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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 $(1,1)$.

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.

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 $\lambda =(1^n)$, 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.

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.