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Reversal preserves the negative-half integral and crystal lattices

Statement

Let $⋆$ be the $Q (q)$ -linear anti-involution of $U_{q}^{-}$ fixing each generator $F_{i}$ , so that for a word $F_{i_{1}} \dots F_{i_{r}}$ one has $(F_{i_{1}} \dots F_{i_{r}})^{⋆} = F_{i_{r}} \dots F_{i_{1}}$ , and let $B$ be the normalized Kashiwara half form of The negative-half bilinear form is symmetric, nondegenerate and integral. Then $⋆$ is a well-defined anti-involution of the Serre quotient, reverses negative words, preserves the divided-power integral form $U_{A}^{-}$ of The Lusztig divided-power integral form $U_{A}$ , commutes with the bar involution $ψ$ , and preserves $B$ : $B (P^{⋆}, Q^{⋆}) = B (P, Q)$ . Consequently $⋆$ preserves the crystal lattice $L (\infty)$ of Crystal lattice polarizations are perfect at $q = 0$ , the integral residue lattice $L_{Z} (\infty) = U_{A}^{-} \cap L (\infty)$ , and the bar lattice $L_{Z}^{-} = U_{A}^{-} \cap ψ (L (\infty))$ of Integral residues of the negative half: freeness, the crystal basis and the form characterization. On the integral crystal residue the induced map permutes $B (\infty) \cup (- B (\infty))$ . Only this signed lattice preservation is asserted, not an unsigned $⋆$ -crystal theorem; the sign is what the gluing of the string-ideal global lifts requires.

Facts & Assumptions

Given: The negative half $U_{q}^{-}$ with generators $F_{i}$ , its divided-power form $U_{A}^{-}$ , the normalized half form $B$ , and the lattices $L (\infty), L_{Z} (\infty)$ .

[F1]

The word Gram formula and norm characterization (Crystal lattice polarizations are perfect at $q = 0$ , The negative-half bilinear form is symmetric, nondegenerate and integral): $B$ is the symmetric normalized form with $B (F_{i} P, Q) = B (P, e_{i}^{'} Q)$ ; its value on two words is a finite sum over color-matching permutations with crossing exponents; $L (\infty) = {P : B (P, P) \in A_{0}} = {P : B (P, L (\infty)) \subseteq A_{0}}$ and $B$ is positive definite on $L (\infty) / q L (\infty)$ with orthonormal basis $B (\infty)$ .

[F2]

Integral residues (Integral residues of the negative half: freeness, the crystal basis and the form characterization): $L_{Z} (\infty)$ is a $Z [q]$ -module whose residue image is $⨁_{b \in B (\infty)} Z b$ , the residue pairing is $Z$ -valued, and the norm- $1$ elements of the integral residue are exactly the signed basis classes $B (\infty) \cup (- B (\infty))$ modulo $q L_{Z}$ .

[F3]

Divided-power form (The Lusztig divided-power integral form $U_{A}$ ): $U_{A}^{-}$ is the $A$ -span of finite products of divided powers $F_{i}^{(r)}$ , graded by $- Q_{+}$ .

Proof

technique · direct

1.1F3algebra

$⋆$ is a well-defined anti-involution. The assignment $F_{i} \mapsto F_{i}$ extended antimultiplicatively is defined on the free algebra, and applying it to a negative Serre relation $\sum_{r} (- 1)^{r} {(\binom{1 - a_{i j}}{r})}_{i} F_{i}^{1 - a_{i j} - r} F_{j} F_{i}^{r} = 0$ and reversing the order of each monomial gives $\sum_{r} (- 1)^{r} {(\binom{1 - a_{i j}}{r})}_{i} F_{i}^{r} F_{j} F_{i}^{1 - a_{i j} - r}$ , which is $(- 1)^{1 - a_{i j}}$ times the same relation after the substitution $r \mapsto 1 - a_{i j} - r$ and the exact symmetry ${(\binom{m}{r})}_{i} = {(\binom{m}{m - r})}_{i}$ of the Gaussian binomials. Hence $⋆$ descends to $U_{q}^{-}$ , fixes $F_{i}$ , and is an anti-involution: it reverses words and its square is the identity.

1.2F3algebra

Integral and bar compatibility. Since $⋆$ fixes each divided power $F_{i}^{(r)}$ (it fixes $F_{i}$ and the scalar $[r]_{i}!$ ), it maps each generator of $U_{A}^{-}$ to a generator, so $⋆ (U_{A}^{-}) = U_{A}^{-}$ . It commutes with $ψ$ because $⋆$ is $Q (q)$ -linear while $ψ$ sends $q \mapsto q^{- 1}$ , and on generators $ψ ⋆ (F_{i}) = ψ (F_{i}) = F_{i} = ⋆ ψ (F_{i})$ .

2.1F1step 1.1algebra

Isometry for the half form. Write $P = F_{i_{1}} \dots F_{i_{r}}$ , $Q = F_{j_{1}} \dots F_{j_{s}}$ as words. Reversing both words, $P^{⋆} = F_{i_{r}} \dots F_{i_{1}}$ , $Q^{⋆} = F_{j_{s}} \dots F_{j_{1}}$ , and the word Gram formula of [F1] expands both $B (P, Q)$ and $B (P^{⋆}, Q^{⋆})$ as sums over bijections between matched colors with the same crossing exponents read backwards; reversing a permutation preserves the weight and the crossing number, so the two sums are termwise equal. Hence $B (P^{⋆}, Q^{⋆}) = B (P, Q)$ on words and, by bilinearity and descent through the Serre ideal, on all of $U_{q}^{-}$ .

3.1F1step 1.2step 2.1

Lattice stability. For $P \in L (\infty)$ one has $B (P^{⋆}, P^{⋆}) = B (P, P) \in A_{0}$ by step 2.1 and the norm characterization of [F1]; hence $P^{⋆} \in L (\infty)$ and $⋆$ preserves $L (\infty)$ . It preserves $U_{A}^{-}$ by step 1.2, so it preserves $L_{Z} = U_{A}^{-} \cap L (\infty)$ ; since it commutes with $ψ$ and $ψ (L_{Z}) = L_{Z}^{-}$ , it also preserves the bar lattice $L_{Z}^{-}$ .

4.1F2step 2.1step 3.1algebra

The signed residue permutation. Steps 2.1–3.1 induce an integral involutive isometry of $L_{Z} / q L_{Z}$ . Each $b$ has residue norm $1$ , so its image also has residue norm $1$ and is $\pm b^{'}$ by [F2]. Involutivity makes this a signed permutation. No representative of $b$ is required to have exact field norm $1$ .

5.1step 1.1step 1.2step 2.1step 3.1step 4.1∎

Conclusion. Steps 1.1–1.2 prove that $⋆$ is an anti-involution preserving $U_{A}^{-}$ and commuting with bar, step 2.1 proves that it is an isometry of the normalized half form, step 3.1 proves preservation of $L (\infty)$ , $L_{Z}$ and $L_{Z}^{-}$ , and step 4.1 gives the signed permutation of the integral crystal residue. No unsigned star-crystal theorem is asserted.

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Sources

Kashiwara, On crystal bases of the q-analogue of universal enveloping algebras (1991) (standard reference, not scraped)