# Todd Trimble Notes on Euler factors

## Gaussians for some classical self-dual groups

### Over the real numbers

Let’s take the classical case of the real numbers first. The topological group of real numbers is a locally compact abelian group which is self-dual under Pontryagin duality, and the Fourier transform is a Hilbert space isomorphism

$L^2(\mathbb{R}, \frac{d x}{\sqrt{2\pi}}) \to L^2(\mathbb{R}, \frac{d x}{\sqrt{2\pi}})$

(this factor $\frac1{\sqrt{2\pi}}$ is the result of “splitting the difference”). This is a special case of Fourier analysis on locally compact abelian groups, which establishes a Hilbert space isomorphism

$\mathcal{F}: L^2(G) \to L^2(\widehat{G})$

more generally for any locally compact abelian group $G$.

Here are two nice things about the standardly normalized Gaussian (details to be recalled in a moment):

• It is carried to itself under Fourier transform;

• A certain parametrized family of Gaussians approximates to the identity (the identity for convolution product, i.e., the Dirac distribution supported at the group identity)

and these two facts can be utilized to prove that the Fourier transform is indeed a Hilbert space isomorphism.

###### Definition

Let $\mathcal{S}$ be the space of smooth functions on $\mathbb{R}$ whose $n^{th}$ order derivatives all vanish rapidly at $\infty$, aka Schwartz space. Define the Fourier transform $\mathcal{F}: \mathcal{S} \to \mathcal{S}$ by

$\mathcal{F}(f)(y) = \frac1{\sqrt{2\pi}} \int_{\mathcal{R}} e^{-i x y} f(x) d x$

This maps Schwartz space to Schwartz space, but will eventually be extended to the Hilbert space completion of $S$.

###### Lemma

$\int_{\mathbb{R}} e^{-x^2/2} d x = \sqrt{2\pi}$.

The classical proof is a famous trick involving squaring the integral, and then changing to polar coordinates.

###### Lemma

For fixed $y$, $\int_{\mathbb{R}} e^{-(x + i y)^2/2} d x = \sqrt{2\pi}$.

This follows from the previous lemma by a contour integration trick. Integrate $e^{-z^2/2}$ over a contour given by the boundary of a rectangle in the complex plane, $[-R, R] \times [0, y]$. The whole contour integral is zero because $e^{-z^2/2}$ is holomorphic inside the rectangle. For large $R$, the contribution of the vertical sides is tiny (on the order of $e^{-R^2/2}$), and the contribution between the horizontal sides is the difference between the integrals of the two lemmas, replacing $\mathbb{R}$ by the interval $[-R, R]$. Taking $R \to \infty$, the difference between the two integrals becomes zero.

###### Lemma

The Gaussian $f(x) = e^{-x^2/2}$ is self-dual under Fourier transform.

###### Proof

The previous integral can be rewritten as

$\int_{\mathbb{R}} e^{- i x y} e^{-x^2/2} e^{y^2/2} d x = \sqrt{2\pi}.$

After a short manipulation, this becomes

$\frac1{\sqrt{2\pi}} \int_{\mathbb{R}} e^{-i x y} e^{-x^2/2} d x = e^{-y^2/2}$

which is what we want.

###### Lemma

The Fourier dual of $f_a(x) = e^{-a x^2/2}$ is $\widehat{f}(y) = \frac1{\sqrt{a}} e^{-y^2/2a}$.

This is just a simple change of variable.

### Over the $p$-adic numbers

For simplicity, let’s work over the $p$-adic completion of the rationals, $\mathbb{Q}_p$. This is a local field (a locally compact but non-discrete field; these have a well-known characterization, involving completions of number fields at places, or function fields of curves over finite fields).

#### Self-duality

Any local field $F$, as an additive topological group, it is isomorphic to its Pontryagin dual $F^\ast$. More exactly, there is an induced map $F \to F^\ast$ induced by the pairing

$F \times F \to S^1: (x, y) \mapsto e^{2\pi i \vert x y\vert}$

where $\vert x\vert$ is the natural valuation for the local field. (Here the $p$-adic valuation of $p^n u$ is $1/p^n$ if $u$ is a unit in the $p$-adic integers.) The assertion is that $F \to F^\ast$ is an isomorphism.

I’m tempted to consider the special case where $F = \widehat{\mathbb{Q}}_p$ from the point of view of the exact sequence of topological groups

$0 \to \widehat{\mathbb{Z}}_p \to \widehat{\mathbb{Q}}_p \to \mathbb{Z}_{(p)}/\mathbb{Z} \to 0$

where the subgroup $\widehat{\mathbb{Z}}_p$ is compact (a pro-group) and is manifestly the Pontryagin-dual of the quotient group $\mathbb{Z}_{(p)}/\mathbb{Z}$ which is discrete (an ind-group). Pontryagin duality assures us that $\mathbb{Z}_{(p)}/\mathbb{Z}$ is the Pontryagin-dual of $\widehat{\mathbb{Z}}_p$. So, the map $F \to F^\ast$ induces a morphism of exact sequences

$\array{ 0 & \to & \widehat{\mathbb{Z}}_p & \to & \widehat{\mathbb{Q}}_p & \to & \mathbb{Z}_{(p)}/\mathbb{Z} & \to & 0 \\ & & \downarrow & & \downarrow & & \downarrow & & \\ 0 & \to & (\mathbb{Z}_{(p)}/\mathbb{Z})^\ast & \to & \widehat{\mathbb{Q}}_{p}^{\ast} & \to & \widehat{\mathbb{Z}}_{p}^{\ast} & \to & 0 }$

for which the first and third vertical arrows are isomorphisms. It follows from the short 5-lemma that the middle vertical arrow is an isomorphism. (The short 5-lemma is well-known for abelian categories; it is slightly less well-known that it holds also in the context of topological abelian groups. See this paper by Borceux and Clementino.)

The same arguments can be adapted to apply to any local completion of any number field.

#### Gaussian

We want an analogue of the Gaussian over $\widehat{\mathbb{Q}}_p$, which again

• Belongs to the Schwartz space (consisting of functions all of whose derivatives vanish rapidly at $\infty$),

• Is self-dual under Fourier transform, and

• Fits into a parametrized family which approximates to the identity.

We will show that the characteristic function of the $p$-adic integers satisfies these requirements.

In the first place, the adeles as a locally compact additive group is also Pontryagin self-dual. We again consider the case over $\mathbb{Q}$, with the remark that everything carries over to adeles over an arbitrary number field.

There are various ways of describing the adeles. Each of the local completions $\widehat{\mathbb{Q}}_p$ at nonarchimedean places has a compact open subring $\widehat{\mathbb{Z}}_p$. One way to define the adeles is to take a restricted direct product

$\mathbb{R} \times \prod_{p}^{\prime} \widehat{\mathbb{Q}}_p \hookrightarrow \mathbb{R} \times \prod_p \widehat{\mathbb{Q}}_p$

consisting of tuples $(r; x_p)$ such that all but finitely many of the $x_p$ belong to $\widehat{\mathbb{Z}}_p$.

(To be included: topology on adeles, local compactness, discrete and cocompact embedding of $\mathbb{Q}$.)

Let $\mathbb{A}_{arch}$ denote the archimedean part of the adeles. Then there is an exact sequence of topological groups

$0 \to \prod_p \widehat{\mathbb{Z}}_p \to \mathbb{A}_{arch} \to \mathbb{Q}/\mathbb{Z} \to 0$

and a morphism of exact sequences

$\array{ 0 & \to & \prod_p \widehat{\mathbb{Z}}_p & \to & \mathbb{A}_{arch} & \to & \mathbb{Q}/\mathbb{Z} & \to & 0 \\ & & \downarrow & & \downarrow & & \downarrow & & \\ 0 & \to & (\mathbb{Q}/\mathbb{Z})^\ast & \to & \mathbb{A}_{arch}^\ast & \to & (\prod_p \widehat{\mathbb{Z}}_p)^\ast & \to & 0 }$

where the first and third vertical arrows are again isomorphisms, and therefore the second is too.

We wish to show that the dual of the discrete group $\mathbb{Q}$ is $\mathbb{A}/\mathbb{Q}$, where $\mathbb{Q}$ is included in the adele ring $\mathbb{A}$. We have an exact sequence

$0 \to (\mathbb{Q}/\mathbb{Z})^\ast \to \mathbb{Q}^\ast \to \mathbb{R}/\mathbb{Z} \to 0$
Revised on October 18, 2011 at 11:47:39 by Todd Trimble