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.)^[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 $\pi :{\mathfrak {g}}\to {\mathfrak {gl}}(V)$ , 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.^[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 $JV\subset V$ 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.^[3]

Proposition — Let ${\mathfrak {g}}$ be a semisimple finite-dimensional Lie algebra over a field of characteristic zero and $x$ an element of ${\mathfrak {g}}$ .^[a]

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).^{[clarification needed]} 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 $K$ .^[4] (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,^[5] there is an associated representation $\Pi$ of $K$ . Integration over $K$ produces an inner product on $V$ for which $\Pi$ is unitary.^[6] 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.^[7]

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,^[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 $C$ 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 ^[9] 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.^[10] 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.^{[clarification needed]}). 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.^{[clarification needed]} $\square$

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

^ Editorial note: this fact is usually stated for a field of characteristic zero, but the proof needs only that the base field be perfect.

^ Hall 2015 Theorem 10.9
^ Jacobson 1979, Ch. II, § 5, Theorem 10.
^ Jacobson 1979, Ch. III, § 11, Theorem 17.
^ Knapp 2002 Theorem 6.11
^ Hall 2015 Theorem 5.10
^ Hall 2015 Theorem 4.28
^ Jacobson 1979, Ch. III, § 7.
^ Hall 2015 Section 10.3
^ Humphreys 1973 Section 6.2
^ Kac 1990, Lemma 9.5.

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.

[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.

[1] Hall 2015 Theorem 10.9

[2] Jacobson 1979, Ch. II, § 5, Theorem 10.

[3] Jacobson 1979, Ch. III, § 11, Theorem 17.

[5] Knapp 2002 Theorem 6.11

[6] Hall 2015 Theorem 5.10

[7] Hall 2015 Theorem 4.28

[8] Jacobson 1979, Ch. III, § 7.

[9] Hall 2015 Section 10.3

[10] Humphreys 1973 Section 6.2

[11] Kac 1990, Lemma 9.5.

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