How to find irreducible representations of $\mathbb{C}S_2$ and $\mathbb{C}S_3$

I just starting to learn representation, so still have a lots of thing that is unclear. And here is a question what I wish to attempt.

Let $S_n$ be the symmetric group and I want to find all the irreducible representations and indecomposable representations of group algebra $\mathbb{C}S_2$ and $\mathbb{C}S_3$?

Something want to make sure, the representation (so is the irreducible representation and indecomposable representation) of a group is the same as the representation its group algebra. And a irreducible representation also is a indecomposable representation.

I am not sure what is a representation. Like for $S_2=\{(1),(12)\}$, is that just $\left( \begin{array}{ccc} 1 & 0 \\ 0 & 1 \end{array} \right),\left( \begin{array}{ccc} 0 & 1 \\ 1 & 0 \end{array} \right)$?

1 Answer

I am not sure what a representation is.

You should be making sure you fix this problem before you consider finding irreducible representations! I will try to motivate and then explain the idea.

Let's get some historical perspective. Originally, groups were conceived of as symmetry groups, sets of transformations that preserve the structure of features of a mathematical object, such as a geometric figure in space or arithmetic equations in a number system. These always are sets of functions which can be composed and inverted. One can then lose the function interpretation of the elements of a group and abstract this to a generic set with an associative binary operation satisfying the same group axioms - this is an abstract group. Finally, if in nature one finds a structure which is naturally a group and whose elements have concrete interpretations (numbers, loops, games), this is a concrete group. Symmetry groups are in particular concrete groups. The distinction between abstract and concrete groups is informal, not formal, by the way.

With the abstraction of groups, we can understand the idea of two groups being "the same" but only having their elements (and the corresponding entries of the Cayley table) relabelled - we say that two groups are isomorphic. In this way they are "essentially the same." Thus, we can speak of different mathematical objects having "the same symmetry" (isomorphic symmetry groups) or otherwise comparing symmetries.

Instead of taking a mathematical object and then attaching a symmetry group to it, one can reverse our perspective by starting with a group and attaching objects for it to "act" on. The fundamental mathematical object (okay, debatable) is a set. The term "group action" is usually reserved for having a group act on a set. This means that every element of $G$ gets an interpretation as an invertible function on a set $X$, in such a way that composition in $G$ corresponds to composition of functions, just as with the idea of a symmetry group. This may be formally described as a group homomorphism $G\to{\rm Perm}(X)$. There is an alternate definition as a map $G\times X\to X$, given the abbreviated notation $(g,x)\mapsto gx$, satisfying "associativity" $(gh)x=g(hx)$.

(Fun fact: if one writes this associativity condition using commutative diagrams, as per the "categorical imperative" to think with diagrams, one may transplant this definition into other categories to define group objects. If one takes group objects in the categories of sets, topological spaces, smooth manifolds or algebraic varieties one gets groups, topological groups, Lie groups and algebraic groups respectively. Anyway, tangential remark.)

Of course, functions on sets hardly captures all there is about symmetry. In different categories there are different types of invertible maps - topological spaces have continuous maps, posets have monotone maps, vector spaces have linear maps - and in general we would want a group to act in the "appropriate" ways on these objects. This gives rise to the most general notion of a group representation: a map $G\to{\rm Aut}(X)$ where $X$ is an object in some category and ${\rm Aut}(X)$ is its automorphism group.

Almost always, by representation we mean group representation (I haven't hinted at any other kind yet), and even more specifically we mean a linear group representation, which uses the category of vector spaces over some field (usually the field $\Bbb R$ because we use it in geometry, or in $\Bbb C$ because the algebra works out so much nicer). In particular, a linear group representation of $G$ is formally a group homomorphism $\rho:G\to{\rm GL}(V)$. This means every element of $G$ gets interpreted as an invertible linear transformation of the vector space $V$.

From there, you can build the idea of isomorphisms of representations, direct sums, indecomposability, irreducibility, semisimplicity, tensor product and so on. I will omit explaining these concepts in this answer. There is also a notion of linear representations of algebras: if $A$ is an algebra over a field, then a representation is an algebra homomorphism $A\to{\rm End}(V)$ for some vector space $V$ over the same field. Thus, every element of $A$ gets interpreted as a linear transformation of $V$, just as with groups, but there's an addition operation too. At this juncture one can prove a complex group representation of $G$ is "essentially the same" as an algebra representation of $\Bbb C[G]$, the complex group algebra. Indeed there is an equivalence of categories. There are a few questions on this site laying around explaining this idea.

(One may also call $V$ an $A$-module, or alternatively define it by a linear map $A\otimes V\to V$ satisfying the aforementioned commutative diagram.) Note that with finite groups, if $|G|$ is invertible in the scalar field then all finite-dimensional representations are semisimple. In particular, indecomposable representations are irreducible. (Maschke's theorem.)

I want to find all the complex irreducible representations of $S_2$ and $S_3$.

Over $\Bbb C$, all irreducible representations of an abelian group are one-dimensional. (This is a consequence of Schur's lemma.) You've given a good representation of $S_2$, but even over $\Bbb R$ it's not irreducible: it fixes the "diagonal" subspaces in the Cartesian plane, $\Bbb R\cdot(1,1)$ and $\Bbb R\cdot(1,-1)$.

For a one-dimensional representation of a group $G$, since ${\rm GL}_1(\Bbb C)$ can be identified with $\Bbb C^\times$ you need to specify a group homomorphism $G\to\Bbb C^\times$. It should be obvious then what the two irreducible representations of $S_2$ are.

Now consider $S_3$. There are automatically two one-dimensional representations $S_3\to\Bbb C^\times$, the trivial representation and the sign representation. (Remember, taking signs of permutations is a group homomorphism $S_n\to\{\pm1\}$.) One can also take the standard representation of $S_3$ by permutation matrices acting on $\Bbb C^3$, but this is not irreducible. Indeed, it has an obvious $G$-invariant subspace of vectors of the form $(z,z,z)$. Indeed, this is the subspace of fixed points, denoted $V^G$ in general.

One thing you learn in the the beginning of representation theory is Weyl's trick: how to construct a $G$-invariant inner product $\langle\cdot,\cdot\rangle_G$ from any choice of inner product $\langle\cdot,\cdot\rangle$. One employs an averaging formula $\langle v,w\rangle_G:=\frac{1}{|G|}\sum_{g\in G}\langle gv,gw\rangle$, which is clearly $G$-invariant and one may check is an inner product. Then, if $W$ is a representation with subrepresentation $V$, one may find a complementary $G$-invariant subspace $U$ so that $W=U\oplus V$ as representations - indeed, one may find such a $U$ as the orthogonal complement of $V$ with respect to $\langle\cdot,\cdot\rangle_G$.

As it turns out, the standard inner product on $\Bbb C^3$ is already $S_3$-invariant. If one takes the orthogonal complement of $(\Bbb C^3)^{S_3}$ one obtains the two-dimensional subrepresentation of points $(x,y,z)\in\Bbb C^3$ satisfying $x+y+z=0$. We can check that it is irreducible too: since it's two-dimensional, if it had a proper nontrivial subrepresentation it would have to be one-dimensional. See if you can continue from there to prove it's irreducible.

This gives altogether three irreducible complex representations of $S_3$, two that have one dimension and one that has two dimensions. (Later, you will learn how to check if you have all of the representations: the sum of the squares of their dimensions is the group's order; in this case we have $1^1+1^2+2^2=3!$) It also showcases a bit of technology, like Schur's lemma, standard representations of symmetric groups, and $G$-invariant inner products yielding $G$-invariant complementary subspaces, which should be covered in introductions to representation theory.

• Thanks for the awesome answer. But what should I really need to write down to say this is a (irreducible) representation? Like in $S_3$ I know that the irreducible representations are trivial, alternating and standard representation. But are they matrices, or homomorphism functions? Thanks. – SamC Aug 6 '15 at 8:47
• @SamC Like I said: a representation is a group homomorphism $G\to{\rm GL}(V)$. If, say, $V$ is $n$-dimensional then picking a basis we can think about it as the coordinate space $F^n$ (where $F$ is the scalar field), and thus think about invertible linear transformations as matrices in ${\rm GL}_n(F)$. In this way, every element of $G$ has a corresponding invertible $n\times n$ matrix associated to it by the representation $G\to{\rm GL}_n(F)$. – whacka Aug 7 '15 at 0:34