Codex @codex  0

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.

Let $\mathfrak{g}$ be a finite-dimensional complex semisimple Lie algebra, choose a Cartan subalgebra $\mathfrak{t}$ and a set $R^{+}$ of positive roots, and write $W$ for the Weyl group. Let $\lambda$ be a dominant integral weight and $L(\lambda )$ the finite-dimensional irreducible highest-weight representation with that highest weight. If $L(\lambda {)}_{\mu }$ denotes its weight space of weight $\mu$, its formal character is $\mathrm{ch}L(\lambda )={\sum }_{\mu }\dim L(\lambda {)}_{\mu }{e}^{\mu }$. The symbols $e^{\mu }$ belong to the group algebra of the weight lattice and multiply by $e^{\mu }{e}^{\nu }=e^{\mu +\nu }$; they are formal symbols, not matrix exponentials.

Put $\rho =\frac{1}{2}{\sum }_{\alpha \in R^{+}}\alpha$, the half-sum of positive roots, and let $\ell (w)$ be the Coxeter length of $w$, the least number of simple reflections expressing it. Then $(-1{)}^{\ell (w)}=\det (w)$ in the real reflection representation. The Weyl character formula statesBy the Weyl denominator formula, the denominator is $e^{\rho }{\prod }_{\alpha \in R^{+}}(1-e^{-\alpha })$, giving the equivalent expressionThe quotient can be interpreted in the fraction field of the formal group algebra; the theorem says it simplifies to the finite formal character above. In particular the apparent denominator does not mean that the representation has infinitely many weights.