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The Fock E- and F-actions on the shape (2,1) at e = 3

Example

At $e = 3$ , charge $0$ , the quantum-group standard vector $∣ 2, 1 ⟩$ has $F_{0} ∣ 2, 1 ⟩ = ∣ 2, 2 ⟩, F_{1} ∣ 2, 1 ⟩ = q^{- 1} ∣ 2, 1, 1 ⟩, F_{2} ∣ 2, 1 ⟩ = ∣ 3, 1 ⟩,$ $E_{0} ∣ 2, 1 ⟩ = 0, E_{1} ∣ 2, 1 ⟩ = q^{- 1} ∣ 1, 1 ⟩, E_{2} ∣ 2, 1 ⟩ = ∣ 2 ⟩ .$ The transported symmetric-group standard vector $M_{2, 1} = J^{- 1} ∣ 2, 1 ⟩$ (the shape is self-conjugate) satisfies $F_{0}^{s y m} M_{2, 1} = M_{2, 2}, F_{1}^{s y m} M_{2, 1} = M_{2, 1, 1}, F_{2}^{s y m} M_{2, 1} = q^{- 1} M_{3, 1},$ $E_{0}^{s y m} M_{2, 1} = 0, E_{1}^{s y m} M_{2, 1} = M_{1, 1}, E_{2}^{s y m} M_{2, 1} = q^{- 1} M_{2},$ because $J$ transposes the outputs and reflects the colors: the operators used are $F_{i}^{s y m} = J^{- 1} F_{- i}^{Q G} J$ and $E_{i}^{s y m} = J^{- 1} E_{- i}^{Q G} J$ of The symmetric-group level-one residue and partition-label convention.

The normalized quantum-group crystal at $2, 1$ has arrows ${\tilde{f}}_{0} \to 2, 2, {\tilde{f}}_{1} \to 2, 1, 1, {\tilde{e}}_{1} \to 1, 1, {\tilde{f}}_{2} = {\tilde{e}}_{2} = 0,$ while the restricted symmetric model at $2, 1$ has instead ${\tilde{f}}_{0} \to 2, 2, {\tilde{f}}_{2} \to 3, 1, {\tilde{e}}_{2} \to 2, {\tilde{f}}_{1} = {\tilde{e}}_{1} = 0.$ The raw vector $2, 1$ is the representative of its crystal residue in each model; an ordinary coefficient such as the $q^{- 1}$ in $F_{1}$ or $F_{2}^{s y m}$ does not refute that, and the Chevalley action of $E_{i}$ , $F_{i}$ is not their normalized Kashiwara action.

Facts & Assumptions

Given: $e = 3$ , charge $0$ , the shape $λ = (2, 1)$ with standard vector $∣ 2, 1 ⟩$ , the Fock node operators and formulas of The node order and the signed exponents $N_{i}^{l}$ and $N_{i}^{r}$ and The Fock space node formulas satisfy the quantum affine relations, and the transport and signature conventions of The symmetric-group level-one residue and partition-label convention.

[F1]

Corners and residues at $(2, 1)$ : the addable nodes are $(1, 3)$ , $(2, 2)$ , $(3, 1)$ with residues $2, 0, 1$ , and the removable nodes are $(1, 2)$ , $(2, 1)$ with residues $1, 2$ (The symmetric-group level-one residue and partition-label convention).

[F2]

Node formulas: for an addable $i$ -node $γ$ of $λ$ , the coefficient of $∣ λ \cup γ ⟩$ in $F_{i} ∣ λ ⟩$ is $q^{N_{i}^{r} (λ, γ)}$ , where $N_{i}^{r}$ counts addable minus removable $i$ -nodes of the smaller diagram (here $λ$ ) strictly to the right of $γ$ (larger content, $γ$ excluded); for a removable $i$ -node $γ$ of $μ$ , the coefficient of $∣ μ ∖ γ ⟩$ in $E_{i} ∣ μ ⟩$ is $q^{- N_{i}^{l}}$ , where $N_{i}^{l}$ counts addable minus removable $i$ -nodes of the smaller diagram $μ ∖ γ$ strictly to the left of $γ$ ; the finite signature selection is defined separately from these coefficients (The node order and the signed exponents $N_{i}^{l}$ and $N_{i}^{r}$ ).

[F3]

The node operators satisfy the affine quantum relations, so $∣ 2, 1 ⟩$ is a vector of a chosen integrable module and the displayed module actions are the ordinary generator actions; ordinary standard coefficients may contain $q^{- 1}$ poles, and normalized Kashiwara lattice stability is a separate statement (The Fock space node formulas satisfy the quantum affine relations).

[F4]

Normalized operators: the transported normalized operators on all partitions are the modular signature operators, and the quantum-group normalized operators act by the reference signature (good $=$ first surviving $-$ , cogood $=$ last surviving $+$ in increasing content order), including zero values (The QG Fock-crystal arrows are the modular good-node operators).

Verification

technique · direct

1.1givenF1F2

Compute the quantum-group Chevalley action at $∣ 2, 1 ⟩$ from [F1] and [F2]. Color $0$ : the only $0$ -node is the addable $(2, 2)$ , whose counts vanish ( $(2, 2)$ is the only residue- $0$ corner), so $F_{0} ∣ 2, 1 ⟩ = ∣ 2, 2 ⟩$ and $E_{0} ∣ 2, 1 ⟩ = 0$ . Color $1$ : the addable $1$ -node is $(3, 1)$ , and in $λ$ the unique residue- $1$ node strictly to its right is the removable $(1, 2)$ , so $N_{1}^{r} = - 1$ and $F_{1} ∣ 2, 1 ⟩ = q^{- 1} ∣ 2, 1, 1 ⟩$ ; the removable $1$ -node is $(1, 2)$ , giving $(1, 1)$ as the smaller diagram, whose residue- $1$ nodes strictly to the left of $(1, 2)$ reduce to the single addable $(3, 1)$ (the removable $(2, 1)$ of that diagram has residue $2$ and is not counted), with addable-minus-removable count $1 - 0 = 1$ , so $N_{1}^{l} = 1$ and $E_{1} ∣ 2, 1 ⟩ = q^{- 1} ∣ 1, 1 ⟩$ . Color $2$ : the addable $2$ -node is $(1, 3)$ , and the only other residue- $2$ node of $(2, 1)$ is the removable $(2, 1)$ , which lies strictly to the left, so $N_{2}^{r} = 0$ and $F_{2} ∣ 2, 1 ⟩ = ∣ 3, 1 ⟩$ ; the removable $2$ -node is $(2, 1)$ , giving the smaller diagram $(2)$ , whose residue- $2$ nodes strictly to the left have count $0$ , so $E_{2} ∣ 2, 1 ⟩ = ∣ 2 ⟩$ .

2.1givenstep 1.1

Transport along $J$ with reflected colors. Since $(2, 1)^{'} = (2, 1)$ , $J M_{2, 1} = ∣ 2, 1 ⟩$ . For $F_{1}^{s y m} = J^{- 1} F_{- 1} J = J^{- 1} F_{2} J$ we get $J^{- 1} ∣ 3, 1 ⟩ = M_{(3, 1)^{'}} = M_{2, 1, 1}$ ; for $F_{2}^{s y m} = J^{- 1} F_{1} J$ we get $q^{- 1} J^{- 1} ∣ 2, 1, 1 ⟩ = q^{- 1} M_{(2, 1, 1)^{'}} = q^{- 1} M_{3, 1}$ ; for $F_{0}^{s y m} = J^{- 1} F_{0} J$ we get $J^{- 1} ∣ 2, 2 ⟩ = M_{2, 2}$ since $(2, 2)^{'} = (2, 2)$ . For the $E$ 's: $E_{1}^{s y m} = J^{- 1} E_{- 1} J = J^{- 1} E_{2} J$ gives $J^{- 1} ∣ 2 ⟩ = M_{(2)^{'}} = M_{1, 1}$ , and $E_{2}^{s y m} = J^{- 1} E_{1} J$ gives $q^{- 1} J^{- 1} ∣ 1, 1 ⟩ = q^{- 1} M_{(1, 1)^{'}} = q^{- 1} M_{2}$ ; $E_{0}^{s y m} = J^{- 1} E_{0} J = 0$ . This is the transported table of the Example statement.

2.2givenF4step 1.1

Normalized crystal arrows. For the quantum-group reading, apply the reference signature: at color $0$ the word is $+$ at $(2, 2)$ , giving ${\tilde{f}}_{0} \to (2, 2)$ ; at color $1$ the increasing-content word is $+$ at $(3, 1)$ then $-$ at $(1, 2)$ , with no cancellable pair, so the first surviving $-$ is $(1, 2)$ (giving ${\tilde{e}}_{1} \to (1, 1)$ ) and the last surviving $+$ is $(3, 1)$ (giving ${\tilde{f}}_{1} \to (2, 1, 1)$ ); at color $2$ the increasing-content word is $-$ at $(2, 1)$ then $+$ at $(1, 3)$ , which cancels, so ${\tilde{f}}_{2} = {\tilde{e}}_{2} = 0$ . For the restricted symmetric model, apply the modular top-to-bottom signature: at color $0$ the word is the single $+$ at $(2, 2)$ , so ${\tilde{f}}_{0} \to (2, 2)$ ; at color $1$ the top-to-bottom word is $-$ at $(1, 2)$ then $+$ at $(3, 1)$ , which cancels, so ${\tilde{f}}_{1} = {\tilde{e}}_{1} = 0$ ; at color $2$ the word is $+$ at $(1, 3)$ then $-$ at $(2, 1)$ , with no cancellable pair, so the first surviving $-$ is $(2, 1)$ (giving ${\tilde{e}}_{2} \to (2)$ ) and the last surviving $+$ is $(1, 3)$ (giving ${\tilde{f}}_{2} \to (3, 1)$ ). These are the two arrow tables of the Example statement, and they correspond under conjugation with negated colors as in [F4].

3.1givenF3F4step 1.1step 2.1step 2.2∎

Raw vectors versus coefficients. By [F4] the standard vectors represent the crystal basis of the $A_{0}$ chart, so $∣ 2, 1 ⟩$ and $M_{2, 1}$ are nonzero modulo $q$ and each represents its crystal residue. The ordinary actions computed in steps 1.1-2.1 have coefficients $1$ and $q^{- 1}$ , and the mismatch with step 2.2 is real: $E_{2} ∣ 2, 1 ⟩ = ∣ 2 ⟩ \neq 0$ while ${\tilde{e}}_{2}^{Q G} (2, 1) = 0$ , and $F_{2} ∣ 2, 1 ⟩ = ∣ 3, 1 ⟩ \neq 0$ while ${\tilde{f}}_{2}^{Q G} (2, 1) = 0$ ; in the symmetric model the corresponding mismatch is in color $1$ : $E_{1}^{s y m} M_{2, 1} = M_{1, 1}$ and $F_{1}^{s y m} M_{2, 1} = M_{2, 1, 1}$ are nonzero, while both color- $1$ crystal arrows vanish. Both color- $2$ crystal arrows in that model are nonzero, as step 2.2 records. Hence the Chevalley coefficients do not define the arrows, exactly as stated.

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Sources

Bernard Leclerc, Fock space representations (complete author-hosted 38-page lecture notes) (standard reference, not scraped)
Alexander Kleshchev, Representation Theory of Symmetric Groups and Related Hecke Algebras, Bull. AMS 47 (2010) 419-481, arXiv:0909.4844 (complete 66-page survey) (standard reference, not scraped)