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Please regard this A COMBINATORIAL CONSTRUCTION FOR SIMPLY–LACED LIE ALGEBRAS on page 7 (it is brief but I hope that page is enought introduction on that topic).

Can I argue, and if so how, that the coxeter group $W$ is finite if the set $R(X)$ is finite (i.e. first $\Longrightarrow$ in Proposition 3.2)? Somewhere I read that a Coxeter group does not neccessarily have to be finite even if the generator set is. Lemma 3.1 states that my $m_{xy}$ (see Coxeter group definition) are 2 or 3 depending on the relation between $x\neq y$ which I already checked and is true.

EDIT: Ok page with paper seems to be down. Updated link.

milkpirate
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  • If all the $m_{i,j}$ are equal to $2$, then all genertors commute and your group is abelian. Since all generators have order 2 and they are finitely many, your group is finite. – Mariano Suárez-Álvarez Mar 30 '16 at 14:03
  • On the other hand, you talk about a set $X$ but you give no indication of what this set is or what the action is, so as it stands your question does not make a lot of sense... – Mariano Suárez-Álvarez Mar 30 '16 at 14:04
  • I am sorry. I made a mistake while explaining my problem. Please have a look again at the original post. – milkpirate Mar 30 '16 at 14:33
  • When you make an error and edit the text, please actually edit the text, do not add «EDIT»> paragraphs! – Mariano Suárez-Álvarez Mar 30 '16 at 15:12
  • Ok did as you asked me to. – milkpirate Mar 30 '16 at 16:08
  • @Mariano Suárez-Alvarez could you have another look at that topic. i think no one else will do anymore bacuse its not new anymore. – milkpirate Apr 02 '16 at 15:14
  • The reason the implication is true is that this is the Weyl group of a root system, rather than an arbitrary Coxeter group. – Tobias Kildetoft Apr 13 '16 at 09:55
  • Sure it is related to the classical root system (https://en.wikipedia.org/wiki/Root_system) but till that point of the paper we "dont know" that. And to be honest I wouldnt know how to transfern the properties of the populations to the ones of the vectors in the classical definition. I was hoping that is possible by "pure" group theory. – milkpirate Apr 13 '16 at 10:09
  • But the graph it started with was one of the classical ones, so we do know this. Anyway, one could certainly also just check that the given ones are in the list of finite Coxeter groups (which are completely classified). – Tobias Kildetoft Apr 13 '16 at 10:17
  • I agree with you we start with classical graphs but we define a totally different way of reflecting things. And just saying these mutation reflections appear to be the same as the usual reflections and thats why everything works out nicely isnt a mathematical argument. We would have to show that $\sigma_x(y)$ is the same as... yeah as what? $(yp_{x_i})_j$ ? It is not that easy. At least its not for me. – milkpirate Apr 13 '16 at 10:34

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