# Monday, August 29

## Continuation of proof and corollaries

:::{.remark}
Recall: we were proving that if $L \leq \liegl(V)$ with $V$ finite dimensional and $L$ consists of nilpotent endomorphisms, then there exists a common eigenvector $v$, so $Lv = 0$.
:::

:::{.proof title="continued"}
We're inducting on $\dim L$ (over all $V$).
Assuming $\dim V > 1$, we showed that proper subalgebras are strictly contained in their normalizers:
\[
K \lneq L \implies K \subsetneq N_L(K)
.\]
Let $K$ be a maximal proper subalgebra of $L$, then $N_L(K) = L$ by maximality and thus $K$ is a proper ideal of $L$.
Then $L/K$ is a Lie algebra of some dimension, which must be 1 -- otherwise the preimage in $L$ under $L\surjects L/K$ would be a subalgebra of $L$ properly between $K$ and $L$.
Thus $K$ is a codimension 1 ideal in $L$.Minimal model program
Choosing any $z\in L\sm K$ yields a decomposition $L = K \bigoplus \FF z$ as vector spaces.
Let $W \da \ts{v\in V\st Kv=0}$, then $W\neq 0$ by the IH.


:::{.claim}
$W$ is an $L\dash$stable subspace.

To see this, let $x\in L, y\in K, w\in W$.
A useful trick:
\[
y.(x.w) = w.(y.w) - [xy].w = 0
,\]
since the first term is zero and $[xy]\in K \normal L$.
:::

Since $z\actson W$ nilpotently, choose an eigenvector $v$ for $z$ in $W$ for the eigenvalue zero.
Then $z.v=0$, so $Lv=0$.
:::

:::{.corollary title="Engel's theorem"}
If all elements of a Lie algebra $L$ are ad-nilpotent, then $L$ is nilpotent as an algebra.
:::

:::{.proof title="?"}
Induct on $\dim L$.
Note that $\ad(L) \leq \liegl(V)$ consists of nilpotent endomorphisms.
Use the theorem to pick $x\in L$ such that $\ad(L).x = 0$, i.e. $[L, x] = 0$, i.e. $x\in Z(L)$ and thus $Z(L)$ is nonzero.
Now $\dim L/Z(L) < \dim L$, and a fortiori its elements are still ad-nilpotent so $L/Z(L)$ is nilpotent.
By proposition 3.2b, $L$ is nilpotent.[^ses_works]

[^ses_works]: 
Note that for arbitrary SESs, the 2-out-of-3 property does not hold for nilpotency, but for the special cases of a quotient by the center it does.

:::

:::{.corollary title="?"}
Let $o\neq L \leq \liegl(V)$ with $\dim V < \infty$ be a Lie algebra of nilpotent endomorphisms (as in the theorem).[^assumption_nilp_endo]
Then $V$ has a basis in which the matrices of $L$ are all strictly upper triangular.

[^assumption_nilp_endo]: 
Note that the assumption is not that $L$ is a nilpotent algebra, but rather the stronger assumption on endomorphisms.

:::

:::{.proof title="?"}
Induct on $\dim V$.
Use the theorem to pick a nonzero $v_1$ with $Lv_1=0$.
Consider $W\da V/\FF v_1$, and view $L \subseteq \Endo(V)$ as a subspace of $\Endo(W)$ -- these are still nilpotent endomorphisms.
By the IH, $W$ has a basis $\ts{\bar v_i}_{2\leq i \leq n}$ with respect to the matrices in $L$ (viewed as a subspace of $\Endo(W)$) are strictly upper triangular.
Let $\ts{v_i} \subseteq V$ be their preimages in $L$; this basis has the desired properties.
This results in a matrix of the following form:

![](figures/2022-08-29_09-38-58.png)

:::

## Chapter 4: Theorems of Lie and Cartan

:::{.remark}
From now on, assume $\FF = \bar\FF$ is algebraically closed and $\characteristic(k) = 0$.
:::

## 4.1: Lie's Theorem

:::{.theorem title="Lie's Theorem"}
Let $L\neq 0$ be a solvable Lie subalgebra of $\liegl(V)$ with $\dim V < \infty$.
Then $V$ contains a common eigenvector for all of $L$.
:::

:::{.proof title="?"}
Induct on $\dim L$.
If $\dim L = 1$, then $L$ is spanned by 1 linearly operator $x$ and over an algebraically closed field, $x$ has at least one eigenvector.
For $\dim L > 1$, take the following strategy:

1. Identify $K\normal L$ proper of codimension 1,
2. By IH, find a common eigenvector for $K$,
3. Verify that $L$ stabilizes "the" subspace of all such common eigenvectors (much harder than before!)
4. In this subspace, find an eigenvector for some $z\in L\sm K$ with $L = K \oplus \FF z$.

Off we go!

**Step 1**:
Since $L$ is solvable, we have $[LL]$ properly contained in $L$.
In $L/[LL]$ choose any codimension 1 subspace -- it is an ideal, which lifts to a codimension 1 ideal $K \subset L$.

**Step 2**:
Since subalgebras of solvable algebras are again solvable, $K$ is solvable.
By the IH, pick a common nonzero eigenvector $v$ for $K$.
There exists a linear map $\lambda: K\to \FF$ such that $x.v = \lambda(x) v$ for all $x\in K$.
Let $W \da \ts{v\in V \st y.v = \lambda(y) v\,\,\forall y\in K}$, which is nonzero.

**Step 3**:
Note $L.W \subseteq W$.
Let $w\in W, x\in L, y\in K$; we WTS $y.(x.w) = \lambda(y)x.w$.
Write
\[
y.(x.w)
&= x.(y.w) - [xy].w \\
&= \lambda(y)(x.w) - \lambda([xy])w
,\]
where the second line follows since $[xy]\in K$.
We then need \( \lambda([xy]) = 0 \) for all $x\in L$ and $y\in K$.
Since $\dim V < \infty$, choose $n$ minimal such that $\ts{w, x.w, x^2.w,\cdots, x^n.w}$ is linearly dependent.
Set $W_i \da \spanof_\FF\ts{w, x.w, \cdots, x^i.w}$, so $W_0 = 0, W_1 = \spanof_\FF\ts{w}$, and so on, noting that 

- $\dim W_n = n$,
- $W_{n+k} = W_n$ for all $k\geq 0$,
- $x.W_n \subseteq W_n$.

:::{.claim}
For all $y\in K$,

\[
y.x^i.w = \lambda(y) x^i.w \mod W_i
.\]

:::

To be continued!
:::