The Haar Measure on the Field of p-Adic Numbers

Introduction

Let $p$ be a prime number. Then the space of $p$-adic numbers $\mathbb{Q}_p$ is a locally compact abelian group. This can be observed through the local basis

$$\overline{B}(x,r)=\{y \in \mathbb{Q}_p:|x-y|_p \le r\},$$

where $|\cdot|_p$ is the $p$-adic norm such that, whenever we write $r=p^mq$ such that $q$ is prime to $p$, we have $|r|_p=p^{-m}$.

We remind the reader that every locally compact abelian group $G$ admits a Haar measure, which is unique up to a scalar multiplication (proof). In this post, we try to find the Haar measure on $\mathbb{Q}_p$, which makes it possible to do harmonic analysis on it. For this reason, in future posts, we also find the dual group of $\mathbb{Q}_p$ as well as the dual measure.

Haar Measure

Let us first recall the basic structure of $\mathbb{Q}_p$. Every element is in the form of Laurent series

$$x=\sum_{j=m}^{\infty}c_jp^j$$

where $m \in \mathbb{Z}$ and $c_j \in \{0,\dots,p-1\}$. The ring of integers $\mathbb{Z}_p$ is exactly the closed disc of radius $1$ at the origin. That is, $\mathbb{Z}_p=\overline{B}(0,1)$ is a compact set. Let $\mu$ be an arbitrary Haar measure on $\mathbb{Q}_p$. Then $\mu(\mathbb{Z}_p)$ is non-zero and finite. We can therefore put

$$m_p(E)=\frac{1}{\mu(\mathbb{Z}_p)}\mu(E).$$

Then in particular $m_p(\mathbb{Z}_p)=1$. This is the canonical Haar measure we are looking for. But it would be hilarious to end the post here. We will give a closer look at it, at least on a $p$-adic level.

Recall that when studying the Lebesgue measure on $\mathbb{R}$ we have encountered some definition in the form of

$$m(E)=\inf\left\{\sum_j\ell(I_j)\right\}$$

where the infimum is taken over all countable collections of open intervals $\{I_j\}$ such that $\bigcup_j I_j \supset E$, and $\ell(I_j)$ is the length of $I_j$. In fact, we can actually write

$$m(E)=\inf\{m(V):V \supset E \text{ is open.}\}$$

On $\mathbb{Q}_p$, we write

$$m_p(E)=\inf\{m_p(V): V \supset E \text{ is open.}\}$$

The point here is how to express $V$. For this reason we need to recall some topology of $\mathbb{Q}_p$.

Some p-adic topology

$\mathbb{Q}_p$ is a separable metric space. Therefore every open set $V$ is a union of open balls.

There is nothing special about this statement. The space has already been equipped with a norm. Besides, as $\mathbb{Q}$ is dense in $\mathbb{Q}_p$, we have nothing to worry about second countability.

Every closed ball of $\mathbb{Q}_p$ is open (hence we call them “balls” thereafter). Every point in the ball is a “centre”. If two balls intersect then one is contained in the other.

This is dramatically different from our understanding of $\mathbb{R}$ or $\mathbb{C}$. Notice that the $p$-adic norm $|\cdot|_p$ only takes the values from $p^k$ with $k \in \mathbb{Z}$ or $0$. For any $r>0$, there exists some $\varepsilon>0$ such that

$$\overline{B}(x,r)=\{y \in \mathbb{Q}_p:\|x-y\|_p \le r\}=\{y \in \mathbb{Q}_p:|x-y|_p<r+\varepsilon\}.$$

The clopenness of balls in $\mathbb{Q}_p$ follows.

Next, recall that $|\cdot|_p$ is non-Archimedean. Consider $y \in \overline{B}(x,r)$. It follows that $|x-y|_p=|y-x|_p \le r$. On the other hand, for any $z \in \overline{B}(x,r)$, we have $|x-z|_p \le r$. Therefore $|y-z|_p \le r$. Hence $\overline{B}(x,r)\subset \overline{B}(y,r)$. Symmetrically we see $\overline{B}(y,r) \subset \overline{B}(x,r)$. Hence they are equal.

Let $\overline{B}(x,r)$ and $\overline{B}(x’,r’)$ be two balls that intersect, and without loss of generality we assume that $r \le r’$. Let $y$ be a point in the intersection, then we see

$$\overline{B}(x',r')=\overline{B}(y,r') \supset \overline{B}(y,r)=\overline{B}(x,r).$$

So far so good. We next try to compute the Haar measure of every ball.

Measure of a ball

Every ball of radius $p^k$ has measure $p^k$ ($k \in \mathbb{Z}$).

First of all notice that $\overline{B}(0,1)=\mathbb{Z}_p$, and we defined $m_p$ so that $m_p(\mathbb{Z}_p)=1$. Therefore every ball of the form $\overline{B}(x,1)$ has measure $1$. Next, notice that $\overline{B}(0,p^k)=p^{-k}\mathbb{Z}_p$ for all $k \in \mathbb{Z}$, it is necessary to unwind $\mathbb{Z}_p$ a little bit more.

We have

$$\mathbb{Z}_p=\coprod_{i=0}^{p-1}(i+p\mathbb{Z}_p)= \coprod_{i=0}^{p-1}\coprod_{j=0}^{p-1}(i+jp+p^2\mathbb{Z}_p)=\cdots$$

Therefore $\mathbb{Z}_p$ is a disjoint union of $p^k$ balls of radius $p^{-k}$ when $k>0$. Hence in this case,

$$m_p(\overline{B}(0,p^{-k}))=p^{-k}m_p(\overline{B}(0,1))=p^{-k}$$

as expected. In other words, for $k<0$, the ball $\overline{B}(0,p^k)$ has measure $p^k$.

For the counterpart, we notice that

$$\begin{aligned} p^{-k}\mathbb{Z}_p &= p^{-k}\coprod_{a_0=0}^{p-1}\dots\coprod_{a_{k-1}=0}^{p-1}(a_0+\dots+a_{k-1}p^{k-1}+p^k\mathbb{Z}_p) \\ &= \coprod_{a_0=0}^{p-1}\dots\coprod_{a_{k-1}=0}^{p-1}(a_0p^{-k}+\dots+a_{k-1}p^{-1}+\mathbb{Z}_p) \end{aligned}$$

which is to say $\overline{B}(0,p^k)=p^{-k}\mathbb{Z}_p$ is a disjoint union of $p^k$ balls of radius $1$. Hence its measure is $p^k$. This concludes our computation of balls in $\mathbb{Q}_p$.

Back to the remaining problem

Now we come back to the definition of $m_p$. Now every open set $V$ can be written in the form

$$V=\bigcup_{i=1}^{\infty}\overline{B}(x_i,p^{m_i}).$$

The union is countable because $\mathbb{Q}_p$ is second countable. By combining intersecting balls, we can assume that the union is also disjoint. It follows that

$$m_p(V)=\sum_{i=1}^{\infty}p^{m_i}.$$

Note: this should be understood in the sense of real series, instead of $p$-adic number, because $m_p$ takes the values in $\mathbb{R}$. So for an arbitrary measurable set, we have

$$m_p(E)=\inf\left\{\sum_{i=1}^{\infty}p^{m_i}:E \subset\bigcup_{i=1}^{\infty}\overline{B}(x_i,p^{m_i})\right\}.$$