# What is the meaning of an “irreducible representation”?

What does it mean to talk about the "irreducible representatives of SO(3)"? I'm struggling to understand the concept of irreducible representations. Could someone give a concrete example for someone unfamiliar with group theory or representation theory?

• Do you mean "irreducible representation" (as in your title) or "irreducible representa tive" (as in the body of your question)? – Arturo Magidin May 13 '11 at 20:33
• @Arturo Magidin: I guess I mean both, since I'm not clear on the distinction. – okj May 13 '11 at 20:41
• You can't mean both. A "representation" of a group is a special kind of homomorphism whose domain is the group. A "representative" of a group would be an element of the group (under some kind of equivalence relation). There is a very standard, clear, definition of "irreducible representation" for any group, and I don't think there is one for "irreducible representative" in general (though, for all I know, there may be one for elements of SO(3)). – Arturo Magidin May 13 '11 at 20:46

A representation of the group $G$ means a homomorphism from $G$ into the group of automorphisms of a vector space $\mathbf{V}$. Essentially, you are trying to interpret each element of $G$ as an invertible linear transformation $\mathbf{V}\to\mathbf{V}$, in order to try to understand the group $G$ by how it "acts on $\mathbf{V}$."

If you have an action $\rho_1$ of $G$ on a vector space $\mathbf{W}$ (that is, one representation), and you have some other action $\rho_2$ of $G$ on another vector space $\mathbf{Z}$ (another representation), then you can use these two actions to construct an action of $G$ on the vector space $\mathbf{W}\oplus\mathbf{Z}$: just let $G$ act on the first coordinate using the old action on $\mathbf{W}$, and let it act on the second coordinate using the old action on $\mathbf{Z}$.

The point to observe, however, is that the action of $G$ on $\mathbf{W}\oplus \mathbf{Z}$ defined this way does not give you any new insights into the structure of $G$: anything you can glean about $G$ from this action, you can learn about $G$ by considering the original actions $\rho_1$ and $\rho_2$. So this new action does not give us anything new.

Conversely, suppose you have one representation $\rho$, with $G$ acting on $\mathbf{V}$, and that there are proper subspaces $\mathbf{W}$ and $\mathbf{Z}$ of $\mathbf{V}$ that satisfy the following properties:

1. $\mathbf{V}=\mathbf{W}\oplus\mathbf{Z}$; and
2. The action of every $g\in G$ on $\mathbf{V}$ maps $\mathbf{W}$ to itself; and
3. The action of every $g\in G$ on $\mathbf{V}$ maps $\mathbf{Z}$ to itself.

Then you can look at the restriction of the action of $G$ on $\mathbf{W}$ to get a representation, and the restriction on $\mathbf{Z}$ to get another representation; and these two representations will give you all the information from the original representation, the same way we had before. The advantage being that since $\mathbf{W}$ and $\mathbf{Z}$ are proper subspaces of $\mathbf{V}$, they have smaller dimension and, presumably, it's easier to understand a subgroup of linear automorphisms for them than for $\mathbf{V}$.

So the moral is that we want to find representations that cannot be "broken up" into smaller ones, because there's no point in trying to understand ones that do break up, we can focus our attention on those that don't, because all the other representations can be built up in terms of the ones that cannot be broken up.

The irreducible representations are precisely the ones that cannot be broken up into smaller pieces (at least for finite groups). Maschke's Theorem says that if you have a representation $\rho$ of a finite group $G$ acting on $\mathbf{V}$, and $\mathbf{W}$ is a subspace of $\mathbf{V}$ such that for all $g\in G$, the image of $\mathbf{W}$ under the action of $g$ is $\mathbf{W}$ itself, then you can find a subspace $\mathbf{Z}$ of $\mathbf{V}$ such that $\mathbf{V}=\mathbf{W}\oplus\mathbf{Z}$ and every $g\in G$ maps $\mathbf{Z}$ to itself (that is, in order to break up $\rho$ into two smaller pieces, it is enough to find a single proper piece on which $\rho$ acts; then you can find a complement for it).

With this in mind, we say:

Let $\rho\colon G\to \mathrm{Aut}(\mathbf{V})$ be a representation of $G$. We say that $\rho$ is irreducible if and only if $\mathbf{V}$ is not the zero vector space, and the only subspaces of $\mathbf{V}$ that are mapped to themselves under the action of every $g\in G$ are $\{\mathbf{0}\}$ and $\mathbf{V}$ itself.

An irreducible representation of $SO(3)$ will be a representation of $SO(3)$ that is irreducible. $SO(3)$ acts naturally on the vector space $\mathbb{R}^3$: it consists of all automorphisms of $\mathbb{R}^3$ that respect the inner product, so this is itself a representation of $SO(3)$ (which is irreducible, because no proper subspace of $\mathbb{R}^3$ is sent to itself by all elements of $SO(3)$).

(Maschke's Theorem holds if the vector space is over a field of characteristic $0$, or if the characteristic does not divide the order of the group; there are similar theorems for certain kinds of infinite groups, but it does not hold for arbitrary infinite groups in general.)

• This is a very helpful answer, thank you. – user27182 Nov 15 '14 at 0:09
• Warning: The concepts irreducible and indecompasable differ in general. – C-Star-W-Star Oct 8 '15 at 17:59

if $\rho:G\to{\rm GL}_n(\mathbb{C})$ is a finite dimensional complex representation of the group $G$, then $\rho$ is irreducible if whenever $\rho(G)(V)\subseteq V$ then $V=\mathbb{C}^n \text{ or } 0$ (there are no non-trivial subspaces of $\mathbb{C}^n$ left invariant by $\rho(G)$).

here's a "visual" example. the real representation of ${\rm SO}_3(\mathbb{R})$ (the elements acting on $\mathbb{R}^3$ by rotations about the origin) is irreducible (there is no line or plane in $\mathbb{R}^3$ that is invariant under all rotations).

• hmm... this is helpful, but I'm still hazy. From my reading I get the feeling that an irreducible representation is a matrix (in the case of SO(3) at least, though it seems that in general they are always tensors), is this correct? If this is the case, what makes irreducible representations (or irreducible representatives) special as compared to other matrices? For example, any member of SO(3) can be represented by a 3x3 matrix of determinant equal to 1. Are these conditions sufficient to say that any matrix that satisfies them is an irreducible representation of SO(3)? – okj May 13 '11 at 20:46
• @okj: An irreducible representation is a map from the group to a group of matrices; under the representation (under the map), each element of the group will map to a matrix. You can think of an irreducible representation as a way to assign to every element of the group (in this case, SO(3)), a particular matrix (linear transformation). The irreducibility refers to the entire map, not to a specific matrix assigned to a specific element. (What may be helping to confuse you is that a representation is a map that sends elements to maps, thinking of matrices as maps) – Arturo Magidin May 13 '11 at 20:55
• @Arturo Magidin: Ok, tell me if I've got this straight. A representation of SO(3) is a map that takes a rotation in $R^3$ and gives it a matrix. If I wanted to I could theoretically define a map that took SO(3) and represented it with 10x10 matrices (presumably it could be a 10x10 matrix whose top-left 3x3 block has determinant 1--i.e. our standard matrix representation for SO(3), and zeros elsewhere). While this could be a faithful representation of SO(3) it is not irreducible, because I can find a line or plane in $R^{10}$ that IS invariant under ALL of my 10x10 matrices. – okj May 13 '11 at 21:17
• @okj: Yes, you could define a map from $SO(3)$ to the automorphism group of a 10-dimensional vector space; that would give you a 10x10 matrix for each element. Whether it would be "a 10x10 whose top-left 3x3 block..." etc. will depend on the map; that's one possibility, but there may be others. Such a representation would indeed not be irreducible, because the subspace spanned by the first three basis vectors would be invariant (it's not a line or a plane, it just has to be a proper subspace of $\mathbf{V}$, and in this case it would be a 3-dim space). – Arturo Magidin May 13 '11 at 23:42