Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 103 (Representation Theory)

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.

The Young lattice, or Young poset, has one vertex for each partition of an integer, including the empty partition, ordered by inclusion of their Young diagrams. A diagram covers another exactly when it adds one box. The Young branching graph is its graded graph: vertices at level $i$ are the partitions of $i$, and edges join diagrams differing by one box.

A descending path from $\lambda ⊢n$ to $(1)$ determines a standard Young tableau: label the successive removed boxes $n,n-1,\dots ,2$, then label the remaining box $1$. Every removed box is a Removable node of a Young diagram, so the resulting labels increase along rows and columns. Conversely, deleting boxes in decreasing label order from a standard Young tableau gives the path. Extending by the unique edge from $(1)$ to the empty diagram yields the usual path-to-tableau correspondence.

For the operator identity, compare the coefficients of each $\mu ⊢i$ in $DU(\lambda )$ and $UD(\lambda )$. If $\mu \ne \lambda$, a common upper cover exists exactly when a common lower cover exists, and each is then unique: the two diagrams differ by exchanging one box, their union is the upper cover and their intersection the lower cover. Since unions and intersections of partition diagrams are again partition diagrams, these off-diagonal coefficients agree.

The coefficient of $\lambda$ in $DU(\lambda )$ is the number of addable nodes of a Young diagram; its coefficient in $UD(\lambda )$ counts the choices of a Removable node of a Young diagram. Along the diagram's boundary the two types of corners alternate, beginning and ending with addable corners. Hence there is exactly one more addable than removable corner. For the empty partition the counts are $1$ and $0$.

Therefore the up and down operators satisfyor $D_{i+1}{U}_i-U_{i-1}{D}_i=I_i$ at level $i$. At level zero, interpret $D_0$ and the lowering contribution as zero. This makes the Young lattice a differential poset.

Use the normal ordering identity for up and down operators, placing every $U$ before every $D$. The commutator from part (a) impliesby induction on $i$. Multiplying the normal-ordered expansion on the left by $D+U$ therefore giveswhere negative indices have coefficient zero and $a_{0,0}(0)=1$.

We prove that $a_{i,j}(\ell )=0$ unless $i,j\ge 0$ and $\ell -i-j=2m$ for some integer $m\ge 0$, and otherwiseThe formula holds at $\ell =0$. For the next step with $\ell +1-i-j=2m$, the three contributions in the recurrence, after factoring out $\ell !/(2^{m}i!j!m!)$, are respectively $i$, $j$, and $2m$. Terms with a negative index or $m-1<0$ are simply absent. Their sum is $i+j+2m=\ell +1$, giving the required $(\ell +1)!$ numerator. Parity and nonnegativity exclude every remaining case. This proves the coefficient formula.

Apply the identity to the empty partition. Since $D(\varnothing )=0$, only terms with $j=0$ survive. To finish at $\lambda ⊢n$, only $U^n$ can contribute, and its coefficient of $\lambda$ is $f_{\lambda }$, by the Young branching graph correspondence with standard Young tableaux. For $\ell =n+2m$,Here $(2m-1)!!=1\cdot 3\cdots (2m-1)$ is the odd double factorial, with $(-1)!!=1$. ThusThe operator $(D+U{)}^{\ell }$ counts oscillating tableaux, allowing both upward and downward steps. A strictly upward path to level $n$ would necessarily have length $n$; the printed operator specification is what determines $N$ when $\ell >n$.

The joint spectrum $\mathrm{Spec}(n)$ consists of the tuples of simultaneous eigenvalues of $X_1,\dots ,X_n$ occurring in irreducible complex $S_n$-modules. A Gelfand–Tsetlin basis is a Gelfand–Tsetlin basis: its vectors lie in the one-dimensional joint eigenspaces selected by successive restrictions. The weight of $v$ is $\alpha =(a_1,\dots ,a_n)$ when $X_{j}v=a_{j}v$ for every $j$.

Define $\mathrm{Cont}(n)$ to be the integer tuples satisfying the content vector of a standard Young tableau conditions: $a_1=0$; for $j>1$, at least one of $a_j-1,a_j+1$ occurs earlier; and between any two occurrences of the same integer $a$, both $a-1$ and $a+1$ occur. We will justify their tableau interpretation in part (iv).

Here is the local spectral rules for Young–Jucys–Murphy elements calculation used in this question. Average a positive definite Hermitian inner product over $S_n$, making the representation unitary. The $X_j$ are commuting self-adjoint operators because transpositions are self-adjoint involutions. Put $s_i=(ii+1)$ and $d=a_{i+1}-a_i$. The group-algebra relations areFor a unit weight vector $v$, the first relation gives $d⟨v,s_{i}v⟩=1$. Thus $d\ne 0$, $|d|\ge 1$, andhas the weight obtained by swapping $a_i,a_{i+1}$, with $\Vert w{\Vert }^2=1-d^{-2}$. If $d=\pm 1$, then $w=0$ and $s_{i}v=\pm v$. If $|d|>1$, then $w\ne 0$, providing an admissible spectral swap in the same irreducible module.

For the proposed three consecutive weights $(a,a+1,a)$, these rules give $s_{i}v=v$ and $s_{i+1}v=-v$. But the braid relation in a Coxeter group would implya contradiction. Hence the proposed tuple is not in $\mathrm{Spec}(n)$. The same argument, with the signs reversed, excludes $(a,a-1,a)$ as well.

Since $X_1=0$, every joint weight has $a_1=0$. In $S_2$, $X_2=(12)$ acts by $1$ on the trivial representation and by $-1$ on the sign representation, which exhaust the irreducible representations. ThusThe content vector of a standard Young tableau conditions similarly require the second coordinate to be a neighbour of $0$, soThese vectors correspond respectively to the single-row tableau of shape $(2)$ and the single-column tableau of shape $(1,1)$.

For $\alpha ,\beta \in \mathrm{Spec}(n)$, define $\alpha \sim \beta$ when their Gelfand–Tsetlin basis lines occur in the same irreducible representation. This is well defined: $G{Z}_n$ contains the centre of $\mathbb{C}{S}_n$, whose primitive central idempotents separate the irreducible modules, and part 1(ii) says these central elements are polynomials in the $X_j$. Consequently the same weight cannot label two different irreducible modules.

For content vectors of standard Young tableaux, an admissible adjacent swap of a standard Young tableau exchanges coordinates $i,i+1$ whose difference has absolute value greater than $1$. Define $\alpha \approx \beta$ as the equivalence relation generated by such swaps within $\mathrm{Cont}(n)$.

Equivalently, after reconstructing the tableaux, $\alpha \approx \beta$ means that the two tableaux have the same final Young diagram. Indeed, an admissible swap exchanges consecutive entries in incomparable cells, preserves the diagram and leaves a standard Young tableau. Conversely, tableaux of a fixed shape are linear extensions of a partially ordered set, with the cells ordered by their row and column requirements; any two linear extensions are connected by adjacent swaps of incomparable elements. We explain this connectivity and its spectral consequence in part (v).

We verify all three content vector of a standard Young tableau conditions using the spectral rules proved in part (i).

The first coordinate is zero by part (ii). For the neighbour condition, suppose some $a_j$ has neither $a_j-1$ nor $a_j+1$ among earlier coordinates. Try moving $a_j$ successively to the left. Each encountered difference is not $\pm 1$; if it is zero, we have an impossible adjacent equality, while otherwise the local spectral rule permits the swap. If $a_j=0$, this procedure must encounter the original first coordinate $0$ and give that contradiction. If $a_j\ne 0$, it moves $a_j$ to the first position, contradicting $X_1=0$. Therefore an earlier neighbour always exists. Starting from $a_1=0$, this also proves inductively that every coordinate is an integer.

For the repeated-value condition, suppose it fails and choose a failing pair of equal coordinates $a$ with the shortest interval. There is no intervening $a$, since a shorter pair would then also fail. There must be at least one intervening neighbour: otherwise the right-hand $a$ could be moved next to the left-hand $a$ by admissible swaps. Suppose, for example, that $a+1$ is missing. There cannot be two occurrences of $a-1$ in the interval: between them the required neighbour $a$ is missing, producing a shorter failing pair. Thus there is exactly one $a-1$. All other intervening values differ from $a$ by more than $1$, so the two copies of $a$ can be moved towards that unique neighbour until the forbidden consecutive pattern $(a,a-1,a)$ appears. Part (i) excludes it. The case in which $a-1$ is missing is identical, using $(a,a+1,a)$. Both neighbours must therefore occur.

For completeness, these conditions really encode standard Young tableaux. If $a$ occurs for the $r$th time, place the next label inThis is the next unfilled cell on diagonal of content $a$. On the first occurrence of a positive diagonal, the earlier-neighbour condition forces its first left predecessor to exist: earlier diagonal values form an interval containing $0$, so a first positive $a$ can only be reached through $a-1$. The negative case uses the predecessor above. On later occurrences, the two neighbours appearing since the previous occurrence ensure that the new cell's left and upper predecessors have already been filled. For diagonal $0$, the same two-neighbour condition supplies both predecessors after the first box. Thus each addition is an addable node of a Young diagram, and the labels form a standard Young tableau.

Conversely, a standard Young tableau starts at $(1,1)$; any later cell has a left or upper predecessor of neighbouring content. Two cells on one diagonal have the cells immediately to the right and below the earlier one between them in the tableau order, supplying both neighbours. Hence its content vector of a standard Young tableau satisfies precisely the stated conditions. The next position on a diagonal is unique, so this construction is a bijection between content vectors of standard Young tableaux and standard Young tableaux.

We have now proved $\boxed{\mathrm{Spec}(n)\subseteq \mathrm{Cont}(n)}$ without assuming the desired spectrum identification.

Use the tableau representation supplied by Young seminormal form, stated explicitly in Question 5, or its normalized Young orthogonal form. It constructs the complex Specht module $S^{\lambda }$ with one basis vector for each standard Young tableau of shape $\lambda$. Its adjacent-transposition matrices imply the content eigenvalues directly.

Indeed, start with $X_1=0$ and use $X_{i+1}=s_i{X}_i{s}_i+s_i$. On an admissible pair $T,R=s_{i}T$, set $a=c_T(i)$, $b=c_T(i+1)$, $d=b-a$, and $q=\sqrt{1-d^{-2}}$. In the orthogonal basis, $X_i$ is $\mathrm{diag}(a,b)$ andMultiplication givesFor a nonadmissible swap, $s_i=\pm 1$ and the next content differs by the same $\pm 1$, giving the same conclusion. Induction therefore proves $X_i{w}_T=c_T(i)w_T$. Every content vector of a standard Young tableau, reconstructed as a standard Young tableau in part (iv), is consequently a spectral weight. Together with the reverse inclusion,

An admissible adjacent swap has nonzero off-diagonal coefficient in Young orthogonal form, and the local spectral calculation in part (i) keeps it inside one irreducible representation. Thus $\alpha \approx \beta$ implies $\alpha \sim \beta$.

Two standard Young tableaux of one shape are connected by admissible swaps. Regard each as a linear extension of a partially ordered set of cells. Move the first cell of the desired extension to the front of the other extension: every cell it crosses is incomparable with it, since otherwise their order would be forced in both extensions. Repeat after fixing that first cell. Each step swaps consecutive labels in incomparable cells. Such cells lie strictly northeast and southwest of each other, so their contents differ by at least $2$ in absolute value. These are exactly admissible swaps.

This also establishes irreducibility of the constructed tableau representations. A submodule is invariant under the commuting $X_j$ and hence under their joint spectral projections; since their weights are distinct, it is a sum of tableau lines. A nonzero off-diagonal coefficient propagates any included line along every admissible edge, and connectivity then includes all lines. Different shapes give nonisomorphic modules, since their content-vector sets are disjoint and content reconstruction determines the shape. Finally, the Robinson–Schensted correspondence gives ${\sum }_{\lambda ⊢n}{f}_{\lambda }^2=n!$, and the sum of squares of irreducible degrees leaves room for no other irreducible representations.

Thus within each irreducible module the weights are exactly the tableaux of one shape, so $\alpha \sim \beta$ implies $\alpha \approx \beta$. Hence the two equivalence relations coincide.

For a cell $x=(i,j)$ in a Young diagram, its hook of a Young diagram consists of $x$, all cells to its right in row $i$, and all cells below it in column $j$. If ${\lambda }^{\prime }$ is the conjugate partition, its hook length isThe hook-length formula is

In the Greene–Nijenhuis–Wilf hook walk, a walk at a noncorner cell chooses uniformly one of the other $h(i,j)-1$ cells in its hook, and stops at a corner. Moves are strictly rightwards or downwards, so termination is certain. A terminal cell must be a Removable node of a Young diagram. For a chosen corner $(\alpha ,\beta )$, its probability is zero unless the start satisfies $a\le \alpha$ and $b\le \beta$.

For a start in that northwest rectangle, defineThe corner conditions are ${\lambda }_{\alpha }=\beta$ and ${\lambda }_{\beta }^{\prime }=\alpha$, so $A_{\alpha }=B_{\beta }=0$, and all earlier $A_i,B_j$ are positive. DefineandThen the closed conditional probability isIn particular, if both inequalities are strict, this isThe separate definitions of $F_{\alpha },G_{\beta }$ handle starts in the terminal row, in the terminal column, or at the corner itself without division by zero.

To prove the formula, the product definitions give the identitieswith the terminal zero identities interpreted as empty sums. For example, summing the $F$ recurrence backwards gives ${\sum }_{r=i+1}^{\alpha }{F}_r={\prod }_{r=i+1}^{\alpha -1}(1+A_r^{-1})$. AlsoThe first-step equation for the hook walk terminal probability isJumps beyond the target rectangle contribute zero. Substituting $p_{i,j}=F_i{G}_j$ proves this equation, and $p_{\alpha ,\beta }=1$ gives its boundary value. Since every step increases a coordinate, backward recursion determines the probability uniquely. This proves the claimed expression.

For $k=1$, the product is empty and equals the identity, the only $1$-cycle in $S_1$. Suppose the cycle-sum identity for Young–Jucys–Murphy elements holds at $k-1$:Multiplication by $X_k={\sum }_{a<k}(ak)$ gives terms $\sigma (ak)$. With permutation products acting right to left, this inserts $k$ immediately after $a$ in the cycle of $\sigma$: the arrows become $a\mapsto k\mapsto \sigma (a)$, with the other arrows unchanged.

Conversely, in any $k$-cycle there is a unique predecessor $a$ of $k$. Delete $k$ and join that predecessor to the successor of $k$; this uniquely recovers $\sigma$ and the factor $(ak)$. Hence every $k$-cycle occurs exactly once, and no other permutation occurs. Induction provesThus ${\xi }_k$ is a conjugacy-class sum in $\mathbb{C}{S}_k$, although it need not be central in $\mathbb{C}{S}_n$ when $k<n$.

Let $T$ be a standard Young tableau of shape $\lambda ⊢n$, and let $\mu$ be the shape formed by its entries $1,\dots ,k$. Since Young–Jucys–Murphy elements act by contents, the eigenvalue of ${\xi }_k$ on its Gelfand–Tsetlin basis vector isIf $\mu$ is not a hook partition, it contains $(2,2)$, whose content is zero, so this product is zero. If $\mu =(a+1,1^b)$ with $a+b+1=k$, its nonzero contents are $1,\dots ,a$ and $-1,\dots ,-b$. Their product is independent of the order of the labels. Therefore the eigenvalue isThe empty product at $k=1$ gives $1$. When $k=n$, $\mu =\lambda$, giving exactly the hook-versus-nonhook distinction suggested by the hint. For $k<n$, it is the intermediate shape $\mu$, rather than the final shape $\lambda$, that controls the answer.

There are $(k-1)!$ $k$-cycles in $S_k$, and each has the same character value of a representation on $S^{\lambda }$. Taking traces in the cycle-sum identity for Young–Jucys–Murphy elements yieldsOnly tableaux whose first $k$ entries form a hook partition contribute.

Write such a shape as ${\mu }_b=(k-b,1^b)$, for $0\le b\le k-1$. There are $f_{{\mu }_b}=\left(\genfrac{}{}{0ex}{}{k-1}{b}\right)$ ways to fill it: choose which $b$ of the labels $2,\dots ,k$ go below the first box. There are $f^{\lambda /{\mu }_b}$ ways to fill the remaining skew diagram with the labels $k+1,\dots ,n$, in standard order. The eigenvalue from part (ii) is $(-1{)}^b(k-b-1)!b!$, so the factoris $1$. Therefore the general character value isHere $f^{\lambda /\mu }$ counts standard skew Young tableaux, and is zero when $\mu$ is not contained in $\lambda$. In particular, for a full $n$-cycle,For $k=1$, the general formula gives $f^{\lambda /(1)}=f_{\lambda }$, the character value at the identity, as it should.

Set $L_i={\lambda }_i+k-i$, for $1\le i\le k$. These are distinct nonnegative integers with $L_1>\cdots >L_k$. Using the falling factorial, the matrix entry can be writtenWhen $k-j>L_i$, the falling factorial vanishes, agreeing with the stipulated convention for a negative factorial argument.

The column polynomials $(X{)}_{k-1},(X{)}_{k-2},\dots ,(X{)}_0$ are monic of successive descending degrees. Subtracting lower-degree combinations of columns turns them into $X^{k-1},X^{k-2},\dots ,1$ without changing the determinant. The Vandermonde determinant therefore givesThe positive sign comes from reversing the usual increasing-degree Vandermonde columns at the same time as writing the differences as $L_i-L_j$.

We now match this denominator with the hook-length formula. The hook lengths in row $i$ are preciselyTo verify the claim, the first hook in the row has length $L_i$. As the column index passes from $j$ to $j+1$, its hook length drops by $1$ plus the number of lower rows ending at column $j$. The skipped lengths are exactly $L_i-L_r$ for those rows: if ${\lambda }_r=j$, then $L_i-L_r={\lambda }_i-j+r-i$. At the right-hand end, lower rows having the same length as row $i$ account for the missing smallest lengths by the same formula. This describes all omitted values, and the remaining count is $L_i-(k-i)={\lambda }_i$.

ConsequentlyCombining the two identities with the hook-length formula proves the determinantal dimension formula

For a standard Young tableau $T$, let $c_T(r)=j-i$ when its entry $r$ lies in cell $(i,j)$. Its content vector of a standard Young tableau is $C(T)=(c_T(1),\dots ,c_T(n))$. We use the following sign convention for axial distance in a Young tableau:This is the axial distance from $j$ to $i$. In particular, write $d=c_T(r+1)-c_T(r)$ for the adjacent transposition $s_r=(rr+1)$.

Here is Young orthogonal form, with one orthonormal basis vector $w_T$ for each standard Young tableau of a fixed shape:If $s_{r}T$ is not standard, omit that term. In this case $r,r+1$ occupy adjacent cells in one row or column, so $d=1$ or $d=-1$, respectively, and the scalar action is $+1$ or $-1$. On a standard pair $T,R=s_{r}T$, $|d|\ge 2$ and the matrix isThis is a real orthogonal involution.

To prove the formula, explicitly use the permitted Young seminormal form theorem: the Specht module has a tableau basis $v_T$ whose vectors are joint eigenvectors of the Young–Jucys–Murphy elements, and whose actions, on an admissible pair oriented so that tableau length increases from $T$ to $R$, areNonadmissible swaps have the scalar actions just described. The theorem supplies these as one globally consistent representation, not as independently chosen two-dimensional blocks.

Average a positive definite Hermitian inner product over $S_n$. Distinct tableau lines are orthogonal because their commuting self-adjoint Young–Jucys–Murphy elements have distinct joint weights. Also $s_r$ is self-adjoint, so the two seminormal equations implyUpon setting $w_T=v_T/\Vert v_T\Vert$, the coefficient from $w_T$ to $w_R$ becomes $\Vert v_R\Vert /\Vert v_T\Vert =\sqrt{1-d^{-2}}$; the reverse coefficient is the same. All phases are consistent because the seminormal basis was global, and the group relations remain true under this change of basis. This proves Young orthogonal form.

For $\lambda =(n)$, there is exactly one standard Young tableau, the row $1,2,\dots ,n$. Its content vector of a standard Young tableau isEvery adjacent distance is $1$, so $s_r{w}_T=w_T$ for every $r$. Thus $\boxed{S^{(n)}\cong {\mathbb{C}}_{\mathrm{triv}}}$ is the trivial representation.

For $\lambda =(1^n)$, there is exactly one standard Young tableau, the column with entries $1,2,\dots ,n$ from top to bottom. Its content vector of a standard Young tableau isEvery adjacent distance is $-1$, so Young orthogonal form gives $s_r{w}_T=-w_T$ for every $r$. Thereforethe sign representation.

Assume $n\ge 2$. The standard Young tableaux of shape $(n-1,1)$ are $T_k$, for $2\le k\le n$: the sole second-row entry is $k$, and the first row isThis lists all of them, since $1$ must occupy the first box and the rest of the first row is forced to increase. Their contents areWrite $w_k=w_{T_k}$. The only nontrivial adjacent swaps exchange the lower entry with $k-1$ or $k+1$. Substituting the content differences into Young orthogonal form yields all generator actions:In the first line at $k=2$, the square-root coefficient is zero and the action is $-w_2$, so no vector $w_1$ is needed. The second line is used only when $k<n$, since the generator $s_n$ does not exist.

The corresponding Specht module is the sum-zero standard representation of the symmetric group:A concrete orthonormal realization isPermuting coordinates gives exactly the actions above, linking this orthogonal tableau basis with the Gelfand–Tsetlin basis found in Question 1. Here the printed $s_j$ are the adjacent Coxeter generators, rather than products of all generators.

A near Young tableau is a filling of a Young diagram by distinct entries from a totally ordered alphabet, increasing along rows and down columns; the entries need not be $1,\dots ,n$. For row insertion of $x$, scan the first row for its leftmost entry larger than $x$. If one exists, replace it by $x$ and insert the displaced entry into the next row by the same rule. Otherwise append $x$ to the row and stop. Continue until a new outer-corner box is appended.