# Does the Weyl group preserve coprimality in Kac-Moody algebras?

Let $$\mathfrak g$$ be a Kac-Moody algebra (symmetric, or hyperbolic, or whatever other assumptions you need) with simple roots $$\alpha_i$$. For $$\alpha$$ a root, write $$\alpha$$ in the basis of simple roots $$\alpha = \sum_i a_i \alpha_i$$. I am interested in when the gcd of all the $$a_i$$ is 1 (in which case I will say $$\alpha$$ is "coprime").

It seems that any real root of $$\mathfrak g$$ is coprime, at least for the cases that I've tested numerically -- namely, $$E_{10}$$, $$E_{11}$$, and $$\begin{bmatrix}2&-3\\-3&2\end{bmatrix}$$. More generally, the Weyl group seems to preserve gcd's in the sense that if $$w$$ is in the Weyl group and $$w\alpha = \sum_i b_i\alpha_i$$, then the gcd of the $$a_i$$ is the same as the gcd of the $$b_i$$. Is this true?

• The Weyl group is generated by simple reflections, so this works, thanks. Jul 2, 2019 at 0:24

Note that, in your notation, if $$d_i \in \mathbb Z$$ are such that $$\sum_i d_i a_i = \gcd_i a_i$$, then also $$\sum_{i \ne j} (d_i + \langle\alpha_j^\vee, \alpha_i\rangle d_j)a_i + d_j(a_j - \sum_i \langle\alpha_j^\vee, \alpha_i\rangle a_i) = \sum_i d_i a_i = \gcd_i a_i$$, so that the simple reflexion $$w$$ in $$\alpha_j$$ preserves gcd's. The general result follows.