= Solution
Take the diagonal <Cartan subalgebra> of $\mathfrak{sl}_3$ and let $\varepsilon_i$ select its $i$th diagonal entry, so $\varepsilon_1+\varepsilon_2+\varepsilon_3=0$ on it. Use <simple roots> $\alpha_1=\varepsilon_1-\varepsilon_2$, $\alpha_2=\varepsilon_2-\varepsilon_3$. The <fundamental weights> are $\omega_1=\varepsilon_1$ and $\omega_2=-\varepsilon_3=\varepsilon_1+\varepsilon_2$. The defining representation $V$ has highest vector $e_1$ and the dual has highest vector $e_3^*$. Thus
$$
\boxed{\operatorname{hw}(V)=\omega_1,\qquad\operatorname{hw}(V^*)=\omega_2.}
$$
Their <crystal basis> diagrams are $B:1\xrightarrow{1}2\xrightarrow{2}3$ and $B^\vee:\bar3\xrightarrow{2}\bar2\xrightarrow{1}\bar1$, with weights $\varepsilon_i$ and $-\varepsilon_i$ respectively. Use the same <crystal tensor-product rule> as in2(d): lowering acts on the first factor when $\varphi_i(u)>\varepsilon_i(v)$; otherwise on the second. Raising uses the first factor when $\varphi_i(u)\ge\varepsilon_i(v)$, so it reverses each lowering arrow. An edge is absent if the selected factor operation is zero.
For $B\otimes B$, abbreviate $i\otimes j$ by $ij$. The complete color-one edge list is $11\to21\to22$, $31\to32$, $13\to23$; the complete color-two list is $21\to31$, $22\to32\to33$, $12\to13$. There are two connected components:
$$
\begin{array}{c|c|c}
\text{highest vertex}&\text{all vertices}&\text{highest weight}\\\hline
11&11,21,22,31,32,33&2\omega_1\\
12&12,13,23&\omega_2
\end{array}
$$
For example $\widetilde f_1(1\otimes1)=2\otimes1$ because $\varphi_1(1)=1>\varepsilon_1(1)=0$, whereas $\widetilde f_1(1\otimes2)=0$ because equality selects the second factor and $\widetilde f_1(2)=0$. This explains which ordered tensor labels lie in each component. At the representation level, the two components are the six-dimensional <symmetric square> and the three-dimensional <exterior square>:
$$
\boxed{V\otimes V\cong L(2\omega_1)\oplus L(\omega_2).}
$$
For $B\otimes B^\vee$, abbreviate $i\otimes\bar j$ by $i\bar j$. The complete color-one edge list is $1\bar3\to2\bar3$, $1\bar2\to2\bar2\to2\bar1$, $3\bar2\to3\bar1$; the complete color-two list is $1\bar3\to1\bar2$, $2\bar3\to3\bar3\to3\bar2$, $2\bar1\to3\bar1$. The isolated vertex is $1\bar1$. Thus the two highest vertices are $1\bar3$ of weight $\varepsilon_1-\varepsilon_3=\omega_1+\omega_2$, and $1\bar1$ of weight zero. Lowering from the former visits all eight other vertices. Consequently
$$
\boxed{V\otimes V^*\cong L(\omega_1+\omega_2)\oplus L(0).}
$$
This also follows concretely from $V\otimes V^*\cong\operatorname{End}(V)=\mathfrak{sl}_3\oplus\mathbb C I$: the first term is the <Adjoint representation>, with highest weight the highest root, and the second is the <trivial Lie algebra representation>. The ordinary invariant vector is $\sum_i e_i\otimes e_i^*$, not the bare tensor suggested by the singleton's crystal label. The explicit edge lists and the diagrams exhibit all eighteen tensor vertices and justify the four highest weights.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2001/iii/paper-5-a2-tensors.png]
{title=All components of the defining A2 tensor square and defining tensor dual crystals, with every lowering edge labeled by its simple-root color}
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