The Weyl dimension formula for a fundamental sl3 module

Given: The Axiom of Choice, the realization of ${\mathfrak{s}\mathfrak{l}}_3$ with simple roots ${\alpha }_1,{\alpha }_2$, positive roots ${\mathrm{\Phi }}^{+}=\{{\alpha }_1,{\alpha }_2,\theta \}$ with $\theta ={\alpha }_1+{\alpha }_2$, the fundamental weight ${\omega }_1$, the Weyl vector $\rho$ and the coroots ${\alpha }_1^{\vee },{\alpha }_2^{\vee },{\theta }^{\vee }$.

The Axiom of Choice is assumed; it enters through the dimension formula and the highest-weight suppliers below (The Axiom of Choice).

In this realization the positive roots are ${\alpha }_1,{\alpha }_2$ and $\theta ={\alpha }_1+{\alpha }_2$, whose coroots in the simply-laced system add: ${\theta }^{\vee }={\alpha }_1^{\vee }+{\alpha }_2^{\vee }$ (Root systems of the classical complex Lie algebras, Classical complex matrix Lie algebras, The root set is a reduced crystallographic root system).

The fundamental weight ${\omega }_1$ is dual to ${\alpha }_1^{\vee }$: $⟨{\omega }_1,{\alpha }_1^{\vee }⟩=1$ and $⟨{\omega }_1,{\alpha }_2^{\vee }⟩=0$, and the Weyl vector satisfies $⟨\rho ,{\alpha }_i^{\vee }⟩=1$ and $\rho ={\alpha }_1+{\alpha }_2$; moreover $⟨\rho ,{\theta }^{\vee }⟩=2$ (Fundamental weights, Integral, dominant, and strictly dominant weights, The root set is a reduced crystallographic root system).

The Weyl dimension formula states $\dim L(\lambda )={\prod }_{\alpha \in {\mathrm{\Phi }}^{+}}(\lambda +\rho ,\alpha )\mathrm{/}(\rho ,\alpha )={\prod }_{\alpha \in {\mathrm{\Phi }}^{+}}⟨\lambda +\rho ,{\alpha }^{\vee }⟩\mathrm{/}⟨\rho ,{\alpha }^{\vee }⟩$ (The Weyl dimension formula).

The defining representation ${\mathbb{C}}^3$ of ${\mathfrak{s}\mathfrak{l}}_3$ has weights ${\epsilon }_1,{\epsilon }_2,{\epsilon }_3$, with ${\epsilon }_1$ the highest weight ${\omega }_1$; it is irreducible, because for a nonzero $v={\sum }_i{c}_i{e}_i$ with $c_j\mathrm{\ne }0$ the matrix units $E_{ij}$ with $i\mathrm{\ne }j$ and the diagonal elements of ${\mathfrak{s}\mathfrak{l}}_3$ produce all three basis vectors $e_1,e_2,e_3$, so the submodule generated by $v$ is everything; hence $L({\omega }_1)\cong {\mathbb{C}}^3$ has dimension $3$ (Classical complex matrix Lie algebras, Highest-weight classification).