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The $i$ -string components of a lattice vector and the single surviving component modulo $q$

Statement

Let $(L, B)$ be a crystal basis of an integrable weight module $M$ in the sense of Crystal lattices, crystal bases and their crystals, let $i \in I$ , $λ \in P$ and $u \in L_{λ}$ , and let $u = \sum_{n \geq 0} F_{i}^{(n)} u_{n}$ be the admissible $i$ -string decomposition of $u$ : the sum is finite, $u_{n} \in \ker E_{i} \cap M_{λ + n α_{i}}$ , and $u_{n} = 0$ whenever $n > ⟨ h_{i}, λ + n α_{i} ⟩$ (Uniqueness of the $i$ -string decomposition of an integrable module). Then:

(a) Components lie in the lattice. $u_{n} \in L \cap M_{λ + n α_{i}}$ for every $n$ .

(b) Lattice string decomposition. $L_{λ} = ⨁_{n \geq 0} {\tilde{f}}_{i}^{n} (L_{λ + n α_{i}} \cap \ker E_{i})$ , the sum being direct because admissible decompositions are unique; equivalently $L = ⨁_{n \geq 0} {\tilde{f}}_{i}^{n} (L \cap \ker E_{i})$ .

(c) Single surviving component. If $u + q L = 0$ in $L_{λ} / q L_{λ}$ , then $u_{n} \in q L$ for every $n$ . If $u + q L \in B_{λ}$ , then there is exactly one index $n_{0}$ with $u_{n_{0}} \notin q L$ , and for it $u_{n_{0}} + q L \in B_{λ + n_{0} α_{i}}$ , $u \equiv F_{i}^{(n_{0})} u_{n_{0}} (m o d q L)$ and $u_{n_{0}} + q L = {\tilde{e}}_{i}^{n_{0}} (u + q L)$ ; in particular $n_{0} = ε_{i} (u + q L)$ (The crystal graph, $ε_{i}$ , $φ_{i}$ and the weight function), and ${\tilde{e}}_{i} u$ and ${\tilde{f}}_{i} u$ are the residues of $F_{i}^{(n_{0} - 1)} u_{n_{0}}$ and $F_{i}^{(n_{0} + 1)} u_{n_{0}}$ .

Everything is a statement about the fixed lattice $L$ and the operators ${\tilde{e}}_{i}, {\tilde{f}}_{i}$ ; no finite-dimensionality, bounded support, uniform nilpotence or choice principle is assumed. Here $F_{i}^{(- 1)} u$ is interpreted as $0$ , and $F_{i}^{(a + 1)} u = 0$ for a top of coroot weight $a$ , so the final formulas include both finite-string endpoints.

Facts & Assumptions

Given: A crystal basis $(L, B)$ of an integrable weight module $M$ (Crystal lattices, crystal bases and their crystals), an index $i \in I$ , a label $λ \in P$ and a vector $u \in L_{λ}$ with admissible $i$ -string decomposition $u = \sum_{n \geq 0} F_{i}^{(n)} u_{n}$ , $u_{n} \in \ker E_{i} \cap M_{λ + n α_{i}}$ , $u_{n} = 0$ for $n > ⟨ h_{i}, λ + n α_{i} ⟩$ (Uniqueness of the $i$ -string decomposition of an integrable module); write $T_{n} : = {\tilde{e}}_{i}^{n} u$ for $n \geq 0$ .

[F1]

Crystal lattice and crystal basis (Crystal lattices, crystal bases and their crystals): $L = ⨁_{μ \in P} L_{μ}$ with $L_{μ} = L \cap M_{μ}$ , ${\tilde{e}}_{i} L \subseteq L$ and ${\tilde{f}}_{i} L \subseteq L$ ; $B = ⨆_{μ} B_{μ}$ is a $Q$ -basis of $L / q L$ with $B_{μ} \subseteq L_{μ} / q L_{μ}$ , ${\tilde{e}}_{i} B \subseteq B \cup {0}$ and ${\tilde{f}}_{i} B \subseteq B \cup {0}$ , and ${\tilde{f}}_{i} b = b^{'}$ if and only if $b = {\tilde{e}}_{i} b^{'}$ for all $b, b^{'} \in B$ .

[F2]

Kashiwara operators (The Kashiwara operators ${\tilde{e}}_{i}$ and ${\tilde{f}}_{i}$ of an integrable module): on the unique admissible decomposition $v = \sum_{n} F_{i}^{(n)} v_{n}$ one has ${\tilde{e}}_{i} v = \sum_{n \geq 1} F_{i}^{(n - 1)} v_{n}$ and ${\tilde{f}}_{i} v = \sum_{n \geq 0} F_{i}^{(n + 1)} v_{n}$ (divided powers), both linear and shifting label weight by $α_{i}$ and $- α_{i}$ ; they preserve $L$ by [F1] and commute with module homomorphisms.

[F3]

Uniqueness of the admissible decomposition (Uniqueness of the $i$ -string decomposition of an integrable module): every vector of an integrable weight module has exactly one finite admissible $i$ -string decomposition as in the Given; additionally $F_{i}^{(n)}$ is injective on $\ker E_{i} \cap M_{μ}$ for $n \leq ⟨ h_{i}, μ ⟩$ and zero for $n > ⟨ h_{i}, μ ⟩$ , and $\ker E_{i} \cap M_{μ} = 0$ when $⟨ h_{i}, μ ⟩ < 0$ .

[F4]

Crystal graph invariants (The crystal graph, $ε_{i}$ , $φ_{i}$ and the weight function): $ε_{i} (b) = \max {n : {\tilde{e}}_{i}^{n} b \neq 0}$ and $φ_{i} (b) = \max {n : {\tilde{f}}_{i}^{n} b \neq 0}$ are finite and nonnegative on $B$ , ${\tilde{e}}_{i}^{n} b \neq 0$ exactly for $0 \leq n \leq ε_{i} (b)$ , and on an existing edge ${\tilde{e}}_{i} b = b^{'}$ one has $ε_{i} (b) = ε_{i} (b^{'}) + 1$ .

Proof

technique · direct

1.1F1F2F3

The tails $T_{n} = {\tilde{e}}_{i}^{n} u$ lie in $L \cap M_{λ + n α_{i}}$ , and $T_{n} = \sum_{k \geq n} F_{i}^{(k - n)} u_{k}$ with ${\tilde{f}}_{i} T_{n + 1} = T_{n} - u_{n}$ for every $n \geq 0$ . Indeed $T_{0} = u \in L_{λ}$ and each ${\tilde{e}}_{i}$ preserves $L$ and raises the label weight by $α_{i}$ by [F2], so $T_{n} \in L_{λ + n α_{i}}$ by [F1]; on the string decomposition ${\tilde{e}}_{i}$ substitutes $F_{i}^{(k)} u_{k} \mapsto F_{i}^{(k - 1)} u_{k}$ by [F2], giving $T_{n} = \sum_{k \geq n} F_{i}^{(k - n)} u_{k}$ , and applying ${\tilde{f}}_{i}$ to $T_{n + 1} = \sum_{k \geq n + 1} F_{i}^{(k - n - 1)} u_{k}$ shifts every divided power up by one, so ${\tilde{f}}_{i} T_{n + 1} = \sum_{k \geq n + 1} F_{i}^{(k - n)} u_{k} = T_{n} - u_{n}$ .

2.1step 1.1F1

Every $u_{n}$ lies in $L$ . For $n \geq 0$ the identity $u_{n} = T_{n} - {\tilde{f}}_{i} T_{n + 1}$ of step 1.1 exhibits $u_{n}$ as a difference of two elements of $L$ : $T_{n}, T_{n + 1} \in L$ by step 1.1 and ${\tilde{f}}_{i} L \subseteq L$ by [F1]; the weight of $u_{n}$ is $λ + n α_{i}$ by the Given, so $u_{n} \in L \cap M_{λ + n α_{i}} = L_{λ + n α_{i}}$ . This proves (a).

2.2step 1.1F1F2

Residue components when $u \in q L$ . Step 1.1 gives $u_{n} = {\tilde{e}}_{i}^{n} u - {\tilde{f}}_{i} {\tilde{e}}_{i}^{n + 1} u$ . Both operators preserve $L$ and are linear, hence preserve $q L$ ; therefore every $u_{n}$ lies in $q L$ .

3.1step 2.1F1F2F3

The lattice string decomposition (b). For a top $v \in L_{λ + n α_{i}} \cap \ker E_{i}$ the vector ${\tilde{f}}_{i}^{n} v$ equals $F_{i}^{(n)} v$ by [F2] applied to the one-term admissible decomposition of $v$ , so $u = \sum_{n} {\tilde{f}}_{i}^{n} u_{n}$ with $u_{n} \in L \cap \ker E_{i}$ by step 2.1, and conversely every ${\tilde{f}}_{i}^{n} (L_{λ + n α_{i}} \cap \ker E_{i})$ lies in $L$ by iterated stability [F1]. For a relation $\sum_{n} {\tilde{f}}_{i}^{n} v_{n} = 0$ with $v_{n} \in L \cap \ker E_{i}$ first discard the top components of coroot weight $a < n$ : their $n$ -th lowering image is $F_{i}^{(n)} v_{n} = 0$ by [F3]; the remaining representatives are admissible, so the relation is an admissible decomposition of $0$ and every remaining $v_{n} = 0$ by the uniqueness half of [F3]. Hence the sum is direct and equals $L_{λ} = ⨁_{μ} L_{μ}$ 's $λ$ -part.

3.2step 1.1step 2.1step 2.2F1F2F4

Residue components when $u + q L = b \in B_{λ}$ . Put $N : = ε_{i} (b)$ . Reducing the identity of step 1.1 gives $u_{n} + q L = {\tilde{e}}_{i}^{n} b - {\tilde{f}}_{i} {\tilde{e}}_{i}^{n + 1} b$ . For $n < N$ the partial-inverse axiom gives ${\tilde{f}}_{i} {\tilde{e}}_{i}^{n + 1} b = {\tilde{e}}_{i}^{n} b$ , so this difference is zero. For $n = N$ it is ${\tilde{e}}_{i}^{N} b \in B$ , and for $n > N$ both terms vanish. Thus exactly $u_{N}$ is outside $q L$ , with residue ${\tilde{e}}_{i}^{N} b$ . Since $F_{i}^{(n)} u_{n} = {\tilde{f}}_{i}^{n} u_{n}$ on a top component, all other summands lie in $q L$ ; consequently $u \equiv F_{i}^{(N)} u_{N} (m o d q L)$ . This proves (c), including $N = 0$ , where raising, rather than lowering, kills the top.

4.1step 3.2step 2.2F1F2F3given∎

The remaining residual formulas and the choice audit. By the previous steps, when $u + q L \in B_{λ}$ every component $u_{n}$ with $n \neq n_{0}$ lies in $q L$ , so ${\tilde{e}}_{i} u = \sum_{n \geq 1} F_{i}^{(n - 1)} u_{n}$ and ${\tilde{f}}_{i} u = \sum_{n \geq 0} F_{i}^{(n + 1)} u_{n}$ are congruent modulo $q L$ to $F_{i}^{(n_{0} - 1)} u_{n_{0}}$ and $F_{i}^{(n_{0} + 1)} u_{n_{0}}$ respectively, with $F_{i}^{(- 1)} u_{0} : = 0$ when $n_{0} = 0$ and $F_{i}^{(a + 1)} u_{n} : = 0$ beyond the string bottom by [F3]; this proves the final clause of (c). Every step used only the fixed lattice $L$ , the finitely supported string decomposition of $u$ and the stated operators, so no finite-dimensionality, bounded support, uniform nilpotence or choice principle is assumed.

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15 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

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)

The ii-string components of a lattice vector and the single surviving component modulo qq