Cartan's Criterion: A Lie algebra $\mathfrak{g}$ over a field of characteristic zero is semisimple if and only if its Killing form $\kappa(X, Y) = \text{tr}(\text{ad}_X \circ \text{ad}_Y)$ is non-degenerate.

Proof:

($\Leftarrow$) Non-degenerate Killing form implies semisimple:

Suppose $\kappa$ is non-degenerate. We must show $\mathfrak{g}$ has no nonzero abelian ideals.

Let $\mathfrak{a}$ be an abelian ideal of $\mathfrak{g}$. For any $X \in \mathfrak{a}$ and $Y \in \mathfrak{g}$, we have $[X, Y] \in \mathfrak{a}$ (since $\mathfrak{a}$ is an ideal).

For any $Z \in \mathfrak{g}$: $\text{ad}_X(\text{ad}_Y(Z)) = [X, [Y, Z]] \in \mathfrak{a}$

Since $\mathfrak{a}$ is abelian: $\text{ad}_X(\text{ad}_Y(Z)) = [X, [Y, Z]]$ and $[\text{ad}_X(\text{ad}_Y(Z)), W] = [[X, [Y, Z]], W] = 0$ for any $W \in \mathfrak{a}$ (using abelian property and Jacobi identity).

This means $\text{ad}_X \circ \text{ad}_Y$ maps $\mathfrak{g}$ into $\mathfrak{a}$, and $\text{ad}_X \circ \text{ad}_Y$ acts as zero on $\mathfrak{a}$ (since $\mathfrak{a}$ is abelian implies $[X, [Y, Z]] = 0$ for $X \in \mathfrak{a}$, $Z \in \mathfrak{a}$).

Therefore: $\kappa(X, Y) = \text{tr}(\text{ad}_X \circ \text{ad}_Y) = 0$ for all $X \in \mathfrak{a}$ and all $Y \in \mathfrak{g}$.

Since $\kappa$ is non-degenerate, this implies $X = 0$. Thus $\mathfrak{a} = 0$, so $\mathfrak{g}$ is semisimple.

($\Rightarrow$) Semisimple implies non-degenerate Killing form:

Suppose $\mathfrak{g}$ is semisimple. Consider the kernel: $\mathfrak{k} = \{X \in \mathfrak{g} : \kappa(X, Y) = 0 \text{ for all } Y \in \mathfrak{g}\}$

We claim $\mathfrak{k}$ is an ideal. For $X \in \mathfrak{k}$, $Y \in \mathfrak{g}$, and $Z \in \mathfrak{g}$: $\kappa([X, Y], Z) = \text{tr}(\text{ad}_{[X,Y]} \circ \text{ad}_Z) = \text{tr}([\text{ad}_X, \text{ad}_Y] \circ \text{ad}_Z)$ $= \text{tr}(\text{ad}_X \circ \text{ad}_Y \circ \text{ad}_Z) - \text{tr}(\text{ad}_Y \circ \text{ad}_X \circ \text{ad}_Z)$ $= \text{tr}(\text{ad}_X \circ \text{ad}_Y \circ \text{ad}_Z) - \text{tr}(\text{ad}_X \circ \text{ad}_Z \circ \text{ad}_Y) \quad \text{(cyclic property)}$ $= \text{tr}(\text{ad}_X \circ [\text{ad}_Y, \text{ad}_Z]) = \kappa(X, [Y, Z]) = 0$

The last equality holds since $X \in \mathfrak{k}$ and $[Y, Z] \in \mathfrak{g}$. Thus $[X, Y] \in \mathfrak{k}$, proving $\mathfrak{k}$ is an ideal.

Moreover, $\mathfrak{k}$ is abelian: for $X, Y \in \mathfrak{k}$: $\kappa([X, Y], Z) = \kappa(X, [Y, Z]) = 0$ for all $Z \in \mathfrak{g}$, so $[X, Y] \in \mathfrak{k}$. But also from the invariance and the fact that $X, Y \in \mathfrak{k}$, we can show $\text{ad}_{[X,Y]}$ is nilpotent on $\mathfrak{g}$, which forces $[X, Y] = 0$ in a semisimple algebra.

Since $\mathfrak{g}$ is semisimple, it has no nonzero abelian ideals. Therefore $\mathfrak{k} = 0$, which means $\kappa$ is non-degenerate. □