# The Unapologetic Mathematician

## Proving the Classification Theorem I

This week, we will prove the classification theorem for root systems. The proof consist of a long series of steps, and we’ll split it up over a number of posts.

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: $V$ is a real inner-product space (of arbitrary dimension), and $\mathfrak{D}=\{\epsilon_1,\dots,\epsilon_n\}$ is a set of $n$ linearly independent unit vectors satisfying $\langle\epsilon_i,\epsilon_j\rangle\leq0$ and $4\langle\epsilon_i,\epsilon_j\rangle^2$ is one of $0$, $1$, $2$, or $3$ for $i\neq j$. For the duration of this proof, we will call such a set of vectors “admissible”. The elements of a base $\Delta$ for a root system $\Phi$, divided by their lengths, are an admissible set.

We construct a Coxeter graph $\Gamma$, 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.

1. 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.
2. 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).

3. The number of pairs of vertices in $\Gamma$ connected by at least one edge is strictly less than $n$ — the number of overall vertices.
4. Let us define

$\displaystyle\epsilon=\sum\limits_{i=1}^n\epsilon_i$

Since the $\epsilon_i$ are linearly independent, $\epsilon$ cannot be zero. Thus we find that

\displaystyle\begin{aligned}0<\langle\epsilon,\epsilon\rangle&=\left\langle\sum\limits_{i=1}^n\epsilon_i,\sum\limits_{j=1}^n\epsilon_j\right\rangle\\&=\sum\limits_{i=1}^n\sum\limits_{j=1}^n\langle\epsilon_i,\epsilon_j\rangle\\&=\sum\limits_{ij}\langle\epsilon_i,\epsilon_j\rangle\\&=n+2\sum\limits_{i

Now, for any pair $\epsilon_i$ and $\epsilon_j$ joined by an edge, we must find that $4\langle\epsilon_i,\epsilon_j\rangle^2$ is $1$, $2$, or $3$, and thus that $2\langle\epsilon_i,\epsilon_j\rangle\leq-1$. Clearly, we can have no more than $n-1$ of these edges if we are to satisfy the above inequality.

5. The graph $\Gamma$ contains no cycles
6. Indeed, if we have a cycle of length $n$ 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 $n$ vertices and at least $n$ edges — the ones involved in the cycle. But this is is impossible by step 2!