The crystal graph, ${\epsilon }_i$, ${\phi }_i$ and the weight function

Finiteness. Let $b\in B$ and let $x\in L$ be a lift of $b$. By [F2] the vector $x$ is a finite sum of string positions $F_i^{(n)}{u}_n$; along each of the finitely many strings both ${\tilde{e}}_i$ and ${\tilde{f}}_i$ move one position and kill the ends, so there are integers $N,M$ with ${\tilde{f}}_i^{N}x=0$ and ${\tilde{e}}_i^{M}x=0$. The operators preserve $L$ and descend to the quotient, so ${\tilde{f}}_i^{N}b=0$ and ${\tilde{e}}_i^{M}b=0$; hence ${\epsilon }_i(b)$ and ${\phi }_i(b)$ are finite.

A highest lift. Let $b_{\mathrm{t}\mathrm{o}\mathrm{p}}:={\tilde{e}}_i^{{\epsilon }_i(b)}b$ and let $y\in L_{\mathrm{w}\mathrm{t}(b_{\mathrm{t}\mathrm{o}\mathrm{p}})}$ be a homogeneous lift of $b_{\mathrm{t}\mathrm{o}\mathrm{p}}$. Since ${\tilde{e}}_i{b}_{\mathrm{t}\mathrm{o}\mathrm{p}}=0$, the element ${\tilde{e}}_{i}y$ lies in $qL$, hence so does ${\tilde{f}}_i{\tilde{e}}_{i}y$; by [F2], $h:=(\mathrm{i}\mathrm{d}-{\tilde{f}}_i{\tilde{e}}_i)(y)=y-{\tilde{f}}_i{\tilde{e}}_{i}y$ is a lift of $b_{\mathrm{t}\mathrm{o}\mathrm{p}}$ that lies in the image of $p_{\mathrm{t}\mathrm{o}\mathrm{p}}=\mathrm{i}\mathrm{d}-{\tilde{f}}_i{\tilde{e}}_i$, that is, in $\ker {\tilde{e}}_i=\ker E_i$; it is nonzero because its class $b_{\mathrm{t}\mathrm{o}\mathrm{p}}$ is nonzero. Thus $h$ is a nonzero $E_i$-highest vector of $M$, and its weight is $\mu :=\mathrm{w}\mathrm{t}(b_{\mathrm{t}\mathrm{o}\mathrm{p}})=\mathrm{w}\mathrm{t}(b)+{\epsilon }_i(b){\alpha }_i$ by [F2].

The negative half. For $U_q^{-}$ and a residue class $b$ of root degree $-\nu$, the operator ${\tilde{e}}_i$ raises the root degree by ${\alpha }_i$ and hence lowers the height $\mathrm{ht}(\nu )$ by one; iterating, ${\tilde{e}}_i^{\mathrm{ht}(\nu )+1}b=0$, so ${\epsilon }_i(b)$ is finite. The operator ${\tilde{f}}_i$ is injective by [F4], so ${\tilde{f}}_i^{n}b\mathrm{\ne }0$ for every $n\ge 0$; the actual lowering-chain length is therefore ${\mathrm{\ell }}_i^{-}=+\mathrm{\infty }$. The formal quantity ${\phi }_i:={\epsilon }_i+⟨h_i,\mathrm{w}\mathrm{t}⟩$ is an integer, well defined on the root-graded crystal, and along a lowering edge it changes by $+1-2=-1$: indeed ${\epsilon }_i$ increases by one and $⟨h_i,\mathrm{w}\mathrm{t}⟩$ decreases by $⟨h_i,{\alpha }_i⟩=2$. It can be negative (for example ${\phi }_i=0$ at the unit, and ${\phi }_i={\epsilon }_i-{\sum }_j{a}_{ij}{\nu }_j$ at degree $-\nu =-{\sum }_j{\nu }_j{\alpha }_j$), and it is not the length of the lowering chain. No integrability or finite-string property of the whole negative half is used.

The component is a finite chain. By [F1] the relation $b\sim b^{\mathrm{\prime }}$ if ${\tilde{f}}_i^{m}b=b^{\mathrm{\prime }}$ or ${\tilde{f}}_i^m{b}^{\mathrm{\prime }}=b$ for some $m\ge 0$ is symmetric and transitive, and each vertex has at most one ${\tilde{f}}_i$-image and at most one ${\tilde{f}}_i$-preimage in $B$; by step 1.1 every class is finite, so the component of $b$ is a finite path. It has a unique top, where ${\tilde{e}}_i$ acts by $0$, and a unique bottom, where ${\tilde{f}}_i$ acts by $0$; writing ${\epsilon }_i$ and ${\phi }_i$ for the distances from the top and to the bottom gives ${\epsilon }_i(b)+{\phi }_i(b)+1=\mathrm{\#}$ of vertices of the component.

The chain has length $a+1$. Put $a:=⟨\mu ,h_i⟩$, so $a\ge 0$ and $h\in H_{\mu }$. By [F3], $F_i^{(n)}h\mathrm{\ne }0$ for $0\le n\le a$ and $F_i^{(n)}h=0$ for $n>a$, so exactly $a+1$ positions of this string are nonzero. Their classes are also nonzero: if $F_i^{(n)}h={\tilde{f}}_i^{n}h\in qL$ for $n\le a$, then applying the lattice-preserving ${\tilde{e}}_i^n$ gives $h\in qL$, contradicting its nonzero top residue. Thus their classes are the vertices reached from $b_{\mathrm{t}\mathrm{o}\mathrm{p}}$ by ${\tilde{f}}_i$ (while nonzero), and by step 2.1 they are precisely the vertices of the component of $b$. Hence the component of $b$ has $a+1$ vertices.

The formulas. If $b$ is the position $n$ of the chain, counted from the top, then ${\epsilon }_i(b)=n$ and, by step 3.1, ${\phi }_i(b)=a-n$. Since $a=⟨\mu ,h_i⟩=⟨h_i,\mathrm{w}\mathrm{t}(b)⟩+2n$ by step 1.2, this gives ${\phi }_i(b)=⟨h_i,\mathrm{w}\mathrm{t}(b)⟩+n={\epsilon }_i(b)+⟨h_i,\mathrm{w}\mathrm{t}(b)⟩$, and also ${\epsilon }_i(b)+{\phi }_i(b)=a=⟨h_i,\mathrm{w}\mathrm{t}(b)⟩+2{\epsilon }_i(b)$, which is the highest rank-one integer of the top label.

Edges. On an existing arrow $b\to {\tilde{f}}_{i}b$, the position increases by one, so ${\epsilon }_i$ increases by $1$ and ${\phi }_i$ decreases by $1$, while the weight decreases by ${\alpha }_i$; conversely, ${\tilde{e}}_i$ moves one position up whenever $b$ is not the top. By step 2.1 the chain has exactly ${\epsilon }_i(b)+{\phi }_i(b)+1$ vertices.

Choice. The data $(L,B)$, the operators and the chain positions are given or uniquely determined; all constructions above use finite sums and the unique string decomposition, so no choice principle is used.