# Why does the dual space consisting of unitary equivalence classes of irreducible representations of compact $G$ not form a group?

I am trying to understand unitary representations of compact groups $$G$$. Equip the class of all unitary irreducible representations of $$G$$ with the usual equivalence relation, unitary equivalence. We define the dual space as the set of equivalence classes with respect to this relation. The dual space cannot ever form a group. I would like to check that my reasoning as to why this is true is correct. This is my reasoning:

We can try to equip this set with binary operation $$[\pi][\pi']:=[\pi\cdot\pi']$$ where $$(\pi\cdot\pi')(x)=\pi(x)\cdot\pi'(x)$$. Then this operation is pointwise multiplication of irreducible representations (which are just group homomorphisms). $$\textbf{But}$$, irreducible representations of compact groups are finite dimensional therefore can be expressed as square matrices, e.g.

$$\pi(x)=n\times n$$ matrix, $$\pi'(x)=m\times m$$ matrix. Then pointwise multiplication (and indeed regular matrix multiplication) cannot be defined unless $$n=m$$. However, in general the dual space will have representations of different dimensions - thus we cannot define a sensible binary operation on them.

This is in contrast with the $$\textit{dual group}$$ for Abelian locally compact groups, because irreducibles are just continuous homomorphisms into $$\mathbb{C}$$ and can therefore be equipped with pointwise multiplication to give group structure.

$$\textbf{One question}$$ comes to mind for me: how do we know there are no other binary operations we might come up with, that would give the dual space a group structure?

• Even if all the irreducible representations of a compact group you considered did have a fixed dimension $n$, there is no reason the operation you mentioned be well defined.
– D_S
May 2, 2021 at 17:00

You ask a soft question, and the answer seems to be "No one has ever come up with one." What operations can you do to two irreducible representations $$\pi$$ and $$\pi'$$ to get another irreducible representation? I can think of none that would produce a group structure.