# The Unapologetic Mathematician

## 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 $A_n$.

February 26, 2010