# Friday, September 02

## 4.2: Jordan-Chevalley decomposition

:::{.remark}
Let $\FF = \bar\FF$ of arbitrary characteristic and $V\in\Vect\slice\FF^\fd$ with $x\in \Endo_\FF(V)$.
The JCF of $x$ is of the form $D+N$ where $D$ is diagonal and $N$ is nilpotent where $D, N$ commute.
Recall $x$ is semisimple (diagonalizable) iff the minimal polynomial of $x$ has distinct roots.
:::

:::{.proposition title="?"}
If $x\in \Endo(V)$,

a. There is a decomposition $x = x_s + x_n$ where $x_s$ is semisimple and $x_n$ is nilpotent.
  This is unique subject to the condition that $x_s, x_n$ commute.

b. There are polynomials $p(T), q(T)$ without constant terms with $x_s = p(x), x_n = q(x)$.
  In particular, $x_s, x_n$ commute with any endomorphism which commutes with $x$.

:::

:::{.lemma title="?"}
Let $x\in \liegl(V)$ with Jordan decomposition $x = x_s + x_n$. 
Then $\ad_x = \ad_{x_s} + \ad_{x_n}$ is the Jordan decomposition of $\ad_x$ in $\Endo(\Endo(V))$.
:::

:::{.proof title="?"}
If $x\in \liegl(V)$ is semisimple then so is $\ad_x$ since the eigenvalues for $\ad_x$ are differences of eigenvalues of $x$.
I.e. if $\ts{v_1,\cdots, v_n}$ is an eigenbasis for $V$ and $x.v_i = a_i v_i$ in this bases, we have $[x e_{ij}] = (a_i - a_j) = e_{ij}$, so $\ts{e_{ij}}$ is an eigenbasis for $\ad_x$.
If $x$ is nilpotent then $\ad_x$ is nilpotent, since $\ad_x(y) = \lambda_x(y) - \rho_x(y)$ where \( \lambda, \rho \) are left/right multiplication, and sums of nilpotents are nilpotent.
One can check $[\ad_{x_s} \ad_{x_n}] = \ad_{[x_s x_n} = 0$ since they commute.
:::

:::{.remark}
One can show that if $L$ is semisimple then $\ad(L) = \Der(L)$, which is used to show that if $L$ is an arbitrary Lie algebra then one has 

- $x = x_s + x_n$,
- $[x_s x_n] = 0$,
- $\ad_{x_s}$ is semisimple and $\ad_{x_n}$ is nilpotent.

This gives a notion of semisimplicity and nilpotency for Lie algebras not of the form $\liegl(V)$.
:::

:::{.lemma title="?"}
Let $U\in \Alg\slice{\FF}^\fd$, then $\Der(U)$ is closed under taking semisimple and nilpotent parts.
:::

:::{.proof title="?"}
Let \( \delta\in \Der(U) \) and write $\delta = \sigma + v$ be the Jordan decomposition of $\delta$ in $\Endo(U)$.
It STS $\sigma$ is a derivation, so for $a\in \FF$ define 
\[
U_a \da \ts{x\in U \st (\delta - a)^k x = 0 \,\,\text{for some } k}
.\]
Note $U = \bigoplus _{a\in \Lambda} U_a$ where $\Lambda$ is the set of eigenvalues of $\delta$, which are also the eigenvalues of $\sigma$ -- this is because $\sigma, v$ are commuting operators, so eigenvalues of $\delta$ are sums of eigenvalues of $s$ and $v$.


:::{.claim}
For any $a,b\in \FF$, $U_a U_b \subseteq U_{a+b}$.
:::

Assuming this, it STS \( \sigma(xy) = \sigma(x)y + x \sigma(y) \) when $x\in U_a, y\in U_b$ where $a,b$ are eigenvalues.
Using that eigenvalues of $\delta$ are also eigenvalues of $\sigma$, since $xy\in U_{a+b}$ by the claim, \( \sigma(xy) = (a+b)xy \) and thus
\[
\sigma(x)y + x \sigma(y) = axy + xby = (a+b)xy
.\]
So \( \sigma\in \Der(U) \).
:::

:::{.proof title="of claim"}
A sub-claim:
\[
(\delta - (a+b) 1) (xy) = \sum_{0\leq i\leq n} {n\choose i} (\delta - aI)^{n-i}x (\delta- b 1)^i y
.\]
:::