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 $𝔤$ be a semisimple Lie algebra over a field of characteristic zero. The theorem states that every finite-dimensional module over $𝔤$ 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 $\pi :𝔤\to 𝔤𝔩(V)$, let $A\subset \mathrm{End}(V)$ be the associative subalgebra of the endomorphism algebra of V generated by $\pi (𝔤)$. 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 $𝔤$-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.

Proposition—Let $𝔤$ be a semisimple finite-dimensional Lie algebra over a field of characteristic zero and $x$ an element of $𝔤$.

1. There exists a unique pair of elements $x_s,x_n$ in $𝔤$ such that $x=x_s+x_n$, $\mathrm{ad}(x_s)$ is semisimple, $\mathrm{ad}(x_n)$ is nilpotent and $[x_s,x_n]=0$.
2. If $\pi :𝔤\to 𝔤𝔩(V)$ is a finite-dimensional representation, then $\pi (x{)}_s=\pi (x_s)$ and $\pi (x{)}_n=\pi (x_n)$, where $\pi (x{)}_s,\pi (x{)}_n$ denote the Jordan decomposition of the semisimple and nilpotent parts of the endomorphism $\pi (x)$.

In short, the semisimple and nilpotent parts of an element of $𝔤$ are well-defined and are determined independent of a faithful finite-dimensional representation.

Proof: First we prove the special case of (i) and (ii) when $\pi$ is the inclusion; i.e., $𝔤$ is a subalgebra of ${𝔤𝔩}_n=𝔤𝔩(V)$. Let $x=S+N$ be the Jordan decomposition of the endomorphism $x$, where $S,N$ are semisimple and nilpotent endomorphisms in ${𝔤𝔩}_n$. Now, ${\mathrm{ad}}_{{𝔤𝔩}_n}(x)$ also has the Jordan decomposition, which can be shown (see Jordan–Chevalley decomposition) to respect the above Jordan decomposition; i.e., ${\mathrm{ad}}_{{𝔤𝔩}_n}(S),{\mathrm{ad}}_{{𝔤𝔩}_n}(N)$ are the semisimple and nilpotent parts of ${\mathrm{ad}}_{{𝔤𝔩}_n}(x)$. Since ${\mathrm{ad}}_{{𝔤𝔩}_n}(S),{\mathrm{ad}}_{{𝔤𝔩}_n}(N)$ are polynomials in ${\mathrm{ad}}_{{𝔤𝔩}_n}(x)$ then, we see ${\mathrm{ad}}_{{𝔤𝔩}_n}(S),{\mathrm{ad}}_{{𝔤𝔩}_n}(N):𝔤\to 𝔤$. Thus, they are derivations of $𝔤$. Since $𝔤$ is semisimple, we can find elements $s,n$ in $𝔤$ such that $[y,S]=[y,s],y\in 𝔤$ and similarly for $n$. Now, let A be the enveloping algebra of $𝔤$; i.e., the subalgebra of the endomorphism algebra of V generated by $𝔤$. 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 $\mathrm{ad}(x)$ is semisimple (resp. nilpotent). This immediately gives (i) and (ii). $◻$

Proofs

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 $𝔤$ is the complexification of the Lie algebra of a simply connected compact Lie group $K$. (If, for example, $𝔤=\mathrm{s}\mathrm{l}(n;ℂ)$, then $K=\mathrm{S}\mathrm{U}(n)$.) Given a representation $\pi$ of $𝔤$ on a vector space $V,$ one can first restrict $\pi$ to the Lie algebra $𝔨$ 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 $𝔤$ is also completely reducible.

Algebraic proof 1

Let $(\pi ,V)$ be a finite-dimensional representation of a Lie algebra $𝔤$ over a field of characteristic zero. The theorem is an easy consequence of Whitehead's lemma, which says $V\to \mathrm{Der}(𝔤,V),v\mapsto \cdot v$ is surjective, where a linear map $f:𝔤\to V$ 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 $L_W\subset \mathrm{End}(V)$ 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 $𝔤$-module given by: for $x\in 𝔤,t\in L_W$,

$x\cdot t=[\pi (x),t]$.

Now, pick some projection $p:V\to V$ onto W and consider $f:𝔤\to L_W$ 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 $[\pi (x),p+t]=0,x\in 𝔤$; that is to say $p+t$ is $𝔤$-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 $W$. $◻$

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 $𝔤$ 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 $𝔤$, 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, $\mathrm{k}\mathrm{e}\mathrm{r}(V_C)$ is an invariant complement to $W$, so that $V$ decomposes as a direct sum of irreducible subspaces:

$V=W\oplus \mathrm{k}\mathrm{e}\mathrm{r}(C_V)$.

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

$V^0=\underset{\lambda \in L}{⨁}{V}_{\lambda }^0$

where $L\subset {𝔥}^{*}$. For each $\lambda \in L$, pick $0\ne v_{\lambda }\in V_{\lambda }$ and $V^{\lambda }\subset V$ the $𝔤$-submodule generated by $v_{\lambda }$ and $V^{\prime }\subset V$ the $𝔤$-submodule generated by $V^0$. We claim: $V=V^{\prime }$. Suppose $V\ne V^{\prime }$. By Lie's theorem, there exists a $𝔟$-weight vector in $V/V^{\prime }$; thus, we can find an $𝔥$-weight vector $v$ such that $0\ne e_i(v)\in V^{\prime }$ 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 $𝔤$-module. Indeed, if it is not simple, then, for some $\mu <\lambda$, $V_{\mu }^0$ contains some nonzero vector that is not a highest-weight vector; again a contradiction. $◻$

Algebraic proof 4

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

External links

• A blog post by Akhil Mathew

References

• Republication of the 1962 original.