Proving the Classification Theorem IV
- Any connected graph takes one of the four following forms: a simple chain, the graph, three simple chains joined at a central vertex, or a chain with exactly one double edge.
- The only possible Coxeter graphs with a double edge are those underlying the Dynkin diagrams , , and .
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.
Here we’ll use the labels on the above graph. We define
As in step 6, we find that and all other pairs of vectors are orthogonal. And so we calculate
And similarly, . We also know that , and so we find
Now we can use the Cauchy-Schwarz inequality to conclude that
where the inequality is strict, since and are linearly independent. And so we find
We thus must have either , which gives us the diagram, or or with the other arbitary, which give rise the the and Coxeter graphs.