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. $\square$

Author

Desvl

Posted on

2022-12-23

Updated on

2022-12-23

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