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 ${\displaystyle {\mathfrak {g}}}$ be a semisimple Lie algebra over a field of characteristic zero. The theorem states that every finite-dimensional module over ${\displaystyle {\mathfrak {g}}}$ is semisimple as a module (i.e., a direct sum of simple modules.)[1]

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 ${\displaystyle \pi$ :{\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

Here is a typical application.[3]

Proposition — Let ${\displaystyle {\mathfrak {g}}}$ be a semisimple finite-dimensional Lie algebra over a field of characteristic zero.[4]

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#Lie algebras) 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). This immediately gives (i) and (ii). ${\displaystyle \square }$

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 ${\displaystyle {\mathfrak {g}}}$ 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.

Algebraic proof 1

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.

Algebraic proof 2

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.

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.[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. ${\displaystyle \square }$

External links

• A blog post by Akhil Mathew

References

1. Hall 2015 Theorem 10.9
2. Jacobson 1962, Ch. II, § 5, Theorem 10. harvnb error: no target: CITEREFJacobson1962 (help)
3. Jacobson 1962, Ch. III, § 11, Theorem 17. harvnb error: no target: CITEREFJacobson1962 (help)
4. Editorial note: this fact is usually stated for a field of characteristic zero, but the proof needs only that the base field be perfect.
5. Knapp 2002 Theorem 6.11
6. Hall 2015 Theorem 5.10
7. Hall 2015 Theorem 4.28
8. Jacobson 1961, Ch. III, § 7. harvnb error: no target: CITEREFJacobson1961 (help)
9. Hall 2015 Section 10.3
10. Humphreys 1973 Section 6.2
11. Kac 1990, Lemma 9.5.

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