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Specializing a quantum highest weight lattice at q=1

Example

Let ${s l}_{2}$ be the rank-one Cartan datum with $λ (h_{1}) = m \in Z_{\geq 0}$ , and let $L (m)$ be the explicit simple module with basis $v_{0}, \dots, v_{m}$ , $K v_{t} = q^{m - 2 t} v_{t}$ , $F v_{t} = [t + 1]_{q} v_{t + 1}$ , $E v_{t} = [m - t + 1]_{q} v_{t - 1}$ (Quantum Verma and simple modules for sl2, Quantum integers, factorials, Gaussian binomials and divided powers at $q_{i}$ ). Let $A = Z [q, q^{- 1}]$ and let $V_{A} = \sum_{t = 0}^{m} A F^{(t)} v_{0} \subseteq L (m)$ be the lattice generated by the highest weight vector under the divided powers of the integral form $U_{A}$ (The Lusztig divided-power integral form $U_{A}$ ). Let $k$ be a field of characteristic $0$ , made an $A$ -algebra by $q \mapsto 1$ , and put $\overset{ˉ}{V} = V_{A} \otimes_{A} k$ .

Then $V_{A}$ is free over $A$ with basis $F^{(t)} v_{0}$ ( $0 \leq t \leq m$ ), and $\overset{ˉ}{V}$ is a $k$ -vector space of dimension $m + 1$ with basis the images ${\overset{ˉ}{v}}_{t}$ on which $\overset{ˉ}{K} = 1, \overset{ˉ}{F} {\overset{ˉ}{v}}_{t} = (t + 1) {\overset{ˉ}{v}}_{t + 1}, \overset{ˉ}{E} {\overset{ˉ}{v}}_{t} = (m - t + 1) {\overset{ˉ}{v}}_{t - 1}, {\overset{ˉ}{T}}_{0, 1} {\overset{ˉ}{v}}_{t} = (m - 2 t) {\overset{ˉ}{v}}_{t},$ with ${\overset{ˉ}{v}}_{- 1} = {\overset{ˉ}{v}}_{m + 1} = 0$ . This is exactly the classical $(m + 1)$ -dimensional simple representation of ${s l}_{2}$ on the coroot basis: ${\overset{ˉ}{T}}_{0, 1}$ is the Cartan generator $h$ acting by the weight $m - 2 t$ , and $\overset{ˉ}{E}, \overset{ˉ}{F}$ are the classical raising and lowering matrices, so the weight multiplicities are $\dim_{k} {\overset{ˉ}{V}}_{μ} = 1$ for $μ = m, m - 2, \dots, - m$ and $0$ otherwise. For $m = 2$ the specialized matrices are $\overset{ˉ}{E} {\overset{ˉ}{v}}_{0} = 0$ , $\overset{ˉ}{E} {\overset{ˉ}{v}}_{1} = 2 {\overset{ˉ}{v}}_{0}$ , $\overset{ˉ}{E} {\overset{ˉ}{v}}_{2} = {\overset{ˉ}{v}}_{1}$ and $\overset{ˉ}{F} {\overset{ˉ}{v}}_{0} = {\overset{ˉ}{v}}_{1}$ , $\overset{ˉ}{F} {\overset{ˉ}{v}}_{1} = 2 {\overset{ˉ}{v}}_{2}$ , $\overset{ˉ}{F} {\overset{ˉ}{v}}_{2} = 0$ , with ${\overset{ˉ}{T}}_{0, 1} = diag (2, 0, - 2)$ , the standard adjoint-type triple.

Facts & Assumptions

Given: The rank-one datum, the simple module $L (m)$ with its string basis, the lattice $V_{A}$ , and the specialization $A \to k$ , $q \mapsto 1$ .

[F1]

$F^{(t)} v_{0} = \frac{1}{[t]!} F^{t} v_{0}$ , the vectors $F^{t} v_{0}$ ( $0 \leq t \leq m$ ) are $Q (q)$ -independent, and the action on the string is $K v_{t} = q^{m - 2 t} v_{t}$ , $F v_{t} = [t + 1]_{q} v_{t + 1}$ , $E v_{t} = [m - t + 1]_{q} v_{t - 1}$ (Quantum Verma and simple modules for sl2, Quantum integers, factorials, Gaussian binomials and divided powers at $q_{i}$ ).

[F2]

Gaussian binomials with nonnegative upper entry lie in $A$ (The quantum Pascal recurrences, the Gauss product formula and Gaussian integrality). Also $[t]! \in A$ and $[t]! \neq 0$ ; the divided powers $E^{(r)}, F^{(r)}$ , the $K^{\pm 1}$ and the $T_{c, t}$ generate $U_{A}$ ; on a weight vector of weight $μ$ the toral binomial $T_{c, t}$ acts by the Gaussian value ${(\binom{μ (h) + c}{t})}_{q}$ , preserving any stable lattice to which that supplier applies (The Lusztig divided-power integral form $U_{A}$ , The q=1 toral specialization identifies the classical Cartan generators (b)).

[F3]

On a $U_{A}$ -stable label-graded lattice the specialization has $\overset{ˉ}{K} = i d$ , ${\overset{ˉ}{T}}_{i, c, t}$ acting by the integer binomials and the classical Cartan commutators; the specialization is defined by base change of the integral form, never by substituting into rational expressions (The q=1 toral specialization identifies the classical Cartan generators (b), Specialization at q=1 recovers the classical enveloping relations (c),(d)).

[F4]

The local deformation lemma Local q=1 deformation proves integrable highest weight universality (a) gives equality of quantum and classical highest-weight characters. Its part (c) identifies the $q = 1$ fibre of the specified ordinary-word lattice $V_{R} = \sum_{w} R F_{w} v_{0}$ , over $R = Q [q]_{(q - 1)}$ , with the classical simple module. In rank one the quantum string is the $(m + 1)$ -dimensional module of Quantum Verma and simple modules for sl2 (a); the actual lattice comparison is checked below.

Verification

technique · check the lattice is preserved by the integral generators with A-valued matrices, tensor with k, and compare the resulting matrices with the classical rank-one simple representation

1.1F1F2F3algebra

The lattice. Put $v_{t} = F^{(t)} v_{0}$ , $0 \leq t \leq m$ . These vectors are $Q (q)$ -independent by [F1], so $V_{A} = ⨁_{t = 0}^{m} A v_{t}$ is free. For $r \geq 0$ , repeated application of the string formulas and division by $[r]!$ give $F^{(r)} v_{t} = {(\binom{t + r}{r})}_{q} v_{t + r}$ if $t + r \leq m$ , and $0$ otherwise; likewise $E^{(r)} v_{t} = {(\binom{m - t + r}{r})}_{q} v_{t - r}$ if $r \leq t$ , and $0$ otherwise. The coefficients are Laurent integral by the Gaussian identities. Also $K^{\pm 1} v_{t} = q^{\pm (m - 2 t)} v_{t}$ , and $T_{c, s} v_{t} = {(\binom{m - 2 t + c}{s})}_{q} v_{t} \in A v_{t}$ , including negative upper labels by the generalized-binomial identity of [F3]'s supplier. Thus every generator of $U_{A}$ preserves $V_{A}$ . Each $v_{t}$ is obtained by a divided power from $v_{0}$ , so $V_{A} = U_{A} v_{0}$ .

1.2F1F2F3algebra

Specialization. Base change of the explicit free basis to $k$ gives $\overset{ˉ}{V} = V_{A} \otimes_{A} k$ has the $k$ -basis ${\overset{ˉ}{v}}_{0}, \dots, {\overset{ˉ}{v}}_{m}$ and dimension $m + 1$ . Substituting $q = 1$ is legitimate here because the action matrices in the basis $F^{(t)} v_{0}$ are known explicitly with entries in $A$ : $\overset{ˉ}{K} = i d$ , $\overset{ˉ}{F} {\overset{ˉ}{v}}_{t} = (t + 1) {\overset{ˉ}{v}}_{t + 1}$ , $\overset{ˉ}{E} {\overset{ˉ}{v}}_{t} = (m - t + 1) {\overset{ˉ}{v}}_{t - 1}$ and ${\overset{ˉ}{T}}_{c, s} {\overset{ˉ}{v}}_{t} = (\binom{m - 2 t + c}{s}) {\overset{ˉ}{v}}_{t}$ , in particular ${\overset{ˉ}{T}}_{0, 1} {\overset{ˉ}{v}}_{t} = (m - 2 t) {\overset{ˉ}{v}}_{t}$ ; this is the operator form of [F3] on the graded lattice.

2.1F2F3F4step 1.1step 1.2algebra∎

Comparison with the classical module. The displayed operators satisfy $[\overset{ˉ}{E}, \overset{ˉ}{F}] = {\overset{ˉ}{T}}_{0, 1}$ , $[{\overset{ˉ}{T}}_{0, 1}, \overset{ˉ}{E}] = 2 \overset{ˉ}{E}$ and $[{\overset{ˉ}{T}}_{0, 1}, \overset{ˉ}{F}] = - 2 \overset{ˉ}{F}$ , including the string ends, so they define a classical ${s l}_{2}$ -module. It is simple: if $x = \sum_{t = 0}^{T} c_{t} {\overset{ˉ}{v}}_{t} \neq 0$ with $c_{T} \neq 0$ , then ${\overset{ˉ}{E}}^{T} x = c_{T} \prod_{j = 1}^{T} (m - j + 1) {\overset{ˉ}{v}}_{0} \neq 0$ in characteristic zero, and ${\overset{ˉ}{F}}^{t} {\overset{ˉ}{v}}_{0} = t! {\overset{ˉ}{v}}_{t}$ generates every basis vector. The empty product covers $T = 0$ and $m = 0$ . This is the simple highest-weight module of weight $m$ . To apply F4 to this lattice, note that $[t]!$ is an $R$ -unit since its value at $q = 1$ is $t! \neq 0$ . Thus $R \otimes_{A} V_{A} = \sum_{t = 0}^{m} R F^{t} v_{0} = V_{R}$ . Since $A \to k$ factors through $R$ , base change of its specified fibre gives precisely $\overset{ˉ}{V}$ ; F4 separately gives the character comparison. The $m = 2$ matrices follow from step 1.2. Freeness here is proved by the explicit string basis, without global freeness or choice.

Depends on

Used by

Direct q=1 substitution in quantum denominators Counterexample

Dependency tree · two levels

34 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

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)