= Solution
Apply the <Weyl character formula> with $\lambda=\rho$. In the <Weyl denominator formula>, replace every formal exponential $e^\mu$ by $e^{2\mu}$. This gives
$$
A_{2\rho}=e^{2\rho}\prod_{\alpha>0}(1-e^{-2\alpha}).
$$
Dividing by $A_\rho=e^\rho\prod_{\alpha>0}(1-e^{-\alpha})$ and cancelling each factor yields
$$
\boxed{\operatorname{ch}L(\rho)=e^\rho\prod_{\alpha>0}(1+e^{-\alpha})
=\prod_{\alpha>0}(e^{\alpha/2}+e^{-\alpha/2}).}
$$
The 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>.
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