# 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.
Proposition — Let ${\mathfrak {g}}$ be a semisimple finite-dimensional Lie algebra over a field of characteristic zero and $x$ an element of ${\mathfrak {g}}$.
* There exists a unique pair of elements $x_{s},x_{n}$ in ${\mathfrak {g}}$ such that $x=x_{s}+x_{n}$, $\operatorname {ad} (x_{s})$ is semisimple, $\operatorname {ad} (x_{n})$ is nilpotent and $[x_{s},x_{n}]=0$.
* If $\pi :{\mathfrak {g}}\to {\mathfrak {gl}}(V)$ 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 ${\mathfrak {g}}$ 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., ${\mathfrak {g}}$ 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). $\square$

## 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 ${\mathfrak {g}}$ 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:
    $V^{0}=\bigoplus _{\lambda \in L}V_{\lambda }^{0}$
where $L\subset {\mathfrak {h}}^{*}$. For each $\lambda \in L$, pick $0\neq v_{\lambda }\in V_{\lambda }$ 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 $V\neq V'$. 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 $0\neq e_{i}(v)\in V'$ 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 $\lambda \geq \mu +\alpha _{i}$; 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 $\mu <\lambda$, $V_{\mu }^{0}$ 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.