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?

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

1 Answer 1


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.


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