The Unapologetic Mathematician

Mathematics for the interested outsider

Proving the Classification Theorem II

We continue with the proof of the classification theorem that we started yesterday.

  1. No more than three edges (counting multiplicities) can be incident on any given vertex of \Gamma.
  2. Consider some vector \epsilon\in\mathfrak{D}, and let \eta_1,\dots,\eta_k be the vectors whose vertices are connected to \epsilon by at least one edge. Since by step 3 the graph \Gamma cannot contain any cycles, we cannot connect any \eta_i and \eta_j, and so we have \langle\eta_i,\eta_j\rangle=0 for i\neq j.

    Since the collection \{\epsilon,\eta_1,\dots,\eta_k\} is linearly independent, there must be some unit vector \eta_0 in the span of these vectors which is perpendicular to each of the \eta_i. This vector must satisfy \langle\epsilon,\eta_0\rangle\neq0. Then the collection \{\eta_0,\eta_1,\dots,\eta_k\} is an orthonormal basis of this subspace, and we can thus write


    and thus


    Now since we can’t have \langle\epsilon,\eta_0\rangle=0, we must find


    and thus


    But 4\langle\epsilon,\eta_i\rangle^2 is the number of edges (counting multiplicities) between \epsilon and \eta_i, and so this sum is the total number of edges incident on \epsilon.

  3. The only connected Coxeter graph \Gamma containing a triple edge is G_2
  4. Indeed, step 4 tells us that no vertex can support more than three incident edges, counting multiplicities. Once we put the triple edge between two vertices, neither of them has any more room for another edge to connect to any other vertices. Thus the connected component cannot grow larger than this.


February 23, 2010 Posted by | Geometry, Root Systems | 3 Comments