In algebra, Weyl's theorem on complete reducibility is a fundamental result in the theory of Lie algebra representations. Let be a semisimple Lie algebra over a field of characteristic zero. The theorem states that every finite-dimensional module over is semisimple as a module
Weyl's theorem implies that the enveloping algebra of a finite-dimensional representation is a semisimple ring in the following way. Given a finite-dimensional Lie algebra representation, let be the associative subalgebra of the endomorphism algebra of V generated by. The ring A is called the enveloping algebra of. If is semisimple, then A is semisimple. Conversely, if A is semisimple, then V is a semisimple A-module; i.e., semisimple as a -module.
Here is a typical application. Proof: First we prove the special case of and when is the inclusion; i.e., is a subalgebra of. Let be the Jordan decomposition of the endomorphism, where are semisimple and nilpotent endomorphisms in. Now, also has the Jordan decomposition, which can be shown to respect the above Jordan decomposition; i.e., are the semisimple and nilpotent parts of. Since are polynomials in then, we see. Thus, they are derivations of. Since is semisimple, we can find elements in such that and similar for. Now, let A be the enveloping algebra of ; i.e., the subalgebra of the endomorphism algebra of V generated by. As noted above, A has zero Jacobson radical. Since, we see that is a nilpotent element in the center of A. But, in general, a central nilpotent belongs to the Jacobson radical; hence, and thus also. This proves the special case. In general, is semisimple when is semisimple. This immediately gives and.
Proofs
Analytic proof
Weyl's original proof was analytic in nature: it famously used the unitarian trick. Specifically, one can show that every complex semisimple Lie algebra is the complexification of the Lie algebra of a simply connected compact Lie group. Given a representation of on a vector space one can first restrict to the Lie algebra of. Then, since is simply connected, there is an associated representation of. Integration over produces an inner product on for which is unitary. Complete reducibility of is then immediate and elementary arguments show that the original representation of is also completely reducible.
Algebraic proof 1
Let be a finite-dimensional representation of a Lie algebra over a field of characteristic zero. The theorem is an easy consequence of Whitehead's lemma, which says is surjective, where a linear map is a derivation if. The proof is essentially due to Whitehead. Let be a subrepresentation. Consider the vector subspace that consists of all linear maps such that and. It has a structure of a -module given by: for, Now, pick some projection onto W and consider given by. Since is a derivation, by Whitehead's lemma, we can write for some. We then have ; that is to say is -linear. Also, as t kills, is an idempotent such that. The kernel of is then a complementary representation to. See also Weibel's homological algebra book.
Algebraic proof 2
is typically proved by means of the quadratic Casimir element of the universal enveloping algebra, and there is also a proof of the theorem that uses the Casimir element directly instead of Whitehead's lemma. Since the quadratic Casimir element is in the center of the universal enveloping algebra, Schur's lemma tells us that acts as multiple of the identity in the irreducible representation of with highest weight. A key point is to establish that is nonzero whenever the representation is nontrivial. This can be done by a general argument or by the explicit formula for. Consider a very special case of the theorem on complete reducibility: the case where a representation contains a nontrivial, irreducible, invariant subspace of codimension one. Let denote the action of on. Since is not irreducible, is not necessarily a multiple of the identity, but it is a self-intertwining operator for. Then the restriction of to is a nonzero multiple of the identity. But since the quotient is a one dimensional—and therefore trivial—representation of, the action of on the quotient is trivial. It then easily follows that must have a nonzero kernel—and the kernel is an invariant subspace, since is a self-intertwiner. The kernel is then a one-dimensional invariant subspace, whose intersection with is zero. Thus, is an invariant complement to, so that decomposes as a direct sum of irreducible subspaces: Although this establishes only a very special case of the desired result, this step is actually the critical one in the general argument.
