## Proving the Classification Theorem IV

We continue proving the classification theorem. The first three parts are here, here, and here.

- 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.

## Proving the Classification Theorem III

Sorry, but what with errands today and work on some long-term plans, I forgot to put this post up this afternoon.

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

- If is a simple chain in the graph , then we can remove these vectors and replace them by their sum. The resulting set is still admissible, and its graph is obtained from by contracting the chain to a single vertex.
- The graph contains no subgraph of any of the following forms:

A simple chain in is a subgraph like

We define

I say that if we replace by in then we still have an admissible set . Any edges connecting to any of the vertices will then connect to the vertex .

It should be clear that is still linearly independent. If there’s no way to write zero as a nontrivial linear combination of the vectors in , then of course there’s no way to do it with the additional condition that the coefficients of all the are the same.

Because we have a simple chain, we find for and otherwise. Thus we calculate (compare to step 2)

and so is a unit vector.

Any can be connected by an edge to at most one of the , or else would contain a cycle (violating step 3). Thus we have either if is connected to no vertex of the chain, or if is connected to by some number of edges. In either case we still find that is , , , or , and so is admissible.

Further, the previous paragraph shows that the edges connecting to any other vertex of the graph are exactly the edges connecting the chain to . Thus we obtain the graph of by contracting the chain in to a single vertex.

Indeed, if one of these graphs occurred in it would be (by step 1) the graph of an admissible set itself. However, if this were the case we could contract the central chain to a single vertex by step 6. We would then find one of the graphs

But each of these graphs has four edges incident on the central vertex, which violates step 4!