Weyl's theorem on complete reducibility

In algebra, Weyl's theorem on complete reducibility is a fundamental result in the theory of Lie algebra representations (specifically in the representation theory of semisimple Lie algebras). Let $$\mathfrak{g}$$ be a semisimple Lie algebra over a field of characteristic zero. The theorem states that every finite-dimensional module over $$\mathfrak{g}$$ is semisimple as a module (i.e., a direct sum of simple modules.)

The enveloping algebra is semisimple
Weyl's theorem implies (in fact is equivalent to) that the enveloping algebra of a finite-dimensional representation is a semisimple ring in the following way.

Given a finite-dimensional Lie algebra representation $$\pi: \mathfrak{g} \to \mathfrak{gl}(V)$$, let $$A \subset \operatorname{End}(V)$$ be the associative subalgebra of the endomorphism algebra of V generated by $$\pi(\mathfrak g)$$. The ring A is called the enveloping algebra of $$\pi$$. If $$\pi$$ is semisimple, then A is semisimple. (Proof: Since A is a finite-dimensional algebra, it is an Artinian ring; in particular, the Jacobson radical J is nilpotent. If V is simple, then $$JV \subset V$$ implies that $$JV = 0$$. In general, J kills each simple submodule of V; in particular, J kills V and so J is zero.) Conversely, if A is semisimple, then V is a semisimple A-module; i.e., semisimple as a $$\mathfrak g$$-module. (Note that a module over a semisimple ring is semisimple since a module is a quotient of a free module and "semisimple" is preserved under the free and quotient constructions.)

Application: preservation of Jordan decomposition
Here is a typical application.

Proof: First we prove the special case of (i) and (ii) when $$\pi$$ is the inclusion; i.e., $$\mathfrak g$$ is a subalgebra of $$\mathfrak{gl}_n = \mathfrak{gl}(V)$$. Let $$x = S + N$$ be the Jordan decomposition of the endomorphism $$x$$, where $$S, N$$ are semisimple and nilpotent endomorphisms in $$\mathfrak{gl}_n$$. Now, $$\operatorname{ad}_{\mathfrak{gl}_n}(x)$$ also has the Jordan decomposition, which can be shown (see Jordan–Chevalley decomposition) to respect the above Jordan decomposition; i.e., $$\operatorname{ad}_{\mathfrak{gl}_n}(S), \operatorname{ad}_{\mathfrak{gl}_n}(N)$$ are the semisimple and nilpotent parts of $$\operatorname{ad}_{\mathfrak{gl}_n}(x)$$. Since $$\operatorname{ad}_{\mathfrak{gl}_n}(S), \operatorname{ad}_{\mathfrak{gl}_n}(N)$$ are polynomials in $$\operatorname{ad}_{\mathfrak{gl}_n}(x)$$ then, we see $$\operatorname{ad}_{\mathfrak{gl}_n}(S), \operatorname{ad}_{\mathfrak{gl}_n}(N) : \mathfrak g \to \mathfrak g$$. Thus, they are derivations of $$\mathfrak{g}$$. Since $$\mathfrak{g}$$ is semisimple, we can find elements $$s, n$$ in $$\mathfrak{g}$$ such that $$[y, S] = [y, s], y \in \mathfrak{g}$$ and similarly for $$n$$. Now, let A be the enveloping algebra of $$\mathfrak{g}$$; i.e., the subalgebra of the endomorphism algebra of V generated by $$\mathfrak g$$. As noted above, A has zero Jacobson radical. Since $$[y, N - n] = 0$$, we see that $$N - n$$ is a nilpotent element in the center of A. But, in general, a central nilpotent belongs to the Jacobson radical; hence, $$N = n$$ and thus also $$S = s$$. This proves the special case.

In general, $$\pi(x)$$ is semisimple (resp. nilpotent) when $$\operatorname{ad}(x)$$ is semisimple (resp. nilpotent). This immediately gives (i) and (ii). $$\square$$

Analytic proof
Weyl's original proof (for complex semisimple Lie algebras) was analytic in nature: it famously used the unitarian trick. Specifically, one can show that every complex semisimple Lie algebra $$\mathfrak{g}$$ is the complexification of the Lie algebra of a simply connected compact Lie group $$K$$. (If, for example, $$\mathfrak{g}=\mathrm{sl}(n;\mathbb{C})$$, then $$K=\mathrm{SU}(n)$$.) Given a representation $$\pi$$ of $$\mathfrak{g}$$ on a vector space $$V,$$ one can first restrict $$\pi$$ to the Lie algebra $$\mathfrak{k}$$ of $$K$$. Then, since $K$ is simply connected, there is an associated representation $$\Pi$$ of $$K$$. Integration over $$K$$ produces an inner product on $$V$$ for which $$\Pi$$ is unitary. Complete reducibility of $$\Pi$$ is then immediate and elementary arguments show that the original representation $$\pi$$ of $$\mathfrak{g}$$ is also completely reducible.

Algebraic proof 1
Let $$(\pi, V)$$ be a finite-dimensional representation of a Lie algebra $$\mathfrak g$$ over a field of characteristic zero. The theorem is an easy consequence of Whitehead's lemma, which says $$V \to \operatorname{Der}(\mathfrak g, V), v \mapsto \cdot v$$ is surjective, where a linear map $$f: \mathfrak g \to V$$ is a derivation if $$f([x, y]) = x \cdot f(y) - y \cdot f(x)$$. The proof is essentially due to Whitehead.

Let $$W \subset V$$ be a subrepresentation. Consider the vector subspace $$L_W \subset \operatorname{End}(V)$$ that consists of all linear maps $$t: V \to V$$ such that $$t(V) \subset W$$ and $$t(W) = 0$$. It has a structure of a $$\mathfrak{g}$$-module given by: for $$x \in \mathfrak{g}, t \in L_W$$,
 * $$x \cdot t = [\pi(x), t]$$.

Now, pick some projection $$p : V \to V$$ onto W and consider $$f : \mathfrak{g} \to L_W$$ given by $$f(x) = [p, \pi(x)]$$. Since $$f$$ is a derivation, by Whitehead's lemma, we can write $$f(x) = x \cdot t$$ for some $$t \in L_W$$. We then have $$[\pi(x), p + t] = 0, x \in \mathfrak{g}$$; that is to say $$p + t$$ is $$\mathfrak{g}$$-linear. Also, as t kills $$W$$, $$p + t$$ is an idempotent such that $$(p + t)(V) = W$$. The kernel of $$p + t$$ is then a complementary representation to $$W$$. $$\square$$

Algebraic proof 2
Whitehead's lemma 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 $$C$$ is in the center of the universal enveloping algebra, Schur's lemma tells us that $$C$$ acts as multiple $$c_\lambda$$ of the identity in the irreducible representation of $$\mathfrak{g}$$ with highest weight $$\lambda$$. A key point is to establish that $$c_\lambda$$ is nonzero whenever the representation is nontrivial. This can be done by a general argument or by the explicit formula for $$c_\lambda$$.

Consider a very special case of the theorem on complete reducibility: the case where a representation $$V$$ contains a nontrivial, irreducible, invariant subspace $$W$$ of codimension one. Let $$C_V$$ denote the action of $$C$$ on $$V$$. Since $$V$$ is not irreducible, $$C_V$$ is not necessarily a multiple of the identity, but it is a self-intertwining operator for $$V$$. Then the restriction of $$C_V$$ to $$W$$ is a nonzero multiple of the identity. But since the quotient $$V/W$$ is a one dimensional—and therefore trivial—representation of $$\mathfrak{g}$$, the action of $$C$$ on the quotient is trivial. It then easily follows that $$C_V$$ must have a nonzero kernel—and the kernel is an invariant subspace, since $$C_V$$ is a self-intertwiner. The kernel is then a one-dimensional invariant subspace, whose intersection with $$W$$ is zero. Thus, $$\mathrm{ker}(V_C)$$ is an invariant complement to $$W$$, so that $$V$$ decomposes as a direct sum of irreducible subspaces:
 * $$V=W\oplus\mathrm{ker}(C_V)$$.

Although this establishes only a very special case of the desired result, this step is actually the critical one in the general argument.

Algebraic proof 3
The theorem can be deduced from the theory of Verma modules, which characterizes a simple module as a quotient of a Verma module by a maximal submodule. This approach has an advantage that it can be used to weaken the finite-dimensionality assumptions (on algebra and representation).

Let $$V$$ be a finite-dimensional representation of a finite-dimensional semisimple Lie algebra $$\mathfrak g$$ over an algebraically closed field of characteristic zero. Let $$\mathfrak b = \mathfrak{h} \oplus \mathfrak{n}_+ \subset \mathfrak{g}$$ be the Borel subalgebra determined by a choice of a Cartan subalgebra and positive roots. Let $$V^0 = \{ v \in V | \mathfrak{n}_+(v) = 0 \}$$. Then $$V^0$$ is an $$\mathfrak h$$-module and thus has the $$\mathfrak h$$-weight space decomposition:
 * $$V^0 = \bigoplus_{\lambda \in L} V^0_{\lambda}$$

where $$L \subset \mathfrak{h}^*$$. For each $$\lambda \in L$$, pick $$0 \ne v_{\lambda} \in V_{\lambda}$$ and $$V^{\lambda} \subset V$$ the $$\mathfrak g$$-submodule generated by $$v_{\lambda}$$ and $$V' \subset V$$ the $$\mathfrak g$$-submodule generated by $$V^0$$. We claim: $$V = V'$$. Suppose $$V \ne V'$$. By Lie's theorem, there exists a $$\mathfrak{b}$$-weight vector in $$V/V'$$; thus, we can find an $$\mathfrak{h}$$-weight vector $$v$$ such that $$0 \ne e_i(v) \in V'$$ for some $$e_i$$ among the Chevalley generators. Now, $$e_i(v)$$ has weight $$\mu + \alpha_i$$. Since $$L$$ is partially ordered, there is a $$\lambda \in L$$ such that $$\lambda \ge \mu + \alpha_i$$; i.e., $$\lambda > \mu$$. But this is a contradiction since $$\lambda, \mu$$ are both primitive weights (it is known that the primitive weights are incomparable.). Similarly, each $$V^{\lambda}$$ is simple as a $$\mathfrak g$$-module. Indeed, if it is not simple, then, for some $$\mu < \lambda$$, $$V^0_{\mu}$$ contains some nonzero vector that is not a highest-weight vector; again a contradiction. $$\square$$

Algebraic proof 4
There is also a quick homological algebra proof; see Weibel's homological algebra book.