Level-One Fock Space and Residue Crystals — Examples

Compute the quantum-group Chevalley action at $\mathrm{\mid }2,1⟩$ 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 $F_0\mathrm{\mid }2,1⟩=\mathrm{\mid }2,2⟩$ and $E_0\mathrm{\mid }2,1⟩=0$. Color $1$: the addable $1$-node is $(3,1)$, and in $\lambda$ the unique residue-$1$ node strictly to its right is the removable $(1,2)$, so $N_1^r=-1$ and $F_1\mathrm{\mid }2,1⟩=q^{-1}\mathrm{\mid }2,1,1⟩$; 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 $E_1\mathrm{\mid }2,1⟩=q^{-1}\mathrm{\mid }1,1⟩$. 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 $F_2\mathrm{\mid }2,1⟩=\mathrm{\mid }3,1⟩$; the removable $2$-node is $(2,1)$, giving the smaller diagram $(2)$, whose residue-$2$ nodes strictly to the left have count $0$, so $E_2\mathrm{\mid }2,1⟩=\mathrm{\mid }2⟩$.

Transport along $J$ with reflected colors. Since $(2,1{)}^{\mathrm{\prime }}=(2,1)$, $J{M}_{2,1}=\mathrm{\mid }2,1⟩$. For $F_1^{\mathrm{s}\mathrm{y}\mathrm{m}}=J^{-1}{F}_{-1}J=J^{-1}{F}_{2}J$ we get $J^{-1}\mathrm{\mid }3,1⟩=M_{(3,1{)}^{\mathrm{\prime }}}=M_{2,1,1}$; for $F_2^{\mathrm{s}\mathrm{y}\mathrm{m}}=J^{-1}{F}_{1}J$ we get $q^{-1}{J}^{-1}\mathrm{\mid }2,1,1⟩=q^{-1}{M}_{(2,1,1{)}^{\mathrm{\prime }}}=q^{-1}{M}_{3,1}$; for $F_0^{\mathrm{s}\mathrm{y}\mathrm{m}}=J^{-1}{F}_{0}J$ we get $J^{-1}\mathrm{\mid }2,2⟩=M_{2,2}$ since $(2,2{)}^{\mathrm{\prime }}=(2,2)$. For the $E$'s: $E_1^{\mathrm{s}\mathrm{y}\mathrm{m}}=J^{-1}{E}_{-1}J=J^{-1}{E}_{2}J$ gives $J^{-1}\mathrm{\mid }2⟩=M_{(2{)}^{\mathrm{\prime }}}=M_{1,1}$, and $E_2^{\mathrm{s}\mathrm{y}\mathrm{m}}=J^{-1}{E}_{1}J$ gives $q^{-1}{J}^{-1}\mathrm{\mid }1,1⟩=q^{-1}{M}_{(1,1{)}^{\mathrm{\prime }}}=q^{-1}{M}_2$; $E_0^{\mathrm{s}\mathrm{y}\mathrm{m}}=J^{-1}{E}_{0}J=0$. This is the transported table of the Example statement.

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].

Raw vectors versus coefficients. By [F4] the standard vectors represent the crystal basis of the $A_0$ chart, so $\mathrm{\mid }2,1⟩$ 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: $E_2\mathrm{\mid }2,1⟩=\mathrm{\mid }2⟩\mathrm{\ne }0$ while ${\tilde{e}}_2^{\mathrm{Q}\mathrm{G}}(2,1)=0$, and $F_2\mathrm{\mid }2,1⟩=\mathrm{\mid }3,1⟩\mathrm{\ne }0$ while ${\tilde{f}}_2^{\mathrm{Q}\mathrm{G}}(2,1)=0$; in the symmetric model the corresponding mismatch is in color $1$: $E_1^{\mathrm{s}\mathrm{y}\mathrm{m}}{M}_{2,1}=M_{1,1}$ and $F_1^{\mathrm{s}\mathrm{y}\mathrm{m}}{M}_{2,1}=M_{2,1,1}$ 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.

The vertex list is the list of $3$-restricted partitions by rank. A partition is $3$-restricted when ${\lambda }_a-{\lambda }_{a+1}<3$ for all $a$ (padded by zeros). Rank $0$: $\mathrm{\varnothing }$. Rank $1$: $(1)$. Rank $2$: the partitions of $2$ are $(2)$ and $(1,1)$, and both satisfy the inequalities ($2-0=2<3$, and $1-1=0$, $1-0=1<3$). Rank $3$: the partitions are $(3)$, $(2,1)$, $(1,1,1)$; $(3)$ fails since $3-0=3$, while $(2,1)$ has $2-1=1<3$ and $(1,1,1)$ has parts differences $0,1<3$. Rank $4$: the partitions are $(4)$, $(3,1)$, $(2,2)$, $(2,1,1)$, $(1,1,1,1)$; $(4)$ fails ($4-0=4\ge 3$) and the other four satisfy $3-1=2<3$, $2-2=0<3$, $2-1=1<3$ together with $1-1=0$, and $0,0,0<3$. Hence the displayed list has $1,1,2,2,4$ shapes, and by [F2] these are precisely the component vertices in ranks $0$-$4$.

The additions from ranks $0$-$2$. Write each signature as the word of signs of the listed nodes and reduce. $\mathrm{\varnothing }$: color $0$ has the single addable node $(1,1)$, word $+$, cogood $(1,1)$, hence ${\tilde{f}}_0\mathrm{\varnothing }=(1)$; colors $1,2$ have no nodes, so those arrows are zero. $(1)$: color $0$ has the removable node $(1,1)$, word $-$ (no cogood, ${\tilde{f}}_0=0$); color $1$ has the addable node $(1,2)$, word $+$, so ${\tilde{f}}_1(1)=(2)$; color $2$ has the addable node $(2,1)$ of residue $-1\equiv 2$, word $+$, so ${\tilde{f}}_2(1)=(1,1)$. $(2)$: color $0$ has no nodes (${\tilde{f}}_0={\tilde{e}}_0=0$); color $1$ has the removable $(1,2)$, word $-$, so ${\tilde{f}}_1(2)=0$ and ${\tilde{e}}_1(2)=(1)$; color $2$ has the addable $(1,3)$ and the addable $(2,1)$, both of residue $2$, in the top-to-bottom word $++$, whose last surviving $+$ is $(2,1)$, so ${\tilde{f}}_2(2)=(2,1)$ and ${\tilde{e}}_2(2)=0$. $(1,1)$: color $0$ has no nodes; color $1$ has the addables $(1,2)$ and $(3,1)$, word $++$, cogood $(3,1)$, so ${\tilde{f}}_1(1,1)=(1,1,1)$; color $2$ has the removable $(2,1)$ of residue $-1\equiv 2$, word $-$, so ${\tilde{e}}_2(1,1)=(1)$ and ${\tilde{f}}_2(1,1)=0$.

The additions from rank $3$ and the vanishing claims. $(2,1)$: color $0$ has the addable $(2,2)$ (residue $0$), word $+$, so ${\tilde{f}}_0(2,1)=(2,2)$; color $1$ has the removable $(1,2)$ and the addable $(3,1)$, top-to-bottom word $-+$, which cancels, so ${\tilde{f}}_1={\tilde{e}}_1=0$ and in particular $(2,1)$ has no residue-$1$ arrow; color $2$ has the addable $(1,3)$ above the removable $(2,1)$, word $+-$ with no cancellable pair, so the last surviving $+$ is $(1,3)$, giving ${\tilde{f}}_2(2,1)=(3,1)$, and the first surviving $-$ is $(2,1)$, giving ${\tilde{e}}_2(2,1)=(2)$. $(1,1,1)$: color $0$ has the addable $(4,1)$ (residue $-3\equiv 0$), word $+$, so ${\tilde{f}}_0(1,1,1)=(1,1,1,1)$; color $1$ has the addable $(1,2)$ above the removable $(3,1)$, word $+-$ with no cancellable pair, so ${\tilde{f}}_1(1,1,1)=(2,1,1)$ and ${\tilde{e}}_1(1,1,1)=(1,1)$; color $2$ has no nodes. These are all the additions from ranks $0$-$3$ inside the window, and every other color at these vertices has just been shown to have no surviving selected sign, hence a zero arrow.

The deletions from rank $4$ and the boundary. $(2,2)$: the only removable node is $(2,2)$, of residue $0$ with word $-$, so ${\tilde{e}}_0(2,2)=(2,1)$, and there is no removable node of residue $1$ or $2$, so the good deletions ${\tilde{e}}_1$ and ${\tilde{e}}_2$ are zero; the addable nodes $(1,3)$ of residue $2$ and $(3,1)$ of residue $1$ are each the unique surviving $+$ of their color, so ${\tilde{f}}_2$ adds $(1,3)$ and ${\tilde{f}}_1$ adds $(3,1)$, both additions leaving rank $5$ and therefore omitted from the displayed window. $(3,1)$: color $2$ has the removables $(1,3)$ and $(2,1)$, both of residue $2$, word $--$, whose first surviving $-$ is $(1,3)$, so ${\tilde{e}}_2(3,1)=(2,1)$; color $1$ has the addable $(3,1)$ of residue $1$ and color $0$ the addables $(1,4)$, $(2,2)$ of residue $0$, all with no surviving $-$ for a good deletion and with cogood additions leaving rank $5$. $(2,1,1)$: color $1$ has the removables $(1,2)$ and $(3,1)$, word $--$ of residue $1$, first surviving $-$ equal to $(1,2)$, so ${\tilde{e}}_1(2,1,1)=(1,1,1)$; color $0$ has the addables $(2,2)$ and $(4,1)$, whose cogood is $(4,1)$ and leaves rank $5$; color $2$ has the single addable $(1,3)$, leaving rank $5$, and no removable. $(1,1,1,1)$: the only removable node is $(4,1)$ of residue $0$, so ${\tilde{e}}_0(1,1,1,1)=(1,1,1)$; the addables $(1,2)$ and $(5,1)$ have residues $1$ and $2$ and lead to rank $5$, so they are omitted rather than declared zero. Finally the conjugate comparison: transposition sends $(3,1)$ to $(2,1,1)$, fixes $(2,1)$, negates residues, so the arrow $(2,1)\stackrel{{\tilde{f}}_2}{\to }(3,1)$ corresponds to the quantum-group arrow ${\tilde{f}}_1$ from $(2,1)$ to $(2,1,1)$ with $f_2$ reflecting to $f_{-2}=f_1$.

The two finite checks. The shape $(4)$ is not $3$-restricted, because ${\lambda }_1-{\lambda }_2=4-0=4\ge 3$; its conjugate is $(1,1,1,1)$ (four columns of height one), not $(1,1,1)$, and it is not $3$-regular because the positive part $1$ occurs four times, and $4\ge 3$. By [F1], therefore, neither $(4)$ nor $(1,1,1,1)$ is a label of the basic component in its respective convention: $(4)$ is not a basic label in the symmetric-group model, and $(1,1,1,1)$ is not an $e$-regular label on the quantum-group side.

$\mathrm{\mid }1111⟩$ lies outside the field basic module. The integrable module of [F2] has the chosen type-one $P$-grading of the convention, with every weight in ${\mathrm{\Lambda }}_0-Q_{+}$. Since the simple roots are independent, a fixed label weight fixes all residue counts $m_i$ and hence the size ${\sum }_i{m}_i$; only finitely many partitions have that size, so its label-weight space is finite-dimensional. The weight ${\mathrm{\Lambda }}_0$ is maximal and its space is the single vacuum line. Thus, with the declared Choice, [F3] applies to $F_{\mathrm{Q}\mathrm{G}}$ with $N_0=M=U_q\mathrm{\mid }\mathrm{\varnothing }⟩=V({\mathrm{\Lambda }}_0)$. Its lattice splitting embeds $(F_0\cap M)\mathrm{/}q(F_0\cap M)$ into $F_0\mathrm{/}q{F}_0$; the corresponding crystal basis is the empty component, precisely the $e$-regular classes by [F2]. If $\mathrm{\mid }1111⟩\in M$, its residue would lie in their span. This contradicts [F5] and the nonregularity in step 1.1. Hence $\mathrm{\mid }1111⟩\mathrm{\notin }M$.

The ambient lower lift is excluded as well. By [F4] $G(1111)=\mathrm{\mid }1111⟩<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>q{F}_0$, so $G(1111)$ has the same residue class as $\mathrm{\mid }1111⟩$; if $G(1111)\in M$, then $G(1111)\in F_0\cap M$ and its residue class would lie in the span of the $e$-regular classes, contradicting the independence of [F5] exactly as in step 2.1. Hence $G(1111)\mathrm{\notin }M$, i.e. the ambient lower global lift is outside the field basic module as well.

Transport to the symmetric-group convention. Applying $J^{-1}$ to the field module $M$ gives $M_{\mathrm{s}\mathrm{y}\mathrm{m}}=J^{-1}M$, and $J^{-1}\mathrm{\mid }1111⟩=M_{(1111{)}^{\mathrm{\prime }}}=M_{(4)}$; since $J^{-1}$ is a linear isomorphism, $M_{(4)}=J^{-1}\mathrm{\mid }1111⟩\mathrm{\notin }{M}_{\mathrm{s}\mathrm{y}\mathrm{m}}$ by step 2.1, and likewise $G_{\mathrm{s}\mathrm{y}\mathrm{m}}(4)=J^{-1}G(1111)\mathrm{\notin }{M}_{\mathrm{s}\mathrm{y}\mathrm{m}}$ by step 3.1. Since also $M_{(4)}\in F_{\mathrm{s}\mathrm{y}\mathrm{m}}$, the transported full Fock space is strictly larger than its cyclic basic module, refuting the false statement.

The full signature remains meaningful on these shapes. At $(4)$ the corners are the removable node $(1,4)$ of residue $3\equiv 0$ and the addable nodes $(1,5)$ of residue $4\equiv 1$ and $(2,1)$ of residue $-1\equiv 2$; the residue-$0$ word is the single $-$, so the modular signature has good node $(1,4)$ and ${\tilde{e}}_0(4)=(3)$, an edge of the full transported crystal. Both $(4)$ and $(3)$ are non-$3$-restricted, so this edge joins two vertices of other components, not a route into the empty component, in agreement with the component identification of [F1]; modular signature statements restricted to restricted labels do not apply to $(4)$ as a basic label.