# Friday, August 26

## 3.2: Engel's theorem

:::{.proposition title="Nilpotent implies solvable"}
Recall $L$ is nilpotent if $L^{n} = 0$ for some $n\geq 0$ (the descending central series) where $L^{i+1} = [LL^{i}]$.
Equivalently, $\prod_{i\leq n} \ad_{x_i} =0$ for any $\ts{x_i}_{i\leq n} \subseteq L$.
Note that $L^{i} \contains L^{(i)}$ by induction on $i$ -- these coincide for $i=0,1$, and one can check
\[
L^{(i+1)} = [L^{(i)} L^{(i)}] \subseteq [L L^{i}] = L^{i+1}
.\]
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:::{.example title="Solvable does not imply nilpotent"}
$\lieb_n$ is solvable but not nilpotent, since $\lieb_n^1 = \lien_n$ but $\lieb_n^2 = \lien_n$ and the series never terminates.
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:::{.example title="?"}
$\lien_n$ is nilpotent, since the number of diagonals with zeros adds when taking brackets $[LL^{i}]$.
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:::{.warnings}
$\lieh$ is also nilpotent, since any abelian algebra is nilpotent.
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:::{.proposition title="?"}
Let $L\in\Lie\Alg\slice\FF$, then

- If $L$ is nilpotent then any subalgebra or homomorphic image of $L$ is nilpotent.
- If $L/Z(L)$ is nilpotent then $L$ is nilpotent.[^lift_series]
- If $L\neq 0$ is nilpotent, then $Z(L) \neq 0$.[^how_to_prove]

[^how_to_prove]: 
If $L^n={n-1} \contains L^n=0$ then $[LL^{n-1}] = 0$ and thus $L^{n-1} \subseteq Z(G)$.

[^lift_series]: 
Lift a series for the quotient, which is eventually in $Z(L)$ since it was zero in the quotient, and then bracketing with $Z(L)$ terminates.

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:::{.exercise title="?"}
Show that if $L/I$ and $I\normal L$ are nilpotent, then $L$ need not be nilpotent.
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:::{.remark}
Distinguish $\Endo(L)$ whose algebra structure is given by associative multiplication and $\liegl(L)$ with the bracket multiplication.
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:::{.definition title="ad-nilpotent"}
An element $x\in L$ is **ad-nilpotent** if $\ad_x \in \Endo(L)$ is a nilpotent endomorphism.
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:::{.remark}
If $L$ is nilpotent then $x\in L$ is ad-nilpotent by taking $x_i = x$ for all $i$.
It turns out that the converse is true:
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:::{.theorem title="Engel's theorem"}
If all $x\in L$ are ad-nilpotent, then $L$ is nilpotent.
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:::{.proof title="?"}
To be covered in an upcoming section.
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:::{.lemma title="?"}
Let $x\in \liegl(V)$ be a nilpotent linear transformation for $V$ finite-dimensional.
Then 
\[
\ad_x: \liegl(V)&\to \liegl(V) \\
y &\mapsto x\circ y - y\circ x
\]
is a nilpotent operator.
:::

:::{.proof title="?"}
Let \( \lambda_x, \rho_x \in \Endo(\liegl(V)) \) be left and right multiplication by $x$, which are commuting nilpotent operators.
The binomial theorem shows that if $D_1, D_2$ are any two commuting nilpotent endomorphisms of a vector space, then $D_1\pm D_2$ is again nilpotent.
But then one can write $\ad_x = \lambda_x - \rho_x$.
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:::{.remark}
If $x\in \liegl(V)$ is nilpotent then so is $\ad_x$.
Conversely, if all $\ad_x$ for $x\in L \leq \liegl(V)$ are nilpotent operators then $L$ is nilpotent by Engel's theorem.
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:::{.warnings}
The converse of the above lemma is not necessarily true:
$x$ being ad-nilpotent does not imply that $x$ is nilpotent.
As a counterexample, take $x=I_n\in \liegl_n(\CC)$, then $\ad_x = 0$ but $x^k=x$ for any $k\geq 1$.
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## 3.3: Proof of Engel's theorem

:::{.remark}
The following is related to the classical linear algebra theorem that commuting operators admit a simultaneous eigenvector:
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:::{.theorem title="?"}
Let $L$ be a Lie subalgebra of $\liegl(V)$ for $V$ finite-dimensional. If $L$ consists of nilpotent endomorphisms, then there exists a nonzero $v\in V$ such that $Lv=0$.
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:::{.proof title="?"}
Proceed by induction on $n = \dim L$ (assuming it holds for *all* vector spaces), where the $n=1$ case is clear -- the characteristic polynomial of such an operator is $f(t) = t^n$, which has roots $t=0$ and every field contains zero.
Once one has an eigenvalue, there is at least one eigenvector.

For $n > 1$, suppose $K \leq L$ is a proper Lie subalgebra.
By hypothesis, $K$ consists of nilpotent elements in $\Endo(V)$, so apply the previous lemma to see that $\ad(K) \subseteq \Endo(L)$ acts by nilpotent endomorphisms of $L$ since they are restrictions to $L$ of nilpotent endomorphisms of $\liegl(V)$.
Since $[KK] \subseteq K$, we can view $\ad(K) \subseteq \Endo(L/K)$ where $L/K$ is a vector space.

By the IH with $V = L/K$, where $\Endo(L/K)$ has smaller dimension, one can find a nonzero $x+K \in L/K$ such that $\ad(K)(x+K)=0$.
Hence one can find an $x\in L\sm K$ such that for all $y\in K$ one has $[yx] \in K$, so $x\in N_L(K)\sm K$.
Thus $K \subsetneq N_L(K)$ is a proper containment.

> To be continued.

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