DOWN-UP ALGEBRAS

Georgia Benkart Tom Roby

Abstract. The algebra generated by the down and up operators on a differential partially ordered set (poset) encodes essential enumerative and structural properties of the poset. Motivated by the algebras generated by the down and up operators on posets, we introduce here a family of infinite-dimensional associative algebras called down-up algebras. We show that down-up algebras exhibit many of the important features of the universal enveloping algebra U(𝔰𝔩_{2}) of the Lie algebra 𝔰𝔩_{2} including a Poincaré-Birkhoff-Witt type basis and a well-behaved representation theory. We investigate the structure and representations of down-up algebras and focus especially on Verma modules, highest weight representations, and category O modules for them. We calculate the exact expressions for all the weights, since that information has proven to be particularly useful in determining structural results about posets.

§1. Introduction

Combinatorial source of down-up algebras.

Assume P is a partially ordered set (poset), and let CP denote the complex vector space whose basis is the set P. For many posets there are two well-defined linear transformations on CP coming from the order relation on P,

d(y) = ∑_{x≺y} x  and  u(y) = ∑_{y≺z} z,

where ≺ denotes the covering relation in the poset. Thus, d(y) is the sum of the elements which y covers, and u(y) is the sum of the elements which cover

1991 Mathematics Subject Classification. 1991 Mathematics Subject Classifications: Primary 16S15, 16S30, 17B35, 17B10 Secondary 81R50, 06A07.

The first author gratefully acknowledges support from National Science Foundation Grant #DMS–9622447 and from the Mathematical Sciences Research Institute, Berkeley. Research at MSRI was supported in part by NSF grant 9701755. The second author gratefully acknowledges support from National Science Foundation Grant #DMS–9353149 and the University of Wisconsin, Madison.

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y. (These are the “down” and “up” operators.) Young’s lattice of all partitions of all nonnegative integers provides an important example of such a poset. If μ = (μ_{1}, μ_{2}, …) and ν = (ν_{1}, ν_{2}, …) are partitions (with parts arranged so μ_{1} ≥ μ_{2} ≥ ⋯ and ν_{1} ≥ ν_{2} ≥ ⋯), then μ ≤ ν if μ_{i} ≤ ν_{i} for each i. The partition ν covers μ if μ ≤ ν and ∑_{i} ν_{i} = 1 + ∑_{i} μ_{i}. In ([St1], [St2]), Stanley found that many of the interesting enumerative and structural properties of Young’s lattice could be deduced from the following relation that the up and down operators satisfy,

du − ud = I,

where I is the identity transformation on CP. Young’s lattice of partitions is not the only example, and Stanley considered more general posets which satisfy the relation du − ud = rI for some fixed positive integer r. Such posets afford a representation of the Weyl algebra generated by u and d/r. Since the Weyl algebra also can be realized as differential operators y ↦ d/dx and x ↦ x (multiplication by x) on C[x], Stanley referred to the posets satisfying du−ud = rI as “r-differential” or simply “differential” when r = 1. Fomin [F] independently defined essentially the same class of posets for r = 1 calling them “Y-graphs”, the terminology inspired by Young’s lattice.

In his study of uniform posets [T], Terwilliger considered finite ranked posets P whose down and up operators satisfy the following relation

d_{i}d_{i+1}u_{i} = α_{i}d_{i}u_{i−1}d_{i} + β_{i}u_{i−2}d_{i−1}d_{i} + γ_{i}d_{i},

where d_{i} and u_{i} denote the restriction of d and u to the elements of rank i. (There is an analogous second relation,

d_{i+1}u_{i}u_{i−1} = α_{i}u_{i−1}d_{i}u_{i−1} + β_{i}u_{i−1}u_{i−2}d_{i−1} + γ_{i}u_{i−1},

which holds automatically in this case because d_{i+1} and u_{i} are adjoint operators relative to a certain bilinear form.) The matrix algebra over the reals generated by these operators together with the idempotent projection maps onto each rank, (referred to as the “incidence algebra” in [T]), is finite-dimensional and in [T] its finite-dimensional simple representations are determined. Terwilliger also discusses many families of examples including the four presented below.

In many classical cases the constants in the above relations do not depend on the rank of the poset. A particular instance of this provides a q-analogue of the notion of differential poset. A partially ordered set whose down and up operators satisfy

(1.1)  \begin{aligned}d^{2}u & = q(q + 1)dud − q^{3}ud^{2} + rd \\ du^{2} & = q(q + 1)udu − q^{3}u^{2}d + ru \end{aligned}

where q and r are fixed complex numbers is said to be “(q, r)-differential”. Examples of (q, r)-differential posets include the posets of alternating forms, quadratic forms, or Hermitian forms over a finite field, as well as Hemmeter’s poset:

(1) [Alt_{q}(n), (n ≥ 2)] The poset of alternating bilinear forms Assume H is an n-dimensional vector space over the field GF (q) of q elements. Consider the set of pairs P = {(U, f) | U is a subspace of H and f is an alternating bilinear form on U} with the following ordering: (U, f) ≤ (V, g) if U is a subspace of V and g|_{U} = f. Then P is a (q, r)- differential poset with r = −q^{n}(q + 1).

(2) [Quad_{q}(n), (n ≥ 2)] The poset of quadratic forms The poset is defined the same way as in the first example but with “quadratic form” replacing “alternating bilinear form”. In this case P is (q, −q^{n+1}(q + 1))-differential.

(3) [Her_{q}(n), (n ≥ 2)] The poset of Hermitian forms In this example, H is taken to be an n-dimensional vector space over GF (q^{2}), and the bilinear forms are “Hermitian”. The result is a poset which is (q^{2}, −q^{2n+1}(q^{2} + 1))-differential.

(4) [Hem_{q}(n), (q odd, n ≥ 2)] Hemmeter’s poset Suppose X^{+} and X^{−} are two copies of the graph of the dual polar space X = C_{n−1}(q). Let Y be the bipartite graph with vertex set X^{+} ∪ X^{−} (forming the bipartition). Two vertices x^{+} ∈ X^{+} and y^{−} ∈ X^{−} are adjacent in Y if and only if x = y or x and y are adjacent in X. The graph Y is a distance regular graph recently discovered by Hemmeter. Now set P = Y and fix u_{0} ∈ Y. Declare u ≤ v if δ(u_{0}, u)+δ(u, v) = δ(u_{0}, v), where δ(u, v) is the distance between u and v. (The choice of u_{0} is irrelevant since Y is vertex transitive.) Then like the poset Alt_{q}(n), P is (q, −q^{n}(q + 1))-differential.

Definition of down-up algebras.

In this paper we study certain infinite-dimensional associative algebras whose generators satisfy relations more general than (1.1). Although the original motivation for our investigations came from posets, we make no assumption about the existence of posets whose down and up operators satisfy our relations.

We say a unital associative algebra A = A(α, β, γ) over the complex numbers C with generators u, d and defining relations

(R1) d^{2}u = αdud + βud^{2} + γd,

(R2) du^{2} = αudu + βu^{2}d + γu, where α, β, γ are fixed but arbitrary elements of C, is a down-up algebra.

It is easy to see that when γ ≠ 0 the down-up algebra A(α, β, γ) is isomorphic to A(α, β, 1) by the map d ↦ d′, u ↦ γu′. Therefore, it would suffice to treat just two cases γ = 0, 1. But rather than divide all our considerations into these cases, it is convenient to retain the notation γ throughout.

Examples of down-up algebras.

If d and u are the down and up operators of a (q, r)-differential poset, and B is the associative algebra they generate, then B satisfies (R1) and (R2) with α = q(q + 1), β = −q^{3}, and γ = r. Thus, B is a homomorphic image of the algebra A(α, β, γ) with these parameters. In an r-differential poset, the relation du − ud = rI is satisfied. Multiplying on the left by d and on the right by d and adding the resulting equations, we get the relation d^{2}u − ud^{2} = 2rd of a (−1, 2r)- differential poset. Thus, the Weyl algebra is a homomorphic image (by the ideal generated by du − ud − r1) of the algebra A(0, 1, 2r). More generally, the q-Weyl algebra is a homomorphic image of the algebra A(0, q^{2}, q+1) (and also of the algebra A(q − 1, q, 2)) by the ideal generated by du − qud − 1. The skew polynomial algebra C_{q}[d, u], or quantum plane (see [M]), is the associative algebra with generators d, u which satisfy the relation du = qud. Therefore, C_{q}[d, u] is a homomorphic image (by the ideal generated by du−qud) of the algebra A(2q, −q^{2}, 0) (also of the algebra A(aq, (1 − a)q^{2}, 0) for any a ∈ C).

Consider a 3-dimensional Lie algebra L over C having a basis x, y, [x, y] such that [x[x, y]] = γx and [[x, y], y] = γy. In the universal enveloping algebra U(L) of L, the relations above become

\begin{matrix}x^{2}y − 2xyx + yx^{2} = γx \\ xy^{2} − 2yxy + y^{2}x = γy. \end{matrix}

Thus, U(L) is a homomorphic image of the down-up algebra A(2, −1, γ) via the mapping ϕ : A(2, −1, γ) → U(L) with ϕ : d ↦ x, ϕ : u ↦ y. In fact, U(L) is isomorphic to A(2, −1, γ). The mapping ψ : L → A(2, −1, γ) with ψ : x ↦ d, ψ : y ↦ u, and ψ : [x, y] ↦ du − ud extends, by the universal property of U(L), to an algebra homomorphism ψ : U(L) → A(2, −1, γ). The two homomorphisms ϕ and ψ are inverses of each other. Every 3-dimensional Lie algebra L over C with L = [L, L] has such a basis (see [J, pp. 13-14]), so its enveloping algebra is a down-up algebra.

There are several noteworthy special cases. The Lie algebra 𝔰𝔩_{2} of 2 × 2 complex matrices of trace zero has a standard basis e, f, h, which satisfies [e, f] = h, [h, e] = 2e, and [h, f] = −2f. From this we see that U(𝔰𝔩_{2}) ≅ A(2, −1, −2). The Heisenberg Lie algebra ℌ has a basis x, y, z where [x, y] = z, and [z, ℌ] = 0. Thus, U(ℌ) ≅ A(2, −1, 0).

In [Sm], S.P. Smith investigated a class of associative algebras that have many of the properties of the enveloping algebra U(𝔰𝔩_{2}) but also many new, interesting aspects. These algebras have a presentation by generators a, b, h and relations [h, a] = a, [h, b] = −b and ab − ba = f(h), where f(h) is a polynomial in h. In the special situation that deg(f) ≤ 1, such an algebra is a homomorphic image of a down-up algebra A(2, −1, γ) for some γ. Algebras related to Smith’s appear in Hodges’ work [H] on primitive factors of U(𝔰𝔩_{2}) and type-A Kleinian singularities.

Woronowicz [Wo] introduced a family of algebras whose presentation is equivalent to having three generators x, y, z which are assumed to satisfy the relations

xz − ζ^{4}zx = (1 + ζ^{2})x zy − ζ^{4}yz = (1 + ζ^{2})y xy − ζ^{2}yx = ζz

for some scalar ζ ≠ ±1, 0. These algebras are down-up algebras with α = ζ^{2}(1 + ζ^{2}), β = −ζ^{6}, and γ = ζ(1 + ζ^{2}). Woronowicz’s algebras fit into a more general construction studied by Jordan in [Jo].

The Lie superalgebra S = 𝔰𝔩(1, 1) = S\overline{0} ⊕ S\overline{1} of 2 × 2 complex matrices, y =

( \begin{matrix}y_{1} & y_{2} \\ y_{3} & y_{4} \end{matrix} ) with supertrace y_{1} − y_{4} = 0 and with multiplication given by the su-

percommutator [x, y] = xy − (−1)^{ab}yx for x ∈ S\overline{a}, y ∈ S\overline{b} has a presentation by generators e, f (which are odd and can be identified with the matrix units e = e_{1,2}, f = e_{2,1}) and relations [e, [e, f]] = 0, [[e, f], f] = 0, [e, e] = 0, [f, f] = 0. Its universal enveloping algebra U(𝔰𝔩(1, 1)) has generators e, f and relations e^{2}f − fe^{2} = 0, ef^{2} − f^{2}e = 0, e^{2} = 0, f^{2} = 0. Thus, U(𝔰𝔩(1, 1)) is a homomorphic image of the down-up algebra A(0, 1, 0) by the ideal generated by the elements e^{2} and f^{2}, which are central in A(0, 1, 0).

Connections with Witten’s deformation of U(𝔰𝔩_{2}).

To provide an explanation of the existence of quantum groups, Witten ([W1], [W2]) introduced a 7-parameter deformation of the universal enveloping algebra U(𝔰𝔩_{2}). Witten’s deformation is a unital associative algebra over a field K (which is algebraically closed of characteristic zero and which could be assumed to be C) and depends on a 7-tuple \underline{ξ} = (ξ_{1}, …, ξ_{7}) of elements of K. It has a presentation by generators x, y, z and defining relations

xz − ξ_{1}zx = ξ_{2}x zy − ξ_{3}yz = ξ_{4}y
yx − ξ_{5}xy = ξ_{6}z^{2} + ξ_{7}z.

We denote the resulting algebra by 𝔚(\underline{ξ}). It can be argued that the following holds:

Theorem 1.2. ([B, Thm. 2.6]) A Witten deformation algebra 𝔚(\underline{ξ}) with

(1.3)  ξ_{6} = 0, ξ_{5}ξ_{7} ≠ 0, ξ_{1} = ξ_{3}, and ξ_{2} = ξ_{4}

is isomorphic to the down-up algebra A(α, β, γ) with α, β, γ given by

(1.4)  α = \frac{1 + ξ_{1}ξ_{5}}{ξ_{5}} ,  β = −\frac{ξ_{1}}{ξ_{5}} ,  γ = −\frac{ξ_{2}ξ_{7}}{ξ_{5}} .

Conversely, any down-up algebra A(α, β, γ) with not both α and β equal to 0 is isomorphic to a Witten deformation algebra 𝔚(\underline{ξ}) whose parameters satisfy (1.3).

A deformation algebra 𝔚(\underline{ξ}) has a filtration, and Le Bruyn ([L1], [L2]) investigated the algebras 𝔚(\underline{ξ}) whose associated graded algebras are Auslander regular. They determine a 3-parameter family of deformation algebras which are called conformal 𝔰𝔩_{2} algebras and whose defining relations are

(1.5)  xz − azx = x  zy − ayz = y  yx − cxy = bz^{2} + z

When c ≠ 0 and b = 0, the conformal 𝔰𝔩_{2} algebra with defining relations given by (1.5) is isomorphic to the down-up algebra A(α, β, γ) with α = c^{−1}(1 + ac), β = −ac^{−1} and γ = −c^{−1}. If b = c = 0 and a ≠ 0, then the conformal 𝔰𝔩_{2} algebra is isomorphic to the down-up algebra A(α, β, γ) with α = a^{−1}, β = 0 and γ = −a^{−1}.

In a recent paper Kulkarni [K1] has shown that under certain assumptions on the parameters a Witten deformation algebra is isomorphic to a conformal 𝔰𝔩_{2} algebra or to a double skew polynomial extension. The precise statement of the result is

Theorem 1.6. ([K1, Thm. 3.0.3]) If ξ_{1}ξ_{3}ξ_{5}ξ_{2} ≠ 0 or ξ_{1}ξ_{3}ξ_{5}ξ_{4} ≠ 0, then 𝔚(\underline{ξ}) is isomorphic to one of the following algebras:

(a) A conformal 𝔰𝔩_{2} algebra with generators x, y, z and relations given by (1.5) for some a, b, c ∈ K.

(b) A double skew polynomial extension (that is, a skew polynomial extension of a

skew polynomial ring) whose generators satisfy (i) xz − zx = x, zy − yz = ζy, yx − ηxy = 0 or (ii) xw = θwx, wy = κyw, yx = λxy

for parameters ζ, η, θ, κ, λ ∈ K.

Kulkarni [K1] studies the simple representations of the conformal sl_{2} algebras and of the skew polynomial algebras in (b) by making essential use of the observation that the conformal 𝔰𝔩_{2} algebras of (1.5) can be realized as hyperbolic rings. Kulkarni applies results of Rosenberg [R] on left ideals of the left spectrum of hyperbolic rings to determine the maximal left ideals for the conformal sl_{2} algebras. Further applications of results of [R] give the left spectrum of the double skew polynomial extensions in (b) of the theorem.

In this work we investigate the structure and representations of the down-up algebras A(α, β, γ). Remarkably down-up algebras have many of the important features of universal enveloping algebras including a Poincaré-Birkhoff-Witt basis theorem and a well-behaved representation theory. Adopting a Lie theoretic approach, we explicitly construct “highest weight” and “lowest weight” representations of the algebras A(α, β, γ) and determine when they are simple. We calculate the exact expressions for all the weights since that information has proven to be especially useful in determining structural results about posets. (In particular, for differential posets Stanley [St1] showed that the spectrum of ud is related to the eigenvalues of the adjacency matrix of certain finite graphs associated with the posets. This gave a new class of graphs whose spectra could be explicitly computed. Many interesting properties associated to random walks on the graph are related to its spectrum.)

In the spirit of the work of [BGG] we investigate two categories of modules (O and O′) for down-up algebras whose simple objects include the highest weight simple modules. Corresponding to each complex number λ, there is a universal infinitedimensional highest weight module V (λ), the so-called Verma module. Each of the weight spaces of V (λ) is 1-dimensional or infinite-dimensional depending on the choice of the parameters α, β, γ, λ. (Roots of unity play a critical role in deciding which occurs.) Its weights are determined by the recurrence relation

λ_{n} = αλ_{n−1} + βλ_{n−2} + γ, λ_{0} = λ, λ_{−1} = 0.

In particular, for the algebra A(1, 1, 0), the weights of V (λ) are given by the Fibonacci sequence λ, λ, 2λ, 3λ, 5λ, 8λ, … for each λ. Besides the examples coming from (q, r)-differential posets and universal enveloping algebras, certain subalgebras of quantized enveloping algebras (quantum groups) are down-up algebras. The highest weight representations that we construct for these subalgebras are not the standard ones.

Our investigations arose from conversations one of the authors (TR) had with Paul Terwilliger. We would like to express our appreciation to him for many helpful suggestions over the course of this project.

§2. Representations of down-up algebras

As before, let A = A(α, β, γ) be a down-up algebra over C with generators u, d and defining relations

(R1) d^{2}u = αdud + βud^{2} + γd,

(R2) du^{2} = αudu + βu^{2}d + γu,

where α, β, γ are fixed but arbitrary elements of C.

Highest weight modules.

Our first objective is to construct “highest weight” representations of downup algebras. Involved in the construction is a certain sequence of elements of C. Starting with λ_{−1} = 0 and an arbitrary λ_{0} = λ ∈ C, define λ_{n} for n ≥ 1 inductively by the following recurrence relation:

(2.1)  λ_{n} = αλ_{n−1} + βλ_{n−2} + γ.

Proposition 2.2. Set λ_{−1} = 0. Let λ_{0} = λ ∈ C be arbitrary, and define λ_{n} for n ≥ 1 as in (2.1). Let V (λ) be the C-vector space having basis {v_{n} | n = 0, 1, 2, … }. Define

(2.3)  \begin{aligned}d · v_{n} & = λ_{n−1}v_{n−1}, n ≥ 1, and d · v_{0} = 0 \\ u · v_{n} & = v_{n+1}. \end{aligned}

Then this action extends to an A(α, β, γ)-module action on V (λ).

Proof. The free associative algebra C⟨u, d⟩ with 1 over C generated by u, d acts on V (λ) by extending (2.3) making V (λ) into an C⟨u, d⟩-module. To show there is an induced action of A, it suffices to verify that

\begin{matrix}(d^{2}u − αdud − βud^{2} − γd)v_{n} = 0 \\ (du^{2} − αudu − βu^{2}d − γu)v_{n} = 0 \end{matrix}

for all n ≥ 0. But this is easy to see since

(d^{2}u − αdud − βud^{2} − γd)v_{n} = λ_{n−1}(λ_{n} − αλ_{n−1} − βλ_{n−2} − γ)v_{n−1},

and

(du^{2} − αudu − βu^{2}d − γu)v_{n} = (λ_{n+1} − αλ_{n} − βλ_{n−1} − γ)v_{n+1} = 0. □

Proposition 2.4.

(a) The A(α, β, γ)-module V (λ) is simple if and only if λ_{n} ≠ 0 for any n.

(b) If m is minimal with the property that λ_{m} = 0, then M(λ) = span_{C}{v_{j} | j ≥ m + 1} is a maximal submodule of V (λ).

(c) Suppose N is a submodule of V (λ) such that N ⊆ span_{C}{v_{j} | j ≥ 1}. Then N ⊆ M(λ).

Proof. Assume N is a nonzero submodule of V (λ), and suppose x = ∑_{k=0}^{n} a_{k}v_{k} ∈ N, where a_{n} ≠ 0. Then

d^{n} · x = λ_{n−1}λ_{n−2} ⋯ λ_{1}λ_{0}a_{n}v_{0} ∈ N.

If the coefficient is nonzero, then v_{0} ∈ N, and since v_{0} generates V (λ), we have N = V (λ) in this case.

If λ_{r} = 0, then span_{C}{v_{j} | j ≥ r + 1} is a proper submodule of V (λ) by (2.3). Thus, V (λ) is simple if and only if λ_{r} ≠ 0 for any r.

Let m be the minimal value such that λ_{m} = 0. Then M(λ) = span_{C}{v_{j} | j ≥ m + 1} is a submodule of V (λ). Suppose N is a submodule of V (λ) properly containing M(λ). We may assume that some y = ∑_{j=0}^{m} b_{j}v_{j} ≠ 0 belongs to N. If n ≤ m is maximal so b_{n} ≠ 0, then d^{n} · y = λ_{n−1}λ_{n−2} ⋯ λ_{1}λ_{0}b_{n}v_{0} ∈ N. By the minimality of m this implies v_{0} ∈ N and hence that N = V (λ). Thus, M(λ) is a maximal submodule of V (λ).

Now assume N is a submodule of V (λ) contained in V_{1} = span_{C}{v_{j} | j ≥ 1}. The submodule N + M(λ) is contained in V_{1}, so it is a proper submodule of V (λ). By the maximality of M(λ), we must have N + M(λ) = M(λ), which implies N ⊆ M(λ). □

Definition 2.5. We will set M(λ) = (0) if V (λ) is simple and otherwise assume M(λ) is defined by Proposition 2.4 (b) using the minimal m such that λ_{m} = 0. Note that when M(λ) ≠ (0), then M(λ) ≅ V (λ_{m+1}).

Example 2.6. Observe for any down-up algebra A = A(α, β, γ) that M(0) = span_{C}{v_{j} | j ≥ 1}, so that V (0)/M(0) is one-dimensional. There is an algebra homomorphism ϵ : A → C with ϵ(u) = 0 = ϵ(d) and ϵ(1) = 1. This gives rise to a one-dimensional A-module, Cv, such that a · v = ϵ(a)v for all a ∈ A, and Cv ≅ V (0)/M(0).

Definition 2.7. A module V for A = A(α, β, γ) is said to be a highest weight module of weight λ if V has a vector y_{0} such that d · y_{0} = 0, du · y_{0} = λy_{0}, and V = Ay_{0}. The vector y_{0} is a maximal vector or highest weight vector of V.

The highest weight module V (λ) is said to be the Verma module of highest weight λ. It is so named because it shares the same universal property as Verma modules for finite-dimensional semisimple complex Lie algebras:

Proposition 2.8. If V is a highest weight module of weight λ of A = A(α, β, γ), then V is a homomorphic image of V (λ).

Proof. Let y_{0} be a maximal vector of V of weight λ. Then V = Ay_{0}. The mapping V (λ) → V with a·v_{0} ↦ a·y_{0} for all a ∈ A is an A-module epimorphism. □

Suppose V is any highest weight A-module of weight λ. Starting with a nonzero maximal vector y_{0} ∈ V of weight λ, define y_{j} inductively by y_{j} = u · y_{j−1} for j ≥ 1. Stop the inductive definition at the first place that a vector y_{m+1} is produced which is a linear combination of the previous vectors. In this case we claim that V = span_{C}{y_{j} | j = 0, 1, …, m}. If no such dependency relation occurs, then we claim that V = span_{C}{y_{j} | j = 0, 1, … }, where the vectors y_{j} are linearly independent. To see these two assertions, first we show that

(2.9)  du · y_{n} = λ_{n}y_{n}

where λ_{0} = λ and the λ_{n}’s satisfy the recurrence relation in (2.1). The proof is by induction on n with the inductive step being

du · y_{n+1} = du^{2} · y_{n} = αudu · y_{n} + βu^{2}d · y_{n} + γu · y_{n}
= αλ_{n}y_{n+1} + βu^{2}du · y_{n−1} + γy_{n+1}
= (αλ_{n} + βλ_{n−1} + γ)y_{n+1} = λ_{n+1}y_{n+1}.

Observe that this implies

(2.10)  d · y_{n} = du · y_{n−1} = λ_{n−1}y_{n−1}

for all n ≥ 1. Since d and u generate A, the relation y_{n+1} = u · y_{n} together with (2.10) shows that V = span_{C}{y_{j} | j = 0, 1, …, m} when a dependency occurs, and V = span_{C}{y_{j} | j = 0, 1, … } otherwise. In both instances the y_{j}’s determine a basis for V.

Weight modules.

If we multiply the relation d^{2}u − αdud − βud^{2} = γd on the left by u and the relation du^{2} − αudu − βu^{2}d = γu on the right by d and subtract the second from the first, the resulting equation is

(2.11)  0 = ud^{2}u − du^{2}d  or  (du)(ud) = (ud)(du).

Therefore, the elements du and ud commute in A = A(α, β, γ). For any basis element v_{n} ∈ V (λ), we have du · v_{n} = λ_{n}v_{n} and ud · v_{n} = λ_{n−1}v_{n}. Using that with n = 0 and λ ≠ 0, it is easy to see that du and ud are linearly independent. Let 𝔥 = Cdu ⊕ Cud.

We say an A-module V is a weight module if V = ∑_{ν∈𝔥^{∗}} V_{ν}, where V_{ν} = {v ∈ V | h · v = ν(h)v for all h ∈ 𝔥}, and the sum is over elements in the dual space 𝔥^{∗} of 𝔥. (Necessarily this is a direct sum.) Any submodule of a weight module is a weight module. If V_{ν} ≠ (0), then ν is a weight and V_{ν} is the corresponding weight space. Each weight ν is determined by the pair (ν′, ν′′) of complex numbers, ν′ = ν(du) and ν′′ = ν(ud), and often it is convenient to identify ν with (ν′, ν′′). In particular, highest weight modules are weight modules in this sense. The basis vector v_{n} of V (λ) is a weight vector whose weight is given by the pair (λ_{n}, λ_{n−1}). We will explore next when two weights (λ_{k}, λ_{k−1}) and (λ_{ℓ}, λ_{ℓ−1}) can be equal. For that we will need to know the values of λ_{n} more precisely.

Proposition 2.12. Assume λ_{−1} = 0, λ_{0} = λ, and λ_{n} for n ≥ 1 is given by the recurrence relation λ_{n} − αλ_{n−1} − βλ_{n−2} = γ. Fix t ∈ C such that

t^{2} = \frac{α^{2} + 4β}{4} .

(i) If α^{2} + 4β ≠ 0, then

λ_{n} = c_{1}r_{1}^{n} + c_{2}r_{2}^{n} + x_{n},

where

r_{1} = \frac{α}{2} + t, r_{2} = \frac{α}{2} − t,

x_{n} = \{ \begin{matrix}(1 − α − β)^{−1}γ if α + β ≠ 1 \\ (2 − α)^{−1}γn if α + β = 1 (necessarily α ≠ 2), \end{matrix}

and ( \begin{matrix}c_{1} \\ c_{2} \end{matrix} ) = \frac{1}{r_{2} − r_{1}} ( \begin{matrix}r_{2} & −1 \\ −r_{1} & 1 \end{matrix} ) ( \begin{matrix}λ − x_{0} \\ αλ + γ − x_{1} \end{matrix} ) .

(ii) If α^{2} + 4β = 0 and α ≠ 0, then

λ_{n} = c_{1}s^{n} + c_{2}ns^{n} + x_{n} where

s = \frac{α}{2}

x_{n} = \{ \begin{matrix}(1 − α − β)^{−1}γ if α + β ≠ 1 \\ 2^{−1}n^{2}γ if α + β = 1 i.e. if α = 2, β = −1, \end{matrix}

and ( \begin{matrix}c_{1} \\ c_{2} \end{matrix} ) = ( \begin{matrix}1 & 0 \\ −1 & 2α^{−1} \end{matrix} ) ( \begin{matrix}λ − x_{0} \\ αλ + γ − x_{1} \end{matrix} ) .

(iii) If α^{2} + 4β = 0 and α = 0, then β = 0 and λ_{n} = γ for all n ≥ 1.

Note in particular, that if α, β are real, then it is natural to take t = (1/2)\sqrt{α^{2} + 4β}.

Proof. These assertions can be shown using standard arguments for solving linear recurrence relations ([Br, Chap. 7]). □

Theorem 2.13. Assume for the algebra A = A(α, β, γ) that λ_{ℓ} = λ_{k} and λ_{ℓ−1} = λ_{k−1} for some ℓ > k. Then

(a) A = A(α, β, γ), where α^{2} + 4β ≠ 0 and α + β ≠ 1, and either (1) γ = 0 and λ_{n} = 0 for all n,

(2) γ ≠ 0, β = 0, and λ_{n} = (1 − α)^{−1}γ for all n, (3)

λ_{n} = (λ − \frac{γ}{1 − α − β} )r^{n} + \frac{γ}{1 − α − β} ,

where r = r_{1} = (1/2)α + t or r = r_{2} = (1/2)α − t (t^{2} = (1/4)(α^{2} + 4β)), and r^{ℓ−k} = 1, or

(4) r_{1}^{ℓ−k} = 1 = r_{2}^{ℓ−k} and

λ_{n} = c_{1}r_{1}^{n} + c_{2}r_{2}^{n} + \frac{γ}{1 − α − β} ,

where c_{1} and c_{2} are as in Proposition 2.12;

(b) A = A(α, 1 − α, 0), where α ≠ 2, and either (1) λ_{n} = 0 for all n, or α = 1 and λ_{n} = λ ≠ 0 for all n, or

(2) λ ≠ 0, (α − 1)^{ℓ−k} = 1, and

λ_{n} = \frac{λ}{2 − α}(1 − (α − 1)^{n+1});

(c) A = A(α, −(1/4)α^{2}, γ) where α ≠ 0, 2, and either (1) γ = 0 and λ_{n} = 0 for all n, or

(2) γ ≠ 0, s = (1/2)α satisfies s^{ℓ−k} = 1, and

λ_{n} = \frac{2γ}{(2 − α)^{2}} ( − αs^{n} + 2);

(d) A = A(2, −1, 0) and λ_{n} = 0 for all n;

(e) A = A(0, 0, γ) and λ_{n} = γ for all n ≥ 1.

Proof. We will divide considerations into cases according to the solutions given for λ_{n} by Proposition 2.12.

Case (a): α + 4β ≠ 0 and α + β ≠ 1.

In this first case r_{1} ≠ r_{2} and

λ_{n} = c_{1}r_{1}^{n} + c_{2}r_{2}^{n} + \frac{γ}{1 − α − β} .

From λ_{ℓ} = λ_{k} and λ_{ℓ−1} = λ_{k−1} we see

\begin{matrix}c_{1}r_{1}^{k}(r_{1}^{ℓ−k} − 1) = −c_{2}r_{2}^{k}(r_{2}^{ℓ−k} − 1) \\ c_{1}r_{1}^{k−1}(r_{1}^{ℓ−k} − 1) = −c_{2}r_{2}^{k−1}(r_{2}^{ℓ−k} − 1). \end{matrix}

Multiplying the second by r_{2} and subtracting gives

c_{1}r_{1}^{k−1}(r_{1}^{ℓ−k} − 1)(r_{1} − r_{2}) = 0.

This shows that c_{1} = 0, r_{1} = 0, or r_{1}^{ℓ−k} = 1. Analogously, c_{2} = 0, r_{2} = 0, or r_{2}^{ℓ−k} = 1.

Suppose initially that 0 = r_{2} = (1/2)α − t. Then r_{1} = (1/2)α + t = α ≠ 0, and t^{2} = (1/4)α^{2} implies β = 0. From Proposition 2.12 we see

(2.14)  λ_{n} = c_{1}α^{n} + \frac{γ}{1 − α} = (λ − \frac{γ}{(1 − α)})α^{n} + \frac{γ}{1 − α}.

When γ = 0, then λ_{n} = λα^{n}, and λ_{ℓ} = λ_{k} forces λ = 0 (Case (1)) or α^{ℓ−k} = 1 (Case (3) with γ = 0). When γ ≠ 0, then either λ_{n} = (1 − α)^{−1}γ for all n (Case (2)) or λ_{n} is as in (2.14) where α^{λ−k} = 1 (Case (3) with β = 0).

When r_{1} = 0 and r_{2} = α, the solutions are precisely the same. So we can assume that neither r_{1} nor r_{2} is 0.

Suppose next that c_{1} = 0. Then λ = λ_{0} = c_{2} + (1 − α − β)^{−1}γ, which gives λ_{n} = (λ − (1 − α − β)^{−1}γ)r_{2}^{n} + (1 − α − β)^{−1}γ. The condition c_{1} = 0 says that

r_{1}λ = (1 − α − β)^{−1}γ(α + β − r_{2}). When c_{2} = 0, then λ = (1 − α − β)^{−1}γ and we have

\frac{r_{1}γ}{1 − α − β} = \frac{(α + β − r_{2})γ}{1 − α − β} .

If γ = 0, then λ = 0 and λ_{n} = 0 for all n (Case (1)). If γ ≠ 0, then α = r_{1} + r_{2} = α + β forces β = 0 and λ_{n} = (1 − α)^{−1}γ (Case (2)). The only other option is that c_{2} ≠ 0 but r_{2}^{ℓ−k} = 1. That gives Case (3). The situation when c_{2} = 0 and c_{1} ≠ 0 also gives Case (3).

Finally, if r_{1}, r_{2}, c_{1}, c_{2} are all nonzero, then r_{1}^{ℓ−k} = 1 = r_{2}^{ℓ−k} and

λ_{n} = c_{1}r_{1}^{n} + c_{2}r_{2}^{n} + \frac{γ}{1 − α − β} ,

for all n (Case (4)).

Case (b): α + 4β ≠ 0 and α + β = 1.

Under these hypotheses, β = 1 − α and 0 ≠ α^{2} + 4(1 − α) = (2 − α)^{2}. Then t^{2} = (1/4)(2 − α)^{2}, and we may assume that t = −(1/2)(2 − α). From that we have r_{1} = α − 1, r_{2} = 1, and

(2.15)  λ_{n} = c_{1}(α − 1)^{n} + c_{2} + \frac{γn}{2 − α}.

When λ_{ℓ} = λ_{k} and λ_{ℓ−1} = λ_{k−1}, then

\begin{matrix}c_{1}(α − 1)^{ℓ} + c_{2} + \frac{γℓ}{2 − α} = c_{1}(α − 1)^{k} + c_{2} + \frac{γk}{2 − α} \\ c_{1}(α − 1)^{ℓ−1} + c_{2} + \frac{γ(ℓ − 1)}{2 − α} = c_{1}(α − 1)^{k−1} + c_{2} + \frac{γ(k − 1)}{2 − α} \end{matrix}

or

(2.16)  \begin{matrix}c_{1}(α − 1)^{k}((α − 1)^{ℓ−k} − 1) = \frac{γ(k − ℓ)}{2 − α} \\ c_{1}(α − 1)^{k−1}((α − 1)^{ℓ−k} − 1) = \frac{γ(k − ℓ)}{2 − α}. \end{matrix}

Therefore,

c_{1}(α − 1)^{k−1}(α − 2)((α − 1)^{ℓ−k} − 1) = 0.

Since α ≠ 2, we see that the left-hand sides of (2.16) are 0, and γ = 0. Solving for c_{1}, c_{2} from Proposition 2.12 we find that c_{1} = (2−α)^{−1}(1−α)λ and c_{2} = (2−α)^{−1}λ. Therefore,

(2.17)  λ_{n} = \frac{λ(1 − α)}{2 − α} (α − 1)^{n} + \frac{λ}{2 − α} 1^{n} = \frac{λ}{2 − α}(1 − (α − 1)^{n+1}).

The condition λ_{ℓ} = λ_{k} implies that λ = 0, and hence all λ_{n} = 0 (Case (1)), or that α = 1 and hence that λ_{n} = λ for all n (Case (1)), or that λ ≠ 0 and (α − 1)^{ℓ−k} = 1 (Case (2)).

Case (c): α^{2} + 4β = 0, α + β ≠ 1 and α ≠ 0.

Here β = −(1/4)α^{2} so that

(2.18)  λ_{n} = c_{1}s^{n} + c_{2}ns^{n} + \frac{γ}{1 − α − β} = c_{1}s^{n} + c_{2}ns^{n} + \frac{4γ}{(2 − α)^{2}} .

It follows from λ_{ℓ} = λ_{k} and λ_{ℓ−1} = λ_{k−1} that

(2.19)  \begin{matrix}c_{1}s^{k}(s^{ℓ−k} − 1) = −c_{2}s^{k}(ℓs^{ℓ−k} − k) \\ c_{1}s^{k−1}(s^{ℓ−k} − 1) = −c_{2}s^{k−1}((ℓ − 1)s^{ℓ−k} − (k − 1)) \end{matrix}

Since s = (1/2)α ≠ 0, we may use these equations to see that

c_{2}(s^{ℓ−k} − 1) = 0 = c_{1}(s^{ℓ−k} − 1).

Now if s^{ℓ−k} = 1, then (2.19) shows that c_{2} = 0, which says that λ = (2 − α)^{−1}(2γ). Therefore, since c_{1} = λ − (2 − α)^{−2}4γ, we have

(2.20)  λ_{n} = (λ − \frac{4γ}{(2 − α)^{2}} )s^{n} + \frac{4γ}{(2 − α)^{2}} = \frac{2γ}{(2 − α)^{2}} ( − αs^{n} + 2),

where s = (1/2)α satisfies s^{ℓ−k} = 1. This gives the solutions in (c). Now if s^{ℓ−k} ≠ 1, then c_{1} = 0 = c_{2} and λ_{n} = (2 − α)^{−2}4γ for all n. Those relations imply (2−α)^{−1}2γ = (2−α)^{−2}4γ, which gives γ = 0 or α = 0. Since α ≠ 0 by assumption, γ = 0 and λ_{n} = 0 for all n.

Case (d): α^{2} + 4β = 0 and α = 2 and β = −1.

The values of λ_{n} are given by λ_{n} = c_{1}s^{n} + c_{2}ns^{n} + (1/2)γn^{2}, which reduces to λ_{n} = c_{1} + c_{2}n + (1/2)γn^{2} because s = α/2 = 1. Solving for c_{1} and c_{2} using Proposition 2.12, we see that c_{1} = λ and c_{2} = λ + γ/2. Therefore,

(2.21)  \begin{aligned}λ_{n} & = c_{1} + c_{2}n + \frac{γn^{2}}{2} = λ + (λ + \frac{γ}{2} )n + \frac{γn^{2}}{2} \\ & = (\frac{γ}{2} n + λ)(n + 1). \end{aligned}

Then

λ_{ℓ} = λ + (λ + \frac{γ}{2} )ℓ + \frac{γℓ^{2}}{2} = λ + (λ + \frac{γ}{2} )k + \frac{γk^{2}}{2} = λ_{k}

implies that

(\frac{γ}{2} )(ℓ^{2} − k^{2}) + (λ + \frac{γ}{2} )(ℓ − k) = 0.

Similarly, from λ_{ℓ−1} = λ_{k−1} we obtain

(\frac{γ}{2} )((ℓ − 1)^{2} − (k − 1)^{2}) + (λ + \frac{γ}{2} )(ℓ − k) = 0,

and those equations combine to give

γ(ℓ − k) = 0.

It must be γ = 0 and λ_{n} = λ(n + 1). But then λ_{ℓ} = λ(ℓ + 1) = λ(k + 1) = λ_{k} shows that λ = 0. We conclude that for the algebra A(2, −1, γ), the only time two weights (λ_{ℓ}, λ_{ℓ−1}) and (λ_{k}, λ_{k−1}) can be equal is if γ = 0 and λ_{n} = 0 for all n.

Case (e): α = β = 0

In this case λ_{n} = γ for all n ≥ 1. □

Corollary 2.22. In V (λ) each weight space is either one-dimensional or infinitedimensional. If an infinite-dimensional weight space occurs, there are only finitely many weights. The cases when V (λ) has an infinite-dimensional weight space are the ones listed in Theorem 2.13.

Submodules of V (λ).

Proposition 2.23. If each weight space of V (λ) is one-dimensional (i.e. if we are not in one of the cases described by Theorem 2.13), then the proper submodules of V (λ) have the form N = span_{C}{v_{j} | j ≥ n} for some n > 0 with λ_{n−1} = 0, and hence they are contained in M(λ).

Proof. Suppose N is a proper submodule of V (λ). There exists some weight ν = (λ_{n}, λ_{n−1}) such that N_{ν} ≠ (0), and we may suppose that n is the least value with that property. Then v_{n} ∈ N, and n is the smallest value for which that is true. Applying u we see that v_{j} ∈ N for all j ≥ n. □

To describe the submodules of V (λ) when its weights are given by Theorem 2.13, it is helpful to observe that if N is a proper submodule of any V (λ), then

(2.24)  I_{N} = {f ∈ C[x] | f(u) · v_{0} ∈ N}

is an ideal of C[x]. There exists a monic polynomial g(x) which generates that ideal. Then dim_{C} V (λ)/N = deg(g) = k, and v_{i} +N for i = 0, 1, …, k −1 is a basis for V (λ)/N.

The situations described by Theorem 2.13 can be grouped into four general types (see 2.17 and 2.20):

(2.25)

(i) γ = 0 and λ_{n} = 0 for all n. (ii) λ_{n} = c for all n where c ∈ C satisfies c − αc − βc = γ.

(iii) λ_{n} = (λ − c)r^{n} + c where c ∈ C and r is a root of x^{2} − αx − β = 0. There is a

least integer p ∈ Z_{>0} such that r^{p} = 1. (iv) λ_{n} = c_{1}r_{1}^{n} + c_{2}r_{2}^{n} + c where r_{1} ≠ r_{2} are roots of the equation x^{2} − αx − β = 0,

and there is some smallest value p ∈ Z_{>0} so that r_{1}^{p} = 1 = r_{2}^{p}.

(i) Assume first that A = A(α, β, 0) and consider the module V (0). Since λ_{n} = 0 for all n, for any g ∈ C[x],

(2.26)  N^{(g)} = span_{C}{f(u)g(u)v_{0} | f ∈ C[x]}

is a submodule of V (0), which is proper if deg(g) ≥ 1. For each ξ ∈ C, the submodule N^{(x−ξ)} is maximal, and every maximal submodule has this form. In particular, M(λ) = N^{(x)}. The quotient module L(0, ξ) =^{def} V (0)/N^{(x−ξ)} is spanned by v_{0} + N^{(x−ξ)} where d · (v_{0} + N^{(x−ξ)}) = 0 and u · (v_{0} + N^{(x−ξ)}) = ξv_{0} + N^{(x−ξ)}.

(ii) Next suppose that A = A(α, β, γ), and the module is V (c) where 0 ≠ c ∈ C is such that c − αc − βc = γ. Then λ_{n} = c ≠ 0 for all n. By Proposition 2.4, V (c) is simple.

(iii) and (iv) These two cases behave similarly, and so we will treat them simultaneously. Assume for A = A(α, β, γ) that the module V (λ) has weights given by

λ_{n} = (λ − c)r^{n} + c where c ∈ C, λ ≠ c, and r^{p} = 1, or
λ_{n} = c_{1}r_{1}^{n} + c_{2}r_{2}^{n} + c where c ∈ C, c_{1}, c_{2} ≠ 0, and r_{1}^{p} = 1 = r_{2}^{p}.

In each instance, we will assume that p is chosen to be minimal. Then λ_{n+p} = λ_{n} for all n ≥ 0, and V (λ) = ∑_{i=0}^{p} V (λ)_{ν_{i}}, where ν_{i} = (λ_{i}, λ_{i−1}). Now ν_{p} = ν_{0} if and only if λ_{p−1} = 0, which happens, for example, if β ≠ 0.

Suppose first that ν_{p} ≠ ν_{0}, so that V (λ) has p + 1 distinct weights. If N is a submodule of V (λ) and N_{ν_{0}} ≠ (0), then N = V (λ). In particular, any proper submodule N is contained in M(λ) by Proposition 2.4, so that M(λ) is the unique maximal submodule. We may assume that N_{ν_{j}} ≠ (0) for some j = 1, …, p. Then (0) ≠ u^{p−j+k}N_{ν_{j}} ⊆ N_{ν_{k}} so that all N_{ν_{k}} ≠ (0). Therefore, for k ≠ p,

(2.27)  {f ∈ C[x] | u^{k}f(u^{p}) · v_{0} ∈ N_{ν_{k}}}

is a nonzero ideal of C[x], and we may assume that g_{k} is its monic generator. Similarly, {f ∈ C[x] | f(u^{p}) · v_{0} ∈ N_{ν_{p}}} is a nonzero ideal, whose monic generator we denote by g_{p}. Then u^{k}g_{k}(u^{p}) · v_{0} ∈ N_{ν_{k}} implies u(u^{k}g_{k}(u^{p}) · v_{0}) ∈ N_{ν_{k+1}}. It follows that

g_{p−1} | g_{p−2} | ⋯ | g_{2} | g_{1} | g_{p} | xg_{p−1}.

Either g_{1} = g_{2} = ⋯ = g_{p} = g and N = N^{(g(x^{p}))} (see (2.26) for the notation), or there is some value m such that g_{p−1} = ⋯ = g_{m+1} = g and g_{m} = ⋯ = g_{1} = g_{p} = xg. Since d(u^{m+1}g(u^{p}) · v_{0}) = λ_{m}u^{m}g(u^{p}) · v_{0}, it must be that λ_{m} = 0 in this case. It is easy to see that λ_{k} ≠ λ_{ℓ} for 1 ≤ k < ℓ ≤ p. Therefore m is the minimal value such that λ_{m} = 0. In this situation, N = N^{(x^{m+1}g(x^{p}))}. As a special case, we have M(λ) = N^{(x^{m+1})}.

In the case that ν_{p} = ν_{0}, the module V (λ) has p distinct weights. Assume N is a proper submodule of V (λ). Arguing as above, we have that N_{ν_{k}} ≠ (0) for each k = 0, 1, …, p − 1. Letting g_{k} be the monic generator for the ideal in (2.27), we have

g_{p−1} | g_{p−2} | ⋯ | g_{2} | g_{1} | g_{0} | xg_{p−1}.

Again there are two possibilities. Either g_{0} = g_{1} = ⋯ = g_{p−1} = g and N = N^{(g(x^{p}))}, or else there is some value m such that g_{p−1} = ⋯ = g_{m+1} = g and g_{m} = ⋯ = g_{1} = g_{0} = xg. Here N = N^{(x^{m+1}g(x^{p}))}, and M(λ) = N^{(x^{m+1})}.

Corollary 2.28. If γ = 0 = λ, then V (λ) has infinitely many maximal proper submodules, each of the form N^{(x−ξ)} = span_{C}{f(u)(u − ξ)v_{0} | f ∈ C[x]} = span_{C}{v_{n} − ξv_{n−1} | n = 1, 2, … } for some ξ ∈ C, and infinitely many onedimensional simple modules L(0, ξ) = V (0)/N^{(x−ξ)}. In all other cases, M(λ) is the unique maximal submodule of V (λ), and there is a unique simple highest weight module of weight λ, L(λ) =^{def} V (λ)/M(λ), up to isomorphism.

Enveloping algebra examples.

Recall that the universal enveloping algebra U(𝔰𝔩_{2}) of 𝔰𝔩_{2} is isomorphic to the algebra A(2, −1, −2), and the universal enveloping algebra U(ℌ) of the Heisenberg Lie algebra ℌ is isomorphic to A(2, −1, 0). For all the algebras A(2, −1, γ), we have computed the values of λ_{n} in the proof of Case (d) of Theorem 2.13, and from (2.21) we see that

λ_{n} = (\frac{γ}{2} n + λ)(n + 1).

In the 𝔰𝔩_{2}-case, the operator h = du−ud is used rather than du. The eigenvalues of h are λ_{n} − λ_{n−1} = λ + nγ = λ − 2n, n = 0, 1, …, (as is customary in the representation theory of 𝔰𝔩_{2}), and V (λ) is simple if and only if λ ∉ Z_{≥0}. The analogous computation in the Heisenberg Lie algebra shows that the central element z = du − ud has constant eigenvalue λ_{n} = λ.

Quantum examples.

Let 𝔤 be a symmetrizable Kac-Moody Lie algebra (for example, a finite-dimensional simple Lie algebra) over C corresponding to the Cartan matrix 𝔄 = (a_{i,j})_{i,j=1}^{n}. Then there are relatively prime integers ℓ_{i} so that the matrix (ℓ_{i}a_{i,j}) is symmetric. Let C(q) be the field of rational functions in the indeterminate q over C. Assume

q_{i} = q^{ℓ_{i}},  and  [m]_{i} = \frac{q_{i}^{m} − q_{i}^{−m}}{q_{i} − q_{i}^{−1}}

for all m ∈ Z_{≥0}. When m ≥ 1, let

[m]_{i}! = ∏_{j=1}^{m}[j]_{i}.

Set [0]_{i}! = 1 and define

[ \begin{matrix}m \\ n \end{matrix} ]_{i} = \frac{[m]_{i}!}{[n]_{i}![m − n]_{i}!}.

Then the quantized enveloping algebra (quantum group) U_{q}(𝔤) of 𝔤 is the unital associative algebra over C(q) with generators E_{i}, F_{i}, K_{i}, K_{i}^{−1} (i = 1, …, n) subject to the relations

(Q1) K_{i}K_{i}^{−1} = K_{i}^{−1}K_{i}, K_{i}K_{j} = K_{j}K_{i}

(Q2) K_{i}E_{j}K_{i}^{−1} = q_{i}^{a_{i,j}}E_{j}

(Q3) K_{i}F_{j}K_{i}^{−1} = q_{i}^{−a_{i,j}}F_{j}

(Q4) E_{i}F_{j} − F_{j}E_{i} = δ_{i,j} \frac{K_{i} − K_{i}^{−1}}{q_{i} − q_{i}^{−1}}

(Q5) ∑_{k=0}^{1−a_{i,j}}(−1)^{k} [ \begin{matrix}1 − a_{i,j} \\ k \end{matrix} ]_{i} E_{i}^{1−a_{i,j}−k}E_{j}E_{i}^{k} = 0 for i ≠ j

(Q6) ∑_{k=0}^{1−a_{i,j}}(−1)^{k} [ \begin{matrix}1 − a_{i,j} \\ k \end{matrix} ]_{i} F_{i}^{1−a_{i,j}−k}F_{j}F_{i}^{k} = 0 for i ≠ j.

Suppose a_{i,j} = −1 = a_{j,i} for some i ≠ j, and consider the subalgebra U_{i,j} generated by E_{i}, E_{j}. In this special case, the quantum Serre relation (Q5) reduces to

\begin{matrix}E_{i}^{2}E_{j} − [2]_{i}E_{i}E_{j}E_{i} + E_{j}E_{i}^{2} = 0 and \\ E_{j}^{2}E_{i} − [2]_{j}E_{j}E_{i}E_{j} + E_{i}E_{j}^{2} = 0. \end{matrix}

Since −ℓ_{i} = ℓ_{i}a_{i,j} = ℓ_{j}a_{j,i} = −ℓ_{j}, the coefficients [2]_{i} and [2]_{j} are equal. The algebra U_{i,j} (with q is specialized to a complex number which is not a root of unity) is isomorphic to A([2]_{i}, −1, 0) by the mapping E_{i} ↦ d, E_{j} ↦ u. The same result is true if the corresponding F ’s are used in place of the E’s. In particular, when 𝔤 = 𝔰𝔩_{3}, the algebra U_{i,j} is just the subalgebra of U_{q}(𝔰𝔩_{3}) generated by the E’s.

Here we compute the values of λ_{n} in the case of the algebra A([2]_{i}, −1, 0). For convenience of notation we write

p = q_{i}.

Then α = [2]_{i} = \frac{p^{2} − p^{−2}}{p − p^{−1}} = p + p^{−1}, β = −1 and γ = 0 so that

α^{2} + 4β = p^{2} + 2 + p^{−2} − 4 = (p − p^{−1})^{2}.

Thus

\begin{aligned}r_{1} & = \frac{p + p^{−1} + p − p^{−1}}{2} = p \\ r_{2} & = \frac{p + p^{−1} − (p − p^{−1})}{2} = p^{−1}, \end{aligned}

\begin{aligned}( \begin{matrix}c_{1} \\ c_{2} \end{matrix} ) & = \frac{1}{p^{−1} − p} ( \begin{matrix}p^{−1} & −1 \\ −p & 1 \end{matrix} ) ( \begin{matrix}λ \\ (p + p^{−1})λ \end{matrix} ) \\ & = \frac{1}{p^{−1} − p} ( \begin{matrix}−pλ \\ p^{−1}λ \end{matrix} ) = \frac{1}{p − p^{−1}} ( \begin{matrix}pλ \\ −p^{−1}λ \end{matrix} ) . \end{aligned}

Therefore

λ_{n} = \frac{pλ}{p − p^{−1}} p^{n} − \frac{p^{−1}λ}{p − p^{−1}} p^{−n}

= (\frac{p^{n+1} − p^{−(n+1)}}{p − p^{−1}} )λ

= [n + 1]_{i}λ.

(q, r)-differential poset examples.

For an algebra arising from a (q, r)-differential poset the case of interest is the algebra A = A(α, β, γ) with defining relations

\begin{aligned}d^{2}u & = q(q + 1)dud − q^{3}ud^{2} + rd, \\ du^{2} & = q(q + 1)udu − q^{3}u^{2}d + ru. \end{aligned}

Here r is an arbitrary element of the field. For this algebra α^{2} + 4β = q^{2}(q − 1)^{2}, so we may assume that t = (1/2)q(q − 1) and r_{1} = 1/2(α + q(q − 1)) = q^{2} and r_{2} = 1/2(α − q(q − 1)) = q. (We will suppose throughout that q ≠ 0, ±1). Then

1 − α − β = 1 − q(q + 1) + q^{3} = q^{2}(q − 1) − (q − 1) = (q − 1)(q^{2} − 1)

so that

x_{n} = (1 − α − β)^{−1}γ = \frac{r}{(q − 1)(q^{2} − 1)}

for all n. We call that common value x. Recall that

( \begin{matrix}c_{1} \\ c_{2} \end{matrix} ) = \frac{1}{r_{2} − r_{1}} ( \begin{matrix}r_{2} & −1 \\ −r_{1} & 1 \end{matrix} ) ( \begin{matrix}λ − x \\ αλ + r − x \end{matrix} ) .

Now

\frac{1}{r_{2} − r_{1}} = \frac{1}{q(1 − q)},

and putting that into the above equation, we get after a bit of calculation

c_{1} = \frac{1}{q(1 − q)}(q(λ − x) − (αλ + r − x))

= \frac{−q}{1 − q} (λ + \frac{r}{q^{2} − 1})

c_{2} = \frac{1}{q(1 − q)}( − q^{2}(λ − x) + (αλ + r − x))

= \frac{1}{1 − q} (λ + \frac{r}{q − 1}).

Thus the eigenvalues of du for this algebra are

\begin{aligned}λ_{n} & = \frac{−q}{1 − q} (λ + \frac{r}{q^{2} − 1})q^{2n} + \frac{1}{1 − q} (λ + \frac{r}{q − 1})q^{n} + \frac{r}{(q − 1)(q^{2} − 1)} \\ & = \frac{q^{n+1} − 1}{q − 1}(q^{n}λ + \frac{q^{n} − 1}{q^{2} − 1}r). \end{aligned}

Fibonacci examples.

For the algebra A(1, 1, 0), the solutions to the associated linear recurrence λ_{n} = λ_{n−1} + λ_{n−2}, λ_{0} = λ, λ_{−1} = 0, (hence the eigenvalues of du on V (λ)) are given by the Fibonacci sequence λ_{0} = λ, λ_{1} = λ, λ_{2} = 2λ, λ_{3} = 3λ, λ_{4} = 5λ, . . .. In this case, the equations in Proposition 2.12 reduce to

λ_{n} = c_{1} (\frac{1 + \sqrt{5}}{2})^{n} + c_{2} (\frac{1 − \sqrt{5}}{2})^{n} ,

where

\begin{aligned}c_{1} & = \frac{λ\sqrt{5}}{5} (\frac{1 + \sqrt{5}}{2}) \\ c_{2} & = −\frac{λ\sqrt{5}}{5} (\frac{1 − \sqrt{5}}{2}) \end{aligned}

or

λ_{n} = λ\frac{\sqrt{5}}{5} ((\frac{1 + \sqrt{5}}{2})^{n+1} − (\frac{1 − \sqrt{5}}{2})^{n+1}) .

Lowest weight modules.

To create lowest weight modules for A = A(α, β, γ) we reverse the roles of d and u. Thus, we assume there is a vector w_{0} such that u · w_{0} = 0, ud · w_{0} = κw_{0} and W = Aw_{0}. When β ≠ 0, the eigenvalues of du are given by the sequence which has κ_{−1} = 0, κ_{0} = κ, an arbitrary complex number, and

(2.29)

βκ_{n} + ακ_{n−1} − κ_{n−2} = −γ, or equivalently
κ_{n} = β^{−1}( − ακ_{n−1} + κ_{n−2} − γ)

for all n ≥ 1.

Proposition 2.30. Let W (κ) be the C-vector space having basis {w_{n} | n = 0, 1, 2, … }.

(a) Assume β ≠ 0, κ_{−1} = 0, and κ_{0} = κ, an arbitrary element of C. Suppose κ_{n} for n ≥ 1 is as in (2.29), and define

(2.31)  \begin{aligned}u · w_{n} & = κ_{n−1}w_{n−1}, n ≥ 1, and u · w_{0} = 0 \\ d · w_{n} & = w_{n+1}. \end{aligned}

Then this action gives W (κ) the structure of a lowest weight A(α, β, γ)-module.

(b) When β = 0 and α ≠ 0, let κ_{n} be defined by

κ_{n} = −γ ∑_{j=1}^{n+1} α^{−j}

for all n ≥ 0. Then W (−γα^{−1}) with the action given by (2.31) is a lowest weight A(α, β, γ)-module.

(c) When γ ≠ 0, the only lowest weight A(0, 0, γ)-module is the 1-dimensional module W = Cw_{0} with d · w_{0} = 0 = u · w_{0}. When γ = 0, set κ_{n} = 0 for all n. Then W (0) with the action given by (2.31) is a lowest weight A(0, 0, 0)-module.

The proof of this result is identical to that of Proposition 2.2, and the following proposition can be shown using standard recurrence relation techniques.

Proposition 2.32. Assume β ≠ 0 and fix t ∈ C such that

t^{2} = \frac{α^{2} + 4β}{4β} .

Let κ_{−1} = 0, κ_{0} = κ ∈ C, and for n ≥ 1, let κ_{n} be given by the recurrence relation βκ_{n} + ακ_{n−1} − κ_{n−2} = −γ.

(i) If α^{2} + 4β ≠ 0, then

κ_{n} = c_{1}r_{1}^{n} + c_{2}r_{2}^{n} + x_{n},

where

r_{1} = −\frac{α}{2β} + t, r_{2} = −\frac{α}{2β} − t,

x_{n} = \{ \begin{matrix}(1 − α − β)^{−1}γ if α + β ≠ 1 \\ (α − 2)^{−1}γn if α + β = 1 and α ≠ 2, \end{matrix}

and ( \begin{matrix}c_{1} \\ c_{2} \end{matrix} ) = \frac{1}{r_{2} − r_{1}} ( \begin{matrix}r_{2} & −1 \\ −r_{1} & 1 \end{matrix} ) ( \begin{matrix}λ − x_{0} \\ αλ + γ − x_{1} \end{matrix} ) .

(ii) If α^{2} + 4β = 0, then α ≠ 0 and

κ_{n} = c_{1}s^{n} + c_{2}ns^{n} + x_{n}

where

s = \frac{α}{2β}

x_{n} = \{ \begin{matrix}(1 − α − β)^{−1}γ if α + β ≠ 1 \\ −\frac{γ}{2} n^{2} if α + β = 1 i.e. if α ≠ 2 and β = −1, \end{matrix}

and ( \begin{matrix}c_{1} \\ c_{2} \end{matrix} ) = ( \begin{matrix}1 & 0 \\ −\frac{2β}{α} & \frac{2β}{α} \end{matrix} ) ( \begin{matrix}κ − x_{0} \\ −\frac{(γ + ακ)}{β} − x_{1} \end{matrix} ) .

Doubly-infinite modules.

Proposition 2.33. Assume β ≠ 0 and let κ, λ be arbitrary but fixed complex numbers. Suppose λ_{0} = λ, λ_{1} = αλ + βκ + γ, and for n ≥ 2, λ_{n} is defined by the recurrence relation λ_{n} = αλ_{n−1} + βλ_{n−2} + γ. Set λ_{−1} = κ and define λ_{n−2} for all n ≤ 0 by the equation λ_{n} = αλ_{n−1} + βλ_{n−2} + γ. The complex vector space V (κ, λ) with basis {v_{n} | n ∈ Z} and with

(2.34)  \begin{matrix}d · v_{n} = λ_{n−1}v_{n−1} \\ u · v_{n} = v_{n+1} \end{matrix}

is an A(α, β, γ)-module.

Proof. As in our previous arguments, we may assume there is an action of the free associative algebra generated by d and u on V (κ, λ) given by (2.34). Then

\begin{aligned}(d^{2}u − αdud − βud^{2} − γd)v_{m} & = λ_{m−1}(λ_{m} − αλ_{m−1} − βλ_{m−2} − γ)v_{m−1} = 0 \\ (du^{2} − αudu − βu^{2}d − γu)v_{m} & = (λ_{m+1} − αλ_{m} − βλ_{m−1} − γ)v_{m} = 0, \end{aligned}

so that there is an induced action by A(α, β, γ). □

If λ_{m} = 0 for any m, then the span of the vectors {v_{n} | n ≥ m+1} is a submodule of V (κ, λ).

Dual modules.

Suppose V is any module for A = A(α, β, γ) and let {v_{a} | a ∈ I} be a basis for V. Assume v_{a}^{∗} ∈ Hom_{C}(V, C) is defined by v_{a}^{∗}(v_{b}) = δ_{a,b} for all a, b ∈ I. Let V^{∗} be the finite dual of V (i.e. the finite linear combinations of the elements v_{a}^{∗}). We will show that V^{∗} carries the structure of an A-module.

The antiautomorphism η

Define a mapping η : A → A by first specifying η(u) = d and η(d) = u and extending this to an antiautomorphism of the free associative algebra generated by u and d. Then

η(d^{2}u) = du^{2} = αudu + βu^{2}d + γu = η(αdud + βud^{2} + γd),

and a similar relation holds for du^{2}. Therefore η induces an antiautomorphism of order 2 on A (which we also denote by η).

Now define

(x · v^{∗})(w) = v^{∗}(η(x) · w)

for all x ∈ A, w ∈ V, and v^{∗} ∈ V^{∗}. It is easy to see that this makes V^{∗} an A-module.

Suppose now that V = V (λ) is a highest weight A-module with basis {v_{n} | n = 0, 1, … }. Then

\begin{aligned}(u · v_{ℓ}^{∗})(v_{n}) & = v_{ℓ}^{∗}(d · v_{n}) = λ_{n−1}v_{ℓ}^{∗}(v_{n−1}) \\ & = \{ \begin{matrix}0 if n ≠ ℓ + 1 \\ λ_{ℓ} if n = ℓ + 1. \end{matrix} \end{aligned}

Therefore u·v_{ℓ}^{∗} = λ_{ℓ}v_{ℓ+1}^{∗}. Similarly, d·v_{ℓ}^{∗} = v_{ℓ−1}^{∗}. If λ_{n} = 0 for some n, then V (λ)^{∗} is not generated by v_{0}^{∗}. But whenever λ_{n} ≠ 0 for all n, then V (λ)^{∗} is a highest weight module. In fact, in that case V (λ)^{∗} ≅ V (λ) under the mapping which sends λ_{n−1} ⋯ λ_{1}λ_{0}v_{n}^{∗} to v_{n}. We summarize what we have shown for highest weight modules and the analogous results for lowest weight and doubly-infinite modules in the following proposition.

Proposition 2.35. If Y is a module for A(α, β, γ), let Y^{∗} denote the finite dual of Y.

(a) If V (λ) is a highest weight module as in Proposition 2.2, then V (λ)^{∗} is an A(α, β, γ)-module (which is a highest weight module isomorphic to V (λ) whenever λ_{n} ≠ 0 for all n) under the action

\begin{matrix}u · v_{n}^{∗} = λ_{n}v_{n+1}^{∗} \\ d · v_{n}^{∗} = v_{n−1}^{∗}. \end{matrix}

(b) If W (κ) is a lowest weight module as in Proposition 2.30, then W (κ)^{∗} is an A(α, β, γ)-module (which is a lowest weight module isomorphic to W (κ) whenever κ_{n} ≠ 0 for all n) under the action

\begin{matrix}u · w_{n}^{∗} = w_{n−1}^{∗} \\ d · v_{n}^{∗} = κ_{n}w_{n+1}^{∗}. \end{matrix}

(c) If V (κ, λ) is a doubly-infinite module as in Proposition 2.35, then V (κ, λ)^{∗} is a doubly-infinite module under the action

\begin{matrix}u · v_{n}^{∗} = λ_{n}v_{n+1}^{∗} \\ d · v_{n}^{∗} = v_{n−1}^{∗}. \end{matrix}

§3. Structural results on down-up algebras

Theorem 3.1. (Poincaré-Birkhoff-Witt Theorem) Assume A = A(α, β, γ) is a down-up algebra over C. Then {u^{i}(du)^{j}d^{k} | i, j, k = 0, 1, … } is a basis of A.

Proof. It is clear from the defining relators f_{1} = d(du)−α(du)d−βud^{2} −γd and f_{2} = (du)u − αu(du) − βu^{2}d − γu that the monomials u^{i}(du)^{j}d^{k}, i, j, k = 0, 1, … are all the irreducible words in A and so span A. We show that they are linearly independent by applying the Diamond Lemma (see [Be]).

We can order the words in the free algebra C⟨d, u⟩ by their total degree in d and u, and then lexicographically with d > u. Two words in C⟨d, u⟩ admit a composition if an end of one is a beginning of the other. We need to argue that whenever the largest words w_{1} = d^{2}u and w_{2} = du^{2} in the relators f_{1} and f_{2} admit a composition say, w_{1} = ab and w_{2} = bc or w_{2} = a′b′ and w_{1} = b′c′, then

\begin{aligned}f_{1}c − af_{2} & = a_{1}t_{1}f_{1}x_{1} + a_{2}t_{2}f_{2}x_{2} or \\ f_{2}c′ − a′f_{1} & = b_{1}y_{1}f_{1}z_{1} + b_{2}y_{2}f_{2}z_{2}, \end{aligned}

where a_{i}, b_{i} ∈ C, t_{i}, x_{i}, y_{i}, z_{i} are words in d, u (possibly empty), t_{i}f_{i}x_{i} < abc, and y_{i}f_{i}z_{i} < a′b′c′ for i = 1, 2. There is exactly one admissible composition, d^{2}u^{2} = d(du)u, of the largest words in the relators, and

\begin{matrix}(d^{2}u − αdud − βud^{2} − γd) & u − d(du^{2} − αudu − βu^{2}d − γu) \\ = βud^{2}u − βdu^{2}d = 0. \end{matrix}

Therefore by the Diamond Lemma, the irreducible words form a basis of A. □

Corollary 3.2. The Gelfand-Kirillov dimension of any down-up algebra A(α, β, γ) is 3.

Proof. Let A^{(n)} = span_{C}{u^{i}(du)^{j}d^{k} | i + 2j + k ≤ n}. The spaces A^{(n)} afford a filtration (0) ⊂ A^{(0)} ⊂ A^{(1)} ⊂ ⋯ ⊂ ∪_{n}A^{(n)} = A(α, β, γ) of the down-up algebra, and A^{(m)}A^{(n)} ⊆ A^{(m+n)} since the defining relations replace the words d^{2}u and du^{2} by words of the same or lower total degree. The number of monomials u^{i}(du)^{j}d^{k} with i + 2j + k = ℓ is (m + 1)(m + 1) if ℓ = 2m and is (m + 1)(m + 2) if ℓ = 2m + 1. Thus, dimA^{(n)} is a polynomial in n with positive coefficients of degree 3, and the Gelfand-Kirillov dimension is given by

GKdim(A(α, β, γ)) = lim sup_{n→∞} log_{n}(dim A^{(n)})

= lim_{n→∞} \frac{ln ( dim A^{(n)})}{ln n}
= 3. □

Recall that the Jacobson radical of a ring R is the intersection of all the annihilators of the simple R-modules. It has many other characterizations, but we will not need them here. A ring whose Jacobson radical is zero is said to be semiprimitive.

Theorem 3.3. If A(α, β, γ) has infinitely many simple Verma modules V (λ), then the intersection of the annihilators of the simple Verma modules is zero. Therefore, such a down-up algebra A(α, β, γ) is semiprimitive.

Proof. Suppose the element a = ∑_{i,j,k} a_{i,j,k}u^{i}(du)^{j}d^{k} annihilates all the simple modules V (λ). Assume there is a least value ℓ so that a_{i,j,ℓ} ≠ 0 for some i, j. Let V (λ) be a simple highest weight module for A(α, β, γ) of weight λ. Applying a to the vector v_{ℓ} in V (λ) we get

0 = a · v_{ℓ} = (λ_{ℓ−1} ⋯ λ_{1}λ_{0}) ∑_{i,j} a_{i,j,ℓ}u^{i}(du)^{j} · v_{0}

= (λ_{ℓ−1} ⋯ λ_{1}λ_{0}) ∑_{i,j} a_{i,j,ℓ}λ_{0}^{j}u^{i} · v_{0}

= (λ_{ℓ−1} ⋯ λ_{1}λ_{0}) ∑_{i,j} a_{i,j,ℓ}λ_{0}^{j}v_{i}.

Recall that a necessary and sufficient condition for V (λ) to be simple is that none of the λ_{n} is zero. The elements v_{i} are linearly independent in V (λ), so combining those results, we see for each i that

0 = ∑_{j} a_{i,j,ℓ}λ_{0}^{j} = ∑_{j} a_{i,j,ℓ}λ^{j}.

Because there are infinitely many values of λ that give simple highest weight modules and only finitely many λ that can be the roots of a nonzero polynomial, we see that the coefficients a_{i,j,ℓ} must all be zero. Since the Jacobson radical is the intersection of the annihilators of all the simple A-modules, it too must be zero. □

Let us examine in detail when a down-up algebra A(α, β, γ) has infinitely many simple Verma modules. Since the Verma module V (λ) is simple if and only if λ_{n} ≠ 0 for all n ≥ 1, the analysis requires the expressions for λ_{n} in Proposition 2.12.

Case 1. α^{2} + 4β ≠ 0. Here λ_{n} = c_{1}r_{1}^{n} + c_{2}r_{2}^{n} + x_{n}, (see Proposition 2.12 (i)), which can be expressed as a polynomial in λ:

λ_{n} = \frac{1}{r_{2} − r_{1}} ((r_{2} − α)r_{1}^{n} + (−r_{1} + α)r_{2}^{n})λ + b_{n},

where r_{2} − α = 1/2α − t − α = −r_{1} and −r_{1} + α = −1/2α − t + α = r_{2}. Thus the coefficient of λ is (r_{2} − r_{1})^{−1}(r_{2}^{n+1} − r_{1}^{n+1}). If this is 0, then r_{1}^{n+1} = r_{2}^{n+1}, which implies (α+τ)^{n+1} = (α−τ)^{n+1} where τ = 2t and τ^{2} = α^{2}+4β. Since τ^{2} = α^{2}+4β ≠ 0, α + τ and α − τ are unequal, and we may suppose one of them — say α + τ ≠ 0.

Then 1 = ((α + τ)^{−1}(α − τ))^{n+1}, so in fact both are nonzero, and the same holds

true if α−τ ≠ 0. We may suppose (α+τ)^{−1}(α−τ) = e^{iθ} where θ = 2πk/(n+1) for some 1 ≤ k ≤ n. Then α − τ = (α + τ)e^{iθ} gives α(e^{iθ} − 1) = −τ(e^{iθ} + 1). Squaring that relation shows that α^{2} = −β(e^{iθ} + e^{−iθ})^{2} = −4β cos^{2}(θ/2). Consequently, λ_{n} is a degree 1 polynomial in λ except when α^{2} = −4β cos^{2}(θ/2) where e^{iθ} is an (n + 1)st root of unity.

Case 2. α^{2} + 4β = 0 and α ≠ 0. In this case λ_{n} = c_{1}s^{n} + c_{2}ns^{n} + x_{n} = ((n + 1)s^{n})λ + b_{n}, where s = α/2. Thus, λ_{n} is a degree 1 polynomial in λ for all n.

Case 3. α^{2} + 4β = 0 and α = 0 (so β = 0). Here λ_{n} = γ for all n ≥ 1.

Proposition 3.4. If

(a) α^{2} + 4β ≠ 0 and α^{2} ≠ −4β cos^{2}(θ/2) for some root of unity e^{iθ},

(b) α^{2} + 4β = 0 and α ≠ 0, or

(c) α = 0 = β and γ ≠ 0,

then A(α, β, γ) has infinitely many simple Verma modules V (λ).

Proof. In the cases described in (a) and (b), each λ_{n} can be regarded as a line λ_{n} = m_{n}λ + b_{n} with nonzero slope m_{n}. Each such line intersects the line y = 0 at one value of λ. Avoiding all those countably many values by choosing λ appropriately, we get infinitely many simple Verma modules V (λ). □

Proposition 3.5. A down-up algebra A = A(α, β, γ) is Z-graded, A = ⊕_{n}A_{n}, where A_{n} = span_{C}{u^{i}(du)^{j}d^{k} | i, j, k = 0, 1, …, and i − k = n}, and A_{0} is a commutative subalgebra.

Proof. The free associative algebra over C generated by d and u can graded by assigning deg(d) = −1 and deg(u) = 1 and extending this by using deg(ab) = deg(a)+deg(b). The relations d^{2}u = αdud+βud^{2} +γd and du^{2} = αudu+βu^{2}d+γu are homogeneous, so the down-up algebra inherits the grading.

To prove that A_{0} is commutative, we first show that u^{i}(du)^{j}d^{i} commutes with du and ud for all i, j. This is true if i+j ≤ 1 by (2.11). Suppose we know the result for all i+j < N, and let i+j = N ≥ 2. Consider (u^{i}(du)^{j}d^{i})(ud). Now if i = 0 this equals (ud)(u^{i}(du)^{j}d^{i}) since ud and du commute. If i = 1, then u(du)^{j}d = (ud)^{j+1}, which commutes with ud. So we may assume that i ≥ 2. Then

(u^{i}(du)^{j}d^{i})(ud) = u^{i}(du)^{j}d^{i−2}(αdud + βud^{2} + γd)d
= αu^{2}(u^{i−2}(du)^{j}d^{i−2})(du)d^{2}+
βu^{2}(u^{i−2}(du)^{j}d^{i−2})(ud)d^{2} + γu^{i}(du)^{j}d^{i} while

(ud)(u^{i}(du)^{j}d^{i}) = u(αudu + βu^{2}d + γu)(u^{i−2}(du)^{j}d^{i})
= αu^{2}(du)(u^{i−2}(du)^{j}d^{i−2})d^{2}+

βu^{2}(ud)(u^{i−2}(du)^{j}d^{i−2})d^{2} + γu^{i}(du)^{j}d^{i}.

These two expressions are equal by our induction hypothesis. Analogously, we consider (u^{i}(du)^{j}d^{i})(du), where we can suppose that i ≥ 1. Then

(u^{i}(du)^{j}d^{i})(du) = u^{i}(du)^{j}d^{i−1}(αdud + βud^{2} + γd)
= α(u^{i}(du)^{j}d^{i})(ud)+

βu(u^{i−1}(du)^{j}d^{i−1})(ud)d + γu^{i}(du)^{j}d^{i} which equals

(du)(u^{i}(du)^{j}d^{i}) = (αudu + βu^{2}d + γu)(u^{i−1}(du)^{j}d^{i})
= α(ud)(u^{i}(du)^{j}d^{i})+
βu(ud)(u^{i−1}(du)^{j}d^{i−1})d + γu^{i}(du)^{j}d^{i}

by what we have already established.

The proof of the proposition will be complete once we show that u^{i}(du)^{j}d^{i} commutes with u^{k}(du)^{ℓ}d^{k} for all i, j, k, ℓ. This is true if i + j ≤ 1 or if i or k is ≤ 1 by our previous considerations, so we assume it is true when i + j + k + ℓ < N and prove it for i + j + k + ℓ = N. We suppose i, k ≥ 2. Now

(u^{i}(du)^{j}d^{i})(u^{k}(du)^{ℓ}d^{k}) = u^{i}(du)^{j}d^{i−2}(αdud + βud^{2} + γd)u^{k−1}(du)^{ℓ}d^{k}
= αu(u^{i−1}(du)^{j}d^{i−1})(ud)(u^{k−1}(du)^{ℓ}d^{k−1})d+

βu^{2}(u^{i−2}(du)^{j}d^{i−2})(ud)(du)(u^{k−2}(du)^{ℓ}d^{k−2})d^{2}+

u(u^{i−1}(du)^{j}d^{i−1})(u^{k−1}(du)^{ℓ}d^{k−1})d,

which can be seen to equal the same expression with the roles of i, j and k, ℓ reversed by induction. □

The center of a down-up algebra.

Let Z(A) = {z ∈ A | za = az for all a ∈ A} denote the center of a down-up algebra A = A(α, β, γ).

Proposition 3.6. Suppose A = ∑_{j∈Z} A_{j} is the Z-grading of A = A(α, β, γ) as in Proposition 3.5, and assume that A has infinitely many simple Verma modules. Then:

(a) The center Z(A) of A is contained in A_{0} = span_{C}{u^{i}(du)^{j}d^{i} | i, j = 0, 1, … }.

(b) If z ∈ Z(A) and V (λ) is any Verma module of A, then z acts as a scalar, say χ_{λ}(z), on V (λ). The mapping χ_{λ} : Z(A) → C is an algebra homomorphism.

(c) A scalar π ∈ C is linked to λ if π = λ_{n+1} for some n ≥ 0 with λ_{n} = 0, where the sequence λ_{1}, λ_{2}, … is constructed using the recurrence relation in (2.1) starting with λ_{−1} = 0 and λ_{0} = λ. If π is linked to λ, then χ_{π} = χ_{λ}.

Proof. Consider the sequence λ = λ_{0}, λ_{1}, λ_{2}, … from (2.1), and let Λ(λ) = {n ∈ Z_{≥0} | λ_{n} = 0}. Suppose v_{0}, v_{1}, … is the canonical basis for V (λ) as in Proposition 2.2, and let z ∈ Z(A). From dz · v_{0} = zd · v_{0} = 0 we see that z · v_{0} = χ_{λ}(z)v_{0} + ∑_{n∈Λ(λ),n>0} a_{n}v_{n} for some scalars a_{n} and χ_{λ}(z). Therefore, when λ ≠ 0 and V (λ) is simple (i.e. when Λ(λ) = ∅) then z · v_{0} = χ_{λ}(z)v_{0}. But za · v_{0} = az · v_{0} = χ_{λ}(z)a · v_{0} for all a ∈ A, and since V (λ) = Av_{0}, this shows that z acts as the scalar χ_{λ}(z) on such Verma modules. Now write z as a sum of its homogeneous components, z = ∑_{j} z_{j}, z ∈ A_{j}. Then since A_{j}v_{n} ⊆ Cv_{j+n} for all j, n, and since zv_{n} = χ_{λ}(z)v_{n} for all n when V (λ) is simple, it must be that z_{j} for j ≠ 0 lies in the annihilator of V (λ). But the intersection of the annihilators of the simple V (λ) modules is zero by Theorem 3.3. Thus, z = z_{0} ∈ A_{0}, as claimed in (a). Once we know that, then it follows for any Verma module V (λ) that z·v_{0} = χ_{λ}(z)v_{0} for some scalar χ_{λ}(z) ∈ C. As before using V (λ) = Av_{0}, we see z acts as χ_{λ}(z) on all of V (λ). As χ_{λ}(zz′)v_{0} = zz′ · v_{0} = z · (z′v_{0}) = χ_{λ}(z)χ_{λ}(z′)v_{0}, it is clear χ_{λ} is an algebra homomorphism. Part (c) is apparent also, since when λ_{n} = 0, then span_{C}{v_{j} | j ≥ n + 1} is a submodule of V (λ) isomorphic to V (λ_{n+1}). □

§4. Category O modules

The representation theory of finite-dimensional complex semisimple Lie algebras centers on a certain important collection of weight modules, the so-called category O modules, whose study was initiated by Bernstein, Gelfand, and Gelfand [BGG]. In this section we define an analogous category of modules for each down-up algebra A = A(α, β, γ). We begin with a general result about weight vectors and then specialize to the modules in O and derive some results about them.

Proposition 4.1. Suppose M is an A-module and m ∈ M is a vector of weight ν = (ν′, ν′′).

(a) Then u · m is a vector of weight μ = (μ′, μ′′) for μ′ = αν′ + βν′′ + γ and μ′′ = ν′.

(b) If β ≠ 0, then d · m is a vector of weight δ = (δ′, δ′′) where δ′ = ν′′ and δ′′ = β^{−1}(ν′ − αν′′ − γ).

Proof. The assertions are evident from the following calculations:

(4.2)

(du)u · m = αu(du) · m + βu(ud) · m + γu · m = (αν′ + βν′′ + γ)u · m
(ud)u · m = u(du) · m = ν′u · m
(du)d · m = d(ud) · m = ν′′d · m

(ud)d · m = β^{−1}(d(du) · m − αd(ud) · m − γd · m) = β^{−1}(ν′ − αν′′ − γ)d · m. □

(4.3) The category O consists of all A-modules M satisfying the following conditions:

(a) M is a weight module relative to 𝔥 = span_{C}{du, ud}, i.e. M = ⨁_{ν} M_{ν} where M_{ν} = {m ∈ M | h · m = ν(h)m for all h ∈ 𝔥};

(b) d acts locally nilpotently on M, so that for each m ∈ M, d^{n} · m = 0 for some n.

(c) M is a finitely generated A-module.

The category O is closed under taking submodules and quotients. It contains all the Verma modules V (λ) and their simple quotient modules L(λ) = V (λ)/M(λ) (and in the case that γ = 0 and λ = 0, the one-dimensional quotients L(0, ξ) =^{def} V (0)/N^{(x−ξ)} (see Corollary 2.28)). Modules in the category O enjoy some of the same basic properties as the category O modules for complex semisimple Lie algebras and for the algebras studied by Smith [Sm].

(4.4) Suppose ω = (ω′, ω′′), where ω′, ω′′ ∈ C. Let Ω(ω) denote the set weights defined inductively starting with ω_{0} = (ω′_{0}, ω′′_{0}), where ω′_{0} = ω′ and ω′′_{0} = ω′′, and proceeding for n ≥ 1 by setting ω_{n} = (ω′_{n}, ω′′_{n}) where

\begin{aligned}ω′_{n} & = αω′_{n−1} + βω′′_{n−1} + γ \\ ω′′_{n} & = ω′_{n−1}. \end{aligned}

Note that ω′_{n} = αω′_{n−1} + βω′_{n−2} + γ for n ≥ 2. When ω = (λ, 0) then ω_{n} = (λ_{n}, λ_{n−1}), where λ_{n} is as in Proposition 2.1. Thus Ω(ω) is just the set of weights of the Verma module V (λ) in the particular instance that ω = (λ, 0).

Lemma 4.5. Suppose β ≠ 0. Assume M is an object in the category O, and let Ω(M) = {ν ∈ 𝔥^{∗} | M_{ν} ≠ (0)} be the set of weights of M. Then there exists a finite set of weights ω^{1}, …, ω^{r} so that Ω(M) ⊆ ∪_{i}Ω(ω^{i}).

Proof. Because M is finitely generated, there exists a finite set of weight vectors m_{1}, …, m_{ℓ} generating M. Then ∑_{j} C[d]m_{j} is finite-dimensional, and moreover it is 𝔥-invariant for 𝔥 = span_{C}{du, ud} by Proposition 4.1. So we can choose a basis y_{1}, …, y_{r} for it such that y_{i} has some weight ω^{i} for each i. Then by Theorem 3.1,

M = ∑_{j} Am_{j} = ∑_{j} C[u]C[du]C[d]m_{j} = ∑_{i} C[u]C[du]y_{i} = ∑_{i} C[u]y_{i}.

The vectors in C[u]y_{i} have weights in Ω(ω^{i}) by Proposition 4.1, so the relation Ω(M) ⊆ ∪_{i}Ω(ω^{i}) is apparent. □

Lemma 4.6. Suppose β ≠ 0. If M is a simple object in the category O, then M ≅ L(λ) for some λ, or γ = 0 and M ≅ V (0)/N^{(x−ξ)} = L(0, ξ) for some ξ ∈ C (as in Corollary 2.28).

Proof. It follows from the local nilpotence of d and from Proposition 4.1 that M contains some highest weight vector v_{0}. The simplicity of M forces M = Av_{0} so that M is a highest weight module. As highest weight modules are quotients of Verma modules, M is as asserted. □

§5. Category O′ modules

In this section we introduce and investigate a more general category of modules for the down-up algebra A = A(α, β, γ). We require throughout that β ≠ 0.

(5.1) The category O′ consists of all A-modules M satisfying the following conditions:

(a) M is a weight module relative to 𝔥 = span_{C}{du, ud}

(b) C[d]m is finite-dimensional for each m ∈ M.

(c) M is a finitely generated A-module.

The modules in O clearly belong to O′, but we will show later that O′ is larger than O by determining the simple modules in O′. Two mappings on weights play a prominent role in these investigations.

Recall (Proposition 4.1) that if a vector m in an A-module has weight ν = (ν′, ν′′) relative to 𝔥, then d · m has weight

(5.2)

δ(ν) = (δ(ν)′, δ(ν)′′)  where
δ(ν)′ = ν′′  and  δ(ν)′′ = β^{−1}(ν′ − αν′′ − γ),

and u · m has weight (5.3)

μ(ν) = (μ(ν)′, μ(ν)′′)  where
μ(ν)′ = αν′ + βν′′ + γ  and  μ(ν)′′ = ν′.

An easy direct computation shows that δ(μ(ν)) = ν and μ(δ(ν)) = ν, (or this can be readily seen from the fact that (du) · m = ν′m and (ud) · m = ν′′m). Observe that the set of weights Ω(ω) in (4.4) alternately can be described as

(5.4)  Ω(ω) = {μ^{k}(ω) | k = 0, 1, … }.

The set of weights of the lowest weight module W (κ) is just {δ^{k}(ω) | k = 0, 1, … }, where ω = (0, κ).

Theorem 5.5. Suppose F is a set of weights such that δ(ω), μ(ω) ∈ F whenever ω ∈ F. Suppose ρ ∈ C is nonzero, and let N(F, ρ) be the C-vector space with basis {v_{ω} | ω ∈ F }.

(a) Define

(5.6)  \begin{aligned}d · v_{ω} & = ρv_{δ(ω)} \\ u · v_{ω} & = ρ^{−1}μ(ω)′′v_{μ(ω)}. \end{aligned}

Then this action extends to give an A(α, β, γ)-module action on N(F, ρ).

(b) If F is generated by any weight ν = (ν′, ν′′) ∈ F under the action of δ and μ, and if ν′ ≠ 0 for any ν ∈ F, then N(F, ρ) is a simple A(α, β, γ)-module.

Proof. Let N(F, ρ) be as in the statement of the theorem, and define an action of the free associative algebra C⟨d, u⟩ on N(F, ρ) using (5.6). Then

(5.7)  \begin{aligned}(du) · v_{ω} & = ρ^{−1}μ(ω)′′d · v_{μ(ω)} = μ(ω)′′v_{ω} = ω′v_{ω} \\ (ud) · v_{ω} & = ρu · v_{δ(ω)} = μ(δ(ω))′′v_{ω} = ω′′v_{ω}. \end{aligned}

Utilizing (5.7) together with (5.2) and (5.3), it is easy to see that

\begin{matrix}(d^{2}u − αdud − βud^{2} − γd) · v_{ω} = ρ(ω′ − αω′′ − βδ(ω)′′ − γ)v_{δ(ω)} = 0 \\ (du^{2} − αudu − βu^{2}d − γu) · v_{ω} = ρ^{−1}μ(ω)′′(μ(ω)′ − αω′ − βω′′ − γ)v_{μ(ω)} = 0. \end{matrix}

Thus, there is an induced action making N(F, ρ) into an A(α, β, γ)-module.

Now suppose F is generated by any weight ν ∈ F under the maps δ and μ defined in (5.2) and (5.3). It follows from (5.7) that N(F, ρ) is a weight module. Let M be a nonzero submodule of N(F, ρ), and assume v = ∑_{ω} a_{ω}v_{ω} is a nonzero vector in M. Since any submodule of a weight module is a weight module, v_{ω} ∈ M if a_{ω} ≠ 0. Now whenever v_{ω} ∈ M for some ω, then d · v_{ω} = ρv_{δ(ω)} and u · v_{ω} = ρ^{−1}μ(ω)′′v_{μ(ω)} both belong to M. Since μ(ω)′′ = ω′ ≠ 0 for any ω ∈ F by assumption, v_{μ(ω)} ∈ F. Because F is generated by ω under δ and μ, M contains v_{τ} for all τ ∈ F. As a consequence, M = N(F, ρ) and N(F, ρ) is a simple A(α, β, γ)-module. □

Suppose now that M is a simple module in the category O′. If M contains a highest weight vector, then M is as in Lemma 4.6. So we may assume henceforth that M has no highest weight vectors. Let m be a nonzero weight vector in M and consider the finite-dimensional space C[d]m. If d^{k} · m = 0 for some k, then d^{k−1} · m is a highest weight vector in M contrary to assumption. It must be that d acts nonnilpotently on C[d]m. Let v be an eigenvector of d in C[d]m corresponding to nonzero eigenvalue, say d · v = ρv, ρ ≠ 0. Write v = ∑_{ω} v_{ω} where v_{ω} belongs to the weight space M_{ω}.

Since d · v = ρv, it must be that d · v has the same number of weight components as v. Equating corresponding components on both sides of the equation gives

(5.8)  d · v_{ω} = ρv_{δ(ω)}.

Thus if S is the C-span of the weight components v_{ω} of v, then dS = S. Now ρu · v = (ud) · v = ∑_{ω} ω′′v_{ω}, which implies that

(5.9)  u · v_{ω} = ρ^{−1}μ(ω)′′v_{μ(ω)}.

Thus, S is an A(α, β, γ)-submodule of M. By simplicity, S = M.

Since there are only finitely many nonzero weight component summands, δ^{ℓ}(ω) = δ^{k}(ω) for some ℓ > k. Applying μ^{k} to both sides, we see δ^{(ℓ−k)}(ω) = ω. Thus, there is some smallest value p so that δ^{p}(ω) = ω and v_{ω}, v_{δ(ω)}, …, v_{δ^{p−1}(ω)} are all nonzero. Their C-span S′ is invariant under d. Using (5.9) and the fact that μ(δ^{i}(ω)) = δ^{i−1}(ω) for i ≥ 1 and μ(ω) = μ(δ^{p}(ω)) = δ^{p−1}(ω), we see that uS′ ⊆ S′ also. Thus, M = S′ by simplicity. The set F = {ω, δ(ω), …, δ^{p−1}(ω)} is δ and μ-invariant. Any weight in it generates the whole set under δ and μ. Now suppose ν is a weight of F and ν′ = μ(ν)′′ = 0. Setting v_{n} = ρ^{n}v_{δ^{n}(ν)} for n = 0, 1, …, p − 1, we have u · v_{0} = 0, d · v_{n} = v_{n+1} for 0 ≤ n ≤ p − 1, and d · v_{n} = ρ^{p}v_{0}. In this case M is a finite-dimensional lowest weight module – a quotient of W (κ) for κ = ν′′. The sole remaining possibility is that ν′ ≠ 0 for any ν ∈ F. In this situation M = N(F, ρ), (and N(F, ρ) is in fact a simple module - compare Theorem 5.5 (b)).

Let us summarize what we have shown:

Theorem 5.10. Assume M is a simple module in the category O′. Then there are three possibilities:

(a) M is a highest weight module, that is, M is isomorphic to L(λ) for some λ or to L(0, ξ) for some ξ ∈ C (when γ = 0).

(b) M is a finite-dimensional lowest weight module with weights ν, δ(ν), …, δ^{p−1}(ν) such that δ^{p}(ν) = ν.

(c) M is isomorphic to N(F, ρ) for some ρ ≠ 0 and some finite set F = {ν, δ(ν), …, δ^{p−1}(ν)} such that δ^{p}(ν) = ν.

We conclude this section by analyzing the special case that the set F is cyclically generated by ω = (λ, κ) under δ and μ, and the weights in

F = {…, μ^{2}(ω), μ^{1}(ω), ω, δ(ω), δ^{2}(ω), … }

are all distinct. Observe that if we define λ_{n} by first setting λ_{0} = λ and λ_{−1} = κ and proceeding by

\begin{aligned}λ_{n} & = αλ_{n−1} + βλ_{n−2} + γ  for n ≥ 1 \\ λ_{−n} & = β^{−1}(λ_{−n+2} − αλ_{−n+1} − γ)  for n ≥ 2, \end{aligned}

then

\begin{aligned}(λ_{n}, λ_{n−1}) & = μ^{n}(ω)  for n ≥ 0 \\ (λ_{−n}, λ_{−n−1}) & = δ^{n}(ω)  for n ≥ 1. \end{aligned}

These are just the weights of the doubly-infinite module V (κ, λ) in Proposition 2.33. It is easy to see using Proposition 2.35 that there is an isomorphism from the dual module V (κ, λ)^{∗} to N(F, 1) taking v_{n}^{∗} ↦ v_{μ^{n}(ω)} for n ≥ 0 and v_{−n}^{∗} ↦ v_{δ^{n}(ω)} for n ≥ 1. Theorem 5.5 shows that if ν′ = λ_{n} ≠ 0 for any ν = (λ_{n}, λ_{n−1}), n ∈ Z, then N(F, 1) ≅ V (κ, λ)^{∗} is simple. This forces V (κ, λ) to be simple as well, for the annihilator in V (κ, λ)^{∗} of any proper submodule of V (κ, λ) would be a proper submodule of V (κ, λ)^{∗}. Consequently, an analogue of the result in Proposition 2.4 holds for the doubly-infinite module V (κ, λ):

Proposition 5.11. Suppose λ_{0} = λ and λ_{−1} = κ and define the sequence λ_{n} as in Proposition 2.33. If λ_{n} ≠ 0 for any n, then the doubly-infinite module V (κ, λ) is simple.

§6. One-dimensional A(α, β, γ)-modules and isomorphisms

Consider a one-dimensional space X = Cx with

d · x = ax  and  u · x = bx.

Then this extends to a module for A = A(α, β, γ) if and only if

0 = (d^{2}u − αdud − βud^{2} − γd) · x = a((1 − α − β)ab − γ)x
0 = (du^{2} − αudu − βu^{2}d − γu) · x = b((1 − α − β)ab − γ)x.

If a = 0 = b, then X = V (0)/M(0). We may suppose that not both a, b are 0, so that X is an A-module if and only if (1 − α − β)ab = γ holds. When γ = 0 and α + β = 1, then any choice of a and b gives X the structure of an A-module, so the one-dimensional A-modules are parametrized by C^{2} in this instance. When γ = 0 and α + β ≠ 1, then either a = 0 or b = 0 must hold. The one-dimensional A-modules are parametrized by {(a, 0) | a ∈ C} ∪ {(0, b) | b ∈ C} in this case. When γ ≠ 0, we may suppose that γ = 1. ¿From (1 − α − β)ab = γ = 1, we see that b = a^{−1}(1 − α − β)^{−1}. Thus, the one-dimensional A-modules are parametrized by C for γ ≠ 0 and α + β ≠ 1, and by a = 0 = b when γ ≠ 0 and α + β = 1.

Now (du) · x = abx = (ud) · x, so that x has weight ν = (ab, ab) for all onedimensional A-modules X. When β ≠ 0, then δ(ν) = (ab, β^{−1}(ab − αab − 1), which equals (ab, ab) when γ = 1. Thus, in the case β ≠ 0, γ ≠ 0 the onedimensional module X is isomorphic to N(F, a) where F = {ν = (ab, ab)} and δ(ν) = ν (compare Theorem 5.10). To summarize we have:

Theorem 6.1. The one-dimensional modules for the down-up algebra A(α, β, γ) are parametrized by

(a) C^{2} when γ = 0 and α + β = 1;

(b) {(a, 0) | a ∈ C} ∪ {(0, b) | b ∈ C} when γ = 0 and α + β ≠ 1;

(c) C when γ ≠ 0 and α + β ≠ 1;

(d) (0, 0) when γ ≠ 0 and α + β = 1.

Corollary 6.2. If the down-up algebras A(α, β, γ) and A(α′, β′, γ′) are isomorphic, then either both α + β and α′ + β′ are equal to 1 or both are different from 1, and either both γ and γ′ are 0 or both are different from 0.

Proof. The result is a simple consequence of Theorem 6.1 and the fact that if ϕ : A(α, β, γ) → A(α′, β′, γ′) is an isomorphism, then any one-dimensional module X = Cx for A(α′, β′, γ′) becomes one for A(α, β, γ) by defining t · x = ϕ(t) · x for all t ∈ A(α, β, γ). □

Open Problems. In conclusion we mention some noteworthy open questions concerning down-up algebras which are not addressed here. Problems (a)-(f) were posed in [B], and since the original version of this paper appeared, several of these problems have been solved.

(a) Determine when the algebra A(α, β, γ) is Noetherian. Kirkman, Musson, and Passman [KMP] have recently proven that A(α, β, γ) is left and right Noetherian if and only if β ≠ 0 if and only if du and ud are algebraically independent.

(b) Find conditions on α, β, γ for A(α, β, γ) to be a domain. When β = 0, then d(du−αud−γ1) = 0 so that A(α, β, γ) has zero divisors for any choice of α, γ ∈ C. It has been shown in [KMP] that A(α, β, γ) is a domain if and only if β ≠ 0 (i.e if and only if the conditions in (a) hold.) It has been proved independently by Kulkarni [K2] (using methods from hyperbolic rings) that β ≠ 0 is a necessary and sufficient condition for A(α, β, γ) to be a domain.

(c) Compute the center of A(α, β, γ). This has been solved in a very recent preprint by Zhao [Z] and announced by Kulkarni [K2].

(d) Determine when A(α, β, γ) is a Hopf algebra. The mapping S(d) = −d, S(u) = −u, which is an algebra antiautomorphism (in fact, the antipode) in the enveloping algebra case, is an algebra antiautomorphism of A(α, β, γ) if and only if β = −1 or β = 1 and α = 0 = γ.

(e) Relate Kulkarni’s presentation of the maximal left ideals in conformal 𝔰𝔩_{2} algebras to the simple modules in category O and category O′. The approach in [K1] using noncommutative algebraic geometry is quite different from the one adopted here and so is the description of the simple modules. Our primary focus here was on determining explicit information about the representations of down-up algebras that could be used in the study of posets in the spirit of the work of Stanley [St1] and Terwilliger [T].

(f) Study the homogenization A[t] of the down-up algebra A = A(α, β, γ), which is the graded algebra generated by d, u, t subject to the relations

\begin{matrix}d^{2}u = αdud + βud^{2} + γdt^{2}, du^{2} = αudu + βu^{2}d + γut^{2}, \\ dt = td, ut = tu. \end{matrix}

Homogenized 𝔰𝔩_{2} is a positively graded Noetherian domain and a maximal order, which is Auslander-regular of dimension 4 and satisfies the Cohen-Macaulay property. Le Bruyn and Smith [LS] have determined the point, line, and plane modules of homogenized 𝔰𝔩_{2} and shown the line modules are homogenizations of the Verma modules.

(g) Investigate indecomposable and projective modules for down-up algebras.

(h) Determine when two down-up algebras are isomorphic. Some information on this problem is derived in Section 6 as a consequence of the determination of the 1-dimensional modules, but we do not address the general problem here.

(i) Investigate prime and primitive ideals of down-up algebras.

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Georgia Benkart, Department of Mathematics, University of Wisconsin, Madison, Wisconsin 53706-1388 benkart@math.wisc.edu

Tom Roby, Department of Mathematics and Computer Science, California State University, Hayward, California 94542-3092 troby@mcs.csuhayward.edu