2. Representation theory: generalities
2.4. Some consequences of Poincar´ e-Birkhoff-Witt
λαLmvα in Vα0,β0, and thus 0 =Lmvα in Vα0,β0 (since λα 6= 0). But since (24) yields
Lmvα =
α−α0− β0
|{z}=1
(m+ 1)
vα−α
| {z }
=v0
= (α−α0−(m+ 1))v0 in Vα0,β0,
this rewrites as 0 = (α−α0−(m+ 1))v0, so that 0 = α−α0 −(m+ 1). But this cannot hold for every m ∈ Z. This contradiction shows that our assumption (that α ∈ Z) was wrong. Thus, α /∈ Z, and our proof of the =⇒ direction is finally done.
Proposition 2.3.3(c) is finally proven.
Proving Proposition 2.3.3 was one part of Homework Set 1 exercise 2; the other was the following:
Proposition 2.3.4. Let α ∈ C and β ∈ C. Then, the Vir-module Vα,β is not irreducible if and only if (α∈Z and β ∈ {0,1}).
We will not prove this; the interested reader is referred to Proposition 1.1 in §1.2 of Kac-Raina.
Remark 2.3.5. Consider the Vir-module Vir (with the adjoint action). Since hCi is a Vir-submodule of Vir, we obtain a Vir-module VirhCi. This Vir-module is isomorphic to V1,−1. More precisely, theC-linear map
VirhCi →V1,−1, Ln7→v−n
is a Vir-module isomorphism. Thus, VirhCi ∼=V1,−1 ∼=Vα,−1 as Vir-modules for every α∈Z (because of Proposition 2.3.3 (a)).
2.4. Some consequences of Poincar´ e-Birkhoff-Witt
We will now spend some time with generalities on Lie algebras and their universal en-veloping algebras. These generalities will be applied later, and while these applications could be substituted by concrete computations, it appears to me that it is better for the sake of clarity to do them generally in here.
Proposition 2.4.1. Let k be a field. Let c be a k-Lie algebra. Let a and b be two Lie subalgebras of c such that a+b=c. Notice that a∩b is also a Lie subalgebra of c.
Let ρ: U(a)⊗U(a∩b)U(b)→U(c) be the k-vector space homomorphism defined by
ρ α⊗U(a∩b)β
=αβ for all α∈U(a) and β ∈U(b)
(this is clearly well-defined). Then, ρ is an isomorphism of filtered vector spaces, of left U(a)-modules and of right U(b)-modules.
Corollary 2.4.2. Let k be a field. Let c be a k-Lie algebra. Let a and b be two Lie subalgebras of c such that a⊕b = c (as vector spaces, not necessarily as Lie algebras). Let ρ : U(a) ⊗k U(b) → U(c) be the k-vector space homomorphism defined by
ρ(α⊗β) = αβ for all α∈U(a) andβ ∈U(b)
(this is clearly well-defined). Then, ρ is an isomorphism of filtered vector spaces, of left U(a)-modules and of right U(b)-modules.
We give two proofs of Proposition 2.4.1. They are very similar (both use the Poincar´ e-Birkhoff-Witt theorem, albeit different versions thereof). The first is more conceptual (and more general), while the second is more down-to-earth.
First proof of Proposition 2.4.1. For any Lie algebra u, we have a k-algebra homo-morphism PBWu :S(u)→gr (U(u)) which sendsu1u2...u`tou1u2...u` ∈gr`(U(u)) for every `∈Nand everyu1, u2, ..., u` ∈u. This homomorphism PBWu is an isomorphism due to the Poincar´e-Birkhoff-Witt theorem.
We can define a k-algebra homomorphism f : gr (U(a)) ⊗gr(U(a∩b)) gr (U(b)) → gr U(a)⊗U(a∩b)U(b)
by f u⊗gr(U(a∩b))v
=u⊗U(a∩b)v ∈grk+` U(a)⊗U(a∩b)U(b)
for any k ∈N, any ` ∈N, anyu∈U≤k(a) andv ∈U≤`(b). This f is easily seen to be well-defined. Moreover, f is surjective39.
It is easy to see that the isomorphisms PBWa:S(a)→gr (U(a)), PBWb :S(b)→ gr (U(b)) and PBWa∩b : S(a∩b) → gr (U(a∩b)) are “compatible” with each other in the sense that the diagrams
S(a)⊗S(a∩b) action ofS(a∩b) onS(a) //
PBWa⊗PBWa∩b ∼=
S(a)
PBWa ∼=
gr (U(a))⊗gr (U(a∩b))
action of gr(U(a∩b)) on gr(U(a)) //gr (U(a))
39To show this, either notice that the image off contains a generating set of gr U(a)⊗U(a∩b)U(b) (because the definition of f easily rewrites as
f α1α2...αk⊗gr(U(a∩b))β1β2...β`
=α1α2...αk⊗U(a∩b)β1β2...β`∈grk+` U(a)⊗U(a∩b)U(b) for anyk∈N, any`∈N, anyα1, α2, ..., αk ∈aandβ1, β2, ..., β`∈b), or prove the more general fact that for anyZ+-filtered algebraA, any filtered rightA-moduleM and any filtered left A-module N, the canonical map
gr (M)⊗gr(A)gr (N)→gr (M⊗AN),
µ⊗gr(A)ν7→µ⊗Aν ∈grm+n(M ⊗AN) (for all µ∈Mm andν∈Nn, for allm, n∈N) is well-defined and surjective (this is easy to prove).
and
S(a∩b)⊗S(b) action ofS(a∩b) onS(b) //
PBWa∩b⊗PBWb ∼=
S(b)
PBWb ∼=
gr (U(a∩b))⊗gr (U(b))
action of gr(U(a∩b)) on gr(U(b)) //gr (U(b)) commute40. Hence, they give rise to an isomorphism
S(a)⊗S(a∩b)S(b)→gr (U(a))⊗gr(U(a∩b))gr (U(b)), α⊗S(a∩b)β 7→(PBWaα)⊗gr(U(a∩b))(PBWbβ). Denote this isomorphism by (PBWa)⊗PBWa∩b (PBWb).
Finally, letσ :S(a)⊗S(a∩b)S(b)→S(c) be the vector space homomorphism defined by
σ α⊗S(a∩b)β
=αβ for all α∈S(a) and β ∈S(b).
This σ is rather obviously an algebra homomorphism. Now, it is easy to see thatσ is an algebra isomorphism41.
40This is pretty easy to see from the definition of PBWu.
41First proof that σis an algebra isomorphism: Since every subspace of a vector space has a comple-mentary subspace, we can find ak-vector subspacedofasuch thata=d⊕(a∩b). Consider such a d.
Sincea=d⊕(a∩b) =d+ (a∩b), the fact thatc=a+brewrites asc=d+ (a∩b) +b
| {z }
=b (sincea∩b⊆b)
=d+b.
Combined with d
=d∩a|{z}
(sinced⊆a)
∩b⊆d∩a∩b= 0 (sinced⊕(a∩b) is a well-defined internal direct sum), this yieldsc=d⊕b.
Recall a known fact from multilinear algebra: Any two k-vector spaces U and V satisfy S(U⊕V)∼=S(U)⊗kS(V) by the canonical algebra isomorphism. Hence,S(d⊕b)∼=S(d)⊗k
S(b).
Buta=d⊕(a∩b) yieldsS(a) =S(d⊕(a∩b))∼=S(d)⊗kS(a∩b) (by the above-quoted fact that any twok-vector spacesU andV satisfyS(U ⊕V)∼=S(U)⊗kS(V) by the canonical algebra isomorphism). Hence,
S(a)⊗S(a∩b)S(b)∼= (S(d)⊗kS(a∩b))⊗S(a∩b)S(b)
∼=S(d)⊗k S(a∩b)⊗S(a∩b)S(b)
| {z }
∼=S(b)
∼=S(d)⊗kS(b)∼=S
d⊕b
| {z }
=c
=S(c).
Thus we have constructed an algebra isomorphism S(a)⊗S(a∩b)S(b)→S(c). If we track down what happens to elements ofd,a∩bandbunder this isomorphism, we notice that they just get sent to themselves, so this isomorphism must coincide withσ(because if two algebra homomorphisms from the same algebra coincide on a set of generators of said algebra, then these two algebra homomorphisms must be identical). Thus,σis an algebra isomorphism, qed.
Second proof that σ is an algebra isomorphism: Define a map τ : c → S(a)⊗S(a∩b)S(b) as follows: For everyc∈c, letτ(c) bea⊗S(a∩b)1 + 1⊗S(a∩b)b, where we have writtencin the form c=a+bwitha∈aandb∈b(in fact, we can writecthis way, becausec=a+b). This mapτ is well-defined, because the value ofa⊗S(a∩b)1 + 1⊗S(a∩b)bdepends only oncand not on the exact values ofaandbin the decompositionc=a+b. (In fact, ifc=a+bandc=a0+b0are two different ways to decompose cinto a sum of an element ofa with an element ofb, thena+b=c=a0+b0,
Now, it is easy to see (by elementwise checking) that the diagram gr (U(a))⊗gr(U(a∩b))gr (U(b))
f
S(a)⊗S(a∩b)S(b)
(PBWa)⊗PBWa∩b(PBWb)
∼=
oo
∼= σ
gr U(a)⊗U(a∩b)U(b)
grρ
,,
S(c)
PBWc
∼=
gr (U(c))
so thata−a0=b0−b, thusa−a0∈a∩b(becausea−a0∈a anda−a0=b0−b∈b), so that a
|{z}
=a0+(a−a0)
⊗S(a∩b)1 + 1⊗S(a∩b)b
= (a0+ (a−a0))⊗S(a∩b)1 + 1⊗S(a∩b)b
=a0⊗S(a∩b)1 + (a−a0)⊗S(a∩b)1
| {z }
=1⊗S(a∩b)(a−a0)
(sincea−a0∈a∩b⊆S(a∩b))
+1⊗S(a∩b)b
=a0⊗S(a∩b)1 + 1⊗S(a∩b)(a−a0)
| {z }
=b0−b
+1⊗S(a∩b)b
=a0⊗S(a∩b)1 + 1⊗S(a∩b)(b0−b) + 1⊗S(a∩b)b
| {z }
=1⊗S(a∩b)((b0−b)+b)
=a0⊗S(a∩b)1 + 1⊗S(a∩b)((b0−b) +b)
| {z }
=b0
=a0⊗S(a∩b)1 + 1⊗S(a∩b)b0.
)
It is also easy to see that τ is a linear map. Thus, by the universal property of the symmetric algebra, the map τ :c→S(a)⊗S(a∩b)S(b) gives rise to a k-algebra homomorphism bτ :S(c)→ S(a)⊗S(a∩b)S(b) that liftsτ.
Anyα∈a satisfies
(bτ◦σ) α⊗S(a∩b)1
=bτ
σ α⊗S(a∩b)1
| {z }
=α1 (by the definition ofσ)
=bτ(α1) =bτ(α) =τ(α) (since bτ liftsτ)
=α⊗S(a∩b)1 + 1⊗S(a∩b)0
by the definition ofτ, sinceα=α+ 0 is a decomposition of αinto a sum of an element ofa with an element ofb
=α⊗S(a∩b)1.
In other words, the map τb◦σ fixes all tensors of the form α⊗S(a∩b)1 with α ∈ a. Similarly, the mapτb◦σfixes all tensors of the form 1⊗S(a∩b)β with β ∈b. Combining the previous two sentences, we conclude that the map mapτb◦σfixes all elements of the set
α⊗S(a∩b)1 | α∈a ∪ 1⊗S(a∩b)β | β ∈b . Thus, there is a generating set of thek-algebraS(a)⊗S(a∩b)S(b) such that the mapbτ◦σfixes all elements of this set (because
α⊗S(a∩b)1 | α∈a ∪
1⊗S(a∩b)β | β∈b is a generating set of the k-algebra S(a)⊗S(a∩b) S(b)). Since this map bτ ◦σ is a k-algebra homomorphism (because bτ andσ arek-algebra homomorphisms), this yields that the mapbτ◦σ is the identity (since ak-algebra homomorphism which fixes a generating set of its domain must be the identity). In other words, we have shown that bτ◦σ= id. A slightly different but similarly simple argument shows that σ◦τb= id. Combiningσ◦τb= id withbτ◦σ= id, we conclude thatbτ is an inverse toσ, so thatσis an algebra isomorphism, qed.
is commutative.42 Hence, (grρ)◦f is an isomorphism, so that f is injective. Since f is also surjective, this yields that f is an isomorphism. Thus, grρ is an isomorphism (since (grρ)◦f is an isomorphism). Sinceρis a filtered map and grρis an isomorphism, it follows thatρis an isomorphism of filtered vector spaces. Hence,ρis an isomorphism of filtered vector spaces, of left U(a)-modules and of right U(b)-modules (since it is clear thatρis a homomorphism ofU(a)-left modules and ofU(b)-right modules). This proves Proposition 2.4.1.
Second proof of Proposition 2.4.1. Let (zi)i∈I be a basis of the k-vector space a∩b.
We extend this basis to a basis (zi)i∈I∪(xj)j∈J of the k-vector spacea and to a basis (zi)i∈I ∪(y`)`∈L of the k-vector space b. Then, (zi)i∈I ∪(xj)j∈J ∪ (y`)`∈L is a basis of the k-vector space c. We endow this basis with a total ordering in such a way that every xj is smaller than every zi, and that every zi is smaller than every y`. By the Poincar´e-Birkhoff-Witt theorem, we have a basis of U(c) consisting of increasing products of elements of the basis (zi)i∈I∪(xj)j∈J∪(y`)`∈L. On the other hand, again by the Poincar´e-Birkhoff-Witt theorem, we have a basis of U(a) consisting of increasing products of elements of the basis (zi)i∈I ∪(xj)j∈J. Note that the zi accumulate at the right end of these products, while the xj accumulate at the left end (because we defined the total ordering in such a way that every xj is smaller than every zi).
Hence, U(a) is a free right U(a∩b)-module, with a basis (over U(a∩b), not overk) consisting of increasing products of elements of the basis (xj)j∈J. Combined with the fact that U(b) is a free k-vector space with a basis consisting of increasing products of elements of the basis (zi)i∈I∪(y`)`∈L (again by Poincar´e-Birkhoff-Witt), this yields thatU(a)⊗U(a∩b)U(b) is a freek-vector space with a basis consisting of tensors of the form
some increasing product of elements of the basis (xj)j∈J
⊗U(a∩b) some increasing product of elements of the basis (zi)i∈I∪(y`)`∈L . The map ρ clearly maps such terms bijectively into increasing products of elements of the basis (zi)i∈I ∪(xj)j∈J ∪(y`)`∈L. Hence, ρ maps a basis of U(a)⊗U(a∩b)U(b) bijectively to a basis of U(c). Thus, ρ is an isomorphism of vector spaces. Moreover, since both of our bases were filtered43, and ρ respects this filtration on the bases, we can even conclude that ρ is an isomorphism of filtered vector spaces. Since it is clear that ρ is a homomorphism of U(a)-left modules and ofU(b)-right modules, it follows that ρ is an isomorphism of filtered vector spaces, of left U(a)-modules and of right U(b)-modules. This proves Proposition 2.4.1.
Proof of Corollary 2.4.2. Corollary 2.4.2 immediately follows from Proposition 2.4.1 (since a⊕b=c yields a∩b= 0, thusU(a∩b) = U(0) =k).
Remark 2.4.3. While we have required k to be a field in Proposition 2.4.1 and Corollary 2.4.2, these two results hold in more general situations as well. For in-stance, Proposition 2.4.1 holds whenever k is a commutative ring, as long as a, b
42In fact, if we follow the pure tensorα1α2...αk⊗S(a∩b)β1β2...β`(withk∈N,`∈N,α1, α2, ..., αk∈a andβ1, β2, ..., β`∈b) through this diagram, we getα1α2...αkβ1β2...β`∈grk+`(U(c)) both ways.
43A basisB of a filtered vector space V is said to befiltered if for every n∈N, the subfamily of B consisting of those elements ofBlying in then-th filtration ofV is a basis of then-th filtration of V.
and a∩b are free k-modules, and a∩b is a direct summand of a as a k-module.
In fact, the first proof of Proposition 2.4.1 works in this situation (because the Poincar´e-Birkhoff-Witt theorem holds for free modules). In a more restrictive situa-tion (namely, whena∩b is a freek-module, and a direct summand of each of a and b, with the other two summands also being free), the second proof of Proposition 2.4.1 works as well. As for Corollary 2.4.2, it holds whenever k is a commutative ring, as long as a and b are free k-modules.
This generality is more than enough for most applications of Proposition 2.4.1 and Corollary 2.4.2. Yet we can go even further using the appropriate generalizations of the Poincar´e-Birkhoff-Witt theorem (for these, see, e. g., P. J. Higgins, Baer Invariants and the Birkhoff-Witt theorem, J. of Alg. 11, pp. 469-482, (1969), http://www.sciencedirect.com/science/article/pii/0021869369900866 ).