By Graham Everest

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12 For all g, h ∈ G, we have χ(g)χ(h) = ˆ χ∈G |G| if g = h 0 if g = h Proof: Note that χ(h−1 ) = χ(h)−1 = χ(h), since χ(h) is on the unit circle. 11 with gh−1 in place of h. ✷ This is the gadget in its ultimate version! As an example, take G = U (Z/5) ∼ = C4 . Here is a table of all the characters. 1 2 4 3 χ0 χ1 χ2 χ3 1 1 1 1 1 i −1 −i 1 −1 1 −1 1 −i −1 i 63 (104) Note that we have written the elements of U (Z/5) in an unusual ordering this is given by 20 , 21 , 22 , 23 . The character values behave likewise.

And in particular, any complex function on G can be written as linear combination of the characters. 14 ˆ Extend χ Given 1 < q ∈ N, consider G := U (Z/q) and a character χ ∈ G. to a function X on N by setting X(n) := χ(n mod q) if n coprime to q 0 otherwise (105) Then the function X is called a Dirichlet character modulo q. Note that this is a slight abuse of language - it is not meant to say that N were a group. However, we will even write χ instead of X for the Dirichlet character associated to χ.

We begin with ∞ s Γ = 2 So, in the domain −x e x ∞ s −1 2 dx = 0 s e−x x 2 0 dx . 99 (57) (s) > 1 + δ, F(s) = π ∞ − 2s n∈N s n−s e−x x 2 0 dx . x (58) Next, replace x by πn2 y in the integral. This means dx/x = dy/y, and we get after some cancelling ∞ 0 s 2 e−πn y y 2 F(s) = n∈N dy . y (59) The interchange of integral and sum is ok, because the series for the zeta function converges uniformly on (s) > 1 + δ. Define 2y e−πn g(y) := = n∈N θ(y) − 1 . 2 (60) Split the integral in equation (59): ∞ F(s) = 1 dy y g(y) + y 1 s 2 s y 2 g(y) 0 dy .