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 $\Phi$.

Thank you in advance

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    $\begingroup$ The confusion might be because you have written the relation between the base and the root system incorrectly. It is either non-negative or non-positive integral linear combinations. $\endgroup$ – Tobias Kildetoft May 16 '20 at 22:53
  • $\begingroup$ 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. $\endgroup$ – square17 May 17 '20 at 13:54
  • $\begingroup$ They are definitely integral combinations of elements from the basis. That is part of the definition of being a basis for a root system. $\endgroup$ – Tobias Kildetoft May 17 '20 at 16:22
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    $\begingroup$ Humphreys does not require the roots of a general root system (as opposed to a crystallographic root system) to be integral linear combinations of the simple roots. He only requires them to be nonnegative linear combinations of the simple roots (sections 1.2 and 1.3). He then defines a crystallographic root system by imposing an additional condition on general root systems (section 2.9), as stated in the question. $\endgroup$ – Ted May 17 '20 at 17:45
  • $\begingroup$ 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 $\Phi$. $\endgroup$ – 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 $\alpha, \beta \in \Delta$.


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