## 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 .
- The only connected Coxeter graph containing a triple edge is

Consider some vector , and let be the vectors whose vertices are connected to by at least one edge. Since by step 3 the graph cannot contain any cycles, we cannot connect any and , and so we have for .

Since the collection is linearly independent, there must be some unit vector in the span of these vectors which is perpendicular to each of the . This vector must satisfy . Then the collection is an orthonormal basis of this subspace, and we can thus write

and thus

Now since we can’t have , we must find

and thus

But is the number of edges (counting multiplicities) between and , and so this sum is the total number of edges incident on .

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.

[...] 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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