In my lecture, the professor constructed a character table for $S_3$.

There are $3$ conjugacy classes, so there are $3$ irreducible characters.

The three characters we used were:

$\bullet$ $\chi_1$, the trivial character.

$\bullet$ $\chi_\rho - \chi_1$, where $\chi_\rho$ denotes the permutation character.

$\bullet$ the determinant of the permutation representation, i.e., the alternating representation.

Is $\chi_\rho - \chi_1$ alsways irreducible? Why?

I know that in this case $\langle \chi_\rho - \chi_1, \chi_\rho - \chi_1 \rangle = 1$, so it is irreducible. I don't see why this would generally be the case, though.

I know that if I just subtract two irreducible characters -- e.g., the alternating character and ($\chi_\rho - \chi_1$) in this character table -- the result isn't irreducible (I don't know if it's even a character?).

  • $\begingroup$ The title asks whether the difference of irreducible characters is always an irreducible character, and the answer is, no. The body asks a very different question, whether the difference between the specific characters $\chi_{\rho}$ and $\chi_1$ is always an irreducible character. $\endgroup$ – Gerry Myerson Mar 2 at 5:44
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    $\begingroup$ I corrected the title $\endgroup$ – Jess Mar 2 at 6:25

Let $S_n$ be the symmetric group, and let $P$ be the permutation representation. As a vector space, $P = \mathbb{C}^n$, with the symmetric group action $\sigma \cdot e_i = e_{\sigma(i)}$ on the standard basis $e_1, \ldots, e_n$. The question is: why is $\chi_P - \chi_1$ always a character, and furthermore why is it irreducible? There are two ways of answering this question, one from a more naive representation-theoretic perspective and the second from a more group-theoretic one.

Whenever we have a permutation representation (we take a group acting on a set, and turn that into a representation by making the set the basis of a vector space), there are two obvious subspaces preserved by the group. The subspaces $$ P_1 = \mathbb{C}(e_1 + \cdots + e_n), \quad P_0 = \{a_1 e_1 + \cdots + a_n e_n \mid a_1 + \cdots + a_n = 0 \}$$ are preserved under the action of $S_n$. Clearly we have $P = P_1 \oplus P_0$, and $P_1$ is isomorphic to the trivial representation, so now we just have to determine whether $P_0$ is irreducible. For this we can do a direct elementary proof.

Let $v = a_1 e_1 + \cdots + a_n e_n \in P_0$ be arbitrary. If we can show that for every $w \in P_1$, there is an element $x \in \mathbb{C}S_n$ of the group algebra such that $xv = w$, then $P_0$ must be irreducible. Note that there is a pair $(i < j)$ such that $a_i \neq a_j$, since $v \in P_0$. Let $(ij) \in S_n$ be the transposition switching $i$ and $j$ and fixing everything else, then $$ (1 - (ij))v = (a_i - a_j) e_i + (a_j - a_i) e_j,$$ which is proportional to $e_i - e_j$. Now we can apply further permutations to show that we can reach the vectors $e_1 - e_2, e_2 - e_3, \ldots, e_{n-1} - e_n$, which form a basis of $P_0$. Hence $P_0$ is irreducible.

Another way of tackling this problem is by calculating characters, which for permutation representations means counting fixed points. To compute $(\chi_P - \chi_1, \chi_P - \chi_1)$ we need to be able to compute the inner products $(\chi_1, \chi_1) = 1$ (which we already know), as well as $(\chi_P, \chi_1)$ and $(\chi_P, \chi_P)$.

To compute $(\chi_P, \chi_1) = \frac{1}{n!} \sum_{\sigma\in S_n} \chi_P(\sigma)$, we first notice that since $P$ is a permutation representation, $\chi_P(\sigma)$ is the number of fixed points $|X^\sigma|$ of $\sigma$ acting on $X = \{1, \ldots, n\}$. By Burnside's Lemma, we have that $(\chi_P, \chi_1) = 1$ since the orbit space $X / S_n$ has only one element. ($S_n$ acts transitively on $X$).

To compute $(\chi_P, \chi_P) = \frac{1}{n!} \sum_{\sigma\in S_n} \chi_P(\sigma)^2$, we can again use Burnside's lemma, this time on the set $Y = X \times X$ with the diagonal action $\sigma \cdot (i, j) = (\sigma(i), \sigma(j))$. The subset of $Y$ fixed by $\sigma$ is of the form $X^\sigma \times X^\sigma$, hence $(\chi_P, \chi_P) = |Y / S_n|$ by Burnside's lemma. Since the action of $S_n$ on $Y$ is transitive (the action of $S_n$ on $X$ is 2-transitive), the orbit space $Y / S_n$ has two elements, the diagonal and everything else. Hence $(\chi_P, \chi_P) = 2$.

We conclude with $$ \begin{aligned} (\chi_P - \chi_1, \chi_P - \chi_1) &= (\chi_P, \chi_P) - 2 (\chi_P, \chi_1) + (\chi_1, \chi_1) \\ &= 2 - 2 + 1 \\ &= 1. \end{aligned}$$

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