= Solution
The PDF uses $V=\mathbb C^{2n+1}$ for $\mathfrak{so}_{2n+1}$; the TeX incorrectly turns this into a power of two. The <crystal of the defining odd-orthogonal representation> is the chain with weights $\varepsilon_1,\ldots,\varepsilon_n,0,-\varepsilon_n,\ldots,-\varepsilon_1$. Label its vertices $1,\ldots,n,0,\bar n,\ldots,\bar1$. Its directed edge colors are
$$
1,2,\ldots,n-1,n,n,n-1,\ldots,2,1.
$$
In particular the short-root string $n\to0\to\bar n$ has two edges of color $n$.
For a full drawing of the <tensor product of crystals> in any rank, take every ordered pair of chain vertices $(a,b)$. Let $\varepsilon_i(b)$ count incoming edges of color $i$ along its string, and let $\varphi_i(a)$ count outgoing ones. We use the convention
$$
\widetilde f_i(a\otimes b)=
\begin{cases}\widetilde f_i(a)\otimes b,&\varphi_i(a)>\varepsilon_i(b),\\
a\otimes\widetilde f_i(b),&\varphi_i(a)\leq\varepsilon_i(b),\end{cases}
$$
omitting an edge when the indicated operator vanishes. This <crystal tensor-product rule> specifies every vertex and every edge; the raising rule uses $\geq$ for its first-factor branch. The picture shows the chain and all 25 tensor vertices for $n=2$, with connected components distinguished by node color.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/iii/paper-102-b2-tensor-crystal.png]
{title=The type-B2 defining crystal and its tensor square}
For $n\geq2$, the three <highest-weight vectors> of the tensor crystal are $1\otimes1$, $1\otimes2$ and $1\otimes\bar1$. They have weights $2\varepsilon_1$, $\varepsilon_1+\varepsilon_2$ and zero. There are no others: a highest tensor vertex must have highest first factor, and the string inequalities then restrict its second factor to these three vertices. Therefore
$$
\boxed{V\otimes V=L(2\varepsilon_1)\oplus L(\varepsilon_1+\varepsilon_2)\oplus L(0).}
$$
The summands are the traceless <symmetric square>, the <exterior square> (the <Adjoint representation>), and the invariant scalar line. Their dimensions are $(2n+1)(n+1)-1$, $n(2n+1)$ and one, summing to $(2n+1)^2$. For $n\geq3$ the first two <highest weights> are $2\omega_1,\omega_2$; for $n=2$ they are $2\omega_1,2\omega_2$. For $n=1$, the three tensor highest vertices are $1\otimes1,1\otimes0,1\otimes\bar1$, giving $L(4\omega_1)\oplus L(2\omega_1)\oplus L(0)$ of dimensions $5,3,1$.
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