# Some details for Zeta functions

Here I’m including proofs and derivations for some technical results used in Ogus’ discussion of Zeta functions.

For $e,n$ positive integers with e fixed, consider the function

$a_n = \begin{cases} 0 \mbox{ if } n/e \not \in \mathbb{Z} \\ e \mbox{ if e divides n} \end{cases}$

Then $a_n = \sum_{\eta \in \mu_e} \eta^n$ where $\mu_e$ is the set of all $e$th roots of unity (this is the first result).  This makes intuitive sense, when n is a multiple of e all the terms in the sum are 1, and in all other cases the roots are permuted and you sum them to get zero.  here’s Ogus’ formal proof:

For brevity a product or sum over $\eta$ means $\eta \in \mu_e$.  So

$\prod_\eta (T - \eta) = T^e -1$ and $\prod_\eta (1 - T\eta) = (1 - T^e)$

The second equality follows because the the leading coefficient should be +1 and all $\eta$ are roots of the rhs.  Note that $T\cdot d(\log f) = \frac{Tdf}{f}$ has the property that $T d(log fg) = Tdf/f + Tdg/g$.   Applying this to

$\prod_\eta (1- \eta T)^{-1} = (1 - T^e)^{-1}$

and comparing sides yeilds

$\sum_\eta \frac{\eta T}{1 - \eta T} = \frac{eT^e}{1 - T^e}$

Expanding as power series gives

$\sum_\eta (\eta T + \eta^2 T^2 + ...) = eT + eT^{2e} + ...$

$\sum_{n = 1}^{\infty} a_n T^n = eT + eT^{2e} + ...$

This gives the first result.

Here’s is the next result.  Let $a_i, b_i \in \mathbb{C}$ be a finite number of complex numbers and set

$Z(T) = \prod_i \frac{1 - a_iT}{1-b_iT}$

Adding zeros if necessary, it can be assumed that there are an equal number of a_i and b_i.  The second result is, as a formal power series,

$Z(T) = \exp (\sum_n \frac{\alpha_n}{n}T^n)$

where $\alpha_n = \sum_i ( b_i^n - a_i^n)$.  This is just some manipulations:

$Td(\log Z(T)) = T d(\log \prod (1 - a_i T) - \log \prod (1 - b_iT))$

$= T d(\sum [\log(1 - a_iT) - \log(1-b_iT)])$

$= T(\sum [ \frac{-a_i}{1 - a_iT} - \frac{-b_i}{1-b_iT}]$

$= \sum_i (b_iT + b_i^2T^2 +.... ) - (a_iT + a_i^2T^2 + ...)$

$=\sum_n \alpha_n T^n$

Or equivalently, $d(\log Z(T)) = \sum_n \alpha_n T^{n-1}$ so integrating gives

$\log Z(T) = \sum_n \frac{a_n}{n}T^n$

and exponentiation gives the second result.