# Wednesday, August 31

## Section 4.1, continuing the proof

:::{.remark}
Recall $\dim L, \dim V < \infty$, $\FF$ is algebraically closed, and $\characteristic \FF = 0$.
For $L \leq \liegl(V)$ solvable, we want a common eigenvector $v\in V$ for $L$.
Steps for the proof:

1. Find a $K\normal L$ proper of codimension 1.
2. Set $W = \ts{v\in V \st x.v = \lambda(x) v \,\forall x\in K}\neq 0$ for some linear \( \lambda: K\to \FF \).
3. Show $L.W \subseteq W$; we needed to show \( \lambda([LK] ) = 0 \).

:::

:::{.proof title="Continued"}
**Step 3**:
Fix $x\in L, w\in W$ and $n$ minimal such that $\ts{x^i w}_{i\leq n}$ is linearly dependent.
For $i\geq 0$ set $W_i = \FF\gens{w, xw, \cdots, x^{i-1}w}$.
Then $\dim W_n = n, W_n = W_{n+i}$ for $i\geq 0$, and $xW_n \subseteq W_n$.


:::{.claim}
For all $y\in K$,
\[
yx^i .w = \lambda(y) x^i w \mod W_i
.\]

:::

:::{.proof title="?"}
This is proved by induction on $i$, where $i=0$ follows from how $W$ is defined.
For $i\geq 1$, use the commuting trick:
\[
yx^i . w 
&= yxx^{i-1}w \\
&= (xy - [xy]) x^{i-1} w \\
&= x(y x^{i-1} w) - [xy]x^{i-1}w \\
&\equiv \lambda(y) x^i w - \lambda([xy])x^{i-1} w \mod W_{i-1} \\
&\equiv \lambda(y) x^i w - \lambda([xy])x^{i-1} w \mod W_{i} \qquad \text{since } W_{i-1} \leq W_i \\
&\equiv \lambda(y) x^i w \mod W_i
.\]

:::

Given this claim, for $i=n$ this says that the matrices of any $y\in K$ with respect to the basis $\ts{x^iw}_{0\leq i \leq n-1}$ is upper triangular with diagonal entries all equal to \( \lambda(y) \).
Thus $\ro{\trace(y)}{W_m} = n \lambda(y)$, and so $[xy]\actson W_n$ with trace $n \lambda([xy])$.
On the other hand, $x,y$ both act on $W_n$ (e.g. by the formula in the claim for $yx^i.w$) and so 
\[
\ro{[xy]}{W_n} = \ro{xy}{W_n} - \ro{yx}{W_n}
,\]
thus $\ro{ \trace([xy])}{W_n} = 0$.
Since $\FF$ is characteristic zero, we have $n \lambda([xy]) = 0 \implies \lambda([xy]) = 0$.

**Step 4**:
By step 1, $L = K \oplus \FF z$ for some $z\in L\sm K$.
Viewing $z: W\to W$ and using $\FF = \bar\FF$, $z$ has an eigenvector $v\in W$.
Since $v\in W$, it is also a common eigenvector for $K$ and thus an eigenvector for $L$ by additivity.
:::

:::{.corollary title="A: Lie's theorem"}
Let $L\leq \liegl(V)$ be a solvable subalgebra, then $L$ stabilizes some flag in $V$.
In particular, there exists a basis for $V$ with respect to which the matrices in $L$ are all upper triangular.
:::

:::{.remark}
Recall that for $V \in\Vect\slice \FF$, a complete flag is an element of
\[
\Fl(V) \da \ts{ 0 = V^0 \subsetneq V^1 \subsetneq \cdots \subsetneq V^n = V \st \dim V^i = i}
.\]
A subalgebra $L$ **stabilizes** a flag if $LV^i \subseteq V^i$ for all $i$, which implies there is a compatible basis (got by extending one vector at a time from a basis for $V^1$) for which $L$ acts by upper triangular matrices.
:::

:::{.proof title="?"}
Use the theorem and induct on $n=\dim V$ as in Engel's theorem -- find a common eigenvector for $V^1$, since $L$ stabilizes one gets an action $L\actson V^i/V^{i-1}$ which is smaller dimension.
Then just lift through the quotient.
:::

:::{.corollary title="B"}
Let $L$ be a solvable Lie algebra, then there exists a chain of ideals 
\[
0 = L_0 \subsetneq L_1 \subsetneq \cdots \subsetneq L_n = L
\]
such that $\dim L_i = i$.
\label{thm:corB}
:::

:::{.proof title="?"}
Consider $\ad L \leq \liegl(L)$.
Apply Lie's theorem: $(\ad L)L_i \subseteq L_i \iff [LL_i] \subseteq L_i$, making $L_i\normal L$ an ideal.
:::

:::{.corollary title="C"} 
Let $L$ be solvable, then $x\in [LL]\implies \ad_L x$ is nilpotent.
Hence $[LL]$ is nilpotent by Lie's theorem.
:::

:::{.proof title="?"}
Find a flag of ideals by \autoref{thm:corB} and let $\ts{x_1,\cdots, x_n}$ be a compatible basis.
Then the matrices $\ts{\ad_x\st x\in L}$ are all upper triangular.
If $x\in [LL]$, without loss of generality $x = [yz]$ for some $y,z\in L$.
Then
\[
\ad_x = [\ad_y \ad_z] = \ad_y\ad_z - \ad_z \ad_y
\]
will be strictly upper triangular (since these are upper triangular and the commutator cancels diagonals) and hence nilpotent.
:::

## Section 4.3

:::{.remark}
We'll come back to 4.2 next time.
For this section, assume $\FF = \bar\FF$ and $\characteristic \FF = 0$.
**Cartan's criterion** for a semisimple $L$ (i.e. $\Rad(L) = 0$) involves the **Killing form**, a certain nondegenerate bilinear form on $L$.
Recall that if $L$ is solvable then $[LL]$ is nilpotent, or equivalently every $x\in [LL]$ is ad-nilpotent.
:::

:::{.lemma title="Checking nilpotency with a trace condition (technical)"}
Let $A \subseteq B$ be subspaces of $\liegl(V)$ (really $\Endo(V)$ as a vector space) with $V$ finite-dimensional.
Let 
\[
M\da\ts{w\in \liegl(V) \st [wB] \subseteq A}
\]
and suppose some $w\in M$ satisfies $\trace(wz) = 0$ for all $z\in M$.
Then $w$ is nilpotent.
:::

:::{.proof title="?"}
Later!
:::

:::{.definition title="Bilinear form terminology"}
A **bilinear form** is a map
\[
\beta(\wait, \wait): L\times L\to \FF
,\]
which is **symmetric** if $\beta(x,y) = \beta(y,x)$ and **associative** if $\beta([xy], z) = \beta(x, [yz])$ for all $x,y,z\in L$.
The **radical** of $\beta$ is
\[
\Rad(\beta) \da \ts{w\in V\st \beta(w, V) = 0}
,\]
and $\beta$ is **nondegenerate** if $\Rad(\beta) = 0$.
:::

:::{.example title="?"}
For $L = \liegl(V)$, take $\beta(x,y)\da \trace(xy)$.
One can check this is symmetric, bilinear, and associative -- associativity follows from the following:
\[
[xy]z &= xyz-yxz\\
x[yz] &= xyz - xzy
.\]
Then note that $y(xz)$ and $(xz)y$ have the same trace, since $\trace(AB) = \trace(BA)$.
:::

:::{.proposition title="?"}
If $\beta$ is associative, then $\Rad(\beta) \normal L$.
:::

:::{.proof title="?"}
Let $z\in \Rad(\beta)$ and $x,y\in L$. 
To see if $[zx]\in \Rad(\beta)$, check
\[
\beta([zx], y) = \beta(z, [xy]) = 0
\]
since $z\in \Rad(\beta)$.
Thus $[zx] \in \Rad(\beta)$.
:::