Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 3 / 19H (Representation Theory)

The Burnside lemma states that for a finite group $G$ acting on a finite set $X$,To prove it, countCounting by $g$ gives ${\sum }_g|X^g|$. Counting by $x$ givesusing the orbit-stabilizer theorem on each orbit $\mathcal{O}$. Dividing by $|G|$ proves the formula.

Let $\pi$ be the character of the permutation representation $\mathbb{C}[X]$. Then $\pi (g)=|X^g|$, and a second application of Burnside's lemma givesFor a two-transitive action there are exactly two orbits on $X\times X$: the diagonal and the ordered pairs of distinct points. Hence $⟨\pi ,\pi ⟩=2$. Transitivity says the trivial representation occurs once. Since the squared multiplicities of the irreducible constituents sum to two, there is exactly one further constituent, with multiplicity one, and it is irreducible and nontrivial. Thusas in the permutation representation of a two-transitive action.

For permutation characters, Burnside lemma on a product givesTwo pairs $(A,B)$ and $(A^{\prime },B^{\prime })$ lie in the same $S_n$-orbit exactly whenWhen $0\le l\le k\le n/2$, every value $0,1,\dots ,l$ is possible because $k+l\le n$. Therefore the symmetric-group subset permutation representation satisfies

For $1\le r\le n/2$, the inclusion map between subset permutation modules embeds $\mathbb{C}[X_{r-1}]$ into $\mathbb{C}[X_r]$, sois a character. Its norm isBy character orthogonality, ${\chi }_r$ is the character of an irreducible representation.

For $r>n/2$, complementation is an $S_n$-equivariant bijection $X_r\cong X_{n-r}$, soConsequentlythe negative of one of the irreducible characters just found, and is not itself the character of a representation.