Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 103 / 1

Work over the complex numbers. The chain $S_0\subset S_1\subset \cdots \subset S_n$ has multiplicity-free restriction: each irreducible representation restricts to a direct sum of pairwise inequivalent irreducible representations. This structural fact can be proved before identifying the branching diagram. Indeed, the permitted Olshanskii centralizer lemma makes $C_{\mathbb{C}{S}_r}(\mathbb{C}{S}_{r-1})$ commutative, since it is generated by the center at the previous level and the commuting element $X_r$. In each irreducible block this is the endomorphism algebra of the restricted module; a repeated summand would give a noncommutative matrix algebra factor. Thus no identification of the branching graph with Young diagrams is being assumed here.

Successively restricting an irreducible representation $V$ gives one-dimensional spaces $V_T$ indexed by paths of irreducible representations from the trivial $S_0$-module to $V$. Choosing one nonzero vector in each $V_T$ gives a Gelfand–Tsetlin basis. Define the Gelfand–Tsetlin algebra as the subalgebra of $\mathbb{C}{S}_n$ acting diagonally in all these bases. The Artin–Wedderburn theorem identifiesUnder this identification the Gelfand–Tsetlin algebra is the direct sum of the full diagonal matrix algebras in the indicated bases, so it is commutative.

To see that it is actually available inside the group algebra, let $e_{\alpha }^{(r)}$ be the central primitive idempotent selecting the irreducible representation $\alpha$ of $S_r$. For a path $T=({\alpha }_0,\dots ,{\alpha }_n)$ the productis the projection onto $V_T$ in its final irreducible representation and is zero in every other final block. These products commute: a center at a higher level commutes with every element at a lower level. The $P_T$ are precisely the diagonal matrix units, and therefore span the proposed algebra.

If an element of the group algebra commutes with every $P_T$, its matrix preserves every $V_T$ and is diagonal in every block. It already belongs to the Gelfand–Tsetlin algebra. Thuswhich proves that it is a maximal commutative subalgebra.

The projections $P_T$ in the Gelfand–Tsetlin algebra satisfyConsequently the map sending a diagonal element to its scalar on each Gelfand–Tsetlin basis vector is a ring isomorphismEach factor is a simple module over this algebra, and the algebra is their direct sum as a module over itself. This proves that $G{Z}_n$ is a semisimple algebra. The argument uses its explicit diagonal structure; being a subalgebra of a semisimple algebra by itself would not suffice.

The Young–Jucys–Murphy elements arewith products of permutations composed from right to left. Fix $i<j$. In $[X_i,X_j]$, all pairs of disjoint transpositions commute. The only terms left, for each $a<i$, areIndeed $(ai)(ij)=(aj)(ai)$ and $(ij)(ai)=(ai)(aj)$. Summing directly provesThus all the Young–Jucys–Murphy elements commute.

The primitive idempotent products used above show that the Gelfand–Tsetlin algebra is generated by the centers at all levels: those products give every diagonal matrix unit, whereas each center acts diagonally. Hence

For the second generation assertion use the permitted Olshanskii centralizer lemma, in the precise formIn particular $Z_r\subseteq \mathrm{alg}(Z_{r-1},X_r)$. Induction puts every $Z_r$ in $\mathrm{alg}(1,X_1,\dots ,X_r)$. Conversely, if $A_r$ is the sum of all transpositions in $S_r$, then $A_r\in Z_r$ and $X_r=A_r-A_{r-1}\in G{Z}_n$. ThereforeHere generation is unital; the zero generator $X_1$ causes no problem.

The deletion projection for a symmetric group removes $n$ from the cycle containing it. A cycle $(a_1\cdots a_{r}n)$ becomes $(a_1\cdots a_r)$, with a singleton understood as a fixed point; all other cycles remain unchanged. This describes a genuine permutation of $\{1,\dots ,n-1\}$ and immediately proves the two identity assertions.

Put $H=S_{n-1}$. For $\sigma ,\theta \in H$ and $k<n$, if $\pi (\theta (k))\ne n$, deletion gives $\sigma \pi (\theta (k))$. If $\pi (\theta (k))=n$, it gives $(\sigma \pi \theta )(n)=\sigma \pi (n)$. These are exactly the two cases in $\sigma {\pi }_n\theta (k)$. Thus

The printed uniqueness assertion has an exceptional case at $n=4$. To locate it, use the two double cosetsAn $H$-bimodule-equivariant set map is determined by $u=\Phi (1)$ and $v=\Phi (t)$. Equivariance requires $u\in Z(H)$ andbecause $H\cap tHt=S_{n-2}$ and $t$ commutes with this subgroup. These conditions are also sufficient: define $\Phi (h)=hu$ and $\Phi (ht{h}^{\prime })=hv{h}^{\prime }$. Changing the double-coset expression only inserts an element of $S_{n-2}$, which commutes with $v$.

For $n\ge 5$, the center of $H$ and this centralizer are both trivial. For the latter, a commuting permutation must fix the unique point outside $\{1,\dots ,n-2\}$ and lie in the center of $S_{n-2}$; that center is trivial when $n-2\ge 3$. Therefore $u=v=1$, giving deletion. For $n=4$, however, $C_{S_3}(S_2)=\{1,(12)\}$. The choiceis well defined and satisfies all the printed equivariance conditions, but sends $(34)$ to $(12)$ rather than to the identity. Uniqueness holds for $n\ge 5$; it is false at $n=4$ as printed. Even requiring the map to restrict to the identity on $H$ does not remove this counterexample.

There is a separate problem with the printed linear-span assertion, when $Z_{((n-1),1)}$ denotes the usual centralizer of a subalgebra. Let $A_{n-1}$ be the sum of all transpositions in $H$. Direct expansion in the group algebra givesThus $Y=X_n^2-2A_{n-1}$ commutes with $H$ and satisfies ${\Pi }_n(Y)=(n-1)1$. For $n\ge 3$ it contains nonzero coefficients on three-cycles involving $n$, whereas every element of $\mathrm{span}\{1,X_n\}$ is supported only on the identity and transpositions. Therefore $Y$ is an explicit counterexample to the asserted equality.

One correct description of the entire preimage is obtained by putting $C=C_{\mathbb{C}{S}_n}(\mathbb{C}H)$. Equivariance gives ${\Pi }_n(C)\subseteq Z(\mathbb{C}H)$, and ${\Pi }_n$ is the identity on that center. It follows thatA useful corrected two-dimensional statement restricts the support to the identity and the transpositions $(in)$: invariance under conjugation by $H$ then forces their coefficients to be equal, yielding exactly $\mathrm{span}\{1,X_n\}$. This additional support restriction is absent from the PDF.