Highest components in the $A_2$ tensor crystal $B({\lambda }_1)\otimes B({\lambda }_1)$

Construct the standard module on the three displayed weight vectors. Set $K_h{u}_j=q^{⟨wt(u_j),h⟩}{u}_j$, $F_1{u}_1=u_2$, $F_2{u}_2=u_3$, $E_1{u}_2=u_1$, $E_2{u}_3=u_2$, and all other $E_i,F_i$ images to zero. The weight shifts give the toral-action relations. The two same-color commutators act diagonally with coroot labels $(1,-1,0)$ and $(0,1,-1)$, respectively, and both cross-color commutators vanish. Each $E_i^2,F_i^2$ is zero, as are $E_i{E}_j{E}_i,F_i{F}_j{F}_i$ for $i\mathrm{\ne }j$, verifying the $A_2$ Serre relations. Thus this is a type-one integrable module, cyclic of highest weight ${\lambda }_1$; the universality in [F1] identifies it as $V({\lambda }_1)$. For each color its nontrivial rank-one string has length one and the third vector is a trivial string, so the Kashiwara operators equal the indicated shifts. The lowering words from $u_1$ are exactly $u_1,u_2,u_3,0$, giving the canonical lattice ${⨁}_j{A}_0{u}_j$ and its three residue vertices. Consequently ${\epsilon }_1=(0,1,0)$ and ${\epsilon }_2=(0,0,1)$, with the stated weights.

Apply the highest-vertex criterion [F3]: ${\epsilon }_1(b_1)=0$ forces $b_1\mathrm{\ne }{u}_2$ and ${\epsilon }_2(b_1)=0$ forces $b_1\mathrm{\ne }{u}_3$, so $b_1=u_1$; then ${\epsilon }_1(b_2)\le ⟨h_1,{\lambda }_1⟩=1$ always holds and ${\epsilon }_2(b_2)\le ⟨h_2,{\lambda }_1⟩=0$ forces $b_2\mathrm{\ne }{u}_3$. Hence the highest vertices are $u_1\otimes u_1$ and $u_1\otimes u_2$, of weights $2{\lambda }_1$ and ${\lambda }_1+({\lambda }_2-{\lambda }_1)={\lambda }_2$, both dominant.

Verify by the two-case rule that $u_1\otimes u_3$ is not highest: ${\phi }_2(u_1)=⟨h_2,wt(u_1)⟩=0<1={\epsilon }_2(u_3)$ by [F2], so ${\tilde{e}}_2$ acts on the second factor and ${\tilde{e}}_2(u_1\otimes u_3)=u_1\otimes u_2\mathrm{\ne }0$.

Enumerate the components with [F4]. The component from $u_1\otimes u_1$ has vertices $(1,1),(2,1),(2,2),(3,1),(3,2),(3,3)$, writing $(j,k)=u_j\otimes u_k$; its nonzero lowering arrows are $(1,1)\stackrel{1}{\to }(2,1)\stackrel{1}{\to }(2,2)$, $(2,1)\stackrel{2}{\to }(3,1)\stackrel{1}{\to }(3,2)$ and $(2,2)\stackrel{2}{\to }(3,2)\stackrel{2}{\to }(3,3)$. The other component is $(1,2)\stackrel{2}{\to }(1,3)\stackrel{1}{\to }(2,3)$. Applying the two comparisons to each listed vertex gives no further nonzero lowering arrows; raising supplies exactly their inverses. Thus the two disjoint components exhaust all nine vertices and have sizes six and three. By F3 their highest weights identify them as $B(2{\lambda }_1)$ and $B({\lambda }_2)$.

The component-multiplicity clause of [F3] gives $V({\lambda }_1)\otimes V({\lambda }_1)\cong V(2{\lambda }_1)\oplus V({\lambda }_2)$. The crystal-basis rank equality gives dimensions six and three from step 3.1, hence $9=6+3$. For the classical standard three-dimensional representation, the symmetric tensors and alternating tensors have dimensions six and three and are complementary under $(1\pm \mathrm{f}\mathrm{l}\mathrm{i}\mathrm{p})\mathrm{/}2$; their highest vectors have weights $2{\lambda }_1$ and ${\lambda }_2$, matching the quantum component calculation.

Choice: clause (a) of the criterion is a finite computation; the identification in step 3.1 uses the uniqueness theorem, whose axiom of choice is declared in this item through [F5] and used only as in its own contract.