# Ch. 8: Root space decompositions (Friday, September 23)

:::{.remark}
Recall that the relations from last time can produce an infinite-dimensional module with basis $\ts{v_0,v_1,\cdots}$.
Note that if \( m\in \ZZ_{\geq 0} \), then \( x.v_{m+1} = ( \lambda- m ) v_m = 0 \).
This says that one can't raise $v_{m+1}$ back to $v_m$, so $\ts{v_{m+1},v_{m+2} \cdots}$ spans a submodule isomorphic to $M(-m-2)$. 
Quotienting yields $L(m) \da M(m) / M(-m-2)$, also called $V(m)$, spanned by $\ts{v_0, \cdots, v_m}$.
Note that $M(-m-2)$ and $L(m)$ are irreducible.
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:::{.remark}
Let $L\in \Lie\Alg\slice\CC^{\fd, \ss}$ for this chapter.
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## 8.1: Maximal toral subalgebras and roots

:::{.remark}
Let $L\ni x = x_s + x_n \in L + L$ be the abstract Jordan decomposition. then if $x=x_n$ for every $x\in L$ then $L$ is nilpotent which contradicts Engel's theorem.
Thus there exists some $x=x_s\neq 0$.
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:::{.definition title="Toral subalgebras"}
A **toral subalgebra** is any nonzero subalgebra spanned by semisimple elements.
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:::{.example title="?"}
The algebraic torus $(\CC\units)^n$ which has Lie algebra $\CC^n$, thought of as diagonal matrices.
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:::{.lemma title="?"}
Let $H$ be a maximal toral subalgebra of $L$.
Any toral subalgebra $H \subseteq L$ is abelian.
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:::{.proof title="?"}
Let $T \leq L$ be toral and let $x\in T$ be a basis element.
Since $x$ is semisimple, it STS $\ad_{T, x} = 0$.
Semisimplicity of $x$ implies $\ad_{L, x}$ diagonalizable, so we want to show $\ad_{T, x}$ has no non-zero eigenvalues.
Suppose that there exists a nonzero $y\in T$ such that $\ad_{T, x}(y) = \lambda y$ for \( \lambda \neq 0 \).
Then $\ad_{T, y}(x) = [yx] = -[xy] = - \ad_{T, x} = - \lambda y\neq 0$, and since $\ad_{T, y}(y) = -[yy] = 0$, $y$ is an eigenvector with eigenvalue zero.
Since $\ad_{T, y}$ is also diagonalizable and $x\in T$, write $x$ as a linear combination of eigenvectors for it, say $x = \sum a_i v_i$. 
Then $\ad_{T, y}(x) = \sum \lambda_i a_i v_i$ and the terms with $\lambda_i = 0$ vanish, and the remaining element is a sum of eigenvectors for $\ad_{T, y}$ with nonzero eigenvalues. $\contradiction$
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:::{.example title="?"}
If $L = \liesl_n(\CC)$ then define $H$ to be the set of diagonal matrices.
Then $H$ is toral and in fact maximal: if $L = H \oplus \CC z$ for some $z\in L\sm H$ then one can find an $h\in H$ such that $[hz] \neq 0$, making it nonabelian, but toral subalgebras must be abelian.
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:::{.definition title="Roots"}
Recall that a commuting family of diagonalizable operators on a finite-dimensional vector space can be simultaneously diagonalized.
Letting $H \leq L$ be maximal toral, this applies to $\ad_L(H)$, and thus there is a basis in which all operators in $\ad_L(H)$ are diagonal.
Set $L_{ \alpha} \da \ts{x\in L \st [hx] = \alpha(h) x\,\,\forall h\in H}$ where \( \alpha: H\to \CC \) is linear and thus an element of $H\dual$.
Note that $L_0 = C_L(H)$, and the set $\Phi\da \ts{\alpha\in H\dual \st \alpha\neq 0, L_\alpha\neq 0}$ is called the **roots** of $H$ in $L$, and $L_\alpha$ is called a **root space**.
Note that $L_0$ is not considered a root space.
This induces a **root space decomposition**
\[
L = C_L(H) \oplus _{ \alpha\in \Phi } L_\alpha
.\]
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:::{.remark}
Note that for classical algebras, we'll show $C_L(H) = H$ and corresponds to the standard bases given early in the book.
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:::{.example title="?"}
Type $A_n$ yields $\liesl_{n+1}(\CC)$ and $\dim H = n$ for $H$ defined to be the diagonal traceless matrices.
Define $\eps_i\in H\dual$ as $\eps_i \diag(a_1,\cdots, a_{n+1}) \da a_i$, the $\Phi \da \ts{\eps_i - \eps_j \st 1\leq i\neq j\leq ,n+1}$ and $L_{\eps_i - \eps_j} = \CC e_{ij}$.
Why: 
\[
[h, e_{ij}] =
\left[ \sum a_k e_{kk}, e_{ij}\right] = a_i e_{ii} e_{ij} - a_j e_{ij} e_{jj} = (a_i - a_j) e_{ij} \da (\eps_i -\eps_j)(h) e_{ij}
.\]

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