# Solvable and Nilpotent Lie Algebras (Wednesday, August 24)

## Humphreys 3.1

:::{.definition title="Derived series"}
Recall that if $L \subseteq \lieg$ is any subset, the **derived algebra** $[LL]$ is the span of $[xy]$ for $x,y\in L$.
This is the analog of needing to take products of commutators to generate a commutator subgroup for groups.
Define the **derived series** of $L$ as 
\[
L^{(0)} = [LL], \quad L^{(1)} = [L^{(0)}, L^{(0)}], \cdots \quad L^{(i+1)} = [L^{(i)} L^{(i)}]
.\]
:::

:::{.proposition title="?"}
These are all ideals.
:::

:::{.proof title="?"}
By induction on $i$ -- it STS that $[x[ab]] \in L^{(i)}$ for $a,b \in L^{(i-1)}$ and $x\in L$.
Use the Jacobi identity and the induction hypothesis that $L^{(i-1)} \normal L$:
\[
[x,[ab]] = [[xa]b] + [a[xb]] \in L^{(i-1)} + L^{(i-1)} \subseteq L^{(i)}
.\]
:::

:::{.definition title="Solvable"}
If $L^{(n)} = 0$ for some $n\geq 01$ then $L$ is called **solvable**.
:::

:::{.remark}
Note that 

- $L$ abelian implies solvable, since $L^{(1)} = 0$.
- $L$ simple implies non-solvable, since this forces $L^{(1)} = L$.

:::

:::{.exercise title="?"}
Let $\mfb \da \mfb_n(\FF)$ be upper triangular matrices, show that $\lieb$ is solvable.

> Use that $[\mfb \mfb] = \mfn$ is strictly upper triangular since diagonals cancel.
> More generally, bracketing matrices with $n$ diagonals of zeros yields matrices with about $2^n$ diagonals of zeros.

:::

:::{.proposition title="?"}
Let $L\in \Lie\Alg\slice\FF$, then

- $L$ solvable implies solvability of all subalgebras and homomorphic images of $L$.
- If $I\normal L$ and $L/I$ are solvable, then $L$ is solvable.
- If $I,J\normal L$ are solvable then $I+J$ is solvable.[^hint_solvable]

[^hint_solvable]: 
Use the third isomorphism theorem.

:::

:::{.exercise title="?"}
Prove these.
:::

:::{.definition title="Radical and semisimple"}
Every $L\in \Lie\Alg\slice\FF^{\fd}$ has a unique maximal solvable ideal, the sum of all solvable ideals, called the **radical** of $L$, denote $\rad(L)$.
$L$ is **semisimple** if $\rad(L) = 0$.
:::

:::{.exercise title="?"}
Prove that any simple algebra is semisimple, and in general $L/\rad(L)$ is semisimple (if nonzero).
:::

:::{.proposition title="?"}
Assume $\liesl_n(\CC)$ is simple, then $R\da \rad(\liegl_n(\CC)) = Z(\lieg) \contains \CC \id_n$ for $\lieg\da\liegl_n(\CC)$.
:::

:::{.proof title="?"}
$\supseteq$: 
Centers are always solvable ideals, since it's abelian and brackets are ideals, and the radical is sum of all solvable ideals.

$\subseteq$: 
Suppose $Z\subsetneq R$ is proper, then there is a non-scalar matrix $x\in R$.
Write $x = aI_n + y$ for $a = \tr(x)/n$ and $0\neq y\in \liesl_n(\CC)$ is traceless.
Consider $I = \gens{x} \normal \liegl_n(\CC)$, i.e. the span of all brackets $[zx]$ for $z\in \lieg$ and their iterated brackets containing $x$, e.g. $[z_1[z_2x]]$.
Note that $[zx]=[zy]$ since $aI_n$ is central.
Since $\liesl_n(\CC)$ is simple, so $\gens{y}_{\liesl_n(\CC)} = \liesl_n(\CC)$ and thus $\liesl_n(\CC) \subseteq I$.
This containment must be proper, since $I \subseteq \rad(\lieg)$ and the latter is solvable, so $I$ must be solvable -- but $\liesl_n(\CC)$ is not solvable.
We can thus choose $x\in I$ such that $x = aI_n + y$ with $a\neq 0$ and $0\neq y \in \liesl_n(\CC)$, so $x-y= aI \in I$ since $y\in I$ because $\liesl_n(\CC) \subseteq I$.
Since $a\neq 0$, we must have $I_n\in I$.
Then $\CC\cdot I_n \subseteq I$, forcing $I = \lieg$ since every matrix in $\liegl_n(\CC)$ is a scalar multiple of the identity plus a traceless matrix.
This contradicts that $I$ is solvable, since $\lieg^{(1)} \da [\lieg\lieg] = \liesl_n(\CC)$.
But $\lieg^{(1)} = \liesl_n(\CC)$, so the derived series never terminates. $\contradiction$
:::

## Humphreys 3.2

:::{.definition title="Lower central series and nilpotent algebras"}
The **descending/lower central series** of $L$ is defined as 
\[
L^0 = L, \quad L^1 = [LL], \quad \cdots L^i = [L, L^{i-1}]
.\]
$L$ is **nilpotent** if $L^n=0$ for some $n$.
:::

:::{.exercise title="?"}
Check that $L^i \normal L$.
:::

:::{.exercise title="?"}
Show that $L$ nilpotent is equivalent to there existing a finite $n$ such that for any set of elements $\ts{x_i}_{i=1}^n$,
\[
(\ad_{x_1} \circ \ad_{x_2} \circ \cdots \circ \ad_{x_n})(y) = 0 \qquad \forall y\in L
.\]
:::