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For the joint spectrum of the Young–Jucys–Murphy elements, adjacent coordinates $a_i,a_{i+1}$ are distinct. If $a_{i+1}-a_i=\pm 1$, the adjacent transposition $s_i$ acts on that line by $\pm 1$. Otherwise interchanging the coordinates gives a spectral vector in the same irreducible representation. These rules follow from $s_i{X}_i{s}_i+s_i=X_{i+1}$ and its two-dimensional eigenspace calculation. Together with the braid relation in a Coxeter group, they exclude the consecutive patterns $(a,a\pm 1,a)$.

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.

Induct on the distance between equal entries $a_i=a_j=a$ of a point of the joint spectrum. Consecutive equal entries are impossible by the local spectral rules for Young–Jucys–Murphy elements. It suffices to consider consecutive occurrences of $a$: proving the assertion for each such pair proves it for any wider pair.

Suppose $a-1$ is missing between them. There can be at most one occurrence of $a+1$ in the interval. Otherwise a pair of consecutive $a+1$ entries has a shorter gap, and the induction hypothesis forces an intervening $a$, contrary to our choice of the pair. Move the two endpoint entries $a$ inward past entries different from $a,a-1,a+1$. Every move is an admissible spectral interchange. If there is no $a+1$, this creates consecutive equal entries. If there is one, this creates $(a,a+1,a)$, already excluded by the braid relation in a Coxeter group. Both are impossible.

Interchanging the roles of $a-1$ and $a+1$ excludes the omission of $a+1$ as well. Therefore

First exclude the short patterns $(a,a+1,a)$ and $(a,a-1,a)$ in the joint spectrum. In the first pattern, the local spectral rules for Young–Jucys–Murphy elements make $s_r$ act by $+1$ and $s_{r+1}$ by $-1$ on the same simultaneous eigenvector. The two sides of the braid relation in a Coxeter group then act by opposite signs. The second pattern gives the same contradiction with signs reversed.

Now induct on the length of a spectral vector. A prefix of length $i-1$ is spectral for $S_{i-1}$: decompose the restricted group representation and retain a nonzero component of the simultaneous eigenvector. If $a_i$ differed by neither $+1$ nor $-1$ from every earlier entry, move it left using the allowed adjacent interchanges. Encountering an equal entry would contradict the distinctness of consecutive eigenvalues. Otherwise it reaches the first position, forcing $a_i=0$. For $i>1$ the original first entry was also zero, so an equal entry would indeed have been encountered. ThusThis argument uses only the stated local rules. In particular, iterating the result from $a_1=0$ also shows that all coordinates are integers.

The exchange condition for a Coxeter group says that if $w=s_1\cdots s_n$ is reduced and $s$ is simple with $\ell (ws)<\ell (w)$, thenfor some $j$. The Matsumoto theorem says that any two reduced expressions for the same element are connected by braid moves. Together they imply the Tits word reduction theorem: a nonreduced word can be transformed by braid moves until two equal adjacent generators can be cancelled.

For the displayed four-armed graph, call the central generator $s$ and the leaves $a,b,c,d$. Different leaves commute, while each leaf $t$ satisfies $sts=tst$. Consider the wordBetween successive occurrences of $s$, the intervening leaf sets alternate between $\{a,b\}$ and $\{c,d\}$. Commuting the two leaves in one block never puts the same leaf on both sides of an $s$, so no length-three braid $tst\leftrightarrow sts$ is ever available. The only possible braid moves are those leaf commutations, and they cannot create adjacent equal letters. Tits reduction therefore shows that $u_N$ is reduced. Since $\ell (u_N)=6N$ is unbounded, the group is infinite.

If $m_{ij}<\infty$, the braid relation in a Coxeter group isFor $m_{ij}=2$ this is commutation; no relation is imposed when $m_{ij}=\infty$. Two reduced expressions are braid equivalent when one can be transformed into the other by finitely many replacements of one side of such a relation by the other.