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Integral residues of the negative half: freeness, the crystal basis and the form characterization

Statement

Let $L_{Z} (\infty) = U_{A}^{-} \cap L (\infty)$ be the integral residue lattice of The integral residue lattices $L_{A} (\infty)$ and $L_{A} (λ)$ , with $L_{Z} / q L_{Z} \to L (\infty) / q L (\infty)$ the canonical injection. Then the image of this injection is exactly $⨁_{b \in B (\infty)} Z b$ ; in particular $L_{Z} / q L_{Z}$ is a free $Z$ -module with basis $B (\infty)$ , and after tensoring with $Q$ it is the full rational crystal space $Q \otimes_{Z} ⨁_{b} Z b = L (\infty) / q L (\infty)$ .

The normalized Kashiwara half form of The negative-half bilinear form is symmetric, nondegenerate and integral takes values on $U_{A}^{-}$ in $K_{Z} = Z [q, q^{- 1}, (1 - q^{2 n})^{- 1} : n \geq 1]$ and on $L_{Z}$ in $A_{Z} = Z [q, (1 - q^{2 n})^{- 1} : n \geq 1]$ ; consequently its residue pairing on $L_{Z} / q L_{Z}$ is $Z$ -valued. In the residue basis $B (\infty)$ the coefficients of an integral element are its integer pairings with the basis vectors, and the norm- $1$ elements of the integral residue $L_{Z} / q L_{Z}$ are exactly $B (\infty) \cup (- B (\infty))$ . The Kashiwara operators ${\tilde{e}}_{i}, {\tilde{f}}_{i}$ preserve $L_{Z}$ , hence induce endomorphisms of the integral residue subgroup. No $A$ -module structure on $L_{Z}$ , no identification of the integral residue with the rational residue as a $Z$ -module, and no unsigned norm- $1$ characterization is asserted; the module integral residue basis is established only in the globalization theorem of this page.

Facts & Assumptions

Given: The negative half $U_{q}^{-}$ with its divided-power form $U_{A}^{-}$ , the crystal lattice $L (\infty)$ and residue basis $B (\infty)$ , and the integral residue lattice $L_{Z} (\infty) = U_{A}^{-} \cap L (\infty)$ .

[F1]

Integral residue lattice (The integral residue lattices $L_{A} (\infty)$ and $L_{A} (λ)$ ): $L_{Z}$ is a $Z [q]$ -module, $M_{A} \cap q L = q L_{Z}$ with $M_{A} = U_{A}^{-}$ , and $L_{Z} / q L_{Z} \to L (\infty) / q L (\infty)$ is injective; also $\overline{L_{Z}} = U_{A}^{-} \cap ψ (L (\infty))$ .

[F2]

Integral skew-string coefficients (Integral skew-string coefficients on the divided-power negative half): the string coefficients of the divided-power half are integral and both Kashiwara operators together with the skew derivations preserve $U_{A}^{-}$ .

[F3]

Crystal basis of the negative half (The crystal basis $B (\infty)$ of the negative half): $(L (\infty), B (\infty))$ is a crystal basis, $B (\infty)$ is a basis of $L (\infty) / q L (\infty)$ , and ${\tilde{e}}_{i}, {\tilde{f}}_{i}$ preserve $L (\infty)$ .

[F4]

Polarization values at $q = 0$ (Crystal lattice polarizations are perfect at $q = 0$ ): the normalized Kashiwara form is the $q = 0$ polarization; $B (P, Q) \in A_{0}$ whenever $P, Q \in L (\infty)$ , and the induced form on $L (\infty) / q L (\infty)$ is the positive definite $Q$ -valued form in which $B (\infty)$ is orthonormal.

[F5]

Half form values (The negative-half bilinear form is symmetric, nondegenerate and integral, The Lusztig-Kashiwara bilinear form on the negative half: the recursion): $B$ is the unique normalized homogeneous form with $B (F_{i} P, Q) = B (P, e_{i}^{'} Q)$ ; its divided-power word denominator bound is $\prod_{i} \prod_{r \geq 1} (1 - q_{i}^{2 r})^{- ν_{i}}$ -admissible, and $B (U_{A}^{-}, U_{A}^{-}) \subseteq K_{Z}$ .

[C1]

Choice: the Axiom of Choice is inherited only through the supplied grand-loop and crystal-basis results (The Axiom of Choice); no additional selection is made in this item, and all stated bases are the supplied ones.

Proof

technique · direct

1.1F2F3

Integral representatives of the crystal basis. By [F2] every divided-power lowering word ${\tilde{f}}_{i_{1}} \dots {\tilde{f}}_{i_{r}} 1$ lies in $U_{A}^{-}$ , and by [F3] it lies in $L (\infty)$ ; hence its class lies in $L_{Z} / q L_{Z}$ and is a nonzero element of $B (\infty)$ , and every $b \in B (\infty)$ is represented by such a word.

1.2F4F5algebra

The value rings. The explicit rank-one divided-word norms $B (F_{i}^{(a)}, F_{i}^{(a)}) = 1 / [a]_{q_{i}^{2}}! = \prod_{r = 1}^{a} (1 + q_{i}^{2} + \dots + q_{i}^{2 (r - 1)})^{- 1}$ and the word Gram expansion of [F5] show that every value on $U_{A}^{-}$ is a $Z [q, q^{- 1}]$ -combination of reciprocal products $\prod (1 - q_{i}^{2 r})^{- 1}$ with bounded $r$ , i.e. lies in $K_{Z}$ ; and every value on $L_{Z} = U_{A}^{-} \cap L (\infty)$ lies in $A_{0} \cap K_{Z}$ by [F4]. The intersection is $A_{Z}$ : clearing denominators in $A_{0} \cap K_{Z}$ writes the value as a Laurent polynomial with denominator $\prod (1 - q^{2 n})^{e}$ of constant term $1$ , regularity at $q = 0$ forces the Laurent polynomial to have no negative powers, and evaluation at $0$ gives an integer. Hence the residue pairing of $L_{Z} / q L_{Z}$ is $Z$ -valued.

2.1F1F4F5step 1.1

Integrality of coefficients against representatives. For $x \in L_{Z}$ with rational residue $\sum_{b} a_{b} b \in L (\infty) / q L (\infty)$ , choose for each $b$ an integral representative $w_{b}$ of $b$ from step 1.1 and pair modulo $q$ : by [F4] the residue form is the positive definite form with orthonormal basis $B (\infty)$ , so $a_{b} = (x, w_{b})$ modulo $q$ , and by [F5] this value lies in $A_{Z}$ ; evaluating the resulting $Z [q]$ -element at $q = 0$ gives an integer because $A_{Z}$ consists of quotients with denominator $\equiv 1 (m o d q)$ . Hence every coefficient $a_{b}$ is an integer and the image of the injection of [F1] is contained in $⨁_{b} Z b$ ; conversely finite integer combinations of the classes of the $w_{b}$ lie in the image, so the image is exactly $⨁_{b} Z b$ .

3.1F4F5step 2.1algebra

Norm-one residue classes. For an integral residue $\overset{ˉ}{x} = \sum_{b} a_{b} b$ , step 2.1 gives $a_{b} \in Z$ , and [F4] gives $B_{0} (\overset{ˉ}{x}, \overset{ˉ}{x}) = \sum_{b} a_{b}^{2}$ . This equals $1$ exactly when $\overset{ˉ}{x} = \pm b$ for one crystal element. This class statement does not characterize the exact norm of its lattice representatives: in rank one $F^{(2)}$ represents a crystal element but has norm $1 / (1 + q^{2})$ . In particular an exact norm- $1$ lattice vector has a signed basis residue, while the converse need not hold.

4.1F2F3step 1.2step 2.1step 3.1∎

Operator stability and conclusion. By [F2] and [F3] the operators ${\tilde{e}}_{i}, {\tilde{f}}_{i}$ preserve both $U_{A}^{-}$ and $L (\infty)$ , hence preserve $L_{Z} = U_{A}^{-} \cap L (\infty)$ and descend to the integral residue subgroup, which by steps 1.1–1.2 is the free $Z$ -module $⨁_{b} Z b$ ; tensoring with $Q$ gives the full rational crystal space by [F3]. Together with steps 2.1 and 3.1 this proves all the assertions, with the stated $Z$ -valued pairing and signed norm- $1$ characterization.

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Sources

Masaki Kashiwara, On crystal bases of the q-analogue of universal enveloping algebras, Duke Mathematical Journal 63 (1991), 465–516 (standard reference, not scraped)
Kyeonghoon Jeong, Seok-Jin Kang and Masaki Kashiwara, Crystal Bases for Quantum Generalized Kac-Moody Algebras, arXiv:math/0305390 (standard reference, not scraped)