7
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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!

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.