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A free $A_{0}$ -lattice generating the module that is not a crystal lattice

Statement refuted

Refuted claim: for a free $A_{0}$ -submodule $L$ of an integrable module that generates it over $Q (q)$ , the two extra crystal-lattice axioms are automatic: the weight decomposition (ii) and the operator stability (iii) would each follow from the other.

This is false. In the two-dimensional simple $U_{q} ({s l}_{2})$ -module $V (ω) = L_{1} (1)$ the lattice $L = A_{0} (v + w) \oplus A_{0} (q v)$ is free of rank $2$ and generates $V (ω)$ , but it violates both (ii) and (iii), so no set $B$ makes $(L, B)$ a crystal basis. Moreover the two axioms are independent, not merely jointly non-automatic: the lattice $L^{'} = A_{0} (q v) \oplus A_{0} w$ satisfies (ii) and violates (iii), while in an explicitly specified rank-one datum with a central toral direction the lattice $A_{0} (u + w) \oplus A_{0} (q u)$ satisfies (iii) and violates (ii). No choice principle is used.

Facts & Assumptions

Given: The type-one datum of ${s l}_{2}$ with $a_{11} = 2$ , $q_{1} = q$ , $P = Z ω$ , $⟨ h_{1}, ω ⟩ = 1$ , $α_{1} = 2 ω$ , and the two-dimensional simple module $V (ω) = L_{1} (1)$ of Divided-power commutation and the simple $U_{q_{i}} ({s l}_{2})$ string modules(iii); and the auxiliary rank-one datum of step 1.3 below.

[F1]

$V (ω)$ has basis $v$ of weight $ω$ and $w : = F_{1} v$ of weight $ω - α_{1}$ , with $E_{1} v = 0$ , $F_{1} v = w$ , $E_{1} w = v$ , $F_{1} w = 0$ , $K_{1} v = q v$ and $K_{1} w = q^{- 1} w$ (Divided-power commutation and the simple $U_{q_{i}} ({s l}_{2})$ string modules, part (iii)).

[F2]

The string decomposition of $v$ is $v = v$ with $v \in \ker E_{1}$ , and that of $w$ is $w = F_{1}^{(1)} v$ ; hence ${\tilde{e}}_{1} v = 0$ , ${\tilde{f}}_{1} v = w$ , ${\tilde{e}}_{1} w = v$ , ${\tilde{f}}_{1} w = 0$ , and $L (ω) = A_{0} v \oplus A_{0} w$ is a crystal lattice with crystal basis the classes of $v$ and $w$ (The Kashiwara operators ${\tilde{e}}_{i}$ and ${\tilde{f}}_{i}$ of an integrable module, Uniqueness of the $i$ -string decomposition of an integrable module, The Kashiwara lattices $L (λ)$ and $L (\infty)$ and their generating sets).

[F3]

A crystal lattice is a free $A_{0}$ -submodule $L \subseteq M$ with (i) $Q (q) \otimes_{A_{0}} L ≅ M$ , (ii) $L = ⨁_{μ} L_{μ}$ with $L_{μ} = L \cap M_{μ}$ free, and (iii) ${\tilde{e}}_{i} L \subseteq L$ , ${\tilde{f}}_{i} L \subseteq L$ ; a crystal basis requires a crystal lattice (Crystal lattices, crystal bases and their crystals).

[F4]

The auxiliary datum: $P = Z ω \oplus Z δ$ , $P^{\lor} = Z h \oplus Z d$ , $⟨ h, ω ⟩ = ⟨ d, δ ⟩ = 1$ and all other pairings zero, $α_{1} = 2 ω$ with simple coroot $h$ and $d_{1} = 1$ ; this is a symmetrizable Cartan datum with $a_{11} = ⟨ α_{1}, h ⟩ = 2$ and a central toral direction $d$ (Symmetrizable Cartan data for quantum groups).

[F5]

For a label module with $E_{1} = F_{1} = 0$ the Kashiwara operators vanish and every $A_{0}$ -submodule is operator stable; $A_{0} = Q [q]_{(q)}$ is a DVR with units the fractions with nonzero value at $0$ (The Kashiwara operators ${\tilde{e}}_{i}$ and ${\tilde{f}}_{i}$ of an integrable module, The local ring $A_{0}$ of rational functions regular at $q = 0$ , The Drinfeld-Jimbo quantized enveloping algebra by generators and relations).

Counterexample

technique · direct

1.1F1F2algebra

The standard lattice. By [F1] and [F2], ${\tilde{e}}_{1} (v + w) = {\tilde{e}}_{1} v + {\tilde{e}}_{1} w = 0 + v = v$ and ${\tilde{e}}_{1} w = v$ , while $L = A_{0} (v + w) \oplus A_{0} (q v)$ is a free $A_{0}$ -module of rank $2$ because $v + w$ and $q v$ are $Q (q)$ -independent; it generates $V (ω)$ because $v = q^{- 1} (q v)$ and $w = (v + w) - v$ .

1.2F2F3algebra

Weight decomposition without operator stability. The lattice $L^{'} = A_{0} (q v) \oplus A_{0} w$ generates $V (ω)$ because $v = q^{- 1} (q v)$ , and it is the direct sum $L_{ω}^{'} \oplus L_{ω - α_{1}}^{'}$ with $L_{ω}^{'} = A_{0} q v$ and $L_{ω - α_{1}}^{'} = A_{0} w$ by distinct weights; but ${\tilde{e}}_{1} w = v$ , and $v = a q v + b w$ with $a, b \in A_{0}$ forces $1 = a q$ and $a = q^{- 1} \notin A_{0}$ , so $v \notin L^{'}$ and $L^{'}$ violates exactly (iii), not (ii).

1.3F4F5algebra

Operator stability without weight decomposition. In the datum of [F4], take one-dimensional modules $M_{0} = Q (q) u$ , $M_{δ} = Q (q) w$ of labels $0, δ$ , with $E_{1} = F_{1} = 0$ , $K_{h} = 1$ on both and $K_{d}$ acting by $1, q$ , respectively. The toral and mixed relations hold, so these are integrable label modules. All Kashiwara operators vanish by [F5]. Hence $L^{''} : = A_{0} (u + w) \oplus A_{0} (q u)$ is operator stable; its displayed generators are independent and span $M_{0} \oplus M_{δ}$ over the field. In $a (u + w) + b q u$ the coordinates are $(a + b q, a)$ , so $L^{''} \cap M_{0} = A_{0} q u$ (set $a = 0$ ), whereas $L^{''} \cap M_{δ} = q A_{0} w$ (set $a = - b q$ ). Their direct sum omits $u + w$ , whose coordinates are both $1 \notin q A_{0}$ . Thus this free generating lattice satisfies (iii) and violates (ii).

2.1F3step 1.1algebra

The lattice is not operator stable. If $v = a (v + w) + b q v$ with $a, b \in A_{0}$ , comparison of the coefficients of $w$ gives $a = 0$ , and then $1 = b q$ , that is $b = q^{- 1} \notin A_{0}$ ; hence $v \notin L$ while ${\tilde{e}}_{1} (v + w) = v$ , so (iii) fails.

2.2F3step 1.1algebra

The lattice is not weight decomposed. The weight pieces are $L_{ω} = L \cap Q (q) v = {a (v + w) + b q v : a = 0} = A_{0} q v$ and $L_{ω - α_{1}} = L \cap Q (q) w = {a (v + w) + b q v : a + b q = 0} = {- b q (v + w) + b q v : b \in A_{0}} = q A_{0} w$ . If $v + w = a q v + b q w$ with $a, b \in A_{0}$ , comparison of the coefficients of $v$ gives $a = q^{- 1} \notin A_{0}$ , so $v + w \notin L_{ω} \oplus L_{ω - α_{1}}$ and $L \neq L_{ω} \oplus L_{ω - α_{1}}$ : (ii) fails too.

3.1F3step 2.1step 2.2

No crystal basis on $L$ . By steps 2.1 and 2.2 the lattice $L$ violates the crystal-lattice axioms (ii) and (iii); since a crystal basis $(L, B)$ requires $L$ to be a crystal lattice by [F3], there is no set $B$ with $(L, B)$ a crystal basis of $V (ω)$ .

4.1step 3.1step 1.2step 1.3∎

Conclusion. The lattice $L$ shows that neither axiom is automatic for a free generating lattice, and steps 1.2 and 1.3 exhibit, respectively, a lattice satisfying (ii) but not (iii) and a lattice satisfying (iii) but not (ii); hence the two axioms are independent in the class of allowed graded quantum data. All maps and vectors are displayed or the zero operators, so no choice principle is used.

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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)

A free A0A0​-lattice generating the module that is not a crystal lattice