# Explicitly decompose $\mathbb{C}^3$ into irreducible representations of $S_3$.

Consider the permutation representation of $S_3$ acting by permuting the elements of a basis of $\mathbb{C}^3$. Explicitly decompose $\mathbb{C}^3$ into irreducible representations.

Can someone please verify my answer? Note: This is NOT homework!

Consider the subspace $W = \operatorname{span} \left\lbrace \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \right\rbrace$ of $\mathbb{C}^3$. Then, $W$ is a one dimensional representation of $S_3$. A proof is as follows: Let $\sigma \in S_3$. Then, $\sigma$ permutes the entries of any vector of the form $\begin{bmatrix} \lambda \\ \lambda \\ \lambda \end{bmatrix}$ to give the same vector. So, $\sigma \cdot w = w$ for any $w \in W$ and $\sigma \in S_3$.

Let $V$ be the two dimensional subspace of $\mathbb{C}^3$ given by $$V = \left\lbrace \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} : x_1+x_2+x_3 = 0 \right\rbrace$$ Then, $V$ is a representation of $S_3$. Note that if the entries of any vector of the above form are permuted, then their sum remains the same.

Then, clearly, $V \cap W = \varnothing$, and $V + W = \mathbb{C}^3$. So, we have $V \oplus W \cong \mathbb{C}^3$ is a decomposition of $S_3$ into irreducible representations.

• How do you know $V$ is irreducible? Other than that, this looks good... – Micah Jul 2 '15 at 2:23

## 1 Answer

You have correctly shown that $\Bbb{C}^3$ is the direct sum of two smaller representations (with one minor exception: $V \cap W$ is the zero vector space, not the empty set). However, you have not given any argument that those representations are actually irreducible. It is clear that $W$ is irreducible, since it's $1$-dimensional, but you should give some argument that $V$ is irreducible.

There are a few ways you might do this, depending on how many theorems you already know:

• A low-tech way of doing this would be to show that $V$ has no $S_3$-invariant subspace, by explicitly computing the subspaces fixed by some elements of $S_3$.
• A high-tech way of doing this would be to compute the character of $V$, and find its Schur inner product with itself. This is probably not actually any easier in this specific case, but it would be for larger groups and higher-dimensional representations...
• Can you prove directly that any invariant subspace of $V$ contains some $e_i - e_j$ with $i\ne j$ ? – Orest Bucicovschi Jul 2 '15 at 12:10