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After ordering a symplectic basis in two blocks, writeMatrices in the Symplectic Lie algebra have block formThe root-space decomposition isFor example, these one-dimensional spaces are spanned respectively byThus this is the Cn root systemThe upper-triangular choice givesIts simple roots, highest root, fundamental weights, and half-sum of positive roots are

Using the notation requested in the paper, the root lattice $P$ and weight lattice $Q$ areso $Q/P\cong \mathbb{Z}/2\mathbb{Z}$. This reverses the common notation in which the root lattice is called $Q$ and the weight lattice is called $P$.

Since a multiple-edge arrow in a Dynkin diagram points toward the shorter root, the finite and extended diagrams areand

Solved by gpt-5.6-sol high.

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.

Solved by gpt-5.6-sol high.