Modular Branching and the Good-Node Crystal — Examples

The $0$-nodes of $\mathrm{\varnothing }$ and $(1)$. The empty partition has the single addable node $(1,1)$ of residue $c-r=0$, so its residue-$0$ word is $(+)$ and the cogood node is $(1,1)$, whence ${\tilde{f}}_0\mathrm{\varnothing }=(1)$; the partition $(1)$ has the single removable node $(1,1)$ of residue $0$, so its residue-$0$ word is $(-)$ with good node $(1,1)$ and ${\tilde{e}}_0(1)=\mathrm{\varnothing }$.

The $1$-nodes of $(1,1)$ and $(1,1,1)$. In $(1,1)$ the residue-$1$ nodes are the addable nodes $(1,2)$ (residue $2-1=1$) and $(3,1)$ (residue $1-3=-2\equiv 1$), ordered $(1,2),(3,1)$ top to bottom; the word is $(+,+)$, the cogood node is the last surviving $+$, namely $(3,1)$, and ${\tilde{f}}_1(1,1)=(1,1,1)$, while adding $(1,2)$ would lead to $(2,1)$. In $(1,1,1)$ the residue-$1$ nodes are the addable node $(1,2)$ and the removable node $(3,1)$; the word $(+,-)$ is already reduced, so the surviving $+$ is $(1,2)$ and the surviving $-$ is $(3,1)$: the cogood node is $(1,2)$ with ${\tilde{f}}_1(1,1,1)=(2,1,1)$, and the good node is $(3,1)$ with ${\tilde{e}}_1(1,1,1)=(1,1)$, in agreement with the inverse property of [F2].

The $2$-nodes of $(1)$ and $(1,1)$. The residue-$2$ nodes of $(1)$: the addable node $(2,1)$ has residue $1-2=-1\equiv 2$, and there is no other, so the word is $(+)$ with cogood node $(2,1)$ and ${\tilde{f}}_2(1)=(1,1)$. For $(1,1)$ the residue-$2$ node is the removable node $(2,1)$ (residue $1-2\equiv 2$), the word is $(-)$, the good node is $(2,1)$ and ${\tilde{e}}_2(1,1)=(1)$.

The $1$-nodes of $(2,1,1)$ and the $2$-nodes of $(2,1,1)$ and $(3,1,1)$. In $(2,1,1)$ the residue-$1$ nodes are the removable nodes $(1,2)$ (residue $2-1=1$) and $(3,1)$ (residue $1-3\equiv 1$), listed $(1,2),(3,1)$: the word $(-,-)$ is reduced, so the good node is the first surviving $-$, namely $(1,2)$, and ${\tilde{e}}_1(2,1,1)=(1,1,1)$. The residue-$2$ nodes of $(2,1,1)$ are the single addable node $(1,3)$ (residue $3-1=2$): the word is $(+)$ with cogood node $(1,3)$ and ${\tilde{f}}_2(2,1,1)=(3,1,1)$. In $(3,1,1)$ the only residue-$2$ boundary node is the removable node $(1,3)$; the interior node $(2,1)$ has the same residue but does not enter the signature; the word is $(-)$ with good node $(1,3)$ and ${\tilde{e}}_2(3,1,1)=(2,1,1)$.

The path and its reverse. Step 1.1, step 2.1, step 1.2 and step 2.2 exhibit, successively, $\mathrm{\varnothing }\to (1)\to (1,1)\to (1,1,1)\to (2,1,1)\to (3,1,1)$ as cogood additions with residues $0,2,1,1,2$, and each reverse step as the deletion of the good node of the same residue; by [F2] this is exactly the assertion that each forward step is a cogood addition and each reverse step a good deletion, and the five residue labels match the displayed path.

All reduced signatures claimed in the example are the ones computed: $(-)$ for $(1)$ at $i=0$ in step 1.1, $(-)$ for $(1,1)$ at $i=2$ in step 2.1, $(+,+)$ for $(1,1)$ and $(+,-)$ for $(1,1,1)$ at $i=1$ in step 1.2, and $(-,-)$ for $(2,1,1)$ at $i=1$, $(+)$ for $(2,1,1)$ at $i=2$ and $(-)$ for $(3,1,1)$ at $i=2$ in step 2.2; in particular $(-,-)$ exhibits two normal nodes of one residue and $(+,-)$ exhibits both signs in a reduced word, and the path still exists as [F3] predicts.

Removable rows and residues. $\lambda =(6,4,4,3,2)$ has consecutive differences $2,0,1,1,2$, so $R(\lambda )=\{1,3,4,5\}$ and $\lambda$ is $\mathrm{\ell }$-restricted for $\mathrm{\ell }=3$ and $\mathrm{\ell }=4$; the residues $i-{\lambda }_i$ are, modulo $3$, $1,2,1,0$ for $i=1,3,4,5$, and modulo $4$, $3,3,1,3$ for the same rows.

Pad by ${\lambda }_6=0$. For $k=1,\dots ,5$, the integers $k-{\lambda }_{k+1}$ are $(-3,-2,0,2,5)$ and $k-{\lambda }_k$ are $(-5,-2,-1,1,3)$. To compute $B_{i,6}$ use indices $i\le k<6$ with the first integer congruent to $i-{\lambda }_i$; for $C_{i,6}$ use $i<k<6$ with the second integer congruent to $i-{\lambda }_i$, then intersect with $R=\{1,3,4,5\}$. For $i=1,3,4,5$, respectively, this gives $(B\cap R,C\cap R)=(\mathrm{\varnothing },\{4\}),(\{4,5\},\mathrm{\varnothing }),(\mathrm{\varnothing },\mathrm{\varnothing }),(\mathrm{\varnothing },\mathrm{\varnothing })$ modulo $3$, and $(\mathrm{\varnothing },\{3,5\}),(\mathrm{\varnothing },\{5\}),(\{5\},\mathrm{\varnothing }),(\mathrm{\varnothing },\mathrm{\varnothing })$ modulo $4$. These are exactly the corrected displayed pairs.

For $\mathrm{\ell }=3$, rows $1,4,5$ pass the lattice test: their demand sets are empty. Row $3$ fails because $\{4,5\}\downarrow ̸\mathrm{\varnothing }$. Hence $R^{normal}=\{1,4,5\}$, with residues $1,1,0$; taking the least normal row in each residue gives $R^{good}=\{1,5\}$. For $\mathrm{\ell }=4$, rows $1,3,5$ pass, while row $4$ fails because $\{5\}\downarrow ̸\mathrm{\varnothing }$. The normal rows all have residue $3$, so $R^{good}=\{1\}$.

$\mathrm{\ell }=3$: decomposability. By [F2] the actual source-labelled restriction $M=D^{\lambda }{\downarrow }_{H(r-1)}$ has the two simple socle factors $D^{\lambda (1)}=D^{(5,4,4,3,2)}$ and $D^{\lambda (5)}=D^{(6,4,4,3,1)}$. Their removed Murphy node contents are $5$ and $-3$, of residues $2$ and $0$ modulo $3$, so [F2]'s central operator has distinct scalars on them. Its minimal polynomial on $M$ has distinct linear factors at those two scalars; the polynomial CRT produces a central primary projector $b$ selecting the first and annihilating the second (other factors are included in $1-b$). Thus $M=bM\oplus (1-b)M$ is a sum of submodules and both are nonzero because they contain the respective simple socle factors. This proves decomposability by actual full-module projections.

$\mathrm{\ell }=4$: indecomposability. By the proved branching input [F2], the socle of $D^{\lambda }{\downarrow }_{H(r-1)}$ is the single simple module $D^{\lambda (1)}=D^{(5,4,4,3,2)}$. By [F3] this makes the restriction indecomposable: a decomposition into two nonzero modules would give two nonzero summand socles. No cell-module self-duality is used.

At $\mathrm{\ell }=4$, rows $1,3,5$ have residue $3$ and are all normal, whereas row $4$ has residue $1$ and fails $\{5\}\downarrow \mathrm{\varnothing }$. At $\mathrm{\ell }=3$, both removable rows of residue $1$ are normal. Thus these tables do not exhibit same-residue removable nodes of different normality; they do show that good rows are the least normal rows of each residue, rather than rows whose residue occurs once. The branching conclusions in steps 4.1–4.2 use the actual general-Hecke supplier and source-label transport [F2], with $\mathrm{\ell }=4$ socle label $D^{(5,4,4,3,2)}$.