# Friday August 30

Review of bilinear forms: let $V = \FF^n$.

**Definition:** 
A bilinear form $\beta: V^2 \to \FF$ can be represented by a matrix $B$ with respect to a basis $\theset{\vector v_i}$ such that
$$
\beta \beta(\sum a_i \vector v_i, \sum b_i \vector v_i) = (a_1 ~a_2 ~\cdots) B (b_1 ~b_2 ~\cdots)
$$


- $\beta$ is *symmetric* iff $\beta(a,b) = \beta(b,a)$.
- $\beta$ is *symplectic* iff $\beta(a, b) = -\beta(b, a)$.
- $\beta$ is *isotropic*  iff $\beta(a, a) = 0$.

For a subspace $U \leq V$, define
$$
U^\perp \coloneqq \theset{\vector v\in V \suchthat \beta(\vector u, \vector v) = \vector 0 ~\forall u\in U}.
$$

> Note: in general, left/right orthogonality are distinguished, but these will be identical when $\beta$ is symmetric/symplectic.

The form $\beta$ is said to be *non-degenerate* iff $V^\perp = 0$ iff $\det B \neq 0$.

Assume $F$ is an algebraically closed field, so $\overline F = F$, and $\mathrm{char} F \neq 2$, then

- If $\beta$ is non-degenerate and symmetric, then $B \sim I_n$
- If $\beta$ is non-degenerate and symplectic, then $B \sim [0, I_{n/2}; I_{n/2}, 0]$.

*Remark:*
$\lieso(n, \FF) = \theset{x\in \liegl(n, F) \suchthat \beta(x(u), v) = -\beta(u, x(v))}$, where $B$ has the matrix $[0, I; I, 0]$ if $n$ is odd, or this matrix with a 1 in the top-left corner if $n$ is odd.

Similarly, $\mathfrak{sp}(2m, \FF)$ can be described this way with the matrix $[0, -I_m; -I_m, 0]$.

**Overview:**
The killing form is defined as $\kappa: \lieg^2 \to \FF$ where $\kappa(x, y) = \mathrm{tr}(\ad_x \circ \ad_y)$.

Then we have **Cartan's Criteria**:

- $\lieg$ solvable $\iff$ $\kappa(x,y) = 0 \forall x\in [\lieg, \lieg], y\in \lieg$.
- $\lieg$ semisimple $\iff \kappa$ is non-degenerate.

Note that if $\lieg$ is semisimple, then $\lieg = \bigoplus_i I_i$ with each $I_i \normal \lieg$ and simple.

## Cartan's Criteria

Some facts:

1. $\kappa$ is symmetric
2. If $\lieg$ is finite dimensional, then $\kappa$ is associative, i.e $\kappa([x,y], z) = \kappa(x, [y, z])$.

*Exercise:* 
Show that if $I \normal \lieg$, then $I^\perp \leq \lieg$ is an ideal.

*Proof of (2):*
In section 4.3, it was shown that $\mathrm{tr}([a,b] \circ c) = \mathrm{tr}(a \circ [b, c])$ for all $a,b,c \in \mathrm{End}(V)$ (provided $V$ is finite dimensional).

So
\begin{align*}
\kappa([x,y], z) &= \mathrm{tr}(\ad_{[x, y]} \circ \ad_z) \\
&= \mathrm{tr}([\ad_x, \ad_y] \circ \ad_z) \\
&= \mathrm{tr}(\ad_x \circ [\ad_y, \ad_z]) \\
&= \mathrm{tr}(\ad_x \circ \ad_{[y, z]} ) \\
&= \mathrm{tr}(x, [y, z]).
.\end{align*}

**Theorem:**
$\lieg$ is semisimple iff $\kappa$ is nondegenerate.

*Proof:*

$\implies$: We want to show that $\lieg^\perp = 0$. Note that $[\lieg^\perp, \lieg^\perp] \subseteq \lieg$, and so for all $x\in [\lieg^\perp, \lieg^\perp]$ and for any $y\in \lieg^\perp$, we have
$$
\kappa(x, y) = \mathrm{tr}(\ad_x \circ \ad_y) = 0
$$
by the const(?) of $\lieg^\perp$. 
This implies $\lieg^\perp$ is solvable.

Using fact (2), we have $\lieg^\perp \normal \lieg$ and thus $\lieg^\perp \subseteq \mathrm{rad}(\lieg)$, which is 0 since because $\lieg$ is semisimple. 
So either $\lieg^\perp = 0$ or $\kappa$ is nondegenerate.

> Used the fact that the radical was a maximal solvable ideal.

$\impliedby$: We want to show that for all $I \normal \lieg$ where $[I, I] = 0$, we have $I^\perp \subseteq \lieg^\perp$.

For $x\in I, y\in \lieg$, we have
$$
(\ad_x \circ \ad_y)^2 = \lieg \mapsvia{\ad_y} \lieg \mapsvia{\ad_x} I \mapsvia{\ad_y} I \mapsvia{\ad_x} 0
$$

And thus $\mathrm{tr}(\ad_x \circ \ad_y) = 0$ and $I \subseteq \lieg^\perp$.


Suppose that $\lieg$ is *not* semisimple. Then there exists a solvable ideal $J \neq 0$ such that the last term $J^i$ in the derived series is an ideal $I \normal \lieg$ such that $[I, I] = 0$, forcing $J^i \subset \lieg^\perp = 0$, which is a contradiction.

## Section 5.2

**Theorem:**
If $\lieg$ is semisimple, then

a. There exist ideals $I_i \normal \lieg$ which are simple Lie algebras satisfying $\lieg = \bigoplus I_i$. Note that $[I_i, I_j] \subseteq I_i \intersect I_j = 0$, since direct summands intersect only trivially.

b. Every simple $I \normal \lieg$ is one of these $I_i$.

c. $\kappa_{I_i} = \restrictionof{\kappa_{\lieg}}{I_i \cross I_i}$, so

![Image](figures/2019-09-06-09:48.png)

*Remark:*
$\lieg$ is semisimple $\iff \lieg = \bigoplus_i I_i$ for some simple Lie algebras $I_i$.

$\impliedby$: For all $i, S \coloneqq \mathrm{rad} \lieg, I_i \normal I_i$ is a solvable ideal. This implies that it is 0, since $I_i$ is simple.

By definition, simple Lie algebras are not abelian.

Supposing that $S = I_i$, we would then have $[S. S] \neq 0$ since $[I_i, I_i] \neq 0$ by definition. But $[S, S] \neq S$ because $S$ is solvable, which says that $S$ is not simple (a contradiction).

Note that $[\mathrm{rad}\lieg, \lieg] \subseteq \bigoplus [\mathrm{rad} \lieg, I_i] = 0$, which forces $\mathrm{rad}\lieg \subseteq Z(\lieg)$. Since $I_i$ is simple, $Z(I_i) = 0$ for all $i$. But $Z(\lieg) = \bigoplus Z(I_i) = 0$, and this forces $\mathrm{rad}(\lieg) \subseteq Z(\lieg) \implies \mathrm{rad}\lieg = 0$. So $\lieg$ is semisimple.

> Next time -- starting the representation theory with $\liesl(2, \FF)$.