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Use the B2 root system conventionThus $V=L_{{\omega }_2}$ is the five-dimensional vector representation of the Special orthogonal Lie algebra ${\mathfrak{so}}_5$. Label its weight verticesThe crystal basis is the colored chainbecause each Kashiwara operator ${\tilde{f}}_i$ subtracts ${\alpha }_i$.

For the tensor product of crystals, write $XY$ for $X\otimes Y$. The complete colored-arrow graph is compactly specified byIts three connected highest-weight components start at $AA$, $AB$, and $AE$. Their vertex sets areTheir highest weights and dimensions identify the ten-vertex component with the exterior square and the other two with the symmetric square. Thereforeof dimensions $10$ and $14+1$, respectively.

The module $L=L_{{\omega }_1}$ is the four-dimensional spin representation. Its weights are $(\pm {\epsilon }_1\pm {\epsilon }_2)/2$, and its crystal is

Every weight of $V$ lies in the root lattice, so every weight of every tensor power $V^{\otimes n}$ also lies in that lattice. But ${\omega }_1=({\epsilon }_1+{\epsilon }_2)/2$ represents the nonzero coset in the quotient of the weight lattice by the root lattice. Consequently no irreducible constituent of $V^{\otimes n}$ can have highest weight ${\omega }_1$, and $L$ never occurs.

The tensor square is the second tensor power $V^{\otimes 2}=V\otimes V$. Over a field of characteristic different from two it decomposes as the direct sum of the symmetric square and exterior square.

For the five-dimensional defining representation of ${\mathfrak{so}}_5$, the symmetric square is the sum of its invariant trace and the traceless representation $V(2{\omega }_2)$, while the exterior square is the Adjoint representation $V(2{\omega }_1)$. Hence

For the defining four-dimensional representation $V$ of the symplectic Lie algebra ${\mathfrak{sp}}_4$, with the short simple root numbered first, its highest weight is ${\omega }_1$. Its tensor square decomposes aswith dimensions $10,5,1$. The first summand is the symmetric square; the other two form the exterior square, split by symplectic contraction of an exterior square.

For $n\ge 2$, the defining Lie algebra representation $V$ of ${\mathfrak{sp}}_{2n}$ has the displayed tensor product decomposition. The crystal of the defining symplectic representation and the crystal tensor-product rule give exactly these three highest vertices, each occurring once. The symmetric square is $L(2{\omega }_1)$ of dimension $n(2n+1)$. The exterior square splits into its invariant inverse-form line and the primitive exterior square $L({\omega }_2)$ of dimension $n(2n-1)-1$. Their dimensions sum to $4n^2$. At $n=1$, the exterior square is only the invariant line, giving $L(2{\omega }_1)\oplus L(0)$.