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.

We continue with the proof of the classification theorem. The first two parts are here and here.

If $\{\epsilon_1,\dots,\epsilon_k\}\subseteq\mathfrak{D}$ is a simple chain in the graph $\Gamma$ , then we can remove these vectors and replace them by their sum. The resulting set is still admissible, and its graph is obtained from $\Gamma$ by contracting the chain to a single vertex.

A simple chain in $\Gamma$ is a subgraph like

We define

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

I say that if we replace $\{\epsilon_1,\dots,\epsilon_k\}$ by $\{\epsilon\}$ in $\mathfrak{D}$ then we still have an admissible set $\mathfrak{D}'$ . Any edges connecting to any of the vertices $\epsilon_i$ will then connect to the vertex $\epsilon$ .

It should be clear that $\mathfrak{D}'$ is still linearly independent. If there’s no way to write zero as a nontrivial linear combination of the vectors in $\mathfrak{D}$ , then of course there’s no way to do it with the additional condition that the coefficients of all the $\epsilon_i$ are the same.

Because we have a simple chain, we find $2\langle\epsilon_i,\epsilon_{i+1}\rangle=-1$ for $1\leq i\leq k-1$ and $\langle\epsilon_i,\epsilon_j\rangle=0$ otherwise. Thus we calculate (compare to step 2)

$\displaystyle\begin{aligned}\langle\epsilon,\epsilon\rangle&=k+\sum\limits_{i<j}2\langle\epsilon_i,\epsilon_j\rangle\\&=k-(k-1)=1\end{aligned}$

and so $\epsilon$ is a unit vector.

Any $\eta\in\Gamma$ can be connected by an edge to at most one of the $\epsilon_i$ , or else $\Gamma$ would contain a cycle (violating step 3). Thus we have either $\langle\eta,\epsilon\rangle=0$ if $\eta$ is connected to no vertex of the chain, or $\langle\eta,\epsilon\rangle=\langle\eta,\epsilon_i\rangle$ if $\eta$ is connected to $\epsilon_i$ by some number of edges. In either case we still find that $4\langle\eta,\epsilon\rangle^2$ is $0$ , $1$ , $2$ , or $3$ , and so $\mathfrak{D}'$ is admissible.

Further, the previous paragraph shows that the edges connecting $\epsilon$ to any other vertex $\eta$ of the graph $\Gamma$ are exactly the edges connecting the chain to $\eta$ . Thus we obtain the graph $\Gamma'$ of $\mathfrak{D}'$ by contracting the chain in $\Gamma$ to a single vertex.

The graph $\Gamma$ contains no subgraph of any of the following forms:

Indeed, if one of these graphs occurred in $\Gamma$ 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!

February 25, 2010 - Posted by John Armstrong | Geometry, Root Systems

4 Comments »

[...] Proving the Classification Theorem IV We continue proving the classification theorem. The first three parts are here, here, and here. [...]

Pingback by Proving the Classification Theorem IV « The Unapologetic Mathematician | February 25, 2010 | Reply
You have \epsilon minus instead of \epsilon = in your definition of the linear combination of a simple chain.

Comment by Robert | February 26, 2010 | Reply
Thanks, fixed.

Comment by John Armstrong | February 26, 2010 | Reply
[...] conclude the proof of the classification theorem. The first four parts of the proof are here, here, here, and [...]

Pingback by Proving the Classification Theorem V « The Unapologetic Mathematician | February 26, 2010 | Reply

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