Proving the Classification Theorem II

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

No more than three edges (counting multiplicities) can be incident on any given vertex of $\Gamma$ .

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

$\displaystyle\epsilon=\sum\limits_{i=0}^k\langle\epsilon,\eta_i\rangle\eta_i$

and thus

$\displaystyle\begin{aligned}1=\langle\epsilon,\epsilon\rangle&=\left\langle\sum\limits_{i=0}^k\langle\epsilon,\eta_i\rangle\eta_i,\sum\limits_{j=0}^k\langle\epsilon,\eta_j\rangle\eta_j\right\rangle\\&=\sum\limits_{i=0}^k\sum\limits_{j=0}^k\langle\epsilon,\eta_i\rangle\langle\epsilon,\eta_j\rangle\langle\eta_i,\eta_j\rangle\\&=\sum\limits_{i=0}^k\sum\limits_{j=0}^k\langle\epsilon,\eta_i\rangle\langle\epsilon,\eta_j\rangle\delta_{i,j}\\&=\sum\limits_{i=0}^k\langle\epsilon,\eta_i\rangle^2\end{aligned}$

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

$\displaystyle\sum\limits_{i=1}^k\langle\epsilon,\eta_i\rangle^2<1$

and thus

$\displaystyle\sum\limits_{i=1}^k4\langle\epsilon,\eta_i\rangle^2<4$

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 $\epsilon$ .

The only connected Coxeter graph $\Gamma$ containing a triple edge is $G_2$

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.

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February 23, 2010 - Posted by John Armstrong | Geometry, Root Systems

3 Comments »

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

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[...] Theorem IV We continue proving the classification theorem. The first three parts are here, here, and [...]

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[...] we conclude the proof of the classification theorem. The first four parts of the proof are here, here, here, and [...]

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