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Classical commutants bound the two generic tensor images

Statement

Let $N \geq 2$ , $n \geq 1$ , $V_{F} = Q (q)^{N}$ , and $M_{F} = V_{F}^{\otimes n}$ . Let $A_{F}$ and $B_{F}$ be respectively the images of the quantum algebra from The type-A quantum algebra and its tensor-action formulas and the normalized Hecke algebra action from The ordered-basis symmetry gives the normalized Hecke tensor action. Let $A_{0}$ be the unital algebra generated by the classical operators $ρ_{n} (a)$ of Classical tensor centralizer from polarization and Newton identities, and let $B_{0}$ be the place-permutation image of $Q [S_{n}]$ . The regular quantum and Hecke matrix families below specialize to generating families for these two rational algebras. Then $\dim_{Q} A_{0} \leq \dim_{F} A_{F}, \dim_{Q} B_{0} \leq \dim_{F} B_{F},$ $\dim_{F} {End}_{B_{F}} (M_{F}) \leq \dim_{Q} A_{0}, \dim_{F} {End}_{A_{F}} (M_{F}) \leq \dim_{Q} B_{0} .$ These are separate image lower bounds and commutant upper bounds; the commuting inclusion needed to turn them into equalities is proved in The quantum and Hecke tensor actions commute.

Facts & Assumptions

Given: The quantum and Hecke actions above, their local relation-descent proofs, and the three local rank/centralizer suppliers below.

[F1]

The specialized classical quantum image satisfies $A_{0} = {End}_{B_{0}} (M_{0})$ (Classical tensor centralizer from polarization and Newton identities).

[F2]

Over $Q$ , the finite permutation image satisfies $B_{0} = {End}_{A_{0}} (M_{0})$ (Finite semisimple matrix density and the double commutant over Q).

[F3]

A finite matrix over $Q [q, q^{- 1}]_{(q - 1)}$ has generic rank at least its rank at $q = 1$ ; this gives both word-image lower bounds and commutant-kernel upper bounds (Finite matrix rank bounds under regular specialization).

Proof

Given: The data in [F1]–[F3].

1.1givenF1F2algebra

Put $R = Q [q, q^{- 1}]_{(q - 1)}$ and $M_{R} = (R^{N})^{\otimes n}$ , a finite free module with fraction-field fiber $M_{F}$ and residue-field fiber $M_{0}$ . All Hecke matrices and all iterated quantum matrices are regular over $R$ by their explicit formulas. For the quantum family, also use $H_{j} = (L_{j} - I) / (q - 1)$ . On a basis word with $m_{j}$ occurrences of $j$ , $L_{j}$ has eigenvalue $q^{m_{j}}$ , so $H_{j}$ is the diagonal matrix with entry $1 + q + \dots + q^{m_{j} - 1}$ if $m_{j} > 0$ and $0$ if $m_{j} = 0$ . Thus $H_{j}$ is regular and specializes to the classical diagonal count operator $ρ_{n} (E_{j j})$ . The generators $E_{i}, F_{i}, H_{j}$ specialize to the adjacent matrix Lie operators and all diagonal count operators, which generate $A_{0}$ by the last conclusion of Classical tensor centralizer from polarization and Newton identities; $L_{j}^{\pm 1}$ specialize to the identity.

2.1givenF1F2F3step 1.1algebra

In a basis of $M_{R}$ , vectorize $X \in End (M_{R})$ . The simultaneous commutator system for the finite family $T_{1}, \dots, T_{n - 1}$ is a finite matrix over $R$ , and its kernel over $F$ is ${End}_{B_{F}} (M_{F})$ while its kernel at $q = 1$ is ${End}_{B_{0}} (M_{0})$ . By [F3] its generic kernel dimension is at most the specialized one, which equals $\dim_{Q} A_{0}$ by [F1]. The same construction for the finite family $E_{i}, F_{i}, L_{j}^{\pm 1}, H_{j}$ has generic kernel ${End}_{A_{F}} (M_{F})$ ; at $q = 1$ the family generates $A_{0}$ , so its specialized kernel is ${End}_{A_{0}} (M_{0})$ , of dimension $\dim_{Q} B_{0}$ by [F2]. This proves both commutant upper bounds independently.

3.1givenF1F2F3step 1.1algebra∎

The algebra $A_{0}$ is finite-dimensional and generated by the specializations at $q = 1$ of the regular quantum family in step 1.1. Words in that finite family span it, so select a finite word basis. The same words in the regular generic matrices have a vectorized-column matrix over $R$ whose specialization has full column rank; [F3] gives their $F$ -linear independence. Hence $\dim_{F} A_{F} \geq \dim_{Q} A_{0}$ . Words in the finite family $T_{1}, \dots, T_{n - 1}$ similarly specialize to a spanning family of $B_{0}$ ; selecting a basis and applying [F3] gives $\dim_{F} B_{F} \geq \dim_{Q} B_{0}$ . Only these chosen regular words are specialized; no arbitrary rational denominator is evaluated at $q = 1$ .

Depends on

Used by

Generic quantum Schur–Weyl mutual centralizers over Q(q) Theorem

Dependency tree · two levels

21 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Alexei Davydov and Alexander Molev, A categorical approach to classical and quantum Schur–Weyl duality (standard reference, not scraped)
Richard Dipper, Stephen Doty and Friederike Stoll, Quantized mixed tensor space and Schur–Weyl duality (standard reference, not scraped)