9

I am just learning a bit about representation theory (a representation being a homomorphism $\rho: G \to GL(V)$ for some complex vector space $V$) and I am trying to understand some examples. I get the representations of the finite cyclic groups and I think I get the representations of the integers. I am wondering about the representations of a group like $\mathbb{R}$ under addition. I know this group is Abelian, and so all the irreducible representations would be $1$-dimensional. If this is too hard I would be happy with a reference.

John Doe
  • 3,529
  • 5
  • 50
  • 89

4 Answers4

8

In the case of an infinite topological group, you probably want to consider only continuous characters. In the compact case, the Peter-Weyl theorem holds, and the situation is similar to the finite case. If you consider arbitrary representations of $\mathbb{R}$, then you can construct some uninteresting and pathological ones from a Hamel basis. In the abelian case, the characters are more or less determined by Pontryagin duality, assuming local compactness. For the particular case of $\mathbb{R}$, it follows that the unitary continuous characters on $\mathbb{R}$ are all of the form $\chi(t) = e^{2\pi i x t}$ for some real $\xi$. If you drop the unitary condition, then you can take $\chi(t) = e^{zt}$ for arbitrary fixed $z\in \mathbb{C}$, more or less by the same argument.

anomaly
  • 26,475
  • Ok, and to clarify, you get one for each real number $\xi$? – John Doe Sep 11 '16 at 19:16
  • Right. If you unravel Pontryagin duality, the content is that the map $\xi \to (t \to e^{2\pi i \xi t})$ is an isomorphism $\mathbb{R} \to \operatorname{Hom}(\mathbb{R}, S^1)$, where the latter is the space of continuous maps $\mathbb{R} \to S^1$. – anomaly Sep 11 '16 at 19:35
  • 3
    A minor point, but if you're looking for all representations you should not restrict to ones with absolute value $1$, so adding in the magnitude you get $\chi(t)=e^{zt}$ for any complex number $z$. – Cyclicduck Oct 29 '20 at 04:55
7

For fun, let's drop the assumption of continuity. $\mathbb{R}$ is a $\mathbb{Q}$-vector space, and so (assuming choice) it has a basis, and hence is an infinite direct sum of copies of $\mathbb{Q}$. The representation theory of $\mathbb{Q}$ is surprisingly complicated: explicitly, the $1$-dimensional representations are given by picking nonzero complex numbers $z_n \in \mathbb{C}^{\times}$ such that

$$z_{nm}^m = z_n$$

for all $i, j$, and then taking

$$\mathbb{Q} \ni \frac{p}{q} \mapsto z_q^p \in \mathbb{C}^{\times}.$$

The obvious examples are given by taking $z_n = \exp \frac{2 \pi i \xi}{n}$ for some $\xi \in \mathbb{C}$ but other examples are possible.

Moreover, the representation theory of $\mathbb{Q}$ isn't semisimple; for example,

$$\mathbb{Q} \ni r \mapsto \left[ \begin{array}{cc} 1 & r \\ 0 & 1 \end{array} \right]$$

is a $2$-dimensional representation which is indecomposable but not irreducible. I don't know off the top of my head how to classify these. And then $\mathbb{R}$ is an infinite direct sum of copies of $\mathbb{Q}$ on top of that...

Qiaochu Yuan
  • 468,795
  • Ah, ok, so you are saying that from these representations of $\mathbb{Q}$ one can find all the representations of $\mathbb{R}$? – John Doe Sep 11 '16 at 20:00
  • @John: no, I'm saying it's at least as hard to classify the representations of $\mathbb{R}$ as the representations of $\mathbb{Q}$, but it could be even harder. Already it's not clear how to classify the representations of $G \times H$ starting from the representations of $G$ and the representations of $H$ unless e.g. $G, H$ are both compact. And $\mathbb{R}$ is an infinite direct sum. – Qiaochu Yuan Sep 11 '16 at 20:11
  • So far what we have is a description of the $1$-dimensional representations of $\mathbb{R}$: if $I$ is a set indexing a basis of $\mathbb{R}$ as a $\mathbb{Q}$-vector space, then $1$-dimensional representations are given by nonzero complex numbers $z_{n, i}, i \in I$ such that $z_{nm, i}^m = z_{n, i}$. – Qiaochu Yuan Sep 11 '16 at 20:13
5

For topological groups like $\mathbb{R}$, the useful objects to study are the continuous representations (or the continuous characters in the case of an abelian group).

It turns out that all continuous characters of $\mathbb{R}$ have the form $\chi(x)=e^{2\pi i\xi x}$ for some $\xi\in\mathbb{R}$. This is generally proved in books that cover Fourier analysis. See for instance Chapter 8 of Folland's Real Analysis.

carmichael561
  • 54,793
2

By the way, though it is true that all irreducibles are 1-dimensional, not all continuous representations split into characters. For example, $$t \mapsto \begin{pmatrix} 1 & t\\ 0 & 1\end{pmatrix}$$ is indecomposable but not irreducible.

C.D.
  • 1,643