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With long-first simple roots ${\alpha }_1={\epsilon }_1-{\epsilon }_2$, ${\alpha }_2={\epsilon }_2$, the Adjoint representation has highest weight ${\epsilon }_1+{\epsilon }_2=2{\Lambda }_2$. Its crystal basis has one vertex $x_{\beta }$ for every root and two distinct zero-weight vertices $z_1,z_2$. The color-one strings are $x_{{\epsilon }_1}\to x_{{\epsilon }_2}$, $x_{{\alpha }_1}\to z_1\to x_{-{\alpha }_1}$, and $x_{-{\epsilon }_2}\to x_{-{\epsilon }_1}$. The color-two strings are $x_{{\epsilon }_1+{\epsilon }_2}\to x_{{\epsilon }_1}\to x_{{\epsilon }_1-{\epsilon }_2}$, $x_{{\epsilon }_2}\to z_2\to x_{-{\epsilon }_2}$, and $x_{-{\epsilon }_1+{\epsilon }_2}\to x_{-{\epsilon }_1}\to x_{-{\epsilon }_1-{\epsilon }_2}$. All other edges vanish. One construction takes the component generated by $1\otimes 2$ in the tensor square of the crystal of the defining odd-orthogonal representation; the crystal tensor-product rule produces exactly these ten vertices. The equivariant isomorphism ${\Lambda }^2{\mathbb{C}}^5\cong {\mathfrak{so}}_5$ identifies its highest weight and representation.

Use directed color-$i$ edges $u\stackrel{i}{\to }v$ for the Kashiwara operator ${\tilde{f}}_i$, so every edge subtracts ${\alpha }_i$ from the weight. The defining crystal of the defining odd-orthogonal representation isThe defining module has highest weight ${\Lambda }_1$ and precisely these five weights, all of multiplicity one. The color-two string through zero has length two, since $⟨{\epsilon }_2,{\alpha }_2^{\vee }⟩=2$. The other indicated strings have length one. Restriction to each simple-root sl2 Lie algebra and the root-string property of a crystal justify the edges; no other weight difference is the relevant simple root.

For the B2 spin crystal $B({\Lambda }_2)$, denote the four weights $\frac{1}{2}(\pm {\epsilon }_1\pm {\epsilon }_2)$ by their sign pairs. Its graph isThe fundamental weight ${\Lambda }_2$ is a minuscule weight: its pairings with all positive coroots are zero or one. Its Weyl orbit has four weights, and the dimension formula gives four, so each occurs once and there are no additional weights. All simple-root strings have length one. Subtracting ${\alpha }_1$ or ${\alpha }_2$ where another weight is present gives exactly this graph.

For the B2 adjoint crystal, write $x_{\beta }$ for the vertex of root weight $\beta$, and use two distinct zero-weight vertices $z_1,z_2$. There are eight root vertices and two zero vertices because the zero-weight space is the two-dimensional Cartan subalgebra. The complete lists of colored strings areEvery arrow in a given row has that row's color; the reverse arrows are the raising operators, and all unlisted lowering operations vanish. In particular the two zero vertices must not be merged.

For a constructive justification, let $V={\mathbb{C}}^5$ with the invariant symmetric form from part(a). The mapis an equivariant isomorphism ${\Lambda }^{2}V\cong {\mathfrak{so}}_5$, identifying the highest weight as ${\epsilon }_1+{\epsilon }_2$. Compute its crystal as the highest-weight component starting at $1\otimes 2$ in $B({\Lambda }_1)\otimes B({\Lambda }_1)$, with labels $1,2,0,\bar{2},\bar{1}$ for the defining chain. The crystal tensor-product rule iswhere ${\epsilon }_i$ counts available raising steps and ${\phi }_i$ counts lowering steps in the color-$i$ string. It generates the following ten vertices:Applying the displayed rule to these labels gives precisely the listed strings, proving the entire diagram rather than only its weight multiplicities. Tensor labels here are crystal vertices in the combinatorial limit, not an assertion that a bare ordinary tensor such as $v_0\otimes v_0$ lies in ${\Lambda }^{2}V$.

Label the weight vertices by $1,\dots ,n,0,\bar{n},\dots ,\bar{1}$, with weights $e_1,\dots ,e_n,0,-e_n,\dots ,-e_1$. The crystal of the defining odd-orthogonal representation is the chainEach arrow of color $i$ subtracts the simple root ${\alpha }_i$. The two arrows through zero form the length-two short-root string.

Here is a complete grid description of the tensor product of crystals, valid for arbitrary $n$. Place the vertex $a\otimes b$ in row $a$, column $b$, in the above chain order; there are $(2n+1{)}^2$ vertices. Let ${\epsilon }_i(a)$ and ${\phi }_i(a)$ count incoming and outgoing steps along the color-$i$ string. For $i<n$, the two strings are $i\to i+1$ and $\overline{i+1}\to \bar{i}$, each of length one. For $i=n$, the string is $n\to 0\to \bar{n}$, with pairs $({\epsilon }_n,{\phi }_n)$ equal to $(0,2),(1,1),(2,0)$. Other vertices have both counts zero.

Using the crystal tensor-product rule, draw every color-$i$ edge byThe first case is a downward arrow to the next row and the second a rightward arrow to the next column; omit an arrow when its indicated factor has no outgoing edge. Thus the rule specifies every vertex and every arrow of the general tensor-square diagram, including the boundary nodes. The figure displays the resulting entire grid for $n=3$, alongside the vector crystal; the generator can draw any positive rank.

The raising rule chooses the first factor when ${\phi }_i(a)\ge {\epsilon }_i(b)$ and the second otherwise. For $n\ge 2$, its only highest vertices areIndeed a highest tensor must have highest first factor, hence $a=1$. For colors $i>1$, the second factor cannot have an incoming edge. For color one, its incoming count must be at most ${\phi }_1(1)=1$. Inspecting the strings leaves precisely $b=1,2,\bar{1}$.

These three connected components encode the Irreducible Lie algebra representationsHere $S_0^{2}V$ is the traceless symmetric square, ${\Lambda }^{2}V$ is the Adjoint representation, and the invariant symmetric form supplies the trivial summand. Their dimensions are $n(2n+3)$, $n(2n+1)$, and one, summing to $(2n+1{)}^2$. In fundamental-weight notation the highest weights are $2{\omega }_1,{\omega }_2,0$ for $n\ge 3$, but $2{\omega }_1,2{\omega }_2,0$ for $n=2$.

For $n=1$ the vector crystal is $1\stackrel{1}{\to }0\stackrel{1}{\to }\bar{1}$. The same tensor rule gives highest vertices $1\otimes 1,1\otimes 0,1\otimes \bar{1}$, of weights $2e_1,e_1,0$, or $4{\omega }_1,2{\omega }_1,0$, respectively. Its summand dimensions are $5,3,1$.

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 ${\epsilon }_1,\dots ,{\epsilon }_n,0,-{\epsilon }_n,\dots ,-{\epsilon }_1$. Label its vertices $1,\dots ,n,0,\bar{n},\dots ,\bar{1}$. Its directed edge colors areIn particular the short-root string $n\to 0\to \bar{n}$ 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 ${\epsilon }_i(b)$ count incoming edges of color $i$ along its string, and let ${\phi }_i(a)$ count outgoing ones. We use the conventionomitting an edge when the indicated operator vanishes. This crystal tensor-product rule specifies every vertex and every edge; the raising rule uses $\ge$ 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.

For $n\ge 2$, the three highest-weight vectors of the tensor crystal are $1\otimes 1$, $1\otimes 2$ and $1\otimes \bar{1}$. They have weights $2{\epsilon }_1$, ${\epsilon }_1+{\epsilon }_2$ 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. ThereforeThe 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 $n\ge 3$ 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 $1\otimes 1,1\otimes 0,1\otimes \bar{1}$, giving $L(4{\omega }_1)\oplus L(2{\omega }_1)\oplus L(0)$ of dimensions $5,3,1$.