The zeta Function, The l Series

The l Series

Here is yet another variation on the dirichlet zeta function. Let m be a positive integer, and let χ be a character taking the group of units mod m, denoted Z_m^*, into the complex plane. Build a function f as follows. If m and n are coprime, reduce n mod m, then apply χ. In other words, f_n = χ(n%m). If m and n are not coprime, let f(n) = 0. Finally, let the l series l_χ,m(s) = ζ(f,s).

Each f_n is 0, or lies on the unit circle. The norms are all 0 or 1. For any s to the right of 1, the terms approach 0 faster than 1/n, and by the integral test, the series is absolutely convergent. Therefore s_a is at most 1.

Because χ respects multiplication, f is a completely multiplicative function. That is, f(a)×f(b) = f(ab).

Review the section on ζ(f,s) when f is multiplicative. with this in mind, l_χ,m is the product, over the primes that do not divide m, of 1 over 1-χ(p)/p^s.

Principal Character

Let χ be the principal character, i.e. the homomorphism that maps Z_m^* to 1.

By convention, l_χ,1(s) = ζ(s).

when m is greater than 1, l_χ,m merely deletes terms from ζ() for which n and m are not relatively prime. With these terms gone, the series remains absolutely convergent to the right of 1, and since f₁, f_m+1, f_2m+1, f_3m+1, etc, are still 1, f does not converge at 1. Therefore s_a and s_c are still 1.

If k is a factor of m, the terms f_n for n divisible by k sum to ζ(s)/k^s, and these are taken away. When m is equal to a prime p, l_χ,p = ζ(s)×(1-1/p^s). The same result holds when χ is the principal character on the units mod p^k. This is because f, and ζ(f,s), have not changed.

Use inclusion exclusion to write l_χ,m as ζ(s) times an expression involving p^-s for the primes p dividing m. Here is an example.

l_χ,693 = ζ(s) × (1 - 3^-s - 7^-s - 11^-s + 21^-s + 33^-s + 77^-s - 231^-s)

Nonprincipal Characters

If χ is not principal, the sum of the image of χ equals 0. This means the partial sums of f are bounded. Multiply by 1/n^s for any positive s, and use abel summation to show convergence. Conversely, when s = 0 the terms do not approach 0. Therefore s_c = 0. Since f_n = 1 when n is 1 mod m, there is no absolute convergence until you drop below the harmonic series, for s > 1. Therefore s_a = 1.