Proving the Classification Theorem I
Our strategy is to determine which Coxeter graphs can arise from actual root systems, and then see what Dynkin diagrams we can get. Since Coxeter graphs ignore relative lengths of roots, we will start by just working with unit vectors whose pairwise angles are described by the Coxeter graph.
So, for the time being we make the following assumptions: is a real inner-product space (of arbitrary dimension), and is a set of linearly independent unit vectors satisfying and is one of , , , or for . For the duration of this proof, we will call such a set of vectors “admissible”. The elements of a base for a root system , divided by their lengths, are an admissible set.
We construct a Coxeter graph , just as before, for every admissible set, and these clearly include the Coxeter graphs that come from root systems. Our task is now to classify all the connected Coxeter graphs that come from admissible sets of vectors.
- If we remove some of the vectors from an admissible set, then the remaining vectors still form an admissible set. The Coxeter graph of the new set is obtained from that of the old one by removing the vertices corresponding to the removed vectors and all their incident edges.
- The number of pairs of vertices in connected by at least one edge is strictly less than — the number of overall vertices.
- The graph contains no cycles
This should be straightforward. The remaining vectors are still linearly independent, and they still have unit length. The angles between each pair are also unchanged. All we do is remove some vectors (vertices in the graph).
Let us define
Since the are linearly independent, cannot be zero. Thus we find that
Now, for any pair and joined by an edge, we must find that is , , or , and thus that . Clearly, we can have no more than of these edges if we are to satisfy the above inequality.
Indeed, if we have a cycle of length in the graph, we could (using our step 1) discard all the vertices not involved in the cycle and still have the graph of an admissible set. But this graph would have vertices and at least edges — the ones involved in the cycle. But this is is impossible by step 2!