# Wednesday, October 12

:::{.remark}
Today: finding bases for root systems.
It's not obvious they always exist, but e.g. in the previous rank 2 examples, $\alpha,\beta$ formed a base.
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:::{.definition title="Height, positive/negative roots"}
Given a base $\Delta \subseteq \Phi$, the **height** of a root $\beta = \sum_{\alpha\in \Delta} k_ \alpha \alpha$ is
\[
\height(\beta) \da \sum k_ \alpha
.\]
If all $k_\alpha \geq 0$, we say $\beta$ is **positive** and write $\beta\in \Phi^+$.
Similarly, $\beta$ is **negative** iff $k_\alpha \leq 0$ for all $\alpha$, and we write $\beta\in \Phi^-$.
This decomposes a root system into $\Phi = \Phi^+ \disjoint \Phi^-$, and moreover $-\Phi^+ = \Phi^-$.
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:::{.remark}
A choice of $\Delta$ determines a partial order on $\Phi$ which extends to $\EE$, where \( \lambda\geq \mu \iff \lambda- \mu \) is a non-negative integer linear combination of elements of $\Delta$.
:::

:::{.lemma title="?"}
If $\Delta \subseteq \Phi$ is a base and $\alpha, \beta\in \Delta$, then 
\[
\alpha\neq \beta\implies ( \alpha, \beta) \leq 0 \text{ and } \alpha- \beta\not\in \Phi
.\]
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:::{.proof title="?"}
We have \( \alpha\neq \pm \beta \) since $\Delta$ is a linearly independent set.
By a previous lemma, if $( \alpha, \beta) > 0$ then $\beta- \alpha \in \Phi$ by a previous lemma. $\contradiction$
:::

:::{.definition title="Regular"}
An element $\gamma\in \EE$ is **regular** iff $\gamma \in \EE\sm \Union_{\alpha\in \Phi} P_ \alpha$ where $P_ \alpha= \alpha^\perp$, otherwise $\gamma$ is **singular**.
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:::{.lemma title="?"}
Regular vectors exist.
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:::{.remark}
The basic fact used is that over an infinite field, no vector space is the *union* of a finite number of proper subspaces.
Note that this is a union, not a sum!

Given a regular vector $\gamma\in \EE$, define
\[
\Phi^+(\gamma) = \ts{ \alpha\in \Phi \st (\alpha, \gamma) > 0 }
,\]
the roots on the positive side of the hyperplane $\alpha^\perp$:

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This decomposes $\Phi = \Phi^+(\gamma) \disjoint \Phi^-(\gamma)$ where $\Phi^-(\gamma) \da - \Phi^+(\gamma)$.
Note that $\gamma$ lies on the positive side of $\alpha^\perp$ for every $\alpha\in \Phi^+(\gamma)$.

:::


:::{.definition title="Decomposable roots"}
A positive root $\beta\in \Phi^+$ is **decomposable** iff $\beta = \beta_1 + \beta_2$ for $\beta_i \in \Phi^+( \gamma)$.
Otherwise $\beta$ is **indecomposable**.
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:::{.theorem title="?"}
There exists a base for $\Phi$.
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:::{.theorem title="?"}
Let \( \gamma\in \EE \) be regular.
Then the set $\Delta(\gamma)$ of all indecomposable roots in $\Phi^+( \gamma)$ is a base for $\Phi$.
Moreover, any base for $\Phi$ arises in this way.
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:::{.proof title="in 5 easy steps"}
\envlist

1. Claim: each root in $\Phi^+( \gamma)$ is in $\ZZ_{\geq 0} \Delta( \gamma)$.
  
  The proof: if not, pick $\beta\in \Phi^+(\gamma)$ which cannot be written this way and choose it such that $(\beta, \gamma)$ is maximal (by finiteness).
  Since $\beta\not\in \Delta( \gamma)$, it is decomposable as $\beta = \beta_1 + \beta_2$ with \( \beta_i \in \Phi^+ \).
  Now \( (\beta, \gamma) = \sum (\beta_i, \gamma) \) is a sum of nonnegative numbers, so $(\beta_i, \gamma) < (\beta, \gamma)$ for $i=1,2$.
  By minimality, \( \beta_i\in \ZZ_{\geq 0 } \Delta(\gamma) \), but then by adding them we get \( \beta\in \ZZ_{\geq 0 } \Delta(\gamma) \). 

2. Claim: if \( \alpha, \beta\in \Delta( \gamma) \) with \( \alpha\neq \beta \) then \( (\alpha, \beta) \leq 0 \).
  Note \( \alpha\neq - \beta \) since $(\alpha, \gamma), (\beta, \gamma) > 0$.
  By lemma 9.4, if $( \alpha, \beta) > 0$ then \( \alpha - \beta\in \Phi \) is a root.
  Then one of \( \alpha- \beta, \beta- \alpha\in \Phi^+( \gamma) \).
  In the first case, \( \beta + (\alpha- \beta ) = \alpha \), decomposing \( \alpha \).
  In the second, \( \alpha + (\beta- \alpha) = \beta \), again a contradiction.

3. Claim: \( \Delta\da \Delta( \gamma) \) is linearly independent.
  Suppose $\sum_{ \alpha\in \Delta} r_ \alpha \alpha = 0$ for some $r_ \alpha \in \RR$.
  Separate the positive terms $\alpha\in \Delta'$ and the remaining $\alpha\in \Delta''$ to write $\eps \da \sum_{ \alpha\in \Delta'}r_ \alpha \alpha = \sum_{ \beta\in \Delta''} t_\beta \beta$ where now $r_ \alpha, t_\beta > 0$.
  Use the two expressions for $\eps$ to write 
  \[
  (\eps, \eps) = \sum _{ \alpha\in \Delta', \beta\in \Delta''} r_ \alpha t_ \beta (\alpha, \beta) \leq 0
  ,\]
  since $r_ \alpha t_ \beta >0$ and $(\alpha, \beta) \leq 0$.
  So $\eps = 0$, since $(\wait, \wait)$ is an inner product.
  Write $0 = (\gamma, \eps) = \sum_{ \alpha\in \Delta'} r_\alpha (\alpha, \delta)$ where $r_ \alpha > 0$ and $(\alpha, \gamma) > 0$, so it must be the case that $\Delta' = \emptyset$.
  Similarly $\Delta'' = \emptyset$, so $r_\alpha = 0$ for all \( \alpha\in \Delta \).

4. Claim: \( \Delta( \gamma) \) is a base for $\Phi$.
  Since $\Phi = \Phi^+(\gamma)\disjoint \Phi^-( \gamma)$, we have B2 by step 1.
  This also implies $\Delta( \gamma)$ is a basis for $\EE$, since we have linear independent by step 3.
  Thus $\ZZ \Delta ( \gamma) \contains \Phi$ and $\RR \Phi = \EE$.

5. Claim: every base of $\Phi$ is $\Delta( \gamma)$ for some regular $\gamma$.
  Given $\Delta$, choose $\gamma\in \EE$ such that $(\gamma, \alpha) > 0$ for all \( \alpha\in \Delta \).
  Then $\gamma$ is regular by B2.
  Moreover $\Phi^+ \subseteq \Phi^+( \gamma)$ and similarly $\Phi^- \subseteq \Phi^-( \gamma)$, and taking disjoint unions yields $\Phi$ for both the inner and outer sets, forcing them to be equal, i.e. $\Phi^{\pm} = \Phi^{\pm}( \gamma)$.
  One can check that $\Delta \subseteq \Delta( \gamma)$ using $\Phi^+ = \Phi^+( \gamma)$ and linear independence of $\Delta$ -- but both sets are bases for $\EE$ and thus have the same cardinality $\ell = \dim \EE$, making them equal.

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