The charge-zero level-one crystal through rank four at e = 3

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 $f_{-2}=f_1$.