Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 103 / 1

Work over $\mathbb{C}$, with the standard inclusions $S_r\subseteq S_n$ fixing the letters $r+1,\dots ,n$. In the group algebra, the Young–Jucys–Murphy elements areThus $X_i$ adds all transpositions involving the newest letter. Conjugating by $S_{i-1}$ merely permutes the summands, so $X_i$ commutes with $\mathbb{C}{S}_{i-1}$. In particular, if $j<i$, then $X_j\in \mathbb{C}{S}_{i-1}$ and $X_i{X}_j=X_j{X}_i$. Hence the whole family is commuting.

Put ${\Omega }_r={\sum }_{1\le a<b\le r}(ab)$, the transposition conjugacy-class sum in $S_r$, with ${\Omega }_1=0$. ThenThis identity will connect the Young–Jucys–Murphy elements with the centres of the successive group algebras.

For an increasing chain of finite groups $G_1\subseteq \cdots \subseteq G_n$, regard every $\mathbb{C}{G}_r$ as a subalgebra of $\mathbb{C}{G}_n$. The Gelfand–Tsetlin algebra is the unital algebra generated by their centres:For this paper the chain is $S_1\subseteq \cdots \subseteq S_n$, so these centres are $Z_1,\dots ,Z_n$. The generators commute: if $r\le s$, an element of $Z_s$ commutes with all of $\mathbb{C}{S}_s$, including $Z_r$. Thus the Gelfand–Tsetlin algebra is commutative. Its joint eigenspaces record successive choices in the restriction of an irreducible representation down the chain.

The relevant centralizer of a subalgebra isEquivalently, it is the fixed subspace of $\mathbb{C}{S}_n$ under conjugation by $S_{n-1}$. We prove the Olshanskii centralizer lemma directly, without assuming multiplicity-free restriction.

A permutation has one cycle containing $n$, of some length $r\ge 1$. Let $\nu$ record the lengths of its other nontrivial cycles, omitting fixed points. Two permutations are conjugate under $S_{n-1}$ exactly when these data $(r,\nu )$ agree: a relabelling of the other letters identifies the distinguished cycles and all remaining cycles. Hence the sums $K_{r,\nu }$ over these conjugation orbits form a basis of $Z_{(n-1,1)}$.

Filter these orbit sums by the number of moved letters other than $n$:If $r=1$, the permutation fixes $n$, and $K_{1,\nu }$ is already a conjugacy-class sum in $Z_{n-1}$. For $r\ge 2$, let $C_{\nu }\in Z_{n-1}$ be the class sum with nontrivial cycle lengths $\nu$. ExpandTerms in which the $r-1$ star transpositions use distinct letters, disjoint from the support of the chosen term of $C_{\nu }$, produce precisely $K_{r,\nu }$, each with coefficient one. Indeed, a cycle through $n$ has a unique expression as an ordered product of star transpositions using each of its other letters once; the remaining cycles uniquely determine the term of $C_{\nu }$.

Every other term moves fewer than $t$ letters outside $n$. A repeated letter uses at most $r-2$ distinct such letters in the star product, and an overlap with the support of $C_{\nu }$ likewise reduces the size of their union. Since the whole product commutes with $S_{n-1}$, these remaining terms are a linear combination of lower-filtered orbit sums. ThusInduction on $t$, starting with the identity, puts every orbit sum in $\mathrm{alg}(Z_{n-1},X_n)$. The reverse inclusion holds because both proposed generators commute with $\mathbb{C}{S}_{n-1}$. ThereforeIn particular this centralizer of a subalgebra is commutative: $Z_{n-1}$ is commutative and $X_n$ commutes with it.

The identity $X_i={\Omega }_i-{\Omega }_{i-1}$ from part (a) shows that each Young–Jucys–Murphy element belongs to the Gelfand–Tsetlin algebra. Conversely, $Z_r$ lies in the centralizer of $\mathbb{C}{S}_{r-1}$ in $\mathbb{C}{S}_r$. Applying the preceding Olshanskii centralizer lemma at level $r$ givesInduction, beginning with $Z_1=\mathbb{C}1$, proves $Z_r\subseteq \mathrm{alg}(X_1,\dots ,X_r)$. ThusHere every generated algebra is unital; $X_1=0$ causes no problem.

To deduce simple branching, use Maschke's theorem: complex representations of finite groups are semisimple modules. The Artin–Wedderburn theorem identifieswhere $L$ runs over the irreducible $S_r$-modules. Write their restrictions asBy Schur lemma, the centralizer in the displayed algebra iswith zero multiplicities omitted. Part (i) makes this algebra commutative, whereas a full matrix algebra of size at least two is noncommutative. Consequently every restriction multiplicity is $0$ or $1$. This proves multiplicity-free restriction at each step, with no use of the branching rule in the centralizer proof.

Let $e_1,\dots ,e_n$ be the coordinate vectors and chooseThese vectors lie in the sum-zero standard representation of the symmetric group. They are mutually orthogonal for the usual Hermitian inner product, with $\Vert v_j{\Vert }^2=j(j+1)$, so they form a basis of its $(n-1)$-dimensional space.

If $i\le j$, every transposition in $X_i$ exchanges two coordinates equal to $1$, so each fixes $v_j$. If $i=j+1$, summing the $j$ transpositions gives coordinate $-1$ in each of positions $1,\dots ,j$ and coordinate $j$ in position $j+1$, hence $X_{j+1}{v}_j=-v_j$. Finally, if $i>j+1$, the $i$th coordinate of $v_j$ is zero. In the sum of the $i-1$ transpositions, any coordinate in its support retains its original value in $i-2$ terms and becomes zero in the remaining term; the $i$th output coordinate is the sum of the original first $i-1$ coordinates, which is zero.

Thus all actions are diagonal in this basis:This supplies an explicit Gelfand–Tsetlin basis of the standard representation.

For $n=4$, take the three vectors from part (iii):Their joint Young–Jucys–Murphy element weights, in the order $(X_1,X_2,X_3,X_4)$, areThey are distinct, and part (ii) says that these operators generate $G{Z}_4$. Therefore a Gelfand–Tsetlin basis isIf an orthonormal basis is desired, use $v_1/\sqrt{2}$, $v_2/\sqrt{6}$, and $v_3/\sqrt{12}$. The corresponding standard Young tableaux have shape $(3,1)$, with the second-row entry respectively $2,3,4$. Their contents agree with the three displayed weights.