# Monday, October 03

## Integrality and Rationality Properties

:::{.claim}
Setup:

- Decompose $L = H \oplus \bigoplus _{ \alpha\in \Phi} L_{ \alpha}$
- Use the isomorphism 
\[
H &\iso H\dual \\
\varphi &\mapsfrom t_{ \varphi}
\]
  to define $(\lambda, \mu) \da \kappa(t_ \lambda, t_ \mu)$ on $H$.

- Choose a basis $\ts{ \alpha_i} \subseteq \Phi \subseteq H\dual$
- For any $\beta \in \Phi$, write \( \beta= \sum c_i \alpha_i \) with $c_i\in \CC$.
Then

\[
c_i\in \QQ
.\]
:::

:::{.proof title="?"}
Write \( ( \beta, \alpha_j) = \sum c_i (\alpha_i, \alpha_j) \) and m
\[
{2 (\beta, \alpha_j) \over (\alpha_j, \alpha_j) }
= \sum c_i {2 (\alpha_i, \alpha_j) \over (\alpha_j, \alpha_j) }
,\]
where the LHS is in $\ZZ$, as is $2( \alpha_i, \alpha_j) \over (\alpha_j, \alpha_j)$.
On the other hand
\[
{2 (\beta, \alpha_j) \over (\alpha_j, \alpha_j) }
= {2 (t_ \beta, t_{\alpha_j} ) \over \kappa(t_{ \alpha_j}, \kappa_{ \alpha_j} ) }
= \kappa(t_ \beta, h_{\alpha_j} )
= \beta(h_{ \alpha_j})
\]
using that \( ( \alpha_j, \alpha_j) = \kappa( t_{ \alpha_j}, t_{ \alpha_j} )\neq 0 \) from before.[^more_gen]
Since $\ts{ \alpha_i}$ is a basis for $H\dual$ and $(\wait, \wait)$ is nondegenerate, the matrix $[ ( \alpha_i, \alpha _j) ]_{1\leq i, j\leq n}$ is invertible.
Thus so is $\left[ 2 ( \alpha_i, \alpha_j) \over (\alpha_j, \alpha_j ) \right]_{1\leq i,j \leq n}$, since it's given by multiplying each column as a nonzero scalar, and one can solve for the $c_i$ by inverting it.
This involves denominators coming from the determinant, which is an integer, yielding entries in $\QQ$.

[^more_gen]: 
More generally,
\[
{2 (\lambda, \alpha)\over (\alpha, \alpha) } = \lambda(h_ \alpha) \qquad\forall \alpha\in \Phi
.\]

:::

:::{.remark}
Given \( \lambda, \mu \in H\dual \) then 
\[
(\lambda, \mu) = \kappa(t_ \lambda, t_\mu) = \Trace(\ad_{t_ \lambda} \circ \ad_{t_\mu} ) = \sum_{ \alpha\in \Phi} \alpha(t_ \lambda) \cdot \alpha(t_\mu)
,\]
using that both ads are diagonal in this basis, so their product is given by the products of their diagonal entries.
One can write this as $\sum_{ \alpha\in \Phi} \kappa(t_ \alpha, t_ \lambda) \kappa(t_ \alpha, t_\mu)$, so we get a formula
\[
( \lambda, \mu ) = \sum_{ \alpha\in \Phi} ( \alpha, \lambda) (\alpha, \mu), \qquad (\lambda, \lambda) = \sum_{ \alpha\in \Phi} (\alpha, \lambda)^2
.\]
Setting $\lambda = \beta$ and dividing by $(\beta, \beta)^2$ yields
\[
{1\over (\beta, \beta)} = \sum_{ \alpha\in \Phi} {(\alpha, \beta)^2 \over (\beta, \beta)^2} \in {1\over 4}\ZZ
,\]
since $(\alpha, \beta)\over (\beta, \beta)\in {1\over 2} \ZZ$.
So $(\beta, \beta)\in \QQ$ and thus $(\alpha, \beta)\in \QQ$ for all \( \alpha, \beta\in \Phi \).
It follows that the pairings \( (\lambda, \mu) \)  on the $\QQ\dash$subspace $\EE_\QQ$ of $H\dual$ spanned by $\ts{ \alpha_i}$ are all rational.
:::

:::{.claim}
$(\wait, \wait)$ on $\EE_\QQ$ is still nodegenerate
:::

:::{.proof title="?"}
If \( \lambda\in \EE_\QQ, ( \lambda, \mu) =0 \forall \mu\in \EE_\QQ \), then $( \lambda, \alpha_i) = 0 \forall i \implies (\lambda, \nu) = 0 \forall \nu\in H\dual \implies \lambda= 0$.
:::

:::{.remark}
Similarly, $(\lambda, \lambda) = \sum_{ \alpha\in \Phi \subseteq \EE_\QQ} ( \alpha, \lambda)^2$ is a sum of squares of rational numbers, and is thus non-negative.
Since $( \lambda, \lambda) = 0 \iff \lambda= 0$, the form on $\EE_\QQ$ is positive definite.
Write $\EE \da \EE_\QQ \tensor_\QQ \RR = \RR\ts{\alpha_i}$, then $(\wait, \wait)$ extends in the obvious way to an $\RR\dash$values positive definite bilinear form on $\EE$, making it a real inner product space.
:::

:::{.theorem title="?"}
Let $L, H, \Phi, \EE\slice\RR$ be as above, then

a. $\Phi$ is a finite set which spans $\EE$ and does not contain zero.
b. If $\alpha\in \Phi$ then $-\alpha\in \Phi$ and thus is the only other scalar multiple in $\Phi$.
c. If \( \alpha, \beta \in \Phi \) then
\[
\beta - \beta(h_ \alpha) \alpha = \beta - {2 (\beta, \alpha) \over ( \alpha, \alpha) } \alpha\in \Phi
,\]
which only depends on $\EE$.
Note that this swaps $\pm \alpha$.
d. If \( \alpha, \beta \in \Phi \) then $\beta(h_\alpha) = {2(\beta, \alpha) \over (\alpha, \alpha)}\in \ZZ$.

Thus $\Phi$ satisfies the axioms of a **root system** in $\EE$.
:::

:::{.example title="?"}
Recall that for $\liesl_3(\CC)$, $\kappa(x,y) = 6 \Trace(xy)$.
Taking the standard basis $\ts{v_i} \da \ts{x_i, h_i, y_i \da x_i^t}$, the matrix $\Trace(v_i v_j)$ is of the form
\[
\mattt 0 0 I 0 A 0 I 0 0\qquad A \da \matt 2 {-1} {-1} 2
.\]
This is far from the matrix of an inner product, but the middle block corresponds to the form restricted to $H$, which is positive definite.
One can quickly check this is positive definite by checking positivity of the upper-left $k\times k$ minors, which here yields $\det(2) = 2, \det A = 4-1 = 3$.
:::

## Part III: Root Systems

## Ch. 9, Axiomatics. 9.1: Reflections in a Euclidean Space

:::{.remark}
Let $\EE$ be a fixed real finite-dimensional Euclidean space with inner product $(\alpha, \beta)$, we consider property (c) from the previous theorem:
\[
\beta - {2( \beta, \alpha) \over (\alpha, \alpha)} \in \Phi \qquad\forall \alpha, \beta\in \Phi
.\]
:::


:::{.definition title="Reflections"}
A **reflection** in $\EE$ is an invertible linear map on an $n\dash$dimensional Euclidean space that fixes pointwise a hyperplane $P$ (of dimension $n-1$) and sending any vector $v\perp P$ to $-v$:

\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2022/Fall/LieAlgebras/sections/figures}{2022-10-03_09-52.pdf_tex} };
\end{tikzpicture}

:::


:::{.remark}
If $\sigma$ is a reflection sending \( \alpha\mapsto - \alpha \), then
\[
\sigma_\alpha(\beta) = \beta - {2( \beta, \alpha) \over (\alpha, \alpha)} \alpha \qquad \forall \beta\in \EE
.\]
One can check that $\sigma_\alpha^2 = \id$.
Some notes on notation:

- Humphreys writes $\inp{ \beta}{ \alpha} \da {2 ( \beta, \alpha) \over (\alpha, \alpha)}$
  This is linear in $\beta$ but not in $\alpha$!
- More modern: $(\beta, \alpha\dual) \da \inp{ \beta}{\alpha}$ where \( \alpha\dual \da {2\alpha\over (\alpha, \alpha)} \) corresponds to $h_\alpha$. 
- Modern notation for the map: $s_\alpha$ instead of $\sigma_\alpha$.

:::