# Why is $\chi_\rho - \chi_1$ Always a Character, and Why is it Irreducible?

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?).

• 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. Mar 2, 2020 at 5:44
• I corrected the title Mar 2, 2020 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}