# Monday October 14

Last time:

Theorem:
If $\Phi$ is irreducible, then the Dynkin diagram is given by $A-G$.

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

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

Thus the graph $\Gamma_A = (V_A, E_A)$ is given by $V_A = A$ and $E_A = \theset{v_i \mapsvia{4\inner{v_i}{v_j}^2} v_j \mid i\neq j}$.


Lemma:

a. If $A$ is admissible, then the number of edges such that $4\inner{v_i}{v_j} \neq 0$ is at most $\abs A - 1$.
b. For every $i$, we have $\deg v_i \leq 3$.
c. If $\Gamma_A$ contains a straight path of length $t$, then the graph $\Gamma'$ obtained by contracting this path is also admissible.

Let $p$ be the point obtained by contracting such a path.

<!--![Image](figures/2019-10-14-09:15.png)-->

Proof of (a):
If $\theset{p_1, \cdots, p_t}$ are linearly independent, then $p\neq 0$.
Thus by positive-definiteness, we have
$0 <_{pd} \inner{p}{p} =_{\# 2} t + \sum_{i < j} 2\inner{p_i}{p_j}$.
Then $t > \sum_{i < j} (-2) \inner{p_i}{p_j} = \sum_{i < j} \sqrt{4\inner{p_i}{p_j}^2}$,
where the quantity in the square root is the number of edges, which is thus greater than or equal to the number of pairs connected.

Proof of (b):
Fix $i$.
Let $u_1 \cdots u_k$ be the vertices in $A$ that are connected to $v_i$ by a single edge.
Then by (a), we have $\inner{u_i}{u_j} = 0$ for all $i\neq j$.

Then the set $\theset{u_1, \cdots, u_k}$ is an orthonormal basis for their span.
Applying Gram-Schmidt, we can write each $v_i = \sum_{j=0}^k \inner{v_i}{u_j} u_j$, where we pick $u_0$ such that the new set $\theset{u_0} \union \theset{u_1, \cdots, u_k}$.
Then $\inner{v_i}{u_0} \neq 0$ for all $i$; otherwise we would have $\theset{u_1, \cdots, u_k, v_i}$ would be linearly dependent, since $v_i = \sum c_i u_i$ from above, which contradicts our initial axiom/assumption.
Then $1 = \inner{v_i}{v_i}$ by A2, which equals $\sum_{j=0}^k \inner{v_i}{u_j}^2 = \inner{v_i}{u_0}^2 + \sum_{j=1}^k \inner{v_i}{u_j}^2$, where the first term is strictly positive.

But then $1 > \sum_{j=1}^k \inner{v_i}{v_j}^2 \geq \frac k 4$ by A3, which then forces $k = \deg v_i \leq 3$.

Proof of (c):
The conditions of A1 are satisfied.
For A2, we have
$$
\inner{p_i}{p_j} = \begin{cases} -\frac 1 2 & \abs{i-j} = 1 \\ 0 & \abs{i-j} > 1 \\ 1 & i=j.\end{cases}
$$

We then have $\inner{p}{p} = t + 2 \sum{i<j} \inner{p_i}{p_j} = t + 2\sum_{i=1}^{t-1} \inner {p_i} {p_{i+1}} = 1$.
Thus $\inner {p} {v_i} = \sum_{j=1}^t \inner {p_j} {v_i} \leq 0$.

For A3, fix $v_i \in A'$.
Then $v_i$ is connected (by a single edge) to at most one point $p_j$, otherwise there would be a cycle.
Thus 
$$
\inner {v_i} {p} = 
\begin{cases} 
\inner {v_i} {p_j} & \text{if $v_i$ is connected to $p_j$} \\ 
0 & \text{else}.
\end{cases}
$$

We thus have $4{\inner {v_i} {p}}^2 = 4{\inner {v_i} {p_j}} \in \theset{0,1,2,3} \mathrm{1}{v_i \sim p_j}$.

## Construction of Root Systems and Automorphisms

We'll start with the construction of types $A-G$.

**Theorem:**
For Dynkin diagrams of type $A-G$, there exists an irreducible root system having the given diagram.

Proof:
By explicit construction.
Fix an orthonormal basis $\theset{\varepsilon_i}$.

**Type $A_\ell$:**
Let
$$
\Phi = \theset{\varepsilon_i - \varepsilon_j \mid 1 \leq i\neq j \leq \ell + 1}
$$

Then $\abs \Phi = \ell^2 + \ell$, and $\Pi = \theset{\alpha_i = \varepsilon_i - \varepsilon_{i+1} \mid 1 \leq i \leq \ell}$.
We then find that $\dim \lieg = \ell^2 + 2\ell$.

> Note that we don't know anything about $\lieg$ yet, but already know its dimension.

Example: $A_2$.
We have $\Pi = \theset{\alpha_1 = \varepsilon_1 - \varepsilon_2, \alpha_2 = \varepsilon_3 - \varepsilon_2}$.
Then $A = (a_{ij})$ with $a_{ij} = \inner {a_i} {a_j\dual}$,
and $\alpha_1\dual = \frac {2\alpha_1}{\inner {\alpha_1} {\alpha_1}} = \frac{2 (\varepsilon_1 - \varepsilon_2)}{\inner{\varepsilon_1 -\varepsilon_2}{\varepsilon_1 - \varepsilon_2}} = \varepsilon_1 - \varepsilon_2 = \alpha_1$.
Doing the computations, it turns out that $\inner {\alpha_1} {\alpha_2\dual} = -1$, $\inner{\alpha_2} {\alpha_1\dual} = -1$, and $\inner {\alpha_i} {\alpha_i\dual} = 2$.

Thus $A = [2, -1; -1, 2]$, which has Dynkin diagram given by:

<!--![Image](figures/2019-10-14-09:44.png)-->

**Type $B_\ell$:**
Recall that these have one "short root":

<!--![Image](figures/2019-10-14-09:45.png)-->

Then $\Phi = \theset{\pm \varepsilon_j, \pm \varepsilon_j \mid 1 \leq i\neq j \leq \ell} \union \theset{\pm \varepsilon_i \mid 1 \leq i \leq \ell}$,
and we have $\Pi = \theset{\alpha_i = \varepsilon_i - \varepsilon_{i-1} \mid 1 \leq i \leq \ell-1} \union \theset{\alpha_\ell \coloneqq \varepsilon_\ell}$.

After carrying out the computation, we have the following Cartan matrix:

<!--![Image](figures/2019-10-14-09:49.png)-->

And $\dim \lieg = 2\ell^2 + \ell$, since $\abs \Phi = 2\ell(\ell-1) + 2\ell = 2\ell^2$.

**Type $D_\ell$:**

<!--![Image](figures/2019-10-14-09:51.png)-->

We obtain $\Phi = \theset{\alpha_i = \varepsilon_i - \varepsilon_{i+1} \mid 1 \leq i \leq \ell-1} \union \theset{\alpha_\ell \coloneqq \varepsilon_{\ell-1} + \varepsilon_\ell}$.
We then find $\inner {\alpha_{\ell-1}} {\alpha_\ell\dual} = 0$ and $\inner {\alpha_{\ell-2}} {\alpha_\ell\dual} = -1$.

**Type $E_\ell$**:
We have $\Pi(E_\ell) = \Pi(D_{\ell - 1}) \union \theset{\alpha_\ell \coloneqq - \frac 1 2 \sum_{i=1}^8 \varepsilon_i}$.

This yields $\abs \Phi = 72, 126, 240$ and $\dim \lieg = 78, 133, 248$, corresponding to $\ell = 6,7,8$.

More results on exceptional Lie Algebras:

<!--![Image](figures/2019-10-14-09:59.png)-->