Weyl's theorem on complete reducibility

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 ${\displaystyle {\mathfrak {g}}}$ is the complexification of the Lie algebra of a simply connected compact Lie group ${\displaystyle K}$.[5] (If, for example, ${\displaystyle {\mathfrak {g}}=\mathrm {sl} (n;\mathbb {C} )}$, then ${\displaystyle K=\mathrm {SU} (n)}$.) Given a representation ${\displaystyle \pi }$ of ${\displaystyle {\mathfrak {g}}}$ on a vector space ${\displaystyle V,}$ one can first restrict ${\displaystyle \pi }$ to the Lie algebra ${\displaystyle {\mathfrak {k}}}$ of ${\displaystyle K}$. Then, since ${\displaystyle K}$ is simply connected,[6] there is an associated representation ${\displaystyle \Pi }$ of ${\displaystyle K}$. Integration over ${\displaystyle K}$ produces an inner product on ${\displaystyle V}$ for which ${\displaystyle \Pi }$ is unitary.[7] Complete reducibility of ${\displaystyle \Pi }$ is then immediate and elementary arguments show that the original representation ${\displaystyle \pi }$ of ${\displaystyle {\mathfrak {g}}}$ is also completely reducible.

Let ${\displaystyle (\pi ,V)}$ be a finite-dimensional representation of a Lie algebra ${\displaystyle {\mathfrak {g}}}$ over a field of characteristic zero. The theorem is an easy consequence of Whitehead's lemma, which says ${\displaystyle V\to \operatorname {Der} ({\mathfrak {g}},V),v\mapsto \cdot v}$ is surjective, where a linear map ${\displaystyle f:{\mathfrak {g}}\to V}$ is a derivation if ${\displaystyle f([x,y])=x\cdot f(y)-y\cdot f(x)}$. The proof is essentially due to Whitehead.[8]

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

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

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

See also Weibel's homological algebra book.

Whitehead's lemma is typically proved by means of the quadratic Casimir element of the universal enveloping algebra,[9] 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 ${\displaystyle C}$ is in the center of the universal enveloping algebra, Schur's lemma tells us that ${\displaystyle C}$ acts as multiple ${\displaystyle c_{\lambda }}$ of the identity in the irreducible representation of ${\displaystyle {\mathfrak {g}}}$ with highest weight ${\displaystyle \lambda }$. A key point is to establish that ${\displaystyle c_{\lambda }}$ is nonzero whenever the representation is nontrivial. This can be done by a general argument [10] or by the explicit formula for ${\displaystyle c_{\lambda }}$.

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

${\displaystyle V=W\oplus \mathrm {ker} (C_{V})}$.

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

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.[11] This approach has an advantage that it can be used to weaken the finite-dimensionality assumptions (on algebra and representation).

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

${\displaystyle V^{0}=\bigoplus _{\lambda \in L}V_{\lambda }^{0}}$

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