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The vectors of a root-system were originally called "roots" because they are the zeros of a characteristic polynomial that comes from the connection of (crystallographic) root-systems to classifying semisimple Lie algbras. From an answer by José Carlos Santos,

It comes from the roots of the characteristic polynomial of an endomorphism. If $\mathfrak g$ is a complex semisimple Lie algbra, $\mathfrak h$ is a Cartan subalgebra and $\alpha\in\mathfrak{h}^*$, then $\alpha$ is a root if, for every $H\in\mathfrak h$, $\alpha(H)$ is an eigenvalue of the endomorphism of $\mathfrak g$ defined by $X\mapsto[H,X]$.

Is there a way to get at this same polynomial just from the definitions of a (crystallographic) root-system without having to talk about the connection to Lie theory?

To provide some specific motivation, I'm reading through a proof of Gabriel's theorem (classifying quivers of finite representation type) where you define the Tits form $q$ for the ADE Dynkin and Euclidean diagrams, and the set of roots associated to that diagram are the nonzero vectors $x \in \mathbb{Z}^n$ for which $q(x) \leq 1$. Since $q(x) \in \mathbb{Z}$ we could say these are the integer roots of the polynomial $q(x)(q(x)-1)$, but I doubt that this generalizes beyond the root systems you get from the ADE Dynkin diagrams. I was looking for a way to justify calling these "roots" without getting into Lie theory, and I thought there might be a justification for the word "root" in terms of general abstract root systems.

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    $\begingroup$ I can recommend Reflection Groups and Coxeter Groups by James Humphreys. I am flipping through it and cannot guarantee that it answers your concerns. On the other hand, I know of no book called Root Systems and Polynomials, so Humphreys is a good start. Plenty of roots, plenty of polynomials. $\endgroup$
    – Will Jagy
    Mar 2, 2018 at 23:17
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    $\begingroup$ @WillJagy Yeah, that's where I started. Humphreys mentions that the term "root" historically comes from Lie theory (section 1.2), and then makes the distinction between general root systems and crystallographic roots systems that relate to Lie theory (section 2.9). I haven't seen the answer to the question spelled out anywhere though. $\endgroup$
    – Mike Pierce
    Mar 3, 2018 at 0:37
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    $\begingroup$ Alright. He's on this site; it appears he is active from time to time. Maybe he will notice this. $\endgroup$
    – Will Jagy
    Mar 3, 2018 at 0:55
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    $\begingroup$ I'm not clear about the motivation for going in this direction, but it's probably feasible at least case-by-case if one first classifies the simple Lie algebras and then contructs them individually in known ways. Anyway, the Bourbaki/Serre notion of abstract root system grew out of early work by Killing and Cartan, with intermediate steps by Jacobson and others. But what is learned by going in the reverse direction? $\endgroup$
    – Jim Humphreys
    Mar 4, 2018 at 0:28
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    $\begingroup$ @Mike: I should clarify that it's probably noit realistic to discuss root systems abstractly without some mention of Lie algebras. The notion of "root" does arise from the characteristic polynomials of adjoint operators and their roots. Without this motivation, why would anyone study "root systems" axiomatically? (On te other hand, the abstract definition allows one to generalize to infinite-dimensional Lie algebras of Kac-Moody type, etc.) $\endgroup$
    – Jim Humphreys
    Mar 14, 2018 at 14:40

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As far as I can tell your question is "can I reconstruct a polynomial from its roots." So, yes, you can consider the product $\prod_{\alpha} (x-\alpha(H))$ as a function on $\mathfrak{h}$ valued in polynomials. This is the same (by the definition of root) as the characteristic polynomial of the adjoint action by $H$ (on $\mathfrak{g}/\mathfrak{h}$; for all of $\mathfrak{g}$, you should multiply by $x^{\mathrm{dim}(\mathfrak{h})}$).

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  • $\begingroup$ I would say (as you clearly meant) that it's the same as the characteristic polynomial, up to multiplication by powers of $x$, by the definition of 'root' and the fact that root spaces are one-dimensional. $\endgroup$
    – LSpice
    Mar 3, 2018 at 21:32
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    $\begingroup$ I may not have made my question clear. I was hoping for a characterization that didn't mention Lie algebras, their Cartan subalgebra, etc, at all. Instead I was hoping for a characterization coming just from the perspective of abstract root systems like are described in Humphrey's Reflection Groups and Coxeter Groups. $\endgroup$
    – Mike Pierce
    Mar 4, 2018 at 1:42

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