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

    \displaystyle\epsilon=\sum\limits_{i=1}^pi\epsilon_i

    \displaystyle\eta=\sum\limits_{j=1}^qj\eta_j

    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

    \displaystyle\begin{aligned}\langle\epsilon,\epsilon\rangle&=\sum\limits_{i=1}^pi^2\langle\epsilon_i,\epsilon_i\rangle+2\sum\limits_{i<j}ij\langle\epsilon_i,\epsilon_j\rangle\\&=\sum\limits_{i=1}^pi^2+\sum\limits_{i=1}^{p-1}i(i+1)2\langle\epsilon_i,\epsilon_{i+1}\rangle\\&=\sum\limits_{i=1}^pi^2-\sum\limits_{i=1}^{p-1}(i^2+i)\\&=p^2-\sum\limits_{i=1}^{p-1}i\\&=p^2-\frac{p(p-1)}{2}\\&=\frac{p(p+1)}{2}\end{aligned}

    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

    \displaystyle\langle\epsilon,\eta\rangle^2=p^2q^2\langle\epsilon_p,\eta_q\rangle^2=\frac{p^2q^2}{2}

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

    \frac{p^2q^2}{2}=\langle\epsilon,\eta\rangle^2<\langle\epsilon,\epsilon\rangle\langle\eta,\eta\rangle=\frac{p(p+1)q(q+1)}{4}

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

    \displaystyle\begin{aligned}\frac{p^2q^2}{2}&<\frac{p(p+1)q(q+1)}{4}\\2p^2q^2&<p(p+1)q(q+1)\\2pq&<(p+1)(q+1)\\2pq&<pq+p+q+1\\pq-p-q+1&<2\\(p-1)(q-1)&<2\end{aligned}

    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.

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February 25, 2010 - Posted by | Geometry, Root Systems

5 Comments »

  1. Whoowee!

    What program do you do your mathwriting in?

    Comment by westwood | February 25, 2010 | Reply

    • WordPress itself supports the \LaTeX writing. For these diagrams I’m using GeoGebra, while for big commutative diagrams I render them in TeXshop and cut/paste the images here.

      Comment by John Armstrong | February 25, 2010 | Reply

  2. Great work! I really enjoy these posts. You should consider joining up with Khan and making some videos of higher level maths.

    You duplicated one line of the equations underneath “As in step 6…and so we calculate”.

    Will you be going over how these tie in with polytopes?

    Comment by Robert | February 26, 2010 | Reply

  3. Thanks, fixed it.

    As it stands, I’m not planning on using the classification immediately. But it provides a nice break in the analysis work for a while.

    Comment by John Armstrong | February 26, 2010 | Reply

  4. [...] Proving the Classification Theorem V Today we conclude the proof of the classification theorem. The first four parts of the proof are here, here, here, and here. [...]

    Pingback by Proving the Classification Theorem V « The Unapologetic Mathematician | February 26, 2010 | Reply


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