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.
- 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
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!