Gaussian Sums of a Dirichlet's Character

Question

In Davenport's chapter 9, They defined

$$\tau(\chi)=\sum_{m=1}^{q} \chi(m) e_q(m)$$ Further if $(n,q)=1$, then we have that $$\chi(n)\tau(\overline{\chi})=\sum_{h=1}^{q} \overline{\chi}(m) e_q(nh)$$ and then they wrote that this gives the desired expression for $\text{ }\chi(n)\text{ }$ provided $\text{ }(n,q)=1$.

My question is, we have $$ \chi(n) = \frac{\sum_{h=1}^{q} \overline{\chi}(m) e_q(nh)}{\tau(\overline{\chi})} $$ but when we are computing the RHS, we need $\overline{\chi}(h)$ so we know what is the value of $\chi(n)$ and so why are we computing $\chi(n)$ with such a complicated expression.

Answer 1

I suspect part of the point might be that on the right side, $n$ is not appearing inside $\chi$-values, and where $n$ does appear it could be a real number, so this equation can serve as a way of defining $\chi(n)$ when $n$ is a real number.

Compare with the equation $$ n! = \int_0^\infty x^ne^{-x}\,dx $$ for $n \geq 0$ in $\mathbf Z$. On the right side, $n$ can be a nonnegative real number, so we can use the right side to define $n!$ when $n$ is a nonnegative real number, e.g., $(1/2)! := \int_0^\infty \sqrt{x}e^{-x}\,dx$.

EDIT: I looked in Davenport Chapter 9, and he explains directly in the second paragraph of that chapter why he wants such a formula for $\chi(n)$:

We need the expression for $\chi(n)$ as a linear combination of imaginary exponentials $e_q(mn)$, which we used earlier in §1 [(4) and (5)]

The formula you ask about was derived when $(n,q) = 1$, but Davenport then immediately shows right afterwards that when $\chi$ is primitive that formula is also true when $(n,q) > 1$: it's true for all integers $n$, no matter what $(n,q)$ is. On pp. 66-67, Davenport uses that formula to show the Gauss sum of each primitive character mod $q$ has absolute value $\sqrt{q}$, and on the next few pages he uses this formula for $\chi(n)$ for all $n$ (when $\chi$ is primitive) in order to obtain the functional equation of the Dirichle $L$-function of $\chi$.

In short, the answer to your question is: be patient. Just read ahead, pay close attention, and you'll see the formula for $\chi(n)$ used!

Answer 2

One way that sums like $\sum_m \chi(m)\psi(nm/q)$ appear (writing $\psi(x)=e^{2\pi ix}$) in the functional equation for Dirichlet $L$-functions, whether done classically or as in Iwasawa-Tate.

The point is that the $\chi$ (or $\overline{\chi}$...) comes out of the indicated sum. Then the Gauss sum, divided by $\sqrt{\pm q}$, is the "$\epsilon$-factor" in the functional equation of the corresponding $L$-function(s). (Yes, for $\chi$ not real-valued, the functional equation relates $L(1-s,\chi)$ and $L(s,\overline{\chi})$ ...)

Yes, as @KCd says, this gives a sort of "integral representation" of $\chi$, though it's not clear to me that this is what extends $\chi$ to an idele class character.

More explicitly: to prove analytic continuation and functional equation of $L(s,\chi)=\sum_n \chi(n)/n^s$, the Riemann-IwasawaTate approach is to use an integral representation of the $L$-function, with suitable Gamma factors: for $\chi$ even (for $\chi$ odd it's slightly different), it is $$ \pi^{-s/2}\Gamma(s/2)L(s) \;=\; {1\over 2}\int_0^\infty y^{s/2} \sum_n\chi(n)e^{-\pi n^2 y}\;dy/y $$ Applying Poisson summation to that linear combination of theta functions inside the integral (to prove the functional equation of that linear combination) spits out the Gauss sum, etc.