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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 $$\frac{1}{N}\sum\limits_{n \leq N} f(n \alpha) \rightarrow \int\limits_{\mathbb{R^N} / \mathbb{Z}^n} f( x) \mathrm{d} x.$$

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By strong approximation, $\mathbb{Q}$ is dense in $\mathbb{A}_{\mathbb{Q}}/\mathbb{Q}_{v}$ for any valuation $v$ on $\mathbb{Q}$ modular.math.washington.edu/books/ant/ant/node108.html –  jspecter Jun 29 '11 at 16:32
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@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
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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

1 Answer 1

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 $\mathbb Q$.

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So, I guess you use something like $q \mapsto (e^{iq}, \pi^{iq})$ and there like. nice example. –  plusepsilon.de 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

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