# Crystallographic root system Coxeter Groups

In Humphreys book in "Reflection Groups and Coxeter Groups" he defines a root system $$\Phi$$ to be crystallographic if it satisfies $$\frac{2(\alpha, \beta)}{(\beta, \beta)} \in \mathbb{Z}$$ $$(\star)$$ for all $$\alpha, \beta \in \Phi$$ and he states that it is enough to require that the ratios be integers when $$\alpha, \beta \in \Delta$$, where $$\Delta$$ is the simple system of a Coxeter group (the elements of $$\Phi$$ are either non-negative or either non-positive linear combinations of the elements of $$\Delta$$ and $$\Delta$$ is a basis for the vector space where the Coxeter group acts).

Having the result $$(\star)$$ for the elements on $$\Delta$$, I don't see how to show this holds for the elements of $$\Phi$$.

• Hello, I agree that I didn't write either non-negative or non-positive positive linear combinations of the elements of $\Delta$, (Already edited the question) but the elements of the root system are not necessarily $\underline{integral}$ linear combinations by its definition, or am I still missing something? Thank you for your help. – square17 May 17 '20 at 13:54
• Yes, for a moment I thought I missed something in the definition of root system. Thank you for clearing that out, but I still don't see clearly why from having $(\star)$ satisfied in $\Delta$ we can deduce it is also satisfied in $\Phi$. – square17 May 17 '20 at 19:11
By the assumption on simple roots, we know that if $$\alpha, \beta \in \Delta$$, then $$s_\alpha(\beta)$$ is an integral linear combination of $$\alpha$$ and $$\beta$$. Since $$\{s_\alpha: \alpha \in \Delta\}$$ generates $$W$$, then for any $$w \in W$$, $$\beta \in \Delta$$, we know that $$w \beta$$ is an integral linear combination of elements of $$\Delta$$. But every root in $$\Phi$$ is of the form $$w \beta$$ (Corollary 1.5) and is thus an integral linear combination of elements of $$\Delta$$.
Now the rest is easy. To show that $$f(\alpha,\beta) := \frac{2(\alpha, \beta)}{(\beta,\beta)} \in \mathbb{Z}$$ for arbitrary $$\alpha, \beta \in \Phi$$, first note that $$f(\alpha,\beta)$$ is invariant under $$W$$, so by replacing $$(\alpha,\beta)$$ by $$(w\alpha, w\beta)$$ for a suitable $$w \in W$$, we may assume that $$\beta \in \Delta$$. Then note that $$f(\alpha, \beta)$$ is linear in $$\alpha$$, so we may assume $$\alpha \in \Delta$$ as well, and we are done by our original assumption that $$f(\alpha,\beta) \in \mathbb{Z}$$ for $$\alpha, \beta \in \Delta$$.