Irreducible representation of $\mathcal{S}_5$ over $\mathbb{C}$ of degree 4

Question

I have come to a point where I need an irreducible representation of $\mathcal{S}_5$ over $\mathbb{C}$ of degree 4.

Can somebody help me to find one and explain how to obtain one?

Answer 1

Consider the permutation representation $V$ obtained by the usual action of $S_5$ on a basis $e_1,\ldots,e_5$ for $\mathbb C^5$. This representation has one obvious invariant subspace, namely the diagonal $$ \Delta : e_1 = e_2 = e_3 = e_4 = e_5 . $$ The 4 dimensional complement $W$ of $\Delta$ is defined by $\sum e_i = 0$. To get an explicit representation of $W$ in terms of matrices, you could use the convenient basis $\{e_i - e_{i+1} : 1 \leq i \leq 4 \}$.

To see that $W$ is irreducible, you can check to see if the character of $W$ has inner product 1 with itself. Once you have an explicit representation of $W$ in terms of matrices, you can calculate this character by hand. Alternately, note that the character $\chi$ of $V$ is given by $$ \chi(g) = |\{x : gx = x\}|, $$ and the character of $W$ is $\chi - 1$.

Whenever you have a group acting doubly transitively on a set of size $n$, this same process will result in an irreducible representation of degree $n-1$. For a proof of this fact, see these notes.

Edit: Let $\phi\colon S_5 \to \operatorname{GL}_2(\mathbb C)$ be the 4 dimensional representation described above. To compute $\phi$ explicitly, let's choose the basis $$ b_i = e_i - e_{i+1}, \quad i=1,2,3,4 $$ for $W$. Then the element $\sigma = (12345)$ acts on this basis like this: \begin{align*} \sigma(b_1) &= e_2 - e_3 = b_2 \\ \sigma(b_2) &= e_3 - e_4 = b_3 \\ \sigma(b_3) &= e_4 - e_5 = b_4 \\ \sigma(b_4) &= e_5 - e_1 = -(b_1+b_2+b_3+b_4) . \end{align*} The element $\tau = (12)$ acts like this: \begin{align*} \sigma(b_1) &= e_2 - e_1 = -b_1 \\ \sigma(b_2) &= e_1 - e_3 = b_1 + b_2 \\ \sigma(b_3) &= e_3 - e_4 = b_3 \\ \sigma(b_4) &= e_4 - e_5 = b_4 . \end{align*} Therefore, with respect to the basis $b_i$, we have $$ \phi(\sigma) = \begin{pmatrix} 0&0&0&-1\\ 1&0&0&-1\\ 0&1&0&-1\\ 0&0&1&-1 \end{pmatrix} , \qquad \phi(\tau) = \begin{pmatrix} -1&1&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{pmatrix}. $$ Since $\sigma,\tau$ generate all of $S_5$, you can get the rest of the $\phi(g), g \in S_5$ from these.