Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 1 / 19G (Representation Theory)

The character of a permutation representation at $g$ equals the number of points fixed by $g$, because a permuted basis vector contributes one to the trace exactly when it is fixed. Thus the character inner product isTo identify the last average, count pairs $(g,z)$ with $gz=z$. Each orbit contributes $|G|$, since its number of elements times their common stabilizer size is $|G|$ by the orbit-stabilizer theorem. ThereforeThe conjugation in the usual character inner product has no effect here because the permutation characters are real-valued.

There is a missing hypothesis in the printed equivalence. A group acting trivially on a two-point set has a one-dimensional zero-sum subspace $S$, hence an irreducible trivial representation, but its action is not doubly transitive. Thus the assertion as written is false. The intended result holds for a transitive action; more generally the trivial two-point action is the only exceptional case under $|X|>1$.

For the intended proof, decompose $\mathbb{C}[X]=1\oplus S$ by separating constant functions from functions with zero sum. If the action is transitive, $⟨1,{\chi }_X⟩=1$. HenceOver the complex numbers, complete reducibility and character orthogonality identify this norm as the sum of squared irreducible multiplicities; it is one precisely when $S$ is irreducible. By part (i), $\Vert {\chi }_X{\Vert }^2$ counts orbits on $X\times X$. Transitivity gives one diagonal orbit, and the norm equals two exactly when all off-diagonal ordered pairs form one orbit, which is double transitivity. This proves both implications with the missing transitivity condition restored.

To check the exact exception, if the action has $r>1$ orbits, then $S$ contains an invariant subspace of dimension $r-1$, from orbitwise constant functions with total sum zero. If $S$ is irreducible this makes $S$ the one-dimensional trivial representation, so $|X|-1=1$ and $|X|=2$. Nontransitivity on two points means both are fixed. Therefore irreducibility is equivalent to double transitivity, except for the trivial two-point action.

The symmetric group is transitive on two-element subsets, giving $⟨1,{\chi }_Y⟩=1$. Its diagonal action on $X\times Y$ has two orbits, according to whether the point lies in the pair. Thus $⟨{\chi }_X,{\chi }_Y⟩=2$, and, since ${\chi }_X=1+{\chi }_S$, the standard zero-sum representation $S$ occurs in $\mathbb{C}[Y]$ once. Its irreducibility follows from the double transitivity of the natural symmetric-group action and the corrected statement of part (ii).

For $n\ge 4$, ordered pairs of two-element subsets have three possible intersection sizes, zero, one and two, each forming a single orbit. Part (i) therefore gives $\Vert {\chi }_Y{\Vert }^2=3$. After removing the one copy each of $1$ and $S$, the remaining character has squared norm $3-1-1=1$, so is itself irreducible. HenceFor $n=3$, taking the complement of a pair is an equivariant bijection $Y\to X$. Thus $\mathbb{C}[Y]\cong 1\oplus S$, with no third constituent. Correspondingly the disjoint-pair orbit is absent and the character norm is two.