Some irreducible complex representation of symmetric groups

Question

Consider the natural action of symmetric groups $S_n$ on $X_i\subset2^{N_n}$, where $N_n=\{1,2,...,n\}$, and $X_i=\{A=\{a_1,...,a_i\}|A\subset N_n\}$ (Basically they are just actions on indices). It induces complex representations of $S_n$ with dimension $\begin{pmatrix}n\\i\end{pmatrix}$, and their characters are denoted by $\chi_i$. We also denote the trivial representation as $\chi_0$. I want to show that $(\chi_i-\chi_{i-1}),1\leq i \leq [\frac{n}{2}]$ are $[\frac{n}{2}]$ irreducible complex representations of $S_n$ but I have got no idea how to prove it. At first I want to find a subspace of $V_i$ isomorphic to $V_{i-1}$, since it works for $i=1$ (Let $V=\{(x_1,...,x_n)|x_1=...=x_n\}\subset V_n$. Then dim$V=1$, and $V_n/V$ is irreducible, with character same as $(\chi_1-\chi_0)$), but it seems a difficult task for $i>1$...so maybe there is a better way to prove this. I think I need some help. Thanks.

Answer 1

We want to show that

$$\langle \chi_i - \chi_{i-1}, \chi_i - \chi_{i-1} \rangle = \langle \chi_i, \chi_i \rangle - 2 \langle \chi_i, \chi_{i-1} \rangle + \langle \chi_{i-1}, \chi_{i-1} \rangle = 1.$$

We'll show this using the fact I described in my comment above that if $\chi_X$ is the character of the permutation representation on a $G$-set $X$, then $\langle \chi_X, \chi_Y \rangle$ is the number of $G$-orbits of the action of $G$ on $X \times Y$. So it suffices to count the number of orbits of the action of $S_n$ on the set of pairs (a subset of $[n] = \{ 1, 2, \dots n \}$ of size $i$, a subset of size $j$), and then to plug in appropriate $i, j$.

I claim that up to the $S_n$ action, a pair of subsets $S, T$ of $[n]$ is determined by the size of its intersection $S \cap T$ and by the size of the complements $S \setminus (S \cap T), T \setminus (S \cap T)$. If the sizes $|S| = i, |T| = j$ are fixed then the only information is the size of the intersection, which can be anywhere from $0$ to $\text{min}(i, j)$. Hence the number of orbits is $\text{min}(i, j) + 1$. This gives

$$\langle \chi_i, \chi_i \rangle = i$$ $$\langle \chi_i, \chi_{i-1} \rangle = i - 1$$ $$\langle \chi_{i-1}, \chi_{i-1} \rangle = i - 1$$

so the inner product we want to compute is

$$i - 2 (i - 1) + (i - 1) = 1$$

as desired.