Weyl character and dimension formulas for sl2

Given: The Axiom of Choice, $\mathfrak{g}={\mathfrak{s}\mathfrak{l}}_2$ with its positive root $\alpha$, fundamental weight $\omega =\alpha \mathrm{/}2$, Weyl group $W=\{1,s\}$, Weyl vector $\rho =\omega$, and the dominant integral weights $m\omega$ with $m\ge 0$.

The Axiom of Choice is assumed; it enters through the character and dimension formulas below (The Axiom of Choice).

$\alpha =2\omega$, $W=\{1,s\}$, $s\omega =-\omega$, $\rho =\omega$ and $\lambda +\rho =(m+1)\omega$; the length of $s$ is $1$ (The Weyl vector rho for a chosen positive system, Integral, dominant, and strictly dominant weights).

$A(\nu )=e^{\nu }-e^{-\nu }$ for every $\nu \in {\mathfrak{h}}^{\ast }$, and the Weyl character formula and Weyl dimension formula read $\mathrm{ch}L(\lambda )=A(\lambda +\rho )A(\rho {)}^{-1}$ and $\dim L(\lambda )={\prod }_{\alpha \in {\mathrm{\Phi }}^{+}}(\lambda +\rho ,\alpha )\mathrm{/}(\rho ,\alpha )$ (The Weyl alternation operator, The Weyl character formula, The Weyl dimension formula).

In the completed ring $\mathcal{R}$ the elements $e^{\pm \omega }$ are invertible with $e^{\omega }{e}^{-\omega }=e^0$, and $e^{\omega }-e^{-\omega }=e^{\omega }(1-e^{-\alpha })$ is invertible with inverse $e^{-\omega }(1-e^{-\alpha }{)}^{-1}$, because $1-e^{-\alpha }$ is invertible (The completed formal character ring, Geometric series are invertible in the completed character ring, The Weyl denominator identity).

For $m\ge 0$ the module $L(m\omega )$ is the finite-dimensional simple module of highest weight $m\omega$ (Highest-weight classification).