## 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.

- The only possible Coxeter graphs with a triple vertex are those of the forms , , , and .

From step 8 we have the labelled graph

Like we did in step 9, we define

These vectors , , and are mutually orthogonal, linearly independent vectors, and that is not in the subspace that they span.

We look back at our proof of step 4 to determine that ; where , , and are the angles between and , , and , respectively. We look back at our proof of step 9 to determine that , , and . Thus we can calculate the cosine

And similarly we find , and . Adding up, we find

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 , , or can be or we don’t have a triple vertex at all. We can also choose which strand is which so that

We can determine from here that

and so we must have , and the shortest leg must be one edge long. Now we have , and so , and must be either or .

If , then the second shortest leg is two edges long. In this case, and . The possibilities for the triple are , , and ; giving graphs , , and , respectively.

On the other hand, if , then the second shortest leg is also one edge long. In this case, there is no more restriction on , and so the remaining leg can be as long as we like. This gives us the family of graphs.

And we’re done! If we have one triple edge, we must have the graph . 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 and the families and , while step 10 narrows down graphs with a triple vertex to , , and , and the family . Finally, if there are no triple vertices or double edges, we’re left with a single simple chain of type .

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