# Monday August 19

## Solvability

> Idea: Define a semisimple Lie algebra

**Definition:**
The derived series for $\lieg$ is given by

\begin{align*}
\lieg^{(0)} = \lieg \\
\lieg^{(1)} = [\lieg^{(0)}, \lieg^{(0)}] \\
\cdots \\
\lieg^{(i+1)} = [\lieg^{(i)}, \lieg^{(i)}]
.\end{align*}

The Lie algebra $\lieg$ is *solvable* if there is some $n$ for which $\lieg^{(n)} = 0$.

*Exercise (to turn in):*
Check that the Lie algebra of upper triangular matrices in $\liegl(n, \FF)$.

*Example:* 
Abelian Lie algebras are solvable

*Example:* 
Simple Lie algebras are *not* solvable.

**Proposition:**
Let $\lieg$ be a Lie algebra, then

1. If $\lieg$ is solvable, then all subalgebras and all homomorphic images of $\lieg$ are also solvable.
2. If $I \normal \lieg$ and both $I$ and $\lieg/I$ are solvable, then so is $\lieg$.
3. If $I, J \normal \lieg$ are solvable, then so is $I+J$.

**Corollary (of part 3 above):**
Any Lie algebra has a unique maximal solvable ideal, which we denote the *radical* $\mathrm{Rad}(\lieg)$.

**Definition:**
A Lie algebra is semisimple if $\mathrm{Rad}(\lieg) = 0$.

*Example:*
Any simple Lie algebra is semisimple.

*Example:* 
Using part (2) above, we can deduce that we can construct a semisimple Lie algebra from *any* Lie algebra: for any $\lieg$, the quotient $\lieg/\mathrm{Rad}(\lieg)$ is semisimple.

## Nilpotency

\begin{align*}
\lieg^{0} = \lieg \\
\lieg^{1} = [\lieg^{0}, \lieg^{0}] \\
\cdots \\
\lieg^{i+1} = [\lieg^{i}, \lieg^{i}]
.\end{align*}

Much like the previous case, we have

*Example:*
Abelian Lie algebras are nilpotent.

*Example:*
Nilpotent Lie algebras are solvable.

*Example:*
The *strictly* upper triangular matrices (with zero on the diagonal) are nilpotent.

1. If $\lieg$ is nilpotent, then all subalgebras and all homomorphic images of $\lieg$ are also nilpotent.
2. If $\lieg/Z(\lieg)$ is nilpotent, then so is $\lieg$.
3. If $\lieg \neq 0$ is nilpotent, then $Z(\lieg) \neq 0$.

**Proposition:**
If $\lieg$ is nilpotent, then $\ad_x \in \mathrm{End}(\lieg)$ is nilpotent for all $x\in \lieg$.

*Proof:*

This is because $\lieg^n = 0 \iff [\lieg, [\lieg, [\lieg, \cdots]]] = 0$, and so for every $x_i, y \in \lieg$ we have $[x_1, [x_2, \cdots [x_n, y]]] = 0$, and so $\ad_{x_1} \circ \ad_{x_2} \circ \cdots \ad_{x_n} = 0$ which implies that $\ad_x^n = 0$ for all $x\in \lieg$.

**Theorem [Engel]:**
If $\ad_x$ is nilpotent for all $x\in \lieg$, then $\lieg$ is nilpotent.

> Remark: This can be confusing if $\lieg$ is a linear algebra, we can consider elements $x \in\lieg$ and ask if it is the case $x$ being nilpotent (as an endomorphism) iff $\lieg g$ is nilpotent? False, a counterexample is $\lieg = \liegl(2, \CC)$, where there exists an $x$ which is *not* nilpotent while $\ad_x$ *is* nilpotent, which contradicts the above theorem.

*Proof:*

We'll first establish a lemma.

**Lemma:**
Let $\lieg \subseteq \liegl(V)$ be a Lie subalgebra for some finite dimensional vector space $V$. If $x$ is nilpotent as an endomorphism on $V$ for all $x\in V$, then there exists a nonzero vector $v\in V$ such that $\lieg v =0$, so $x\in \lieg \implies x(v) = 0$.

*Proof of lemma*:
Use induction on $\dim \lieg$, splitting into two separate base cases:

- Case $\dim \lieg = 0$, then $\lieg = \theset{0}$.
- Case $\dim g = 1$, left as an exercise.

Inductive step: Let $A$ be a maximal proper subalgebra and define $\phi: A \to \liegl(\lieg/A)$ where $a \mapsto (x + A \mapsto [a, x] + A)$. We need to check that $\phi$ is a homomorphism, this just follows from using the Jacobi identity.

We also need to show that $\im \phi \leq \liegl(\lieg/a)$ is a Lie subalgebra, and $\dim \im \phi < \dim \lieg$. The claim is that $\phi(a) \in \mathrm{End}(\lieg/A)$ is nilpotent for all $a\in A$. By the inductive hypothesis, there is a nonzero coset $y + A \in \lieg/A$ such that $(\im \phi) \cdot (y+A) = A$. Since $y\not\in A$, then $\phi(a)(y+A) = A$ for all $a\in A$, and so $[a,y]\in A$.

We want to show that $A$ is a subalgebra of codimension 1, and $A \oplus F_y \leq \lieg$ is a Lie subalgebra.
This is because $[a_1 + c_1y, a_2 + c_2 y] = [a_1, a_2] + c_2[a_1, y] - c_2[a_2, y] + c_1c_2[y, y]$. The last term is zero, the middle two terms are in $A$, and because $A$ is closed under the bracket, the first term is in $A$ as well.

But then $A \oplus F_y$ is a larger subalgebra than $A$, which was maximal, so it must be everything. So $A \oplus F_y = \lieg$. So $A \normal \lieg$ because $[a_1, a_2 + cy]$ is in $A, A \oplus F_y = \lieg$ respectively, and this equals $[a_1, a_2] + c[a_1, y]$, where both terms are in $A$.

> Proof to be continued.