U.U.D.M. Project Report 2019:17

Examensarbete i matematik, 30 hpHandledare: Volodymyr MazorchukExaminator: Denis GaidashevJuni 2019

Department of MathematicsUppsala University

Category O for Takiff 𝖘𝖑2Christoffer Söderberg

Category O for Takiff sl2

Christoffer Soderberg

June 3, 2019

Abstract. We study basic properties of a natural analogue of the BGGcategory O for the non-semi-simple Takiff Lie algebra of sl2. The mainemphasis is on trying to understand the structure of Verma modules and theGabriel quiver for the blocks of this category.

1 Introduction

The Takiff algebra is defined to be a Lie algebra over a truncated polynomial ring. Takiffalgebras were used by Takiff in 1971 to describe, in some cases, the ring of polynomialover an algebra invariant under the adjoint action.

In this thesis we take the construction of the original Takiff algebra of sl2 and studythe natural analogue of the category O, or sometimes called category BGG. For theoriginal definition of the category O we will need a semi-simple Lie algebra, but in ourcase it turns out that the Takiff algebra of sl2 is not semi-simple, and therefore we haveto alter our definition of category O to ensure that the Verma modules are in fact inthis category. This is due to the fact that the study of Verma modules is a center piecewhen studying the category O. In our case we no longer require that the objects areweight modules, we instead take them to be generalized weight modules.

The main focus of this paper lies in the study of Verma modules. One of the reasonswhy the Verma modules are so important is because when we later trying to decomposeour category in terms of the simples objects, we will need some theory for the simpleobjects to compute everything explicitely. This is where the Verma modules comes in.It turns out that every simple object is a quotient of some Verma module. Not only dowe compute the block decomposition for our category, but we also compute the Gabrielquivers for all the different blocks.

The outline for this paper is as follows: In section 2 we define our Takiff algebra andfind its triangular decomposition. In section 3 we define the universal enveloping algebratogether with some of its properties. Section 4 studies the Casimir elements in our

1

universal enveloping algebra for Takiff sl2. In section 5 we define the thick category Oand proving some basic properties of this category. Section 6 introduces Verma modulesand deducing a formula for all submodules of the Verma modules. In section 7 wewill derive the Hasse diagrams for our Verma modules. In section 8 we study the blockdecomposition for our category. Lastly, section 9 we will introduce all the Gabriel quiversfor all of the different blocks in our category.

2 The Takiff algebra of sl2

Definition 2.1. The generelized Takiff algebra, or sometimes called the truncated cur-rent Lie algebra, g of the Lie algebra sl2 is defined as

g = sl2 ⊗C C[x]/C〈xn+1〉 , (1)

with the Lie bracket defined as [a⊗ xi, b⊗ xj] = [a, b]⊗ xi+j.

We get the original Takiff algebra by setting n = 1, which will be the main focus in thispaper. Since sl2 is of complex dimension 3 we get that g is of complex dimension 6. TheLie algebra g will have the basis {f, h, e, f , h, e}, where f , h and e is the standard basisfor sl2 and f = f ⊗ x, h = h⊗ x and e = e⊗ x. Note that the subalgebra span{f , h, e}is a non-zero abelian ideal, implying that g is not semi-simple.

Let g0 = span{f, h, e} and g1 = span{f , h, e}. Then g = g0 ⊕ g1. With this our gwill in fact be a super Lie algebra. Following [2], for i ∈ Z2 we can define ∆i as the

set α ∈ h∗0 such that α 6= 0 and g(α)i 6= 0. Here, using the notation in [2], g

(α)i is the

maximal subspace of g such that h− α(h) acts locally nilpotent. Setting ∆ = ∆0 ∪∆1.The elements in ∆ will be the roots of g. Now the root lattice Q is the subgroup of h∗

generated by ∆.

Note that Q = 2Zh∗. Thus taking the linear map l : Q → Z where 2nh∗ 7→ 2n weget a triangular decomposition of g as g = n+ ⊕ n0 ⊕ n−, where n+ =

⊕l(α)>0 g

(α),

n0 =⊕

l(α)=0 g(α) and n− =

⊕l(α)<0 g

(α). Note that n0 = h. We will denote b = h⊕ n+.

3 Universal enveloping algebra

Definition 3.1. Let g be a Lie algebra over C. The universal enveloping algebra of g isdefined as the quotient

U(g) = T (g)/I(g), (2)

where T (g) is the tensor algebra of g and I(g) the two-sided ideal of T (g) generated byall elements of the form xy − yx− [x, y], where x and y are elements of g.

2

Definition 3.2. A filtered ring is a ring S with a family {Fn|n = 0, 1, 2, . . . } of subgroupsof S+ such that

1. for each i, j, FiFj ⊂ Fi+j

2. for i < j, Fi ⊂ Fj

3.⋃Fn = S

The family {Fn} is called a filtration of S.

Note that U(g) is an extension of the subring C, with generators {f, h, e, f , h, e}. Withthis we can define a filtration for U(g) in the following way. An element in U(g) of theform

cx1x2 . . . xn (3)

where c ∈ C and xi ∈ {f, h, e, f , h, e} for all i = 1, 2, . . . , n is called a word of length n.Let us define Fi as the additive subgroup of U(g) consisting of words of length i or less,F0 = C. This family {Fi} is a filtration of U(g).

Definition 3.3. A graded ring is a ring T together with a family {Tn|n = 0, 1, 2, . . . }of subgroups of T+ such that

1. TiTj ⊂ Ti+j, and

2. T = ⊕nTn, as abelian groups.

The family {Tn} is called a grading of T . A non-zero element of Tn is said to be homo-geneous of degree n.

Let S be a filtered ring. Then one can construct a graded ring T in the following way.Define Tn = Fn/Fn−1 and T = ⊕Tn. It is enough to consider the homogeneous ele-ments when defining the multiplication. Let a ∈ Fn\Fn−1, then a have degree n anda = a + Fn−1 is the leading term of a. Now let b be of degree m. We can defineab = ab + Fn+m−1. This is well-defined and makes T into a graded ring called gr(S),which is denoted by the associated graded ring of S.

The filtration discussed above leads to the associated graded ring gr(U(g)).

Theorem 3.4. If S is a filtered ring and gr(S) is left or right Noetherian then S is leftor right Noetherian respectively.

3

Proof. Let I ⊂ S be a right ideal of S. Then we can define the right ideal gr(I) of gr(S)by setting (gr(I))n = (I + Fn−1) ∩ Fn/Fn−1, and

gr(I) =⊕n

(gr(I))n. (4)

Note that (I +Fn−1)∩Fn/Fn−1 ' (I ∩Fn)/(I ∩Fn−1). If a ∈ (gr(I))n and r ∈ (gr(I))mthen

ar = ar + Fn+m−1 ∈ (I + Fn−1) ∩ Fn/Fn−1. (5)

This implies that gr(I) is in fact a right ideal of gr(S). Similarly, every left ideal I ⊂ Sone can define a left ideal gr(I) ⊂ gr(U(g)).

Hence if gr(S) does not have any strictly ascending chain of left or right ideals, neitherhas S.

Proposition 3.5. The associated graded ring gr(U(g)) is a commutative C-algebra,generated over C by the set {f, h, e, f , h, e}.

Proof. The relation xixj − xjxi = [xi, xj] in U(g) shows that

xixj − xjxi = [xi, xj] + F1 (6)

in F2/F1. The fact that g is a Lie algebra implies that [xi, xj] ∈ F1, forcing xixj = xjxi.In other words, gr(U(g)) is commutative. The rest follows from the definition of thefiltration of U(g).

Corollary 3.6. Let g be a Lie algebra. If g is finite dimensional then U(g) is a Noethe-rian ring.

Proof. It is known that every finitely generated commutative algebra over C is Noethe-rian. Thus by Proposition 3.5 we have that gr(U(g)) is Noetherian. By Theorem 3.4U(g) is Noetherian.

The following theorem will be useful when we are working with universal algebras.

Theorem 3.7. (Poincare–Birkhoff–Witt theorem) The set of monomials

{fk1 fk2hk3hk4ek5 ek6|k1, k2, . . . , k6 ∈ Z>0} (7)

forms a basis for U(g).

4

4 Casimir elements of U(g)

Following [8]. Let Eij for i, j = 1, 2 be the standard basis for gl2 and let E(k)ij = Eij⊗xk,

where k = 0, 1, be the basis for g2 = gl2 ⊗C C[x]/C〈xn+1〉 . Then the elements

θ1 = E(0)11 + E

(0)22 − 2

θ2 = E(1)11 + E

(1)22

θ3 = E(0)11 E

(1)22 + E

(1)11 (E

(0)22 − 2)− E(0)

21 E(1)12 − E

(1)21 E

(0)12

θ4 = E(1)11 E

(1)22 − E

(1)21 E

(1)12

(8)

generate Z(g2).

Proposition 4.1. The elements

c1 =1

4h2 +

1

2h(

1

2h− 2)− f e− f e

c2 =1

4h2 − f e

(9)

generate Z(g).

Proof. To get the elements in Z(g) from Z(g2) we can set E11 + E22 = 0 since we want

the trace to be zero. From this we have that E(k)11 = 1

2h⊗xk, E(k)

22 = 12h⊗xk, E(k)

12 = e⊗xk

and E(k)21 = f ⊗ xk.

5 Defining the Thick Category O

Definition 5.1. The category O for g is defined as the full subcategory of U-mod con-sisting of modules M on which the action of U(b) is locally finite in the sense thatdimU(b)v <∞ for all v ∈M .

The first thing we need to prove are some basic properties of O.

Proposition 5.2. Let M ∈ O. If N ⊂M is a submodule, then N ∈ O.

Proof. Let N ⊂ M be a submodule. By definition of M is finitely generated. Let{m1,m2, . . . ,mn} generate M . Then for all i = 1, 2, . . . , n there exists a subalgebraWi ⊂ U(g) such that Wi(mi) = U(mi) ∩ N . Now by corollary 3.6 our subring Wi isfinitely generated. From this we can conclude that

⋃ni=1 Gen(Wi)mi generate N and

that |⋃ni=1 Gen(Wi)mi| <∞.

Proposition 5.3. Let M ∈ O. If N ⊂M then M/N ∈ O.

Proof. Let {m1,m2, . . . ,mn} be the generating set for M . Then M/N ∈ U -mod followsfrom the fact that {m1 +N,m2 +N, . . . ,mn +N} generates M/N .

5

To show that U(b) acts locally finite on M/N we need to show that dimU(b)(v+N) <∞ for all v + N ∈ M/N . Since N is closed under the action of U(g) we get thatU(b)(v + N) = {w + N |w ∈ U(b)v}. In other words, U(b)(v + N) = U(b)v/N . Whichhas lower dimension than U(b). Thus dimU(b)(v +N) < dimU(b)v <∞.

Proposition 5.4. Let 0f−→ X → Y

g−→ Z → 0 be a short exact sequence. If X,Z ∈ Othen Y ∈ O.

Proof. Let {x1, x2, . . . , xn} and {z1, z2, . . . , zm} be the generating set for X and Z respec-tively. Exactness implies that g is surjective. Thus there exists yi ∈ Y such that g(yi) =zi for all i = 1, 2, . . . ,m. Now I claim that {f(x1), f(x2), . . . , f(xn), y1, y2, . . . , ym} is agenerating set for Y . Let v ∈ Y . Now

g(v) =m∑i=1

rizi (10)

where ri ∈ U(g) for all i = 1, 2, . . . ,m. Using the fact that g is a homomorphism we canrewrite the right hand to

m∑i=1

rizi =m∑i=1

rig(yi) = g

(m∑i=1

riyi

). (11)

Clearly we have that g (v −∑m

i=1 riyi) = 0, and thus v−∑m

i=1 riyi ∈ ker g. Exactness atY yields that v −

∑mi=1 riyi ∈ Im f . Hence there exists sk ∈ U(g) for all k = 1, 2, . . . , n

such that f (∑n

k=1 skxk) = v −∑m

i=1 riyi. Using the fact that f is an homomorphismand solving for v gives us

v =n∑k=1

skf(xk) +m∑i=1

riyi. (12)

Hence Y is generated by {f(x1), f(x2), . . . , f(xn), y1, y2, . . . , ym}.

Left to show that for any v ∈ Y we have that dimU(b)v < ∞. By above we can writev =

∑nk=1 skf(xk) +

∑mi=1 riyi. Then

dimU(b)v ≤ dimU(b)n∑k=1

skf(xk) + dimU(b)m∑i=1

riyi. (13)

Here f and g will induce an isomorphisms between

U(b)n∑k=1

skxk ' U(b)n∑k=1

skf(xk) (14)

and the inequality

dimU(b)m∑i=1

riyi ≤ dimU(b)m∑i=1

rizi (15)

6

respectively. Since X,Z ∈ O we have that the left hand side in the above expressionsare finite, which implies that dimU(b)v <∞.

Definition 5.5. A non-empty full subcategory T of an abelian category A (of modulesover a ring R) is a Serre subcategory if for any exact sequence

0→M →M ′ →M ′′ → 0 (16)

then M ′ is in T if and only if M and M ′′ are in T .

Corollary 5.6. The category O is a Serre subcategory of U(g)-mod.

Proof. It follows directly from Proposition 5.2, Proposition 5.3 and Proposition 5.4.

Definition 5.7. Let λ ∈ C and M ∈ O. The generalized weight space Mλ is defined as

Mλ = {v ∈M |(h− λ(h))kv = 0,∀h ∈ h, k � 0} (17)

A g-module M is called a generalized weight module if M =⊕

λMλ. We will denotethe support of M by Supp(M), which is the set of all λ ∈ C such that Mλ 6= 0. Thefollowing two propositions will give us a clue on how the generalized weight modulesbehave in our objects of O.

Remark 5.8. By the PBW-theorem each generator v for an object M ∈ O can be writtenas v =

∑ni=1 vi where each vi ∈ Mλi for some λi ∈ C. Now since all the generalized

weight spaces are invariant under the action of h we can use powers of h− λi(h) for alli = 1, . . . , n to prove that all vi ∈M for all i = 1, . . . , n.

Proposition 5.9. Let M ∈ O. There exists λ1, λ2, . . . , λk such that

Supp(M) ⊂n⋃i=1

(λi + 2Z). (18)

Proof. Let {m1,m2, . . . ,mn} be the generating set for M . By the remark 5.8 we canassume that all mi ∈Mλi for some λi ∈ C. The roots in g implies that

h(emi) = (λi + 2)emi

h(emi) = (λi + 2)emi

h(fmi) = (λi − 2)fmi

h(fmi) = (λi − 2)fmi.

(19)

This immediately yields that Supp(M) ⊂⋃ni=1(λi + 2Z).

Proposition 5.10. Let M ∈ O. For all λ ∈ C we have that dimMλ <∞.

7

Proof. Let m = {m1,m2, . . . ,mn} be the generating set for M . By the definition ofM ∈ O we have that dimU(b)(mi) < ∞, which implies that dimU(b)(m) < ∞. Withthis we can write M = U(n−)U(b)(m). Without loss of generality we can assume thatm is a basis for U(b)(m). Thus M = U(n−)(m). Let U(n−)µ−λi = {faf b|a+ b = µ−λ}.Clearly dimU(n−)µ−λi <∞ and together with the fact that

Mµ =n∑i=1

U(n−)µ−λimi, (20)

shows that dimMµ <∞.

Proposition 5.11. For all M,N ∈ O, the dimension dim Hom(M,N) <∞.

Proof. Let {m1,m2, . . . ,mn} be the generating set for M . A homomorphism f : M → Nmust satisfy that if v ∈ M and u ∈ U(g) then f(uv) = uf(v), since it is a mod-ule homomorphism. This implies that f is completely defined by what it does on thegenerating set. Using Remark 5.8 let λ1 = h(m1). By Proposition 5.10 we have thatdimNλi < ∞ for all i = 1, 2, . . . , n. Thus dim Hom(U(g)(mi), Nλi) < ∞. This is truefor all i = 1, 2, . . . , n, hence dim Hom(M,N) <∞.

Proposition 5.12. If M = U〈v1, v2, . . . , vn〉, where v1, v2, . . . , vn ∈M , and if for everyi = 1, 2, . . . , n the dimension dimU(b)vi <∞ then M ∈ O.

Proof. By the remark 5.8 we can assume that all vi ∈ Mλi for some λi ∈ C. Sincee and e raises the weight by 2 and that all Mλ are h-invariant and h-invariant wehave that dimU(g)vi < ∞ implies that there exists Ni ∈ Z such that if λ ≥ λi + Ni

then U(g)vi ∩ Mλ = 0. Let N = inf{λi + Ni|i = 1, 2, . . . , n}. Then if λ ≥ N thenM ∩Mλ = 0 since M is finitely generated. Using Proposition 5.10 we can conclude thatdimU(g)v <∞ if v ∈Mλ for some λ ∈ C.

Now let v ∈M be an arbitrary element. We can write v =∑m

i=1wi for some wi ∈Mλwi.

Now

dimU(g)v = dimU(g)m∑i=1

wi ≤ dimm∑i=1

U(g)wi <∞ (21)

since dimU(g)wi <∞ for all i = 1, 2, . . . ,m

6 Verma Modules

For λ ∈ h∗, define Cλ to be the one-dimensional b-module with generator vλ such that

n+vλ = 0, Hvλ = λ(H)vλ, ∀H ∈h. (22)

Definition 6.1. The Verma module is defined as follows:

∆(λ) = IndgbCλ∼= U(g)⊗U(b) Cλ. (23)

8

From the definition and the PBW-theorem it is clear that {f if j ⊗ vλ|i, j ∈ Z≥0} formsa basis for ∆(λ). Let us begin by writing down how the basis for g acts on ∆(λ). Foran arbitrary basis vector for ∆(λ) we get that

hf if j ⊗ vλ = (λ(h)− 2i− 2j)f if j ⊗ vλhf if j ⊗ vλ = λ(h)f if j ⊗ vλ − 2if i−1f j+1 ⊗ vλef if j ⊗ vλ = i(λ(h)− i− 2j + 1)f i−1f j ⊗ vλ + jλ(h)f if j−1 ⊗ vλef if j ⊗ vλ = iλ(h)f i−1f j ⊗ vλ − i(i− 1)f i−2f j+1 ⊗ vλ.

(24)

With some abusive notation let vλ be 1⊗ vλ.

Lemma 6.2. For any λ ∈ h∗, ∆(λ) ∈ O.

Proof. It is finitely generated since vλ generates ∆(λ).

Left to show that the action of U(b) is locally finite. Let v ∈ ∆(λ). We can write

v =m∑p=1

∑i,i≥0i+j=k

aijfif j ⊗ vλ, (25)

where aij ∈ C and m ∈ Z≥0. Let W be a vector space with the basis

{f if j ⊗ vλ|0 ≤ i+ j ≤ m}. (26)

Clearly W is finite dimensional and also v ∈ W . From (24) we get that W is U(b)-invariant. Thus dimU(b)(v) ≤ dimW <∞.

The goal of this section is to examine the submodules of ∆(λ) for all λ ∈ h∗.

Lemma 6.3. Let T ∈ Matn×n(C) and let W ∈ Cn be a non-zero subspace of Cn suchthat T (W ) ⊂ W . Then W contains a weight vector of T .

Proof. Let V ∈ Matn×r(C) such that the columns of V form a basis for W . As a T -invariant subspace we have that TV = V U , where ∈ Matr×r(C). A consequence ofthe fundamental theorem of algebra we know that every linear operator must have anon-zero weight vector. Let x be a non-zero weight vector for U with weight λ. NowUx = λx and acting with V gives us TV x = V Ux = λV x. Hence V x is a weight vectorof T and we also have that V x 6= 0 because V is linearly independent, also V x ∈ W bythe construction of V .

Note that ∆(λ)λ−2kh∗ = span{f if j ⊗ vλ|i + j = k}. With this we can write our Vermamodules as ∆(λ) =

⊕k∈N ∆(λ)λ−2kh∗ .

Lemma 6.4. If λ ∈ C and N ⊂ ∆(λ) is a non-zero submodule, then there exists anon-zero v ∈ N such that hv = λ(h)v.

9

Proof. Let N be a non-zero sub-module of ∆(λ). Then there exists µ ∈ h∗ such thatNµ 6= 0. From Proposition 5.10 we have that dimNµ < ∞. From the relations (24) wehave that Nλ is h-invariant. Using Lemma 6.3 we get that there exists a weight vectorv of h. Last thing we need to check is that the weight for v must be λ(h). This followsfrom the fact that the action of h onto ∆(λ)λ−2kh∗ in matrix form is

[h] =

λ(h)−2k λ(h)

−2(k − 1) λ(h). . . . . .

−2 λ(h)

, (27)

in the basis {fk ⊗ vλ, fk−1f ⊗ vλ, . . . , fk ⊗ vλ}. This is true for all k ∈ N, thus forcingthe weight to be λ(h).

Furthermore, a weight vector for h in ∆(λ)λ−2kh∗ is a linear combination of elements ofthe form fk ⊗ vλ. With this we can prove our first main result.

Theorem 6.5. ∆(λ) is simple if and only if λ(h) 6= 0.

Proof. Let N ⊂ ∆(λ) be a non-zero submodule. By Lemma 6.4 there exists a weightvector v ∈ N of h. This weight vector will be on the form

v = p(f)⊗ vλ, (28)

where p(t) ∈ C[t] with degree l ∈ Z because how h acts on ∆(λ)λ−2kh∗ . Acting with el

on v yields

elv = l(l + 1)

2λ(h)lplvλ, (29)

where pl is the leading coefficient of p(t). By assumption λ(h) 6= 0 implies that vλ ∈ N ,but vλ generates ∆(λ), thus N = ∆(λ). Hence ∆(λ) is simple.

Theorem 6.6. If λ(h) = 0 and λ(h) 6∈ Z≥0 then all submodules of ∆(λ) are of the formU(n−)(f j ⊗ vλ) for some j ∈ Z≥0.

Proof. Let N ⊂ ∆(λ) be a non-zero submodule. By Lemma 6.4 we have a weight vectorv ∈ N for h. As before we can write it as v = p(f) ⊗ vλ where p(t) ∈ C[t] with degreel ∈ Z. We also have that there exists an a ∈ N such that p(t) = tap(t), where p(t) has aconstant term. Let us now act with h on v.

hv =l∑

k=a

pk(λ(h)− 2k)fk ⊗ vλ (30)

Where the pk’s are the coefficients in p(t). Since λ(h) 6∈ Z we have that λ(h) − 2k 6= 0for all k ∈ Z. By taking

10

v − 1

λ(h)− 2lhv (31)

we get a polynomial with

deg(v − 1

λ(h)− 2lhv) < l. (32)

Inductively we can reduce our weight vector to a weight vector of the form fa⊗vλ. Hencefa ⊗ vλ ∈ N . Let j = inf{a|fa ⊗ vλ ∈ N}. We want to show that f j ⊗ vλ generates N .Assume that it does not generate N . Then we can assume that there exists

w 6∈ U(g)(f j ⊗ vλ). (33)

Since all the terms with higher exponents over f than j lies in U(g)(f j ⊗ vλ) we canrewrite w as

w =a−1∑k=0

qk(f)fk ⊗ vλ (34)

where qk ∈ C[t] for all k = 0, . . . , a − 1. Now let b = sup{deg qk|k = 0, . . . , a − 1}.By assumption we have that λ(h) = 0 which implies that acting with e only reducesthe exponent on f . Making ebw a weight vector for h since all the terms has a zeroexponent on f . By above we can reduce this weight vector to a weight vector of theform f g ⊗ vλ where g < j. This contradicts the minimality of j. Now (24) shows thatN = U(n−)(f j ⊗ vλ).

Theorem 6.7. If λ(h) = 0 and λ(h) ∈ Z≥0, then for all submodules N ⊂ ∆(λ) we haveone of the following cases

1. N = U(n−)(f j ⊗ vλ) for some j ∈ Z≥0.

2. N = U(g)(fλ(h)−2k+1fk ⊗ vλ) for some k ∈ Z≥0, and λ(h)− 2k + 1 > 0.

3. N = U(g){f j ⊗ vλ, fλ(h)−2k+1fk⊗ vλ} for some j, k ∈ Z≥0 where λ(h)− 2j+ 1 > 0and λ(h)− 2k + 1 > 0.

Proof. Let N ⊂ ∆(λ) be a non-zero submodule. By Lemma 6.4 there exists a weightvector for h in N . If we do the same procedure as in the proof of Theorem 6.6 we havethat there exists j ∈ Z≥0 such that f j⊗vλ, and f i⊗vλ 6∈ N for all i < j, where i ∈ Z≥0.If N = U(n−)(f j ⊗ vλ) we are in the first case.

Note that efnfm ⊗ vλ = 0 when n = λ(h) − 2m + 1. Then for any v ∈ N there existsα ∈ N such that we can reduce fλ(h)+1v to the form

11

eαfλ(h)+1v =∑p=1

λ(h)−2p+1>0

apfλ(h)−2p+1fp ⊗ vλ +

∑λ(h)−2p+1≤0

bpfp ⊗ vλ, (35)

where ap, bp ∈ C. Also note that

ehfλ(h)−2p+1fp ⊗ vλ =

2(λ(h)− 2p+ 1)(λ(h)− 2p)(4p+ 1)fλ(h)−2(p+1)+1fp+1 ⊗ vλ.(36)

By (36) we can reduce eαfλ(h)+1v to a weight vector of h. With this and the argumentsfrom Theorem 6.6 we can assume that eαfλ(h)+1v is

eαfλ(h)+1v =∑p≥0

λ(h)−2p+1>0

apfλ(h)−2p+1fp ⊗ vλ (37)

All the terms in that sum lives in different weight spaces, thus as in proof of The-orem 6.6, using h we can show that fλ(h)−2p+1fp ⊗ vλ ∈ N whenever ap 6= 0. Letk = inf{p|fλ(h)−2p+1fp ⊗ vλ ∈ N}. If N is generated by fλ(h)−2k+1fk ⊗ vλ we are in thesecond case.

If neither of the first cases, then assume that there exists v ∈ N such that no term in vlies in U(g)(fλ(h)−2k+1fk ⊗ vλ, f j ⊗ vλ). Assume that fλ(h)+1v − w 6= 0 where w is thepart of fλ(h)+1v that lies in U(g)(fλ(h)−2k+1fk⊗vλ). Doing the same procedure as abovewe arrive at a contradiction with the minimality of k. Thus v =

∑jp=k apf

cp fp ⊗ vλ forsome cp ∈ C, where λ(h)− 2p + 1 < 0. But now by the same reasoning as in the proofof Theorem 6.6, we arrive at a contradiction with the minimality of j.

Definition 6.8. Let λ ∈ h∗. Define L(λ) = ∆(λ)/max(∆(λ)), where

max(∆(λ)) =∑

N⊂∆(λ)Nλ=0

N. (38)

Lemma 6.9. The submodule max(∆(λ)) is a unique maximal submodule of ∆(λ).

Proof. The construction gives us that max(∆(λ))λ = 0, since it is a sum of submodulesN that satisfies Nλ = 0. This shows that max(∆(λ)) 6= ∆(λ). Let M ⊂ ∆(λ) suchthat max(∆(λ)) ⊂ M . If Mλ 6= 0 then vλ ∈ M . Since vλ generates ∆(λ) we have thatM = ∆(λ).

If Mλ = 0, then M ⊂ max(∆(λ)) by definition and thus M = max(∆(λ)). This provesmaximality of max(∆(λ)).

Let us now prove uniqueness. Assume that M ⊂ ∆(λ) is maximal. Then Mλ = 0, thusM ⊂ max(∆(λ)) by the definition of max(∆(λ)). Since M is maximal and the fact thatmax(∆(λ)) 6= ∆(λ) implies that M = max(∆(λ)).

12

Corollary 6.10. For all λ ∈ h∗ the simple quotient L(λ) is unique.

Proof. The submodule max(∆(λ)) is unique and therefore L(λ) is unique.

Proposition 6.11. Let M ∈ O. If M is simple then M ' L(λ) for some λ.

Proof. From Proposition 5.9 we have that there exists a highest weight λ. The weightspace is finite dimensional from Proposition 5.10. By the fact that it is the highestweight vector implies that eMλ = eMλ = 0. In other words, every element in Mλ isa weight vector for e and e with the weights being 0. We also have that Mλ is an h-invariant subspace and using the fact that [h, h] = 0, we know that there exists a commonweight vector v ∈ Mλ for h and h. Note that U(b)(v) = U(h)(v) = Cv. Furthermore

U(b)(v) ' Cλ. Now using the fact that (IndU(g)U(b),Res

U(g)U(b)) is a pair of adjoint functors,

where ResU(g)U(b) is defined as the restriction of U(g) onto U(b), yields

0 6= HomU(b)(Cλ,ResU(g)U(b)M) = HomU(g)(Ind

U(g)U(b)Cλ,M), (39)

but IndU(g)U(b)Cλ = ∆(λ). Every image of a homomorphism is a submodule. Thus every

non-zero f : ∆(λ) → M is surjective since M is simple. Taking the quotient with thekernel yields that L(λ) 'M .

7 Hasse diagrams for Verma modules

Proposition 7.1. The Hasse diagram for ∆(λ), where λ(h) 6= 0, is just a point.

Proof. Follows immediately from the fact that ∆(λ) is simple.

Proposition 7.2. The Hasse diagram for ∆(λ), where λ(h) = 0 and λ(h) 6∈ Z≥0, is

∆(λ) U(g)(f ⊗ vλ) U(g)(f 2 ⊗ vλ) U(g)(f 3 ⊗ vλ) · · ·(40)

Proof. Since all submodules are of the form U(g)(f j ⊗ vλ) and that

· · · ⊂ U(g)(f 3 ⊗ vλ) ⊂ U(g)(f 2 ⊗ vλ) ⊂ U(g)(f ⊗ vλ) ⊂ U(g)(vλ) (41)

we have our desired Hasse diagram.

Lemma 7.3. Let ξ ∈ h∗/Q such that λ(h) = 0 for all λ ∈ ξ. Then there is an inclusion

∆(λ− 2h∗)→ ∆(λ) (42)

for all λ ∈ ξ.

Proof. Since λ(h) = 0 we have that ef ⊗ vλ is zero and thus U(n+)f ⊗ vλ = 0. Thisimplies that there is an inclusion from ∆(λ − 2h∗) to ∆(λ) defined as vλ−2h∗ 7→ fvλ,which extends to a homomorphism.

13

Let mλ ∈ Z≥0 be the biggest number such that λ(h)− 2mλ + 1 > 0.

Proposition 7.4. The Hasse diagram for ∆(λ), where λ(h) = 0 and λ(h) ∈ Z≥0, is

∆(λ)

U(g){fλ(h)+1, f} ∆(λ− 2h∗)

......

. . .

U(g){fλ(h)+1, fmλ} U(g)(fλ(h)−2+1fmλ) · · · ∆(λ− 2mλ)

U(g)(fλ(h)+1) U(g)(fλ(h)−2+1) · · · U(g)(fλ(h)−2mλ+1) · · ·

(43)

where the height of this diagram is λ(h) + 1.

Proof. We will prove this inductively. We will start with the case where λ(h) = 0. FromLemma 7.3 we have that ∆(λ−2h∗) ⊂ ∆(λ), and thus from Proposition 7.2 we can writeout its diagram. Using Theorem 6.7 we have two submodules left, namely U(g)(f ⊗ vλ)and ∆(λ). Here it is easy to see that ∆(λ− 2h∗) ⊂ U(g)(f ⊗ vλ) ⊂ ∆(λ). So we arriveat the following Hasse diagram

∆(λ)

U(g)(f ⊗ vλ) ∆(λ− 2) · · ·(44)

which is the desired Hasse diagram.

Using induction on λ(h) we only need to look at the submodules that are not a submoduleof ∆(λ − 2h∗). Using Theorem 6.7 we know that a submodule N ⊂ ∆(λ) where N 6⊂∆(λ− 2h∗) then N is one of the following three cases

1. N = ∆(λ),

2. N = U(g){fλ(h)+1 ⊗ vλ, f j ⊗ vλ} for some j ∈ N where λ(h)− 2j + 1 > 0,

14

∆(λ)

U(g){fλ(h)+1, f} ∆(λ− 2h∗)

......

. . .

U(g){fλ(h)+1, fmλ} U(g)(fλ(h)−2+1fmλ) · · · ∆(λ− 2mλ)

U(g)(fλ(h)+1) U(g)(fλ(h)−2+1) · · · U(g)(fλ(h)−2mλ+1) · · ·

(47)

3. N = U(g)(fλ(h)+1 ⊗ vλ).

Note that

U(g)(fλ(h)+1 ⊗ vλ) ⊂ U(g){fλ(h)+1 ⊗ vλ, f ⊗ vλ} ⊂ U(g){fλ(h)+1 ⊗ vλ, f 2 ⊗ vλ}⊂ · · · ⊂ U(g)(fλ(h)+1 ⊗ vλ).

(45)

and that

U(g){fλ(h)−2+1f ⊗ vλ, f j ⊗ vλ} ⊂ U(g){fλ(h)+1 ⊗ vλ, f j ⊗ vλ}U(g)(fλ(h)−2+1f ⊗ vλ) ⊂ U(g)(fλ(h)+1 ⊗ vλ).

(46)

Hence the induction on λ(h) implies that the Hasse diagram for ∆(λ) is (47) which isthe desired form.

8 Block decomposition

In this section we will study the block decomposition of the category O.

First we want to introduce a duality in O. Let σ : g→ g be defined as

σ(e) = −f σ(e) = −fσ(h) = h σ(h) = h

σ(f) = e σ(f) = e

(48)

15

.This is clearly an involution. Let M ∈ O, using that M =

⊕λ∈hMλ we define the

duality functor � as

M� =⊕λ∈h

M∗λ (49)

such that if g ∈M�, x ∈M and a ∈ g then a · g(x) = g(σ(a) · x).

If M ∈ O and let m be a generating set for M . Then m� is a generating set for M�,furthermore the definition of σ implies that dimU(b)v <∞ where v ∈ M�. Indeed forall Mλ there exists a N ∈ Z≥0 such that Mλ+2nh∗ = 0 for all n ≥ N . Thus for all M∗

λ

there exists a N ∈ Z≥0 such that M∗λ+2nh∗ = 0 for all n ≥ N . Since every element is a

linear combination of elements in different M∗λ we have dimU(b)v <∞ where v ∈M�.

If f : M → N , then we define �(f) : N� →M� by α 7→ α ◦ f . This makes the dualityfunctor � a contravariant and involutive functor on O.

Proposition 8.1. The duality functor � : O → O is exact.

Proof. It is enough to prove that short exact sequences maps to short exact sequences.Let

0→Mf−→ X

g−→ N → 0 (50)

be a short exact sequence. We want to show that the sequence

o→ N��(g)−−→ X�

�(f)−−→M� → 0 (51)

is exact.

Exactness at N�: Since �(g)(α) = α◦g, then exactness at N� comes from the fact thatg is an epimorphism.

Exactness at X�: The composition �(f) � (g) = g ◦ f ◦ = 0 by exactness of (50),therefore Im(�(g)) ⊂ ker(�(f)). Now let α ∈ ker(�(f)). Define β ∈ N� such thatβ(n) = α(x), where g(x) = n. Then �(g)(β) = β ◦ g = α. Hence it is exact at X�.

Exactness at M�: Since �(f)(α) = α ◦ f , then exactness at M� comes from the factthat f is an monomorphism.

Thus 51 is exact.

Proposition 8.2. Let M ∈ O, then Supp(M) = Supp(M�).

Proof. Let λ, µ ∈ Supp(M) and α ∈M∗λ . Then

(h− µ(h))α = 0 ⇐⇒ λ = µ (52)

16

.Thus M�

λ 6= 0 and λ ∈ Supp(M�). The fact that this functor is involutive implies thatSupp(M) = Supp(M�).

By Corollary 6.10 and Proposition 6.11 the simple objects in our category are fully de-termined by its highest weight and its support. Thus by Proposition 8.2 the dualityfunctor will preserve the isomorphism classes of the simple objects.

We will decompose O using its simple objects. There is a one-to-one correspondencebetween simple objects L(λ) ∈ O and pairs (λ(h), λ(h)) up to isomorphism of the L(λ).Now consider the smallest equivalence relation ∼ on simple object in O such that ∼e⊂∼,where ∼e is defined as

L ∼e L′ ⇐⇒ Ext1O

(L,L′) 6= 0. (53)

With this relation we can decompose the category such that two simple objects are inthe same direct summand if they are equivalent under ∼.

First of all, morphisms in O must preserve the action of Casimir elements. Thus takingtwo object M and M ′ with highest weights λ and λ′ respectively. Then using c2 inProposition 4.1 we have the following decomposition of O

O =⊕

ξ∈h∗/Q

O[ξ], (54)

as Serre categories, where O[ξ] be the full subcategory of O consisting of all M such thatSupp(M) ⊂ ξ. In order for us to understand this decomposition we must understandO[ξ] for all ξ ∈ h/Q.

Proposition 8.3. Let ξ ∈ h∗/Q such that λ(h) = 0 for all λ ∈ ξ. The category O[ξ] isindecomposable in the sense of Serre categories.

Proof. The inclusion map from Lemma 7.3 together with the fact that ∆(λ) is inde-composable implies that L(λ) and L(λ − 2h∗) lies in the same indecomposable directsummand of O[ξ] for any λ ∈ ξ.

Lemma 8.4. Let M ∈ O. Then M is projective if and only if Ext1O

(M,N) = 0 for all

N ∈ O.

Proof. Given an exact short sequence 0→ X → L→ N → 0. The functor Ext1O

inducesby definition a long exact sequence

0→ HomU(g)(M,K)→ HomU(g)(M,L)→ HomU(g)(M,N)→ Ext1O

(M,K)→ . . . (55)

Now M is projective if and only if HomU(g)(M, ) is exact. Since ExtiO

(M,N) = 0 for

all i = 1, 2, . . . and all N ∈ O, by the definition of Ext1O

, we have that HomU(g)(M, ) is

exact if and only Ext1O

(M,N) = 0 for all N ∈ O.

17

Proposition 8.5. If λ(h) = 0 then ∆(λ) is not projective.

Proof. Let ∇(λ+h∗) be the dual module of ∆(λ+h∗). Firstly, ∆(λ+h∗) has L(λ+h∗)as simple top, which implies that ∇(λ+h∗) has L(λ+h∗) as simple socle. Now considerthe following short exact sequence

0→ L(λ+ h∗)f−→ ∇(λ+ h∗)

g−→ coker(f)→ 0. (56)

Here we have that L(λ+ h∗) is not the simple socle for coker(f), thus

HomU(g)(∆(λ), coker(f)) 6= 0. (57)

Since λ is the highest weight of ∆(λ) we get that HomU(g)(∆(λ), L(λ + h∗)) = 0. AlsoL(λ + h∗) is the socle of ∇(λ + h∗) which yields that HomU(g)(∆(λ),∇(λ + h∗)) = 0.From this we can conclude that HomU(g)(∆(λ), ) is not exact, in other words, ∆(λ) isnot projective.

Lemma 8.6. For two simple objects ∆(λ),∆(µ) ∈ O where λ 6= µ, λ(h) 6= 0 andµ(h) 6= 0 we have

Ext1O

(∆(λ),∆(µ)) = 0. (58)

Proof. Since we have the adjunction between Res and Ext it is enough to show thatExt1

O(Cλ,Res

U(g)U(b)∆(µ)) = 0. With the use of the isomorphism Cλ → C0 where vλ 7→

v0 + λ(h) we only need to show that Ext1O

(C0,ResU(g)U(b)∆(µ)) = 0. Now if Res

U(g)U(b)∆(µ) is

injective, then it would automatically imply the desired result. Taking an element fromExt1

O(∆(λ),∆(µ)), that is a short exact sequence of the form

0→ ∆(µ)→ X → ∆(λ)→ 0. (59)

Now if ResU(g)U(b)∆(µ) is injective then every short exact sequence of that form will split,

that is, it will exists an element in x ∈ Xµ such that n+x = 0. Now if n+x = 0 then

n+hx = 0. Thus it will be enough to show that ResU(g)U(b)∆(µ) is injective as an n+-module.

Consider the module C[e, e] which is a n+-module. This is a free module in the categoryof C[e, e]-fgmod, which is the caregory of finitely graded C[e, e]-modules, and whereC[e, e]i is the set of all homogeneous polynomials of degree i. Since it is a free module itis also a projective module. The duality � can be extended to a duality in C[e, e]-fgmodin the standard way, therefore we will denote it by the same symbol. Now the dualmodule C[e, e]�, which is by definition

C[e, e]� =∞⊕

i=−∞

(C[e, e]−i)∗, (60)

an injective module. We would like to show that C[e, e]� is isomorphic to ResU(g)U(b)∆(µ)

as n+-modules. Since C is the socle for both ResU(g)U(b)∆(µ) and C[e, e]�, we have the

following diagram

18

C ResU(g)U(b)∆(µ)

C[e, e]�

∃φ (61)

The morphism φ exists because C[e, e]� is injective. It is easy to see that the dimensionof (C[e, e]−i)

∗ and δ(µ)µ(h)−2i are the same, which implies that φ is an isomorphism.

Hence ResU(g)U(b)∆(µ) is injective.

Corollary 8.7. We can decompose Oh6=0 in the following way

O[ξ] =⊕λ∈ξ

O[ξ]λ (62)

where O[ξ]λ is the Serre subcategory generated by ∆(λ).

Proof. By Theorem 6.5 we have that L(λ) = ∆(λ). Then the decomposition followsfrom Lemma 8.6.

9 Gabriel Quivers

In this section we are going to compute the Gabriel quivers for the blocks in O.

Computing Gabriel quivers for our blocks, it is convenient to introduce the full subcat-egory O[ξ, λ] of O[ξ] consisting of all modules M of the form Mλ+2ih∗ = 0 for all i ∈ Z≥0

for some λ ∈ ξ. The reason we introduce these subcateories is that from the definitionwe have that O[ξ, λ] ↪→ O[ξ, λ + 2h∗], and thus we can compute the quiver for O[ξ] in-ductively using O[ξ, λ]. Hence the problem is reduce to computing the quiver for O[ξ, λ].

Lemma 9.1.

1. If λ(h) 6= 0, then Ext1O

(L(λ), L(λ)) ∼= C2.

2. If λ(h) = 0 and λ(h) 6∈ Z≥0, then Ext1O

(L(λ), L(λ)) ∼= C2.

3. If λ(h) = 0 and λ(h) ∈ Z≥0, then Ext1O

(L(λ), L(λ)) = 0.

19

Proof. Consider the non-split short exact sequence

0→ ∆(λ)→M → ∆(λ)→ 0 (63)

Let m1 ∈ ∆(λ)λ be the elements that generates ∆(λ) and let m2 ∈ Mλ be the elementthat maps to m1 under the quotient map. With this dimMλ = 2 and has the basis{m1,m2}. Since the sequence does not split and the fact that Mλ is a generalized weightspace implies that

[h]Mλ=

[λ(h) α

0 λ(h)

][h]Mλ

=

[λ(h) β

0 λ(h)

], (64)

where α, β ∈ C. This is due to the fact that (H − λ(H))mi = 0 for all i = 1, 2 and allH ∈ h together with the fact that dimU(b)mi < ∞ for all i = 1, 2. Since α and β arefree parameters, we have that dim Ext1

O(∆(λ),∆(λ)) ≤ 2. Let us now examine for which

values on α and β gives isomorphic extensions.

Consider the following diagram

0 ∆(λ) M ∆(λ) 0

0 ∆(λ) N ∆(λ) 0

f1 g1

γ

f2 g2(65)

where

[h]Mλ=

[λ(h) α1

0 λ(h)

][h]Mλ

=

[λ(h) β1

0 λ(h)

][h]Nλ =

[λ(h) α2

0 λ(h)

][h]Nλ =

[λ(h) β2

0 λ(h)

].

(66)

By the commutativity of the first square we get that

γ(m1) = γ(f1(vλ)) = f2(vλ) = n1. (67)

Similarly for the other square we have that γ(m2) = n2. The fact that γ is a homomor-phism implies that α1 = α2 and β1 = β2. Indeed

hγ(m2) = α2n1 + λ(h)n2

hγ(m2) = β2n1 + λ(h)n2

(68)

but on the other hand we have

hγ(m2) = γ(hm2) = γ(α1m1 + λ(h)m2) = α1n1 + λ(h)n2,

hγ(m2) = γ(hm2) = γ(β1m1 + λ(h)m2) = β1n1 + λ(h)n2.(69)

20

Now 1 follows from Theorem 6.5.

Now if λ(h) = 0 and λ 6∈ Z≥0, then recall that we can view L(λ) as a sl2 module viag � g/sl2. Consider the short exact sequence

0→ L(λ)→M → L(λ)→ 0, (70)

where M is constructed in a similar way as before. Note that since L(λ) is a quotient of∆(λ) we have that M is a quotient of ∆(V ). In other words, there exists a short exactsequence of the form

0→ K → ∆(V )→M → 0. (71)

If we let α 6= 0, β = 0, then we can pick

K = U(n−)⊗ C〈f ⊗m1, f ⊗m2〉, (72)

which will give us a non-trivial extension. To find the another non-trivial extension wewill set α = 0, β 6= 0. It is easy to see that K has to be generated by elements in∆(V )λ−2h∗ . We find the highest weight vectors in ∆(V )λ−2h∗ by acting with n+ on anarbitrary element.

e(af ⊗m1 + bf ⊗m1 + cf ⊗m2 + df ⊗m2) = aλ(h)m1 + αbm2 + cλ(h)m2 = 0

e(af ⊗m1 + bf ⊗m1 + cf ⊗m2 + df ⊗m2) = αam2 = 0(73)

where a, b, c, d ∈ C. In this system of equation we will have two free parameters c andd, which implies that there exists a two dimensional subspace of ∆(V )λ−2h∗ which isgenerated highest weight vectors. Taking K to be generated by these weight vectorsgives us another non-trivial extension of L(λ) with itself. These two extensions are notisomorphic by the same argument as in the first case. This shows 2.

In the third case where λ(h) = 0 and λ(h) ∈ Z≥0 we also have that dimL(λ) < ∞.Since L(λ) can be viewed as an sl2 module we can use the Weyl theorem that says thatevery finite dimensional sl2 module is semi-simple. This shows 3.

Theorem 9.2. The subcategory O[ξ]λ, where λ(h) 6= 0, has the following Gabriel quiver

•

a b

(74)

Proof. The result follows from the fact that the subcategory has only one simple objectL(λ) and the first case in Lemma 9.1.

Let us consider the following quivers

21

· · · 6 4 2 0

a a

c

d

b

a

c

d

b

a

c

d

b b

(75)

and

· · · −4 −2 0 2 4 · · ·a a

c

d

b

a

c

d

b

a

c

d

b

a

c

d

b

a

c

d

b b

(76)

which we will denote ∞Q and ∞Q∞ respectively.

Lemma 9.3. The quiver ∞Q is the Gabriel quiver for O[ξ, λ], where λ(h) = 0 andλ(h) 6∈ Z for all λ ∈ ξ.

Proof. The simples in O[ξ, λ] are L(λ−2ih∗) for all i ∈ Z≥0. Let us first assign L(λ−2ih∗)to the vertex i in ∞Q. We have to show that if j < i, then

Ext1O

(L(λ− 2ih∗), L(λ− 2jh∗)) =

{C, if i = j + 1

0, otherwise. (77)

To prove this consider the short exact sequence

0→ L(λ− 2jh∗)→ X → L(λ− 2ih∗)→ 0. (78)

Here X must have highest weight λ − 2i and thus must be a quotient of ∆(λ − 2ih∗).By Theorem 6.6 and Lemma 7.3 we have that the radical of ∆(λ− 2ih∗) is unique andis equal to Rad(∆(λ− 2ih∗)) = ∆(λ − 2(i + 1)h∗). This together with the fact that∆(λ− 2ih∗) is indecomposable for all i ∈ Z≥0 ≥ 0 implies (77).

Now using the fact that the duality � is exact and preserves the category O[ξ, λ] andisomorphism classes of simple modules we get that

Ext1O

(L(λ− 2ih∗), L(λ− 2jh∗)) ∼= Ext1O

(L(λ− 2jh∗), L(λ− 2ih∗)). (79)

It remains to show that Ext1O

(L(λ − 2ih∗), L(λ − 2ih∗)) ∼= C for all i ∈ Z≥0, but thisfollows from the second case in Lemma 9.1.

Theorem 9.4. The Gabriel quiver for O[ξ], where ξ is non-integral and λ(h) = 0 forall λ ∈ ξ, is ∞Q∞.

Here ξ being integral means that λ(h) ∈ Z for all λ ∈ ξ.

Proof. This follows directly from Lemma 9.3 and by taking the inductive limit.

22

It remains to compute the Gabriel quiver for when ξ is integral and λ(h) = 0 for allλ ∈ ξ. When ξ is integral recall that the action of e in ∆(λ) looks like

ef if j ⊗ vλ = i(λ(h)− i− 2j + 1)f i−1f j ⊗ vλ. (80)

This implies that we have an sl2 extension of L(λ) with L(r · λ), in the sense thatg � g/sl2 ∼= sl2. Here L(r · λ) = L(−λ− 2). It can also be seen by the following shortexact sequence

0→ L(r · λ)→ ∆(λ)/∆(λ− 2)→ L(λ)→ 0. (81)

Now consider the following two quivers

0 2 4 · · ·

−2 −4 −6 · · ·

a

s a

a

b

s

a

b

s

b

a

c

d

tb

a

c

d

b

t

a

c

d

b

t

b

(82)

and

1 3 5 · · ·

−1 −3 −5 −7 · · ·

b

a

s

a

b

s

a

b

s

b

b

c

d

a

b

c

d

a

t

b

c

d

a

t

b

c

d

a

t

a

(83)

which we will denote by Γeven and Γodd respectively. Let us also define Γevenn (Γodd

n ) asthe full subquiver of Γeven (Γodd) which consist of all vertices up to n.

Lemma 9.5. The Gabriel quiver for O[ξ, λ], where ξ is integral and λ(h) = 0 for allλ ∈ ξ, is:

1. Γevenn when λ(h) ∈ 2Z for all λ ∈ ξ,

2. Γoddn when λ(h) ∈ 2Z + 1 for all λ ∈ ξ.

23

Proof. Let us start with the first case. First we assign L(λ− 2ih∗) to the vertex n− 2i.We want to show that if i < j and λ− 2ih∗ 6= 0 then

Ext1O

(L(λ− 2ih∗), L(λ− 2jh∗)) =

C, if i = j + 1

C, if r · (λ− 2ih∗) = λ− 2jh∗

0, otherwise

(84)

Consider the short exact sequence

0→ L(λ− 2jh∗)→ X → L(λ− 2ih∗)→ 0. (85)

Since i < j, X must have λ− 2ih∗ as its highest weight, therefore X must be a quotientof ∆(λ− 2ih∗). By Theorem 6.7, the radical of ∆(λ− 2ih∗) must be either

∆(λ− 2(i+ 1)h∗) or U(g){f ⊗ vλ, fλ(h)−2i−1 ⊗ vλ}. (86)

If Rad(∆(λ−2ih∗)) = ∆(λ−2(i+1)h∗), then Rad(∆(λ−2(i+1)h∗)) = ∆(λ−2(i+2)h∗).Now using the fact that the radical is unique for Verma modules we have that

Ext1O

(L(λ− 2ih∗), L(λ− 2(i+ 1)h∗)) ∼= C. (87)

If Rad(∆(λ−2ih∗)) = U(g){f ⊗vλ, fλ(h)−2i−1⊗vλ}, then by Proposition 7.4 the moduleU(g){f ⊗ vλ, fλ(h)−2i−1 ⊗ vλ} has two maximal submodules namely

∆(λ− 2(i+ 1)h∗) and U(g){f 2 ⊗ vλ, fλ(h)−2i−1 ⊗ vλ}. (88)

The first one yields

Ext1O

(L(λ− 2ih∗), L(λ− 2jh∗)) ∼= C, r · (λ− 2ih∗) = λ− 2jh∗ (89)

and the second one yields

Ext1O

(L(λ− 2ih∗), L(λ− 2(i+ 1)h∗)) ∼= C. (90)

This shows 84. When λ − 2ih∗ = 0, then r · (λ − 2ih∗) = λ − 2(i + 1)h∗ and with theargument above we get that

Ext1O

(L(λ− 2ih∗), L(λ− 2(i+ 1)h∗)) ∼= C2. (91)

Same as the previous theorem. By using that the duality � preserves the isomorphismclasses we get the reversed cases for free. This together with Lemma 8.6 proves the firstcase. The second case follows by the same arguments.

Theorem 9.6. The Gabriel quiver for O[ξ, λ], where ξ is integral and λ(h) = 0 for allλ ∈ ξ, is:

1. Γeven when λ(h) = 0 and λ(h) ∈ 2Z for all λ ∈ ξ,

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2. Γodd when λ(h) = 0 and λ(h) ∈ 2Z + 1 for all λ ∈ ξ.

Proof. This follows by Lemma 9.5 and taking the inductive limit.

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References

[1] Joseph Bernstein, Izrail Moiseevich Gel’fand, and Sergei Izrail’evich Gel’fand. Cat-egory of g-modules. Funktsional’nyi Analiz i ego Prilozheniya, 10(2):1–8, 1976.

[2] Ivan Dimitrov, Olivier Mathieu, and Ivan Penkov. On the structure of weight mod-ules. Transactions of the American Mathematical Society, 352(6):2857–2869, 2000.

[3] Brendan Dubsky, Rencai Lu, Volodymyr Mazorchuk, and Kaiming Zhao. Categoryo for the schrodinger algebra. Linear Algebra and its Applications, 460:17–50, 2014.

[4] Peter Gabriel. Unzerlegbare darstellungen i. manuscripta mathematica, 6(1):71–103,Mar 1972.

[5] James E. Humphreys. Representations of Semisimple Lie Algebras in the BGG cat-egory O, volume 94. American Mathematical Society, 2008.

[6] J. C. Robson J. C. McConnell. Noncommutative Noetherian Rings, volume 30. J.Wiley and Sons, Feb 1988.

[7] Volodymyr Mazorchuk. Lectures on SL2(c)-Modules. Imperial College Pr, Dec 2009.

[8] Alexander Molev. Casimir elements for certain polynomial current lie algebras. 1996.

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