Dual crystals reverse arrows and weights: two small computations

(A) Apply ${\tilde{f}}_1(b^{\vee })=({\tilde{e}}_{1}b{)}^{\vee }$ and ${\tilde{e}}_1(b^{\vee })=({\tilde{f}}_{1}b{)}^{\vee }$ of [F1] to $b=v_0,v_1$: ${\tilde{e}}_1{v}_0=0$ gives ${\tilde{f}}_1(v_0^{\vee })=0$; ${\tilde{e}}_1{v}_1=v_0$ gives ${\tilde{f}}_1(v_1^{\vee })=v_0^{\vee }$; ${\tilde{f}}_1{v}_0=v_1$ gives ${\tilde{e}}_1(v_0^{\vee })=v_1^{\vee }$; ${\tilde{f}}_1{v}_1=0$ gives ${\tilde{e}}_1(v_1^{\vee })=0$. So the only arrow is $v_1^{\vee }\stackrel{1}{\to }{v}_0^{\vee }$ and the values $({\epsilon }_1,{\phi }_1)=({\phi }_1,{\epsilon }_1)$ are $(1,0)$ at $v_0^{\vee }$ and $(0,1)$ at $v_1^{\vee }$; weights are negated by [F1]. Comparing with [F2], the map $v_1^{\vee }\mapsto v_0$, $v_0^{\vee }\mapsto v_1$ preserves arrows, weights and $\epsilon \mathrm{/}\phi$, so $B(1{)}^{\vee }\cong B(1)$ with highest vertex $v_1^{\vee }$.

(B) By [F1] the arrows become $u_3^{\vee }\stackrel{2}{\to }{u}_2^{\vee }\stackrel{1}{\to }{u}_1^{\vee }$, with weights ${\lambda }_2,{\lambda }_1-{\lambda }_2,-{\lambda }_1$ along that chain. The pairs $({\epsilon }_1,{\phi }_1)$ at $(u_1^{\vee },u_2^{\vee },u_3^{\vee })$ are $(1,0),(0,1),(0,0)$, and the color-two pairs are $(0,0),(1,0),(0,1)$. Construct a module on $w_1,w_2,w_3$ with these successive weights: set $F_2{w}_1=w_2$, $F_1{w}_2=w_3$, $E_2{w}_2=w_1$, $E_1{w}_3=w_2$, all other $E_i,F_i$ images zero, and $K_h{w}_j=q^{⟨wt(w_j),h⟩}{w}_j$. These maps satisfy [F6]: the weight shifts give the toral relations, the same-color commutators have diagonal entries $(0,1,-1)$ and $(1,-1,0)$, matching $(K_i-K_i^{-1})\mathrm{/}(q_i-q_i^{-1})$, and the cross-color commutators vanish. Each $E_i^2,F_i^2$ and each alternating triple $E_i{E}_j{E}_i,F_i{F}_j{F}_i$ with $i\mathrm{\ne }j$ vanishes, proving both $A_2$ Serre relations. The module is nonzero, integrable and generated by its highest vector $w_1$ of dominant weight ${\lambda }_2$, so F6 identifies it with $V({\lambda }_2)$. Each color has one length-one string and one trivial string; [F7] therefore makes the canonical lattice ${⨁}_j{A}_0{w}_j$ with residue crystal $B({\lambda }_2)$. The map $u_3^{\vee }\mapsto w_1$, $u_2^{\vee }\mapsto w_2$, $u_1^{\vee }\mapsto w_3$ preserves all crystal data. To check the module-dual description as well, realize the standard module by $F_1{u}_1=u_2$, $F_2{u}_2=u_3$, $E_1{u}_2=u_1$, $E_2{u}_3=u_2$, the given weights and zero remaining images. The same relation check with colors exchanged identifies this cyclic integrable module with $V({\lambda }_1)$ by [F6]. For its linear dual basis $u_j^{\ast }$, [F8] gives the displayed $w$-actions on $w_1=u_3^{\ast }$, $w_2=-q_1{u}_2^{\ast }$, $w_3=q_1^2{u}_1^{\ast }$ (here $q_1=q_2$). Thus $V({\lambda }_2)$ is the dual three-dimensional module, without a crystal-component uniqueness or AC premise.

(C) The swap sends $(v_j\otimes v_k{)}^{\vee }$ to $v_k^{\vee }\otimes v_j^{\vee }$. On the right the three-element lowering chain is $(v_1^{\vee }\otimes v_1^{\vee })\to (v_0^{\vee }\otimes v_1^{\vee })\to (v_0^{\vee }\otimes v_0^{\vee })$, with weights $2{\lambda }_1,0,-2{\lambda }_1$; each next raising reverses the preceding arrow by [F4]. The remaining vertex $v_1^{\vee }\otimes v_0^{\vee }$ has weight zero and both operators vanish there: the tie ${\phi }_1(v_1^{\vee })={\epsilon }_1(v_0^{\vee })=1$ selects the bottom second factor for lowering and the top first factor for raising. This is the image of the isolated dual vertex $(v_0\otimes v_1{)}^{\vee }$. The reversed three-element chain is $B(2)$ and the isolated vertex is $B(0)$ by [F5], so both sides have the claimed decomposition.

The arrow and tensor calculations are finite applications of [F1]; the canonical identification in (B) uses the explicit relation check, highest-weight module universality [F6] and lowering-word construction [F7]. No choice principle or general crystal-component uniqueness theorem is invoked.