# Wednesday August 21

Last time: we had a theorem that said that if $\lieg \in \liegl(V)$ and every $x\in\lieg$ is nilpotent, then there exists a nonzero $v \in V$ such that $\lieg v = 0$.

We proceeded by induction on the dimension of $V$, constructing $\im \phi \subseteq \liegl(\lieg/A)$, and showed that $\lieg = A \oplus Fy$. Now consider
$$
W = \theset{v\in V \suchthat Av = 0},
$$

which is $\lieg\dash$invariant, so $\lieg(W) \subseteq W$, or for all $a\in A, x\in \lieg, v\in W$, we have $a\actson x(v) = 0$. This is true because $a\actson x = x\circ a + [a, x] \in \liegl(V)$. But $V$ is killed by any element in $A$, and both of these terms are in $A$. In particular, the $y$ appearing in $Fy$ also satisfies $y \in W$. Consider $\restrictionof{y}{W} \in \mathrm{End}(w)$, and we want to apply the inductive hypothesis to $F \restrictionof{y}{W} \subseteq \liegl(V)$.

We need to check that $\restrictionof{y}{W} \in \mathrm{End}(V)$, which is true exactly because $y$ is nilpotent.
So we can construct a nonzero $v\in W \subset V$ such that $y(v) = 0$, and so $\lieg v = 0$.

**Claim:** 
$\phi(a) \in \mathrm{End}(\lieg/A)$ is nilpotent. 

Each $a\in A\subset \lieg$ is nilpotent by assumption.
Define the maps for left multiplication by $a$, $m_\ell: x \mapsto ax$, and the right multiplication $m_r: x \mapsto xa$.
These are nilpotent, and since $m_\ell, m_r$ commute, the difference $m_\ell - m_r$ is nilpotent, and this is exactly $\ad_a$. But then $\phi(a)$ is nilpotent.

> Good proof for using all of the definitions!

Now we can see what the consequences of having such a nonzero vector are. This theorem implies Engel's theorem, which says that if $\ad_x \in \mathrm{End}(\lieg)$ is nilpotent for every $x\in \lieg$, then $\lieg$ is nilpotent.

*Proof:* 

By induction on dimension. The base case is easy. For the inductive step, the previous theorem applies to $\ad g \subset \liegl(\lieg)$. So we can produce the nonzero $v\in \lieg$ such that $\ad \lieg v = 0$. Then $[x, v] = 0$ for all $x\in\lieg$, so either $v\in Z(\lieg)$ or $Z(\lieg) \neq 0$. In either case, $\lieg / Z(\lieg)$ has smaller dimension. Since $\ad_x$ is nilpotent, so is $\ad_x + Z(\lieg)$, and so $\lieg/Z(\lieg)$ is nilpotent.
By an earlier proposition, since the quotient is nilpotent, so is the total space. $\qed$

Let $\mathfrak{N}(F)$ be the subalgebra of $\liegl(F)$ consisting of strictly upper triangular matrices. We have a corollary: if $\lieg \subset \liegl(n, F)$ is a Lie subalgebra such every $x\in \lieg$ is nilpotent as an endomorphism of $F$, then the matrices of $\lieg$ with respect to some bases of in $\mathfrak{N}(n, F)$.

The proof is by induction on $n$, where the base case is easy. For the inductive step, we use the previous theorem to get a $v_1$ such that $x(v_1) = 0$ for all $x\in \lieg$. Let $\overline V = F^n/Fv_1 \cong F^{n-1}$, and define $\phi: \lieg \to \liegl(\overline V)$ where $x \mapsto (\overline y \mapsto \overline{y(x)})$.

Then $\im \phi \leq \liegl(n-1, F)$ as a subalgebra, and every $\phi(x)\in \mathrm{End}(F^{n-1})$ is nilpotent, since $x$ was nilpotent on the larger space. But (see notes) then $x$ can be written as a strictly upper-triangular matrix.

## Chapter 2: Semisimple Lie Algebras

We now assume $\mathrm{char} ~F = 0$ and $\overline F = F$.

**Theorem:**
If $\lieg$ is a solvable Lie subalgebra of $\liegl(V)$ for some finite dimensional $V$, then $V$ contains a common eigenvector for a $x\in\lieg$, i.e. a $\lambda: \lieg \to F, x \mapsto \lambda(x)$ such that $x(v) = \lambda(x) v$ for all $x\in\lieg$.

*Proof:* 
We will use induction on the dimension of $\lieg$. For the inductive step:

**Claim 1:** 
There is an ideal $A\normal \lieg$ such that $\lieg = A \oplus Fy$ for some $y\neq 0$, so $A$ is a subalgebra of a solvable Lie algebra $\lieg$ and thus solvable itself. By hypothesis, we can produce a $w \in V\setminus\theset{0}$, and thus a functional $\lambda: A \to F$ such that $aw = \lambda(a) w$ for all $a\in A$. So we define
$$
V_\lambda = \theset{v\in V \suchthat av = \lambda(a)v \forall a\in A}
$$

where $w\in V_\lambda$.

**Claim 2:** 
$y(V_\lambda) \subseteq V_\lambda$, or $\restrictionof{y}{V_\lambda}\in\mathrm{End}(V_\lambda)$.

Thus $F(\restrictionof{y}{V_\lambda}) \leq \liegl(V_\lambda)$ is a Lie algebra of dimension 1, and thus solvable. By the inductive hypothesis, we can find a $v\in V_\lambda$ and some $\mu \in F$ such that $y(v) = \mu v$. An arbitrary element $x\in\lieg$ can be written as $x = a + cy$ for some $a\in A, c\in F$ and it acts by $x(v) = a(v) + cy(v) = \lambda(a) v + c\mu v = (\lambda(a) + c)v \in V_\lambda$.