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Integral skew-string coefficients on the divided-power negative half

Statement

Let $U_{q}^{-}$ be the negative half of the Drinfeld–Jimbo algebra with its Kashiwara skew derivations $e_{i}^{'}, e_{i}^{''}$ (Kashiwara's skew derivations $e_{i}^{'}$ and $e_{i}^{''}$ on the negative half), let $A = Z [q, q^{- 1}]$ and let $U_{A}^{-}$ be the divided-power half generated by the $F_{i}^{(r)}$ (The Lusztig divided-power integral form $U_{A}$ ). Then:

(i) $e_{i}^{'}$ preserves $U_{A}^{-}$ : $e_{i}^{'} (U_{A}^{-}) \subseteq U_{A}^{-}$ .

(ii) Define, for $x \in U_{q}^{-}$ , $P_{i} x : = \sum_{n \geq 0} (- 1)^{n} q_{i}^{- n (n - 1) / 2} F_{i}^{(n)} (e_{i}^{'})^{n} x,$ the sum being finite on each $x$ because $e_{i}^{'}$ is locally nilpotent, and the powers $(e_{i}^{'})^{n}$ being ordinary, not divided, powers of the derivation. Then $P_{i}$ is a well-defined $Q (q)$ -linear map with $e_{i}^{'} P_{i} = 0$ that fixes $\ker e_{i}^{'}$ pointwise and annihilates $F_{i} U_{q}^{-}$ ; hence $U_{q}^{-} = \ker e_{i}^{'} \oplus F_{i} U_{q}^{-}$ and $P_{i}$ is the projection onto $\ker e_{i}^{'}$ along $F_{i} U_{q}^{-}$ .

(iii) Every $x \in U_{A}^{-}$ has a unique finite expansion $x = \sum_{n \geq 0} F_{i}^{(n)} x_{n}, e_{i}^{'} x_{n} = 0,$ with $x_{n} = q_{i}^{n (n - 1) / 2} P_{i} (e_{i}^{'})^{n} x \in U_{A}^{-}$ for every $n$ ; the sum is finite because $(e_{i}^{'})^{n} x = 0$ for large $n$ . Consequently the string operators ${\tilde{e}}_{i} (\sum_{n} F_{i}^{(n)} x_{n}) : = \sum_{n \geq 1} F_{i}^{(n - 1)} x_{n}$ and ${\tilde{f}}_{i} (\sum_{n} F_{i}^{(n)} x_{n}) : = \sum_{n \geq 0} F_{i}^{(n + 1)} x_{n}$ preserve $U_{A}^{-}$ .

No global $A$ -basis and no freeness of $U_{A}^{-}$ over $A$ is assumed or asserted.

Facts & Assumptions

Given: The negative half $U_{q}^{-} = ⨁_{ν \in Q_{+}} U_{q}^{-} [- ν]$ , the maps $e_{i}^{'}$ , $e_{i}^{''}$ of Kashiwara's skew derivations $e_{i}^{'}$ and $e_{i}^{''}$ on the negative half, the divided powers $x^{(m)} = x^{m} / [m]_{i}!$ , and the $A$ -algebra $U_{A}^{-}$ .

[F1]

$e_{i}^{'} (F_{j}) = δ_{i j}$ , $e_{i}^{'}$ is $Q (q)$ -linear, and $e_{i}^{'}$ satisfies the twisted Leibniz rule $e_{i}^{'} (x y) = e_{i}^{'} (x) y + q_{i}^{⟨ wt (x), h_{i} ⟩} x e_{i}^{'} (y)$ for homogeneous $x$ , with the operator identity $e_{i}^{'} F_{j} = q_{i}^{- a_{i j}} F_{j} e_{i}^{'} + δ_{i j}$ ; in particular $e_{i}^{'} (F_{i} y) = q_{i}^{- 2} F_{i} e_{i}^{'} (y) + y$ and $e_{i}^{'} (F_{j} y) = q_{i}^{- a_{i j}} F_{j} e_{i}^{'} (y)$ for $j \neq i$ (Kashiwara's skew derivations $e_{i}^{'}$ and $e_{i}^{''}$ on the negative half).

[F2]

$e_{i}^{'}$ is locally nilpotent: for every homogeneous $x$ of degree $- ν$ and every $k > ht (ν)$ one has $(e_{i}^{'})^{k} x = 0$ (Kashiwara's skew derivations $e_{i}^{'}$ and $e_{i}^{''}$ on the negative half (d)).

[F3]

Every word in the $F_{j}$ of degree $- ν$ has exactly $ν_{i}$ letters $F_{i}$ ; $U_{A}^{-}$ consists of the finite $A$ -linear combinations of the products $F_{j_{1}}^{(r_{1})} \dots F_{j_{m}}^{(r_{m})}$ of divided powers, and these are $Q (q)$ -linear combinations of words (The Lusztig divided-power integral form $U_{A}$ , Positive, negative and toral quantum subalgebras and their root gradings).

[F4]

Divided-power multiplication: $F_{i}^{(n)} F_{i}^{(m)} = \frac{F_{i}^{n + m}}{[n]_{i}! [m]_{i}!} = {(\binom{n + m}{n})}_{i} F_{i}^{(n + m)}$ , and the Gaussian coefficients are Laurent polynomials in $q_{i}$ with integer coefficients (Quantum integers, factorials, Gaussian binomials and divided powers at $q_{i}$ , The quantum Pascal recurrences, the Gauss product formula and Gaussian integrality (iv)).

[F5]

Gauss product identity and its inversion: for $N \geq 0$ , $\prod_{j = 0}^{N - 1} (1 + q_{i}^{2 j} z) = \sum_{r = 0}^{N} q_{i}^{r (N - 1)} {(\binom{N}{r})}_{i} z^{r}$ ; hence $\sum_{r = 0}^{N} (- 1)^{r} q_{i}^{r (N - 1)} {(\binom{N}{r})}_{i} = 0$ for $N \geq 1$ , and applying the field involution $q_{i} \mapsto q_{i}^{- 1}$ (which fixes every Gaussian coefficient) gives $\sum_{r = 0}^{N} (- 1)^{r} q_{i}^{- r (N - 1)} {(\binom{N}{r})}_{i} = 0$ for $N \geq 1$ , while the sum equals $1$ for $N = 0$ (The quantum Pascal recurrences, the Gauss product formula and Gaussian integrality (iii), Quantum integers, factorials, Gaussian binomials and divided powers at $q_{i}$ ).

Verification

Proof technique: compute $e_{i}^{'}$ and its powers on divided-power words, use the resulting explicit q-binomial sum for the projector, and read the decomposition and the integrality of its coefficients off that computation.

1.1givenF1F4algebra

The deletion rule. From $e_{i}^{'} F_{i} = q_{i}^{- 2} F_{i} e_{i}^{'} + 1$ and $e_{i}^{'} (1) = 0$ , induction on $r \geq 1$ gives $e_{i}^{'} (F_{i}^{r}) = (\sum_{s = 0}^{r - 1} q_{i}^{- 2 s}) F_{i}^{r - 1} = q_{i}^{- (r - 1)} [r]_{i} F_{i}^{r - 1}$ . Dividing by the nonzero $[r]_{i}!$ yields $e_{i}^{'} (F_{i}^{(r)}) = q_{i}^{- (r - 1)} F_{i}^{(r - 1)}$ . For $j \neq i$ , the relation $e_{i}^{'} F_{j} = q_{i}^{- a_{i j}} F_{j} e_{i}^{'}$ and $e_{i}^{'} (1) = 0$ give $e_{i}^{'} (F_{j}^{r}) = 0$ , hence $e_{i}^{'} (F_{j}^{(r)}) = 0$ .

2.1step 1.1F1F3algebra

$e_{i}^{'}$ preserves $U_{A}^{-}$ : by [F3] it suffices to check a divided-power word $w = F_{j_{1}}^{(r_{1})} \dots F_{j_{m}}^{(r_{m})}$ , by induction on $m$ using the twisted Leibniz rule of [F1] and the deletion rule of step 1.1: $e_{i}^{'} (w) = e_{i}^{'} (F_{j_{1}}^{(r_{1})}) w^{'} + q_{i}^{⟨ wt (F_{j_{1}}^{(r_{1})}), h_{i} ⟩} F_{j_{1}}^{(r_{1})} e_{i}^{'} (w^{'})$ , where $w^{'}$ is the restricted word, the first factor is $0$ or $q_{i}^{- (r_{1} - 1)} F_{i}^{(r_{1} - 1)} \in U_{A}^{-}$ by step 1.1, the scalar $q_{i}^{⟨ - r_{1} α_{j_{1}}, h_{i} ⟩} = q_{i}^{- r_{1} a_{i j_{1}}} \in A$ , and $e_{i}^{'} (w^{'}) \in U_{A}^{-}$ by induction; products of divided powers lie in $U_{A}^{-}$ . By [F1] linearity extends this to all of $U_{A}^{-}$ .

2.2step 1.1F1algebra

Powers of the derivation on a string: if $u \in \ker e_{i}^{'}$ then for $0 \leq n \leq m$ , $(e_{i}^{'})^{n} F_{i}^{(m)} u = q_{i}^{- n m + n (n + 1) / 2} F_{i}^{(m - n)} u$ . For $n = 0$ this is clear; for the induction step, $e_{i}^{'} (F_{i}^{(m - n)} u) = e_{i}^{'} (F_{i}^{(m - n)}) u + q_{i}^{- 2 (m - n)} F_{i}^{(m - n)} e_{i}^{'} (u) = q_{i}^{- (m - n - 1)} F_{i}^{(m - n - 1)} u$ , where the second summand vanishes because $e_{i}^{'} u = 0$ ; multiplying by $q_{i}^{- n m + n (n + 1) / 2}$ gives the exponent $- (n + 1) m + (n + 1) (n + 2) / 2$ as required.

3.1step 2.2givenalgebra

The projector: $P_{i}$ is well defined by [F2], and $Q (q)$ -linear. Consecutive terms cancel in $e_{i}^{'} P_{i}$ : applying $e_{i}^{'}$ to the $n$ -th term gives $(- 1)^{n} q_{i}^{- n (n - 1) / 2} [e_{i}^{'} (F_{i}^{(n)}) (e_{i}^{'})^{n} + q_{i}^{- 2 n} F_{i}^{(n)} (e_{i}^{'})^{n + 1}] = (- 1)^{n} q_{i}^{- n (n - 1) / 2} [q_{i}^{- (n - 1)} F_{i}^{(n - 1)} (e_{i}^{'})^{n} + q_{i}^{- 2 n} F_{i}^{(n)} (e_{i}^{'})^{n + 1}]$ by steps 1.1, 2.2, and the contribution of the second part of the $n$ -th term cancels the first part of the $(n + 1)$ -st term because $- n (n - 1) / 2 - 2 n = - n (n + 1) / 2 - (n)$ ; the $n = 0$ contribution of the second part is $(e_{i}^{'})^{1}$ and equals the first part of the $n = 1$ term, so $e_{i}^{'} P_{i} = 0$ . If $e_{i}^{'} x = 0$ then $(e_{i}^{'})^{n} x = 0$ for $n \geq 1$ , so $P_{i} x = F_{i}^{(0)} (e_{i}^{'})^{0} x = x$ : $P_{i}$ fixes $\ker e_{i}^{'}$ pointwise.

3.2step 2.2F4F5algebra

$P_{i}$ annihilates $F_{i} \ker e_{i}^{'}$ : for $u \in \ker e_{i}^{'}$ and $m \geq 1$ , step 2.2 gives $(e_{i}^{'})^{n} (F_{i}^{(m)} u) = q_{i}^{- n m + n (n + 1) / 2} F_{i}^{(m - n)} u$ for $n \leq m$ , and $(e_{i}^{'})^{n} (F_{i}^{(m)} u) = q_{i}^{- n (m - 1)} F_{i}^{(m - 1)} (e_{i}^{'} u) = 0$ for $n > m$ (the surviving term of the Leibniz expansion at the first step where the derivative reaches $u$ ); hence $P_{i} (F_{i}^{(m)} u) = \sum_{n = 0}^{m} (- 1)^{n} q_{i}^{- n (n - 1) / 2} F_{i}^{(n)} q_{i}^{- n m + n (n + 1) / 2} F_{i}^{(m - n)} u = \sum_{n = 0}^{m} (- 1)^{n} q_{i}^{- n m + n} {(\binom{m}{n})}_{i} F_{i}^{(m)} u = [\sum_{n = 0}^{m} (- 1)^{n} q_{i}^{- n (m - 1)} {(\binom{m}{n})}_{i}] F_{i}^{(m)} u$ , using [F4] for $F_{i}^{(n)} F_{i}^{(m - n)}$ . By [F5] the bracket is $1$ for $m = 0$ and $0$ for $m \geq 1$ ; hence $P_{i}$ fixes $\ker e_{i}^{'}$ and annihilates $F_{i}^{(m)} \ker e_{i}^{'}$ for every $m \geq 1$ .

4.1step 2.2step 3.1step 3.2F2algebra

Decomposition and uniqueness: for $x \in U_{q}^{-}$ let $m$ be maximal with $(e_{i}^{'})^{m} x \neq 0$ (finite by [F2]) or $m = 0$ if $e_{i}^{'} x = 0$ , and put $x_{m} : = q_{i}^{m (m - 1) / 2} P_{i} (e_{i}^{'})^{m} x$ ; then $e_{i}^{'} x_{m} = q_{i}^{m (m - 1) / 2} e_{i}^{'} P_{i} (e_{i}^{'})^{m} x = 0$ by step 3.1, and $x^{'} : = x - F_{i}^{(m)} x_{m}$ satisfies $(e_{i}^{'})^{m} x^{'} = (e_{i}^{'})^{m} x - q_{i}^{- m (m - 1) / 2} x_{m} = 0$ by step 2.2 and step 3.1, so $ord (x^{'}) \leq m - 1$ . Descending induction on $m$ therefore writes $x = \sum_{n} F_{i}^{(n)} x_{n}$ with $e_{i}^{'} x_{n} = 0$ and $x_{n} = q_{i}^{n (n - 1) / 2} P_{i} (e_{i}^{'})^{n} x$ : at each stage the coefficient of the remainder agrees with the claimed formula because $P_{i}$ annihilates the subtracted strings $F_{i}^{(m - n)} x_{m}$ by step 3.2, so $P_{i} (e_{i}^{'})^{n} x^{'} = P_{i} (e_{i}^{'})^{n} x$ for $n < m$ . If also $x = \sum_{n} F_{i}^{(n)} y_{n}$ with $e_{i}^{'} y_{n} = 0$ , subtract the two finite expansions and put $z_{n} : = x_{n} - y_{n}$ . If some $z_{n}$ is nonzero, let $N$ be its largest index. Applying $(e_{i}^{'})^{N}$ to $0 = \sum_{n} F_{i}^{(n)} z_{n}$ gives $0 = q_{i}^{- N (N - 1) / 2} z_{N}$ by step 2.2, a contradiction. Thus every $z_{n} = 0$ , proving uniqueness.

5.1step 2.1step 3.1step 3.2step 4.1F2F3algebra∎

Integrality and conclusion: each $x_{n} = q_{i}^{n (n - 1) / 2} P_{i} (e_{i}^{'})^{n} x$ lies in $U_{A}^{-}$ , because $e_{i}^{'}$ preserves $U_{A}^{-}$ by step 2.1, every $(e_{i}^{'})^{n} x$ is therefore in $U_{A}^{-}$ , the coefficients $q_{i}^{\pm m (m - 1) / 2}$ and $(- 1)^{m}$ lie in $A$ , and each $F_{i}^{(m)}$ is a generator of $U_{A}^{-}$ ; hence the finite sum $P_{i} (\cdot)$ of the definition lies in $U_{A}^{-}$ on $U_{A}^{-}$ . Consequently every string coefficient $x_{n}$ of an $x \in U_{A}^{-}$ is integral. For the projection claim, apply step 4.1 to the expansion of an arbitrary $y$ : $F_{i} y = \sum_{n \geq 1} [n]_{i} F_{i}^{(n)} y_{n - 1}$ decomposes with all string coefficients in $\ker e_{i}^{'}$ and zero $n = 0$ coefficient, so $P_{i} (F_{i} y) = 0$ by the formula; hence $U_{q}^{-} = \ker e_{i}^{'} \oplus F_{i} U_{q}^{-}$ with $P_{i}$ the projection along $F_{i} U_{q}^{-}$ , and $x_{0} = P_{i} x \in U_{A}^{-}$ whenever $x \in U_{A}^{-}$ . Finally the string operators of (iii) are well defined by the uniqueness of the expansion, and they preserve $U_{A}^{-}$ because they only shift the divided-power labels $n \mapsto n \pm 1$ of the integral coefficients $x_{n} \in U_{A}^{-}$ . No global $A$ -basis and no freeness of $U_{A}^{-}$ was used anywhere.

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Sources

Masaki Kashiwara, On crystal bases of the $q$-analogue of universal enveloping algebras (1991) (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)