# Friday October 11

Recall from last time the Dynkin diagrams.
If $\Phi$ is irreducible, then its diagram is one of the following:

![Image](figures/2019-10-11-09:14.png)\

**Definition:**
A subset $A = \theset{v_1, \cdots, v_n} \subseteq E$ is *admissible* iff

1. $A$ is linearly independent.
2. $\inner{v_i}{v_i} = 1$ for all $i$, and $\inner{v_i}{v_j} \leq 0$ if $i\neq j$.
3. $s_{ij} = 4\inner{v_i}{v_j}^2 \in \theset{0,1,2,3}$ if $i\neq j$.

Define a graph $\Gamma_A = (V_A, E_A)$ where $V_A = A$ and $E_A= \theset{s_{ij} \suchthat i\neq j}$.
If $\Pi = \theset{\alpha_1, \cdots, \alpha_\ell} \subseteq \Phi$ is a base, then $A \coloneqq \theset{v_i = \frac{\alpha_i}{\sqrt{\inner{\alpha_i}{\alpha_i}}}}$.

**Lemma:**

a. If $A$ is admissible, then $\# \theset{(v_i, v_j) \in E_A \suchthat 4\inner{v_i}{v_j}^2 \neq 0} \leq \abs{A} - 1$, and $\Gamma_A$ contains no graph cycles.
b. $\deg V_i \leq 3$ for all $i$.
c. If $\Gamma_A$ contains a path $p_1 \to \cdots \to p_t$, then $A' \coloneqq \theset{p} \union A\setminus \theset{p_1, \cdots, p_t}$ where $p \coloneqq \sum p_i$. Moreover, $\Gamma_{A'}$ is obtained from $\Gamma_A$ by contracting this path onto $p$.

*Proof of theorem:*

Assume the lemma holds.
Let $\Gamma$ be the Coxeter diagram of $\Phi$; then $\Phi$ is connected.

**Case 1:** 
$\Gamma$ has a triple edge.
But then both vertices on this edge have degree 3, so this is the maximal number of edges between them.
But since $\Gamma$ must be connected, this is everything.

**Case 2:** 
$\Gamma$ has no triple edges but some double edge.
We will first show that $\Gamma$ has only one double edge.

Suppose otherwise; then $\Gamma$ has at least two double edges occurring. 
Without loss of generality (e.g. by taking a subgraph), these are connected by a path of single edges.
By the lemma, we can contract this path to get an admissible subset.
But then there is a vertex of degree 4, contracting $\deg V_i \leq 3$ for all $i$.

Now we'll show that $\Gamma$ has no branching point, i.e. a vertex of degree exactly 3.
If this occurs, then a double edge is connected to such a vertex by a path.
Contracting this path yields a vertex of degree 4, again a contradiction.

By these two statements, $\Gamma$ has the general form:

\begin{align*}
\Gamma = v_1 \to \circ \to \cdots \to v_p \to\to w_q \to \circ \to \cdots \to w_1
.\end{align*}

Let $v = \sum i v_i$ and $w = \sum i w_i$, then $\inner{v}{v} = \frac{1}{2} p(p+1)$, and $\inner{w}{w} = \frac 1 2 q(q+1)$.
Note that $\inner{v_i}{w_j} = -1/\sqrt 2$ if $i=p$ and $j=q$, and 0 otherwise.

Thus $\inner{v}{w} = \cdots = \frac 1 2 p^2q^2$.
By Cauchy-Schwarz, this is strictly less than $\inner{v}{v} \inner{w}{w} = \frac 1 4 p(p+1)q(q+1)$.
We then obtain $(p-1)(q-1) < 2$.
Supposing wlog that $p \geq q$, we have either $p=q=2$, in which case we get $\circ \to \circ \to\to \circ \to \circ$.
Otherwise $q=1$, and we get $\circ \to \cdots \to \circ \to\to \circ$.

**Case 3:** 
$\Gamma$ has only single edges.
We want to show $\Gamma$ has only one branching point, i.e. a vertex of degree 3.
If it has 2, we can contract the intermediate path to get a vertex of degree 4.
So we have the following situation:

![Image](figures/2019-10-11-09:41.png)\

Define $x = \sum i x_i, y = \sum i y_i, w = \sum i w_i$, and $\hat w, \hat x, \hat y$ to be their normalization.
Then $B = \theset{b_i} \coloneqq \theset{\hat w, \hat y, \hat y, z}$ is orthonormal and linearly independent, so we can apply Gram-Schmidt.
This yields a $z' \neq 0$ such that 
$$
z = \sum \inner{z}{\hat b_i} \hat b_i
$$

In particular, $\inner{z}{z'}z' \neq 0 z'$, otherwise $z$ is a linear combination of the $x_i, y_i, w_i$.
Thus $\inner{z}{\hat w}^2 + \inner{z}{\hat x}^2 + \inner{z}{\hat y}^2 > 1$.
We can compute $\inner{z}{\hat w} = \frac{-q/2}{\sqrt{\frac 1 2 q(q+1)}}$, and so $\inner{z}{\hat w}^2 = \frac q {2(q+1)}$.

From this, we can obtain $\frac 1 {q+1} + \frac 1 {r+1} + \frac 1 {p+1} > 1$.
We can assume $p \geq q \geq r \geq 1$, since these correspond to the lengths of paths in the above image.
This allows us to do some case-by-case analysis.

Using this, we find $\frac 3 {r+1} > 1$, and so $r=1$ must hold.
Similarly, $\frac 2 {q+1} > \frac 1 2$, which forces $q = \in \theset{1, 2}$.

Supposing $r=q=1$, then we get type $D_\ell$ because $p$ can be anything.
Supposing otherwise that $r=1, q=2, p \in \theset{2,3,4}$, we get type $E$.