty} \tau(n) \exp \left(2 \pi i n \frac{z+l}{p}\right)+p^{11} \sum_{n=1}^{\infty} \tau(n) \exp \left(2 \pi i pnz\right) \\ &=\sum_{n=1}^{\infty}\left(\frac{1}{p} \sum_{l=0}^{p-1} e^{2 \pi i n \frac{l}{p}}\right) \tau(n) q^{\frac{n}{p}}+p^{11} \sum_{n=1}^{\infty} \tau(n) q^{p n}\\ &= \sum_{n=1}^{\infty} \tau(p n) q^{n}+p^{11} \sum_{n=1}^{\infty} \tau(n) q^{p n}\\ &= \sum_{n=1}^{\infty}\left(\tau(p n)+p^{11} \tau\left(\frac{n}{p}\right)\right) q^{n} \end{aligned}$

$\displaystyle \tau(p n)+p^{11} \tau\left(\frac{n}{p}\right)=\tau(p) \tau(n)$

$\displaystyle \tau(m n)=\tau(m) \tau(n), \quad \text {if} ~(m, n)=1$

$\displaystyle \tau\left(p^{k+1}\right)=\tau(p) \tau\left(p^{k}\right)-p^{11} \tau\left(p^{k-1}\right)$

$\displaystyle \sum_{k=0}^{\infty} \tau\left(p^{k}\right) x^{k}=\frac{1}{1-\tau(p) u+p^{11} x^{2}}$

$\displaystyle \begin{aligned} \sum_{n=1}^{\infty} \tau(n) n^{-s} &=\prod_{p}\left(\sum_{k=0}^{\infty} \tau\left(p^{k}\right) p^{-k s}\right) \\ &=\prod_{p}\left(1-\tau(p) p^{-s}+p^{11-2 s}\right)^{-1} \end{aligned}$

Below, we prove the third statement about the congruences of ${\tau(n).}$
Congruence proof: The main idea is to relate ${\Delta}$ to Eisenstein series.

$\displaystyle E_{k}(z)=\frac{1}{2} \sum_{(c, d)=1} \frac{1}{(c z+d)^{k}}$

$\displaystyle E_{k}(z)=1-\frac{2 k}{B_{k}} \sum_{n=1}^{\infty} \sigma_{k-1}(n) q^{n}$

$\displaystyle E_{k}\left(\frac{a z+b}{c z+d}\right)=(c z+d)^{k} E_{k}(z)$

$\displaystyle \left(\begin{array}{ll} a & b \\ c & d \end{array}\right) \in S L_{2}(\mathbb{Z})$

$\displaystyle E_{6}(z)=1-504 \sum_{n=1}^{\infty} \sigma_{5}(n) q^{n}$

$\displaystyle E_{12}(z)=1+\frac{65520}{691} \sum_{n=1}^{\infty} \sigma_{11}(n) q^{n}$

$\displaystyle \frac{E_{12}-E_{6}^{2}}{\Delta}$

is ${S L_{2}(\mathbb{Z})}$ invariant and

$\displaystyle \frac{E_{12}-E_{6}^{2}}{\Delta} \rightarrow \frac{65520}{691}+1008 ~~\text{as}~~ q\rightarrow 0$

$\displaystyle E_{12}-E_{6}^{2}=\frac{65520+1008 \cdot 691}{691} \Delta$

$\displaystyle \implies \sigma_{11}(n) \equiv \tau(n) \bmod 691$

$\displaystyle \tau(n) =\sigma_{11}(n)+\frac{691}{756}\left(-\sigma_{11}(n)+\sigma_{5}(n)-252 \sum_{m=1}^{n-1} \sigma_{5}(m) \sigma_{5}(n-m)\right)$

Let

$\displaystyle \begin{aligned} F(z) &=q \prod_{n=1}^{\infty}\left(1-q^{n}\right)^{2}\left(1-q^{11 n}\right)^{2} \\ &=\sum_{n=1}^{\infty} c(n) q^{n} \end{aligned}$

Looking at ${\mod 11}$ , we see that

$\displaystyle \begin{aligned} F &=q \prod_{n=1}^{\infty}\left(1-q^{n}\right)^{2}\left(1-q^{11 n}\right)^{2} \\ & \equiv q \prod_{n=1}^{\infty}\left(1-q^{n}\right)^{2}\left(\left(1-q^{n}\right)^{11}\right)^{2} \bmod 11 \\ & \equiv \Delta \bmod 11 \end{aligned}$

Therefore

$\displaystyle c(n) =\tau(n) \mod 11$

The second statement about the bounds is a much harder result- proved by Deligne using machinery of Riemann Hypothesis over finite fields.

https://archive.org/stream/proceedingsofcam1920191721camb#page/n141/mode/2up

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