# Soft question about irreducible representations

So I'm a beginner in representation theory and I'm a bit confused about the 'point' of irreducible representations. From what I understand, irreducible representations are important because they allow us to understand decompose the action of a linear transformation on a vector space. Here are my two questions:

1.) How does that help us to understand the group?

2.) Are we looking to find irreducible representations of each group element?

## 1 Answer

(1) Irreducible representations are more about understanding representations than they necessarily are about understanding groups. You said it yourself: they are about decomposing representations into their constituent parts, like natural numbers being factored into products of prime numbers, or molecules decomposed as a collection of atoms.

By themselves groups are just sets with binary operations satisfying some properties - and from this vantage point alone that doesn't inspire much confidence in their interestingness, since in a certain sense 'most' algebraic structures are boring and unimaginably complicated - but the defining properties of the group operation are abstracted from the idea of symmetry, and this motivation is resurrected in the idea of a group action. It is debatable to what extent it is an exaggeration to say the whole point of groups is to act on things. Since space is an important concept, in particular vector spaces, it makes sense to study linear group actions in their own right, i.e. representations.

That said, it's true that representation theory has been applied to group theory, combinatorics, quantum physics, chemistry and other areas. For instance, see

(BTW there are, actually, naturally occuring groups which are not just symmetry groups:

(2) The phrase "irreducible representation of each group element" makes no sense to me. Perhaps you could make this question more intelligible. If you mean find the value of $\rho(g)$ for all $g\in G$, where $\rho:G\to\mathrm{GL}(V)$ is an irreducible representation, then these elements are only defined up to pointwise conjugation of $\rho$ by linear transformations. Generally, the work involved pertaining to irreducible representations is finding ways to describe them, classify them, and calculate their characters (since most things you want to know about representations may be calculated practically using just the character table).

• Thanks for the response! What I mean by the second question is that if $\rho: G -> GL(V)$ is a an irreducible representation, does it take all elements of G to a diagonal matrix or just certain elements of G? For instance, I know that with $G = SU(2)$ you can restrict the group to the subgroup $U(1)$ to find irreducible representations $\rho_{U(1)}$, but I'm assuming that it doesn't map the rest of $SU(2)$ to diagonal matrices. – hijasonno Aug 12 '16 at 14:41
• @hijasonno The elements of $GL(V)$ are linear transformations. To represent them as matrices, you need to pick an ordered basis of $V$. Whether a particular $\rho(g)$ is represented by a diagonal matrix depends on which ordered basis you choose. For "almost all" ordered bases, none of the $\rho(g)$s (except for central elements $g\in Z(G)$ because of Schur's lemma, assuming $V$ is a complex irrreducible representation) will be represented by diagonal matrices. Each $\rho(g)$ will be diagonalisable, but the whole set of $\rho(g)$s will not be simultaneously diagonalizable unless $\dim V=1$. – arctic tern Aug 12 '16 at 14:54
• Also, how in the world did you expect anyone to interpret question (2) you've written that way? You didn't even use the word "diagonal"... – arctic tern Aug 12 '16 at 14:57
• Yeah I knew while I writing it that it was a very poor way of wording it but it was also compounded by my limited understanding of irreducible representations. Sorry about that. So when we ask ourselves to "classify all irreducible representations of a group" are we asking ourselves to find all ordered bases such that we can diagonalize each $\rho(g)$? – hijasonno Aug 12 '16 at 21:20