Representations of the Heisenberg algebra A

2.2.1. General remarks

Consider the oscillator algebra (aka Heisenberg algebra)A=ha_i | i∈Zi+hKi. Recall that

[ai, aj] =iδi,−jK for any i, j ∈Z; [K, a_i] = 0 for any i∈Z.

Let us try to classify the irreducible A-modules.

Let V be an irreducible A-module. Then, V is countably-dimensional (by Remark 2.1.2, since U(A) is countably-dimensional), so that by Corollary 2.1.3, the endomor-phism K |_V is a scalar (because K is a central element of A and thus also a central element of U(A)).

If K |_V= 0, then V is a module over the Lie algebra ACK = ha_i | i∈Zi. But since ha_i | i∈Ziis an abelian Lie algebra, irreducible modules over ha_i | i∈Ziare 1-dimensional (again by Corollary 2.1.3), so thatV must be 1-dimensional in this case.

Thus, the case whenK |_V= 0 is not an interesting case.

Now consider the case when K |_V=k 6= 0. Then, we can WLOG assume thatk = 1, because the Lie algebra A has an automorphism sending K to λK for any arbitrary λ6= 0 (this automorphism is given bya_i 7→λa_i for i >0, and a_i 7→a_i for i≤0).

We are thus interested in irreducible representations V of A satisfying K |_V= 1.

These are in an obvious 1-to-1 correspondence with irreducible representations of U(A)(K −1).

Proposition 2.2.1. We have an algebra isomorphism

ξ:U(A)(K−1)→D(x₁, x₂, x₃, ...)⊗C[x₀],

whereD(x₁, x₂, x₃, ...) is the algebra of differential operators in the variables x₁,x₂, x3, ... with polynomial coefficients. This isomorphism is given by

ξ(a−i) = x_i for i≥1;

ξ(a_i) = i ∂

∂xi

for i≥1;

ξ(a₀) = x₀.

Note that we are sloppy with notation here: Since ξ is a homomorphism from U(A)(K −1) (rather than U(A)), we should write ξ(a_−i) instead of ξ(a_−i), etc..

We are using the same letters to denote elements ofU(A) and their residue classes in U(A)(K −1), and are relying on context to keep them apart. We hope that the reader will forgive us this abuse of notation.

Proof of Proposition 2.2.1. It is clear¹³ that there exists a unique algebra homomor-phism ξ:U(A)(K−1)→D(x₁, x₂, x₃, ...) satisfying

ξ(a−i) =x_i fori≥1;

ξ(a_i) =i ∂

∂xi

for i≥1;

ξ(a₀) =x₀.

It is also clear that this ξ is surjective (since all the generators x_i, ∂

∂x_i and x₀ of the algebraD(x₁, x₂, x₃, ...)⊗C[x₀] are in its image).

In the following, a map ϕ : A → N (where A is some set) is said to be finitely supported if all but finitely many a ∈ A satisfy ϕ(a) = 0. Sequences (finite, infinite, or two-sided infinite) are considered as maps (from finite sets, N or Z, or occasionally other sets). Thus, a sequence is finitely supported if and only if all but finitely many of its elements are zero.

If A is a set, then N^Afin will denote the set of all finitely supported maps A →N. By the easy part of the Poincar´e-Birkhoff-Witt theorem (this is the part which states that the increasing monomials span the universal enveloping algebra¹⁴), the family¹⁵

→

Y

i∈Z

aⁿ_iⁱ ·K^m

!

(...,n−2,n−1,n0,n1,n2,...)∈N^Zfin, m∈N

is a spanning set of the vector space U(A). Hence, the family

→

Y

i∈Z

aⁿ_iⁱ

!

(...,n−2,n−1,n0,n1,n2,...)∈N^Z_fin

13from the universal property of the universal enveloping algebra, and the universal property of the quotient algebra

14The hard part says that these increasing monomials are linearly independent.

15Here,

→

Q

i∈Z

aⁿ_iⁱ denotes the product...aⁿ₋₂⁻²aⁿ₋₁⁻¹aⁿ₀⁰aⁿ₁¹aⁿ₂².... (This product is infinite, but still has a value since only finitely manyni are nonzero.)

is a spanning set ofU(A)(K−1), and since this family maps to a linearly indepen-dent set under ξ (this is very easy to see), it follows that ξ is injective. Thus, ξ is an isomorphism, so that Proposition 2.2.1 is proven.

Definition 2.2.2. Define a vector subspace A₀ of A by A₀ =ha_i | i∈Z{0}i+ hKi.

Proposition 2.2.3. This subspace A₀ is a Lie subalgebra of A, and Ca₀ is also a Lie subalgebra of A. We haveA =A0⊕Ca0 as Lie algebras. Hence,

U(A)(K −1) =U(A₀⊕Ca₀)(K−1)∼= (U(A₀)(K−1))

| {z }

∼=D(x1,x2,x3,...)

⊗C[a₀]

| {z }

∼=C[x0]

(since K ∈ A₀). Here, the isomorphism U(A₀)(K −1) ∼= D(x₁, x₂, x₃, ...) is defined as follows: In analogy to Proposition 2.2.1, we have an algebra isomorphism

ξe:U(A₀)(K−1)→D(x₁, x₂, x₃, ...) given by

ξe(a−i) = x_i for i≥1;

ξe(a_i) = i ∂

∂x_i for i≥1.

The proof of Proposition 2.2.3 is analogous to that of Proposition 2.2.1 (where it is not completely straightforward).

2.2.2. The Fock space

From Proposition 2.2.3, we know that

U(A₀)(K−1)∼=D(x₁, x₂, x₃, ...)⊆End (C[x₁, x₂, x₃, ...]).

Hence, we have a C-algebra homomorphism U(A₀) → End (C[x₁, x₂, x₃, ...]). This makesC[x₁, x₂, x₃, ...] into a representation of the Lie algebraA₀. Let us state this as a corollary:

Corollary 2.2.4. The Lie algebra A₀ has a representation F = C[x₁, x₂, x₃, ...]

which is given by

a−i 7→x_i for every i≥1;

a_i 7→i ∂

∂x_i for every i≥1, K 7→1

(where “a−i 7→x_i” is just shorthand for “a−i 7→(multiplication by x_i)”). For every µ∈C, we can upgrade F to a representation F_µ of A by adding the condition that a₀ |_Fµ=µ·id.

Definition 2.2.5. The representationF ofA₀introduced in Corollary 2.2.4 is called the Fock module or the Fock representation. For every µ ∈ C, the representation F_µ ofA introduced in Corollary 2.2.4 will be called the µ-Fock representation of A.

The vector space F itself is called the Fock space.

Let us now define some gradings to make these infinite-dimensional spaces more manageable:

Definition 2.2.6. Let us grade the vector space A by A = L

n∈Z

A[n], where A[n] = ha_ni for n 6= 0, and where A[0] = ha₀, Ki. With this grading, we have [A[n],A[m]]⊆ A[n+m] for alln∈Zand m∈Z. (In other words, the Lie algebra A with the decomposition A = L

n∈Z

A[n] is a Z-graded Lie algebra. The notion of a

“Z-graded Lie algebra” that we have just used is defined in Definition 2.5.1.) Note that we are denoting the n-th homogeneous component of A by A[n] rather than A_n, since otherwise the notationA₀ would have two different meanings.

Definition 2.2.7. We grade the polynomial algebraF by setting deg (x_i) =−i for each i. Thus, F = L

n≥0

F [−n], where F [−n] is the space of polynomials of degree

−n, where the degree is our degree defined by deg (x_i) = −i (so that, for instance, x²₁+x₂ is homogeneous of degree−2). With this grading, dim (F [−n]) is the number p(n) of all partitions of n. Hence,

X

n≥0

dim (F [−n])qⁿ =X

n≥0

p(n)qⁿ= 1

(1−q) (1−q²) (1−q³)· · · = 1 Q

i≥1

(1−qⁱ) in the ring of power series Z[[q]].

We use the same grading for F_µ for every µ∈ C. That is, we define the grading on F_µ by F_µ[n] =F [n] for every n ∈Z.

Remark 2.2.8. Some people prefer to grade F_µ somewhat differently from F: namely, they shift the grading for F_µ by µ²

2 , so that deg 1 = −µ²

2 in F_µ, and gen-erally F_µ[z] = F

µ² 2 +z

(as vector spaces) for every z ∈ C. This is a grading by complex numbers rather than integers (in general). (The advantage of this grading is that we will eventually find an operator whose eigenspace to the eigenvalue n is F_µ[n] =F

µ² 2 +n

for every n∈C.) With this grading, the equality P

n≥0

dim (F [−n])qⁿ = 1 Q

i≥1

(1−qⁱ) rewrites as P

n∈C

dim (F_µ[−n])qⁿ⁺

µ²

2 = q^µ2 Q

i≥1

(1−qⁱ), if we allow power series with complex

ex-ponents. We define a “power series” ch (F_µ) by

ch (F_µ) = X

n∈C

dim (F_µ[−n])qⁿ⁺

µ²

2 = q^µ2 Q

i≥1

(1−qⁱ).

But we will not use this grading; instead we will use the grading defined in Definition 2.2.7.

Proposition 2.2.9. The representation F is an irreducible representation of A₀. Lemma 2.2.10. For every P ∈F, we have

P (a₋₁, a₋₂, a₋₃, ...)·1 = P inF.

(Here, the term P(a−1, a−2, a−3, ...) denotes the evaluation of the polynomial P at (x1, x2, x3, ...) = (a−1, a−2, a−3, ...). This evaluation is a well-defined element of U(A₀), since the elements a−1, a−2, a−3, ...of U(A₀) commute.)

Proof of Lemma 2.2.10. For every Q ∈ F, let multQ denote the map F → F, R 7→ QR. (In Proposition 2.2.1, we abused notations and denoted this map simply by Q; but we will not do this in this proof.) Then, by the definition of ξ, we have ξ(a_−i) = mult (x_i) for every i≥1.

Since we have defined an endomorphism multQ∈ EndF for every Q ∈F, we thus obtain a map mult :F →EndF. This map mult is an algebra homomorphism (since it describes the action of F on theF-module F).

Let P ∈F. Since ξ is an algebra homomorphism, and thus commutes with polyno-mials, we have

ξ(P (a₋₁, a₋₂, a₋₃, ...))

=P (ξ(a−1), ξ(a−2), ξ(a−3), ...) = P(mult (x1),mult (x2),mult (x3), ...) (since ξ(a−i) = mult (x_i) for everyi≥1)

= mult



P(x₁, x₂, x₃, ...)

| {z }

=P





since mult is an algebra homomorphism, and thus commutes with polynomials

= multP.

Thus,

P(a−1, a−2, a−3, ...)·1 = (multP) (1) =P ·1 =P.

This proves Lemma 2.2.10.

Proof of Proposition 2.2.9. 1) The representation F is generated by 1 as a U(A₀ )-module (due to Lemma 2.2.10). In other words,F =U(A0)·1.

2) Let us forget about the grading on F which we defined in Definition 2.2.7, and instead, once again, define a grading on F by deg (x_i) = 1 for every i ∈ {1,2,3, ...}.

Thus, the degree of a polynomialP ∈F with respect to this grading is what is usually referred to as the degree of the polynomial P.

If P ∈ F and if α·x^m₁¹x^m₂²x^m₃ ³... is a monomial in P of degree degP, with α 6= 0, this yields that for every nonzero P ∈ F, the representation F is generated by P as a U(A₀)-module (since F = U(A₀)· 1

|{z}

∈U(A₀)·P

⊆ U(A₀)·U(A₀)· P = U(A₀)·P).

Consequently, F is irreducible. Proposition 2.2.9 is proven.

Proposition 2.2.11. Let V be an irreducible A₀-module on which K acts as 1.

Assume that for any v ∈ V, the space C[a1, a2, a3, ...]·v is finite-dimensional, and the a_i with i >0 act on it by nilpotent operators. Then, V ∼=F asA₀-modules.

Before we prove this, a simple lemma:

16Proof. LetP∈F. Letα·x^m₁¹x^m₂²x^m₃³... be a monomial inP of degree degP, withα6= 0. When we apply the differential operator ∂_x^m₁¹

m1! gets mapped toαby the differential operator ∂_x^m₁¹

m1!

Lemma 2.2.12. Let V be an A₀-module. Let u ∈ V be such that a_iu = 0 for all i > 0, and such that Ku = u. Then, there exists a homomorphism η : F → V of A₀-modules such that η(1) =u. (This homomorphism η is unique, although we won’t need this.)

We give two proofs of this lemma. The first one is conceptual and gives us a glimpse into the more general theory (it proceeds by constructing an A0-module Ind^A0

CK⊕A⁺₀ C, which is an example of what we will later call a Verma highest-weight module in Definition 2.5.14). The second one is down-to-earth and proceeds by direct construction and computation.

First proof of Lemma 2.2.12. Define a vector subspaceA⁺₀ ofA₀byA⁺₀ =ha_i | i positive integeri.

It is clear that the internal direct sum CK ⊕ A⁺₀ is well-defined and an abelian Lie subalgebra of A₀. We can make C into an CK⊕ A⁺₀

-module by setting Kλ =λ for every λ ∈C;

aiλ= 0 for every λ∈C and every positive integeri.

Now, consider theA₀-module Ind^A0

CK⊕A⁺₀ C=U(A₀)⊗_U(CK⊕A⁺₀)C. Denote the element 1⊗_U(^CK⊕A+0) 1∈U(A₀)⊗_U(^CK⊕A+0)C of this module by 1.

We will now show the following important property of this module:

For anyA₀-module T, and any t ∈T satisfying (a_it = 0 for alli >0) and Kt=t,

Once this is proven, we will (by considering η_F,1) show that Ind^A_CK⊕A+

0 C∼= F, so this property will translate into the assertion of Lemma 2.2.12.

Proof of (14). Letτ :C→T be the map which sends every λ∈Ctoλt∈T. Then, τ is C-linear and satisfies

τ(Kλ) for every λ∈C and every positive integeri.

Thus,τis a CK⊕ A⁺₀

-module map. In other words,τ ∈Hom

CK⊕A⁺₀

C,Res^A0

CK⊕A⁺₀ T . By Frobenius reciprocity, we have

Hom_A0

under this isomorphism is an A0 -module map η_{T ,t}: Ind^A_CK⊕A⁰ +

(by the proof of Frobenius reciprocity)

= 1t=t.

Hence, there exists a homomorphism η_{T ,t}: Ind^A0

CK⊕A⁺₀ C→T of A₀-modules such that η_{T ,t}(1) = t. This proves (14).

It is easy to see that the element 1∈F satisfies (a_i1 = 0 for all i >0) andK1 = 1.

Thus, (14) (applied to T = F and t = 1) yields that there exists a homomorphism η_F,1 : Ind^A0

CK⊕A⁺₀ C → F of A₀-modules such that η_F,1(1) = 1. This homomorphism η_F,1 is surjective, since

F =U(A₀)· 1

|{z}

=η_F,1(1)

(as proven in the proof of Proposition 2.2.9)

=U(A₀)·η_F,1(1) =η_F,1(U(A₀)·1) since η_F,1 is an A₀-module map

⊆Imη_F,1.

Now we will prove that this homomorphism η_F,1 is injective.

In the following, a mapϕ:A→N(whereAis any set) is said to befinitely supported if all but finitely manya∈A satisfyϕ(a) = 0. Sequences (finite, infinite, or two-sided infinite) are considered as maps (from finite sets, N or Z, or occasionally other sets).

Thus, a sequence is finitely supported if and only if all but finitely many of its elements are zero.

If A is a set, then N^Afin will denote the set of all finitely supported maps A →N. By the easy part of the Poincar´e-Birkhoff-Witt theorem (this is the part which states that the increasing monomials span the universal enveloping algebra), the family¹⁸





→

Y

i∈Z{0}

aⁿ_iⁱ·K^m





(...,n−2,n−1,n1,n2,...)∈N^Zfin^{0}, m∈N

is a spanning set of the vector space U(A₀).

Hence, the family









→

Y

i∈Z{0}

aⁿ_iⁱ·K^m



⊗_U(CK⊕A⁺₀) 1





(...,n−2,n−1,n1,n2,...)∈N^Zfin^{0}, m∈N

is a spanning set of the vector space U(A0)⊗_U(^CK⊕A+0)C= Ind^A_CK⊕A⁰ + 0 C.

Let us first notice that this family is redundant: Each of its elements is contained in the smaller family









→

Y

i∈Z{0}

aⁿ_iⁱ



⊗_U(^CK⊕A+0) 1





(...,n−2,n−1,n1,n2,...)∈N^Z_fin^{0}

.

18Here,

→

Q

i∈Z{0}

aⁿ_iⁱ denotes the product...aⁿ₋₂⁻²aⁿ₋₁⁻¹aⁿ₁¹aⁿ₂².... (This product is infinite, but still has a value since only finitely manyni are nonzero.)

19 Hence, this smaller family is also a spanning set of the vector space Ind^A0

CK⊕A⁺₀ C. This smaller family is still redundant: Every of its elements corresponding to a sequence (..., n−2, n−1, n₁, n₂, ...)∈N^Zfin^{0} satisfying n₁+n₂+n₃+... >0 is zero²⁰, and zero elements in a spanning set are automatically redundant. Hence, we can replace

19This is because any sequence (..., n₋₂, n₋₁, n1, n2, ...)∈N^Z{0}fin and anym∈Nsatisfy

(by repeated application ofK1=1)

sinceK^m∈U CK⊕ A⁺₀ only finitely manyn_i are nonzero.

There exists some positive integer`satisfyingn`>0 (sincen1+n2+n3+... >0). Letjbe the greatest such `(this is well-defined, since only finitely manyni are nonzero).

Sincejis the greatest positive integer`satisfyingn`>0, it is clear thatjis the greatest integer

`satisfyingn`>0. In other words,aⁿ_j^j is the rightmost factor in the product

→

this smaller family by the even smaller family and we still have a spanning set of the vector space Ind^A0

CK⊕A⁺₀ C.

Clearly, sequences (..., n−2, n−1, n₁, n₂, ...)∈N^Zfin^{0} satisfyingn₁ =n₂ =n₃ =...= 0 are in 1-to-1 correspondence with sequences (..., n−2, n−1) ∈ N{...,−3,−2,−1}

fin . Hence, we can reindex the above family as follows:



So we have proven that the family



21. Since the family vec-tor space F (in fact, this family consists of all monomials of the polynomial ring C[x₁, x₂, x₃, ...] =F), we thus conclude that η_F,1 sends a spanning family of the vector space Ind^A0

CK⊕A⁺₀ C to a basis of the vector space F. Thus, η_F,1 must be injective²². Altogether, we now know that η_F,1 is a surjective and injective A0-module map.

Thus,η_F,1 is an isomorphism of A₀-modules.

Now, apply (14) to T =V andt =u. This yields that there exists a homomorphism η_V,u : Ind^A0 (because eachai with negativei acts onF by multiplication withx_−i)

=

fin . We thus have shown that

every (..., n₋₂, n₋₁) ∈ N{...,−3,−2,−1}

22Here we are using the following trivial fact from linear algebra: If a linear mapϕ:V →W sends a spanning family of the vector spaceV to a basis of the vector spaceW (as families, not just as sets), then this mapϕmust be injective.

Now, the composition η_V,u ◦η⁻¹_F,1 is a homomorphism F → V of A₀-modules such that

η_V,u◦η⁻¹_F,1

(1) =η_V,u η⁻¹_F,1(1)

| {z }

=1 (sinceη_F,1(1)=1)

=η_V,u(1) =u.

Thus, there exists a homomorphism η : F → V of A0-modules such that η(1) = u (namely, η=η_V,u◦η⁻¹_F,1). This proves Lemma 2.2.12.

Second proof of Lemma 2.2.12. Letη be the map F →V which sends every polyno-mialP ∈F =C[x1, x2, x3, ...] to P (a−1, a−2, a−3, ...)·u∈V. ²³ This map ηis clearly C-linear, and satisfies η(F) ⊆ U(A₀)·u. In order to prove that η is an A₀-module homomorphism, we must prove that

η(aiP) = aiη(P) for every i∈Z{0} and P ∈F (15) and that

η(KP) = Kη(P) for every P ∈F. (16)

First we show that

Kv=v for every v ∈U(A₀)·u. (17)

Proof of (17). SinceK lies in the center of the Lie algebra A₀, it is clear that K lies in the center of the universal enveloping algebra U(A₀). Thus, Kx = xK for every x∈U(A₀).

Now let v ∈U(A₀)·u. Then, there exists somex∈U(A₀) such thatv =xu. Thus, Kv=Kxu=x Ku

|{z}

=u

=xu=v. This proves (17).

Proof of (16). Since K acts as the identity on F, we have KP =P for everyP ∈F. Thus, for everyP ∈F, we have

η(KP) =η(P) = Kη(P)

since (17) (applied to v =η(P) ) yields Kη(P) =η(P) (because η(P)∈η(F)⊆U(A₀)·u)

.

This proves (16).

Proof of (15). Let i ∈Z{0}. If i < 0, then (15) is pretty much obvious (because in this case, a_i acts as x−i on F, so that a_iP =x−iP and thus

η(a_iP) =η(x_−iP) = (x_−iP) (a₋₁, a₋₂, a₋₃, ...)·u=a_iP (a₋₁, a₋₂, a₋₃, ...)·u

| {z }

=η(P)

=a_iη(P)

for every P ∈ F). Hence, from now on, we can WLOG assume that i is not < 0.

Assume this. Then, i≥0, so that i >0 (since i∈Z{0}).

In order to prove the equality (15) for all P ∈ F, it is enough to prove it for the case when P is a monomial of the form x_{`1</sub>x<sub>`2}...x_`m for some m ∈ N and some

23Note that the term P(a₋₁, a₋₂, a₋₃, ...) denotes the evaluation of the polynomial P at (x1, x2, x3, ...) = (a₋₁, a₋₂, a₋₃, ...). This evaluation is a well-defined element of U(A0), since the elementsa₋₁,a₋₂,a₋₃,... ofU(A0) commute.

(`<sub>1</sub>, `₂, ..., `_m)∈ {1,2,3, ...}^m. ²⁴ In other words, in order to prove the equality (15), it is enough to prove that

η(a_i(x_{`1</sub>x<sub>`2}...x_{`m</sub>)) = a<sub>i</sub>η(x<sub>`1}x_{`2</sub>...x<sub>`m}) for every m∈N and every (`<sub>1</sub>, `₂, ..., `_m)∈ {1,2,3, ...}^m. (18)

Thus, let us now prove (18). In fact, we are going to prove (18) by induction over m. The induction base is very easy (using a_i1 = i ∂

∂xi

1 = 0 and a_iu= 0) and thus left to the reader. For the induction step, fix some positive M ∈ N, and assume that (18) is already proven form =M −1. Our task is now to prove (18) for m =M.

So let (`1, `2, ..., `M) ∈ {1,2,3, ...}<sup>M</sup> be arbitrary. Denote by Q the polynomial x<sub>`2</sub>x_{`3</sub>...x<sub>`M}. Then,x_{`1</sub>Q=x<sub>`1}x_{`2</sub>x<sub>`3}...x_{`M</sub> =x<sub>`1}x_{`2</sub>...x<sub>`M}.

Since (18) is already proven for m = M − 1, we can apply (18) to M − 1 and (`2, `3, ..., `M) instead of m and (`1, `2, ..., `m). We obtain η(ai(x`2x`3...x`M)) = aiη (x<sub>`2</sub>x_{`3</sub>...x<sub>`M}). Sincex_{`2</sub>x<sub>`3}...x_`M =Q, this rewrites as η(a_iQ) =a_iη(Q).

Since any x∈ A₀ and y ∈ A₀ satisfy xy= yx+ [x, y] (by the definition ofU(A₀)), we have

a_ia−`1 =a−`1a_i+ [a_i, a−`1]

| {z }

=iδi,−(−`1)K

=a−`1a<sub>i</sub>+i δi,−(−`1)

| {z }

=δi,`1

K =a−`1a<sub>i</sub>+iδ<sub>i,`1</sub>K.

On the other hand, by the definition of η, every P ∈ F satisfies the two equalities η(P) =P (a−1, a−2, a−3, ...)·u and

η(x_{`1</sub>P) = (x<sub>`1}P) (a−1, a−2, a−3, ...)

| {z }

=a−`1·P(a−1,a−2,a−3,...)

·u=a−`₁·P (a−1, a−2, a−3, ...)·u

| {z }

=η(P)

=a_−`1 ·η(P). (19)

Sincea_i acts onF asi ∂

∂x_i, we havea_i(x_`1Q) = i ∂

∂x_i (x_`1Q) anda_iQ=i ∂

∂x_iQ. Now,

a_i





x_{`1</sub>x<sub>`2}...x_`M

| {z }

=x`1Q





=a_i(x_`1Q) = i ∂

∂x_i (x_`1Q) = i ∂

∂x_ix_`1

Q+x_`1 ∂

∂x_iQ

(by the Leibniz rule)

=i ∂

∂x_ix_`1

| {z }

=δ_i,`1

Q+x_`1 ·i ∂

∂x_iQ

| {z }

=aiQ

=iδ_{i,`1</sub>Q+x<sub>`1} ·a_iQ=x_{`1</sub> ·a<sub>i</sub>Q+iδ<sub>i,`1}Q,

so that

η(a_i(x_{`1</sub>x<sub>`2}...x_{`M</sub>)) =η(x<sub>`1} ·a_iQ+iδ_{i,`1</sub>Q) = η(x<sub>`1} ·a_iQ)

| {z }

=a−`1·η(aiQ) (by (19), applied toP=aiQ)

+iδ_i,`1η(Q)

=a−`1 ·η(aiQ)

| {z }

=aiη(Q)

+iδi,`1η(Q) = a−`1 ·aiη(Q) +iδi,`1η(Q).

24This is because such monomials generateF as a C-vector space, and because the equality (15) is linear inP.

Compared to

a_iη





x_{`1</sub>x<sub>`2}...x_`M

| {z }

=x`1Q





=a_i η(x_`1Q)

| {z }

=a−`1·η(Q) (by (19), applied toP=Q)

= a_ia−`1

| {z }

=a−`1ai+iδi,`1K

·η(Q)

= (a−`1a<sub>i</sub>+iδ<sub>i,`1</sub>K)·η(Q) = a−`1 ·a<sub>i</sub>η(Q) +iδ<sub>i,`1</sub> Kη(Q)

| {z }

=η(Q)

(by (17), applied tov=η(Q) (sinceη(Q)∈η(F)⊆U(A0)·u))

=a_{−`1</sub> ·a<sub>i</sub>η(Q) +iδ<sub>i,`1}η(Q),

this yields η(a_i(x_{`1</sub>x<sub>`2}...x_{`M</sub>)) =a<sub>i</sub>η(x<sub>`1}x_{`2</sub>...x<sub>`M}). Since we have proven this for every (`<sub>1</sub>, `₂, ..., `_M) ∈ {1,2,3, ...}^M, we have thus proven (18) for m = M. This completes the induction step, and thus the induction proof of (18) is complete. As we have seen above, this proves (15).

From (15) and (16), it is clear that η isA₀-linear (sinceA₀ is spanned by the a_i for i∈Z{0} and K). Sinceη(1) =u is obvious, this proves Lemma 2.2.12.

Proof of Proposition 2.2.11. Pick some nonzero vectorv ∈V. LetW =C[a₁, a₂, a₃, ...]·

v. Then, by the condition, we have dimW < ∞, and a_i : W → W are commuting nilpotent operators²⁵. Hence, T

i≥1

Kerai 6= 0 ²⁶. Hence, there exists some nonzero u ∈ T

i≥1

Kera_i. Pick such a u. Then, a_iu = 0 for all i > 0, and Ku = u (since K acts as 1 on V). Thus, there exists a homomorphism η : F →V of A₀-modules such that η(1) = u (by Lemma 2.2.12). Since both F and V are irreducible and η 6= 0, this yields thatη is an isomorphism. This proves Proposition 2.2.11.

2.2.3. Classification of A₀-modules with locally nilpotent action of C[a₁, a₂, a₃, ...]

Proposition 2.2.13. LetV be any A0-module having a locally nilpotent action of C[a₁, a₂, a₃, ...]. (Here, we say that the A₀-module V has a locally nilpotent action of C[a₁, a₂, a₃, ...] if for anyv ∈V, the space C[a₁, a₂, a₃, ...]·v is finite-dimensional, and the ai with i > 0 act on it by nilpotent operators.) Assume that K acts as 1 on V. Assume that for every v ∈ V, there exists some N ∈ N such that for every n ≥ N, we have a_nv = 0. Then, V ∼= F ⊗U as A₀-modules for some vector space U. (The vector space U is not supposed to carry any A0-module structure.)

Remark 2.2.14. From Proposition 2.2.13, we cannot remove the condition that for every v ∈V, there exists some N ∈N such that for every n ≥N, we have anv = 0.

In fact, here is a counterexample of how Proposition 2.2.13 can fail without this condition:

25Of course, when we writeai:W →W, we don’t mean the elementsai ofA0 themselves, but their actions on W.

26Here, we are using the following linear-algebraic fact:

If T is a nonzero finite-dimensional vector space over an algebraically closed field, and if b1, b2, b3, ...are commuting linear maps T →T, then there exists a nonzero common eigenvector of b1, b2, b3, .... If b1, b2, b3, ... are nilpotent, this yields T

i≥1

Kerbi 6= 0 (since any eigenvector of a nilpotent map must lie in its kernel).

LetV be the representation C[x₁, x₂, x₃, ...] [y](y²) of A₀ given by a−i 7→x_i for every i≥1;

a_i 7→i ∂

∂x_i +y for every i≥1, K 7→1

(where we are being sloppy and abbreviating the residue class y ∈ C[x₁, x₂, x₃, ...] [y](y²) by y, and similarly all other residue classes). We have an exact sequence

0 ^//F ^{i //}V ^{π //}F ^//0 of A₀-modules, where the mapi:F →V is given by

i(P) =yP for every p∈F =C[x₁, x₂, x₃, ...],

and the map π : V → F is the canonical projection V → V(y) ∼= F. Thus, V is an extension of F by F. It is easily seen that V has a locally nilpotent action of C[a₁, a₂, a₃, ...]. But V is not isomorphic to F ⊗U as A₀-modules for any vector spaceU, since there is a vectorv ∈V satisfyingV =U(A₀)·v (for example,v = 1), whereas there is no vector v ∈ F ⊗U satisfying F ⊗U = U(A₀)·v if dimU > 1, and the case dimU ≤1 is easily ruled out (in this case, dimU would have to be 1, so that V would be ∼=F and thus irreducible, and thus the homomorphisms i and π would have to be isomorphisms, which is absurd).

Before we prove Proposition 2.2.13, we need to define the notion of complete coflags:

Definition 2.2.15. Letk be a field. LetV be a k-vector space. LetW be a vector subspace of V. Assume that dim (VW)<∞. Then, acomplete coflag from V to W will mean a sequence (V0, V1, ..., VN) of vector subspaces of V (with N being an integer) satisfying the following conditions:

- We haveV₀ ⊇V₁ ⊇...⊇V_N.

- Every i∈ {0,1, ..., N}satisfies dim (VVi) =i.

- We haveV₀ =V and V_N =W.

(Note that the conditionV₀ =V is superfluous (since it follows from the condition that every i ∈ {0,1, ..., N} satisfies dim (VVi) = i), but has been given for the sake of intuition.)

We will also denote the complete coflag (V₀, V₁, ..., V_N) by V = V₀ ⊇ V₁ ⊇ ... ⊇ VN =W.

It is clear that if k is a field,V is a k-vector space, andW is a vector subspace of V satisfying dim (VW)<∞, then a complete coflag fromV to W exists.²⁷

27In fact, it is known that the finite-dimensional vector space VW has a complete flag (F0, F1, ..., FN); now, if we let p be the canonical projection V → VW, then

p⁻¹(FN), p⁻¹(FN−1), ..., p⁻¹(F0)

is easily seen to be a complete coflag fromV to W.

Definition 2.2.16. Let k be a field. Let V be a k-algebra. Let W be a vector subspace of V. Let i be an ideal of V. Then, an i-coflag from V to W means a complete coflag (V₀, V₁, ..., V_N) from V to W such that

every i∈ {0,1, ..., N −1} satisfiesi·V_i ⊆V_i+1.

Lemma 2.2.17. Let k be a field. Let B be a commutative k-algebra. Let I be an ideal of B such that thek-vector spaceBI is finite-dimensional. Let i be an ideal of B. Let M ∈N. Then, there exists an i-coflag from B toi^M +I.

Proof of Lemma 2.2.17. We will prove Lemma 2.2.17 by induction over M: Induction base: Lemma 2.2.17 is trivial in the case whenM = 0, because i⁰

|{z}

=B

+I = B+I =B. This completes the induction base.

Induction base: Letm ∈N. Assume that Lemma 2.2.17 is proven in the case when M =m. We now must prove Lemma 2.2.17 in the case whenM =m+ 1.

Since Lemma 2.2.17 is proven in the case when M = m, there exists an i-coflag (J₀, J₁, ..., J_K) from B to i^m+I. This i-coflag clearly is a complete coflag from B to i^m+I.

Since

dim (i^m+I) i^m+1+I

≤dim B i^m+1+I

because (i^m+I) i^m+1+I

injects intoB i^m+1+I

≤dim (BI) since B i^m+1+I

is a quotient ofBI

<∞ (since BI is finite-dimensional), there exists a complete coflag (U₀, U₁, ..., U_P) from i^m+I to i^m+1+I.

Since (U₀, U₁, ..., U_P) is a complete coflag fromi^m+I toi^m+1+I, we haveU₀ =i^m+I, and each of the vector spacesU0, U1, ..., UP contains i^m+1+I as a subspace.

Also, every i ∈ {0,1, ..., P} satisfies U_i ⊆ i^m +I (again since (U₀, U₁, ..., U_P) is a complete coflag from i^m+I toi^m+1+I).

Since (J0, J1, ..., JK) is a complete coflag fromB toi^m+I, while (U0, U1, ..., UP) is a complete coflag from i^m+I toi^m+1+I, it is clear that

(J₀, J₁, ..., J_K, U₁, U₂, ..., U_P) = (J₀, J₁, ..., JK−1, U₀, U₁, ..., U_P)

is a complete coflag fromB toi^m+1+I. We now will prove that this complete coflag (J₀, J₁, ..., J_K, U₁, U₂, ..., U_P) = (J₀, J₁, ..., JK−1, U₀, U₁, ..., U_P)

actually is an i-coflag.

In order to prove this, we must show the following two assertions:

Assertion 1: Everyi∈ {0,1, ..., K−1} satisfiesi·J_i ⊆J_i+1. Assertion 2: Everyi∈ {0,1, ..., P −1} satisfies i·Ui ⊆Ui+1.

Assertion 1 follows directly from the fact that (J₀, J₁, ..., J_K) is an i-coflag.

Assertion 2 follows from the fact thati· U_i

|{z}

⊆i^m+I

⊆i·(i^m+I)⊆ i·i^m

| {z }

=i^m+1

+ i·I

|{z}

⊆I (sinceIis an ideal)

⊆

i^m+1+I ⊆U_i+1(because we know that each of the vector spacesU₀,U₁,...,U_P contains i^m+1+I as a subspace, so that (in particular) i^m+1+I ⊆U_i+1).

Hence, both Assertions 1 and 2 are proven, and we conclude that (J₀, J₁, ..., J_K, U₁, U₂, ..., U_P) = (J₀, J₁, ..., JK−1, U₀, U₁, ..., U_P)

is an i-coflag. This is clearly an i-coflag from B to i^m+1 +I. Thus, there exists an i-coflag from B toi^m+1+I. This proves Lemma 2.2.17 in the case when M =m+ 1.

The induction step is complete, and with it the proof of Lemma 2.2.17.

Proof of Proposition 2.2.13. Letv ∈V be arbitrary. Let I_v ⊆C[a₁, a₂, a₃, ...] be the

annihilator ofv. Then, the canonicalC-algebra mapC[a₁, a₂, a₃, ...]→End (C[a₁, a₂, a₃, ...]·v) (this map comes from the action of theC-algebraC[a₁, a₂, a₃, ...] on C[a₁, a₂, a₃, ...]·v)

gives rise to an injective map C[a₁, a₂, a₃, ...]I_v → End (C[a₁, a₂, a₃, ...]·v). Since this map is injective, we have dim (C[a₁, a₂, a₃, ...]I_v)≤dim (End (C[a₁, a₂, a₃, ...]·v))<

∞ (since C[a₁, a₂, a₃, ...]·v is finite-dimensional). In other words, the vector space C[a₁, a₂, a₃, ...]I_v is finite-dimensional.

LetW be theA₀-submodule ofV generated byv. In other words, letW =U(A₀)·v.

LetW be theA₀-submodule ofV generated byv. In other words, letW =U(A₀)·v.

Im Dokument Pavel Etingof: 18.747: Infinite-dimensional Lie algebras (Spring term 2012 at MIT) (Seite 26-53)