Topics in Abstract Algebra/Lie algebras

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Let be a vector space. is called a Lie algebra if it is equipped with the bilinear operator , denoted by , subject to the properties: for every

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

(ii) is called the Jacobi identity.

Example: For , define , the cross product of and . The known properties of the cross products show that is a Lie algebra.

Example: Let . A member of is called a derivation. Define . Then .

Theorem Let be a finite-dimensional vector space.

  • (i) If is a Lie algebra consisting of nilpotent elements, then there exists such that for every .
  • (ii) If is solvable, then there exists a common eigenvalue .

Theorem (Engel) is nilpotent if and only if is nilpotent for every .
Proof: The direct part is clear. For the converse, note that from the preceding theorem that is a subalgebra of . Thus, is nilpotent and so is .

Theorem is solvable if and only if is nilpotent.
Proof: Suppose is solvable. Then is a subalgebra of . Thus, . Hence, is nilpotent, and so is nilpotent. For the converse, note the exact sequence:

Since both and are solvable, is solvable.

3 Therorem (Weyl's theorem) Every representation of a finite-dimensional semisimple Lie algebra:

is completely reducible.
Proof: It suffices to prove that every -submodule has a -submodule complement. Furthermore, the proof reduces to the case when is simple (as a module) and has codimension one. Indeed, given a -submodule , let be the subspace consisting of elements such that is a scalar multiplication. Since any commutator of elements is zero (that is, multiplication by zero), it is clear that has dimension 1. may not be simple, but by induction on the dimension of , we can assume that. Hence, has complement of dimension 1, which is spanned by, say, . It follows that is the direct sum of and the kernel of . Now, to complete the proof, let be a simple -submodule of codimension 1. Let be a Casimir element of . It follows that is the direct sum of and the kernel of . (TODO: obviously, the proof is very sketchy; we need more details.)