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The Coxeter–Dynkin diagrams tell us that in a spherical Coxeter simplex most of the dihedral angles are right. Say among $\tfrac{n{\cdot}(n+1)}2$ dihedral angles we can have at most $n$ angles which are not right.

Is it possible to see this statement without classification?

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  • $\begingroup$ Anton: Did you read Vinberg's paper on hyperbolic Coxeter groups? He useda probabilistic argument of Stanley and the key was that in signature (n,1) most angles are right. You might've able to use the same argument in the finite case. $\endgroup$
    – Misha
    Commented Aug 16, 2013 at 22:39
  • $\begingroup$ @Misha, yes I did. He use this fact in the proof and that is one of the reasons I asked the question. $\endgroup$
    – Anton Petrunin
    Commented Aug 19, 2013 at 1:01
  • $\begingroup$ Maybe sort of a moral explanation could be something like there cannot be too many pairwise noncommuting involutions in a Coxeter group of relevant type... $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Aug 27, 2014 at 7:51
  • $\begingroup$ @Anton: For the "spherical" case it's simplest here to think just about finite Coxeter groups (crystallographic or not). Your reference to Dynkin points in other directions such as root systems in Lie algebras. Both classifications (Coxeter graphs, Dynkin diagrams) are similar in the finite case, but not identical. $\endgroup$
    – Jim Humphreys
    Commented Aug 27, 2014 at 13:08

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I'd be very surprised if one can do better than just reading part of the proof of the classification, which is a beautiful combinatorial argument already, in my opinion.

For example, in the proof of the classification of irreducible root systems given in Fulton and Harris (of Theorem 21.11), part (iv) says that you can collapse a long string of nodes in the diagram if the endpoints are connected to any other nodes. This will inductively force diagrams with more than one set of multiple lines (i.e. non-right angles) to be quite small.

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    $\begingroup$ Probably you right, but I wanted to be surprised :) $\endgroup$
    – Anton Petrunin
    Commented Aug 15, 2013 at 20:22
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Let the root system be $v_1$, …, $v_n$ with all elements normalized to be length $1$. So $\langle v_i, v_i \rangle =1$, we have $\langle v_i, v_j \rangle \leq - \cos (\pi/3) = -1/2$ for at least $n$ pairs $(i,j)$, and we have $\langle v_i, v_j \rangle \leq 0$ for all $i \neq j$.

But then $$\left\langle \sum_{i=1}^n v_i, \sum_{i=1}^n v_i \right\rangle \leq n + 2 n (-1/2) =0,$$ a contradiction.

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  • $\begingroup$ +1, I think this is a standard way to show that spherical Coxeter diagrams have no loops. $\endgroup$
    – S. Carnahan ♦
    Commented Aug 27, 2014 at 0:09
  • $\begingroup$ Yes, I also think this is standard. (Although I seem to remember arguing that a graph with cycles has a chordless cycle when I saw this proof presented, and then looking just at the roots in the cycle, and that is evidently not necessary.) $\endgroup$
    – David E Speyer
    Commented Aug 27, 2014 at 0:11
  • $\begingroup$ Terminological aside: In the graph-theoretic literature, a "chordless cycle" is also known as an "induced cycle". The latter is perhaps more common, at least more common than one might initially think. $\endgroup$
    – François G. Dorais
    Commented Aug 27, 2014 at 2:40

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