Representation theory - how to show a subrepresentation is irreducible? Consider the permutation group $S_3$ that permutes $\{1,2,3\}$. Let $V = \mathbb{C}^3$ be a vector space. We can have a representation of $S_3$ using $3\times 3$ matrices.
In the lecture I am following, it is then shown that $V = V_1\oplus V_2$ where $V_1 = span\{(1,1,1)^T\}$ and $V_2 = span\{(a,b,c)^T \vert\  a+b+c = 0\}$. How can one show that these are irreducible subrepresentations? For $V_1$, it is one dimensional so it follows but for $V_2$, how can one see that there is no subspace that is invariant under $S_3$?
Is there a general technique to know that one has decomposed the representation into a direct sum of subrepresentations?
 A: In general (for an arbitrary group), $V_2$ may or may not be irreducible. Here, you can check $V_2$ is irreducible by computing its chatacter:
There are three conjugacy classes of $S_3$, represented by $1$, $(123)$, and $(12)$. The character of $V_2$ has value $2$, $-1$, and $0$ on these classes, respectively. The inner product of it with itself is
$$\frac{1}{6}(2^2 \times 1+(-1)^2 \times 2 + 0^2 \times 3) = 1$$
So it is irreducible.
A: If you haven't been introduced to characters yet, then here is a more direct approach for low dimensional problems.
note: I assume you are using the standard (injective) representation of $S_3$ with permutation matrices. (Otherwise e.g. you could have a trivial 3-d representation where every permutation is mapped to $I_3$.)
Note that $S_3$ is not an abelian group, thus its (injectively mapped) representation isn't either.  If $V_2$ is a direct sum of two 1-d subpaces (i.e. if $V_2$ is reducible), then all 6 of your permutation matrices are simultaenously diagonalizable
$\implies$ the representation is commutative $\implies S_3$ is abelian, a contradiction.
