Topics in Abstract Algebra/Lie algebras

Let $$V$$ be a vector space. $$(V, [, ])$$ is called a Lie algebra if it is equipped with the bilinear operator $$V \times V \to V$$, denoted by $$[, ]$$, subject to the properties: for every $$x, y, z \in V$$ (ii) is called the Jacobi identity.
 * (i) [x, x] = 0
 * (ii) [x, [y, z]] + [y, [z, x]] + [z, [x, y]] = 0

Example: For $$x, y \in \mathbf{R}^3$$, define $$[x, y] = x \times y$$, the cross product of $$x$$ and $$y$$. The known properties of the cross products show that $$(R^3, [,])$$ is a Lie algebra.

Example: Let $$\operatorname{Der}(V) = \{ D \in \operatorname{Ext}(V) : D(x y) = (Dx) y + x Dy \}$$. A member of $$\operatorname{Der}(V)$$ is called a derivation. Define $$[x, y] = xy - yx$$. Then $$[x, y] \in \operatorname{Der}(V)$$.

Theorem Let $$V$$ be a finite-dimensional vector space.
 * (i) If $$\mathfrak{g} \subset \mathfrak{gl}_k(V)$$ is a Lie algebra consisting of nilpotent elements, then there exists $$v \in V$$ such that $$x(v) = 0$$ for every $$x \in \mathfrak{g}$$.
 * (ii) If $$\mathfrak{g}$$ is solvable, then there exists a common eigenvalue $$v \in V$$.

Theorem (Engel) $$\mathfrak{g}$$ is nilpotent if and only if $$\operatorname{ad}(x)$$ is nilpotent for every $$x \in \mathfrak{g}$$.

Proof: The direct part is clear. For the converse, note that from the preceding theorem that $$\operatorname{ad}(\mathfrak{g})$$ is a subalgebra of $$\mathfrak{n}_k$$. Thus, $$\operatorname{ad}(\mathfrak{g})$$ is nilpotent and so is $$\mathfrak{g}$$. $$\square$$

Theorem $$\mathfrak{g}$$ is solvable if and only if $$[\mathfrak{g}, \mathfrak{g}]$$ is nilpotent.

Proof: Suppose $$\mathfrak{g}$$ is solvable. Then $$\operatorname{ad}[\mathfrak{g}, \mathfrak{g}]$$ is a subalgebra of $$\mathfrak{b}_k$$. Thus, $$\operatorname{ad}[\mathfrak{g}, \mathfrak{g}] \subset \mathfrak{n}_k$$. Hence, $$\operatorname{ad}[\mathfrak{g}, \mathfrak{g}]$$ is nilpotent, and so $$[\mathfrak{g}, \mathfrak{g}]$$ is nilpotent. For the converse, note the exact sequence:
 * $$0 \longrightarrow [\mathfrak{g}, \mathfrak{g}] \longrightarrow \mathfrak{g} \longrightarrow \mathfrak{g} / {[\mathfrak{g}, \mathfrak{g}]} \longrightarrow 0$$

Since both $$[\mathfrak{g}, \mathfrak{g}]$$ and $$\mathfrak{g} / [\mathfrak{g}, \mathfrak{g}]$$ are solvable, $$\mathfrak{g}$$ is solvable. $$\square$$

3 Therorem (Weyl's theorem) Every representation of a finite-dimensional semisimple Lie algebra:
 * $$\mathfrak g \to \operatorname{End}(V)$$

is completely reducible.

Proof: It suffices to prove that every $$\mathfrak g$$-submodule has a $$\mathfrak g$$-submodule complement. Furthermore, the proof reduces to the case when $$W$$ is simple (as a module) and has codimension one. Indeed, given a $$\mathfrak g$$-submodule $$W$$, let $$E \subset \operatorname{Hom}(V, W)$$ be the subspace consisting of elements $$f$$ such that $$f|_W$$ is a scalar multiplication. Since any commutator of elements $$f \in E$$ is zero (that is, multiplication by zero), it is clear that $$E / [E, E]$$ has dimension 1. $$E$$ may not be simple, but by induction on the dimension of $$E$$, we can assume that. Hence, $$E$$ has complement of dimension 1, which is spanned by, say, $$f$$. It follows that $$V$$ is the direct sum of $$W$$ and the kernel of $$f$$. Now, to complete the proof, let $$W$$ be a simple $$\mathfrak g$$-submodule of codimension 1. Let $$c$$ be a Casimir element of $$\mathfrak g \to \operatorname{End}(V)$$. It follows that $$V$$ is the direct sum of $$W$$ and the kernel of $$c$$. $$\square$$ (TODO: obviously, the proof is very sketchy; we need more details.)