= Solution
For a finite-support <formal character>, use the same pullback rule $(wu)(\xi)=u(w^{-1}\xi)$; equivalently $w(e^\lambda)=e^{w\lambda}$. This specifies the action on the denominator, which is a finite delta sum rather than a <polynomial> function. The whole infinite-support completion need not be Weyl-stable, so no such claim is required.
A simple <reflection> $s_\alpha$ permutes the positive roots other than $\alpha$ and sends $\alpha$ to $-\alpha$. In the product defining $D$, all factors are therefore permuted except the $\alpha$ factor, whose sign reverses. Thus $s_\alpha D=-D$. Simple reflections generate $W$, and each has <determinant> $-1$, so
$$
\boxed{wD=(\det w)D.}
$$
Back to article page
