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 eth 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.



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