# Friday, September 30

## 8.4

:::{.proposition title="A"}
$\alpha\in \Phi\implies \dim L_{\pm \alpha} = 1$, and $\alpha, \lambda \alpha\in \Phi\implies \lambda = \pm 1$.
:::

:::{.proof title="of proposition A"}
Consider $M \da \bigoplus \bigoplus _{ c\in \CC} L_{c \alpha} \leq L\in\mods{\liesl( \alpha)}$.
Write \( \liesl(\alpha) = \gens{ x_ \alpha, h_ \alpha, y_ \alpha} \), we decomposed $M = K \oplus \liesl(\alpha) \oplus W$ where $\ker \alpha \leq H$ and $W \intersect H = 0$.
WTS: $W = 0$. So far, we've shown that if $L(s) \subseteq W$ for $s\in \ZZ_{\geq 0}$ (which guarantees finite dimensionality), then $s$ can't be even -- otherwise it has a weight zero eigenvector, forcing it to be in $H$, but $W \intersect H = 0$.

Aside: $\alpha\in \Phi \implies 2\alpha\not\in \Phi$, since it would have weight $(2\alpha)(h_ \alpha) = 2\alpha h_ \alpha = 4$, but weights in irreducible modules have the same parity as the highest weight and no such weights exist in $M$ (only $0, \pm 2$ in $K \oplus \liesl_2(\alpha)$ and only odd in $W$).
Suppose $L(s) \subseteq W$ and $s\geq 1$ is odd. Then $L(s)$ has a weight vector for $h_ \alpha$ of weight 1.
This must come from $c=1/2$, since $(1/2) \alpha (h_ \alpha) = (1/2) 2 = 1$, so this is in $L_{\alpha/2}$.
However, by the aside, if $\alpha\in \Phi$ then $\alpha/2\not\in\Phi$.

Thus it $W$ can't contain any odd roots or even roots, so $W = 0$.
Note also that $L_{\pm\alpha}\not\subset K \oplus W$, forcing it to be in $\liesl( \alpha)$, so $L_{ \alpha} = \gens{x_ \alpha}$ and $L_{- \alpha} = \gens{y _{\alpha} }$.
:::

:::{.proposition title="B"}
Let \( \alpha, \beta\in \Phi \) with \( \beta\neq \pm \alpha \) and consider $\beta + k \alpha$ for $n\in \ZZ$.

a. $\beta(h_ \alpha) \in \ZZ$.
b. $\exists r,q\in \ZZ_{\geq 0}$ such that for $k\in \ZZ$, the combination $\beta + k \alpha\in \Phi \iff k \in [-r, q] \ni 0$.
  The set $\ts{ \beta + k \alpha \st k\in [-r, q]} \subseteq \Phi$ is the **$\alpha\dash$root string through $\beta$**.
c. If \( \alpha + \beta\in \Phi \) then \( [L_ \alpha L_ \beta ] = L_{ \alpha + \beta} \).
d. \(\beta- \beta(h_ \alpha) \alpha\in \Phi \).

:::

:::{.proof title="?"}
Consider 
\[
M \da \bigoplus _{k\in \ZZ} L_{ \beta + k \alpha} \leq L \quad\in\mods{\liesl( \alpha)}
.\]

a. $\beta(h _{\alpha} )$ is the eigenvalues of $h_ \alpha$ acting on $L_ \beta$.
  But by the lemma, $\beta(h_ \alpha)\in \ZZ$.

b. By the previous proposition, $\dim L_{ \beta+ k \alpha} = 1$ if nonzero, and the weight of $h_\alpha$ acting on it is $\beta( h _{\alpha} ) + 2k$ all different for distinct $k$.
  By $\liesl_2\dash$representation theory, we know the multiplicities of various weight spaces as the sum of dimensions of the zero and one weight spaces, and thus $M$ is a single irreducible $\liesl(\alpha)\dash$module.
  So write $M - L(d)$ for some $d\in \ZZ_{\geq 0}$, then $h_ \alpha\actson M$ with eigenvalues $\ts{d,d-2,\cdots, -d+2, -d}$.
  But $h_ \alpha\actson M$ with eigenvalues $\beta( h_ \alpha) + 2k$ for those $k\in \ZZ$ with $L_{\beta + k \alpha}\neq 0$.
  Since the first list is an unbroken string of integers of the same parity, thus the $k$ that appear must also be an unbroken string.
  Define $r$ and $q$ by setting $d = \beta(h_\alpha) + 2q$ and $-d =\beta( h_ \alpha ) - 2r$ to obtain $[-r, q]$.
  Adding these yields $0 = 2\beta( h_ \alpha) + 2q-2r$ and $r-q = \beta(h_ \alpha)$.

d. Let $M\cong L(d) \in \mods{\liesl(\alpha)}$ and $x_\beta \in L_ \beta\smz \subseteq M$ with $x_ \alpha\in L_{ \alpha}$.
  If $[x_ \alpha x_ \beta] = 0$ then $x_ \beta$ is a maximal $\liesl(\alpha)\dash$vector in $L(d)$ and thus $d = \beta(h_ \alpha)$.
  But $\alpha + \beta\in \Phi \implies \beta)(h_ \alpha) + 2$ is a weight in $M$ bigger than $d$, a contradiction.
  Thus $\alpha + \beta\in \Phi \implies [x_ \alpha x_ \beta] \neq 0$.
  Since this bracket spans and $\dim L_{ \alpha + \beta} = 1$, so $[L_ \alpha L_ \beta] = L_{ \alpha + \beta}$.

e. Use that $q\geq 0, r\geq 0$ to write $-r \leq -r+q \leq q$.
  Then \[
  \beta - \beta(h_ \alpha) \alpha = \beta - (r-q) \alpha = \beta + (-r+q \alpha)\da \beta + \ell\alpha
  \] where $\ell\in [-r, q]$.
  Thus $\beta + \ell\alpha\in \Phi$ is an unbroken string by (b).

:::

:::{.question}
Is it true that $\bigoplus_{k\in \ZZ} L_{\beta+ k \alpha} = \bigoplus _{c\in \CC} L_{\beta + c\alpha}$?
The issue is that $c\in \ZZ + {1\over 2}$ is still possible.
:::

## 8.5: Rationality properties of $\Phi$

:::{.remark}
Recall that $\kappa$ restrict to a nondegenerate bilinear form on $H$ inducing $H\dual\iso H$ via $\phi\mapsto t_\phi$ where $\kappa(t_\phi, \wait) = \phi(\wait)$.
Transfer to a nondegenerate symmetric bilinear form on $H\dual$ by $(\lambda, \mu) \da \kappa(t_\lambda, t_\mu)$.
By prop 8.3 we know $H\dual$ is spanned by $\Phi$, so choose a $\CC\dash$basis $\ts{ \alpha_1,\cdots, \alpha_n} \subseteq \Phi$.
Given $\beta\in\Phi$, write $\beta = \sum c_i \alpha_i$ with $c_i\in \CC$.
:::

:::{.claim}
$c_i \in \QQ$ for all $i$!
:::