# Does the Divisor Function $\sigma(n)$ have analogues for other Fuchsian groups?

I have been reading about the divisor function $\sigma = 1 \ast 1$ and proved an elementary identity:

$$\Big[\sum_{d|n} \sigma_0(d)\Big]^2 = \sum_{d|n} \sigma_0(d)^3$$

Here $\sigma_0 = \sum_{d|n} 1$ counts the divisors of $n$. Could be the first of many such kind of identity? I wouldn't even be sure where to look for identities for $\sigma_k$.

Perhaps you could prove such a result with Eisenstein series:

$$G_k(z) = - \frac{B_k}{k} + \sum_{n=1}^\infty \sigma_{k-1}(n)q^n = \frac{(2\pi i)^k}{(k-1)!} \times \frac{1}{2}\sum \frac{1}{(m \tau + n)^k}$$

However $G_0$ is not a meaningful Eisenstein series.

Could the divisor function have analogues for other Fuchsian groups? The Ramanujan sum

$$\sum_{(a,q)=1} e^{2\pi i \frac{a}{q}} = \mu(q)$$

related to the Mobius function. This certainly has analogues for other Fuchsian groups.

• The stated identity seems to be false when $n=p$ is prime; I get $(2+p)^2$ on the left and $1+(1+p)^3$ on the right. – David E Speyer Aug 19 '15 at 16:47
• @DavidSpeyer I meant the divisor counting function, as in this question – john mangual Aug 19 '15 at 17:16
• Oh, I see. In that case, there is a simple explanation: Both sides are multiplicative, so any identity which holds for $n=p^a$ holds for all $n$. But $\sum_{d|p^a} \sigma_0(d)^k$ is a polynomial in $a$ and of course there are going to be algebraic identities between various polynomials in $a$. Your relation is $\left( \sum_{b=0}^a b \right)^2 = \sum_{b=0}^a b^3$. Write down any other polynomial relation between Faulhaber polynomials en.wikipedia.org/wiki/… and it will also be a polynomial relation between the functions $\sum_{d|n} \sigma_0(d)^k$. – David E Speyer Aug 19 '15 at 17:28
• Sorry, above is silly: It would only work for a relation of the form (product of Faulhaber's) = (other product of Faulhaber's). Otherwise, the reduction to prime powers breaks. – David E Speyer Aug 19 '15 at 17:33

Looks similar to the elementary formula $(\sum_{k=1}^n k)^2=\sum_{k=1}^n k^3$. Here's why. The function $\sigma_0$ is multiplicative, so \begin{aligned} \sum_{d\mid n} \sigma_0(d) &= \prod_{p^{e_p}\|n} (1+\sigma_0(p)+\sigma_0(p^2)+\cdots+\sigma_0(p^{e_p}))\\ &= \prod_{p^{e_p}\|n} (1+2+\cdots+(e_p+1))\\ &= \prod_{p^{e_p}\|n} \frac{(e_p+1)(e_p+2)}{2}.\\ \end{aligned} Similarly, since $\sigma_0^3$ is multiplicative, \begin{aligned} \sum_{d\mid n} \sigma_0(d)^3 &= \prod_{p^{e_p}\|n} (1+\sigma_0(p)^3+\sigma_0(p^2)^3+\cdots+\sigma_0(p^{e_p})^3)\\ &= \prod_{p^{e_p}\|n} (1^3+2^3+\cdots+(e_p+1)^3)\\ &= \prod_{p^{e_p}\|n} \left(\frac{(e_p+1)(e_p+2)}{2}\right)^2.\\ \end{aligned} Voila, no need for Eisenstein series! And similar results are easily proven.
If you are looking for identities involving $\sigma_k$, you could try looking at Duke's article "When is the product of two Hecke eigenforms an eigenform?". For instance, the fact that $E_8=E_4^2$ immediately implies that $$\sigma_7(n)=\sigma_3(n)+120\sum_{m=1}^{n-1}\sigma_3(m)\sigma_3(n-m)$$ by simply computing Fourier coefficients. This work has been extended by a number of authors. (Duke only considered eigenforms for $\mathrm{SL}_2(\mathbb Z)$.) For instance, Emmons and Lanphier (in "Products of an arbitrary number of Hecke eigenforms") considered products of an arbitrary number of eigenforms and Johnson (in "Hecke eigenforms as products of eigenforms") considered eigenforms of higher level (e.g. eigenforms for the Fuchsian group $\Gamma_1(N)$).