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Rank-one tensor strings and their crystal limit rule

Statement

Let $i \in I$ , write $z = q_{i}$ , and let $V (a)$ , $V (b)$ be the simple rank-one modules of nonnegative integral highest weights $a, b$ for the subalgebra generated by $E_{i}, F_{i}, K_{i}^{\pm 1}$ , with divided bases $v_{j} = F_{i}^{(j)} v_{0}$ $(0 \leq j \leq a)$ and $w_{k} = F_{i}^{(k)} w_{0}$ $(0 \leq k \leq b)$ (Divided-power commutation and the simple $U_{q_{i}} ({s l}_{2})$ string modules). Consider the tensor product with the coproduct $Δ E_{i} = E_{i} \otimes K_{i}^{- 1} + 1 \otimes E_{i}$ , $Δ F_{i} = F_{i} \otimes 1 + K_{i} \otimes F_{i}$ (The Drinfeld–Jimbo formulas define a Hopf algebra), its Kashiwara operators ${\tilde{e}}_{i}, {\tilde{f}}_{i}$ (Uniqueness of the $i$ -string decomposition of an integrable module) and the $Z$ -graded $Q (q)$ -vector space $V (a) \otimes V (b)$ with the tensor lattice $L : = ⨁_{j = 0}^{a} ⨁_{k = 0}^{b} A_{0} (v_{j} \otimes w_{k})$ over the local ring $A_{0}$ (The local ring $A_{0}$ of rational functions regular at $q = 0$ ). Then:

(i) $L$ is a crystal lattice: it is a free $A_{0}$ -module generating $V (a) \otimes V (b)$ over $Q (q)$ , it respects the weight grading, and ${\tilde{e}}_{i} L \subseteq L$ , ${\tilde{f}}_{i} L \subseteq L$ .

(ii) Modulo $q$ one has, for all $0 \leq j \leq a$ , $0 \leq k \leq b$ , ${\tilde{f}}_{i} (v_{j} \otimes w_{k}) = {\begin{cases} v_{j + 1} \otimes w_{k} & if a - j > k, \\ v_{j} \otimes w_{k + 1} & if a - j \leq k, \end{cases}$ and ${\tilde{e}}_{i} (v_{j} \otimes w_{k}) = {\begin{cases} v_{j - 1} \otimes w_{k} & if a - j \geq k, \\ v_{j} \otimes w_{k - 1} & if a - j < k, \end{cases}$ where terms leaving the divided-basis ranges are $0$ . In particular ${\tilde{f}}_{i} b = b^{'}$ if and only if $b = {\tilde{e}}_{i} b^{'}$ for residues $b, b^{'}$ of the product basis (the partial-inverse property).

(iii) The nonzero residue strings are the chains $(0, t), (1, t), \dots, (a - t, t), (a - t, t + 1), \dots, (a - t, b)$ , $0 \leq t \leq \min (a, b)$ , and they partition the rectangle $0 \leq j \leq a$ , $0 \leq k \leq b$ .

This is a finite, choice-free calculation, independent of the grand loop and of global integral freeness.

Facts & Assumptions

Given: The index $i$ , $z = q_{i}$ , nonnegative integers $a, b$ , the rank-one modules $V (a), V (b)$ with divided bases, and the tensor product $V (a) \otimes V (b)$ with the coproduct of the statement.

[F1]

$V (a)$ has basis $v_{0}, \dots, v_{a}$ with $K_{i} v_{j} = z^{a - 2 j} v_{j}$ , $F_{i} v_{j} = [j + 1]_{z} v_{j + 1}$ , $E_{i} v_{j} = [a - j + 1]_{z} v_{j - 1}$ , and divided powers act by $F_{i}^{(m)} v_{j} = {(\binom{j + m}{m})}_{z} v_{j + m}$ , $E_{i}^{(m)} v_{j} = {(\binom{a - j + m}{m})}_{z} v_{j - m}$ (Divided-power commutation and the simple $U_{q_{i}} ({s l}_{2})$ string modules, parts (i)-(iii)); the same holds for $V (b)$ with basis $w_{k} = F_{i}^{(k)} w_{0}$ .

[F2]

$Δ E_{i} = E_{i} \otimes K_{i}^{- 1} + 1 \otimes E_{i}$ and $Δ F_{i} = F_{i} \otimes 1 + K_{i} \otimes F_{i}$ define the Hopf coproduct, so the tensor product is a module with $Δ E_{i}, Δ F_{i}$ nilpotent and $K_{i}$ diagonalisable with the sum of the two weights (The Drinfeld–Jimbo formulas define a Hopf algebra).

[F3]

On each $V (a)$ there is a symmetric bilinear $σ$ -contravariant form $B_{a}$ , normalized by $B_{a} (v_{0}, v_{0}) = 1$ , with the divided-string norms $B_{a} (F_{i}^{(r)} v_{0}, F_{i}^{(r)} v_{0}) = z^{r (a - r)} {(\binom{a}{r})}_{z} = : d (a, r)$ for $0 \leq r \leq a$ , and $B_{a} (F_{i}^{(n)} h, F_{i}^{(n)} h) = d (m, n) B_{a} (h, h)$ for every highest vector $h$ of highest weight $m$ (Cartan projection constructs the Verma contravariant form and identifies its radical).

[F4]

${ord}_{z} [m]_{z} = 1 - m$ for $m \geq 1$ , and ${ord}_{z} {(\binom{N}{R})}_{z} = - R (N - R)$ with leading coefficient $1$ for $0 \leq R \leq N$ ; elements of $Q (q)$ of nonnegative order lie in $A_{0}$ , and $d (N, R) = z^{R (N - R)} {(\binom{N}{R})}_{z}$ lies in $1 + q A_{0}$ , hence is a unit of $A_{0}$ (Quantum integers, factorials, Gaussian binomials and divided powers at $q_{i}$ , The quantum Pascal recurrences, the Gauss product formula and Gaussian integrality, The local ring $A_{0}$ of rational functions regular at $q = 0$ ).

[F5]

Every element of the tensor product has a unique admissible $i$ -string decomposition, and the Kashiwara operators ${\tilde{e}}_{i}, {\tilde{f}}_{i}$ act on the string of a highest vector $h$ of length $A$ by ${\tilde{e}}_{i} F_{i}^{(n)} h = F_{i}^{(n - 1)} h$ and ${\tilde{f}}_{i} F_{i}^{(n)} h = F_{i}^{(n + 1)} h$ , with $F_{i}^{(- 1)} h = F_{i}^{(A + 1)} h = 0$ ; $M = ⨁$ of its string spaces (Uniqueness of the $i$ -string decomposition of an integrable module).

[F6]

Define the symmetric bilinear product form $B (x \otimes y, x^{'} \otimes y^{'}) : = B_{a} (x, x^{'}) B_{b} (y, y^{'})$ . The single-factor forms of [F3] (Cartan projection constructs the Verma contravariant form and identifies its radical) make the product basis orthogonal with diagonal Gram entries $d (a, j) d (b, k)$ . Its tensor contravariance is established locally in step 1.1 from the generator values $σ (K_{i}) = K_{i}$ , $σ (E_{i}) = z F_{i} K_{i}^{- 1}$ , $σ (F_{i}) = z^{- 1} K_{i} E_{i}$ and the coproduct [F2] (The Drinfeld–Jimbo formulas define a Hopf algebra); it is not asserted by the single-module Cartan-projection lemma.

Proof

technique · direct. Construct the rank-one highest vectors of the tensor product by an explicit recurrence, use the $\mathbb Q$ positivity of the residue form to identify the tensor lattice, compute the residues of the divided strings and read off the crystal rule; all steps are finite and choice-free

1.1F1F2F3algebra

Tensor contravariance and divided coproduct. Since $Δ$ is an algebra map, $Δ (σ (E_{i})) = z Δ (F_{i}) Δ (K_{i}^{- 1}) = z F_{i} K_{i}^{- 1} \otimes K_{i}^{- 1} + 1 \otimes z F_{i} K_{i}^{- 1} = (σ \otimes σ) Δ (E_{i})$ , and $Δ (σ (F_{i})) = z^{- 1} Δ (K_{i}) Δ (E_{i}) = z^{- 1} K_{i} E_{i} \otimes 1 + z^{- 1} K_{i} \otimes K_{i} E_{i} = (σ \otimes σ) Δ (F_{i})$ ; the equality for $K_{i}^{\pm 1}$ is immediate. Both $Δ σ$ and $(σ \otimes σ) Δ$ reverse products, so equality on generators proves equality on the rank-one algebra. Factor contravariance then gives $B (Δ (X) (x \otimes y), x^{'} \otimes y^{'}) = B (x \otimes y, (σ \otimes σ) Δ (X) (x^{'} \otimes y^{'})) = B (x \otimes y, Δ (σ (X)) (x^{'} \otimes y^{'}))$ for all $X$ ; extend by bilinearity to all tensor vectors. This proves the tensor contravariance of the product form. From $Δ F_{i} = F_{i} \otimes 1 + K_{i} \otimes F_{i}$ and the $q$ -commutation $(K_{i} \otimes F_{i}) (F_{i} \otimes 1) = z^{- 2} (F_{i} \otimes 1) (K_{i} \otimes F_{i})$ , expanding the $n$ -th power gives $Δ F_{i}^{(n)} = \sum_{s = 0}^{n} z^{- s (n - s)} F_{i}^{(n - s)} K_{i}^{s} \otimes F_{i}^{(s)}$ : the coefficient of $(F_{i} \otimes 1)^{n - s} (K_{i} \otimes F_{i})^{s}$ in $(Δ F_{i})^{n}$ is $z^{- s (n - s)} {(\binom{n}{s})}_{z}$ , and division by $[n]_{z}!$ cancels the Gaussian binomials as $[n]_{z}! = {(\binom{n}{s})}_{z} [s]_{z}! [n - s]_{z}!$ .

1.2F1F2algebra

The highest vectors. For $0 \leq t \leq \min (a, b)$ define $c_{0} = 1$ and $c_{l + 1} = - c_{l} z^{b - 2 t + 2 l + 2} \frac{[b - t + l + 1]_{z}}{[a - l]_{z}}$ for $0 \leq l < t$ ; all denominators are nonzero since $l < t \leq a$ , and set $h_{t} : = \sum_{l = 0}^{t} c_{l} v_{l} \otimes w_{t - l}$ . Substituting $Δ E_{i}$ and comparing the coefficient of $v_{l - 1} \otimes w_{t - l}$ for $1 \leq l \leq t$ gives $c_{l} [a - l + 1]_{z} z^{2 (t - l) - b} + c_{l - 1} [b - t + l]_{z}$ , which is zero by the recurrence; the coefficients of $v_{- 1} \otimes w_{t}$ and $v_{t} \otimes w_{- 1}$ vanish because those basis vectors do not exist. Hence $Δ E_{i} h_{t} = 0$ , and $h_{t}$ has weight $a + b - 2 t$ since $K_{i}$ acts on $v_{l} \otimes w_{t - l}$ by $z^{a - 2 l} z^{b - 2 (t - l)} = z^{a + b - 2 t}$ .

1.3F4F6algebra

The norm criterion. For $x = \sum_{j, k} α_{j, k} v_{j} \otimes w_{k}$ one has $B (x, x) = \sum_{j, k} α_{j, k}^{2} d (a, j) d (b, k)$ , so $x \in L$ implies $B (x, x) \in A_{0}$ because $d (a, j) d (b, k) \in A_{0}$ ; conversely, if $B (x, x) \in A_{0}$ but $x \notin L$ , take the minimal $N \geq 1$ with $q^{N} x \in L$ and let $\overline{x} \neq 0$ be its residue in $L / q L$ . Then $B (q^{N} x, q^{N} x) = q^{2 N} B (x, x) \in q A_{0}$ , while on residues the form $\overline{B}$ is diagonal with all entries congruent to $1$ modulo $q$ , so $\overline{B} (\overline{x}, \overline{x})$ is the sum of the squares of the rational residues of the coordinates of $\overline{x}$ , a positive rational, contradicting $\overline{B} (\overline{x}, \overline{x}) = 0$ . Hence $L = {x : B (x, x) \in A_{0}}$ .

2.1F4F6step 1.2

Valuations. By [F4], ${ord}_{z} c_{l + 1} - {ord}_{z} c_{l} = (b - 2 t + 2 l + 2) + (1 - (b - t + l + 1)) - (1 - (a - l)) = a - t + 1$ for every $l$ , so the orders telescope to ${ord}_{z} c_{l} = l (a - t + 1) > 0$ for $l \geq 1$ , with ${ord}_{z} c_{0} = 0$ ; hence $c_{l} \in A_{0}$ , $c_{0} = 1$ and $c_{l} \in q A_{0}$ for $l > 0$ . Therefore $h_{t} \equiv v_{0} \otimes w_{t} (m o d q L)$ , and by [F6] $B (h_{t}, h_{t}) = \sum_{l} c_{l}^{2} d (a, l) d (b, t - l) \equiv d (b, t) \equiv 1 (m o d q)$ , since $d (a, l) \in 1 + q A_{0}$ and the $l > 0$ terms lie in $q A_{0}$ .

2.2F1step 1.1step 1.2

Coefficients. Put $u_{t, n} : = F_{i}^{(n)} h_{t}$ and write $u_{t, n} = \sum_{j, k} γ_{j, k} v_{j} \otimes w_{k}$ ; by weight, $γ_{j, k} = 0$ unless $j + k = t + n$ . Applying $Δ F_{i}^{(n)} = \sum_{s} z^{- s (n - s)} F_{i}^{(n - s)} K_{i}^{s} \otimes F_{i}^{(s)}$ of step 1.1 to $h_{t}$ and using [F1] gives the finite sum $γ_{j, k} = \sum_{l} c_{l} z^{- s (n - s) + s (a - 2 l)} {(\binom{j}{l})}_{z} {(\binom{k}{t - l})}_{z}$ with $s = k - t + l$ , over the integers $l$ with $0 \leq l \leq \min (t, j)$ and $0 \leq s \leq n$ (the symmetry of the Gaussian binomials rewrites the divided-power coefficients).

3.1F3F4F5step 2.1step 1.3

The tensored strings lie in $L$ . Let $A : = a + b - 2 t$ and $u_{t, n} : = F_{i}^{(n)} h_{t}$ for $0 \leq n \leq A$ . By F5, $F_{i}^{(n)} h_{t} \neq 0$ for $0 \leq n \leq A$ ; the span of these $A + 1$ vectors is generated by $h_{t}$ and stable under $E_{i}, F_{i}, K_{i}$ , so it is the simple rank-one module of highest weight $A$ by [F1], and the form restricted to it is a nonzero contravariant form, hence by [F3] a multiple of the normalized form, with factor $B (h_{t}, h_{t})$ ; hence $B (u_{t, n}, u_{t, n}) = d (A, n) B (h_{t}, h_{t}) \in A_{0}$ by step 2.1 and [F4]. The norm criterion of step 1.3 therefore gives $u_{t, n} \in L$ .

4.1step 3.1step 2.1step 2.2F4algebra

Integrality of the residue. Every summand in step 2.2 is an integer Laurent series in $z$ : its factors $c_{l}$ , $[m]_{z}$ and ${(\binom{N}{R})}_{z}$ are quotients of Laurent polynomials whose denominator factors $[m]_{z}$ have leading coefficient $\pm 1$ , so their inverses are Laurent series with integer coefficients with finitely many negative terms. Since $u_{t, n} \in L$ by step 3.1, each $γ_{j, k}$ lies in $A_{0}$ ; an element of $A_{0}$ that is also an integer Laurent series has an integer residue, and the sum of the squares of the residues is $1$ because $B (u_{t, n}, u_{t, n}) \equiv 1 (m o d q)$ by step 2.1 and [F4], while the Gram entries are congruent to $1$ .

5.1F4step 2.2step 4.1

Distinguished coefficient. If $n \leq a - t$ , take $(j, k) = (n, t)$ . The summand with $l = 0$ is $c_{0} z^{0} {(\binom{n}{0})}_{z} {(\binom{t}{t})}_{z} = 1$ ; a summand with $l \geq 1$ has $s = l$ and valuation $l (a - t + 1) + l (a - 2 l) - l (n - l) - l (t - l) - l (n - l) = l (2 a - 2 n - 2 t + 1) + l^{2} \geq l + l^{2} > 0$ by the additivity of ${ord}_{z}$ in [F4] (here $a - t - n \geq 0$ ). Hence $γ_{n, t} \equiv 1 (m o d q)$ .

5.2F4step 2.2step 4.1algebra

Second distinguished coefficient. If $n \geq a - t$ , take $(j, k) = (a - t, n - a + 2 t)$ . The summand with $l = 0$ is $z^{t (n - a + t)} {(\binom{n - a + 2 t}{t})}_{z}$ , of valuation $t (n - a + t) - t (n - a + t) = 0$ and leading coefficient $1$ by [F4]; a summand with $l \geq 1$ and ${(\binom{j}{l})}_{z} \neq 0$ , so $l \leq a - t$ , has $s = n - a + t + l$ and valuation $l (a - t + 1) - s (n - s) + s (a - 2 l) - l (j - l) - (t - l) (k - t + l) = l + l^{2} > 0$ , where the exponent identity uses $k - t + l = s$ . Hence $γ_{a - t, n - a + 2 t} \equiv 1 (m o d q)$ .

6.1step 4.1step 5.1step 5.2

The residues are the chain positions. By steps 5.1-5.2 the distinguished coefficient of $u_{t, n}$ has residue $1$ . Since all residues are integers and their squares sum to $1$ by step 4.1, every other coefficient has residue $0$ . Hence the residue of $u_{t, n}$ is the single basis element $v_{j} \otimes w_{k}$ with $(j, k) = (n, t)$ when $n \leq a - t$ and $(j, k) = (a - t, n - a + 2 t)$ when $n \geq a - t$ .

7.1step 6.1algebra

Partition into chains. Fixing $t$ and letting $n$ run over $0, \dots, a + b - 2 t$ , the positions of step 6.1 are $(0, t), (1, t), \dots, (a - t, t), (a - t, t + 1), \dots, (a - t, b)$ . These are the chains of the statement, and they partition the rectangle $0 \leq j \leq a$ , $0 \leq k \leq b$ : a cell $(j, k)$ lies on the chain of $t = \min (k, a - j)$ , and then it is the position $n = j$ if $k = t$ and $n = a - t + k - t$ if $k > t$ (then $j = a - t$ ), which determines the position uniquely. The total number of vectors $u_{t, n}$ is $\sum_{t = 0}^{\min (a, b)} (a + b - 2 t + 1) = (a + 1) (b + 1)$ .

8.1step 3.1step 6.1step 7.1algebra

$A_{0}$ -basis. The matrix expressing the family ${u_{t, n}}$ in the product basis reduces modulo $q$ to a permutation matrix by steps 6.1-7.1, hence has determinant congruent to $\pm 1$ modulo $q$ , a unit of $A_{0}$ ; the adjugate identity over $A_{0}$ makes the matrix invertible over $A_{0}$ . Since ${u_{t, n}} \subseteq L$ by step 3.1 and both families have $(a + 1) (b + 1)$ members, ${u_{t, n}}$ is an $A_{0}$ -basis of $L$ .

9.1step 7.1step 8.1F5algebra∎

The crystal rule. By [F5] the operators act on the strings by ${\tilde{e}}_{i} F_{i}^{(n)} h_{t} = F_{i}^{(n - 1)} h_{t}$ and ${\tilde{f}}_{i} F_{i}^{(n)} h_{t} = F_{i}^{(n + 1)} h_{t}$ , with the endpoint terms zero; since the strings form the $A_{0}$ -basis of $L$ by step 8.1, both operators preserve $L$ and shift the residues along the chains of step 7.1. Reading off which tensor factor changes gives the two displayed rules: along $(n, t) \mapsto (n + 1, t)$ the first factor moves exactly when $a - (n) > t$ , that is $a - j > k$ , and along $(a - t, k) \mapsto (a - t, k + 1)$ the second factor moves, which is the case $a - j \leq k$ ; the ${\tilde{e}}_{i}$ -rule is read in the reverse direction, where the first factor moves exactly when $a - j \geq k$ . Terms leaving the ranges are $0$ . For residues $b = v_{j} \otimes w_{k}$ and $b^{'}$ : if $a - j > k$ then ${\tilde{f}}_{i} b = v_{j + 1} \otimes w_{k}$ and $a - (j + 1) \geq k$ , so ${\tilde{e}}_{i} b^{'} = b$ ; if $a - j \leq k$ then ${\tilde{f}}_{i} b = v_{j} \otimes w_{k + 1}$ and $a - j < k + 1$ , so ${\tilde{e}}_{i} b^{'} = b$ ; the converse implications follow by the same two evaluations of the displayed rules. Hence ${\tilde{f}}_{i} b = b^{'}$ if and only if $b = {\tilde{e}}_{i} b^{'}$ . No choice is used, and neither the grand loop nor any freeness over $Z [q, q^{- 1}]$ enters.

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Sources

Masaki Kashiwara, On crystal bases of the q-analogue of universal enveloping algebras, Duke Math. J. 63 (1991), no. 2, 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)
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)