Residue-node reading orders and the conjugation dictionary

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.