A Thesis entitled
On weight modules over sl2
by Hanh Nguyen
Submitted to the Graduate Faculty as partial fulfillment of the requirements for the Master of Arts in Mathematics
Dr. Akaki Tikaradze, Committee Chair
Dr. Gerard Thompson, Committee Member
Dr. Mao-Pei Tsui, Committee Member
Dr. Patricia R. Komuniecki, Dean College of Graduate Studies
The University of Toledo August 2013 Copyright 2013, Hanh Nguyen
This document is copyrighted material. Under copyright law, no parts of this document may be reproduced without the expressed permission of the author. An Abstract of
On weight modules over sl2 by Hanh Nguyen
Submitted to the Graduate Faculty as partial fulfillment of the requirements for the Master of Arts in Mathematics The University of Toledo August 2013
An important class of representations of sl2 consists of weight modules. The goal of this thesis is to give a self-contained exposition of the classification of simple weight modules over C.
iii Acknowledgments
I would like to express my deepest gratitude to my advisor Professor Akaki
Tikaradze. Without his guidance and persistent help, this thesis would not have been possible.
I would also like to offer special thanks to my loving family, especially my husband
Trieu Le for his encouragement and help.
Last but not least, I would like to thank my professors and friends for all the support.
iv Contents
Abstract iii
Acknowledgments iv
Contents v
1 Basic Definitions and Introduction 1
2 Universal Enveloping Algebras 9
3 Verma modules and their quotients 11
3.1 sl2-modules which have a maximal or a minimal vector ...... 13 3.2 Complete Reducibility ...... 15
4 sl2 -modules which have neither a maximal or a minimal vector 19
References 24
v Chapter 1
Basic Definitions and Introduction
We start by recalling few basic notions related to Lie algebras.
Definition 1.1. A vector space g over C, with an operation g × g → g denoted (x, y) 7→ [x, y] and called the bracket or commutator of x and y, is called a Lie algebra over C if the following satisfied:
(L1) The bracket operation is bilinear.
(L2)[ x, x] = 0 for all x in g
(L3)[ x, [y, z]] + [y, [z, x]] + [z, [x, y]] = 0 for all x, y, z in g
Remark 1.2. (L3) is called the Jacobi identity.
Remark 1.3. (L3) shows that g is not associative.
Remark 1.4. Applying (L1) and (L2), we obtain
0 = [x + y, x + y] = [x, x + y] + [y, x + y]
= [x, x] + [x, y] + [y, x] + [y, y]
= [x, y] + [y, x].
It follows that [x, y] = −[y, x] for all x, y in g. Hence [·, ·] is skew symmetric.
1 Example 1.5. Any vector space V over any field is a Lie algebra with the trivial
bracket operation: [x, y] = 0 for all x, y in V .
Example 1.6. General linear algebra. Let V be a vector space over C. Let End(V ) denote the set of all linear transformations from V into itself. Then End(V )
is an associative ring with respect to the usual composition operation of linear trans-
formations. Now we define a new operation [x, y] = xy − yx, then End(V ) becomes a
Lie algebra over C. We write gl(V ) for End(V ) and call it the general linear algebra. Below we check that the bracket [x, y] satisfies all three axioms (L1),(L2) and (L3)
in the definition.
(L1) Since the composition of linear maps is bilinear, this axiom is satisfied.
(L2) For all x in gl(V ), we have [x, x] = xx − xx = 0.
(L3) For all x, y, z in gl(V ), we compute
[x, [y, z]] + [y, [z, x]] + [z, [x, y]] = [x, yz − zy] + [y, zx − zx] + [z, xy − yx]
= x(yz − zy) − (yz − zy)x + y(zx − xz)
− (zx − xz)y + z(xy − yx) − (xy − yx)z
= xyz − xzy − yzx + zyx + yzx − yxz
− zxy + xzy + zxy − zyx − xyz + yxz
= 0.
Thus, axiom (L3) is also satisfied.
Definition 1.7. A subspace K is a subalgebra of g if and only if K is closed under
the bracket operation, that is, for every x, y ∈ K,[x, y] ∈ K.
Example 1.8. Let V be a finite dimensional vector space. Let sl(V ) be the set of
all linear transformations on V with trace zero. Then sl(V ) is in fact a subalgebra
2 of gl(V ). First, using the identity Tr(x + y) = Tr(x) + Tr(y) which holds for all
x, y ∈ gl(V ), we infer that sl(V ) is a linear subspace of gl(V ). Second, it follows from
linearity of the trace and the identity Tr(xy) = Tr(yx) that
Tr([x, y]) = Tr(xy − yx) = Tr(xy) − Tr(yx) = 0.
Thus, [x, y] ∈ sl(V ), which shows that sl(V ) is closed under the bracket operation.
We call sl(V ) the special linear algebra over V .
We now discuss the dimension of sl(V ) in the case V has finite dimension. Write
dim(V ) = n + 1. Because sl(V ) is a proper subspace of gl(V ) (note that the identity
transformation does not belong to sl(V )) and dim(gl(V )) = (n + 1)2, we see that
2 dim(sl(V )) 6 (n + 1) − 1. On the other hand, we may exhibit this number of
linearly independent matrices of trace zero by taking all ei,j (i 6= j) along with
hi = ei,i − ei+1,i+1(1 6 i 6 n). It can be checked that these matrices are linearly 2 independent. We have (n + 1) − (n + 1) such ei,j matrices and n such hi matrices. This shows that the number of matrices we have is (n+1)2 −(n+1)+n = (n+1)2 −1.
We conclude that dim(sl(V )) = (n + 1)2 − 1.
2 For an example, consider the case when V = C . We use sl2 to denote sl(V ). With the above notation, we have
0 1 0 0 1 0 e1,2 = , e2,1 = , h1 = e1,1 − e2,2 = . 0 0 1 0 0 −1
Put e := e1,2, f := e2,1, h := h1. Then it follows from the above discussion that the collection {e, f, h} is a basis of sl(2, C). Direct calculations show
[e, f] = h, [h, e] = 2e, [h, f] = 2f.
3 Definition 1.9. Let g and g0 be Lie algebra over C. A linear transformation φ : g → g0 is called a homomorphism between g and g0 if φ([x, y]) = [φ(x), φ(y)] for all x, y in g.
Definition 1.10. Let g be a Lie algebra. A representation of g is a homomorphism
φ : g → gl(V ), where V is a vector space over C. Here V can be finite or infinite dimensional.
Definition 1.11. A vector space V over C, endowed with an operation g × V → V (denoted (x, v) 7→ x · v or just xv) is called a g-module if for all x, y ∈ g, all v, w ∈ V , and a, b ∈ F , the following conditions are satisfied:
(M1)( ax + by) · v = a(x · v) + b(y · v),
(M2) x · (av + bw) = a(x · v) + b(x · w),
(M3)[ x, y] · v = x · (y · v) − y · (x · v).
Clearly above two notions are the same. It will be convenient to use the language of representations along with the language of modules.
Example 1.12. Let g be a Lie algebra over C. Define Der(g) to be the set of all linear maps D : g → g which satisfy the Leibniz rule
D([x, y]) = [D(x), y] + [x, D(y)] for all x, y ∈ g.
Then Der(g) is a subspace of gl(g). Each linear map in Der(g) is called a derivation.
We now define a special type of derivations. Each x ∈ g defines a map adx : g → g by adx(y) = [x, y] for all y ∈ g. We claim that adx ∈ Der(g). Indeed, adx is linear. For any α, β ∈ F and y, z ∈ g, we have
adx(αy + βz) = [x, αy + βz] = α[x, y] + β[x, z] = α adx(y) + β adx(z).
4 Now we show that adx satisfies the Leibniz rule. For all y, z ∈ g,
adx([y, z]) = [x, [y, z]] = −[z, [x, y]] − [y, [z, x]] (from Jacobi identity)
= [[x, y], z] + [y, [x, z]]
= [adx(y), z] + [y, adx(z)].
Consider the map ad defined by
ad : g −→ gl(g)
x 7→ adx
We claim that ad is a homomorphism of Lie algebras. Indeed, ad is linear by the linearity of the Lie bracket. We now show that ad preserves the Lie bracket, that is, ad[x,y] = [adx, ady] = adx ◦ ady − ady ◦ adx for all x, y ∈ g. For each z ∈ g,
ad[x,y](z) = [[x, y], z] = −[z, [x, y]
= [x, [y, z]] + [y, [z, x]] (by Jacobi identity)
= [x, [y, z]] − [y, [x, z]]
= adx([y, z]) − ady([x, z])
= adx(ady(z)) − ady(adx(z))
= [adx, ady](z).
Thus, (g, ad) is a representation of g.
Definition 1.13. Let V,W be g-modules. A homomorphism of g-modules is a linear map φ : V → W such that φ(x · v) = x · φ(v) for all x ∈ F and v ∈ V . When φ is a vector space isomorphism then we call it an isomorphism of g-modules.
Definition 1.14. Let V be a g-module. A g-submodule of V is a subspace L of V
5 which is closed under the action of g, i.e. for any x ∈ g, we have x · L ⊂ L.
Definition 1.15. A g-module V is called irreducible or simple if it has only trivial g-submodules (V and 0). We do not regard a zero dimensional vector space as an irreducible g-module. The module V is called completely reducible if it can be written as a direct sum of irreducible g-submodules.
In particular, a representation of sl2 is defined as follow.
Definition 1.16. We already know that sl2 is a 3-dimensional Lie algebra with the basis {e, f, h} and the brackets are [h, e] = 2e,[h, f] = −2f,[e, f] = h. Let V be any sl2-module then for any v ∈ V we have
h · (e · v) − e · (h · v) = 2e · v.
h · (f · v) − f · (h · v) = −2f · v.
e · (f · v) − f · (e · v) = h · v.
Definition 1.17 (Weight sl2-modules). An sl2-module is a weight module if the action of h on V is diagonalizable with finite dimensional eigenspaces. This yields a
decomposition of V as the direct sum of subspaces Vλ = {v ∈ V | h · v = λv}, λ ∈ C
and dim(Vλ) < ∞. If Vλ 6= {0} then Vλ is called a weight space of h in V . For any v ∈ V − {0}, if h · v = λv then v is called a weight vector of weight λ. The set of all
weight of V will be denoted by wt(V ).
Definition 1.18. Let Vλ be a weight space of h in V . A nonzero weight vector v in
Vλ is called a maximal vector if e · v = 0.
Definition 1.19. Let Vλ be a weight space of h in V . A nonzero weight vector v in
Vλ is called a minimal vector if f · v = 0.
Proposition 1.20. Let Vλ be a weight space of a sl2-module. For any v ∈ Vλ we have:
6 (a) e · v ∈ Vλ+2.
(b) f · v ∈ Vλ−2.
Proof. Since v ∈ Vλ, we have h · v = λv. Now consider
[h, e] · v = h · (e · v) − e · (h · v) (by M3)
=⇒ 2e · v = h · (e · v) − λe · v
=⇒ h · (e · v) = (λ + 2)e · v.
Thus, e · v ∈ Vλ+2. Similarly, we consider
[h, f] · v = h · (f · v) − f · (h · v) (by M3)
=⇒ − 2f · v = h · (f · v) − λf · v
=⇒ h · (f · v) = (λ − 2)f · v.
Hence, f · v ∈ Vλ−2.
Proposition 1.21. Let V be a finite dimensional sl2-module. Then there exists a maximal vector and a minimal vector.
Proof. By the Fundamental Theorem of Algebra, every polynomial in C has a root. This implies that the characteristic polynomial of h on V has at least a root. As a result, the action of h on V has at least one eigenvector. Let v be a nonzero vector such that h·v = λv. If the sequence of vectors v, e·v, e2 ·v, e3 ·v, . . . has infinite nonzero elements then it forms an infinite sequence of linearly independent eigenvectors of V because each nonzero vector has different weight by the Proposition 1.20. Since V is
finitely dimensional, there must exist a positive integer m such that em · v 6= 0 and em+1 · v = e · (em · v) = 0. Hence, em · v is a maximal vector of V .
7 Similarly, let us look at the the sequence of vectors v, f · v, f 2 · v, f 3 · v, . . . By the
same reason, there must be a positive integer n such that f n · v 6= 0 and f n+1 · v =
f · (f n · v) = 0. Hence, f n · v is a minimal vector of V .
Proposition 1.22. A simple sl2-module V which has at least one weight vector is a
weight sl2-module.
Proof. Let W be the direct sum of all weight spaces of V then W 6= {0} because V
has at least one weight vector. It is clear that W is closed under the actions of e, f, h.
Thus, W a sl2-submodule of V . However V is simple, V = W . Therefore, V is a
simple weight sl2-module.
The following theorem follows from the above results.
Theorem 1.23. Any finitle simple sl2-module is a weight module with a maximal vector and a minimal vector.
The goal of the thesis is to classify all simple weight sl2-modules. We can divide
the problem of classification of simple weight sl2-modules into two cases:
Case 1: Simple weight sl2-modules with either a maximal vector or a minimal vector.
Case 2: Simple weight sl2-modules without any maximal or minimal vectors.
8 Chapter 2
Universal Enveloping Algebras
Definition 2.1. Let g be a Lie-algebra over C . The Universal Enveloping Algebra U(g) is defined as a quotient of the free algebra T (g) generated by g by the relations xy − yx = [x, y], x, y ∈ g.
Let us look at some examples of Universal Enveloping Algebras.
Example 2.2. Let g = span{x1} be a one dimensional Lie-algebra over C. Then g has only bracket [x1, x1] = 0. By the above definition, U(g) is generated by only x1.
Hence, U(g) is the polynomial algebra F [x1] in one variable x1.
Example 2.3. Let g = sl2. Then U(g) is a free algebra generated by e, f, h satisfying the relations ef − fe = h, hf − fh = −2f, he − eh = 2e.
U(sl2) = C < e, f, h > / < ef − fe = h, hf − fh = −2f, he − eh = 2e > .
Although the following fundamental theorem of Poincare-Birkhoff-Witt holds for arbitrary Lie algebras, we will only state it for the case when g is finite dimensional.
Theorem 2.4 (PBW Theorem). Let g be a finite dimensional Lie-algebra over C and let {x1, x2, . . . , xn} be an ordered basis of g. Then the Universal Enveloping Algebra
a1 a2 an U(g) has a basis given by {x1 x2 . . . xn |ai ≥ 0}
9 Proof. The original PBW Theorem and its proof can be found in [1, pp. 92–93].
L i j k Let g = sl2 then U(g) can be written as U(sl2) = Ch e f . It also means that
U(sl2) is as the same size as polynomial algebra in three variables. Now we look at the relation between g-modules and U(g)-modules.
Proposition 2.5. Let g be a finite dimensional Lie-algebra. Then there is a bijec-
tive correspondence between g-modules and U(g)-modules. Furthermore, under this
correspondence a g-module is simple if only if it is a simple U(g)-modules.
The proof is a tautology. From now on we will freely switch between g-modules,
g-representations and (left) Ug-modules, as these notions are identical.
Proposition 2.6. Given V be a simple g-module then V is isomorphic to U(g)/I for
some left maximal ideal of I of Ug.
Proof. Since V is a simple g-module then V is generated by some vector m ∈ V . Let
us define the map
φ : U(g) → V
a 7→ ma
We can prove that φ is a homomorphism with the ker φ = Ann(m). Thus, V '
U(g)/Ann(m).
10 Chapter 3
Verma modules and their quotients
From now on in this chapter we put g = sl2(C). At first we will compute some commutators in Ug, which will be used repeatedly. We will often make use of the
Leibnitz rule for commutators in associative algebras: [a, bc] = [a, b]c + b[a, c].
Theorem 3.1. The following formulas hold in Ug(n ≥ 0)
[e, f n+1] = (n + 1)(h + n)f n = (n + 1)f n(h − n)
[f, en+1] = −(n + 1)(h − n)en
[h, en+1] = 2(n + 1)en+1
[h, f n+1] = −2(n + 1)f n+1.
Proof. We proceed by the induction on n. The case n = 0 is trivial.
[e, f n+1] = hf n + f[e, f n]
= hf n + fn(h + n − 1)f n−1
= (n + 1)hf n + 2nf n + n(n − 1)f n = (n + 1)(h + n)f n.
11 [h, en+1] = [h, e]en + e[h, en] = 2en+1 + 2nen+1 = 2(n + 1)en+1. The rest of the formulas are verified in a similar fashion.
Lemma 3.2. Let V be an g-module and let v be a maximal vector of weight λ. Then
(a) h · f iv = (λ − 2i)f iv,
(b) f · f iv = f i+1v,
(c) e · f iv = i(λ − i + 1)f i−1v.
Proof. Using Theorem 3.1, we have
h(f iv) = λ(f iv) + (−2i)f iv = (λ − 2i)f iv.
Next,
ef iv = f iev + [e, f i]v = if i−1(h − i + 1)v = i(λ − i + 1)f i−1v.
Now using the above lemma, we have the following.
Proposition 3.3. Let V be a simple weight g-module with maximal vector v0. Then
i V = span{f · v0 : i ≥ 0}.
P i Proof. Let us put W = i≥0 Cf v. Then it follows from the above Lemma that W is a submodule of V. Moreover, 0 6= W since v ∈ W. Therefore, W = V.
12 3.1 sl2-modules which have a maximal or a mini-
mal vector
Definition 3.4. Let λ ∈ C. We define Mλ as a quotient of Ug by the left ideal
Ug(h − λ) + Uge. We will denote by vλ the image of 1 in Mλ.
n n It follows immediately from the PBW theorem that Mλ = ⊕n≥0Cf vλ, f vλ 6= 0.
We note that wt(Mλ) = λ − 2Z+ and dim Mλ(λ − 2n) = 1 for any n ∈ Z+.
Lemma 3.5. Let V be a simple weight sl2-module with a maximal vector x of weight
λ. Then there is a unique surjective homomorphism φ : Mλ → V of sl2-modules such
that φ(vλ) = v.
i i Proof. We define φ on the basis of Mλ by φ(f vλ) = f · x for all i ≥ 0 and extend to
all of Mλ by linearity. One can check that φ is an sl2-homomorphism. The surjectivity of φ follows from Theorem 3.3.
Lemma 3.6. If λ ∈ Z+, then maximal vectors of Mλ (up to a nonzero constant
λ+1 multiple) are eλ, f eλ. Otherwise only maximal vector of Mλ is eλ.
n n−1 n Proof. We have ef vλ = n(λ − n + 1)f vλ. Thus, f eλ, n > 0 is a maximal vector if an only if n = λ + 1.
Theorem 3.7. Let V be a simple weight sl2 module with a maximal vector of weight
λ. Then V is a quotient of Mλ. We have two cases:
Case 1: λ∈ / Z+ (the set of all non-negative integers) then Mλ is simple and V is iso-
morphic to Mλ.
Case 2: λ ∈ Z+ then Mλ has a unique proper submodule N which is isomorphic to Mλ−2.
We have V ' Mλ/N, which has dimension λ + 1.
13 Proof. Let N ⊂ Mλ be a submudule, then wt(N) ⊂ λ − 2Z contains a largest weight,
denote it by µ. Let v be a nonzero vector in Nµ. Then Nµ+2 = 0, in particular ev = 0.
Hence, if λ∈ / Z+, then v ∈ Cvλ. Therefore, Mλ = Ugvλ = Ugv ⊂ N. Thus, Mλ
is simple. Now suppose that λ ∈ Z+. Then without loss of generality µ = −λ and
2λ 2λ 0 v = f vλ. Thus, we have shown that any submodule of Mλ contains Usl2f vλ = M . On the other hand, since N(−λ − 2 + i) = 0 for all i ≥ 1, we see that N ⊂ M 0.
0 We claim that M’ is isomorphic to M−λ−2. Indeed, since M is generated by
a maximal vector of weight −λ − 2, it is isomorphic to a quotient of M−λ−2. But
0 −λ − 2 ∈/ Z+. So M−λ−2 is simple. Therefore, M is isomorphic to M−λ−2. Put
λ+1 i Lλ = Mλ/Ugf vλ. Clearly f vλ form a basis of L(λ), 0 ≤ i ≤ λ. In particular, dim L(λ) = λ + 1. We claim that L(λ) is simple. Indeed, we have that L(n) =
M(λ)/M 0 and M 0 is the unique nontrivial submodule of M(λ). Therefore, L(λ) is
simple.
As a consequence, we obtain the following classification of finite dimensional simple
sl2-module.
Theorem 3.8. Let V be a finite dimensional simple sl2-module and let λ = dim V −1. (1.) Relative to the action of h, V can be written as a direct sum of weight spaces:
V = Vλ ⊕ Vλ−2 ⊕ · · · V−λ+2 ⊕ V−λ, where dim Vµ = 1 for each µ = λ, λ − 2,..., −λ. (2.) V has, up to a nonzero scalar multiple, a unique maximal vector and a unique minimal vector.
i (3.) Let v0 be a maximal vector of V then V has a special basis as {f · vi : 0 ≤ i ≤ dim V − 1} and sl2 acts on V by the following rules
h · vi = (λ − 2i)vi
e · vi = i(λ − i + 1)vi−1
14 f · vi = vi+1.
In particular, for each possible finite dimension, there exists one irreducible sl2- module (up to isomorphism).
So far we have showed that given a simple weight sl2-module V with a maximal vector of weight λ then we have
Case 1: If λ ∈ Z+ then V is a finite dimensional simple weight sl2 module with dim V = λ + 1.
Case 2: If λ∈ / Z+ then V is isomorphic to Mλ.
Similarly, by the same fashion we can prove that given a simple sl2 module with a minimal vector of weight λ then we have
Case 1: If λ ∈ Z− then V is a finite dimensional simple sl2 module with dim V = −λ + 1.
Case 2: If λ∈ / Z+ then V is isomorphic to Nλ where Nλ is defined as a quotient of Ug by the left ideal Ug(h − λ) + Ugf.
3.2 Complete Reducibility
We have classified finite dimensional simple weight sl2-modules. Using the classifi-
cation We will show complete reducibility of finite dimensional representations of sl2. This is a particular case of the Weyl’s complete reducibility theorem for semi-simple
complex Lie algebras.
2 First of all we introduce the Casimir element ∆ = h + 4fe + 2h ∈ Usl2.
Lemma 3.9. The Casimir element ∆ belongs to the center of Usl2.
and claim that ∆ is in the center of U(sl2).
15 Proof. It suffices to check that ∆ commutes with e, f and h. We have
[h, ∆] = [h, h2 + 4fe + 2h] = [h, h2] + 4[h, fe] + 2[h, h]
= 4[h, f]e + 4f[h, e] = 4(−2fe) + (4f)(2e) = 0.
[e, ∆] = [e, h2 + 4fe + 2h] = [e, h2] + 4[e, fe] + 2[e, h]
= h[e, h] + [e, h]h + 4[e, f]e + 2[e, h] = −2he − 2eh + 4he − 4e
= 2[h, e] − 4e = 4e − 4e = 0.
[f, ∆] = [f, h2 + 4fe + 2h] = [f, h2] + 4[f, fe] + 2[f, h]
= [f, h]h + h[f, h] + 4f[f, e] + 2[f, h] = 2fh + 2hf − 4fh + 4f
= 2[h, f] + 4f = −4f + 4f = 0.
Thus ∆ belongs to the center of U(sl2).
Now let V be a finite dimensional sl2-module. Consider ∆ as a linear map from V into V . We define
λ m V = v ∈ V | (∆ − λI) v = 0 for some m ∈ Z+ .
We claim that V λ is a submodule of V .
Proof. We want to prove that V λ is closed under the actions of e, f, and h. Indeed,
λ m for any v ∈ V , we have (∆ − λI) v = 0 for some m ∈ Z+. Since ∆ commutes with e, f, and h, we have
(∆ − λI)me · v = e(∆ − λI)mv = 0.
16 Hence, e · v belongs to V λ. Similarly, we see that f · v and h · e belong to V λ.
From linear algebra we know that V = L V λ.
Let V1 be a finite dimensional simple sl2-module with highest weight n ≥ 1. We
know that dim V1 = n + 1. Let v be a maximal vector. Then e · v = 0 and h · v = nv. We then have
∆v = (h2 + 4fe + 2h)v = h2 + 4fev + 2hv = n2v + 0 + 2nv = ((n + 1)2 − 1)v.
Suppose V2 is another finite dimensional simple weight sl2-module with the highest
weight m ≥ 0 and hence dim V2 = m + 1. The above calculation shows that the maximal vector is an eigenvector of ∆ with eigenvalue (m + 1)2 − 1.
If (n + 1)2 − 1 = (m + 1)2 − 1, then we have (n − m)(m + n + 2) = 0. Since n
and m are positive integers, we see that n = m. Therefore ∆ has distinct eigenvalues
for each irreducible representation of sl2. Hence it can be used to separate different irreducible representations.
Theorem 3.10. Let V be a finite dimensional sl2-module. Then V is isomorphic to
a direct sum of simple sl2-modules.
We will proceed by induction on dimV. We may assume that V = V λ, that is, ∆
has only one generalized eigenvalue λ ∈ C.. We can always find a submodule N of V such that V/N is simple. For example, we can take N to be the submodule of V of
highest dimension.
By the induction hypothesis, N = L L(n), λ = (n + 1)2 − 1, m ≥ 0 and N is
isomorphic to a direct sum of copies of L(n), since dim N < dim V and ∆ acts on V
with only one eigenvalue. Thus we have the following exact sequence
p 0 −→ N −→ V −−−−→ L(n) −→ 0.
17 We will prove that the above exact sequence is split. Let 0 6= v ∈ L(n) be a
minimal vector, thus h · v = nv and f · v = 0. Let v0 ∈ V be a generalized eigenvector of weight n with respect to h such that p(v0) = v. Then en+1v0 has generalized weight
n + 2 and fv0 has generalized weight −n − 2. But the set of eigenvalues of h on V is
n, n − 2, ··· , −n. Thus, en+1v0 = 0 and fv0 = 0.
Thus, using Theorem 3.1.
0 = [f, en+1]v0 = −(n + 1)(h − n)env0.
Thus env0 has weight n. Since p(env0) = enp(v0) 6= 0 in L(n). Thus env0 6∈ Ker p =
L(n)m.
n 0 0 n 0 Let us consider the module generated by e v ,N = Usl2(f v ). Since dim V < ∞,
0 N’ is a nonzero finite dimensional quotient of the Verma module Mn, therefore N is
0 isomorphic to L(n). But since N 6⊂ N the restriction p|N : N → L(n) must be an isomorphism since both N 0,L(n) are simple modules. So the exact sequence is split
and we are done.
18 Chapter 4
sl2 -modules which have neither a maximal or a minimal vector
In this section we will complete the classification of simple weight modules.
Definition 4.1. Let λ, µ ∈ C. Then the quotient of Usl2 by the ideal Usl2(h − λ) +
Usl2(fe − µ) will be denoted by Wλ,µ.
Remark 4.2. It is straightforward to see that Wλ,µ=span{vi, wi : i ≥ 0} where the action of e, f, h is given as follows
v0 = w0
h · v0 = λv0
fe · v0 = µv0
h · vi = (λ + 2i)vi
e · vi = vi+1
f · vi = {(1 − i)λ + µ − i(i − 1)}vi−1
h · w0 = λw0
fe · w0 = µw0
19 h · wi = (λ − 2i)wi
e · wi = {(iλ + µ − i(i − 1)}wi−1
f · wi = wi+1
It is also clear that set of all weights of Wλ,µ is {λ + 2Z}.
Lemma 4.3. Let M be a simple weight sl2-module. Then for any λ ∈ wt(M), there is µ ∈ C such that M is a quotient of Wλ,µ.
Proof. First of all, we consider the action h and ef on M. We see that h and ef are commuting because
[ef, h] = e[f, h] + [e, h]f By Leibniz rule
= e(−2f) + 2ef
= 0.
Given λ ∈ wt(M), let Mλ = {v ∈ M|h · v = λv}. Let v be any vector in Mλ, we have h · v = λv∗. Applying ef to ∗, we have
ef(h · v) = ef(λv)
h(ef · v) = λ(ef · v) because ef, h are commuting
Thus Mλ is invariant under ef. Thus, there is 0 6= w ∈ Mλ and µ ∈ C such that ef · w = µv. Thus, w is the common eigenvector of h, ef. Since M is a simple sl2-module, it is generated by w. So, M is isomorphic to Usl2(C)/Ann(w). But
Usl2(h − λ) + Usl2(fe − µ) ⊂ Ann(w). Therefore, M is a quotient of Wλ,µ.
Lemma 4.4. Let M be a simple weight sl2-module which has neither a maximal vector or a minimal vector. Then M is isomorphic to Wλ,µ for some λ, µ ∈ C.
20 Proof. By the above lemma, for each λ ∈ wt(M) there is µ ∈ C such that M is a
quotient of Wλ,µ. The weight space of M is a subset of the weight space of Wλ,µ. Now
we are going to prove that in fact wt(M) = wt(Wλ,µ) for some special λ, µ.
Suppose that wt(M) * wt(Wλ,µ). Then there is β ∈ wt(Wλ,µ) such that either (β + 2) ∈/ wt(M) or (β − 2) ∈/ wt(M).
Case 1:(β + 2) ∈/ wt(M). Take any vector v ∈ Mβ then e · v ∈ Mβ+2 = 0. Thus, e · v = 0. That means v is a maximal vector of M.
Case 2:(β−2) ∈/ wt(M). Similarly, take any vector v ∈ Mβ then f·v ∈ Mβ−2 = 0. Thus, f · v = 0. That means v is a minimal vector of M.
Since, by the hypothesis of the lemma, M does not have a maximal or a minimal
vector. Thus, both cases can not occur. Therefore, we have wt(M) = wt(Wλ,µ).
That means M is isomorphic to Wλ,µ. Moreover, M is simple sl2-module, Wλ,µ is also
a a simple sl2-module. Therefore, M is isomorphic to sl2-module for some special λ, µ.
Theorem 4.5. The sl2 module Mλ,µ is simple if only if Mλ,µ does not have any maximal vector or minimal vector.
Proof. Suppose Wλ,µ has a maximal vector or a minimal vector then we want to
prove that Wλ,µ is not simple. Let 0 6= v ∈ Wλ,µ be a maximal vector of Wλ,µ. Then v = ciwi for some positive integer i and h · v = (λ − 2i)v. Let W be the vector space
generated by v. Then we can prove that W is a submodule of Wλ,µ. The set of all
weights of W is less than or equal to λ − 2i. However, the set of all weights of Wλ,µ
is 2Z + λ. Hence, W 6= Wλ,µ. It means Wλ,µ is not simple.
Conversely, suppose Wλ,µ is not simple. Let {0} 6= N ( Wλ,µ be a proper sub-
module of Wλ,µ. Then we want to prove Wλ,µ has a maximal or a minimal vector. The proof here is similar with the proof of the Lemma 4.4.
Hence, the proof is complete.
21 We now are going to find the conditions for λ, µ such that Wλ,µ is simple. By the
Theorem 4.5, it means we have to find λ, µ for Wλ,µ does not have a maximal or a minimal vector.
Suppose Wλ,µ has a maximal vector v. Then v = cnwn for some n ≥ 0, cn 6= 0
and wn 6= 0. We have
0 = e · v
= cne · wn
= cn[µ + (1 − n)λ − n(n − 1)]wn−1.
Hence, µ + (1 − n)λ − n(n − 1) = 0 ⇒ µ = n(n − 1) + (n − 1)λ .
Suppose Wλ,µ has a minimal vector u. Then u = cmvm for some m ≥ 0, cm 6= 0
and vm 6= 0. We have
0 = f · u
= cmf · vm
= cm[mλ + µ − m(m − 1)]vm−1.
Hence, mλ + µ − m(m − 1) = 0 ⇒ µ = m(m − 1) − mλ.
Therefore, Wλ,µ is simple, if and only if µ 6= n(n − 1) + (n − 1)λ and µ 6= m(m − 1) − mλ for any nonnegative integers n, m.
To summarize, we have the following theorem.
Theorem 4.6. Let V be simple weight sl2-module which does not have a maximal or a minimal vector and let λ ∈ wt(V ). Then V is isomorphic to Wλ,µ for some µ such that µ 6= [n(n − 1) + (n − 1)λ] and µ 6= [m(m − 1) − mλ] for any nonnegative integers
m, n.
∼ Now given Wλ,µ, we have to find all the Wλ0,µ0 such that Wλ,µ = Wλ0,µ0 .Let φ be
22 an isomorphism between Wλ,µ and Wλ0,µ0 . Then we have
φ : Wλ,µ → Wλ0,µ0
0 v0 7→ vn or
0 v0 7→ wn
For some positive integer n.
0 Suppose φ(v0) = vn . By the definition of Wλ,µ, we have h · v0 = λv0(∗). Applying φ to (∗), we have
φ(h · v0) = φ(λv0)
h · (φ(v0)) = φ(λv0)
0 0 h · vn = λvn
0 0 (λ + 2n)vn = λvn.
Hence, λ = λ0 + 2n. we are now going to find the condition for µ and µ0. Let
0 us consider φ(fe · v0) = φ(µv0) = µφ(v0) = µvn . But we also have φ(fe · v0) =
0 0 0 0 0 0 fe · (φv0)) = fe · vn = f · (e · vn ) = f · vn+1 = [−nλ + µ − n(n + 1)]vn . Therefore, µ = −nλ0 + µ0 − n(n + 1) .
0 0 Now suppose φ(v0) = wn . Similarly, by direct calculations we have λ = λ + 2n and µ = nλ0 + µ0 − n(n + 1).
∼ 0 0 0 Therefore, Wλ,µ = Wλ0,µ0 when λ = λ +2n, n ∈ Z+ and µ = (n+1)λ +µ −n(n+1),
0 0 0 or λ = λ + 2n, n ∈ Z− and µ = nλ + µ − n(n + 1).
23 References
[1] James E. Humphreys, Introduction to Lie algebras and representation theory,
Graduate Texts in Mathematics, vol. 9, Springer-Verlag, New York, 1978.
24