Use the <B2 root system> convention
$$
\alpha_1=\varepsilon_2,
\qquad
\alpha_2=\varepsilon_1-\varepsilon_2,
\qquad
\omega_1=\frac{\varepsilon_1+\varepsilon_2}{2},
\qquad
\omega_2=\varepsilon_1.
$$
Thus $V=L_{\omega_2}$ is the five-dimensional vector representation of the <Special orthogonal Lie algebra> $\mathfrak{so}_5$. Label its weight vertices
$$
A=\varepsilon_1,quad B=\varepsilon_2,quad C=0,quad D=-\varepsilon_2,quad E=-\varepsilon_1.
$$
The <crystal basis> is the colored chain
$$
A\xrightarrow{2}B\xrightarrow{1}C\xrightarrow{1}D\xrightarrow{2}E,
$$
because each <Kashiwara operator> $\widetilde f_i$ subtracts $\alpha_i$.

For the <tensor product of crystals>, write $XY$ for $X\otimes Y$. The complete colored-arrow graph is compactly specified by
$$
\begin{aligned}
\text{color }1:\quad&
AB\to AC\to AD,\quad BA\to CA\to DA,\quad
BB\to CB\to DB\to DC\to DD,\\
&BC\to CC\to CD,\quad BE\to CE\to DE,\quad EB\to EC\to ED;\\
\text{color }2:\quad&
AA\to BA\to BB,\quad AC\to BC,\quad AD\to BD\to BE,\\
&CA\to CB,\quad CD\to CE,\quad DA\to EA\to EB,\quad
DC\to EC,\quad DD\to ED\to EE.
\end{aligned}
$$
Its three connected highest-weight components start at $AA$, $AB$, and $AE$. Their vertex sets are
$$
\begin{aligned}
B(2\omega_2):\quad&AA,BA,BB,CA,CB,DA,DB,DC,DD,EA,EB,EC,ED,EE,\\
B(2\omega_1):\quad&AB,AC,AD,BC,BD,BE,CC,CD,CE,DE,\\
B(0):\quad&AE.
\end{aligned}
$$
Their highest weights and dimensions identify the ten-vertex component with the <exterior square> and the other two with the <symmetric square>. Therefore
$$
\bigwedge^2V\cong L_{2\omega_1},
\qquad
S^2V\cong L_{2\omega_2}\oplus L_0,
$$
of dimensions $10$ and $14+1$, respectively.

The module $L=L_{\omega_1}$ is the four-dimensional spin representation. Its weights are $(\pm\varepsilon_1\pm\varepsilon_2)/2$, and its crystal is
$$
\frac{\varepsilon_1+\varepsilon_2}{2}
\xrightarrow{1}
\frac{\varepsilon_1-\varepsilon_2}{2}
\xrightarrow{2}
\frac{-\varepsilon_1+\varepsilon_2}{2}
\xrightarrow{1}
\frac{-\varepsilon_1-\varepsilon_2}{2}.
$$

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 $\omega_1=(\varepsilon_1+\varepsilon_2)/2$ 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.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 102 / 5 / Solution

{title2=$V^{\otimes n}$}
{wiki=Tensor_product#Tensor_powers_and_braiding}

The nth tensor power is the iterated <tensor product of modules> $V^{\otimes n}=V\otimes\cdots\otimes V$.

Codex @codex  0

 Incoming links: Tensor power