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Weyl's theorem on complete reducibility

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 ${\displaystyle \pi :{\mathfrak {g}}\to {\mathfrak {gl}}($V)} , let ${\displaystyle A\subset \operatorname {End} ($V)} be the associative subalgebra of the endomorphism algebra of V generated by ${\displaystyle \pi ({\mathfrak {g}})}$. The ring A is called the enveloping algebra of ${\displaystyle \pi }$. If ${\displaystyle \pi }$ is semisimple, then A is semisimple.^[2] (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 ${\displaystyle JV\subset V}$ implies that ${\displaystyle JV=0}$. 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 ${\displaystyle {\mathfrak {g}}}$-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[edit]

Here is a typical application.^[3]

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

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

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

Proofs[edit]

Analytic proof[edit]

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 ${\displaystyle K}$.^[4] (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,^[5] 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.^[6] 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.

Algebraic proof 1[edit]

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 derivationif${\displaystyle f([x,y])=x\cdot f($y)-y\cdot f(x)} . The proof is essentially due to Whitehead.^[7]

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 ${\displaystyle W}$. ${\displaystyle \square }$

Algebraic proof 2[edit]

Whitehead's lemma is typically proved by means of the quadratic Casimir element of the universal enveloping algebra,^[8] 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 ${\displaystyle c_{\lambda }}$isnonzero whenever the representation is nontrivial. This can be done by a general argument ^[9] 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.

Algebraic proof 3[edit]

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.^[10] 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.^{[clarification needed]}). 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.^{[clarification needed]} ${\displaystyle \square }$

Algebraic proof 4[edit]

This section needs expansion. You can help by adding to it. (July 2024)

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

External links[edit]

• Ablog post by Akhil Mathew

References[edit]

• Hall, Brian C. (2015). Lie Groups, Lie Algebras, and Representations: An Elementary Introduction. Graduate Texts in Mathematics. Vol. 222 (2nd ed.). Springer. ISBN 978-3319134666.
• Humphreys, James E. (1973). Introduction to Lie Algebras and Representation Theory. Graduate Texts in Mathematics. Vol. 9 (Second printing, revised ed.). New York: Springer-Verlag. ISBN 0-387-90053-5.
• Jacobson, Nathan (1979). Lie algebras. New York: Dover Publications, Inc. ISBN 0-486-63832-4. Republication of the 1962 original.
• Kac, Victor (1990). Infinite dimensional Lie algebras (3rd ed.). Cambridge University Press. ISBN 0-521-46693-8.
• Knapp, Anthony W. (2002), Lie Groups Beyond an Introduction, Progress in Mathematics, vol. 140 (2nd ed.), Boston: Birkhäuser, ISBN 0-8176-4259-5
• Weibel, Charles A. (1995). An Introduction to Homological Algebra. Cambridge University Press.