# Representations of abelian groups

A classical result is

Theorem: Let $$G$$ be a abelian group and $$(V, \rho)$$ be a irreducible representation of $$G$$ over a algebraically closed field $$k$$. If $$V$$ is finite dimensional (more generally, if $$\mathrm{dim}_k V < |k|)$$ then $$\mathrm{dim}_k V = 1$$.

This is essencially Schur/Dixmier's lemma, the proof is based at $$\rho(g)$$ having a eingenvalue, $$\forall g \in G$$. But what if i don't suppose anything about the dimension of $$V$$?

 Question:

Are every irreducible $$G$$-module finite dimensional? Maybe the unitary ones? Do I really need to go topological world and suppose $$G$$ compact abelian and consider only strongly continuous representations or something like this?

 An idea:

$$G$$-modules are equivalent to $$k[G]$$-modules and if $$G$$ is abelian then this is a commutative algebra over $$k$$. If $$G$$ is finite, then $$k[G]$$ is finite dimensional over $$k$$ and $$V = k[G]v$$ for $$v \in V \setminus 0$$, we are done. More generally, this proves that all irredutible representations of a finite group are finite dimensional.

In abelian case, $$k[G]^*$$ is also a commutative cocommutative Hopf algebra. A $$k[G]$$-module $$V$$ gives a $$k[G]^*$$-comodule $$V^*$$ and simple $$k[G]^*-$$comodules are finite dimensional over $$k$$ (see here). But dualizing transforms injections into surjections, I don't know if still $$V^*$$ is a simple comodule.

It is not true in the infinite-dimensional case. An irreducible representation of an abelian group $$A$$ over a field $$k$$ is the same thing as a simple module over the commutative $$k$$-algebra $$k[A]$$. Since $$k[A]$$ is commutative, simple modules can be identified with quotients by maximal ideals. If $$m$$ is such a maximal ideal, then $$k[A]/m$$ is a field extension of $$k$$.

But it is possible for $$k[A]/m$$ to be a nontrivial field extension of $$k$$, even if $$k$$ is algebraically closed, if $$A$$ is large enough. For example we can arrange to have $$k[A]/m \cong k(t)$$ by choosing $$A = k(t)^{\times}$$, together with the natural map $$k[k(t)^{\times}] \to k(t)$$.

On the other hand, the following can be salvaged. I've seen this called Dixmier's lemma but I'm not sure it's the same statement other people call Dixmier's lemma.

(Dixmier's?) Lemma: If $$V$$ is an irreducible representation of a group $$G$$ over an algebraically closed field $$k$$ such that $$\dim V < |k|$$, then $$\text{End}_G(V) = k$$.

Proof. Schur's lemma still holds in the sense that $$\text{End}_G(V)$$ is always a division algebra. Since $$k$$ is algebraically closed, in order for this division algebra to contain elements other than the elements of $$k$$ it must contain a transcendental element. So we observe that $$k(t)$$ itself has dimension at least the cardinality of $$k$$ since the rational functions $$\frac{1}{t - a}, a \in k$$ are linearly independent. Hence if $$\dim V < |k|$$ then $$\text{End}_G(V)$$, which has dimension at most $$|V|^2$$, also has dimension less than $$|k|$$ (which must be infinite), so it cannot contain a transcendental element. $$\Box$$

Corollary: If $$k$$ is an algebraically closed field and $$A$$ is an abelian group, then every irreducible representation of $$A$$ of dimension strictly less than $$|k|$$ is $$1$$-dimensional. In particular, if $$A$$ has cardinality strictly less than $$k$$, then every irreducible representation of $$A$$ is $$1$$-dimensional.

Proof. If $$V$$ is an irreducible representation of an abelian group $$A$$ then every $$a \in A$$ acts by endomorphisms on $$V$$, so the lemma $$\text{End}_A(V) \cong k$$ gives that every $$a \in A$$ acts by a scalar in $$k$$. Hence $$V$$ must be $$1$$-dimensional. $$\Box$$

So e.g. irreducible representations of countable abelian groups over $$\mathbb{C}$$ are $$1$$-dimensional (but not over $$\overline{\mathbb{Q}}$$!). This is true even though the eigenvalue argument is a priori unavailable in this setting: for example, in the regular representation of $$\mathbb{Z}$$ (the action of $$k[x, x^{-1}]$$ on itself) no nonzero element of the group has any eigenvalues.

• Thank you! This $k(t)^\times$ is really simple. Oct 12, 2022 at 17:07