A congruence for the Ramanujan Tau function.

Here is a fun problem from Apostol’s “Modular Functions and Dirichlet Series in Number Theory” that involves one of Ramanujan’s conjectures.

$\textbf{Problem: }$ : Let $\Delta(z) = q \prod_{n=1}^\infty(1-q^n)^{24} = \sum_{n=1}^\infty \tau(n)q^{n}, \quad q=e^{2\pi i z}.$

Prove that $\tau(n) \equiv \sigma_{11}(n) \mod 691.$

$\textbf{Solution:}$ First we note $\Delta$ is a cusp form of weight 12. Then we write out the $q$ -expansions for $E_6$ , $E_6^2$ , and $E_{12}$ .

$E_6 = 1+ \dfrac{2\cdot 6}{B_6} \sum_{n=1}^\infty \sigma_5(n) q^n = 1-(42)(12) \sum_{n=1}^\infty \sigma_5(n) q^n$

$E_6^2= 1- (2)(12)(42) \sum_{n=1}^\infty \sigma_5(n) q^n+ (42)^2(12)^2 \sum_{n=1}^\infty \sum_{m=1}^{n-1}\sigma_5(n)\sigma(n-m)q^n$

$E_{12} = 1 - \dfrac{2(12)}{B_{12}} \sum_{n=1}^\infty \sigma_{11}(n) q^n = 1 - \dfrac{2(12)}{\dfrac{-691}{2730}} \sum_{n=1}^\infty \sigma_{11}(n) q^n = 1+ \dfrac{(24)(2730)}{691} \sum_{n=1}^\infty \sigma_{11}(n) q^n.$

$\Delta = a E_{12} + bE_6^2$ . Since $\Delta$ has no constant terms, $a+b=0$ . So $a=-b$ .

We will also use the fact that $\tau(1)=1$ .

The coefficient of $q$ in $E_{12}(z)$ is $\dfrac{(224)(2730)}{691}$ .

The coefficient of $q$ in $E_6^2(z)$ is $-(2)(12)(42).$

So $\tau(1) = 1$ and

$\tau(1) = \alpha\left(\dfrac{(224)(2730)}{691}\right) - \alpha(-(2)(12)(42)) = \alpha \left(\dfrac{762048}{691}\right)$ .

So $\alpha = \dfrac{691}{762048}=\dfrac{691}{(24)(756)(42)}$ .

Comparing coefficients of $q$ for $\Delta = a E_{12} - a E_6^2$ , and

$\tau(n) = \alpha \left[\dfrac{(24)(2730)}{(691)} \sigma_{11}(n) +(24)(42)\sigma_5(n) -(12)^2(42)^2 \sum_{m=1}^{n-1} \sigma_5(m)\sigma_5(n-m) \right]$

$\alpha \cdot \dfrac{(24)(2730)}{691} = \dfrac{691}{(24)(756)(42)} \cdot \dfrac{(24)(2730)}{691} = \dfrac{65}{756}.$

$\alpha \cdot (24)(42) = \dfrac{691}{756}$

$\alpha \cdot (12)^2(42)^2 = \dfrac{691}{756} \cdot (12)(24) = \frac{691}{3}.$ So

$\tau(n)= \dfrac{65}{756} \sigma_{11}(n) +\dfrac{691}{756} \sigma_5(n) - \dfrac{691}{3} \sum_{m=1}^{n-1} \sigma_6(m) \sigma_5(n-m).$

$\tau(n) \equiv \dfrac{65}{756} \sigma_{11}(n) \mod 691 \equiv \dfrac{65+691}{756} \sigma_{11}(n) \equiv \dfrac{756}{756} \sigma_{11}(n) \equiv \sigma_{11}(n) \mod 691.$

A congruence for the Ramanujan Tau function.

Written by bs8277

Leave a Reply Cancel reply