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The unmodified integral torus has a quantum-binomial basis

Statement

Let $(I, A, D, P, P^{\lor}, q)$ be a finite symmetrizable Cartan datum (Symmetrizable Cartan data for quantum groups), put $A = Z [q, q^{- 1}]$ (The Laurent polynomial ring as the principal localisation of Z[t] at t), let $L : = \sum_{i \in I} Z (d_{i} h_{i}) \subseteq P^{\lor}$ , and let $U_{A}^{0}$ be the $A$ -subalgebra of $U_{q}^{0}$ generated by the $K_{h}^{\pm 1}$ and the toral quantum binomials $T_{i, c, t}$ (The Lusztig divided-power integral form $U_{A}$ ).

(a) One simple torus. Fix $i \in I$ and write $z : = q_{i}$ , $K : = K_{i}$ , $T_{t} : = T_{i, 0, t}$ for $t \geq 0$ . Then for all $r, s \geq 0$ , $c \in Z$ and $t \geq 1$ , $T_{r} T_{s} = \sum_{j = 0}^{\min (r, s)} z^{r s} K^{- j} {(\binom{r + s - j}{r})}_{z} {(\binom{r}{j})}_{z} T_{r + s - j}, T_{c + 1, t} = z^{t} T_{c, t} + K^{- 1} z^{- c + t - 1} T_{c, t - 1},$ where ${(\binom{m}{r})}_{z}$ is the Gaussian binomial of Quantum integers, factorials, Gaussian binomials and divided powers at $q_{i}$ ; all coefficients lie in $A$ . Moreover the $A$ -subalgebra of $U_{q}^{0}$ generated by $K^{\pm 1}$ and all $T_{c, t}$ equals the $A$ -span of ${K^{δ} T_{t} : δ \in {0, 1}, t \geq 0}$ , and that set is an $A$ -basis.

(b) Full torus. Fix a $Z$ -basis of $P^{\lor}$ , let $J$ be the finite set of coordinates occurring in the finitely many vectors $d_{i} h_{i}$ , and let $D : = ⨁_{j \in J} Z e_{j}$ be the corresponding coordinate direct summand. For every coset of $L$ in $P^{\lor}$ choose the representative $c$ whose $D$ -component has minimal sum of coordinate absolute values, ties broken lexicographically, and whose complementary component is the coset's common complementary component. Then $U_{A}^{0} = ⨁_{c} K_{c} \cdot A -span {\prod_{i \in I} K_{i}^{δ_{i}} T_{i, 0, t_{i}} : δ_{i} \in {0, 1}, t_{i} \geq 0},$ the sum over the chosen representatives, so the displayed elements form an $A$ -basis of $U_{A}^{0}$ .

Both statements are proved directly on the unmodified generators over the non-PID ring $A$ : no integral PBW or canonical-basis input, no inference from torsion-freeness to freeness, and no modified- $_{A} \dot{U}$ result is used.

Facts & Assumptions

Given: The Cartan datum with finitely many indices, the Laurent ring $A$ , the subgroup $L = \sum_{i} Z (d_{i} h_{i}) \subseteq P^{\lor}$ , and the toral binomials $T_{i, c, t} = \prod_{s = 1}^{t} (K_{i} q_{i}^{c - s + 1} - K_{i}^{- 1} q_{i}^{- c + s - 1}) / (q_{i}^{s} - q_{i}^{- s})$ for $i \in I$ , $c \in Z$ , $t \geq 0$ .

[F1]

$A = Z [q, q^{- 1}]$ is a subring of $Q (q)$ containing $z^{\pm m}$ and ${(\binom{m}{r})}_{z}$ for all $m, r \geq 0$ : the Gaussian binomials are Laurent polynomials in $q_{i}$ with integer coefficients (The Laurent polynomial ring as the principal localisation of Z[t] at t, The quantum Pascal recurrences, the Gauss product formula and Gaussian integrality (iv)), and $q_{i}^{\pm 1} \in A$ (Quantum integers, factorials, Gaussian binomials and divided powers at $q_{i}$ ).

[F2]

Each $T_{i, c, t}$ is a Laurent polynomial in $K_{i}^{\pm 1}$ with coefficients in $Q (q)$ , all $T_{i, c, t}$ commute with one another and with every $K_{h}$ , $T_{i, c, 0} = 1$ , $T_{i, c, 1} = (K_{i} q_{i}^{c} - K_{i}^{- 1} q_{i}^{- c}) / (q_{i} - q_{i}^{- 1})$ , and $T_{i, c, t} = T_{i, c, t - 1} (K_{i} q_{i}^{c - t + 1} - K_{i}^{- 1} q_{i}^{- c + t - 1}) / (q_{i}^{t} - q_{i}^{- t})$ (The Lusztig divided-power integral form $U_{A}$ , Divided-power multiplication, toral binomials and the $E_{i}^{(a)} F_{i}^{(b)}$ reordering identity (ii)).

[F3]

$U_{q}^{0} ≅ Q (q) [P^{\lor}]$ as algebras, with $K_{h} K_{h^{'}} = K_{h + h^{'}}$ and $K_{0} = 1$ ; the $K_{h}$ for $h$ in a $Z$ -basis of $P^{\lor}$ are algebraically independent, and $U_{q}^{0}$ is commutative (Triangular decomposition of a quantized enveloping algebra (ii)).

[F4]

The simple coroots $h_{i}$ are linearly independent over $Q$ in $P^{\lor} \otimes_{Z} Q$ , and all $d_{i} > 0$ ; hence $L = ⨁_{i} Z (d_{i} h_{i}) ≅ Z^{n}$ and each $d_{i} h_{i}$ has finite support in the fixed basis of $P^{\lor}$ , so the set $J$ of coordinates occurring in the $d_{i} h_{i}$ is finite and $L \subseteq D$ (Symmetrizable Cartan data for quantum groups, Free abelian group on a set).

[F5]

For every nonzero integer $t$ , $q^{t} \neq 1$ : after replacing $t$ by $∣ t ∣$ , equality would make $q$ a root of $T^{∣ t ∣} - 1$ . Thus $z^{s} - z^{- s} \neq 0$ for $s \geq 1$ (Quantum integers, factorials, Gaussian binomials and divided powers at $q_{i}$ ).

Verification

technique · evaluate both sides of the product identity at the infinitely many toral points $K=z^n$, using the Gaussian Vandermonde identity proved from the weighted-subset model of the ordinary Gaussian binomial; deduce the shift and stability identities by direct factorization; then span all shifts by the stable pair $\{K^{\delta}T_t\}$ and read off independence from the extremal exponents

1.1F2F5algebra

The simple torus and its evaluation. Fix $i$ and write $z = q_{i}$ , $K = K_{i}$ , $T_{t} = T_{i, 0, t}$ . By [F2], $T_{t} = \prod_{s = 1}^{t} (K z^{1 - s} - K^{- 1} z^{s - 1}) / (z^{s} - z^{- s})$ , a Laurent polynomial in $K$ whose support is contained in $[- t, t]$ : indeed each factor equals $K z^{1 - s} (1 - K^{- 2} z^{2 s - 2})$ , so $T_{t} = K^{t} \prod_{s = 1}^{t} z^{1 - s} (1 - K^{- 2} z^{2 s - 2}) / \prod_{s = 1}^{t} (z^{s} - z^{- s})$ , a $Q (q)$ -combination of $K^{t - 2 l}$ for $0 \leq l \leq t$ , and the two extreme coefficients (of $K^{t}$ and of $K^{- t}$ ) are nonzero by [F5]. Evaluating at $K = z^{n}$ for an integer $n \geq 0$ gives $T_{t} (z^{n}) = \prod_{s = 1}^{t} [n - s + 1]_{z} / [s]_{z}$ with $[m]_{z} = (z^{m} - z^{- m}) / (z - z^{- 1})$ , which equals the Gaussian binomial ${(\binom{n}{t})}_{z}$ when $t \leq n$ , and equals $0$ when $t > n$ (the factor $s = n + 1$ has numerator $z^{0} - z^{0} = 0$ ); accordingly we use the convention ${(\binom{n}{t})}_{z} = 0$ for $t > n$ or $t < 0$ . Finally, if a Laurent polynomial in $K$ with coefficients in $Q (q)$ vanishes at $K = z^{n}$ for infinitely many $n$ , it is zero: after multiplying by a power of $K$ it becomes a polynomial whose infinitely many pairwise distinct roots $z^{n}$ force all coefficients to vanish (the $z^{n}$ are distinct because $q$ is transcendental by [F5]).

1.2F1algebra

The Gaussian Vandermonde. For an indeterminate $Q$ and $m, r \geq 0$ put $[m]_{Q} : = 1 + Q + \dots + Q^{m - 1}$ and $B (m, r) : = [m]_{Q}! / ([r]_{Q}! [m - r]_{Q}!)$ , so that with $Q = z^{2}$ one has $B (m, r) = {(\binom{m}{r})}_{z^{2}}$ in the notation of [F1]. We first record the weighted-subset model: if $i n v (S) : = # {(a, b) : a < b, a \notin S, b \in S}$ for $S \subseteq {1, \dots, m}$ , then $B (m, r) = \sum_{S \subseteq {1, \dots, m}, ∣ S ∣ = r} Q^{i n v (S)}$ . Indeed both sides equal $1$ for $m = r = 0$ and $0$ for $r < 0$ or $r > m$ ; splitting the subsets according to whether $m$ lies in $S$ gives the recurrence $B (m, r) = B (m - 1, r) + Q^{m - r} B (m - 1, r - 1)$ for the right-hand side, and the same recurrence holds for the left-hand side because $B (m - 1, r) + Q^{m - r} B (m - 1, r - 1) = \frac{[m - 1]_{Q}!}{[r]_{Q}! [m - r]_{Q}!} ([m - r]_{Q} + Q^{m - r} [r]_{Q}) = \frac{[m - 1]_{Q}! [m]_{Q}}{[r]_{Q}! [m - r]_{Q}!} = B (m, r)$ , using $[m - r]_{Q} + Q^{m - r} [r]_{Q} = [m]_{Q}$ . Splitting a subset $S \subseteq {1, \dots, n}$ with $∣ S ∣ = s$ into $A : = S \cap {1, \dots, r}$ , $∣ A ∣ = j$ , and $B^{'} : = S \cap {r + 1, \dots, n}$ , the inversion count splits as $i n v (S) = i n v (A) + i n v (B^{'}) + (r - j) (s - j)$ , because a crossing inversion $a < b$ with $a \in {1, \dots, r} ∖ A$ , $b \in B^{'}$ can always be realized, while pairs $(b, a)$ with $b \in {r + 1, \dots, n}$ , $a \leq r$ are never inversions; hence for $0 \leq r \leq n$ , $B (n, s) = \sum_{j = 0}^{\min (r, s)} Q^{(r - j) (s - j)} B (r, j) B (n - r, s - j) .$

1.3F1F2algebra

The shift and stability identities. Put $N_{s} (c) = K z^{c - s + 1} - K^{- 1} z^{- c + s - 1}$ . Since $N_{s} (c + 1) = N_{s - 1} (c)$ for $s \geq 2$ , factoring the first $t - 1$ common factors reduces the shift to $N_{1} (c + 1) - z^{t} N_{t} (c) = K^{- 1} z^{- c + t - 1} (z^{t} - z^{- t})$ , proving $T_{c + 1, t} = z^{t} T_{c, t} + K^{- 1} z^{- c + t - 1} T_{c, t - 1}$ . The product definition also gives $(z^{t + 1} - z^{- t - 1}) T_{t + 1} = T_{t} (K z^{- t} - K^{- 1} z^{t})$ . Rearranging yields $K^{- 1} T_{t} = z^{- 2 t} K T_{t} - z^{- t} (z^{t + 1} - z^{- t - 1}) T_{t + 1}$ ; multiplying by $K z^{2 t}$ yields $K^{2} T_{t} = z^{2 t} T_{t} + z^{t} (z^{t + 1} - z^{- t - 1}) K T_{t + 1}$ . Every coefficient is in $A$ .

2.1step 1.2F1algebra

Normalization to the symmetric Gaussian binomials. By the identification of Quantum integers, factorials, Gaussian binomials and divided powers at $q_{i}$ , ${(\binom{m}{r})}_{z} = z^{- r (m - r)} {(\binom{m}{r})}_{z^{2}} = z^{- r (m - r)} B (m, r)$ for $0 \leq r \leq m$ . First multiply step 1.2 by $B (n, r)$ and use the factorial identity $B (n, r) B (n - r, s - j) = B (n, r + s - j) B (r + s - j, r)$ . Then multiply by $z^{- r (n - r)} z^{- s (n - s)}$ and substitute $B (m, r) = z^{r (m - r)} {(\binom{m}{r})}_{z}$ in each term, the powers of $z$ on the right-hand side combine to $2 (r - j) (s - j) + (r + s - j) (n - r - s + j) + r (s - j) + j (r - j) - r (n - r) - s (n - s) = r s - j n,$ as a direct expansion shows; in the resulting identity ${(\binom{n}{r})}_{z} {(\binom{n}{s})}_{z} = \sum_{j = 0}^{\min (r, s)} z^{r s - j n} {(\binom{n}{r + s - j})}_{z} {(\binom{r + s - j}{r})}_{z} {(\binom{r}{j})}_{z}$ both sides are zero when $n < r$ or $n < s$ (then $r + s - j \geq \max (r, s) > n$ for every $j$ , so the factor ${(\binom{n}{r + s - j})}_{z}$ vanishes), and the identity therefore holds for all $n, r, s \geq 0$ under the convention of step 1.1.

3.1step 1.1step 2.1F1F2algebra

The product identity. With notation as in step 1.1, we claim $T_{r} T_{s} = \sum_{j = 0}^{\min (r, s)} z^{r s} K^{- j} {(\binom{r + s - j}{r})}_{z} {(\binom{r}{j})}_{z} T_{r + s - j}$ for all $r, s \geq 0$ . Both sides lie in $Q (q) [K^{\pm 1}]$ : the left side by [F2], the right side because the coefficients are in $A \subseteq Q (q)$ and $T_{r + s - j}$ is a Laurent polynomial in $K$ [F2]. Evaluating at $K = z^{n}$ and using step 1.1, the left side becomes ${(\binom{n}{r})}_{z} {(\binom{n}{s})}_{z}$ and the right side becomes $\sum_{j} z^{r s - j n} {(\binom{r + s - j}{r})}_{z} {(\binom{r}{j})}_{z} {(\binom{n}{r + s - j})}_{z}$ , and these agree for every $n \geq 0$ by step 2.1. By the evaluation principle of step 1.1 the two Laurent polynomials coincide, proving the identity; its coefficients lie in $A$ by [F1].

4.1step 3.1step 1.3F1F2algebra

Spanning and the simple-torus basis. Let $M : = A -span {K^{δ} T_{t} : δ \in {0, 1}, t \geq 0}$ . First, $K^{m} T_{t} \in M$ for all $m \in Z$ , $t \geq 0$ : for $m \geq 2$ , multiply the second identity of step 1.3(b) by $K^{m - 2}$ to get $K^{m} T_{t} = z^{2 t} K^{m - 2} T_{t} + z^{t} (z^{t + 1} - z^{- t - 1}) K^{m - 1} T_{t + 1}$ , which reduces $∣ m ∣$ and lies in $M$ by induction on $∣ m ∣$ from the base $m = 0, 1$ ; for $m = - 1$ use the first identity of step 1.3(b) directly; for $m \leq - 2$ , multiply it by $K^{m + 1}$ to get $K^{m} T_{t} = z^{- 2 t} K^{m + 2} T_{t} - z^{- t} (z^{t + 1} - z^{- t - 1}) K^{m + 1} T_{t + 1}$ , again reducing $∣ m ∣$ . Second, all $T_{c, t} \in M$ by induction on $t$ (the case $t = 0$ gives $T_{c, 0} = 1$ ): for $c \geq 0$ the shift identity $T_{c + 1, t} = z^{t} T_{c, t} + K^{- 1} z^{- c + t - 1} T_{c, t - 1}$ of step 1.3 (with unit coefficient $z^{t} \in A^{\times}$ ) iterates upward from $T_{0, t} = T_{t} \in M$ , and for $c \leq - 1$ its other solution $T_{c, t} = z^{- t} T_{c + 1, t} - K^{- 1} z^{- c - 1} T_{c, t - 1}$ iterates downward from $c = 0$ . Third, $M$ is an $A$ -subalgebra: $(K^{m} T_{r}) (K^{n} T_{s}) = K^{m + n} T_{r} T_{s}$ expands by step 3.1 into the terms $z^{r s} {(\binom{r + s - j}{r})}_{z} {(\binom{r}{j})}_{z} K^{m + n - j} T_{r + s - j}$ , which lie in $M$ by the first part. Since $M$ contains $K^{\pm 1}$ and every $T_{c, t}$ , it contains the generated subalgebra; conversely $K^{δ} T_{t}$ is a product of the generators $K$ and $T_{0, t}$ , so $M$ equals the $A$ -subalgebra generated by $K^{\pm 1}$ and all $T_{c, t}$ , which is the simple-torus part of $U_{A}^{0}$ . Fourth, ${K^{δ} T_{t}}$ is $A$ -independent: if a finite combination with coefficients $(a_{δ, t})$ is zero and not all coefficients vanish, let $N : = \max {t : a_{0, t} \neq 0 or a_{1, t} \neq 0}$ ; the exponent $N + 1$ of $K$ occurs only in $K T_{N}$ (each $T_{t}$ has support in $[- t, t]$ by step 1.1, and $K^{δ} T_{t}$ with $t \leq N$ can reach exponent $N + 1$ only for $(δ, t) = (1, N)$ ), and its coefficient of $K^{N + 1}$ is the nonzero extreme coefficient of $T_{N}$ , so $a_{1, N} = 0$ ; then the exponent $- N$ occurs only in $T_{N}$ ( $K^{1} T_{t}$ has support in $[1 - t, 1 + t]$ with $1 - t > - N$ for $t \leq N$ ), forcing $a_{0, N} = 0$ ; descending on $N$ gives all coefficients zero. Hence ${K^{δ} T_{t}}$ is an $A$ -basis of $M$ , as claimed in (a).

5.1step 4.1F3F4algebra∎

The full torus. By [F3], $U_{q}^{0} = Q (q) [P^{\lor}]$ with basis-like multiplication $K_{h} K_{h^{'}} = K_{h + h^{'}}$ ; by [F4], $L = ⨁_{i} Z (d_{i} h_{i})$ is free of rank $∣ I ∣$ and the $K_{d_{i} h_{i}} = K_{i}$ are algebraically independent. Let $B$ be the $A$ -subalgebra of $Q (q) [L]$ generated by the $K_{i}^{\pm 1}$ and all $T_{i, c, t}$ ; since $Q (q) [L] = ⨂_{i} Q (q) [K_{i}^{\pm 1}]$ and the generators with distinct $i$ lie in distinct tensor factors, step 4.1 applied in each factor gives $B = A -span {\prod_{i} K_{i}^{δ_{i}} T_{i, 0, t_{i}} : δ_{i} \in {0, 1}, t_{i} \geq 0}$ with the displayed products forming an $A$ -basis. In $Q (q) [P^{\lor}] = ⨁_{c} K_{c} Q (q) [L]$ , the direct sum running over the chosen representatives (whose existence and explicitness: the $D$ -components of a fixed coset form a coset of $L$ in the finitely generated free group $D$ , and the elements of minimal coordinate absolute-value sum are finitely many by boundedness of the coordinates, so the lexicographic tie-break is a finite comparison; the complementary component is determined by the coset because $L \subseteq D$ by [F4]), every product of generators of $U_{A}^{0}$ lies in $⨁_{c} K_{c} B$ because $K_{h_{1}} \dots K_{h_{m}} = K_{\sum h_{l}}$ and the $T$ -products lie in $B$ ; conversely $K_{c} \in U_{A}^{0}$ and $B \subseteq U_{A}^{0}$ (each $K_{i}^{δ_{i}} T_{i, 0, t_{i}}$ is a product of generators). Hence $U_{A}^{0} = ⨁_{c} K_{c} B$ , and the union of $K_{c}$ times the $A$ -basis of $B$ over the representatives is an $A$ -basis of $U_{A}^{0}$ , proving (b). The construction uses only finitely many explicit coordinate comparisons per coset and no choice function on an infinite family; no integral PBW basis, canonical basis or modified-algebra input is used, and no freeness is inferred from torsion-freeness over the non-PID ring $A$ .

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Sources

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)
Kyeonghoon Jeong, Seok-Jin Kang and Masaki Kashiwara, Crystal Bases for Quantum Generalized Kac-Moody Algebras, arXiv:math/0305390 (standard reference, not scraped)