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.


1 Answer 1


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.

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

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