Selberg Identity for Kloosterman Sums, Multiplicativity of Ramanujan Tau

The Ramanujan Tau function is defined by

$\displaystyle \Delta(z)=\sum_{n \geq 1} \tau(n) q^{n}=q \prod_{n \geq 1}\left(1-q^{n}\right)^{24}=\eta(z)^{24}.$

The sequence ${\tau(n)}$ looks like

$\displaystyle \begin{aligned} &1,-24,252,-1472,4830,-6048,-16744,84480,-113643,-115920,534612,-370944,-577738,\\ &401856,1217160,987136,-6905934,2727432,10661420,-7109760,-4219488,-12830688,\cdots \end{aligned}$

Ramanujan observed multiplicativity property

$\displaystyle \begin{aligned} &\tau(m n)=\tau(m) \tau(n) \text { if } (m, n)=1 \\ &\tau\left(p^{r+1}\right)=\tau(p) \tau\left(p^{r}\right)-p^{11} \tau\left(p^{r-1}\right) \text { for } p \text { prime and } r>0 . \end{aligned}$

And these both formulae together can be written as

$\displaystyle \tau(m)\tau(n)=\sum_{d \mid(m, n)} d^{11} \tau\left(\frac{m n}{d^{2}}\right).$

Standard way to prove these properties by the use of Hecke operators.

By the Petersson trace formula applied to holomorphic cusp forms on ${SL_2(\mathbb Z)}$ we get

$\displaystyle \tau(m) \tau(n)= \frac{\|\Delta\|^{2} }{ 4 \pi^{13} \Gamma(11)} (m n)^{\frac{11}{2}}\left(\delta_{m, n}+2 \pi \sum_{c>0} \frac{S(m, n ; c)}{c} J_{11}\left(\frac{4 \pi \sqrt{m n}}{c}\right)\right)$

where

$\displaystyle \|\Delta \|^2 =\langle \Delta \rangle =\int_{\mathbb H /SL_2(\mathbb Z)} y^{12} \left|\Delta(z)\right|^2 \frac{dxdy}{y^2}$

is the Petersson norm of ${\Delta(z)}$ ,

$\displaystyle S(m, n ; c)=\sum_{a d \equiv 1(\bmod c)} e\left(\frac{m a+n d}{c}\right)$

is the Kloosterman sum and the

$\displaystyle J_{\nu}(x)=\sum_{\ell=0}^{\infty} \frac{(-1)^{\ell}}{\ell ! \Gamma(\ell+1+\nu)}\left(\frac{x}{2}\right)^{\nu+2 \ell}$

is the Bessel function of first kind.

In particular we have,

$\displaystyle \tau(n)= \frac{\|\Delta\|^{2} }{ 4 \pi^{13} \Gamma(11)} (n)^{\frac{11}{2}}\left(\delta_{n, 1}+2 \pi \sum_{c>0} \frac{S(1, n ; c)}{c} J_{11}\left(\frac{4 \pi \sqrt{n}}{c}\right) \right).$

This is an amazing formula, this complicated sum of Kloosterman sums twisted with Bessel functions turn out to be an integer! (up to the constants in the front)

Proof of multiplicativity:

We start by taking ${\frac{mn}{d^2}}$ in the formula for ${\tau(n)}$

$\displaystyle \tau(\frac{mn}{d^2}) =\frac{\|\Delta\|^{2} }{ 4 \pi^{13} \Gamma(11)} (\frac{mn}{d^2})^{\frac{11}{2}}\left(\delta_{\frac{mn}{d^2}, 1}+2 \pi \sum_{c>0} \frac{S(1, \frac{mn}{d^2} ; c)}{c} J_{11}\left(\frac{4 \pi \sqrt{\frac{mn}{d^2}}}{c}\right) \right).$

and compute $\sum_{d|(m,n)} d^{11} \tau(\frac{mn}{d^2})$ using this expression.

Comparing the expression for ${\tau(m) \tau(n)}$ and ${\tau(mn)}$ , we need to prove the following.

$\displaystyle \delta_{m,n} =\sum_{d|(m,n)} \delta_{\frac{mn}{d^2}, 1}$

and

$\displaystyle S(m, n ; c) =\sum_{d|(m,n, c)} dS\left(1, \frac{mn}{d^2} ; c/d\right)$

Note there is a change of variables ${c \rightarrow dc}$ . The ${c}$ in the RHS of the above sum corresponds to ${dc}$ in the Petterson expansion for ${\tau(\frac{mn}{d^2}).}$

The first equation is easy to check. The only way ${\delta_{\frac{mn}{d^2}, 1}}$ is non-zero is if ${m=n=d.}$

Let’s verify the second identity. It was proved by Selberg.

Proof of Selberg’s Identity:

In the case where ${(m, c)=(n, c)=1}$ , replacing ${a}$ by ${na}$ in the sum we get

$\displaystyle S(m, n ; c)=\sum_{a d \equiv 1(\bmod c)} e\left(\frac{m a+n d}{c}\right) =\sum_{a d \equiv 1(\bmod c)} e\left(\frac{mn a+ d}{c}\right) = S(1, mn, c)$

And we are done in this case because the RHS has only one term corresponding to ${d=1}$ which is exactly ${S(1, mn, c)=S(mn, 1, c).}$ In fact we just used ${(n, c)=1}$ .

But it’s enough to assume ${(m, n, c)=1}$ as seen by the following:

Let ${(c, n)= n_1, (c, m)=m_1}$ and write ${c=n_1m_1c_1}$ . We have ${(m_1, n_1)=1}$ because ${(m, n, c)=1.}$

Now if ${c_1}$ , ${n_1}$ and ${n_2}$ all have no common factors between each other, choosing ${a_1}$ such

$\displaystyle a_1 = {m}a \mod n_1, a_1 =nd \mod m_1, a_1 =nd \mod c_1$ we get

$\displaystyle a_1 +mnd_1 = ma+ nd \mod c$

which implies ${S(m, n ; c) =S(1, mn, c)}$ .

If ${c_1}$ has a common with with ${n_1}$ or ${m_1}$ , let say with ${n_1}$ , then ${c}$ and ${n}$ are divisible by some prime twice, that is ${p^2 \mid c, n}$ but ${p}$ doesn’t divide ${m_1}$ . In this case the Kloosterman sums ${S(m, n ; c)}$ and ${S(mn, 1 ; c)}$ are both zero. We can prove it from the following:

If ${q=q_1q_2}$ where ${(q_1, q_2)=1}$ are coprime integers we write ${a=x_{1} q_{2}+x_{2} q_{1}}$ where ${ x_2 = a\overline q_1 \mod q_2, x_1 = a\overline q_2 \mod q_2}$ . So we get the following twisted multiplicativity in the modulus.

$\displaystyle S(m, n ; q)=S\left(m, n \overline{q_{2}}^{2} ; q_{1}\right) S\left(m, n \overline{q_{1}}^{2} ; q_{2}\right).$

Twisted multiplicativity allows us to reduce to the computations to prime powers. Because ${p^2\mid c}$ , we will need the following expression to write sums with ${c=p^k}$ in terms of sums with lower modulus ${p^{k-1}.}$

If ${c=p^k}$ , where ${k\ge 2}$ , we write ${a = x + p^{k-1}x_1}$ where ${a=x \bmod p^{k-1}}$ and ${1\le x_1<p}$ to get

$\displaystyle S(m, n, p^k) = \sum_{ \substack{x \bmod p^{k-1}\\ (x, p)=1} } e\left(\frac{m x+n \bar{x}}{p^{k}}\right) \sum_{x_1=1}^{p} e\left(\frac{x_1\left(m-n \bar{x}^{2}\right)}{p}\right)$

So we can see that by reducing to the case of prime powers, the inner sum is zero when ${p^2}$ divides ${n}$ and doesn’t divide ${m}$ both in the case of ${S(m, n, c)}$ and ${S(mn, 1, c)}$ .

So we proved that ${S(m, n ; c) =S(1, mn, c)}$ and we are done because in this case ${(m, n, c)=1}$ there is only one term in the Selberg identity.

Next prove the Selberg identity for the case ${c=p^k.}$

The identity reduces to

$\displaystyle S(m, n ; p^{k})=S(mn, 1 ; p^{k})+\delta_{p|(m,n)} ~p S\left(\frac{m}{p}, \frac{n}{p} ; p^{k-1}\right)$

If ${p}$ doesn’t divide ${m}$ or ${n}$ , we can change variable and shift the ${n}$ to get ${S(m, n ; p^{k})=S(mn, 1 ; p^{k})}$ just like before. So assume ${p}$ divides both ${m}$ and ${n}$ and so we need to prove

$\displaystyle S(m, n ; p^{k})=S(mn, 1 ; p^{k})+ p S\left(\frac{m}{p}, \frac{n}{p} ; p^{k-1}\right)$

For instance if ${k=1}$ and ${p|(m,n)}$ , we have

$\displaystyle S(m, n ; p) =p-1, \quad S(mn, 1 ; p)=-1, S\quad \left(\frac{m}{p}, \frac{n}{p} ; 1\right)=p.$

For ${k \ge 2}$ , the expression in terms of lower modulus shows that ${S(mn, 1 ; p^{k})=0}$ because ${p^2|mn}$ . Now the equation reduces to

$\displaystyle S(m, n ; p^{k})=p S\left(\frac{m}{p}, \frac{n}{p} ; p^{k-1}\right)$

which is obvious since each term in the sum on LHS repeats ${p}$ times.

We now move onto the general case. We basically use that ${S(m, n ; c) =S(1, mn, c)}$ when ${(m, n, c)=1.}$

Let assume ${(m, n, c)= g}$ , so we have ${(m/g, n/g, c/g)=1}$

The RHS equals

$\displaystyle \sum_{d|g} dS\left(1, \frac{mn}{d^2} ; \frac{c}{d}\right) =\sum_{g=dd'} dS\left(1, \frac{n}{g}\frac{m}{g} d'^2 ; \frac{c}{g}d'\right)$

Let us denote ${m'=m/g, n'=n/g, c'=c/g}$ , so we have

. $\displaystyle RHS =\sum_{g=dd'} dS\left(1, m'n' d'^2 ; c'd'\right)$

If ${(c', d')\neq 1}$ , then the modulus ${c'd}$ is divisible by some ${p^2}$ , using the formula for prime-power moduli mentioned above we get ${S\left(1, \frac{n}{g}\frac{m}{g} d'^2 ; \frac{c}{g}d'\right)=0}$ .

So assume ${(c', d')= 1.}$

By the twisted multiplicativity we get

$\displaystyle S\left(1, m'n'd'^2 ; c'd'\right) =S\left(1, m'n' ; c' \right) S\left(1, 0 ;d' \right) = \mu(d')S\left(1, m'n' ; c' \right) .$

We now use that ${(m', n', c')=1}$ to change ${S\left(1, m'n' ; c' \right) }$ to ${S(m', n', c')}$ , and get

$\displaystyle S\left(1, m'n'd'^2 ; c'd'\right) =\mu(d')S(m', n', c') 1_{(c', d')=1}.$

$\displaystyle RHS =\sum_{g=dd'} d \mu(d')S(m', n', c')1_{(c', d')=1}.$

We apply twisted multiplicativity in reverse to ${c', d'}$ to get

$\displaystyle RHS =\sum_{g=dd'} 1_{(c', d')=1} d \mu(d') S(m', n', c') =\left( \sum_{\substack{g=dd_1d_2\\ d_1|c'}} \mu(d_1) \mu(d_1d_2)d\right) S(m', n', c')$

$\displaystyle LHS= S(m, n ; c) =S(m'g, n'g; c'g)$

Now ${a \mod c}$ reduce to a coprime residue modulo ${c'}$ . So ${S(m'g, n'g; c'g)}$ is just ${ S(m', n', c') }$ repeated a few number of times. In fact the number of times it repeats is the number of times ${a \mod c}$ reduces to a fixed class mod ${c'.}$ This is ${\phi(g)}$ number of times if ${g}$ is coprime to ${c'}$ . But in general it repeats exactly

$\displaystyle \left( \sum_{\substack{g=dd_1d_2\\ d_1|c'}} \mu(d_1) \mu(d_1d_2)d\right) =(g, c')\phi\left(\frac{g}{(g, c')}\right)$

number of times and we see that ${LHS=RHS.}$ (We can verify the above identity by looking at prime power by mutliplicativity of the quantities involved)

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