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For the defining representation of the special linear Lie algebra ${\mathfrak{sl}}_3$, the crystal basis is $1\stackrel{1}{\to }2\stackrel{2}{\to }3$, with weights ${\epsilon }_i$ restricted to trace-zero diagonal matrices. Its dual is $\bar{3}\stackrel{2}{\to }\bar{2}\stackrel{1}{\to }\bar{1}$, with weights $-{\epsilon }_i$. Using the crystal tensor-product rule with lowering on the first factor when ${\phi }_i(a)>{\epsilon }_i(b)$, the tensor square has highest vertices $1\otimes 1$ and $1\otimes 2$, of highest weights $2{\omega }_1,{\omega }_2$, generating components of six and three vertices. The defining tensor dual has highest vertices $1\otimes \bar{3}$ and $1\otimes \bar{1}$, generating components of eight and one vertices and weights ${\omega }_1+{\omega }_2,0$. These encode the symmetric square plus exterior square, and the trace-free endomorphisms plus scalar identity, respectively. A crystal vertex labelled by a tensor pair need not represent that same undeformed ordinary simple tensor; the actual invariant vector is ${\sum }_i{e}_i\otimes e_i^{\ast }$.

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.

For the B2 root system with ${\alpha }_1={\epsilon }_1-{\epsilon }_2$ long and ${\alpha }_2={\epsilon }_2$ short, the spin fundamental weight is ${\Lambda }_2=({\epsilon }_1+{\epsilon }_2)/2$. Its four-dimensional highest-weight representation has the minuscule weight orbit $\frac{1}{2}(\pm {\epsilon }_1\pm {\epsilon }_2)$, each of multiplicity one. The crystal basis is the chain $++\stackrel{2}{\to }+-\stackrel{1}{\to }-+\stackrel{2}{\to }--$, where signs denote the coordinates. Each Kashiwara operator subtracts its indicated simple root; the sl2 Lie algebra strings have length one. This fixes every edge and gives a connected highest-weight graph with the correct character.

The defining representation of ${\mathfrak{so}}_{2n+1}$ has weights ${\epsilon }_1,\dots ,{\epsilon }_n,0,-{\epsilon }_n,\dots ,-{\epsilon }_1$. Its crystal basis is a chain on these vertices, with lowering-edge colors $1,\dots ,n-1,n,n,n-1,\dots ,1$. The short-root string has length two through zero. With the crystal tensor-product rule, for $n\ge 2$ the tensor square has highest vertices $1\otimes 1$, $1\otimes 2$ and $1\otimes \bar{1}$, corresponding to the traceless symmetric square, the exterior square and the trivial Lie algebra representation. Their dimensions are $(2n+1)(n+1)-1$, $n(2n+1)$ and one.

For the symplectic Lie algebra ${\mathfrak{sp}}_{2n}$, the defining crystal basis is the chain $1,2,\dots ,n,\bar{n},\dots ,\bar{1}$, with weights ${\epsilon }_1,\dots ,{\epsilon }_n,-{\epsilon }_n,\dots ,-{\epsilon }_1$. Its successive lowering-edge colors are $1,2,\dots ,n-1,n,n-1,\dots ,1$. Every step subtracts its color's simple root. Applying the crystal tensor-product rule to two copies gives a complete grid description of the tensor-square crystal. For $n\ge 2$, its only highest vertices are $1\otimes 1$, $1\otimes 2$, and $1\otimes \bar{1}$, of highest weights $2{\omega }_1$, ${\omega }_2$, and zero. For $n=1$, the middle vertex is absent and only the other two highest vertices remain.

The seven-dimensional irreducible representation of the G2 root system has highest weight ${\omega }_1=2{\alpha }_1+{\alpha }_2$. Its crystal basis is the chain of weights $2{\alpha }_1+{\alpha }_2,{\alpha }_1+{\alpha }_2,{\alpha }_1,0,-{\alpha }_1,-{\alpha }_1-{\alpha }_2,-2{\alpha }_1-{\alpha }_2$, with successive edge colors $1,2,1,1,2,1$. Each step subtracts its color's simple root. The nonzero weights are the six short roots, each of multiplicity one; the zero weight also has multiplicity one.

Write ${\epsilon }_i(b)$ and ${\phi }_i(b)$ for the numbers of raising and lowering steps of color $i$ available from a vertex of a crystal basis. One convention for the tensor product of crystals isThe raising operator uses the first factor for ${\phi }_i(a)\ge {\epsilon }_i(b)$ and the second otherwise. The different strictness makes raising and lowering inverse along every edge. An edge is absent if the chosen Kashiwara operator vanishes. Stating the convention is essential when identifying the highest-weight vectors as ordered tensor vertices.

On a crystal, ${\tilde{f}}_i$ follows an edge of color $i$ and ${\tilde{e}}_i$ follows that edge backwards.

These maps are the raising and lowering operations of a crystal basis.

Take the diagonal Cartan subalgebra of ${\mathfrak{sl}}_3$ and let ${\epsilon }_i$ select its $i$th diagonal entry, so ${\epsilon }_1+{\epsilon }_2+{\epsilon }_3=0$ on it. Use simple roots ${\alpha }_1={\epsilon }_1-{\epsilon }_2$, ${\alpha }_2={\epsilon }_2-{\epsilon }_3$. The fundamental weights are ${\omega }_1={\epsilon }_1$ and ${\omega }_2=-{\epsilon }_3={\epsilon }_1+{\epsilon }_2$. The defining representation $V$ has highest vector $e_1$ and the dual has highest vector $e_3^{\ast }$. ThusTheir crystal basis diagrams are $B:1\stackrel{1}{\to }2\stackrel{2}{\to }3$ and $B^{\vee }:\bar{3}\stackrel{2}{\to }\bar{2}\stackrel{1}{\to }\bar{1}$, with weights ${\epsilon }_i$ and $-{\epsilon }_i$ respectively. Use the same crystal tensor-product rule as in2(d): lowering acts on the first factor when ${\phi }_i(u)>{\epsilon }_i(v)$; otherwise on the second. Raising uses the first factor when ${\phi }_i(u)\ge {\epsilon }_i(v)$, so it reverses each lowering arrow. An edge is absent if the selected factor operation is zero.

For $B\otimes B$, abbreviate $i\otimes j$ by $ij$. The complete color-one edge list is $11\to 21\to 22$, $31\to 32$, $13\to 23$; the complete color-two list is $21\to 31$, $22\to 32\to 33$, $12\to 13$. There are two connected components:For example ${\tilde{f}}_1(1\otimes 1)=2\otimes 1$ because ${\phi }_1(1)=1>{\epsilon }_1(1)=0$, whereas ${\tilde{f}}_1(1\otimes 2)=0$ because equality selects the second factor and ${\tilde{f}}_1(2)=0$. This explains which ordered tensor labels lie in each component. At the representation level, the two components are the six-dimensional symmetric square and the three-dimensional exterior square:

For $B\otimes B^{\vee }$, abbreviate $i\otimes \bar{j}$ by $i\bar{j}$. The complete color-one edge list is $1\bar{3}\to 2\bar{3}$, $1\bar{2}\to 2\bar{2}\to 2\bar{1}$, $3\bar{2}\to 3\bar{1}$; the complete color-two list is $1\bar{3}\to 1\bar{2}$, $2\bar{3}\to 3\bar{3}\to 3\bar{2}$, $2\bar{1}\to 3\bar{1}$. The isolated vertex is $1\bar{1}$. Thus the two highest vertices are $1\bar{3}$ of weight ${\epsilon }_1-{\epsilon }_3={\omega }_1+{\omega }_2$, and $1\bar{1}$ of weight zero. Lowering from the former visits all eight other vertices. ConsequentlyThis also follows concretely from $V\otimes V^{\ast }\cong \mathrm{End}(V)={\mathfrak{sl}}_3\oplus \mathbb{C}I$: the first term is the Adjoint representation, with highest weight the highest root, and the second is the trivial Lie algebra representation. The ordinary invariant vector is ${\sum }_i{e}_i\otimes e_i^{\ast }$, not the bare tensor suggested by the singleton's crystal label. The explicit edge lists and the diagrams exhibit all eighteen tensor vertices and justify the four highest weights.

Use the ordered alphabetwith weights $\mathrm{wt}(i)={\epsilon }_i$ and $\mathrm{wt}(\bar{i})=-{\epsilon }_i$. The crystal of the defining symplectic representation is the chainThe label of an arrow is its simple root color; a Kashiwara operator ${\tilde{f}}_i$ follows an arrow of color $i$ and lowers the weight by ${\alpha }_i$.

Here is a grid description of the entire tensor product of crystals, valid for every $n$. Put one vertex $a\otimes b$ at row $a$ and column $b$, in the alphabet order above. Let $P_i$ be the set of chain positions with outgoing color $i$, and $Q_i$ the positions with incoming color $i$:Their indicators are ${\phi }_i$ and ${\epsilon }_i$. Use the crystal tensor-product ruleAn arrow is omitted if the selected factor has no outgoing arrow of that color. Thus the complete grid has a downward color-$i$ arrow when $a\in P_i$ and $b\notin Q_i$, and a rightward color-$i$ arrow when $b\in P_i$ and $a\notin P_i$. There are no other arrows. These two coordinate rules specify every vertex and every edge of the requested general-rank drawing. The following figure displays the full grid at $n=3$, including all components, together with the standard chain; increasing $n$ uses exactly the same grid construction.

For completeness, highest vertices can be read from the raising rule, which selects the first factor when ${\phi }_i(a)\ge {\epsilon }_i(b)$. If $a$ has an incoming edge of color $i$, then ${\phi }_i(a)=0$; either the raising operator acts nontrivially on $a$, or ${\epsilon }_i(b)=1$ and it acts nontrivially on $b$. Thus a highest vertex must have $a=1$. At $a=1$, all ${\phi }_i(a)$ vanish except ${\phi }_1(a)=1$. To be highest, the incoming color of $b$, if any, must consequently be $1$. For $n\ge 2$, the possibilities are exactlyTheir weights are $2{\epsilon }_1=2{\omega }_1$, ${\epsilon }_1+{\epsilon }_2={\omega }_2$, and zero. Each occurs once. Connected highest-weight components of the crystal basis describe the irreducible representations in the tensor product of Lie algebra representations, soThis is the tensor-square decomposition of the defining symplectic representation. The symmetric square is $L(2{\omega }_1)$; the exterior square is the direct sum of its invariant inverse-form line and the primitive exterior square $L({\omega }_2)$. Their dimensions are respectively $n(2n+1)$, $n(2n-1)-1$, and one, summing to $4n^2=\dim (V\otimes V)$. At $n=1$, the chain has two vertices and the highest tensor vertices are only $1\otimes 1$ and $1\otimes \bar{1}$: the decomposition is $L(2{\omega }_1)\oplus L(0)$, of dimensions three and one.

Choose a short simple root ${\alpha }_1$ and a long simple root ${\alpha }_2$. The G2 root system has positive rootsThe two hexagons formed by the short and long roots give the usual twelve-root diagram. The fundamental weights areThe second is the highest root, so the irreducible module $L({\omega }_2)$ is the Adjoint representation of a Lie algebra.

The seven weights of $L({\omega }_1)$ are zero and the six short roots, each with multiplicity one. Its crystal, with arrows denoting the lowering operators ${\tilde{f}}_i$, is the chain

Applying the Weyl dimension formula to $\lambda =n_1{\omega }_1+n_2{\omega }_2$ gives

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 product crystal uses the $i$-string data ${\epsilon }_i$ and ${\phi }_i$ to decide on which tensor factor a Kashiwara operator acts. Its connected components are highest-weight crystals and encode the decomposition of the tensor product representation.

This tensor-product operation combines one crystal basis with another, with factor selection governed by their string parameters.