Proving the Classification Theorem V
- 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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