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

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 $\pi: \mathfrak \to \mathfrak(V)$, let $A \subset \operatorname(V)$ 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. (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.

''Proof'': First we prove the special case of (i) and (ii) when $\pi$ is the inclusion; i.e., $\mathfrak g$ is a subalgebra of $\mathfrak_n = \mathfrak(V)$. Let $x = S + N$ be the Jordan decomposition of the endomorphism $x$, where $S, N$ are semisimple and nilpotent endomorphisms in $\mathfrak_n$. Now, $\operatorname_(x)$ also has the Jordan decomposition, which can be shown (see Jordan–Chevalley decomposition) to respect the above Jordan decomposition; i.e., $\operatorname_(S), \operatorname_(N)$ are the semisimple and nilpotent parts of $\operatorname_(x)$. Since $\operatorname_(S), \operatorname_(N)$ are polynomials in $\operatorname_(x)$ then, we see $\operatorname_(S), \operatorname_(N) : \mathfrak g \to \mathfrak g$. Thus, they are derivations of $\mathfrak$. Since $\mathfrak$ is semisimple, we can find elements $s, n$ in $\mathfrak$ such that , S= , s y \in \mathfrak and similarly for $n$. Now, let ''A'' be the enveloping algebra of $\mathfrak$; i.e., the subalgebra of the endomorphism algebra of ''V'' generated by $\mathfrak g$. As noted above, ''A'' has zero Jacobson radical. Since , 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 $\operatorname(x)$ is semisimple (resp. nilpotent). 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 $\mathfrak$ is the complexification of the Lie algebra of a simply connected compact Lie group $K$. (If, for example, $\mathfrak=\mathrm(n;\mathbb)$, then $K=\mathrm(n)$.) Given a representation $\pi$ of $\mathfrak$ on a vector space $V,$ one can first restrict $\pi$ to the Lie algebra $\mathfrak$ of $K$. Then, since $K$ is simply connected, there is an associated representation $\Pi$ of $K$. Integration over $K$ produces an inner product on $V$ for which $\Pi$ is unitary. Complete reducibility of $\Pi$ is then immediate and elementary arguments show that the original representation $\pi$ of $\mathfrak$ 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 $V \to \operatorname(\mathfrak g, V), v \mapsto \cdot v$ is surjective, where a linear map $f: \mathfrak g \to V$ is a derivation if f( , y = x \cdot f(y) - y \cdot f(x). The proof is essentially due to Whitehead.

Let $W \subset V$ be a subrepresentation. Consider the vector subspace $L_W \subset \operatorname(V)$ 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 $\mathfrak$-module given by: for $x \in \mathfrak, t \in L_W$,
:x \cdot t = pi(x), t/math>.
Now, pick some projection $p : V \to V$ onto ''W'' and consider $f : \mathfrak \to L_W$ given by f(x) = , \pi(x)/math>. 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; that is to say $p + t$ is $\mathfrak$-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, 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 $\mathfrak$ 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 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 $\mathfrak$, 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, $\mathrm(V_C)$ is an invariant complement to $W$, so that $V$ decomposes as a direct sum of irreducible subspaces:
:$V=W\oplus\mathrm(C_V)$.

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. 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 $\mathfrak b = \mathfrak \oplus \mathfrak_+ \subset \mathfrak$ be the Borel subalgebra determined by a choice of a Cartan subalgebra and positive roots. Let $V^0 = \$. Then $V^0$ is an $\mathfrak h$-module and thus has the $\mathfrak h$-weight space decomposition:
:$V^0 = \bigoplus_ V^0_$
where $L \subset \mathfrak^*$. For each $\lambda \in L$, pick $0 \ne v_ \in V_$ and $V^ \subset V$ the $\mathfrak g$-submodule generated by $v_$ and $V' \subset V$ the $\mathfrak g$-submodule generated by $V^0$. We claim: $V = V'$. Suppose $V \ne V'$. By Lie's theorem, there exists a $\mathfrak$-weight vector in $V/V'$; thus, we can find an $\mathfrak$-weight vector $v$ such that $0 \ne e_i(v) \in V'$ 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 $\lambda \ge \mu + \alpha_i$; 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.). Similarly, each $V^$ is simple as a $\mathfrak g$-module. Indeed, if it is not simple, then, for some $\mu < \lambda$, $V^0_$ contains some nonzero vector that is not a highest-weight vector; again a contradiction. $\square$

Algebraic proof 4

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

External links

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blog post
by Akhil Mathew

References

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* Republication of the 1962 original.
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* {{cite book , last=Weibel , first=Charles A. , author-link=Charles Weibel , title=An Introduction to Homological Algebra , year=1995 , publisher=Cambridge University Press

Lie algebras
Theorems in representation theory