= Solution
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. Thus
$$
\boxed{\{a_i-1,a_i+1\}\cap\{a_1,\ldots,a_{i-1}\}\ne\varnothing.}
$$
This argument uses only the stated local rules. In particular, iterating the result from $a_1=0$ also shows that all coordinates are <integers>.
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