0
$\begingroup$

Can someone please provide me with a reference for the fact that the Weyl group $W$ associated to a root system $\Phi$ can be realised as a Coxeter group?

This means that a Weyl group $W$ has a presentation of the form $\langle s_i \mid s_i^2, (s_is_j)^{n_{ij}}\rangle.$

This is normally done by choosing a set of positive roots $\Phi^+$ and a set of simple roots $\Sigma=\lbrace \alpha_i \rbrace \subset \Phi^+$ and mapping the generators of the free group $\langle s_i\rangle$ to the $s_{\alpha_i}$, the reflections across the hyperplane perpendicular to each $\alpha_i$. If we choose the $n_{ij}$ to be the order of $s_{\alpha_i}s_{\alpha_j}$ then clearly this map factors to the quotient.

The difficulty is in showing that the map is injective. If we happen to have $s_{i_1}\dots s_{i_n}$ and $s_{j_i}\dots s_{j_m}$ mapping to the same element $w$, we must show that the former elements are the same in the quotient of the free group. I have a vague understanding that this can be done by looking at the two induced 'galleries' connecting the fundamental domain (Weyl chamber)) $C$ with $wC$.

I hope someone can provide a reference with all the details of the argument in the preceeding paragraph (preferably not Bump's Lie Groups 2nd Ed.).

$\endgroup$
2
$\begingroup$

A proof can be found in Section 1.9 of James E. Humphreys’ Reflection groups and Coxeter groups. However, this proof does not use Weyl chambers but instead argues that any relation $s_{\alpha_{i_1}} \dotsm s_{\alpha_{i_n}} = 1$ in $W$ can be deduced from the given relations $(s_{\alpha_i} s_{\alpha_j})^{n_{ij}} = 1$.

(Disclaimer: I’ve never read through the proof myself, so I can’t say how well this works.)


I think that another source is Bourbaki’s Lie Groups and Lie Algebras, Chapter VI, §1, no. 5, Theorem 2, part (vii) (page 166 in my edition). I seems that (in true Bourbaki fashion) one has to follow the rabbit hole given references to the earlier chapter to assemble the complete proof. But at least a quick glance at the first reference (Chapter V, §3, no. 2, Theorem 1) suggests that this proof uses Weyl chambers.

$\endgroup$
  • $\begingroup$ Thanks! Will take a look at these proofs. $\endgroup$ – BetaTumSeNaHoPaega Jun 15 at 15:30

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.