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Residue sequences need not separate when e is at most n

Statement refuted

The hypothesis $e > n$ in Separated Hecke residues give seminormal idempotents and bases cannot be dropped. Take $F = C$ , $q = - 1$ (so $e = e (C, - 1) = 2$ , The quantum characteristic of an invertible Hecke parameter), $n = 3$ and $λ = (2, 1)$ . The two distinct standard tableaux $s = \begin{matrix} 1 & 2 \\ 3 \end{matrix}$ and $t = \begin{matrix} 1 & 3 \\ 2 \end{matrix}$ have contents ${content}_{s} (1), {content}_{s} (2), {content}_{s} (3) = 0, 1, - 1$ and ${content}_{t} (1), {content}_{t} (2), {content}_{t} (3) = 0, - 1, 1$ , and the extended quantum-integer identity gives $[1]_{- 1} = 1 = [- 1]_{- 1}$ , since $[- 1]_{- 1} = - (- 1)^{- 1} [1]_{- 1} = 1$ . Hence $([{content}_{s} (k)]_{q})_{k = 1, 2, 3} = (0, 1, 1) = ([{content}_{t} (k)]_{q})_{k = 1, 2, 3} (q = - 1) :$ the q-content sequences collide although $s \neq t$ , so the separation conclusion of the theorem fails. The difference of the two putative separating eigenvalues at $k = 2$ is $[- 1]_{- 1} - [1]_{- 1} = 0$ . Thus interpolation cannot distinguish the two tableaux: a formula that excludes equal values may still define a polynomial, but it is not a separating projector for these two labels. The attempted generator formula has axial distance $ρ = 2$ and $[ρ]_{- 1} = 0$ . In fact $L_{2}$ has a nontrivial Jordan block on this cell, so this cell has no simultaneous JM-eigenbasis. This explicit example shows that the uniform separation hypothesis $e > n$ cannot be dropped; it does not claim that every individual cell at every parameter with $e \leq n$ lacks a seminormal basis. At the boundary $e = n = 2$ , still over $C$ with $q = - 1$ , the row and column tableaux both have tuple $(0, 1)$ . The global Lagrange products in the theorem are consequently $1$ for both labels, so they are not orthogonal. Thus even the proposed replacement $e \geq n$ fails for the full global-idempotent statement.

Facts & Assumptions

Given: The field $C$ , $q = - 1 \in C^{\times}$ , the primary case $n = 3$ and the boundary test $n = 2$ , the standard tableaux $s, t$ of shape $(2, 1)$ above, and the q-integers of Hecke Jucys-Murphy elements and q-contents.

[F1]

(The quantum characteristic of an invertible Hecke parameter) $e (C, - 1) = 2$ : the least $m \geq 1$ with $[m]_{- 1} = 0$ is $m = 2$ , since $[2]_{- 1} = 1 + q = 0$ and $[1]_{- 1} = 1 \neq 0$ .

[F2]

(Hecke Jucys-Murphy elements and q-contents) The q-content value of an entry is $[content]_{q}$ with $[m]_{q} = 1 + q + \dots + q^{m - 1}$ for $m \geq 0$ and $[- m]_{q} = - q^{- m} [m]_{q}$ for $m > 0$ ; at $q = - 1$ one has $[1]_{- 1} = 1$ and $[- 1]_{- 1} = - (- 1)^{- 1} [1]_{- 1} = 1$ .

[F3]

(Separated Hecke residues give seminormal idempotents and bases) Under the hypothesis $e > n$ the q-content sequence determines the standard tableau, and the denominators of the spectral projectors are nonzero.

Counterexample

1.1given

Contents of the two tableaux: for $s$ with rows $(1, 2)$ and $(3)$ the nodes are $(1, 1), (1, 2), (2, 1)$ for $1, 2, 3$ , giving contents $0, 1, - 1$ ; for $t$ with rows $(1, 3)$ and $(2)$ the nodes are $(1, 1), (2, 1), (1, 2)$ for $1, 2, 3$ , giving contents $0, - 1, 1$ .

2.1step 1.1F1F2

Collision: by [F1] and [F2], $[2]_{- 1} = 0$ and $[1]_{- 1} = [- 1]_{- 1} = 1$ ; hence $[{content}_{s} (1)]_{q} = 0 = [{content}_{t} (1)]_{q}$ , $[{content}_{s} (2)]_{q} = 1 = [{content}_{t} (2)]_{q}$ and $[{content}_{s} (3)]_{q} = 1 = [{content}_{t} (3)]_{q}$ , so the two q-content sequences coincide even though $s \neq t$ .

3.1step 2.1F1F3

Interpolation and axial distance. The eigenvalue difference that would be needed to distinguish $s$ and $t$ at $k = 2$ is $1 - 1 = 0$ . Equal-value exclusion in a polynomial product does not make its literal denominators zero; it instead loses the factor that could distinguish these tuples. No polynomial in the JM operators can provide the two distinct joint-eigenvalue labels used in the separated-range construction. For the swap of $2$ and $3$ , $ρ = 1 - (- 1) = 2$ and $[2]_{- 1} = 0$ , so the displayed separated-range generator formula is undefined here.

4.1givenF2step 2.1step 3.1

Verify the failure of a JM-eigenbasis on this cell directly. Put $m_{s} = 1 + T_{1}$ and $m_{t} = (1 + T_{1}) T_{2}$ modulo the higher row-shape ideal. The quadratic relation gives $m_{s} T_{1} = q m_{s}$ . The higher row symmetrizer is $1 + T_{1} + T_{2} + T_{1} T_{2} + T_{2} T_{1} + T_{1} T_{2} T_{1}$ , so its vanishing in this cell gives $m_{t} T_{1} = - m_{s} - m_{t}$ . At $q = - 1$ , since $L_{2} = q^{- 1} T_{1} = - T_{1}$ , it follows that $m_{s} L_{2} = m_{s}$ and $m_{t} L_{2} = m_{s} + m_{t}$ . Thus $L_{2} - I$ is nonzero and $(L_{2} - I)^{2} = 0$ on this two-dimensional cell. It is not diagonalizable over $C$ , so a simultaneous JM-eigenbasis cannot exist on this cell. Together with step 2.1 this supplies the claimed concrete boundary counterexample, without making a universal claim about all cells outside the separated range.

5.1givenF2F3step 2.1∎

Check the boundary case $n = e = 2$ . The row and column tableaux have contents $(0, 1)$ and $(0, - 1)$ , whose q-content tuples both equal $(0, 1)$ at $q = - 1$ by [F2]. These are all standard tableaux of size two, so the theorem's global sets are $R_{1} = {0}$ and $R_{2} = {1}$ . For either label every inner product in its Lagrange formula is empty, giving $F_{r o w} = F_{c o l u m n} = 1$ . Their product is $1 \neq 0$ in the nonzero algebra $H_{C, - 1} (S_{2})$ , which has basis $1, T_{1}$ . Hence they are not orthogonal, disproving replacement of $e > n$ by $e \geq n$ in the global theorem. This supplements the same-shape collision and Jordan block at size three.

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Sources

Andrew Mathas, Hecke algebras and Schur algebras of the symmetric group, complete prepublication manuscript (1998; expanded published edition: AMS University Lecture Series 15, 1999); complete 79-page manuscript, SHA-256 8cfeb611a95a3ff25cb95ad22bfd3a68475818301c34b04d131a8480a1b997e0 (standard reference, not scraped)