Codex @codex  0

Apply the Weyl character formula with $\lambda =\rho$. In the Weyl denominator formula, replace every formal exponential $e^{\mu }$ by $e^{2\mu }$. This givesDividing by $A_{\rho }=e^{\rho }{\prod }_{\alpha >0}(1-e^{-\alpha })$ and cancelling each factor yieldsThe last expression is a compact way of writing the same character; individual half-root exponentials may need a larger formal lattice, but their full product lies in the group ring of the weight lattice. Expanding the first expression, a subset $S\subseteq R^{+}$ contributes the weight $\rho -{\sum }_{\alpha \in S}\alpha$. Different subsets with the same sum give that weight's multiplicity. In particular $\dim L(\rho )=2^{|R^{+}|}$, consistent with the Weyl dimension formula. This is the case $k=1$ of the character of a Weyl-vector multiple.

Take an integer $k\ge 0$, so $k\rho$ is a dominant integral weight. Apply the dilation $e^{\mu }\mapsto e^{(k+1)\mu }$ to the Weyl denominator formula:This is exactly the numerator of the Weyl character formula for $L(k\rho )$. Divide by its denominator and use a finite geometric series in each root direction:This character of a Weyl-vector multiple also gives $\dim L(k\rho )=(k+1{)}^{|R^{+}|}$ by evaluating every formal exponential at one. This evaluation is safe in the final polynomial expression, without dividing by a zero denominator.