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 $\backslash pi:\; \backslash mathfrak\{g\}\; \backslash to\; \backslash mathfrak\{gl\}(V)$, let $A\; \backslash subset\; \backslash operatorname\{End\}(V)$ be the associative subalgebra of the endomorphism algebra of V generated by $\backslash pi(\backslash mathfrak\; g)$. The ring A is called the enveloping algebra of $\backslash pi$. If $\backslash pi$ is semisimple, then A is semisimple.{{harvnb|Jacobson|1979|loc=Ch. II, § 5, Theorem 10.}} (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 $JV\; \backslash subset\; V$ 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 $\backslash mathfrak\; g$-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.)

Here is a typical application.{{harvnb|Jacobson|1979|loc=Ch. III, § 11, Theorem 17.}}

{{math_theorem|name=Proposition|math_statement=Let $\backslash mathfrak\; g$ be a semisimple finite-dimensional Lie algebra over a field of characteristic zero and $x$ an element of $\backslash mathfrak\; g$.{{efn|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. There exists a unique pair of elements $x\_s,\; x\_n$ in $\backslash mathfrak\; g$ such that $x\; =\; x\_s\; +\; x\_n$, $\backslash operatorname\{ad\}(x\_s)$ is semisimple, $\backslash operatorname\{ad\}(x\_n)$ is nilpotent and $[x\_s,\; x\_n]\; =\; 0$.
2. If $\backslash pi\; :\; \backslash mathfrak\{g\}\; \backslash to\; \backslash mathfrak\{gl\}(V)$ is a finite-dimensional representation, then $\backslash pi(x)\_s\; =\; \backslash pi(x\_s)$ and $\backslash pi(x)\_n\; =\; \backslash pi(x\_n)$, where $\backslash pi(x)\_s,\; \backslash pi(x)\_n$ denote the Jordan decomposition of the semisimple and nilpotent parts of the endomorphism $\backslash pi(x)$.

In short, the semisimple and nilpotent parts of an element of $\backslash mathfrak\; g$ 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 $\backslash pi$ is the inclusion; i.e., $\backslash mathfrak\; g$ is a subalgebra of $\backslash mathfrak\{gl\}\_n\; =\; \backslash mathfrak\{gl\}(V)$. Let $x\; =\; S\; +\; N$ be the Jordan decomposition of the endomorphism $x$, where $S,\; N$ are semisimple and nilpotent endomorphisms in $\backslash mathfrak\{gl\}\_n$. Now, $\backslash operatorname\{ad\}\_\{\backslash mathfrak\{gl\}\_n\}(x)$ also has the Jordan decomposition, which can be shown (see Jordan–Chevalley decomposition) to respect the above Jordan decomposition; i.e., $\backslash operatorname\{ad\}\_\{\backslash mathfrak\{gl\}\_n\}(S),\; \backslash operatorname\{ad\}\_\{\backslash mathfrak\{gl\}\_n\}(N)$ are the semisimple and nilpotent parts of $\backslash operatorname\{ad\}\_\{\backslash mathfrak\{gl\}\_n\}(x)$. Since $\backslash operatorname\{ad\}\_\{\backslash mathfrak\{gl\}\_n\}(S),\; \backslash operatorname\{ad\}\_\{\backslash mathfrak\{gl\}\_n\}(N)$ are polynomials in $\backslash operatorname\{ad\}\_\{\backslash mathfrak\{gl\}\_n\}(x)$ then, we see $\backslash operatorname\{ad\}\_\{\backslash mathfrak\{gl\}\_n\}(S),\; \backslash operatorname\{ad\}\_\{\backslash mathfrak\{gl\}\_n\}(N)\; :\; \backslash mathfrak\; g\; \backslash to\; \backslash mathfrak\; g$. Thus, they are derivations of $\backslash mathfrak\{g\}$. Since $\backslash mathfrak\{g\}$ is semisimple, we can find elements $s,\; n$ in $\backslash mathfrak\{g\}$ such that $[y,\; S]\; =\; [y,\; s],\; y\; \backslash in\; \backslash mathfrak\{g\}$ and similarly for $n$. Now, let A be the enveloping algebra of $\backslash mathfrak\{g\}$; i.e., the subalgebra of the endomorphism algebra of V generated by $\backslash mathfrak\; g$. 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, $\backslash pi(x)$ is semisimple (resp. nilpotent) when $\backslash operatorname\{ad\}(x)$ is semisimple (resp. nilpotent).{{clarify|why?|date=October 2020}} This immediately gives (i) and (ii). $\backslash square$

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 $\backslash mathfrak\{g\}$ is the complexification of the Lie algebra of a simply connected compact Lie group $K$.{{harvnb|Knapp|2002}} Theorem 6.11 (If, for example, $\backslash mathfrak\{g\}=\backslash mathrm\{sl\}(n;\backslash mathbb\{C\})$, then $K=\backslash mathrm\{SU\}(n)$.) Given a representation $\backslash pi$ of $\backslash mathfrak\{g\}$ on a vector space $V,$ one can first restrict $\backslash pi$ to the Lie algebra $\backslash mathfrak\{k\}$ of $K$. Then, since $K$ is simply connected,{{harvnb|Hall|2015}} Theorem 5.10 there is an associated representation $\backslash Pi$ of $K$. Integration over $K$ produces an inner product on $V$ for which $\backslash Pi$ is unitary.{{harvnb|Hall|2015}} Theorem 4.28 Complete reducibility of $\backslash Pi$ is then immediate and elementary arguments show that the original representation $\backslash pi$ of $\backslash mathfrak\{g\}$ is also completely reducible.

Let $(\backslash pi,\; V)$ be a finite-dimensional representation of a Lie algebra $\backslash mathfrak\; g$ over a field of characteristic zero. The theorem is an easy consequence of Whitehead's lemma, which says $V\; \backslash to\; \backslash operatorname\{Der\}(\backslash mathfrak\; g,\; V),\; v\; \backslash mapsto\; \backslash cdot\; v$ is surjective, where a linear map $f:\; \backslash mathfrak\; g\; \backslash to\; V$ is a derivation if $f([x,\; y])\; =\; x\; \backslash cdot\; f(y)\; -\; y\; \backslash cdot\; f(x)$. The proof is essentially due to Whitehead.{{harvnb|Jacobson|1979|loc=Ch. III, § 7.}}

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

:$x\; \backslash cdot\; t\; =\; [\backslash pi(x),\; t]$.

Now, pick some projection $p\; :\; V\; \backslash to\; V$ onto W and consider $f\; :\; \backslash mathfrak\{g\}\; \backslash to\; L\_W$ given by $f(x)\; =\; [p,\; \backslash pi(x)]$. Since $f$ is a derivation, by Whitehead's lemma, we can write $f(x)\; =\; x\; \backslash cdot\; t$ for some $t\; \backslash in\; L\_W$. We then have $[\backslash pi(x),\; p\; +\; t]\; =\; 0,\; x\; \backslash in\; \backslash mathfrak\{g\}$; that is to say $p\; +\; t$ is $\backslash mathfrak\{g\}$-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 $W$. $\backslash square$

Whitehead's lemma is typically proved by means of the quadratic Casimir element of the universal enveloping algebra,{{harvnb|Hall|2015}} Section 10.3 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 $c\_\backslash lambda$ of the identity in the irreducible representation of $\backslash mathfrak\{g\}$ with highest weight $\backslash lambda$. A key point is to establish that $c\_\backslash lambda$ is nonzero whenever the representation is nontrivial. This can be done by a general argument {{harvnb|Humphreys|1973}} Section 6.2 or by the explicit formula for $c\_\backslash lambda$.

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 $\backslash mathfrak\{g\}$, 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, $\backslash mathrm\{ker\}(V\_C)$ is an invariant complement to $W$, so that $V$ decomposes as a direct sum of irreducible subspaces:

:$V=W\backslash oplus\backslash mathrm\{ker\}(C\_V)$.

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.{{harvnb|Kac|1990|loc=Lemma 9.5.}} 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 $\backslash mathfrak\; g$ over an algebraically closed field of characteristic zero. Let $\backslash mathfrak\; b\; =\; \backslash mathfrak\{h\}\; \backslash oplus\; \backslash mathfrak\{n\}\_+\; \backslash subset\; \backslash mathfrak\{g\}$ be the Borel subalgebra determined by a choice of a Cartan subalgebra and positive roots. Let $V^0\; =\; \backslash \{\; v\; \backslash in\; V\; |\; \backslash mathfrak\{n\}\_+(v)\; =\; 0\; \backslash \}$. Then $V^0$ is an $\backslash mathfrak\; h$-module and thus has the $\backslash mathfrak\; h$-weight space decomposition:

:$V^0\; =\; \backslash bigoplus\_\{\backslash lambda\; \backslash in\; L\}\; V^0\_\{\backslash lambda\}$

where $L\; \backslash subset\; \backslash mathfrak\{h\}^*$. For each $\backslash lambda\; \backslash in\; L$, pick $0\; \backslash ne\; v\_\{\backslash lambda\}\; \backslash in\; V\_\{\backslash lambda\}$ and $V^\{\backslash lambda\}\; \backslash subset\; V$ the $\backslash mathfrak\; g$-submodule generated by $v\_\{\backslash lambda\}$ and $V\text{'}\; \backslash subset\; V$ the $\backslash mathfrak\; g$-submodule generated by $V^0$. We claim: $V\; =\; V\text{'}$. Suppose $V\; \backslash ne\; V\text{'}$. By Lie's theorem, there exists a $\backslash mathfrak\{b\}$-weight vector in $V/V\text{'}$; thus, we can find an $\backslash mathfrak\{h\}$-weight vector $v$ such that $0\; \backslash ne\; e\_i(v)\; \backslash in\; V\text{'}$ for some $e\_i$ among the Chevalley generators. Now, $e\_i(v)$ has weight $\backslash mu\; +\; \backslash alpha\_i$. Since $L$ is partially ordered, there is a $\backslash lambda\; \backslash in\; L$ such that $\backslash lambda\; \backslash ge\; \backslash mu\; +\; \backslash alpha\_i$; i.e., $\backslash lambda\; >\; \backslash mu$. But this is a contradiction since $\backslash lambda,\; \backslash mu$ are both primitive weights (it is known that the primitive weights are incomparable.{{clarify|More details should be given on this matter|date=September 2020}}). Similarly, each $V^\{\backslash lambda\}$ is simple as a $\backslash mathfrak\; g$-module. Indeed, if it is not simple, then, for some , $V^0\_\{\backslash mu\}$ contains some nonzero vector that is not a highest-weight vector; again a contradiction.{{clarify|More explanation|date=September 2020}} $\backslash square$

{{expand section|date=July 2024}}

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

• A [http://amathew.wordpress.com/2010/01/31/weyls-theorem-on-complete-reducibility/ blog post] by Akhil Mathew

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{{Reflist|20em}}

• {{cite book |last=Hall |first=Brian C. |title=Lie Groups, Lie Algebras, and Representations: An Elementary Introduction|edition = 2nd|series= Graduate Texts in Mathematics|volume=222 |year=2015 |publisher=Springer|isbn=978-3319134666}}
• {{cite book | last=Humphreys | first=James E. | author-link=James E. Humphreys | title=Introduction to Lie Algebras and Representation Theory | edition=Second printing, revised | series=Graduate Texts in Mathematics | volume=9 | publisher=Springer-Verlag | location=New York | year=1973 | isbn=0-387-90053-5 | url-access=registration | url=https://archive.org/details/introductiontoli00jame }}
• {{cite book |author-link=Nathan Jacobson |author-last=Jacobson |author-first=Nathan |title=Lie algebras |publisher=Dover Publications, Inc. |place=New York |year=1979 |isbn=0-486-63832-4}} Republication of the 1962 original.
• {{cite book|first=Victor|last=Kac|author-link=Victor Kac|title=Infinite dimensional Lie algebras|edition= 3rd |publisher= Cambridge University Press |year=1990 |isbn=0-521-46693-8 |url=https://books.google.com/books?id=kuEjSb9teJwC&q=Victor%20G.%20Kac&pg=PP1}}
• {{citation|last=Knapp|first=Anthony W.|author-link=Anthony W. Knapp|title=Lie Groups Beyond an Introduction|edition= 2nd|series=Progress in Mathematics|volume=140|publisher=Birkhäuser|place= Boston|year= 2002|isbn=0-8176-4259-5}}
• {{cite book |last=Weibel |first=Charles A. |author-link=Charles Weibel |title=An Introduction to Homological Algebra |year=1995 |publisher=Cambridge University Press }}

Category:Lie algebras

Category:Theorems in representation theory