# Representation theory of the additive group of the rationals?

What do the finite-dimensional continuous complex representations of the additive group $\mathbb{Q}$ with the usual topology look like? With the discrete topology? Which representations are indecomposable? Irreducible?

The only ones I can think of are of the form $t \mapsto e^{tA}$ for some $A \in \mathcal{M}_n(\mathbb{C})$. I would be willing to believe that these are the only ones in the first case, but I'm less sure in the second case.

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I guess you know, that all irreducible representation of a locally compact abelian group are characters, and that the category of finite dimensional representation of a locally compact abelian group is semisimple. – plusepsilon.de Jun 29 '11 at 16:08
@Marc: Dear Marc, I just noticed the last sentence of this comment. It is not true in general that finite dimensional reps. of a locally compact ab. gp. are semisimple, if the group is not compact. E.g. if $G = \mathbb Z$, then the cat. of fin. dim. $G$-reps. is just the cat. of finitely generated torsion $\mathbb C[T,T^{-1}]$-modules, which is not semisimple. (Concretely, invertible matrices can have non-trivial Jordan blocks.) If we restrict to unitary representations, then we do get a semi-simple category. Regards, – Matt E Aug 3 '13 at 4:46

One way to think of $\mathbb Q$ is the direct limit over positive integers $n$ of $\frac{1}{n} \mathbb Z$. Thus giving a character of $\mathbb Q$ is the same as giving an element in the projective limit of the character groups of $\frac{1}{n}\mathbb Z$. In particular, if we restrict to unitary characters, we find that $\mathbb Q^{\vee}$ is the projective limit of circle groups $S^1$ under the $n$th power maps. This object is (I think) called a solenoid; to number theorists it is better known as the adele class group $\mathbb A/\mathbb Q$. (Here and throughout I am using the discrete topology; if one instead considers the induced topology from $\mathbb R$, then, as Robin explains, one just gets characters of $\mathbb R$.)

The exact sequence $0 \to \hat{\mathbb Z} \to \mathbb Q^{\vee} \to S^1 \to 0$ in Pete's answer arises from the map taking the solenoid to the base $S^1$; the fibres of this map are copies of $\hat{\mathbb Z}$.

If we wanted not necessarily unitary characters, we would instead get the projective limit of copies of $\mathbb C^{\times}$ under the $n$th power maps. Since $\mathbb C^{\times} = \mathbb R_{> 0} \times S^1$, and since $\mathbb R_{> 0}$ is uniquely divisible, this projective limit is simply $\mathbb R_{> 0}$ times the solenoid.

On a slightly tangential note, let me remark that the relationship with the adeles is important (e.g. it is the first step in Tate's thesis):

Since the adeles are the (restricted) product of $\mathbb R$ and each $\mathbb Q_p$, and since these are all self-dual, it is easy to see that $\mathbb A$ is self-dual.

One then has the exact sequence $$0 \to \mathbb Q \to \mathbb A \to \mathbb A/\mathbb Q \to 0$$ which is again self-dual (the duality swaps $\mathbb Q$ and the solenoid $\mathbb A/\mathbb Q$).

One should compare this with the exact sequence $$0 \to \mathbb Z \to \mathbb R \to \mathbb R/\mathbb Z = S^1 \to 0.$$ This is again self-dual ($\mathbb R$ is self-dual, and duality swaps the integers and the circle).

This brings out the important intuition that the adeles are to $\mathbb Q$ as $\mathbb R$ is to $\mathbb Z$.

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+1: this is certainly what I should have said. – Pete L. Clark Nov 17 '10 at 9:27

In the usual topology, each continuous representation of $(\mathbb{Q},{+})$ will extend to a continuous representation of $(\mathbb{R},{+})$ so you'll only get the obvious ones.

In the discrete topology there are non-obvious representations even in dimension one. Consider the map $\alpha:\mathbb{Q}\to \mathbb{Q}_p\to\mathbb{Q}_p/\mathbb{Z}_p \cong \mathbb{Z}[1/p]/\mathbb{Z}$. Then $t\mapsto\exp(2\pi i\alpha(t))$ is a character of $\mathbb{Q}$, but not of the form $t\mapsto e^{at}$.

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Taking Pontrjagin duals of the short exact sequence of discrete abelian groups

$0 \rightarrow \mathbb{Z} \rightarrow \mathbb{Q} \rightarrow \mathbb{Q}/\mathbb{Z} \rightarrow 0$

gives

$0 \rightarrow \hat{\mathbb{Z}} \rightarrow \mathbb{Q}^{\vee} \rightarrow S^1 \rightarrow 0$,

-- here $\hat{\mathbb{Z}}$ is the profinite completion of $\mathbb{Z}$ -- so this classifies the one-dimensional "unitary" continuous representations of $\mathbb{Q}$. I believe that the exact sequence above is split; of all places, this came up at a dinner party I attended last week. (Added: no, it is not split as a sequence of topological groups or even as a sequence of groups: see Matt E's comment below.)

N.B.: If you ask a number theorist (broadly construed) what the "usual topology" on $\mathbb{Q}$ is, she will say that it is the discrete topology. This is the topology it inherits from the topology on the adele ring $A_{\mathbb{Q}}$.

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Thanks. Could you be a little more explicit about the classification? I'm not very familiar with exact sequences. – Qiaochu Yuan Nov 16 '10 at 22:23
@Qiaochu: It should be pretty clear that $\hat{\mathbb{Z}}$ is a subgroup of the Pontrjagin dual group of $\mathbb{Q}$, and the exactness means that the quotient is isomorphic to the circle group $S^1$. To say that it's split means that there is a subgroup of the Pontrjagin dual which maps isomorphically onto $S^1$ via the quotient map, whence an internal direct product decomposition $\mathbb{Q}^{\vee} \cong \hat{\mathbb{Z}} \times S^1$. Equivalently, there exists a section to the quotient map. But I'm not 100% sure this is true; I'd have to think about it a bit... – Pete L. Clark Nov 16 '10 at 22:28
Dear Pete, There is not a section. To see that there is no continuous section, one can use double duality, and note that the original exact sequence is not split. To see that there is no section at all, continuous or not, one notes that mult. by any non-zero $n \in \mathbb Z$ is a bijection on $\mathbb Q$, hence also on $\mathbb Q^{\vee}$. Thus the latter space is torsion free, while $S^1$ contains a lot of torsion elements. – Matt E Nov 17 '10 at 6:50
Dear Matt, Thanks for your comment. I wish you were at the dinner party... – Pete L. Clark Nov 17 '10 at 9:23