I am studying the development of Tate's Thesis in Lang's Algebraic Number Theory and have a conceptual question.

The setting: Let $k=\mathbb{Q}_p$. Let $\mu$ be the unique Haar measure giving $\mu(\mathbb{Z}_p)=1$; note that $\mu(p\mathbb{Z}_p)=1/p=|p|$, so that in general if $E\subset k$ is a Borel set and $a\in k^\times$, then $\mu(aE)=|a|\mu(E)$. A quasi-character $c$ of the multiplication group $k^\times$ is a continuous homomorphism into $\mathbb{C}$. Since $k^\times = U\times P$, where $U$ is the unit group of $\mathbb{Z}_p$, and $P$ is the cyclic group generated multiplicatively by $p$ (thus $P\cong \mathbb{Z}$), $c$ can be written as $c(a)=\chi(a)|a|^s$, where $\chi$ is the character of $k^\times$ obtained by projecting to $U$ and then applying $c$'s restriction to $U$, and $s$ is the complex number determined (up to a multiple of $2\pi i/\log p$) by the equation $c(p)=1/p^s$.

Let $\lambda:k\rightarrow \mathbb{Q}/\mathbb{Z}$ be defined by reducing a $p$-adic number modulo $\mathbb{Z}_p$ and embedding the resulting sum of powers of $p^{-1}$ in $\mathbb{Q}/\mathbb{Z}$, and define the Fourier transform of an $L^1$ function $k\rightarrow \mathbb{C}$ by

$$\hat f(y) = \int_k f(x)e^{-2\pi i\lambda(xy)}d\mu(x)$$

Now, for a given quasi-character $c$, define $\hat c$ by $\hat c(a) = |a|[c(a)]^{-1}$ for $a\in k^\times$.

Finally, define a "local zeta-function" that takes as parameters a function $f:k\rightarrow \mathbb{C}$ that is sufficiently integrable, and a quasi-character $c$ of $k^\times$, as follows:

$$\zeta(f,c) = \int_{k^\times} f(a)c(a)|a|^{-1}d\mu(a)$$

(The $|a|^{-1}d\mu(a)$ is the Haar measure on $k^\times$.)

The question: In this setting, Lang proves that for any two sufficiently integrable functions $f,g$, and any quasi-character $c$, the equation

$$\zeta(f,c)\zeta(\hat g,\hat c) = \zeta(\hat f,\hat c)\zeta(g,c)$$

holds. This equation is eventually going to be used to prove the classical functional equation of the classical $\zeta$-functions.

It is kind of a miraculous formula: it implies that the ratio $\zeta(f,c)/\zeta(\hat f,\hat c)$ is independent of the function $f$ and is actually just a function of $c$. Lang calls it $\rho(c)$.

I follow the proof step-by-step but I am missing the forest for the trees. So my question is this:

What is the ratio $\rho(c)=\zeta(f,c)/\zeta(\hat f,\hat c)$? What is it telling us about the quasi-character $c$? Why, morally, is it independent of $f$?

Apologies that this question is not more precise. An example of a more precise question whose answer would advance my understanding of what I am going for here is

  • Writing $c(a)=\chi(a)|a|^s$ for a character $\chi$ of $k^\times$ and fixing the character $\chi$ and the field $k=\mathbb{Q}_p$, $\rho$ becomes a complex function of a complex number. It ought to be some very nice function because its definition is made of such natural components. Is it? Once we are given $p,\chi$, can we write down $\rho(s)$ in terms of familiar functions like $\Gamma$, $\exp$, etc.?

Thanks in advance.


The function $\rho(s)$ is explicitly described in Tate's thesis, in each of the possible cases.

In the archimedean case, it is a gamma factor (or a ratio of gamma factors, evaluated at $s$ and $1-s$).

In the non-archimedean case, it is equal to the square root of the discriminant, times the ratio of the Euler factor evaluated at $s$ and $1-s$, times the "root number" of the functional equation (itself the product of a Gauss sum and the conductor of $c$).

I would write everything down but it's already nicely compiled in section 2.5 of Tate's thesis.

As for the independence of the ratio, it is indeed a surprising thing. The equation $\zeta(f, c) \zeta(\hat g, \hat c) = \zeta(\hat f, \hat c) \zeta(g, c)$ expresses the self-adjointness of $\hat{\: }$ under an appropriate pairing (the double integral appearing in the proof). It is no coincidence that functions which transform nicely under $\hat{\: }$ end up playing a special role: their $\zeta$-functions are essentially the classical zeta-functions.

I really recommend reading Tate's thesis directly. It has aged very well and it is a marvelous piece of mathematics!

  • 2
    $\begingroup$ +1 There is no better exposition than Tate's thesis itself. $\endgroup$ – Marc Palm Sep 26 '13 at 14:43

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