# The Pontryagin Dual group of Q_p

# Introduction

Let $G$ be a locally compact abelian group (for example, $\mathbb{R}$, $\mathbb{Z}$, $\mathbb{T}$, $\mathbb{Q}_p$). Then every irreducible unitary representation $\pi:G \to U(\mathcal{H}_\pi)$ is one dimensional, where $\mathcal{H}_\pi$ is a non-zero Hilbert space, in which case we take it as $\mathbb{C}$. It follows that $\pi(x)(z)=\xi(x)z$ for all $z \in \mathbb{C}$ where $\xi \in \operatorname{Hom}(G,\mathbb{T})$, viewing $\mathbb{T}$ as the unit circle in the complex plane. Such homomorphisms are called (unitary) **characters**, and we denote all characters of $G$ by $\widehat{G}$, call it the Pontryagin dual group. It should ring a bell about representation theory in finite groups. For convenience, instead of $\xi(x)$, we often write $\langle x,\xi \rangle$. We also write $\langle x,\xi\rangle\langle y,\xi \rangle=\langle x+y ,\xi\rangle$, and the following examples will remind the reader the reason.

Some easily accessible examples are:

- $\widehat{\mathbb{R}} \cong \mathbb{R}$, with $\langle x,\xi \rangle = e^{2\pi i \xi x}$.
- $\widehat{\mathbb{T}} \cong \mathbb{Z}$, with $\langle z, n \rangle = z^n$.
- $\widehat{Z} \cong \mathbb{T}$, with $\langle n,z \rangle = z^n$.
- $\widehat{\mathbb{Z}/k\mathbb{Z}} \cong \mathbb{Z}/k\mathbb{Z}$, with $\langle m,n\rangle =e^{2\pi i m n / k}$.

# The Dual of p-adic Field

But we want to show that

It is broken down into several steps. But it shall be clear that $\mathbb{Q}_p$ is a topological group with respect to addition.

## Step 1 - Find the simplest character

Every $p$-adic number $x \in \mathbb{Q}_p$ can be written in the form

where $m \in \mathbb{Z}$, $x_j \in \{1,2,\dots,p-1\}$ for all $j$. We immediately define

and claim that $\xi_1$ is a character. Notice that the right hand side is always well-defined, because all summands when $j \ge 0$ contributes nothing as $\exp(2\pi i x_jp^j)=1$. That is to say, the right hand side can be understood as a finite product: when $m \ge 0$, i.e. $x \in \mathbb{Z}_p$, the pairing $\langle x, \xi \rangle = 1$; when $m<0$ however, $\langle x,\xi_1 \rangle = \exp\left( 2\pi i \sum_{j=m}^{-1}x_jp^j\right)$. Therefore it is legitimate to write

From this it follows immediately that

The function $\xi_1$ is continuous because it is continuous on $\mathbb{Z}_p$, being constant. Therefore it is safe to say that $\xi_1$ is a character with kernel $\mathbb{Z}_p$.

A quick thought would be, generating all characters out of $\xi_1$, something like $\xi_p$, $\xi_{1+p+p^2+\dots}$. But that might lead to a nightmare of subscripts. Instead, we try to discover as many as possible. For any $y \in \mathbb{Q}_p$, we define

In other words, $\xi_y$ is defined by $x \mapsto \langle xy,\xi_1\rangle$. Since multiplication is continuous, we see immediately that $\xi_y$ is a character, not very more complicated than $\xi_1$. We will show that this is all we need. To do this, we need to *characterise* all characters. Characters have the same image but their kernels differ. That is where we attack the problem.

## Step 2 - Study the kernels of characters

For $\xi_y$ above, notice that $\langle x,\xi_y\rangle=1$ if and only if $xy \in \ker\xi_1=\mathbb{Z}_p$, i.e. $|xy|_p \le 1$. Therefore

We expect that all characters are of the form $\xi_y$. Therefore their kernels shall be like $\ker\xi_y$ naturally. Notice that for fixed $y$, we have $|y|_p=p^m$ for some $m \in \mathbb{Z}$. As a result $\ker\xi_y = \overline{B}(0,p^{-m})$. For this reason we have the following (more obscure) argument

Lemma 1.If $\xi \in \widehat{\mathbb{Q}}_p$, there exists an integer $k$ such that $\overline{B}(0,p^{-k}) \subset \ker\xi$.

*Proof.* Since $\xi$ is continuous, $\langle 0,\xi\rangle=1$ on the circle, there exists $k$ such that $\overline{B}(0,p^{-k}) \subset \xi^{-1}\{z \in \mathbb{T}:|z-1| < 1\}$ (this is to say the right hand side is an open set). But $\overline{B}(0,p^{-k})$ is a group (as $|\cdot|_p$ is non-Archimedean), therefore it maps into a subgroup of $\mathbb{T}$, which can only be $\{1\}$. $\square$

We cannot say the kernel of $\xi$ is exactly of the form $\overline{B}(0,p^{-k})$ yet, but we have a way to formalise them now. If $\overline{B}(0,p^{-k}) \subset \ker\xi$ for all $k$, then $\xi=1$ is the unit in $\widehat{\mathbb{Q}}_p$. Otherwise, for each $\xi$, there is a smallest $k_0$ such that $\overline{B}(0,p^{-k_0})\subset \ker\xi$ but $\overline{B}(0,p^{-k}) \not \subset \ker\xi$ whenever $k<k_0$. In another way around, we have $\langle p^{k_0-1},\xi\rangle \ne1$ but $\langle p^k,\xi\rangle=1$ whenever $k \ge k_0$. As one may guess, such $k_0$ subjects to the “size” of $\xi$. For convenience we study the case when $k_0=0$ first.

Lemma 2 (“Fourier series”).Suppose for given $\xi \in \widehat{\mathbb{Q}}_p$, $\langle 1,\xi \rangle = 1$ but $\langle p^{-1},\xi \rangle \ne 1$. There is a sequence $(c_j)$ taking values in $\{0,1,\dots,p-1\}$ such that $\langle p^{-k},\xi \rangle=\exp\left(2\pi i\sum_1^k c_{k-j}p^{-j}\right)$ for all $k=1,2,\dots$. In particular, $c_0 \ne 0$.

*Proof.* Put $\omega_k=\langle p^{-k},\xi\rangle$. Then $\omega_0=1$ but $\omega_k \ne 1$ for all $k \ge 1$. Since

each $\omega_{k+1}$ is a $p$-th root of $\omega_{k}$, and in particular $\omega_1$ is a $p$-th root of unity. There exists $c_0 \in \{1,\dots,p-1\}$ such that

and the overall formula for $\omega_k$ follows from induction. $\square$

One would guess that for the corresponding $k_0$, the “size” of $\xi$ should be $p^{k_0}$. This looks realistic, but will be tedious. Right now we still only study the case when $k_0=0$.

Lemma 3.Notation being in lemma 2, there exists $y \in \mathbb{Q}_p$ with $|y|_p=1$ such that $\xi = \xi_y$.

*Proof.* From lemma 2 we obtain a series $y=\sum_{j=0}^{\infty}c_jp^j$ with $c_0 \ne 0$. Then in particular $|y|_p=1$. By expanding the term, we see

It follows that $\langle x,\xi \rangle = \langle x,\xi_y \rangle$ for all $x \in \mathbb{Q}_p$. $\square$

Now we are ready to conclude our observation of the dual group.

## Step 3 - Realise the dual group

Theorem.The map $\Lambda:y \mapsto \xi_y$ is an isomorphism of topological groups. Hence $\mathbb{Q}_p \cong \widehat{\mathbb{Q}}_p$.

*Proof.* First of all we study the algebraic isomorphism. First of all if $\xi_y=1$, then

Hence the map $\Lambda$ is injective. To show that $\Lambda$ is surjective, fix $\xi \in \widehat{\mathbb{Q}}_p$. By the comment below lemma 1, there is a smallest integer $k$ such that $\langle p^j,\xi \rangle = 1$. Then one considers the character $\eta$ defined by

It satisfies the condition in lemma 3, therefore there exists $z \in \mathbb{Q}_p$ such that $\eta=\xi_z$, and it follows that $\xi=\xi_{p^{-k}z}$.

Next we show that $\Lambda$ is a homeomorphism. Observe the following sets

ranging over $\ell \ge 1$ and $k \in \mathbb{Z}$. These sets constitute a local base at $1$ for $\widehat{\mathbb{Q}}_p$. We need to show that it corresponds to a local base of $\mathbb{Q}_p$ under the map $\Lambda$:

The image of the set $\{x:|x|_p \le p^k\}$ under $\xi_1$ is $\{1\}$ if $k \le 1$ and is the group of $p^k$-th roots of unity if $k>0$, and hence is contained in $\{z:|z-1|<\ell^{-1}\}$ if and only if $k \le 0$. It follows that $\xi_y \in N(\ell,k)$ if and only if $|y|_p \le p^{-k}$, i.e., $y \in \overline{B}(0,p^{-k})$. We are done. $\square$

The Pontryagin Dual group of Q_p