Is $\operatorname{Stab}(\lambda)$ generated by the simple reflections it contains, for $\lambda\in A_0$?

Question

For a finite Weyl group, the stabilizer of an element in the fundamental domain is generated by the simple reflections of the Weyl group that is contains. Does the same still hold for the closure of the fundamental alcove of an affine Weyl group?

Answer 1

Yes, let $W_a$ be the affine Weyl group, and $S_a$ its set of simple reflections. Let $G=\operatorname{Stab}(\lambda)$ for $\lambda\in\overline{A_\circ}$, and let $G'\leq G$ be the subgroup generated by the simple reflections $G$ contains. I claim $G=G'$.

Suppose $w\in G$. If $\ell(w)=0$ or $1$, it's clear $w\in G'$. Suppose $\ell(w)>1$. Pick $s\in S_a$ such that $\ell(ws)<\ell(w)$. By the standard properties of the length function, this means the corresponding hyperplane $H_s$ separates $A_\circ$ and $w^{-1}A_\circ$.

Since $\lambda=w^{-1}\lambda$, by virtue of $w\in G$, we have $\lambda\in w^{-1}\overline{A_\circ}$. There are two main cases. If $s=s_\alpha$ is just a usual simple reflection, we must have $(\lambda,\alpha)\leq 0$, and if $s=s_{\tilde{\alpha},1}$ is the reflection corresponding affine hyperplane corresponding to the highest root $\tilde{\alpha}$ translated by $\frac{1}{2}\tilde{\alpha}^\vee$, we must have $(\lambda,\tilde{\alpha})\geq 1$.

But $\lambda\in\overline{A_\circ}$, so $(\lambda,\alpha)\geq 0$ and $(\lambda,\tilde{\alpha})\leq 1$. Necessarily $(\lambda,\alpha)=0$, or $(\lambda,\tilde{\alpha})=1$. In either case, $$ s_\alpha(\lambda)=\lambda $$ or $$ s_{\tilde{\alpha},1}(\lambda)=\lambda-((\lambda,\tilde{\alpha})-1)\tilde{\alpha}^\vee=\lambda. $$ So whatever $s$ is, $s\lambda=\lambda$, so $s\in G'$. But $ws\lambda=w\lambda=\lambda$, so $ws\in G$, so by induction, $ws\in G'$, thus $w=wss\in G'$, whence $G=G'$.