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

I know that there is classification of local fields, but here is a closely related question: Can the additive group of $\mathbb{Q}$ be a proper dense subgroup of a locally compact abelian group, whose topology is complete, other than the p adic numbers or the reals? I think of this question more as a collection, and I guess I will have to try out various examples here.

1.Example by MattE: Consider $\alpha =(\alpha_1, \dots, \alpha_n) \in \mathbb{R}^n$ linearly independent over $\mathbb{Q}$, then the map $q \mapsto q \alpha:= ( q \alpha_1, \dots, q \alpha_n)$ becomes dense in the $n$ torus, i.e. $\mathbb{R}^n / \mathbb{Z}^n$, actually even more it becomes equidistributed in the following sense $$\frac{1}{N}\sum\limits_{n \leq N} f(n \alpha) \rightarrow \int\limits_{\mathbb{R^N} / \mathbb{Z}^n} f( x) \mathrm{d} x.$$

share|cite|improve this question
By strong approximation, $\mathbb{Q}$ is dense in $\mathbb{A}_{\mathbb{Q}}/\mathbb{Q}_{v}$ for any valuation $v$ on $\mathbb{Q}$ – jspecter Jun 29 '11 at 16:32
@Theo: One of those is indecomposable, the other not! – user641 Jun 29 '11 at 16:33
@Steve: Thanks, silly me. – t.b. Jun 29 '11 at 16:37
@late_learner : Good point... – Joel Cohen Jun 29 '11 at 16:47
As for groups in which $\mathbb{Q}$ is a lattice here's a stupid example. Consider any compact group $C$ and $\mathbb{Q}$ in the discrete topology. Then $\mathbb{Q}$ is a lattice in the locally compact group $\mathbb{Q} \times C.$ – jspecter Jun 29 '11 at 16:59
up vote 10 down vote accepted

It can surely be embedded densely in many such groups. E.g. it can be embedded into $(S^1)^n$ for any $n$. (Here $S^1$ is the circle group.)

To see this, choose an element $\alpha$ in $(S^1)^n$ whose powers are dense in $(S^1)^n$.

Now for inductively, for each integer $m$, choose $\alpha_m$ such that $\alpha_m^m = \alpha$, in a compatible way (i.e. so that if $m' = d m,$ then $\alpha_{m'}^d = \alpha_m$). Then the $\alpha_m$ together generate a copy of $\mathbb Q$ inside $(S^1)^n$, which will be dense.

(A little more succintly, I am using the fact that $(S^1)^n$ is divisible, hence injective, to extend the embedding $\mathbb Z \hookrightarrow (S^1)^n$ to an embedding $\mathbb Q \hookrightarrow (S^1)^n$.)

Another way to think about this example, when $n = 2$ say, is that we take a line with irrational slope in $(S^1)^2$; this gives a dense copy of $\mathbb R$, which contains inside it a dense copy of $\mathbb Q$.

share|cite|improve this answer
So, I guess you use something like $q \mapsto (e^{iq}, \pi^{iq})$ and there like. nice example. – Jun 29 '11 at 17:28
Couldn't you just extend the embedding $f : \mathbb{Z} \to (S^1)^n$ by denoting $\alpha = (e^{i \theta_1}, \ldots, e^{i \theta_n})$ and setting $f(q) = (e^{i q\theta_1}, \ldots, e^{i q\theta_n})$ ? – Joel Cohen Jun 29 '11 at 17:54
@Joel: Dear Joel, Yes, you're right! (I just bashed out the first thing that came to mind.) Thanks, and best wishes, – Matt E Jun 29 '11 at 20:11

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.