Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What does the Haar measure on $\hat{\mathbf{Z}} = \prod_p \mathbf{Z}_p$ look like?

Does it bear any relation to the "upper density" of a subset $S\subset\mathbf{N}$ defined by $m(S) = \limsup_{n\to\infty} |S\cap\{1,\ldots,n\}|/n$?

share|cite|improve this question
up vote 2 down vote accepted

Since $\widehat{Z}$ and $\mathbb{Z}_p$ are compact, we will assume unit volume. This is standard, but you can change at the Haar measure at any set of finite places.

As you will know any the topology on $\widehat{Z}$ is that of an infinite product of space (se Tychonoffs theorem), in particular any open set looks like

$$O = O_{p_1} \times O_{p_2} \times .... \times O_{p_k} \times \prod\limits_{p \neq p_j} \mathbb{Z}_p$$

and the measure of $O$ is precisely

$$ \mu(O) = \mu_{p_1}(O_{p_1}) \cdots \mu_{p_k}(O_{p_k}).$$

Now for any open set $O$ the set $O \cap \mathbb{Z}$ has finite index in $\mathbb{Z}$, i.e. an ideal, so the upper density will be $index^{-1}$, so here there are actually equal.

For any open set $X \subset \widehat{Z}$, the measure $\mu(X)$ is precisely the upper density of $S =X \cap \mathbb{Z}$ in the standard parametrization.

For measurable set is does not work, since the upper density is obviously not a measure.

I am not too familiar with the upper density, so I cannot go the other way;(

share|cite|improve this answer
Note that you know a Borel measure, if you know the measure of all open sets. – Mar 24 '12 at 14:22
Hold on, ideals don't have trivial upper density. For instance, $2\mathbf{Z}$ has upper density $1/2$. – Sean Eberhard Mar 24 '12 at 14:43
I meant "not interesting" = "trivial", not that it has denisty $1$. – Mar 24 '12 at 14:45
So is it is true, then, that if $O\subset\hat{\mathbf{Z}}$ is open then $\mu(O)$ is just the upper density (or rather just density: we don't need the 'sup') of $O\cap\mathbf{Z}$, isn't it? – Sean Eberhard Mar 24 '12 at 14:53
Yes, but how do you go the other direction? – Mar 24 '12 at 14:55

A way to construct an Haar measure on $\hat{\Bbb Z}$ is to take the product $\prod dz_p$ where $dz_p$ is an Haar measure on $\Bbb Z_p$ normalized so that $\int_{\Bbb Z_p}\,dz_p=1$.

This measure actually extends to an Haar measure on the full ring of adeles $\Bbb A=\Bbb Q\Bbb R\hat{\Bbb Z}$.

Any other Haar measure on $\hat{\Bbb Z}$ (or $\Bbb A$) is a scalar multiple of this one.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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