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Direct q=1 substitution in quantum denominators

Statement refuted

False claim: if $x \in U_{A}$ is written as a rational expression in $q$ (with numerator and denominator in $U_{A}$ ), then the operator induced by $x$ on the $q = 1$ specialization of a $U_{A}$ -stable lattice is computed by substituting $q = 1$ into the numerator and denominator separately.

The rule fails already for the first toral element: $T_{0, 1} = (K - K^{- 1}) / (q - q^{- 1})$ has numerator and denominator both vanishing under separate substitution, while the actual specialized operator is the classical coroot action on the weight lattice.

Facts & Assumptions

Given: The rank-one quantum group $U_{q} ({s l}_{2})$ , the integral form $U_{A}$ over $A = Z [q, q^{- 1}]$ , and the rank-one simple modules.

[F1]

$T_{0, 1} = \frac{K - K^{- 1}}{q - q^{- 1}}$ is a generator of the integral form $U_{A}$ ; on a weight vector of weight $μ$ it acts by the Gaussian value $[μ (h_{1})]_{q} = \frac{q^{μ (h_{1})} - q^{- μ (h_{1})}}{q - q^{- 1}}$ , and at $q = 1$ this value is the integer $μ (h_{1})$ (The Lusztig divided-power integral form $U_{A}$ , The q=1 toral specialization identifies the classical Cartan generators (b), Quantum integers, factorials, Gaussian binomials and divided powers at $q_{i}$ ).

[F2]

In the two-dimensional simple module $L (1)$ with basis $v_{0}, v_{1}$ one has $K v_{0} = q v_{0}$ , $K v_{1} = q^{- 1} v_{1}$ , $F v_{0} = v_{1}$ , $F v_{1} = 0$ , $E v_{0} = 0$ , $E v_{1} = v_{0}$ ; in particular $T_{0, 1}$ acts as $diag (1, - 1)$ (Quantum Verma and simple modules for sl2 (a) at $m = 1$ ).

[F3]

The string formula $E_{i} F_{i}^{(b)} = F_{i}^{(b)} E_{i} + \frac{F_{i}^{(b - 1)}}{q^{b} - q^{- b}} \sum_{s = 0}^{b - 1} (q^{- 2 s} K - q^{2 s} K^{- 1})$ , and on a weight vector $w$ of weight $μ$ with $E_{i} w = 0$ it gives $E_{i} F_{i}^{(b)} w = [μ (h_{i}) - b + 1]_{q} F_{i}^{(b - 1)} w$ , while separate substitution into the displayed fraction gives $0 / 0$ (Divided-power commutation and the simple $U_{q_{i}} ({s l}_{2})$ string modules (ii), The Lusztig divided-power integral form $U_{A}$ ).

[F4]

Base change $U_{A} \otimes_{A} k$ with $q \mapsto 1$ gives well-defined images of integral elements. The classical Cartan and Serre relations hold in its positive quotient by all $K_{h} - 1$ , and hence as operators on every $U_{A}$ -stable label-graded lattice, where ${\overset{ˉ}{K}}_{h} = i d$ (Specialization at q=1 recovers the classical enveloping relations (b),(c)). The full unquotiented specialization retains other toral characters, as stated in (d). The rank-one lattice of Specializing a quantum highest weight lattice at q=1 meets the grading and stability hypotheses and has the stated classical coroot action.

Counterexample

1.1F1algebra

The element and its two evaluations. Let $x = T_{0, 1} = \frac{K - K^{- 1}}{q - q^{- 1}}$ . By [F1], $x \in U_{A}$ and on a weight vector of weight $μ$ it acts by the scalar $[μ (h_{1})]_{q}$ , whose value at $q = 1$ is the integer $μ (h_{1})$ . The separate substitution $K \mapsto 1$ , $q \mapsto 1$ into the written expression gives $0 / 0$ : the numerator $K - K^{- 1}$ and the denominator $q - q^{- 1}$ both specialize to $0$ and no value is defined. This already refutes the displayed rule.

1.2F1F2algebra

Explicit witness on $L (1)$ . By [F2], in the two-dimensional module $L (1)$ the element $K$ acts as $diag (q, q^{- 1})$ and $x = T_{0, 1}$ acts as $diag ([1]_{q}, [- 1]_{q}) = diag (1, - 1)$ , a nonzero operator. Substituting $q = 1$ into the matrix of $K$ gives the identity operator $diag (1, 1)$ , so the operator induced by $x$ is not obtained by substituting $q = 1$ into the numerator and denominator of the rational expression: the naive substitution loses the entire Cartan action.

1.3F1F3algebra

Further witnesses. (i) The Gaussian integers: $[m]_{q} = \frac{q^{m} - q^{- m}}{q - q^{- 1}}$ has numerator and denominator vanishing at $q = 1$ for every $m \neq 0$ , while the correct specialization is the integer $m$ ; for example ${(\binom{2}{1})}_{q} = [2]_{q} = (q^{2} - q^{- 2}) / (q - q^{- 1})$ gives $0 / 0$ by separate substitution, whereas its Laurent expression $q + q^{- 1}$ specializes to $2$ . (ii) The divided-power commutation coefficient $\frac{1}{q^{b} - q^{- b}} \sum_{s = 0}^{b - 1} (q^{- 2 s} K - q^{2 s} K^{- 1})$ of [F3]: its numerator and denominator both vanish at $q = 1$ , yet on a weight vector $w$ of weight $μ$ with $E_{i} w = 0$ its action is the finite value $[μ (h_{i}) - b + 1]_{q} \mapsto μ (h_{i}) - b + 1$ given by the string formula. Each of these elements lies in $U_{A}$ and has a well-defined specialized operator through the integral form.

2.1F1F2F3F4algebra∎

The correct statement. By [F4], specialization is the base change $U_{A} \otimes_{A} k$ with $q \mapsto 1$ : it is defined on elements of the integral form (and on operator matrices on graded lattices, whose entries are Laurent polynomials), and the cancellations in Steps 1.1-1.3 are performed inside $U_{A}$ before any reduction. On the label-graded lattice $A v_{0} \oplus A v_{1}$ of $L (1)$ , the action factors through the positive quotient by $K - 1$ , and $T_{0, 1}$ induces the classical coroot operator $diag (1, - 1)$ by [F4]. The coroot identification is made in that quotient or on these lattice operators; it is not asserted in the full unquotiented algebra. No choice is used; every witness is an explicit finite matrix computation.

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Sources

Richard Borcherds, Mark Haiman, Theo Johnson-Freyd, Nicolai Reshetikhin and Vera Serganova, Berkeley Lectures on Lie Groups and Quantum Groups (book-length lecture notes, last updated 18 January 2024) (standard reference, not scraped)
Math 263C Quantum Groups (Stanford, 2022), lecture slides 'Lusztig's quantum group. Tilting modules. The Fusion Category in Action' (standard reference, not scraped)
George Lusztig, Introduction to Quantum Groups (Progress in Mathematics 110, Birkhauser, 1993; second printing 1994; author-hosted scan) (standard reference, not scraped)