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> [!definition] Definition. ([[derived and central series of a Lie algebra]])
> Let $\mathfrak{g}$ be a [[Lie algebra]]. Let $[I,J]=\text{span}\{ [i,j] \}_{i \in I}^{j \in J}$; this is an [[ideal of a Lie algebra|ideal]].[^1] The **derived subalgebra** [[Lie subalgebra|of]] $\mathfrak{g}$ is defined to be $[\mathfrak{g}, \mathfrak{g}]$. Note that $[\mathfrak{g}, \mathfrak{g}]=0$ iff $\mathfrak{g}$ is [[abelian Lie algebra|abelian]].
>
Now, fix notation $\mathfrak{g}^{(0)}=\mathfrak{g}^{0}=\mathfrak{g}$, and then inductively define ideals $\begin{align}
\mathfrak{g}^{(n)}&:=[\mathfrak{g}^{(n-1)}, \mathfrak{g}^{(n-1)}] \\
\mathfrak{g}^{n} &:= [\mathfrak{g}, \mathfrak{g}^{n-1}].
\end{align}$
The **derived series** of $\mathfrak{g}$ is $\mathfrak{g} \supset \overbrace{ \mathfrak{g}^{(1)} }^{ [\mathfrak{g}, \mathfrak{g}] } \supset \overbrace{ \mathfrak{g}^{(2)} }^{ \big[ [\mathfrak{g}, \mathfrak{g}], [\mathfrak{g}, \mathfrak{g}] \big] } \supset \mathfrak{g}^{(3)} \supset\cdots $
The **central series** of $\mathfrak{g}$ is $\mathfrak{g} \supset \underbrace{ \mathfrak{g}^{1} }_{ [\mathfrak{g}, \mathfrak{g}] } \supset \underbrace{ \mathfrak{g}^{2} }_{ \big[ \mathfrak{g}, [\mathfrak{g, \mathfrak{g}}] \big] } \supset \mathfrak{g}^{3} \supset \cdots$
Note that number of derived series brackets grows exponentially, while the number of central series brackets grows linearly.
^definition-1
> [!note] Mnemonic.
> Everything in the derived series is derived subalgebra.
^note
> [!definition] Definition. (Nilpotent, Solvable)
> We call $\mathfrak{g}$ **solvable** if its derived series terminates: $\mathfrak{g}^{(n)}=0$ for some $n \geq 1$.
>
We call $\mathfrak{g}$ **nilpotent** if its central series terminates: $\mathfrak{g}^{n}=0$ for some $n \geq 1$.
>
$\mathfrak{g}^{(n)} \subset \mathfrak{g}^{n}$, so nilpotent $\implies$ solvable.[^2]
^definition-2
> [!basicexample]
> - [ ] check as a direct computation using elementary matrices: $[\mathfrak{gl}_{n}, \mathfrak{gl}_{n}]=\mathfrak{sl}_{n}$
> - Any [[abelian Lie algebra]] is nilpotent (hence solvable), because $\mathfrak{g}^{1}=[\mathfrak{g}, \mathfrak{g}]=0$.
> - The [[strictly triangular matrix|Lie algebra of strictly upper triangular matrices]] $\mathfrak{n}_{n} \subset \mathfrak{gl}_{n}$ is nilpotent (hence solvable), as $\mathfrak{g}^{n}=0$
> - The [[upper-triangular matrix|upper triangular Lie algebra]] $\mathfrak{b}_{n} \subset \mathfrak{gl}_{n}$ is solvable, but not nilpotent if $n \geq 2$.
^basic-example
- [ ] see also handwritten notes from lecture 6
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####
[^1]: Check using Jacobi identity.
[^2]: The converse is false! For a counterexample, let $\mathfrak{g}$ be the two-dimensional [[Lie algebra]] with [[basis]] $x,y$ and bracket determined by $[x,y]=y$. Then $\mathfrak{g}$ is solvable but not nilpotent.
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#### References
> [!backlink]
> ```dataview
> TABLE rows.file.link as "Further Reading"
> FROM [[]]
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> GROUP BY Tag
> ```
> [!frontlink]
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> FROM outgoing([[]])
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> GROUP BY Tag
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