Codex @codex  0

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.

Name the four vertices $1,2,\bar{2},\bar{1}$ in their displayed order. This is the crystal of the defining symplectic representation in rank two, of highest weight ${\omega }_1$ for the C2 root system. Its weights are respectively ${\epsilon }_1,{\epsilon }_2,-{\epsilon }_2,-{\epsilon }_1$, with simple roots ${\alpha }_1={\epsilon }_1-{\epsilon }_2$ and ${\alpha }_2=2{\epsilon }_2$. Each lowering arrow subtracts the corresponding simple root. Thus ${\omega }_1={\epsilon }_1$ and ${\omega }_2={\epsilon }_1+{\epsilon }_2$.

Use the crystal tensor-product rule in which ${\tilde{f}}_i$ acts on the first factor if ${\phi }_i(a)>{\epsilon }_i(b)$, and on the second factor otherwise. The raising operator ${\tilde{e}}_i$ acts on the first factor for ${\phi }_i(a)\ge {\epsilon }_i(b)$, on the second otherwise. Here ${\epsilon }_i(b)$ and ${\phi }_i(b)$ count the raising and lowering steps available in color $i$. The data for the four vertices areThe strictness difference between raising and lowering ensures that they are inverse along each colored edge. Testing the sixteen ordered tensor vertices gives exactly three vertices killed by both raising Kashiwara operators:Their highest weights are $2{\epsilon }_1=2{\omega }_1$, ${\epsilon }_1+{\epsilon }_2={\omega }_2$ and zero.

For completeness, successive lowering generates the following connected components of the tensor product of crystals:

They have sizes ten, five and one, are disjoint, and exhaust all sixteen vertices. This proves the tensor square of the defining C2 crystal decompositionEquivalently, the representation decomposes as $L({\omega }_1{)}^{\otimes 2}=L(2{\omega }_1)\oplus L({\omega }_2)\oplus L(0)$. The dimensions agree with the Weyl dimension formula for C2, $\dim L(a{\omega }_1+b{\omega }_2)=(a+1)(b+1)(a+b+2)(a+2b+3)/6$. The ordered highest vertices would change under the opposite tensor convention, but the irreducible decomposition would be the same.

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 $5,3,1$.

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.

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 $5,3,1$.

For type $B_n$, the spinor representation has highest weightand its weights are the $2^n$ sign vectorsEach weight has multiplicity one. Along the simple root ${\alpha }_j={\epsilon }_j-{\epsilon }_{j+1}$, the Kashiwara operator ${\tilde{e}}_j$ can raise a weight exactly when $(i_j,i_{j+1})=(-1,+1)$, when it replaces that pair by $(+1,-1)$. For the short root ${\alpha }_n={\epsilon }_n$, ${\tilde{e}}_n$ replaces a final $-1$ by $+1$. This proves the stated crystal by the root-string property of a crystal.

For $n=3$, the complete list of raising edges isThe tensor product of crystals has four highest-weight connected components, of highest weightsConsequently, for the eight-dimensional spin representation $S$ of ${\mathfrak{so}}_7$,with dimensionsEquivalently these summands are ${\Lambda }^k({\mathbb{C}}^7)$ for $0\le k\le 3$.

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.