18.745: LECTURE 23

PROFESSOR: VICTOR KAČ SCRIBES: DAVID MEYER AND CHRISTOPHER DAVIS

From now on, let 𝔤 be an arbitrary finite-dimensional Lie Algebra over an algebraically closed field 𝔽 of characteristic 0.

Theorem. (Weyl Complete Reducibility) If 𝔤 is semisimple, then any finite-dimensional 𝔤-module V is completely reducible, ie: for any submodule U there exists a complementary submodule U′ so that V = U ⊕ U′.

Corollary. Any finite-dimensional 𝔤-module is isomorphic to a direct sum of irreducible 𝔤-modules.

Theorem. (Levi) Let 𝔤 be any finite-dimensional Lie Algebra over 𝔽, and R(𝔤) the radical of 𝔤. Then there exists a semisimple subalgebra 𝔰 of 𝔤 such that 𝔤 = 𝔰+R(𝔤). (As previously noted, 𝔰 ∩ R(𝔤) = {0}).

Theorem. (Mal’cev) In Levi’s theorem, if 𝔰_{1} is another semisimple subalgebra of 𝔤 such that 𝔤 = 𝔰_{1} + R(𝔤), then there exists an automorphism ϕ of 𝔤 such that ϕ(𝔰) = 𝔰_{1}.

Recall that a “projector” of a vector space V to a subspace U ⊂ V is a linear operator P on V such that P (V) = U and P (u) = u for all u ∈ U. Note that V = U ⊕ Ker(P) for any projector P.

Given a projector P_{0} of V to U, any other projector of V to U is of the form P = P_{0} + A where A(V) ⊂ U and A(U) = 0.

Proof. (Weyl’s Theorem) Consider End V as a 𝔤-module, with: a ∗ A = [a, A] = aA − Aa, for a ∈ 𝔤, A ∈ End V. Notational Remark: since V is a 𝔤-module, we are identifying elements of 𝔤 with elements of End V. Hence, “aA” and “Aa” represent composition in End V, while a ∗ A represents the 𝔤-module action defined above.

Let M be the subspace of End V consisting of all A such that A(V) ⊂ U and A(U) = 0. This is a submodule of End V:

(a ∗ A)V = (aA − Aa)V ⊂ a(A(V)) + A(a(V)) ⊂ a(U) + A(V) ⊂ U
(a ∗ A)U = (aA − Aa)U ⊂ a(A(U)) + A(a(U)) ⊂ a(0) + A(U) = 0

Pick an arbitrary projector P_{0} : V ⟶ U, and define the following 1-cocycle:

f : 𝔤 ⟶ M; f(a) = a ∗ P_{0} = aP_{0} − P_{0}a

It is clear f maps into M since a and P_{0} commute on U, and:

(aP_{0} − P_{0}a)V ⊂ a(P_{0}(V)) + P_{0}(a(V)) ⊂ a(U) + P_{0}(V) ⊂ U

f is also clearly a 1-cocycle, since:

f([a, b]) = [a, b] ∗ P_{0} = a ∗ (b ∗ P_{0}) − b ∗ (a ∗ P_{0}) = a ∗ f(b) − b ∗ f(a).

Date: December 2nd, 2004.

(the second equality is just the defining property of a 𝔤-module)

Thus, by the main theorem of Lecture 22, f is trivial: there exists A ∈ M such that f(a) = a∗A for all a ∈ 𝔤. Equivalently: a∗P_{0} = a∗A, or a∗(P_{0} −A) = 0. Let P = P_{0} − A. Observe that P is just a projector of V to U, and we have a ∗ P = 0, ie, aP = P a, for all a ∈ 𝔤. Hence, Ker(P) is a-invariant for all a ∈ 𝔤, and thus Ker(P) is a 𝔤-submodule of V. Since V = U ⊕ Ker(P), Ker(P) is a 𝔤-submodule of V complementary to U, as desired. □

Proof. (Levi’s Theorem, using induction on dim 𝔤) If Rad(𝔤) is not abelian, consider 𝔤̄ = 𝔤/[Rad(𝔤), Rad(𝔤)]. Since dim(𝔤̄) < dim(𝔤), by the inductive assumption there exists semisimple 𝔰̄ ⊂ 𝔤̄ such that 𝔤̄ = 𝔰̄ + Rad(𝔤̄). Hence 𝔤 = 𝔤_{1} + Rad(𝔤), where 𝔤_{1} is the preimage of 𝔰̄ in 𝔤, and dim(𝔤_{1}) < dim(𝔤). Now apply the inductive assumption to 𝔤_{1}, so that we can write 𝔤_{1} = 𝔰 + Rad(𝔤_{1}), with 𝔰 semisimple. Thus 𝔤 = 𝔰 + (Rad(𝔤_{1}) + Rad(𝔤)). Note that Rad(𝔤_{1}) + Rad(𝔤) is an ideal of 𝔤: Write an arbitrary element of Rad(𝔤_{1})+Rad(𝔤) as a+b, where a ∈ Rad(𝔤_{1}) and b ∈ Rad(𝔤). Write an arbitrary element of 𝔤 as c + d, where c ∈ 𝔤_{1} and d ∈ Rad(𝔤). Then

[a + b, c + d] = [a, c + d] + [b, c + d]
= [a, c] + [a, d] + [b, c + d],

We have [a, c] ∈ Rad(𝔤_{1}) because Rad(𝔤_{1}) is an ideal of 𝔤_{1}, [a, d], [b, c + d] ∈ Rad(𝔤) because Rad(𝔤) is an ideal. Furthermore, Rad(𝔤_{1}) + Rad(𝔤) is solvable, because (Rad(𝔤_{1}) + Rad(𝔤))^{(n)} ⊆ Rad(𝔤_{1})^{(n)} + Rad(𝔤), which follows from the fact that Rad(𝔤_{1}) is a sub-algebra and Rad(𝔤) is an ideal of 𝔤. Hence, if Rad(𝔤_{1})^{(n)} = 0, and Rad(𝔤)^{(m)} = 0, then (Rad(𝔤_{1}) + Rad(𝔤))^{(m+n)} = 0. Hence, we have a solvable ideal of 𝔤 which contains Rad(𝔤), so we conclude that our ideal is in fact Rad(𝔤), yielding the equality

𝔤 = s + Rad(𝔤).

What remains is to consider the case when Rad(𝔤) is abelian. Consider the 𝔤-module structure End 𝔤 defined by:

a ∗ m = (ad a)m − m(ad a)

for a ∈ 𝔤, m ∈ End 𝔤. Further, consider the following submodule of End 𝔤:

M̃ = {m ∈ End 𝔤|m(𝔤) ⊂ Rad(𝔤) and m(Rad(𝔤)) = 0}

Let M = M̃/R̃ where R̃ = {ad(a)|a ∈ Rad(𝔤)}. Note that since Rad(𝔤) is abelian, it acts trivially on M. Hence, M is actually an 𝔰-module. (where 𝔰 = 𝔤/Rad(𝔤))

Exercise 23.1: Show that M̃ is a submodule of End(𝔤), and R̃ is a submodule of M̃.

Proof. M̃ is clearly a subspace of End(𝔤). Given a ∈ 𝔤 and m ∈ M̃:

(a ∗ m)(𝔤) ⊂ (ad a)m(𝔤) + m((ad a)𝔤) ⊂ (ad a)Rad(𝔤) + m(𝔤) ⊂ Rad(𝔤)

(a∗m)(Rad(𝔤)) ⊂ (ad a)m(Rad(𝔤))+m(ad a)(Rad(𝔤)) ⊂ (ad a)0+m(Rad(𝔤)) = 0 Hence a ∗ m ∈ M̃. So M̃ is a submodule of End(𝔤). R̃ is also clearly a subspace of End(𝔤). Given ad a ∈ R̃, with a ∈ Rad(𝔤):

(ad a)(𝔤) = [a, 𝔤] ⊂ Rad(𝔤)
(ad a)(Rad(𝔤)) = [a, Rad(𝔤)] = 0

Hence R̃ is a subspace of M̃. To see that it is a submodule, note that for ad a ∈ R̃ and b ∈ 𝔤:

b ∗ (ad a) = (ad b)(ad a) − (ad a)(ad b) = [ad b, ad a] = ad([b, a]) ∈ R̃
□

Let P_{0} be a projector of 𝔤 to Rad 𝔤, and consider the following 1-cocycle:

f : s → M
a → (ad ã)P_{0} − P_{0}(ad ã),

where ã is any preimage of a under the map 𝔤 → 𝔤/Rad 𝔤.

Exercise 23.2: The map f above is a well-defined 1-cocycle.

Proof. Let b ∈ Rad 𝔤.

(ad ã)P_{0} − P_{0}(ad ã) − ((ad ã + b̃)P_{0} − P_{0}(ad ã + b̃)) =
(ad b̃)P_{0} − P_{0}(ad b̃) = ad − b̃,

since we are assuming Rad 𝔤 is abelian, and since P_{0} acts as the identity on Rad 𝔤. Hence, the map f is well-defined, since the target of our map is M̃/R̃.

It remains to make sure that our map is indeed a 1-cocycle. In the notation above, it’s clear that [ã, b̃] maps to [a, b]. So, on one hand,

f([a, b]) = (ad [ã, b̃])P_{0} − P_{0}(ad [ã, b̃])

and on the other hand

af(b) − bf(a) = (ad a)f(b) − f(b)(ad a) − (ad b)f(a) + f(a)(ad b)
= (ad a)((ad b̃)P_{0} − P_{0}(ad b̃)) − ((ad b̃)P_{0} − P_{0}(ad b̃))(ad a)

− (ad b)((ad ã)P_{0} − P_{0}(ad ã)) + ((ad ã)P_{0} − P_{0}(ad ã))(ad b)

= (ad [ã, b̃])P_{0} − P_{0}(ad [ã, b̃]).
□

By the Fundamental Theorem on Cohomology, which may be applied because s is semi-simple, there exists m ∈ M such that a_{m} = (ad ã)P_{0} − P_{0}(ad ã). But, simply writing out a_{m} and subtracting, this means that

(1)  (ad ã)(P_{0} − m̃) − (P_{0} − m̃)(ad ã) = ad r_{a},

where m̃ is a preimage of m in M̃, and r_{a} ∈ Rad 𝔤.

Consider the projector P = P_{0} − m̃ of 𝔤 to Rad 𝔤.

Case 1: If all r_{a} = 0, then equation 1 above implies that ad ã commutes with P for all a ∈ 𝔤. In this case, let s = Ker P. Because P is a projector, s ∩ Rad 𝔤 = 0 and 𝔤 = s + Rad 𝔤. So we get that 𝔤 = s ⊕ Rad 𝔤 as vector spaces. This is in fact a direct sum of ideals: that Rad 𝔤 is an ideal is clear, and s is an ideal because, by the commutativity mentioned above, for every a ∈ 𝔤, s_{1} ∈ s,

0 = [a, P (s_{1})] = P ([a, s_{1}]).

Case 2: Now assume that not all r_{a} = 0. Let 𝔤_{1} = {a ∈ 𝔤 | P (ad a) = (ad a)P.} This is a subalgebra by the first properties of the adjoint representation. It is a proper subalgebra by our assumption, and equation 1 together with the facts that (ad r_{a})P = 0 since Rad 𝔤 is abelian and P (ad r_{a}) = ad r_{a} since P is a projector, implies

P (ad a − r_{a}) = (ad a − r_{a})P.

Hence, 𝔤 = 𝔤_{1} + Rad 𝔤.

Applying the inductive assumption, we can find a subalgebra s of 𝔤_{1} such that 𝔤_{1} = s + Rad 𝔤_{1}, a direct sum of vector spaces. Hence, 𝔤 = s + Rad 𝔤 + Rad 𝔤_{1}. As in the proof of Weyl’s Theorem, this is equivalent to 𝔤 = s + Rad 𝔤.

□

Proof. (Mal’cev’s Theorem) Suppose the radical is abelian, and, from Levi’s Theorem, we have 𝔤 = s ⊕ Rad 𝔤. Let P_{s} and P_{r} be the canonical projectors of 𝔤 onto s and Rad 𝔤, respectively. Let a_{r}, b_{r} ∈ Rad 𝔤, and a_{s}, b_{s} ∈ s. Say a = a_{r} + a_{s}, b = b_{r}+b_{s}. Such decompositions exist and are unique, because of our decomposition of 𝔤. Because Rad 𝔤 is an ideal, P_{s}([a, b]) = [s_{1}, s_{2}], and thus P_{s} is a Lie algebra homomorphism. Furthermore, by again considering the decomposition a = a_{r} + a_{s}, it’s clear that P_{r} + P_{s} = 1.

Let s_{1} ⊕ Rad 𝔤 = 𝔤 be another Levi decomposition.

By the adjoint representation, since it is an ideal, Rad 𝔤 is an s_{1} module. Define the 1-cocycle f : s_{1} → Rad 𝔤 by the formula f(a) = P_{r}(a).

Exercise 23.3: Check that f is indeed a 1-cocycle.

Proof. Write decompositions of two elements a, b ∈ s_{1} as above (in particular, a_{s} ∈ s, not s_{1}). Then

P_{r}([a_{r} + a_{s}, b_{r} + b_{s}]) = [a_{r}, b_{r}] + [a_{s}, b_{r}] + [a_{r}, b_{s}]
= [a_{s}, b_{r}] + [a_{r}, b_{s}]

since we are assuming Rad 𝔤 to be abelian. This clearly equals

[a_{r} + a_{s}, P_{r}(b_{r} + b_{s})] − [b_{r} + b_{s}, P_{r}(a_{r} + a_{s})],

and so f is a 1-cocycle. □

By the Fundamental Theorem, this cocycle is trivial. So, there exists r ∈ Rad 𝔤 such that P_{r}(a) = [a, r], for any a ∈ s_{1}. So we have that P_{r} = −ad r on s_{1}.

Since Rad 𝔤 is abelian, (ad r)^{2} = 0. Hence exp (ad r) = 1 + (ad r), and, from Exercise 9.2, that must be an automorphism of 𝔤. Call it σ.

Let a ∈ s_{1}. We have

σ(a) = (1 + ad r)a = (1 − P_{r})a = P_{s}(a) ∈ s.

Hence, we have an automorphism σ of 𝔤 such that σ(s_{1}) ⊆ s. An automorphism is, of course, injective, and by the vector space decompositions 𝔤 = s ⊕ Rad 𝔤 and 𝔤 = s_{1} ⊕ Rad 𝔤, we have that the dimensions of s and s_{1} are equal. Hence, σ(s_{1}) = s.

Exercise 23.4: As in the proof of Levi’s Theorem, reduce Mal’cev’s Theorem in the case of non-abelian radical Rad 𝔤 to the case where Rad 𝔤 is abelian.

Proof. We take a slightly different route, following Jacobson, [Ja62]. Let 𝔑 denote the sum of all the nilpotent ideals of 𝔤. That 𝔑 is a nilpotent ideal is obvious. We will prove inductively that for every i ≥ 1, there is an automorphism A_{i} of 𝔤 such that A_{i}(s_{1}) ⊆ s + 𝔑^{(i)}. Considering dimensions and the fact that 𝔑 is solvable, we will be able to conclude Mal’cev’s Theorem.

Considering the equation

P_{r}([a_{r} + a_{s}, b_{r} + b_{s}]) = [a_{r}, b_{r}] + [a_{s}, b_{r}] + [a_{r}, b_{s}],

we see that P_{r}([a, b]) ∈ [𝔤, Rad 𝔤] for each a, b ∈ s_{1}.

Lemma 1. With notation as above, we have that [𝔤, Rad 𝔤] ⊆ 𝔑.

Proof. We know from lecture five that [Rad 𝔤, Rad 𝔤] ⊆ 𝔑. By Engel’s Theorem, this is equivalent to ad [a_{r}, b_{r}] being nilpotent for a_{r}, b_{r} ∈ Rad 𝔤. Using the Jacobi identity, and the fact that Rad 𝔤 is an ideal, we get that ad [a, b_{r}] is nilpotent for any a ∈ 𝔤. □

Lemma 2. As s_{1} is semi-simple, we have [s_{1}, s_{1}] = s_{1}.

Proof. This follows from the structure theorem: the result holds for a simple Lie algebra t because it contain no non-trivial proper ideals, and thus [t, t] of such an ideal is either 0 or the whole simple algebra, and as t is not abelian, it must be the latter. The result continues to hold under passage to direct sums. □

Now, consider any a ∈ s_{1}. We have just seen that we may write a = [a_{1}, a_{2}], for some a_{1}, a_{2} ∈ s_{1}. Hence, from what we proved above, P_{r}(a) ∈ 𝔑. We know that P_{r} + P_{s} = 1, so we have that s_{1} ⊆ s + 𝔑^{(1)}. This provides the first step in our induction, with the identity automorphism playing the role of A_{1}.

Simplify notation by assuming that s_{1} ⊆ s + 𝔑^{(k)} (i.e., if necessary, replace s_{1} with the isomorphic Lie Algebra A_{k}(s_{1})). Because s is semi-simple, s ∩ Rad 𝔤 = ∅, and hence P_{r}(s_{1}) ⊆ 𝔑^{(k)}.

We can make 𝔑^{(k)} into a s_{1}-module by defining a · z = [P_{s}(a), z]. We check that this is indeed a module:

[a, b] · z = [P_{s}([a, b]), z]
= [[P_{s}(a), P_{s}(b)], z]
= −[[P_{s}(b), z], P_{s}(a)] − [[z, P_{s}(a)], P_{s}(b)]
= [P_{s}(a), [P_{s}(b), z]] + [P_{s}(b), [z, P_{s}(a)]]
= [P_{s}(a), [P_{s}(b), z]] − [P_{s}(b), [P_{s}(a), z]]
= a · (b · z) − b · (a · z),

as required. As 𝔑^{(k+1)} is an ideal, we can consider also 𝔑^{(k)}/𝔑^{(k+1)} as an s_{1}- module, with action a · \overline{z} = \overline{[P_{s}(a), z]}.

From what we saw two paragraphs ago, [P_{r}(a), P_{r}(b)] ∈ 𝔑^{(k+1)}, for a, b ∈ s_{1}. So,

\overline{P_{r}([a, b])} = \overline{[a_{s}, b_{r}]} − \overline{[b_{s}, a_{r}]} = a · \overline{b_{r}} − b · \overline{a_{r}}.

Define the map f : a → \overline{P_{r}(a)}. The map is linear because the projector is linear, and the above equation says precisely that f is a one-cocycle. Hence, by the fundamental theorem on cohomology, there is some element \overline{z} ∈ 𝔑^{(k)}/𝔑^{(k+1)} such that, for any a ∈ s_{1}, we have \overline{P_{r}(a)} = a · \overline{z} = \overline{[P_{r}(a), z]}, where z is any lift of \overline{z}.

6 PROFESSOR: VICTOR KAČ SCRIBES: DAVID MEYER AND CHRISTOPHER DAVIS

Let A = exp(ad z). Again, from exercise 9.2, this is an automorphism. We have

A(a) = a + [z, a] + \frac{1}{2!}[z, [z, a]] + ⋯
≡ a + [z, a] mod 𝔑^{(k+1)}
≡ a_{r} + a_{s} + [z, a_{r}] + [z, a_{s}] mod 𝔑^{(k+1)}

and we know a_{s} ∈ 𝔑^{(k)} and \overline{a_{r}} = \overline{[a_{r}, z]}, so

≡ a_{s} mod 𝔑^{(k+1)}.

Thus, we have that A(s_{1}) ⊆ s + 𝔑^{(k+1)}, and we proceed by induction.

□

As already mentioned, the proof is now complete, when we consider that the dimensions of s and s_{1} must be the same. □

References

[Ja62] N. Jacobson, Lie Algebras, Interscience Publishers, 1962.