Topics in Abstract Algebra/Lie algebras

From Wikibooks, open books for an open world
Jump to: navigation, search

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

  • (i) [x, x] = 0
  • (ii) [x, [y, z]] + [y, [z, x]] + [z, [x, y]] = 0

(ii) is called the Jacobi identity.

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.)

References[edit]