The Unapologetic Mathematician

Mathematics for the interested outsider

Proving the Classification Theorem IV

We continue proving the classification theorem. The first three parts are here, here, and here.

  1. Any connected graph \Gamma takes one of the four following forms: a simple chain, the G_2 graph, three simple chains joined at a central vertex, or a chain with exactly one double edge.
  2. This step largely consolidates what we’ve done to this point. Here are the four possible graphs:

    The labels will help with later steps.

    Step 5 told us that there’s only one connected graph that contains a triple edge. Similarly, if we had more than one double edge or triple vertex, then we must be able to find two of them connected by a simple chain. But that will violate step 7, and so we can only have one of these features either.

  3. The only possible Coxeter graphs with a double edge are those underlying the Dynkin diagrams B_n, C_n, and F_4.
  4. Here we’ll use the labels on the above graph. We define



    As in step 6, we find that 2\langle\epsilon_i,\epsilon_{i+1}\rangle=-1=2\langle\eta_j,\eta_{j+1}\rangle and all other pairs of vectors are orthogonal. And so we calculate


    And similarly, \langle\eta,\eta\rangle=\frac{q(q+1)}{2}. We also know that 4\langle\epsilon_p,\eta_q\rangle^2=2, and so we find


    Now we can use the Cauchy-Schwarz inequality to conclude that


    where the inequality is strict, since \epsilon and \eta are linearly independent. And so we find


    We thus must have either p=q=2, which gives us the F_4 diagram, or p=1 or q=1 with the other arbitary, which give rise the the B_n and C_n Coxeter graphs.


February 25, 2010 Posted by | Geometry, Root Systems | 5 Comments

Proving the Classification Theorem III

Sorry, but what with errands today and work on some long-term plans, I forgot to put this post up this afternoon.

We continue with the proof of the classification theorem. The first two parts are here and here.

  1. If \{\epsilon_1,\dots,\epsilon_k\}\subseteq\mathfrak{D} is a simple chain in the graph \Gamma, then we can remove these vectors and replace them by their sum. The resulting set is still admissible, and its graph is obtained from \Gamma by contracting the chain to a single vertex.
  2. A simple chain in \Gamma is a subgraph like

    We define


    I say that if we replace \{\epsilon_1,\dots,\epsilon_k\} by \{\epsilon\} in \mathfrak{D} then we still have an admissible set \mathfrak{D}'. Any edges connecting to any of the vertices \epsilon_i will then connect to the vertex \epsilon.

    It should be clear that \mathfrak{D}' is still linearly independent. If there’s no way to write zero as a nontrivial linear combination of the vectors in \mathfrak{D}, then of course there’s no way to do it with the additional condition that the coefficients of all the \epsilon_i are the same.

    Because we have a simple chain, we find 2\langle\epsilon_i,\epsilon_{i+1}\rangle=-1 for 1\leq i\leq k-1 and \langle\epsilon_i,\epsilon_j\rangle=0 otherwise. Thus we calculate (compare to step 2)


    and so \epsilon is a unit vector.

    Any \eta\in\Gamma can be connected by an edge to at most one of the \epsilon_i, or else \Gamma would contain a cycle (violating step 3). Thus we have either \langle\eta,\epsilon\rangle=0 if \eta is connected to no vertex of the chain, or \langle\eta,\epsilon\rangle=\langle\eta,\epsilon_i\rangle if \eta is connected to \epsilon_i by some number of edges. In either case we still find that 4\langle\eta,\epsilon\rangle^2 is 0, 1, 2, or 3, and so \mathfrak{D}' is admissible.

    Further, the previous paragraph shows that the edges connecting \epsilon to any other vertex \eta of the graph \Gamma are exactly the edges connecting the chain to \eta. Thus we obtain the graph \Gamma' of \mathfrak{D}' by contracting the chain in \Gamma to a single vertex.

  3. The graph \Gamma contains no subgraph of any of the following forms:
  4. Indeed, if one of these graphs occurred in \Gamma it would be (by step 1) the graph of an admissible set itself. However, if this were the case we could contract the central chain to a single vertex by step 6. We would then find one of the graphs

    But each of these graphs has four edges incident on the central vertex, which violates step 4!

February 25, 2010 Posted by | Geometry, Root Systems | 4 Comments