Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm supposed to show that each Complex finite dimensional irreducible representation of an abelian group is one dimensional.

For any map $\phi: V \rightarrow V$ it holds that $\phi(\rho(g)v) = \rho(g) \phi(v)$. Also since the group $\rho(h) \rho(g) v = \rho(g) \rho(h) v$. From a previous exercise I know that $\phi = \lambda \cdot id_V$ for some $\lambda \in \mathbb{C}$. This transforms the previous equation into $\lambda \cdot id_V \cdot (\rho(g)v) = \rho(g) \lambda \cdot id_V \cdot v$ which implies that $\lambda \cdot id_V \cdot (\rho(g)v) =\lambda \cdot \rho(g) \cdot v$. Now I'm not quite sure how to bring into play that $G$ is abelian. Could someone give me a hint?

Cheers!

share|improve this question
add comment

2 Answers

Let us sort out things a bit.

Let $\rho: G \to \operatorname{GL}(V)$ be an irreducible representation of any group $G$.

You have seen that if $\varphi : V \to V$ commutes with all $\rho(g)$, for $g \in G$, that is $$ \varphi (\rho(g) v) = \rho(g) \varphi (v)\tag{comm} $$ for all $g \in G$ and $v \in V$, then $\varphi = \lambda \operatorname{id}_{V}$ for some $\lambda \in \mathbf{C}$.

Now if $G$ is abelian we have $\rho(x) \rho(g) = \rho(g) \rho(x)$ for all $g, x \in G$, so that $\varphi = \rho(x)$ satisfies (comm).

It follows that for any $x \in G$ there is $\lambda \in \mathbf{C}$ such that $$ \rho(x) = \lambda \operatorname{id}_{V}, $$ so all $\rho(x)$ are scalars, and then leave every subspace invariant.

Since the representation is irreducible, $V$ must have dimension $1$ then.

share|improve this answer
add comment

As an alternative approach:
Since all homomorphisms from an abelian group $G$ to $C^*$ are irreducible characters of $G$, and there are $|G|$ many of these, by dint of orthogonality relations, we conclude that they are all irreducible characters of $G$, and hence all irreducible representations of $G$ are of dimension $1$.
Inform me of any error. Thanks.

share|improve this answer
    
+1 This would be my approach when $G$ is finite. –  Alexander Gruber Apr 9 '13 at 1:11
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.