Level-One Fock Space and Residue Crystals

For every $a\ge 1$ the multiplicity of the part $a$ in ${\lambda }^{\mathrm{\prime }}$ equals ${\lambda }_a-{\lambda }_{a+1}$. Indeed the columns of $[\lambda ]$ of height at least $a$ are exactly the first ${\lambda }_a$ columns, because column $j$ has height $\mathrm{\#}\{r:{\lambda }_r\ge j\}\ge a$ if and only if $j\le {\lambda }_a$; hence the columns of height exactly $a$ number ${\lambda }_a-{\lambda }_{a+1}$, and these heights are precisely the parts of ${\lambda }^{\mathrm{\prime }}$ by [F1].

Transposition of nodes: $(r,c)\in [\lambda ]$ if and only if $(c,r)\in [{\lambda }^{\mathrm{\prime }}]$, so $(r,c)\mapsto (c,r)$ is a bijection from the nodes of $\lambda$ onto the nodes of ${\lambda }^{\mathrm{\prime }}$. It preserves addability and removability: deleting $(r,c)$ from $[\lambda ]$ leaves the diagram of a partition $\mu$ if and only if deleting $(c,r)$ from $[{\lambda }^{\mathrm{\prime }}]$ leaves the diagram of ${\mu }^{\mathrm{\prime }}$, and the same argument with diagrams enlarged by one node gives addability. It negates the residue, $\mathrm{res}(c,r)=r-c=-(c-r)=-\mathrm{res}(r,c)$ in $I=\mathbb{Z}\mathrm{/}e\mathbb{Z}$.

Distinct corners of a partition have distinct contents. Let $x=(r,c)\mathrm{\ne }y=(s,d)$ be addable or removable nodes with equal content $c-r=d-s$; by [F2] each is of the form $(a,{\lambda }_a)$ (removable) or $(a,{\lambda }_a+1)$ (addable, the new bottom row included via ${\lambda }_{k+1}=0$). If $r=s$, two distinct corners must be the removable and addable endpoints of that row, whose contents differ by $1$, so they cannot have equal content. Interchange the nodes if needed to take $r<s$. If $x$ and $y$ are both removable with $r<s$, then ${\lambda }_r\ge {\lambda }_s$ while ${\lambda }_r-r={\lambda }_s-s$ forces ${\lambda }_r-{\lambda }_s=r-s<0$, a contradiction. If both are addable with $r<s$, then ${\lambda }_r\ge {\lambda }_s$ while equal contents force ${\lambda }_r-{\lambda }_s=r-s<0$, again a contradiction. If $x$ is removable and $y$ addable with $r<s$, equal contents give ${\lambda }_r-{\lambda }_s=r+1-s$; for $s\ge r+2$ the right side is at most $-1$, so ${\lambda }_r<{\lambda }_s$, contradicting ${\lambda }_r\ge {\lambda }_s$, and for $s=r+1$ it gives ${\lambda }_r={\lambda }_s$, which contradicts the addability of $(s,{\lambda }_s+1)$, that is ${\lambda }_{s-1}={\lambda }_r>{\lambda }_s$ (here $s\ge 2$ because $r\ge 1$). If $x$ is addable with smaller row $r$ than the removable $y=(s,{\lambda }_s)$, then ${\lambda }_r\ge {\lambda }_s$ while ${\lambda }_r+1-r={\lambda }_s-s$ forces ${\lambda }_s-{\lambda }_r=s+1-r\ge 2$, contradicting ${\lambda }_r\ge {\lambda }_s$. All possible pairs are covered, so equal contents force $x=y$.

Consequently $\lambda$ is $e$-restricted, i.e. ${\lambda }_a-{\lambda }_{a+1}<e$ for all $a$, if and only if the part $a$ occurs in ${\lambda }^{\mathrm{\prime }}$ fewer than $e$ times, i.e. ${\lambda }^{\mathrm{\prime }}$ is $e$-regular: step 1.1 identifies the two lists of integers compared with $e$. For $\lambda =\mathrm{\varnothing }$, where ${\lambda }_a=0$ for all $a$, both conditions are the empty family of inequalities and both hold; for $a>k$ one has ${\lambda }_a-{\lambda }_{a+1}=0<e$, so the restrictedness condition is finite, and $m_i(\mathrm{\varnothing })=0$ for all $i$.

Step 1.2 also gives the weight bookkeeping: the transposition bijects the residue-$i$ nodes of $\lambda$ onto the residue-$(-i)$ nodes of ${\lambda }^{\mathrm{\prime }}$, so $m_i(\lambda )=m_{-i}({\lambda }^{\mathrm{\prime }})$ for every $i$. Hence $\mathrm{wt}(\mathrm{\mid }{\lambda }^{\mathrm{\prime }}⟩)={\mathrm{\Lambda }}_0-{\sum }_j{m}_j({\lambda }^{\mathrm{\prime }}){\alpha }_j={\mathrm{\Lambda }}_0-{\sum }_j{m}_{-j}(\lambda ){\alpha }_j=\tau (\mathrm{wt}(M_{\lambda }))$: applying the transposition of labels to the weight is the reflection $i\mapsto -i$ on the coefficients. In particular the relabelled basis vector $M_{\lambda }$ is a weight vector of weight $\mathrm{wt}(M_{\lambda })$ for the transported toral action, since $K_h^{\mathrm{s}\mathrm{y}\mathrm{m}}{M}_{\lambda }=J^{-1}{K}_{\tau h}^{\mathrm{Q}\mathrm{G}}J{M}_{\lambda }=q^{⟨\tau h,\mathrm{wt}(\mathrm{\mid }{\lambda }^{\mathrm{\prime }}⟩)⟩}{M}_{\lambda }=q^{⟨h,\tau \mathrm{wt}(\mathrm{\mid }{\lambda }^{\mathrm{\prime }}⟩)⟩}{M}_{\lambda }=q^{⟨h,\mathrm{wt}(M_{\lambda })⟩}{M}_{\lambda }$, using that $\tau$ preserves the pairing and exchanges $h_j$ with $h_{-j}$ (The cyclic affine type A Cartan datum and its residues).

The transported operators act by node moves with reflected colors. For any partition $\mu$, the QG operator $E_{-i}^{\mathrm{Q}\mathrm{G}}$ sends $\mathrm{\mid }\mu ⟩$ to a linear combination of $\mathrm{\mid }\mu \setminus \gamma ⟩$ over the removable nodes $\gamma$ of $\mu$ of residue $-i$, and $F_{-i}^{\mathrm{Q}\mathrm{G}}$ sends $\mathrm{\mid }\mu ⟩$ to a combination of $\mathrm{\mid }\mu \cup \beta ⟩$ over the addable nodes $\beta$ of $\mu$ of residue $-i$ (The level-one partition Fock space and node residues); conjugating by $J$ and using step 1.2, $E_i^{\mathrm{s}\mathrm{y}\mathrm{m}}$ sends $M_{\lambda }$ ($J{M}_{\lambda }=\mathrm{\mid }{\lambda }^{\mathrm{\prime }}⟩$) to the $J^{-1}$-images of the results of removing residue-$(-i)$ removable nodes of ${\lambda }^{\mathrm{\prime }}$, that is, to combinations of $M_{\lambda \setminus \gamma }$ over the removable nodes $\gamma$ of $\lambda$ of residue $i$, and likewise $F_i^{\mathrm{s}\mathrm{y}\mathrm{m}}$ adds residue-$i$ nodes of $\lambda$. Thus at the level of nodes the symmetric-group model uses the same residue labels as the modular signature of the Definition, not their negatives.

At quantum characteristic $e=0$ the colors form the set $I=\mathbb{Z}$ and, by [F2] and step 1.3, each residue class contains at most one node of $\lambda$: an $i$-node is a corner of content exactly $i$, and there is at most one such corner. Hence the residue word of every color has length at most one, no cancellation can occur, and for every nonempty $\lambda$ the removable corner $(k,{\lambda }_k)$ of its last row is the good node of its own residue, so repeated good deletions carry $\lambda$ to $\mathrm{\varnothing }$; consult the $e=0$ case of the survey's level-one example, under which every level-one partition is a crystal label. No finite affine datum, charge or weight formula is asserted here.

The conjugation $C(\lambda )={\lambda }^{\mathrm{\prime }}$ is a bijection from the $e$-restricted partition class onto the $e$-regular class by step 2.1, and it is not an involution of the $e$-restricted class: at $e=3$ it sends the restricted shape $(1,1,1)$ to the non-restricted shape $(3)$, and at $e=2$ it sends the restricted shape $(1,1,1,1)$ to the non-restricted shape $(4)$; the same-model sign (Mullineux) involution of the survey instead preserves the restricted label set and is computed by negating the residues of a good-node path inside it. In the group regime the published dictionary for conjugate-sign duality of simple labels is exactly the stated identification $D(\lambda )\cong D^{{\lambda }^{\mathrm{\prime }}}\otimes \mathrm{sgn}$ (p-regular and p-restricted labels under transpose and sign).

We prove the order claim by a case distinction on the two types. Let $x=(r,c)$, $y=(s,d)$ be corners of the same residue with $r<s$. Case both removable: $c={\lambda }_r$, $d={\lambda }_s$ and ${\lambda }_r\ge {\lambda }_s$; if ${\lambda }_r={\lambda }_s$ then ${\lambda }_{r+1}\ge {\lambda }_s={\lambda }_r$ (weak decrease, and $r+1\le s$), contradicting ${\lambda }_r>{\lambda }_{r+1}$ from [F1], so $c={\lambda }_r>{\lambda }_s=d$. Case both addable: $c={\lambda }_r+1$, $d={\lambda }_s+1$, and if ${\lambda }_r={\lambda }_s$ then ${\lambda }_{s-1}={\lambda }_r={\lambda }_s$ (as $r\le s-1$, weak decrease gives ${\lambda }_{s-1}\le {\lambda }_r$, while ${\lambda }_{s-1}\ge {\lambda }_s$), contradicting the addability condition ${\lambda }_{s-1}>{\lambda }_s$ of [F1]; hence $c={\lambda }_r+1>{\lambda }_s+1=d$. Case $x$ removable and $y$ addable: $c={\lambda }_r$ and $d={\lambda }_s+1$; addability of $y$ gives ${\lambda }_{s-1}>{\lambda }_s$. If $s\ge r+2$, then ${\lambda }_r\ge {\lambda }_{s-1}\ge {\lambda }_s+1=d$; equality ${\lambda }_r=d$ would force ${\lambda }_{r+1}\ge {\lambda }_{s-1}\ge d={\lambda }_r$ (as $r+1\le s-1$), contradicting removability ${\lambda }_r>{\lambda }_{r+1}$, so $c={\lambda }_r>d$. If $s=r+1$, then ${\lambda }_r={\lambda }_{s-1}>{\lambda }_s=d-1$, so ${\lambda }_r\ge d$; equality ${\lambda }_r=d$ would give $\mathrm{cont}(x)-\mathrm{cont}(y)=(c-d)+(s-r)=1$, and equal residues would force $e\mid 1$, impossible for $e\ge 2$; hence $c>d$ in this case too. Case $x$ addable and $y$ removable: $c={\lambda }_r+1$ and $d={\lambda }_s\le {\lambda }_r$, so $c>d$. In every case $c>d$, and then $\mathrm{cont}(x)-\mathrm{cont}(y)=(c-d)+(s-r)>0$: both the column index and the content strictly decrease along increasing row index.

The reduction of [F2] is independent of the order of its replacements. Call a pair $(j,i)$ with $j<i$ a valid pair of a word $w\in \{-,0,+{\}}^n$ if $w_j=-$, $w_i=+$ and all entries strictly between are $0$, and reduce $w$ greedily by scanning left to right, pushing each encountered $-$ and, at each $+$, cancelling it with the most recently pushed uncancelled $-$ if there is one, leaving the $+$ as a survivor otherwise; the scan is deterministic. If $(j,i)$ is a valid pair, then the scan of $w$ cancels the $+$ at $i$ with the $-$ at $j$: no entry strictly between $j$ and $i$ is nonzero, so at time $i$ the position $j$ is still uncancelled and is the most recent uncancelled $-$. Consequently passing from $w$ to the word $w^{\mathrm{c}}$ that is $w$ with positions $j,i$ set to $0$ yields the same scan pairing on all remaining positions: the stack contents before $j$ and after $i$ agree, and no decision is taken at $j$ or $i$ in $w^{\mathrm{c}}$. By induction on the number of replacements, every word reachable from $w$ by replacements has the same scan survivors as $w$, so every maximal reduction ends in the scan reduced word and that word is unique; in particular $\epsilon$, $\phi$ and the surviving nodes of [F2] are well defined.

Two distinct $i$-nodes of $\lambda$ share neither a row nor a column. Same row: the only corners in a row $r$ are the removable node $(r,{\lambda }_r)$ and, when it is addable, the node $(r,{\lambda }_r+1)$, whose columns differ by $1$; equal residues would force $e\mid 1$, impossible for $e\ge 2$, and a node is not repeated. Same column $d$: if both nodes are removable, $(r,{\lambda }_r)$ and $(s,{\lambda }_s)$ with ${\lambda }_r={\lambda }_s=d$ and $r<s$, then ${\lambda }_{r+1}\ge {\lambda }_s={\lambda }_r$ contradicts ${\lambda }_r>{\lambda }_{r+1}$; if both are addable, $(r,{\lambda }_r+1)$ and $(s,{\lambda }_s+1)$ with ${\lambda }_r={\lambda }_s=d-1$ and $r<s$, then ${\lambda }_{s-1}\le {\lambda }_r={\lambda }_s$ and ${\lambda }_{s-1}\ge {\lambda }_s$ give ${\lambda }_{s-1}={\lambda }_s$, contradicting addability at row $s$; if $(r,{\lambda }_r)$ is removable and $(s,{\lambda }_s+1)$ addable with ${\lambda }_r={\lambda }_s+1$ and $r<s$, then removability gives ${\lambda }_{r+1}\le {\lambda }_s$, while $s\ge r+2$ would give ${\lambda }_{r+1}\ge {\lambda }_{s-1}>{\lambda }_s$, that is ${\lambda }_{r+1}\ge {\lambda }_s+1$, a contradiction; hence $s=r+1$, and then the two contents differ by exactly $1$, because $\mathrm{cont}(s,{\lambda }_s+1)={\lambda }_s+1-s={\lambda }_r-(r+1)=\mathrm{cont}(r,{\lambda }_r)-1$, so equal residues would again force $e\mid 1$, impossible; finally the opposite mixed orientation is impossible because addability of $(r,d)$ would give ${\lambda }_r=d-1<d={\lambda }_s\le {\lambda }_r$ for $r<s$.

Conjugation reverses the reading order and preserves the signs. By [F4] transposition sends the $i$-nodes of $\lambda$ bijectively onto the $(-i)$-nodes of ${\lambda }^{\mathrm{\prime }}$, preserving addable/removable type, and negates contents. If $x_1,\dots ,x_n$ is the top-to-bottom $i$-node list of $\lambda$, then by step 1.1 the contents $\mathrm{cont}(x_m)$ strictly decrease in $m$, so the contents of the images $x_m^{\mathrm{\prime }}$ strictly increase in $m$; by step 1.1 applied to ${\lambda }^{\mathrm{\prime }}$, increasing content is exactly bottom to top on the ${\lambda }^{\mathrm{\prime }}$ corner list. Hence $x_1^{\mathrm{\prime }},\dots ,x_n^{\mathrm{\prime }}$ is the bottom-to-top $(-i)$-node list of ${\lambda }^{\mathrm{\prime }}$, and its sign word equals the sign word of $x_1,\dots ,x_n$ because transposition preserves the type of every node.

By step 2.2 the two words being reduced are identical: the modular signature of $\lambda$ at residue $i$ lists $x_1,\dots ,x_n$ top to bottom, and the reference order of [F3] applied to ${\lambda }^{\mathrm{\prime }}$ at residue $-i$ lists $x_1^{\mathrm{\prime }},\dots ,x_n^{\mathrm{\prime }}$ by increasing content, which is the same sequence of signs. Both rules cancel $-+$ and select the first surviving $-$ as good and the last surviving $+$ as cogood, so their survivor patterns coincide position by position, and the selected node of $\lambda$ is the transposed node of the selected node of ${\lambda }^{\mathrm{\prime }}$: the good (removable) $i$-node of $\lambda$ transposes to the first-surviving-$-$, hence good, removable $(-i)$-node of ${\lambda }^{\mathrm{\prime }}$, and the cogood (addable) $i$-node transposes to the last-surviving-$+$, hence cogood, addable $(-i)$-node of ${\lambda }^{\mathrm{\prime }}$. The survivor counts also agree, ${\epsilon }_i(\lambda )={\epsilon }_{-i}({\lambda }^{\mathrm{\prime }})$ and ${\phi }_i(\lambda )={\phi }_{-i}({\lambda }^{\mathrm{\prime }})$ in the two readings.

Reversing a word exchanges the two descriptions. Cancelling a $-+$ pair of a word $w$ amounts to cancelling the corresponding $+-$ pair of the reversed word $w^{\mathrm{R}}$; the first surviving $-$ of $w$ is the last surviving $-$ of $w^{\mathrm{R}}$, and the last surviving $+$ of $w$ is the first surviving $+$ of $w^{\mathrm{R}}$. Applying this to the increasing-content $(-i)$-signature of ${\lambda }^{\mathrm{\prime }}$ rewrites the dictionary of step 3.1 in the decreasing-content form: cancel $+-$, delete at the last surviving $-$ and add at the first surviving $+$. Both forms describe one and the same reduction and select the same actual nodes of ${\lambda }^{\mathrm{\prime }}$, and by step 3.1 the same nodes of $\lambda$ after transposition.

No hypothesis on $\lambda$ beyond the corner combinatorics was used: steps 1.1 and 2.1 hold for every partition, and steps 1.2-4.1 are computations on the resulting sign words. The modular and reference readings therefore agree as signature computations on all partitions, and only the later restriction of the two models to their basic label sets selects a connected component from them.

For every partition $\mu$ and every color $i$, the selection operations of the two readings correspond under transposition. By [F2] the top-to-bottom $i$-node word of $\mu$ equals the increasing-content $(-i)$-node word of ${\mu }^{\mathrm{\prime }}$ position by position, with the same signs, so the reduced survivor words coincide; consequently ${\epsilon }_i(\mu )={\epsilon }_{-i}({\mu }^{\mathrm{\prime }})$ and ${\phi }_i(\mu )={\phi }_{-i}({\mu }^{\mathrm{\prime }})$, and if ${\epsilon }_i(\mu )>0$ then the good $i$-node of $\mu$ is the transposed node of the good $(-i)$-node of ${\mu }^{\mathrm{\prime }}$, while if ${\phi }_i(\mu )>0$ then the cogood $i$-node of $\mu$ is the transposed node of the cogood $(-i)$-node of ${\mu }^{\mathrm{\prime }}$. In particular ${\epsilon }_i(\mu )=0$ if and only if ${\epsilon }_{-i}({\mu }^{\mathrm{\prime }})=0$.

There is no naming-polarity swap between the two conventions: in the modular reading the good node is by definition removable and the cogood node addable ([F1]), in the $p$-restricted operator definition the good node is removable and the cogood node addable ([F3]), and in the Fock-space reading the good node is the surviving $-$, hence removable, while the cogood node is the surviving $+$, hence addable ([F4]); the normalized Fock operators remove the good node and add the cogood node ([F5]). Hence both dictionaries compare a deletion with a deletion and an addition with an addition, and the only change is the color reflection $i\mapsto -i$.

The transported crystal operators on all partition residues are given by the modular signature rule. The field operators $J^{-1}{\tilde{e}}_{-i}^{\mathrm{Q}\mathrm{G}}J$ and $J^{-1}{\tilde{f}}_{-i}^{\mathrm{Q}\mathrm{G}}J$ preserve $(F_0{)}_{\mathrm{s}\mathrm{y}\mathrm{m}}$ by [F5]; let $b_{\mu }:=M_{\mu }+q(F_0{)}_{\mathrm{s}\mathrm{y}\mathrm{m}}$. The induced map of residue spaces sends $b_{\mu }$ to $\mathrm{\mid }{\mu }^{\mathrm{\prime }}⟩+q{F}_0$. By [F4]-[F5] its raising image is the residue of the partition obtained by deleting the good $(-i)$-node of ${\mu }^{\mathrm{\prime }}$, or zero when no such node exists. Step 1.1 identifies that node with the transpose of the modular good $i$-node $A$ of $\mu$, so ${\tilde{e}}_i^{\mathrm{s}\mathrm{y}\mathrm{m}}{b}_{\mu }=b_{\mu \setminus A}$ when ${\epsilon }_i(\mu )>0$, and zero otherwise. The same argument gives ${\tilde{f}}_i^{\mathrm{s}\mathrm{y}\mathrm{m}}{b}_{\mu }=b_{\mu \cup B}$ for the modular cogood node $B$, and zero when ${\phi }_i(\mu )=0$. These are equalities of crystal residues: on raw standard vectors they assert congruences modulo $q(F_0{)}_{\mathrm{s}\mathrm{y}\mathrm{m}}$, including at zero arrows, rather than exact field-vector equalities.

Weights transform by the reflection. By [F1] the transported weight satisfies $\mathrm{wt}(M_{\mu })={\mathrm{\Lambda }}_0-{\sum }_i{m}_i(\mu ){\alpha }_i$ and $m_i(\mu )=m_{-i}({\mu }^{\mathrm{\prime }})$; hence $\mathrm{wt}(\mathrm{\mid }{\mu }^{\mathrm{\prime }}⟩)={\mathrm{\Lambda }}_0-{\sum }_j{m}_j({\mu }^{\mathrm{\prime }}){\alpha }_j={\mathrm{\Lambda }}_0-{\sum }_j{m}_{-j}(\mu ){\alpha }_j=\tau (\mathrm{wt}(M_{\mu }))$, and the weight of a crystal arrow is preserved by the node moves of step 2.1 because those moves change one residue count by one on each side consistently.

Restricted case. Let $\lambda$ be $e$-restricted, so that ${\lambda }^{\mathrm{\prime }}$ is $e$-regular by [F1] and hence a vertex of the empty component of the full Fock crystal by [F5], on which the operators ${\tilde{e}}^{\mathrm{Q}\mathrm{G}}$, ${\tilde{f}}^{\mathrm{Q}\mathrm{G}}$ of [F5] act. By step 2.1 applied at $\mu =\lambda$, the modular operator ${\tilde{e}}_i$ on $\lambda$ (delete the good $i$-node, zero if ${\epsilon }_i(\lambda )=0$) is intertwined with ${\tilde{e}}_{-i}^{\mathrm{Q}\mathrm{G}}$ on ${\lambda }^{\mathrm{\prime }}$: $C{\tilde{e}}_i\lambda ={\tilde{e}}_{-i}^{\mathrm{Q}\mathrm{G}}C\lambda$, and likewise $C{\tilde{f}}_i\lambda ={\tilde{f}}_{-i}^{\mathrm{Q}\mathrm{G}}C\lambda$, both sides being zero together by the count agreement of step 1.1. Since transposition is a bijection from the $e$-restricted onto the $e$-regular labels and [F5] identifies the latter precisely as the vertex set of the empty component, the conjugation of the modular restricted graph is exactly that empty component, with colors negated; for $e=p$ prime the modular side is the operator system of [F3], and in general it is the reading fixed in [F1].

Finally, the arrows are not read off from ordinary Chevalley coefficients. The selection is the signature reduction of [F4], and [F4]-[F5] record that a pole or a unique Chevalley term does not determine the crystal arrow; the transported operator ${\tilde{e}}_i^{\mathrm{s}\mathrm{y}\mathrm{m}}$ of step 2.1 is the normalization inherited from ${\tilde{e}}_{-i}^{\mathrm{Q}\mathrm{G}}$, not the transported Chevalley operator $E_i^{\mathrm{s}\mathrm{y}\mathrm{m}}=J^{-1}{E}_{-i}^{\mathrm{Q}\mathrm{G}}J$, whose coefficients (such as $q^{-1}$ from the formulas of Leclerc (38)-(39)) differ from the arrow data.