# Wednesday September 4

## 4.3: Cartan's Criterion

**Lemma:**
Let $A\subset B$ be two subspaces of $\liegl(V)$ where $\dim V < \infty$.
Set $M = \theset{x \in \liegl(V) \suchthat [x, B] \subset A}$.
Suppose that $x\in M$ satisfies $\Tr(xy) = 0$ for all $y\in M$.
Then $x$ is nilpotent.

*Proof:*
Let $x = s+n$ (where $s = x_s$ and $n = x_n$) be be the Jordan decomposition of $x$.
Fix a basis $v_1 \cdots v_m$ of $V$ relative to which $s$ has matrix $\mathrm{diag}(a_1 \cdots a_m)$.
Let $E$ be the vector subspace of $F$ over the prime field $Q$ spanned by the eigenvalues $a_1 \cdots a_m$.
We have to show that $s = 0$, or equivalently that $E= 0$, since $E$ has finite $Q\dash$dimension by construction.
It will suffice to show that the dual space $E\dual$ is 0, i.e. that any linear functional $f: E \to Q$ is zero.

Given $f$, let $y$ be the element of $\liegl(V)$ whose matrix is given by $\mathrm{diag{(f(a_1), \cdots f(a_m))}}$.
If $\theset{e_{ij}}$ is a basis of $\liegl(V)$, then $\ad_s(e_{ij}) = (a_i - a_j)e_{ij}$ and $\ad_y(e_{ij}) = (f(a_i) - f(a_j)) e_{ij}$.

Now let $r(t) \in F[t]$ be a polynomial with no constant term, satisfying $r(a_i - a_j) = f(a_i) - f(a_j)$ for all pairs $i,j$.
The existence of such $r(t)$ follows from Lagrange interpolation, and the fact that if $a_i = a_j$ then $0 = r(a_j) - r(a_i) = r(a_i - a_j) = r(0)$, so $r$ has no constant term. 
Thus there is no ambiguity in the assigned values, since $a_i -a_j = a_j - a_l$ would imply (by linearity of $f$) that $f(a_i) - f(a_j) = f(a_k) - f(a_l)$.
Thus $\ad_y = r(\ad_s)$.

> Note that Lagrange Interpolation is a special case of the Chinese Remainder Theorem for polynomials. If all $x_i$s are distinct, then $p_i(x) = x - x_i$ are all pairwise coprime. Then dividing $\frac{p(x)}{p_i(x)} = p(x_i)$. So letting $A_1 \cdots A_k$ be constants in $k$, there is a unique polynomial of degree less than $k$ such that $p(x_i) = A_i$. Thus there is a polynomial $p(x)$ such that $p(x) = A_i \mod p_i(x)$, and $p(x_i) = A_i$.

Now $\ad_s$ is the semisimple part of $\ad_x$.
By lemma A of 4.2, $\ad_s$ can be written as a polynomial in $\ad_x$ without a constant term.
Therefore $\ad_y$ is also a polynomial in $\ad_x$ without constant term.
By hypothesis, $\ad_x$ maps $B$ into $A$, so we have $\ad_y(B) \subset A$, and so $y\in M$.
Using the hypothesis of the lemma, $\Tr(xy) = 0$, and so $\sum a_i f(a_i) = 0$.
The left side is a $Q\dash$linear combination of elements of $E$.
Applying $f$, we obtain $\sum f(a_i)^2 = 0$. 
But the numbers $f(a_i)$ are rational, so this forces all of them to be zero.
Finally, $f$ must be identically 0 because the $a_i$ span $E$. $\qed$

> Note that $\Tr([x,y]z) = \Tr(x [y, z])$. To verify this, write $[x,y]z = xyz - yxz$ and $x[y,z] = xyz - xzy$, then use the fact that $\Tr(y(xz)) = \Tr((xz)y)$.

**Theorem (Cartan's Criterion):**
Let $L \leq \liegl(V)$ be a subalgebra with $V$ finite dimensional.
Suppose $\Tr(xy) = 0$ for all $x \in [L, L]$ and $y\in L$.
Then $L$ is solvable.

*Proof:*
It suffices to show that $[L, L]$ is nilpotent, or just that all $x\in [L, L]$ are nilpotent endomorphisms.
We apply the above lemma, with $V$ as given, $A = [L, L]$, and $B = L$, so $M = \theset{x\in\liegl(V)\suchthat [x, L] \subset [L, L]}$.
We have $L \subset M$.
Our hypothesis is that $\Tr(xy) = 0$ for all $x\in [L, L]$ and $y\in L$.
To use the lemma to reach the desired conclusion, we need a stronger result: that $\Tr(xy) = 0$ for $x\in [L, L]$ and $y\in M$.

If $[x,y]$ is a generator of $[L, L]$ and $z\in M$, then $\Tr([x, y]z) = \Tr(x[y,z]) = \Tr([y,z]x)$.
By definition of $M$, $[y, z] \in [L, L]$, so the right side is 0 by hypothesis.

**Corollary:**
Let $L$ be a Lie algebraic such that $\Tr(\ad_x \circ \ad_y) = 0$ for all $x\in [L, L], y\in L$. 
Then $L$ is solvable.

*Proof:* 
Apply the theorem to the adjoint representation of $L$.
We then get $\ad L$ is solvable.
Since $\ker \ad = Z(L)$ is also solvable, $L$ itself is solvable.

## Killing Form

### Criterion for Semisimplicity

Let $L$ be any lie algebra. 
If $x,y \in L$, then define $\kappa(x, y) = \Tr(\ad_x \circ \ad_y)$. 
Then $k$ is a symmetric bilinear form on $L$, called the **killing form**.

**Theorem:**
$\lieg$ is solvable $\iff \kappa(x, y) = 0$ for all $x\in [\lieg, \lieg], y\in \lieg$.

*Proof:*

$\impliedby$: By Cartan's Criterion.

$\implies$: Exercise.

*Example:*
The killing form of $\liesl(2, F)$.

We have
\begin{align*}
x = \left(\begin{array}{ll}{0} & {1} \\ {0} & {0}\end{array}\right) \\
h = \left(\begin{array}{rr}
1 & 0 \\
0 & -1
\end{array}\right) \\
y = \left(\begin{array}{rr}
0 & 0 \\
1 & 0
\end{array}\right)
.\end{align*}

Then $\ad_h = \mathrm{diag}(2, 0, -2)$, and
\begin{align*}
\ad_x = \left(\begin{array}{ccc}{0} & {-2} & {0} \\ {0} & {0} & {1} \\ {0} & {0} & {0}\end{array}\right) \\
\ad_y = =\left(\begin{array}{ccc}{0} & {0} & {0} \\ {-1} & {0} & {0} \\ {0} & {2} & {0}\end{array}\right)
.\end{align*}

and thus $k$ has the matrix
\begin{align*}
\left(\begin{array}{lll}{0} & {0} & {4} \\ {0} & {8} & {0} \\ {4} & {0} & {0}\end{array}\right)
.\end{align*}

where $k_{ij} = \kappa(x_i, x_j)$ where $x_i$ is a basis of $L$.