The Unapologetic Mathematician

Mathematics for the interested outsider

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.

  1. The only possible Coxeter graphs with a triple vertex are those of the forms D_n, E_6, E_7, and E_8.
  2. From step 8 we have the labelled graph

    Like we did in step 9, we define

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

    \displaystyle\eta=\sum\limits_{j=1}^{q-1}i\eta_j

    \displaystyle\zeta=\sum\limits_{k=1}^{r-1}i\zeta_k

    These vectors \epsilon, \eta, and \zeta are mutually orthogonal, linearly independent vectors, and that \psi is not in the subspace that they span.

    We look back at our proof of step 4 to determine that \cos(\theta_1)^2+\cos(\theta_2)^2+\cos(\theta_3)^3<0; where \theta_1, \theta_2, and \theta_3 are the angles between \psi and \epsilon, \eta, and \zeta, respectively. We look back at our proof of step 9 to determine that \langle\epsilon,\epsilon\rangle=\frac{p(p-1)}{2}, \langle\eta,\eta\rangle=\frac{q(q-1)}{2}, and \langle\zeta,\zeta\rangle=\frac{r(r-1)}{2}. Thus we can calculate the cosine

    \displaystyle\begin{aligned}\cos(\theta_1)^2&=\frac{\langle\epsilon,\psi\rangle^2}{\langle\epsilon,\epsilon\rangle\langle\psi,\psi\rangle}\\&=\frac{(p-1)^2\langle\epsilon_{p-1},\psi\rangle^2}{\langle\epsilon,\epsilon\rangle}\\&=\frac{(p-1)^2\frac{1}{4}}{\frac{p(p-1)}{2}}\\&=\frac{p-1}{2p}\\&=\frac{1}{2}\left(1-\frac{1}{p}\right)\end{aligned}

    And similarly we find \cos(\theta_2)^2=\frac{1}{2}(1-\frac{1}{q}), and \cos(\theta_3)^2=\frac{1}{2}(1-\frac{1}{r}). Adding up, we find

    \displaystyle\begin{aligned}\frac{1}{2}\left(3-\frac{1}{p}-\frac{1}{q}-\frac{1}{r}\right)&<1\\3-\frac{1}{p}-\frac{1}{q}-\frac{1}{r}&<2\\\frac{1}{p}+\frac{1}{q}+\frac{1}{r}&>1\end{aligned}

    This last inequality, by the way, is hugely important in many areas of mathematics, and it’s really interesting to find it cropping up here.

    Anyway, now none of p, q, or r can be 1 or we don’t have a triple vertex at all. We can also choose which strand is which so that

    \displaystyle\frac{1}{p}\leq\frac{1}{q}\leq\frac{1}{r}\leq\frac{1}{2}

    We can determine from here that

    \displaystyle\frac{3}{2}\geq\frac{3}{r}\geq\frac{1}{p}+\frac{1}{q}+\frac{1}{r}>1

    and so we must have r=2, and the shortest leg must be one edge long. Now we have \frac{1}{p}+\frac{1}{q}>\frac{1}{2}, and so \frac{2}{q}>\frac{1}{2}, and q must be either 2 or 3.

    If q=3, then the second shortest leg is two edges long. In this case, \frac{1}{p}>\frac{1}{6} and p<6. The possibilities for the triple (p,q,r) are (3,3,2), (4,3,2), and (5,3,2); giving graphs E_6, E_7, and E_8, respectively.

    On the other hand, if q=2, then the second shortest leg is also one edge long. In this case, there is no more restriction on p, and so the remaining leg can be as long as we like. This gives us the D_n family of graphs.

And we’re done! If we have one triple edge, we must have the graph G_2. If we have a double edge or a triple vertex, we can have only one, and we can’t have one of each. Step 9 narrows down graphs with a double edge to F_4 and the families B_n and C_n, while step 10 narrows down graphs with a triple vertex to E_6, E_7, and E_8, and the family D_n. Finally, if there are no triple vertices or double edges, we’re left with a single simple chain of type A_n.

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February 26, 2010 - Posted by | Geometry, Root Systems

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