3
$\begingroup$

Let $\Phi$ be a root system in the following sense: (1) $\Phi \subset \mathbb{R}^n$ consists of a finite number of nonzero vectors, (2) for each $\alpha \in \Phi$, $\Phi \cap \mathbb{R} \alpha = \{-\alpha,\alpha\}$ and (3) for each $\alpha \in \Phi$, $s_\alpha \Phi = \Phi$ where $s_\alpha$ is the reflection in $\alpha$.

Let $\Delta \subset \Phi$ be a simple system. Then, for $\alpha \neq \beta$ in $\Delta$, we can show that $(\alpha,\beta) \leq 0$.

Humpfreys (in Reflection Groups and Coxeter Groups) proves this by contradiction, in a weird way. I want to know why my simple argument below does not work (since if it did, he would just do it this way).

Argument: Suppose for a contradiction that $(\beta,\alpha) \geq 0$. Then $s_\alpha \beta = \beta - c\alpha$ where $c = \dfrac{2(\beta,\alpha)}{(\alpha,\alpha)} \geq 0$. Then since $\Delta$ is a basis of $\text{span}_\mathbb{R} \Phi$, every element of $\Phi$ has a unique representation as a linear combination of $\Delta$ with coefficients of all the same sign. This is already a contradiction to $s_\alpha \beta = \beta - c\alpha$.

Where is my misunderstanding?

I would like to emphasise that these root systems are not crystallographic, as one sees with Lie Algebras.

$\endgroup$
  • 1
    $\begingroup$ The negation of $(\beta,\alpha) \leq 0$ is that $(\beta, \alpha) > 0$, so you should change those weak inequalities to strict ones in your argument. The argument works nonetheless. $\endgroup$ – Hugh Denoncourt Jun 3 '14 at 18:45
5
+100
$\begingroup$

Your argument is correct. There are two potential misunderstandings.

The first involves his choice to have this fact appear as a corollary to the main theorem of the section. It is common for authors to choose to apply the main theorem of a section to derive corollaries that really have simple self-contained proofs of their own like the one you gave. He wants to show the power of the section's main theorem, so he uses it to derive the obtuse-angle fact as a corollary.

But I suspect your real confusion is this: The $\Delta$ of the proof of the main theorem is a set he constructs that may or may not be simple, and as such, has to be proven to be simple. Your argument begins with $\Delta$ as a simple system, which is fine for the obtuse-angle fact, but no good for proving the main theorem.

He is trying to prove that every positive root system $\Pi$ in $\Phi$ contains a unique simple system $\Delta$. For the existence portion of the proof, he lets $\Delta \subseteq \Pi$ be a minimal set subject to the requirement that each root in $\Pi$ is a non-negative linear combination of roots in $\Delta$. (The set $\Pi$ is such a set, but is probably not minimal.) He can't assume that $\Delta$ is itself a simple system, because then the argument would be circular. To prove $\Delta$ is a simple system, he has to show that $\Delta$ is a linearly independent set, which he does by proving that for this (not necessarily simple) $\Delta$, the obtuse-angle property holds. Since every simple system is contained in some positive root system $\Pi$, and this constructed $\Delta$ is the unique simple system contained in $\Pi$, it follows that the obtuse-angle property holds for arbitrary simple systems.

$\endgroup$
  • $\begingroup$ There still seems to be a mistake at the end of the proof in the book though (last line of the proof on page 9). Shouldn't the cases to consider at the end of the proof be $c-c_\alpha > 0$ and $c-c_\alpha \leq 0$ and not $c+c_\alpha > 0$ and $c+c_\alpha \leq 0$ as he states? $\endgroup$ – Daugmented Sep 29 '15 at 10:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.