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Home , Adjoint representation, Verma module, Weyl character formula

U.U.D.M. Project Report 2019:18

Verma Modules, The and Embedding Theorems

Aban Zehra Husain

Examensarbete i matematik, 30 hp Handledare: Volodymyr Mazorchuk Examinator: Denis Gaidashev Juni 2019

Department of Uppsala University

Uppsala University

Department of Mathematics

Masters Thesis

Verma Modules, The Weyl Character Formula and Embedding Theorems

Supervisor: Author: Prof. Dr. Volodymyr Aban Zehra Husain Mazorchuk

2019 Contents

1 Preliminaries 1 1.1 Lie Algebras ...... 1 1.2 Modules ...... 3 1.3 The and the ...... 4 1.4 The Universal Enveloping Algebra and the Poincaré Birkhoff Witt Theorem . . . 6 1.5 Complete Reducibility Theorem ...... 8 1.6 Weight and weight modules ...... 10 1.7 Induced Representations ...... 11

2 The Weyl Character Formula 11 2.1 Verma Modules ...... 11 2.2 The Weyl Character Formula ...... 15 2.2.1 The Weyl Dimension Formula ...... 17 2.3 Examples ...... 18 2.3.1 g = sl2 ...... 18 2.3.2 g = sl3 ...... 19 2.3.3 The Natural Representation ...... 19 2.3.4 The Adjoint Representation ...... 20

3 Homomorphisms Between Verma Modules 20 3.1 Submodules and Subquotients of Verma Modules ...... 21 3.2 Homomorphisms Between Verma Modules ...... 22 3.3 Existence of Embeddings (Integral Case) ...... 23 3.4 Existence of Embeddings (General Case) ...... 25 3.5 Antidominant Weights and Simple Verma Modules ...... 25 3.6 The example of M(ρ) ...... 26 3.6.1 The chains corresponding to α and β ...... 26 3.6.2 The chains containing both α and β ...... 27 Aban Husain 1

1 Preliminaries

Unless otherwise specified, all vector spaces are over C.

1.1 Lie Algebras Definition 1.1. A g equipped with a bilinear operation [·, ·] (sometimes called a Lie bracket) such that: (i) [x, x] = 0 for all x ∈ g; (ii) [x, [y, z]] + [y, [z, x]] + [z, [x, y]] = 0 for all x, y, z ∈ g. is a . If A is an associative algebra over a field, a Lie bracket may be defined by the [x, y] = xy − yx for all x, y ∈ A, where (x, y) 7→ xy is the multiplicative operation on A. Then (A, [·, ·]) is called the underlying Lie Algebra of A, and A is referred to as an enveloping algebra. Every Lie algebra may be embedded into the underlying Lie algebra for some associative algebra. Definition 1.2. A vector subspace g0 of g is a Lie subalgebra if [x, y] ∈ g0 for all x, y ∈ g0. Definition 1.3. An ideal is a subalgebra g0 of g such that [x, y] ∈ g0 for all x ∈ g and y ∈ g0. If g0 is an ideal, the bracket operation defines a bracket operation on g/g0, so g/g0 is a Lie algebra, called the quotient Lie algebra. If g contains no proper nonzero ideals it is termed a and called a semi-simple Lie algebra if g decomposes into simple components. Let V be a vector space and End(V ) be the algebra of of V . The underlying Lie algebra of End(V ) is called gl(V ). Let V be finite dimensional with dimension n. Given a fixed of V , gl(V ) may be identified with the algebra of n × n matrices and denoted gln. The canonical basis of gln in given by eij (the n × n matrix with 1 in the (ij) position and 0 elsewhere). The subalgebras of gl(V ) are referred to as the linear Lie algebras. Example. Let V be a vector space of dimension n. The special linear algebra is the subalge- bra of gln consisting of endomorphisms with 0, denoted by sln. As trace is independent of basis, this subalgebra is well defined. Given that Tr(xy) = Tr(yx) and Tr(x+y) = Tr(x)+Tr(y), clearly sln is a Lie subalgebra of gln. A basis of sln may be formed by matrices of the form eij 2 for i 6= j and eii − ei+1,i+1 for 1 ≤ i < n, thus having dimension n − 1. Example. Let V be a vector space of odd dimension n = 2k + 1 with a bilinear form on V defined by the matrix 1 0 0  s = 0 0 Ik 0 Ik 0 where Ik is the k × k identity matrix. The special orthogonal algebra is the subalgebra of gln t consisting of endomorphisms x such that sx = −x s, and is denoted son. If x is partitioned as s is, then an equivalent condition is that x must be of the form   0 b1 b2 t x = −b2 m n  t t −b1 p −m

t for arbitrary 1 × k matrices b1, b2 and k × k matriced m, n, p with the added constraints n = −n and pt = −p. A basis may be constructed by restricting to each of the associated partitions. The matrices e1,k+i+1 − ei+1,1 and e1,i+1 − ek+i+1,1 for 1 ≤ i ≤ k form b2 and b1. Corresponding to m, n and p respectively, the matrices ei+1,j+1 − ek+i+1,k+j+1 for 1 ≤ i 6= j ≤ k, ei+1,k+j+1 − ej+1,k+i+1 for 1 ≤ i < j ≤ k and ek+i+1,j+1 − ek+j+1,i+1 for 1 ≤ i < j ≤ k form a partial basis. With the addition of diagonal matrices of the form eii − ek+i,k+i for 2 ≤ i ≤ k + 1 yields a basis 2 of son with dimension 2k + k. Aban Husain 2

In the case of an even dimensional vector space, the orthogonal algebra is constructed similarly, where the bilinear form on V is of the form  0 I  s = k . Ik 0

Let g be a Lie algebra, and define an operation (x, y) 7→ −[x, y]. Then g equipped with this new bracket operation becomes another Lie algebra called the opposite to g. Definition 1.4. Let g, g0 be Lie algebras. A φ : g → g0 is a homomorphism of Lie algebras if φ([x, y]) = [φ(x), φ(y)] for all x, y ∈ g. Definition 1.5. Let X,Y be subsets of g. The set of elements of X that commute, i.e. the set of elements x such that [x, y] = 0, with all elements y of Y is called the centralizer of Y in X. The centralizer of g in g is an ideal and referred to as the of g. Definition 1.6. The normalizer n of a subalgebra h in g is the subalgebra consisting of all x ∈ g such that [x, h] ∈ h. The subalgebra h is an ideal of n. Definition 1.7. Let g be a Lie algebra. The descending central series of g is the chain of subalgebras:

g = g1 ⊇ = [g, g1] ⊇ g3 = [g, g2] ⊇ · · · ⊇ gn = [g, gn−1] ⊇ ..., and the derived series is

g = g(0) ⊇ g(1) = [g(0), g(0)] ⊇ g(2) = [g(1), g(1)] ⊇ · · · ⊇ g(n) = [g(n−1), g(n−1)] ⊇ ....

For any g, the derived subalgebra [g, g] is nonzero, and hence [g, g] = g given that g is a sum of simple Lie algebras. Definition 1.8. A Lie algebra g is nilpotent if it satisfies one of the following equivalent conditions:

k (i) there exists k ∈ Z≥0 such that g = 0;

(ii) there exists k ∈ Z≥0 such that [x1, [x2, [..., [xk−1, xk],..., ]]] = 0 for all x1, . . . , xk ∈ g; (iii) there exists a decreasing series of ideals of g :

g1 ⊇ g2 ⊇ · · · ⊇ gn

such that g1 = g, gn = 0 and [g, gi] ⊆ gi+1 for i < n. Definition 1.9. A Lie algebra g is solvable if it satisfies one of the following equivalent condi- tions:

(k) (i) there exists k ∈ Z≥0 such that g = 0; (ii) there exists a decreasing series of ideals of g :

g1 ⊇ g2 ⊇ · · · ⊇ gn

such that g1 = g, gn = 0 and [gi, gi] ⊆ gi+1 for i < n. By definition of the descending central series and derived series, g(k) ⊂ gn+1. Hence if the descending central series terminates, so must the derived series, i.e. every nilpotent subalgebra is solvable.

Example. The subalgebra of gln consisting of strictly upper triangular matrices is nilpotent, and thus solvable. Conversely, the subalgebra consisting of upper triangular matrices is solvable only. Aban Husain 3

1.2 Modules Definition 1.10. Let V be a vector space. A homomorphism % : g → gl(V ) is termed a representation of g in V . The dimension of the representation % is the dimension of the space V , and V is referred to as a g-. The labels representation and module will be used interchangably. For simplicity, %(x)v is denoted xv unless specification is necessary.

Example. Let g be any Lie algebra. Then g functions as a g-module via the action %(x)y = [x, y]. This is termed the adjoint representation. The %(x) is denoted ad x.

n Example. Given g = sln, the natural representation is the representation of sln in C such that each x ∈ sln acts by normal on vectors. This representation holds for any other Lie subalgebra of gln. Definition 1.11. Given representations %, %0 of g in V,V 0; a linear map φ : V → V 0 is a g-homomorphism if %0(x)φ(v) = φ(%(x)v) for all x ∈ g. Definition 1.12. Two representations %, %0 of g in V,V 0 are equivalent or isomorphic if there exists a g-isomorphism from V to V 0. The equivalence class [V ] is generally referred to as a representation instead a class of representations.

Definition 1.13. Let %i be representations of g in a family of vector spaces (Vi)i∈I and V = L L i∈I Vi. Then % : g → gl(V ), defined by %(x) = i∈I %i(x), is a representation of g in V called the direct sum of representations. If all the representations %i are equivalent to a representation π of g in W , then % is a multiple of π and V is a multiple of W .

Definition 1.14. Given a representation % of g in V and a vector subspace W ⊂ V , W is stable under % if %(W ) ⊂ W . Given a stable subspace W , % W (x) = %(x) W is a subrepresentation of % and W is a submodule. Definition 1.15. Given a submodule W ⊂ V , with representation % on V , let π be the endomor- phism of V/W induced by %. Then π is a representation termed the quotient representation and V/W is a quotient module. Definition 1.16. A representation % of g in V is simple or irreducible if V contains no proper nonzero submodules. Lemma 1.17 (Schur’s Lemma, [Ser05]). If % is an irreducible representation of g in V , and ψ ∈ End(V ) commutes with %(x) for all x ∈ g, then ψ acts via scalar multiplication. Definition 1.18. A representation % is semi-simple or completely reducible if any of the following equivalent conditions are satisfied: (i) % is a direct sum of simple representations;

(ii) V is a direct sum of simple submodules; (iii) for each submodule W of V , there exists a complement W 0 of W that is a submodule of V . Every finite dimensional representation of a semisimple Lie algebra is completely reducible. Given that the proof of this statement requires a few more definitions, it will be given later. Under the adjoint representation, clearly the ideals of g yield the submodules. Thus any simple (resp. semi-simple) Lie algebra g is a simple (resp. semi-simple) module under the adjoint representation. Aban Husain 4

1.3 The Cartan Subalgebra and the Root System Definition 1.19. An endomorphism u of a finite dimensional vector space V is diagonalizable if there exists a basis B of V such that the matrix of u is diagonal with respect to B Definition 1.20. An endomorphism u of a finite dimensional vector space V is triangulariz- able if any of the following three equivalent conditions hold:

(i) there exists a basis B of V such that the matrix of u is upper triangular with respect to B; (ii) there exists a basis B of V such that the matrix of u is lower triangular with respect to B; Definition 1.21. A nilpotent subalgebra h of g, such that h equals its normalizer in g, is called a Cartan subalgebra.

Theorem 1.22. Let h be a nilpotent subalgebra of g. Consider the adjoint representation of h in g. Assume ad x is triangularizable for all x ∈ h, and define for λ ∈ h∗:

gλ = {x ∈ g :[h, x] = λ(h)x}.

Then:

L λ (i) g = λ∈h∗ g ; (ii) [gλ, gµ] ⊆ gλ+µ; (iii) g0 is a subalgebra of g such that h ⊆ g0;

(iv) h is a Cartan subalgebra if and only if h = g0; (v) g0 is a Cartan subalgebra if it is nilpotent. Definition 1.23. Let h be a Cartan subalgebra of g. If ad x is triangularizable for all x ∈ h, then h is a splitting Cartan subalgebra. If h is a splitting Cartan subalgebra of g, set of λ ∈ h∗ such that gλ 6= 0 is called the root system of g with respect to h. The set of simple roots ∆ ∈ Φ is a base for Φ and a basis for h∗ Definition 1.24. Let g be a semi-simple Lie algera and h a splitting Cartan subalgebra of g. The pair (g, h) is a split semi-simple Lie algebra and the set of roots of g for this Cartan subalgebra is denoted Φ(g, h). Theorem 1.25. Let (g, h) be a split semi-simple Lie algebra with correspodning set of roots Φ = Φ(g, h). Then:

L α α (i) g = h ⊕ α∈Φ g where dim g = 1 for each α ∈ Φ; α −α (ii) If α ∈ Φ then −α ∈ Φ, and hα = [g , g ] is a one dimensional vector subspace of h; (iii) Φ spans the vector space h∗;

α −α (iv) For any α ∈ Φ, sα = hα ⊕ g ⊕ g is a subalgebra of g isomorphic to sl2. Given a basis of h such that the adjoint representation is triangularizable, let (·, ·) be the standard inner product on the vector space h∗ with respect to the dual basis. Define the operator h·, ·i : ∗ ∗ (β,α) h × h → k by hβ, αi = 2 (α,α) .

Example. Let g = sln and eij the n×n matrix with 1 in the (i, j) position and 0 elsewhere. Let h be the subalgebra of diagonal matrices spanned by eii − ei+1,i+1. Then h is a splitting Cartan ∗ subalgebra of g. With respect to h, the set of roots of g is of the form i − j, where i ∈ h α maps eii to 1 and ejj = 0 for j 6= i. For α = i − j, the root space g is the one dimensional space spanned by eij. Aban Husain 5

α Example. Let (g, h) be a split semi-simple Lie algebra, and α a root of g. For nonzero xα ∈ g , −α ∗ choose x−α ∈ g such that α(hα) = 2 for hα = [xα, x−α]. Then for any λ ∈ h , the operator hλ, αi is equivalent to λ(hα). Let V be the subalgebra of g spanned by xα, x−α and hα. Let φ be a homomorphism from sl2 to V defined by 1 0  0 1 0 0 7→ h , 7→ x , 7→ x . 0 −1 α 0 0 α 1 0 −α

Then φ is an isomorphism, so for any α ∈ Φ, the subalgebra sα = gα ⊕hα ⊕g−α may be identified α with sl2, where h is the one dimensional subspace spanned by hα. For any representation % of g, the restriction % sα may be identified with an sl2 representation. Proposition 1.26. Let (g, h) be a split semi-simple Lie algebra with root system Φ. Let ∆ be a base for Φ. Given α, β ∈ Φ:

(i) hβ, αi ∈ Z; (ii) the only roots that are scalar multiples of α are α and −α; P (iii) β = α∈∆ nαα where all nα ∈ Z≥0 or all nα ∈ Z≤0. Definition 1.27. For a split semi-simple Lie algebra (g, h) and root system Φ, fix a base ∆. Then: P (i) a root β = α∈∆ nαα is positive (resp. negative) if nα ∈ Z≥0 (resp. nα ∈ Z≤0); (ii) n = L gα is termed the positive root space, where Φ is the set of positive roots, + α∈Φ+ + and similarly n = L gα is termed the negative root space; − α∈Φ−

(iii) the decomposition g = h ⊕ n+ ⊕ n− is called the triangular decomposition;

(iv) b+ = h ⊕ n+ is termed the Borel subalgebra with respect to h and ∆. If h∗0 ⊂ h∗ is a subspace of codimension one such that h∗0 does not contain any α ∈ Φ, then h∗0 splits h∗ into two connected components containing positive and negative roots. In particular, as any subspace of codimension one is the of a linear functional λ on h∗, the set of positive roots with respect to such a functional is

Φ+ = {α | λ(α) > 0}.

Example. Let g = sln and h the Cartan subalgebra of g consisting of diagonal matrices. Define a total order ≺ on {1, . . . , n} and let s(i) be the successor of i with respect to the order ≺. Then the set of positive roots consists of the roots i − j for i ≺ j and ∆ = {i − s(i)}. For such an order, the Borel subalgebra is of the form M b = h ⊕ gα.

α=i−j i≺j i,j∈{1,...,n}

The Borel subalgebras of sln with respect to h are in one-to-one correspondence with the total orders on {1, . . . , n}.

∗ Definition 1.28. Let σα be the reflection of h across the hyperplane orthogonal to α, defined by λ 7→ λ − hλ, αiα. The W of a split semi-simple Lie algebra (g, h) is the group generated by σα for all α ∈ Φ. Proposition 1.29. Given a basis ∆ for a root system Φ, the Weyl group W is generated by the simple reflections σα for α ∈ ∆. Additionally W acts transitively on Φ. Let l(w) be the size of the smallest set of simple reflections such that their product forms w for w ∈ W and set sgn(w) = −1l(w). Aban Husain 6

1.4 The Universal Enveloping Algebra and the Poincaré Birkhoff Witt Theorem All statements and proofs follow those provided in [Dix96].

Let g be a Lie algebra and T (g) the tensor algebra, i.e.

∞ M T (g) = T k(g) k=0 where T k(g) = g⊗k. In particular T 0(g) = k and T 1(g) = g. With the product being tensor multiplication, T (g) is an associative algebra. Let J ⊂ T be the two-sided ideal generated by all terms of the form x ⊗ y − y ⊗ x − [x, y] L∞ k for all x, y ∈ g. Given the generators of J, we have J ⊂ T+(g) where T+(g) = k=1 T (g). Definition 1.30. The associative algebra U(g) = T (g)/J is called the universal enveloping algebra such that the mapping σ composed of the canonical mappings g → T (g) and T (g) → U(g) yields σ(x)σ(y) − σ(y)σ(x) = σ([x, y]) for all x, y ∈ g. The center of the universal enveloping algebra is denoted Z(g) and in the case of a commutative enveloping algebra, U(g) = S(g). 0 0 Let U (g) be the canonical of T (g) in U(g) and U+(g) the image of T+(g). Given that 0 T (g) = T (g) ⊕ T+(g) and J ⊂ T+(g), we have

0 U(g) = U (g) ⊕ U+(g) where U 0(g) = k. Hence U(g) is generated by the image of 1 and g in U(g). Lp p Denote the canonical image of k=0 T (g) in U(g) by Up(g). Fix a basis x1, . . . , xn for g and denote the canonical image xi in U(g) by yi. For any finite sequence I let yI = yi1 . . . yip for I = (i1, . . . , ip). Define the ordering i ≤ I if i ≤ i1, . . . , ip.

Lemma 1.31. Let a1, . . . , ap ∈ g, the mapping σ be the composition of the canonical mappings g → T (g) and T (g) → U(g) and π a permutation of {1, . . . , p}. Then:

σ(a1) . . . σ(ap) − σ(aπ(1)) . . . σ(aπ(p)) ∈ Up−1(g).

Additionally the yI generate the vector space Up(g) for all increasing seqences I of length ≤ p. Proof. The permutation is generated by transpositions of the form (j, j + 1), so it is sufficient to prove this for transpositions of this form. Given that

σ(a1) . . . σ(ap) − σ(a1) . . . σ(aj+1)σ(aj) . . . σ(ap) = σ(a1) . . . σ([aj, aj+1]) . . . σ(ap), the identity holds true. Thus for every increasing sequence I of length ≤ p and any permutation π, yπ(I) is a linear combination of yJ for increasing sequences J of length ≤ p. The vector space Up(g) is therefore generated by the elements yI for all increasing sequences I of length ≤ p.

Lemma 1.32. Let P be the algebra of polynomials k[z1, . . . , zn] and Pi the set of elements with degree ≤ i. Set zI = zi1 . . . zip for I = (i1, . . . , ip). There exists a unique linear mapping fp : g ⊗ Pp → P , for each p ∈ Z≥0, which satisfies the following conditions:

(i) fp(xi ⊗ zI ) = zizI for i ≤ I, zI ∈ Pp;

(ii) fp(xi ⊗ zI ) − zizI ∈ Pq for zI ∈ Pq, q ≤ p;

(iii) fp(xi ⊗ fp(xj ⊗ zJ )) = fp(xj ⊗ fp(xi ⊗ zJ )) + fp([xi, xj] ⊗ zj) for zJ ∈ Pp−1. Aban Husain 7

Additionally

fp = fp−1. g⊗Pp−1

Proof. Given p = 0, for all i ∈ {1, . . . , n}, the map f0 must be defined as f0(xi ⊗ 1) = zi to satisfy condition ((i)). In this case, the conditions ((ii)) and ((iii)) follow immediately. We shall proceed by induction. Assume there exists some p, such that fk is unique and satisfies conditions

((i)), ((ii)) and ((iii)) for all k < p. If there exists such an fp, then fp = fp−1. Thus it is g⊗Pp−1 sufficient to show that fp−1 has only one linear extension fp on g ⊗ Pp such that the conditions of the lemma are satisfied. As P is a commutative algebra, we may take I to be an increasing sequence of p elements and define fp(xi ⊗ zI ). If i ≤ I, then the choice of fp is restricted to fp(xi ⊗ zI ) = zizI by ((i)). If i 6≤ I, then I can be written as (j, J) where J is an increasing sequence of p − 1 elements and j < 1, j ≤ J. Then zjzJ = fp−1(xj ⊗ zJ ), and following from conditions ((i)) and ((iii)): fp(xi ⊗ zI ) = fp(xi ⊗ fp−1(xj ⊗ zJ ))

= fp(xj ⊗ fp−1(xi ⊗ zJ )) + fp−1([xi, xj] ⊗ zJ ).

As zJ ∈ Pp−1 and ((ii)) holds true for fp−1, we have

fp−1(xi ⊗ zJ ) − zizJ = w ∈ Pp−1 and then fp(xi ⊗ zI ) = fp(xj ⊗ fp−1(xi ⊗ zJ )) + fp−1([xi, xj] ⊗ zJ )

= fp(xj ⊗ (zizJ + w)) + fp−1([xi, xj] ⊗ zJ )

= zjzizJ + fp−1(xj ⊗ w) + fp−1([xi, xj] ⊗ zJ ).

Hence define fp(xi ⊗ zI ) = zizI for i ≤ I and as above for I = (j, J) with j < i and j ≤ J. This is a unique linear extension of fp−1 to g ⊗ Pp such that the conditions ((i)) and ((ii)) hold by definition. Hence we must show that fp defined as such satisfies ((iii)), i.e.

fp(xi ⊗ fp(xj ⊗ zJ )) = fp(xj ⊗ fp(xi ⊗ zJ )) + fp([xi, xj] ⊗ zj) for arbitrary i, j and zJ ∈ Pp−1. If j < i and j ≤ J, then the condition holds from the definition of fp. As the bracket operation is anti-symmetric, the condition holds for i < j and i ≤ J as well. If i = j then the condition is trivially satisfied. Assume that i, j 6≤ J. Let J = (k, K) such that k < i and k < j. By the induction hypothesis, as zK ∈ Pp−2, fp(xj ⊗ zJ ) = fp(xj ⊗ fp(xk ⊗ zK ))

= fp−1(xj ⊗ fp−1(xk ⊗ zK ))

= fp−1(xk ⊗ fp−1(xj ⊗ zK )) + fp−1([xj, xk] ⊗ zK )

= fp(xk ⊗ fp(xj ⊗ zK )) + fp([xj, xk] ⊗ zK ).

As zK ∈ Pp−2, condition ((ii)) yields fp(xj ⊗ zK ) = zjzK + w for some w ∈ Pp−2. Then

fp(xi ⊗ fp(xj ⊗ zJ )) = fp(xi ⊗ fp(xk ⊗ (zjzK + w))) + fp(xi ⊗ fp([xj, xk] ⊗ zK )).

For the first term on the right hand side of the equality, condition ((iii)) holds by the definition of fp as k < j, k ≤ K, and by the induction hypothesis as w ∈ Pp−2. The condition holds for the second term as zK ∈ Pp−2. This yields fp(xi ⊗ fp(xj ⊗ zJ )) = fp(xi ⊗ fp(xk ⊗ fp(xj ⊗ zK )) + fp(xi ⊗ fp([xj, xk] ⊗ zK ))

= fp(xk ⊗ fp(xi ⊗ fp(xj ⊗ zK )) + fp([xi, xk] ⊗ fp(xj ⊗ zK ))

+ fp([xj, xk] ⊗ fp(xi ⊗ zK )) + fp([xi, [xj, xk]] ⊗ zK ). Aban Husain 8

Interchanging i, j and the using the properties of the bracket operation fp(xi ⊗ fp(xj ⊗ zJ )) − fp(xj ⊗ fp(xi ⊗ zJ ))

= fp(xk ⊗ [fp(xi ⊗ fp(xj ⊗ zK )) − fp(xj ⊗ fp(xi ⊗ zK ))]) + fp([xi, [xj, xk]] ⊗ zK )

− fp([xj, [xi, xk]] ⊗ zK )

= fp(xk ⊗ fp([xi, xj] ⊗ zK )) + fp([xi, [xj, xk]] ⊗ zK ) + fp([xj, [xk, xi]] ⊗ zK )

= fp([xi, xj] ⊗ fp(xk ⊗ zK )) + fp([xk, [xi, xj]] ⊗ zK ) + fp([xi, [xj, xk]] ⊗ zK )

+ fp([xj, [xk, xi]] ⊗ zK )

= fp([xi, xj] ⊗ fp(xk ⊗ zK ))

= fp([xi, xj] ⊗ zJ ).

Hence fp(xi ⊗ fp(xj ⊗ zJ )) = fp(xj ⊗ fp(xi ⊗ zJ )) + fp([xi, xj] ⊗ zj) for all i, j and zJ ∈ Pp−1, so condition ((iii)) holds for fp.

Lemma 1.33. The yI form a basis for the vector space U(g) for every increasing sequence I. Proof. Combining the maps from the previous lemma for all p yields a bilinear man f : g×P → P such that f(xi, yI ) = yiyI for i ≤ I and

f(xi, f(xj, zJ )) = f(xj, f(xi, zJ )) + f([xi, xj], zJ ).

This defines a representation % of g on P such that ϕ(xi)zI = zizI for i ≤ I. By universality of of U(g) there is a homomorphism φ : U(g) → End(P ) such that φ(yi)zI = zizI for i ≤ I. By induction, if I = (i1, . . . , ip) is a increasing sequence then

φ(yI ) = φ(yi1 . . . yip ) · 1 = zi1 . . . zip = zI .

As φ is a homomorphism, given that the zI for each increasing sequence I are linearly indepen- dent, so are the yI . The yI , for every increasing sequence I generate U(g) and are linearly independent. Hence they form a basis for the vector space U(g).

Theorem 1.34 (Poincare-Birkhoff-Witt). Let (x1, . . . , xn) be a basis for g. Then a basis for v1 v2 vn U(g) can be formed by the set of elements x1 x2 . . . xn where v1, . . . , vn ∈ Z≥0.

Corollary 1.35. Assume that g has a triangular decomposition g = n− ⊕ h ⊕ n+, such that (y1, . . . , yn), (h1, . . . , hn) and (x1, . . . , xn) are bases for n−, h and n+ respectively. Then a basis a1 an b1 bn c1 cn for U(g) is formed by the set of elements y1 . . . yn h1 . . . hn x1 . . . xn where ai, bi, ci ∈ Z≥0, and U(g = U(n−)U(h)U(n+).

1.5 Complete Reducibility Theorem Theorem 1.36 ([Ser05]). If g is semisimple, and % is a finite dimensional representation of g, then % is completely reducible.

Proof. The proof is completed in steps, following from proof given in [Ser05]. (i) Let g be a semisimple Lie algebra and % a finite dimensional representation of g in V . We may assume that % is a faithful representation, i.e. injective. If not we may quotient g with ker(%), which does not affect the reducibility of the representation. Define the bilinear form B%(x, y) = TrV %(x)%(y). As trace is invariant of basis, this form is well defined. Given that % is injective, it is quite clear that B% is nondegenerate (i.e. if B%(x, y) = 0 for all y, then x = 0). Aban Husain 9

(ii) Let (xi) be a set basis of g, and B a nondegenerate bilinear form. As B is nondegenerate, there exists a dual basis (yi) of g with respect to B, in other words B(xi, yj) = δij where δij = 1 if i = j, and 0 otherwise. Define the element X b = xiyj ∈ U(g). i,j

This element is contained in the center of U(g) and is called the corresponding to B.

(iii) Denote the Casimir element corresponding to B% by c%. As c% lies in the center of U(g), it commutes with %(g), hence acts as an endomorphism of V as a g-module. Additionally, given the definition of B%, we have X X Tr(%(c%)) = Tr(%(xi)%(yj)) = Tr(B%(xi, yi)) = dim g. i,j i

(iv) Assume % is an irreducible representation. Then c% acts via scalar multiplication on V . Given that Tr(c%) = dim g,%(c%) is zero if and only if g = 0. Hence c% acts via nonzero scalar multiplication on V .

(v) Define the exact sequence of g-modules

0 −→ W −→ V −→ C −→ 0 where g acts trivially on C. Due to the semisimplicity of g, as [g, g] = g and C is one dimensional, g must act trivially. We want to show that the sequence splits, i.e. there is a one dimensional submodule of V complementary to W that maps to C.

(vi) We may reduce the case of the above short exact sequence to W being a simple submodule. As W is a submodule of codimension one in V , for any X ⊂ W , we have W/X is a submodule of codimension one in V/X. Hence there is some W˜ /X ⊂ V/X which is a complement to W/X in V/X. Then W ⊕ W˜ = V . Thus we have the following short exact sequence

0 → X → W˜ → C → 0 which yields by construction X˜ ⊂ W˜ such that X ⊕ X˜ = W.˜ Then

X˜ ⊕ W = X˜ ⊕ W ⊕ X = W ⊕ W˜ = V, so X˜ is a complementary subspace in V to W .

(vii) We may assume g acts faithfully on W . If not, let a ⊂ g be the subalgebra that acts trivially on W . As g acts trivially on C, and the homomorphisms of the short exact sequence are g homomorphisms, a(V ) ⊂ W . Since a(W ) = 0, we have [a, a] acting trivially on V . Since g is semisimple, so is a, hence [a, a] = a. As a ⊂ Ker(g → End(V )) we have g/a acts on V and acts faithfully on W . Given that g/a retains its semisimplicity, we may assume that g acts on W faithfully.

(viii) Assume % is a representation of g in W such that % acts faithfully and W is simple. Then the Casimir element c% corresponding to B% acts as a g endomorphism on V where, c%(V ) ⊂ W as c% acts trivially on C. Assuming that g is nonzero we have c%(W ) = W . Then Ker(c%) ⊂ V is a complement of W in V that is stable under g.

(ix) Let 0 → W → V → W˜ → 0 Aban Husain 10

be a short exact sequence of arbitrary g-modules. Viewing Homg(V,W ) as a g-module, define the following submodules:

V = {φ ∈ Hom(V,W ): φ W ∈ C}

W = {φ ∈ Hom(V,W ): φ W = 0}. Thus there is a short exact sequence of g-modules

0 → W → V → C → 0. From above there exists some element φ ∈ V invariant under g that is mapped to 1. Hence ker(φ) ⊂ V is invariant under g, and is therefore a complementary submodule to W . This yields the decomposition V = W ⊕ ker(φ), hence the sequence 0 → W → V → W˜ → splits.

1.6 Weight and weight modules Definition 1.37. Let % be a representation of g on V with splitting Cartan subalgebra h. For ∗ λ ∈ h , a weight space is a subspace Vλ of V given by:

Vλ = {v ∈ V : hv = λ(h)v ∀h ∈ h}. ∗ For λ ∈ h such that Vλ is non-zero, λ is termed a weight of V and any non-zero v ∈ Vλ is a weight vector of weight λ. Definition 1.38. Given a weight λ of a g-module with splitting Cartan subalgebra h and root system Φ, it is an integral weight if hλ, αi ∈ Z for all α ∈ Φ. A weight is dominant integral with respect to some base ∆ if hλ, αi ∈ Z≥0 for all α ∈ ∆, or equivalently all α ∈ Φ+. Definition 1.39. The set of fundamental weights of a split semi-simple Lie algebra (g, h) with root system Φ and base ∆ = {α1, . . . , αn} is the set ω1, . . . , ωn such that hωi, αji = δij where ( 1, if i = j, δij = 0, if i 6= j. Clearly the integral weights are combinations of fundamental weights with integer coefficients. The set of integral weights is called the weight lattice. Definition 1.40. Let V be a module of g. If V admits a decomposition into weight spaces, i.e. L V = λ∈h∗ Vλ, then V is a weight module Lemma 1.41. Let V be an arbitrary module of a split semisimple Lie algebra (g, h). Then L L λ∈h∗ Vλ is a submodule of V . If V is finite dimensional, then V = λ∈h∗ Vλ, i.e. V is a weight module. Definition 1.42. Let V be a weight module of g and v+ a weight vector with weight λ. If + + n+v = 0 then v is called a maximal vector and and λ is a highest weight. Definition 1.43. Let V be a module of a split semi-simple Lie algebra (g, h) with a chosen base ∆ ⊂ Φ. Let b+ = h ⊕ n+ be the Borel subalgebra corresponding to ∆. If V is generated by a maximal vector v+ of weight λ, then it is termed a b-highest weight module with highest weight λ. Lemma 1.44. Let V be a b-highest weight module for a Lie algebra g. Then: (i) V is a weight module; (ii) the weights of V are of the form λ − P n α where n ∈ ; α∈Φ+ α α Z≥0

(iii) dimVλ = 1. Lemma 1.45. Let V,W be irreducible b-highest weight modules of g with the same highest weight λ. Then V and W are isomorphic as g-modules. Proposition 1.46. Let V be a finite dimensional, irreducibe weight module of g. Then every weight of V is integral. Aban Husain 11

1.7 Induced Representations Definition 1.47. Let W be an h-module and h be a subalgebra of g. Given that U(g) is a right U(h)-module, we may form a left U(g)-module U(g) ⊗U(h) W called the g-module induced by W , denoted ind(W, g). If the representation of h corresponding to W is %, then we may refer to the representation of g induced by %, denoted ind(%, g). Definition 1.48. Given a finite dimensional vector space E with a vector subspace F stable under some endomorphism u. Define trE/F u = tru − tr(u|F ). Let h be a subalgebra of g. Define 1 θ (x) = tr ad x g,h 2 g/h g for all x ∈ h. Definition 1.49. Let % be a representation of h in W and h a subalgebra of g. Define

∼ % (x) = %(x) + θg,h(x) · 1 for all x ∈ h. Then %∼ is a representation of h in W and ind(%∼, g) = ind∼(%, g) the twisted representation of g induced by % and the corresponding module, ind∼(W, g), is termed the twisted g-module induced by the h-module W .

2 The Weyl Character Formula

Given a split semi-simple Lie algebra (g, h) and λ ∈ h∗, one may construct a representation of g such that λ is the highest weight with respect to a given Borel subalgebra. Additionally a simple highest weight module may be constructed for any λ ∈ h∗. In the case of dominant integral weights, one has a finite dimensional simple highest weight module. For such a module, the dimensions of the module and each of its weight spaces may be combinatorically determined via computation by the Weyl character formula.

2.1 Verma Modules Let (g, h) be a split semi-simple Lie algebra with root system Φ and λ ∈ h∗. For a chosen base ∆ and corresponding Borel subalgebra b = h ⊕ n+, define a one dimensional representation τλ of b by τλ(h + n) = λ(h) for h ∈ h and n ∈ n+. Let % be the twisted representation of g induced by τλ: ∼ % = ind (τλ, g). From the definition of the twisted induced module: 1 θ (x) = tr ad x, g,b 2 g/b g

1 so θg,b(x) = tr adg(x) for x ∈ b. As adgn is nilpotent for n ∈ n+ and adgh = −2ρ(h) for 2 n− n− h ∈ h, where ρ is the half sum of positive roots, θg,b(h + n) = −ρ(h). Hence

∼ τλ (x) = τλ(x) + θg,b(x) = τλ−ρ.

The representation % can be seen as the representation of g induced by τλ−ρ. Definition 2.1. The g-module M(λ), corresponding to the representation %, is a b-highest weight module with highest weight λ − ρ, called the Verma module, i.e:

M(λ) = U(g) ⊗U(b) C where C is the module associated with the representation τλ−ρ. Aban Husain 12

Proposition 2.2. Let M(λ) be a Verma module, α ∈ Φ and xα ∈ gα. Let α1, . . . , αn ∈ Φ+ be pairwise distinct roots and B(µ) the number of families (nα)α∈Φ+ ⊂ Z≥0 such that µ = P n α. Then: α∈Φ+ α L (i) M(λ) = µ∈h∗ M(λ)µ. (ii) All weights µ of M(λ) are given by X µ = λ − ρ − nαα

α∈Φ+

where nα ∈ Z≥0, and dimM(λ)µ = B(λ − ρ − µ).

(iii) For each weight µ: X M(λ) = xp1 . . . xpn ⊗ . µ −α1 −αn C p1,...,pn∈Z≥0 λ−δ−p1α1−···−pnαn=µ

(iv) Additionally:

M(λ)λ−ρ = 1 ⊗ C M(λ) = U(n−)M(λ)λ−ρ

U(n+)M(λ)λ−ρ = 0.

Proof. To prove (iv), as M(λ) = U(g)⊗U(b) C, it is generated by 1⊗1. Clearly M(λ)λ−ρ = 1⊗C, and n+(1 ⊗ 1) = 0 by definition of τλ−ρ, so U(n+)M(λ)λ−ρ = 0. By the Poincaré-Birkhoff-Witt Theorem, U(g) = U(n−)U(h)U(n+), hence

M(λ) = U(n−)M(λ)λ−ρ.

For (iii), if h ∈ h then h(xp1 . . . xpn ⊗ 1) = [h, xp1 . . . xpn ] ⊗ 1 + xp1 . . . xpn h ⊗ 1 −α1 −αn −α1 −αn −α1 −αn = (−p α − · · · − p α )(h)(xp1 . . . xpn ⊗ 1) 1 1 n n −α1 −αn + (λ − ρ)(h)(xp1 . . . xpn ⊗ 1) −α1 −αn = (λ − ρ − p α − · · · − p α )(h)(xp1 . . . xpn ⊗ 1). 1 1 n n −α1 −αn Both (i) and (ii) follow directly from (iii). Definition 2.3. Let V be a g-module. The central character of V is a Lie algebra homomor- phism χ : Z(g) → C such that for all x ∈ Z(g) and v ∈ V the action of x is defined by xv = χ(x)v. Proposition 2.4. Let V be a g-module generated by a weight vector v of weight λ, such that n+v = 0. Then: (i) There exist a unique homomorphism φ : M(λ + ρ) → V such that φ(1 ⊗ 1) = v and φ is surjective.

(ii) V = U(n−)v. (iii) Every endomorphism of V as a g-module is scalar. (iv) V has a central character. (v) The homomorphism φ is bijective if and only if V 6= 0 and u acts injectively on V for each nonzero u ∈ U(n−). Aban Husain 13

Proof. Both assertions (i) and (ii) follow from the construction of M(λ + ρ) as an induced representation and as v generates V as a g-module. Assume ϕ is an endomorphism of V as a g-module. Then hϕ(v) = ϕ(hv) = λ(h)ϕ(v).

As Vλ is one dimensional, and ϕ(v) ∈ Vλ, there exists some k ∈ C such that ϕ(v) = kv. Since v generates V , for any u ∈ U(g) ϕ(uv) = u(ϕ(v)) = kuv. Any element in Z(g) acts on V as an endomorphism, hence assertion (iv) follows. Finally, if φ is an isomorphism then clearly each u ∈ U(n−) acts injectively on V . If not, there must exist some nonzero u ∈ U(n−) such that φ(u ⊗ 1) = 0. Thus

uv = uφ(1 ⊗ 1) = φ(u ⊗ 1) = 0 so unless V = 0, u does not act injectively.

Let the central character of M(λ) = χλ. Proposition 2.5. Let λ, µ ∈ h∗. Then the following statements are equivalent:

(i) χλ = χµ (ii) λ ∈ W µ. Proposition 2.6. Let M(λ) be a Verma module of g. Every proper submodule of M(λ) is P contained in M(λ)+ = µ6=λ−ρ M(λ)µ. Additionally there exists a maximal proper submodule K and the g-module M(λ)/K = L(λ) is irreducible. L Proof. Let W be a submodule of M(λ) distinct from M(λ). Given that M(λ) = µ∈h∗ M(λ)µ, we have M W = W ∩ M(λ)µ and W ∩ M(λ)λ−ρ = 0 µ∈h∗ as dim M(λ)λ−ρ = 1. Therefore W must be contained in M(λ)+, so the sum of all proper submodules of M(λ) is contained in M(λ)+. The sum K of all proper submodules of M(λ) is thus maximal and L(λ) = M(λ)/K is simple.

Proposition 2.7. If V is a simple g-module such that there exists a nonzero v ∈ Vλ−ρ annihilated by n+, then V is isomorphic to L(λ). Proof. Let φ : M(λ) → V be the unique surjective homomorphism such that φ(1 ⊗ 1) = v. V is simple and V ' M(λ)/ker φ, thus ker φ = K so

V = M(λ)/K = L(λ).

Proposition 2.8. Let V be a finite dimensional simple g-module. Then (i) There exists unique λ ∈ h∗ such that V is isomorphic to L(λ + ρ).

(ii) λ is a dominant integral weight and dim Vλ = 1. (iii) If µ is a weight of V , then µ = λ − P n α for n ∈ and (µ, µ) ≤ (λ, λ). α∈Φ+ α α Z≥0 Proof. (i) As V is finite dimensional, there must exist some weight λ such that λ + α is not a weight of V for all positive roots α. Additionally, as V is simple, it is generated by v ∈ Vλ. Thus V is isomorphic to L(λ + ρ).

α −α (ii) Let sα = hα ⊕ g ⊕ g be the subalgebra of g isomorphic to slα where α ∈ Φ+. If % is the

representation of g in V , then % is isomorphic to a representation of sl2. As V is finite sα dimensional and xαVλ = 0, we have λ(hα) = hλ, αi ∈ Z≥0 and Vλ is one dimensional. Aban Husain 14

(iii) Let µ be a weight of V . There exists some w ∈ W such that wµ is a dominant integral weight. Since every element in the Weyl orbit of µ is a weight of V , and the inner product on h∗ is stable under w, we may assume µ to be dominant integral. As λ is the highest weight, λ − µ is a sum of positive roots with positive integer coefficients. Hence hλ + µ, λ − µi ≥ 0 so:

0 ≤ (λ + µ, λ − µ) = (λ, λ) + (µ, λ) − (λ, µ) − (µ, µ) = (λ, λ) − (µ, µ),

and the assertion follows.

Lemma 2.9. Let A be an associative algebra with an underlying Lie algebra defined by the bracket [x, y] = xy − yx. Let x, y, h ∈ A be elements such that [h, y] = −2y and [x, y] = h. For m ∈ Z≥0: [x, ym] = m(h + m − 1)ym−1 = mym−1(h − m + 1). Proof. For m = 0, 1 the equality is obvious. Assuming the identity holds for m, then

[x, ym+1] = xym+1 − ym+1x = xyym − yxym + yxym − yymx = [x, y]ym + y[x, ym] = hym + ym(h + m − 1)ym−1 = hym + m(h + m − 1)ym + [y, m(h + m − 1)ym−1] = hym + m(h + m − 1)ym + 2mym = (m + 1)(h + m)ym

Proposition 2.10. Let λ ∈ h∗ and α ∈ ∆. Assume m = hλ, αi is a positive integer. Given the −α canonical generator v = 1 ⊗ 1 of M(λ), for nonzero x−α ∈ g , let V be the submodule of M(λ) 0 m generated by v = xα v. Then V is isomorphic to M(σαλ). 0 Proof. As v 6= 0 and σαλ = λ − mα, we have

0 v ∈ M(λ)λ−ρ−mα = M(λ)σαλ−ρ.

β α β β β If β ∈ ∆ with β 6= α, then [g , g ] = 0 and g v = 0 as g ⊂ n+. Hence if xβ ∈ g then

0 m m m xβv = xβx−αv = [xβ, x−α]v + x−αxβv = 0.

α Let xα ∈ g be such that [xα, x−α] = hα, where hλ, αi = λ(hα). Following from the statement of (2.9)

0 m xαv = xαx−αv m m = [xα, x−α]v + x−αxαv m−1 = mx−α (hα − m + 1)v m−1 = mx−α (λ(hα) − ρ(α) − m + 1)v = 0.

0 Hence n+v = 0. Additionally as nonzero u ∈ U(n−) acts injectively on M(λ), it does too on any submodule. Therefore V is isomorphic to M(σαλ).

Lemma 2.11. Let V be a be a g-module generated by an element v of Vλ such that n+v = 0. α m For α ∈ Φ, let xα in g . For all α ∈ ∆, assume x−αv = 0 for sufficiently large m. Then V is a finite dimensional, simple module. Aban Husain 15

Lemma 2.12. Let λ ∈ h∗ be a dominant integral weight, K the maximal proper submodule of −α M(λ+ρ) and v = 1⊗1. For all α ∈ ∆, let x−α be a nonzero element of g and mα = hλ, αi+1. Then X mα X mα K = U(g)x−α v = U(n−)x−α v α∈∆ α∈∆ and K has finite codimension.

Proof. As λ is a dominant integral weight, for each α ∈ ∆, mα = hλ + ρ, αi is a positive integer. mα Thus for each α, the submodule generated by x−α v is

mα Vα = U(n+)x−α v ' M(σα(λ + ρ)) P and is distinct from M(λ + ρ). Thus α∈∆ Vα is distinct from M(λ + ρ), and thereby contained m P in K. For sufficiently large m and each α ∈ ∆, we have x−αv = 0 in M(λ + ρ)/ α∈∆ Vα. P mα Hence the quotient is simple and finite dimensional, so K = α∈∆ U(n−)x−α v and has finite codimension. Theorem 2.13. There is a one-to-one correspondence between the set of dominant integral weights and the set of classes of finite dimensional, simple g-modules.

Proposition 2.14. Let V be a finite dimensinal g-module, and w0 ∈ W such that w0(∆) = −∆. Then

∗ ∗ P ∗ ∗ (i) For any λ ∈ h the orthogonal subspace of Vλ in V is µ6=−λ Vµ , so V−λ can be identified with the dual of Vλ. (ii) If V is simple and λ is its highest weight, then V ∗ is a simple module and has highest weight −w0λ.

2.2 The Weyl Character Formula Let M(λ) be the Verma module of a split semi-simple Lie algebra (g, h) with root system Φ and base ∆. Define Q+ to be the set of linear combinations of Φ+ with nonnegative integer coefficients. Additionally define a partial order on weights by λ ≥ µ if λ − µ ∈ Q+ h∗ h∗ Denote the additive group of mappings from h∗ to Z by Z , the subgroup Z[h∗] ⊂ Z of maps h∗ with finite support, and the subgroup Zhh∗i ⊂ Z of maps with support contained in a finite ∗ union of sets v − Q+ for v ∈ h . Definition 2.15. Let V be a weight module of g such that each weight space is finite dimensional. h∗ The character of V is a map in Z defined by:

Ch V (µ) = dim Vµ.

h∗ For any λ ∈ h∗, let eλ ∈ Z be the map with support {λ}, which takes the value 1 at λ. Then ∗ P λ ∗ ∗ all elements of Zhh i are of the form λ∈h∗ cλe with cλ ∈ Z for all λ ∈ h . Then Zhh i may be viewed as a ring with multiplication defined by

X λ X 0 µ X X 0 v cλe cµe = cλcµe , λ∈h∗ µ∈h∗ v∈h∗ λ+µ=v λ,µ∈h∗

h∗ specifically eλeµ = eλ+µ. The Weyl group W operates on Z by w(eλ) = ewλ. Lemma 2.16. Let V and W be g-modules. Then:

(i) if V has a character and W is a submodule of V , both W and V/W have characters, and

Ch V = Ch W + Ch V/W ; Aban Husain 16

(ii) if Ch V, Ch W ∈ Zhh∗i, then the g-module V ⊗ W has a character, and Ch(V ⊗ W ) = (Ch V )(Ch W ).

Lemma 2.17. Let X wρ ∗ d = sgn(w)e ∈ Z[h ] w∈W and X −γ ∗ K = B(γ)e ∈ Zhh i. γ∈Q+ As an element of Zhh∗i, d is invertible, and Ke−ρd = 1. Proposition 2.18. Let λ ∈ h∗. The g-module M(λ) has a character, and

−1 λ ∗ Ch M(λ) = d e ∈ Zhh i P wρ for d = w∈W sgn(w)e . ∗ Proof. For any γ ∈ h , dimM(λ)γ = B(λ − ρ − γ), so X Ch M(λ) = B(γ)eλ−ρ−γ . γ∈h∗

As Ke−ρd = 1, and Ch M(λ) = eλ−ρK, we have Ch M(λ) = d−1eλ.

∗ Lemma 2.19. Let M be an arbitrary g-module with central character χλ0 , for some λ0 ∈ h , ∗ and character in Zhh i. Let DM be the set of λ ∈ W λ0 such that Supp(Ch M)∩{λ−δ+Q+}= 6 ∅. Then we have X Ch M = nλCh M(λ),

λ∈DM where nλ ∈ Z for each λ ∈ DM .

Proof. If M = 0, then Supp(Ch M) = DM = ∅. Hence assume M 6= 0. Let µ−ρ ∈ Supp Ch M be a maximal element with respect to the the Borel subalgebra defined by ∆. Assume dimMµ−ρ = m. Let V be the submodule of M generated by any v ∈ Mµ−ρ. Then there exists a unique surjective homomorphism φ such that φ(1⊗1) = v. Thus there must exist some g-homomorphism m m ϕ :(M(µ)) → M such that (M(µ)µ−ρ) is mapped bijectively into Mµ−ρ. As such, the central character of M(µ) must be χλ0 , so µ ∈ W λ0 and µ ∈ DM . Let kerϕ = L and cokerϕ = N. Then there exists an exact sequence of homomorphisms 0 → L → (M(µ))m → M → N → 0. Given that Ch V = Ch W +Ch V/W for any g-module V and submodule W , and Ch (M(µ)m = m Ch M(µ), we have Ch M = m Ch M(µ) − Ch L + Ch N.

As N is the co-kernel of ϕ, Supp Ch N ⊂ Supp Ch M, so DN ⊂ DM . Since µ − ρ is a maximal weight with respect to ∆, {µ − ρ + Q+} ∩ Supp Ch M = µ − ρ. Hence DN ( DM given that µ − ρ 6∈ Supp Ch N. If λ ∈ DL, then {λ − ρ + Q+} must intersect with Supp Ch M(µ), as L is a submodule of m (M(µ)) . Thus µ − ρ ∈ {λ − ρ + Q+}, and it follows that DL ⊂ DM . As L is the kernel of ϕ, m m which maps (M(µ)µ−ρ) bijectively, L ∩ (M(µ)µ−ρ) = 0. Therefore µ 6∈ DL, so DL ( DM . ∗ Given that Ch M ∈ Zhh i, the set DM must be finite. Hence DL and DN are subsets of DM with strictly smaller cardinalities. As both L, N have central character χλ0 , induction on the cardinality of the sets DL and DN shows Ch M to be a Z-linear combination of characters of M(λ) for λ ∈ W λ0. Aban Husain 17

Theorem 2.20 (The Weyl Character Formula). Let V be a finite dimensional, irreducible g- module with highest weight λ. Then X X sgn(w)ewρCh V = sgn(w)ew(λ+ρ). w∈W w∈W

Proof. As V is a finite dimensional highest weight module, it is isomorphic to L(λ+ρ) and has a −1 µ finitely supported character. Additionally, it has central character χλ+ρ. Since Ch M(µ) = d e we have X −1 w(λ+ρ) Ch V = nwd e , nw ∈ Z. w∈W P w(λ+ρ) Therefore dCh V = w∈W nwe . For any w ∈ W, dimVµ = dimVwµ, so Ch V is stable under the action of W . On the other hand, wd = sgn(w)d, so dCh V is an alternating function. Hence X dCh V = a sgn(w)ew(λ+ρ) w∈W λ+ρ for some a ∈ Z. The weight space Vλ is one dimensional, so the coefficient of e must be 1 in dCh V , and so a = 1. This yields the character formula X X sgn(w)ewρCh V = sgn(w)ew(λ+ρ). w∈W w∈W

Corollary 2.21. Let µ ∈ h∗ and V a simple finite dimensional module with highest weight λ. Then the multiplicity of µ as weight of V is X sgn(w)B(w(λ + ρ) − (µ + ρ)). w∈W Proof. By the Weyl Character formula:

X ChV = d−1 sgn(w)ew(λ+ρ) w∈W X = Ke−ρ sgn(w)ew(λ+ρ) w∈W X X = B(γ)e−γ−ρ sgn(w)ew(λ+ρ).

γ∈Q+ w∈W

Thus ChV = P sgn(w) P B(γ)ew(λ+ρ)−γ−ρ. Applying this to µ yields w∈W γ∈Q+ X Ch(V )(µ) = sgn(w)B(w(λ + ρ) − (µ + ρ)). w∈W

2.2.1 The Weyl Dimension Formula Theorem 2.22. Let V be a finite dimensional irreducible module of (g, h) with highest weight λ. Then Π hλ + ρ, αi dimV = α∈Φ+ . Πα∈Φ+ hρ, αi ∗ ∗ µ µ Proof. Let α ∈ Φ+ and define fα : Zhh i → Zhh i by linearly extending f(e ) = hµ, αie . Given that µ+µ0 0 µ+µ0 µ+µ0 0 µ+µ0 fα(e ) = hµ + µ , αie = hµ, αie + hµ , αie Aban Husain 18

the map fα is a derivation. As fα and fβ commute for α 6= β, let f = Πα∈Φ+ fα. This is a differential operator but no longer a derivation. As previously defined, Ke−ρd = 1, thus we have

X wρ −ρ α d = sgn(w)e = e Πα∈+ (e − 1). w∈W

Let v : Z[h∗] → Z be the homomorphism mapping g to the sum of its values. Thus v(Ch V ) = dimV . Using the product form of d, apply f and v consecutively, in that order, to dChV . As fα is a derivation, fα(dCh V ) = fα(d)Ch V + dfα(Ch V ). α 0 Additionally v(e − 1) = 0 for all α ∈ Φ+, so for any proper subset Φ ⊂ Φ+ we have

α v(Πα∈Φ0 fα(e − 1)) = 0. This yields v(f(dCh V )) = v(f(d))v(Ch V ). To compute f(v(d)), we use use the linearity of P wρ ρ rho f and v, and d = w∈W sgn(w)e . For e , clearly v(f(e )) = Πα∈Φ+ hρ, αi. The Weyl group acts transitively on the root system and preserves the usual inner product (·, ·) on h∗, so hwρ, αi = hρ, w−1αi. Additionally the number of positive roots mapped to negative roots by −1 l(w) w ∈ W is l(w) = l(w ), sgn(w) = (−1) , and hρ, −αi = −hρ, αi for any α ∈ Φ+. Thus

wρ l(w) v(f(e )) = (−1) Πα∈Φ+ hρ, αi. Hence

X l(w) v(f(d)) = sgn(w)(−1) Πα∈Φ+ hρ, αi w∈W X l(w) l(w) = (−1) (−1) Πα∈Φ+ hρ, αi w∈W

= |W |Πα∈Φ+ hρ, αi

P w(λ+ρ) The right hand side of the Weyl dimension formula is the sum w∈W sgn(w)e . From w(λ+ρ) l(w) above, v(f(e )) = (−1) Πα∈Φ+ hλ + ρ, αi, which yields

X w(λ+ρ) v(f( sgn(w)e )) = |W |Πα∈Φ+ hλ + ρ, αi. w∈W Hence applying f and consequently v to both sides of the Weyl character formula results in the dimension formula Π hλ + ρ, αi dimV = α∈Φ+ . Πα∈Φ+ hρ, αi

2.3 Examples

2.3.1 g = sl2

Let h be the Cartan subalgebra of sl2 consisting of diagonal matrices, with corresponding roots 1 − 2 and 2 − 1. Given the normal linear ordering on the Z>0, the half sum of positive roots 1−2 becomes ρ = 2 which is also the fundamental weight. Thus the dominant integral weights are in one-to-one correspondence with nonnegative integers. Let λ = mρ and V be a finite dimensional, irreducible representation of sln with highest weight λ. The Weyl character formula yields

ρ −ρ X µ+ρ µ−ρ (e − e )Ch V = cµ(e − e ) µ∈h∗ = e(m+1)ρ − e−(m+1)ρ. Aban Husain 19

Given that (e(m+1)ρ − e−(m+1)ρ)(µ) is nonzero if and only if µ = (m + 1)ρ, −(m + 1)ρ we have the the following equalities: cmρ − c(m+2)ρ = 1

c−(m+2)ρ − c−mρ = −1

cnρ − c(n+2)ρ = 0, n 6= m, −(m + 2).

Thus for all integers i, j, c(m+2i+1)ρ = c(m+2j+1)ρ. By the initial condition, V is finite di- mensional. Therefore there are only finitely many weight spaces, so c(m+2i+1)ρ = 0 for all integers i. Additionally c(m+2i)ρ = c(m+2j)ρ and c−(m+2i)ρ = c−(m+2j)ρ for i, j ≥ 1. Hence c(m+2i)ρ = c−(m+2i)ρ = 0 for all i ≥ 1 from the finite dimensionality of V . Finally the formula results in c−mρ = c−(m−2)ρ = ··· = c(m−2)ρ = cmρ. As λ = mρ is the highest weight of V , its weight space must be one dimensional. Hence the the character of V is given by

Ch V = emρ + e(m−2)ρ + . . . e−(m−2)ρ + e−mρ.

2.3.2 g = sl3

As before, let h be the Cartan subalgebra of sl3 consisting of diagonal matrices. With the usual ordering on Z>0, the positive roots are α = 1 − 2, β = 2 − 3 and γ = 1 − 3 with base ∆ = {α, β}. Thus the half sum of the positive roots is ρ = γ and the fundamental weights are ω1 = 1, ω2 = −3. The Weyl group of sl3 is

W = {1, σα, σβ, σασβ, σβσα, σασβσα}.

∗ While h is two dimensional, the weights 1, 2, and 3 are constrained by 1 + 2 + 3 = 0, so ω1, ω2 function as fundamental weights. Any dominant integral weight is of the form λ = m1ω1 + m2ω2 = (m1, m2). Then

hω1 + ω2, αi = 1, hω1 + ω2, βi = 1 hω1 + ω2, γi = 2 and

h(m1 + 1)ω1 + (m2 + 1)ω2, αi = m1 + 1, h(m1 + 1)ω1 + (m2 + 1)ω2, βi = m2 + 1,

h(m1 + 1)ω1 + (m2 + 1)ω2, γi = m1 + m2 + 2.

Therefore the Weyl dimension formula in the case of sl3 becomes (m + 1) (m + 1) (m + m + 2) dimV = 1 2 1 2 . 1 1 2

2.3.3 The Natural Representation

Let V be the natural representation of sl3. The highest weight is λ = (1, 0) = 1. From the Weyl dimension formula (1 + 1) (0 + 1) (1 + 0 + 2) dim V = 1 1 2 = 3.

The dimension of a weight space is invariant under the Weyl orbit of a weight, and dim Vλ = 1 as λ is the highest weight. The Weyl orbit of λ is as follows

σα(1) = 2, σβ(1) = 1, σβσα(1) = 3, σασβ(1) = 2, σασβσα(1) = 3.

As dimV = 3, the Weyl orbit of 1 is the set of weights of V , and

Ch(V ) = e1 + e2 + e3 . Aban Husain 20

Alternatively, the character of V may be computed with the Weyl character formula, given by

(eρ − eα − eβ + e−α + e−β − eρ)Ch(V ) = e21−3 − e22−3 − e21−2 + e23−2 + e22−1 − e23−1 .

Applying 2 + ρ to both sides yields

c2 − c2+ρ−α − c2+ρ−β + c2+ρ+α + c2+ρ+β − c2+2ρ = 0.

Given that λ = 1 is the highest weight of V , clearly 2 + ρ − α = 22 − 3, 2 + ρ + α = 21 − 3, 2 + ρ + β = 22 − 22 + 1, 2 + 2ρ = 2 + 21 − 23 are not weights of V . Thus all that remains is

c2 − c1 = 0 and so c2 = 1. Similarly, applying 3 + ρ to both sides yields c3 = 1. As dimV = 3, we obtain the same result Ch(V ) = e1 + e2 + e3 .

Of the set of weights of V , there are 1, 2 = 1 − α, 3 = 1 − α − β. Thus the lowest weight of ∗ ∗ V is 3, so −3 is the highest weight of V , i.e. V ' L(0, 1).

2.3.4 The Adjoint Representation

The finite dimensional irreducible representation of sl3 with highest weight λ = (1, 1) is the adjoint representation. Then 1 + 1 1 + 1 1 + 1 + 2 dim(V ) = 1 1 2 = 8.

The highest weight of V is γ = ρ = 1 − 3. As the Weyl group acts transitively on the set of roots, dimVµ = 1 for all µ ∈ Φ. The Weyl orbit of ρ is as follows

σα(ρ) = β, σβ(ρ) = α, σασβ(ρ) = −α, σβσα = −β, σασβσα(ρ) = −ρ.

Given that λ + ρ = 2ρ : X sgn(w)ew(λ+ρ) = e2ρ − e2β − e2α + e−2α + e−2β − e−2ρ w∈W and X (µ+ρ) (µ+β) (µ+α) (µ−α) (µ−β) (µ−ρ) dCh(V ) = cµ(e − e − e + e + e − e ). µ∈h∗ Applying both sides of the Weyl character formula to ρ results with

dCh(V )(ρ) = c0 − cρ−β − cρ−α + cρ+α + cρ+β − c2ρ = 0.

The highest weight of V is ρ, so ρ + α, ρ + β and 2ρ are not weights of V . As ρ − α and ρ − β are both roots of sl3, cρ−α = cρ−β = 1. Hence c0 = 2 and

Ch(V ) = 2e0 + eρ + eα + eβ + e−α + e−β + e−ρ.

3 Homomorphisms Between Verma Modules

All theorems are taken from [Dix96] with the proofs supplemented by [Hum08]. Aban Husain 21

3.1 Submodules and Subquotients of Verma Modules Unless otherwise specified, g will denote a finite dimensional, semisimple Lie algebra.

Definition 3.1. Let % be a representation of g in V . A series (V0,...,Vn) of submodules of V such that V = V0 ⊃ V1 ⊃ · · · ⊃ Vn = 0 is termed a composition series. If Vi/Vi+1 is simple for each i ∈ {0, . . . , n − 1}, then the series is termed a Jordan-Hölder series. Definition 3.2. A ring is called a Noetherian ring if each every chain of ascending ideals terminates eventually. A module is termed a Noetherian module if it satisfies the following equivalent conditions: (i) every chain of ascending submodules terminates eventually; (ii) every submodule is finitely generated. While it will not be proven here, we shall take from [MR01] that the universal enveloping algebra of a finite dimensional Lie algebra is a Noetherian ring. Proposition 3.3. Let λ ∈ h∗ and M(λ) the associated Verma module of the split semi-simple Lie algebra (g, h). Then: (i) M(λ) has a Jordan-Hölder series;

(ii) Every simple subquotient of M(λ) is isomorphic to L(µ) for some µ ∈ W λ ∩ (λ − Q+). Proof. (ii) Let N and N 0 be submodules of M(λ) such that N 0 ⊂ N and N/N 0 is simple. Given that the decomposition of M(λ) into weight spaces is inherited by quotients, we have

0 M 0 N/N = (N/N )µ. µ∈h∗ The set of weights of N/N 0 must be contained in the set of weights of M(λ). Hence the weights 0 of N/N are contained in λ − ρ − Q+. Thus there must exist some µ ∈ λ − Q+ such that µ − ρ 0 0 is a weight of N/N but µ − ρ + α is not for all α ∈ ∆. Then for some nonzero v ∈ (N/N )µ−ρ, 0 0 0 the module N/N is generated by v and n+v = 0. Thus N/N ' L(µ). Conversely, as N/N is a subquotient of M(λ), the two share a central character. Hence, as χλ = χµ, µ must lie in the Weyl orbit of λ. Therefore the simple subquotient N/N 0 is isomorphic to L(µ), where µ ∈ W λ ∩ (λ − Q+).

(i) Given that M(λ) is generated by a single element and U(g) is a Noetherian ring, M(λ) is a Noetherian U(g)-module. Thus every nonzero submodule N of M(λ) contains a maximal submodule N 0 such that N/N 0 is simple. If M(λ) has no Jordan-Hölder series, then there would exist an infinite decreasing sequence of submodules (N0,N1,... ) such that Ni/Ni+1 is simple. As a direct consequence of (ii), infinitely many of these subquotients would be isomorphic to L(µ) for some µ ∈ W λ ∩ (λ − Q+). This would imply that µ would be a weight of infinite multiplicity in M(λ). Given that every weight space of M(λ) must be finite dimensional, there must exist a Jordan-Hölder series of M(λ). Proposition 3.4. Let λ, µ ∈ h∗. If M(µ) is isomorphic to a submodule of M(λ) then µ ≤ λ and µ ∈ W (λ).

Proof. If M(µ) is isomorphic to a submodule of M(λ) then µ − ρ ∈ λ − ρ − Q+, so λ − µ ∈ Q+ and µ ≤ λ. Additionally, as a submodule, M(µ) and M(λ) share the same central character, so µ ∈ W λ. Lemma 3.5. Let R be a left Noetherian ring (i.e. each chain of ascending left ideals terminates eventually). Assume x ∈ R is not a right zero divisor. The ideal Rx must intersect nontrivially with each left ideal I. In particular, if R has no right zero divisors, then any two nonzero left ideals must intersect nontrivially. Aban Husain 22

Proof. Assume I ⊂ R is a nontrivial left ideal, x ∈ R is not a right divisor and I ∩ Rx = 0. Hence x 6∈ I, so there is a chain of ideals

0 ⊂ I ⊂ I + Ix ⊂ I + Ix + Ix2 ⊂ ... where each inclusion is proper, in contradiction to the Noetherian property of R. If the the sequence terminates then there must be some n such that there exists nonzero ai ∈ I where

2 n a0 + a1x + a2x + ··· + anx = 0.

n−1 This implies that a0 = −(a1 + a2x + . . . anx )x and therefore

n−1 a0 = −(a1 + a2x + . . . anx )x ∈ I ∩ Rx = 0.

n−1 Since x is not a right zero divisor, this forces a1 + a2x + . . . anx = 0. Inductively, ai = 0 for all i ∈ {0, . . . , n}, yielding a contradiction. Hence the intersection of I and Rx must be nontrivial. Proposition 3.6. Let λ ∈ h∗ and M(λ) the associated Verma module. Then (i) there exists a smallest nonzero submodule V in M(λ); (ii) the module V is isomorphic to M(µ) for some µ ∈ h∗.

∗ Proof. For any λ ∈ h ,M(λ) and U(n−) are isomorphic as n−-modules when the latter is equipped with the left regular representation. From above, any two nonzero left ideals of U(n−) must intersect nontrivially. Given that any g-submodule of M(λ) is also a n−-submodule, any two such nonzero submodules must also intersect nontrivially. As M(λ) has a Jordan-Hölder series, it must contain a minimal submodule V . This is the unique smallest submodule.

As V is the smallest nonzero submodule in the Jordan-Hölder series, V/0 is simple and therefore isomorphic to L(µ) for some µ. As a submodule of M(λ) however, U(n−) acts injectively on V . Hence V ' M(µ) = L(µ).

3.2 Homomorphisms Between Verma Modules Theorem 3.7. Let λ, µ ∈ h∗. Then (i) every nonzero homomorphism ϕ : M(µ) → M(λ) is injective;

(ii) the space Homg(M(µ),M(λ)) has dimension at most one.

Proof. (i) Let vµ and vλ be the canonical generators of M(µ) and M(λ) respectively and assume ϕ : M(µ) → M(λ) is a nonzero homomorphism of g-modules. Given that M(µ) and M(λ) are isomorphic to U(n−) as left U(n−) modules, there must exist some u ∈ U(n−) such that 0 ϕ(vµ) = uvλ. Assume that ϕ is not injective. Then there exists nonzero u ∈ U(n−) such that 0 0 u vµ 6= 0 but ϕ(u vµ) = 0. Then

0 0 0 ϕ(u vµ) = u ϕ(vµ) = (u u)vλ = 0.

0 As u is nonzero by assumption, and U(n−) contains no right zero divisors, u = 0. This yields

ϕ(M(µ)) = ϕ(U(n−)vµ) = U(n−)ϕ(vµ) = U(n−)uvλ = 0, so ϕ is trivial. Hence every nonzero g-homomorphism must be injective.

(ii) Let ϕ1, ϕ2 : M(µ) → M(λ) be g-homomorphisms such that ϕ1(M(µ)) = ϕ2(M(µ)). Then there must exist some automorphism φ of M(µ) such that ϕ2 = ϕ1 ◦ φ. However all automor- phisms of M(µ) act via scalar multiplication, hence ϕ1 and ϕ2 are linearly dependent. If M(µ) is simple and both ϕ1 and ϕ2 are nonzero, then both homomorphisms map M(µ) to the unique sim- ple submodule of M(λ), so ϕ1(M(µ)) = ϕ2(M(µ)). Finally we have the general case. As M(µ) Aban Husain 23 contains a simple submodule isomorphic to L(ϑ) = M(ϑ), there exists an injective homomor- phism ψ : M(ϑ) → M(µ). As M(ϑ) is simple, we have already shown that Homg(M(ϑ),M(λ)) has dimension at most one. Define the linear mapping

f : Homg(M(µ),M(λ)) → Homg(M(ϑ),M(λ)) by ϕ 7→ ϕ ◦ ψ. Assume f(ϕ) = 0. Then ϕ = 0 as ψ(M(ϑ)) 6= 0 and we have already shown that all nonzero homomorphisms are injective. Hence f is injective and

dim Homg(M(µ),M(λ)) ≤ 1.

3.3 Existence of Embeddings (Integral Case) ∗ Proposition 3.8. Let λ ∈ h be a dominant integral weight and w ∈ W . If w = σn . . . σ1 is a reduced decompostion of w, where σi is the reflection associated with αi ∈ ∆, let

λ0 = λ, λ1 = σ1λ0, . . . , λi = σiλi−1, . . . , λn = wλ.

Then:

(i) λ0 ≥ λ1 ≥ · · · ≥ λn and hλi, αi+1i ∈ Z>0 for all i ∈ {0, . . . , n − 1};

(ii) M(λ0) ⊃ M(λ1) ⊃ · · · ⊃ M(λn) = M(wλ). Lemma 3.9. Let a be a with universal enveloping algebra U(a). Let x ∈ a, u ∈ U(a) and p a nonnegative integer. Then there exists l ∈ Z≥0 such that xlu ∈ U(a)xp.

Proof. Denote left (resp. right) multiplication by x on U(a) by lx (resp. rx). The adjoint action of a on itself extends to U(a) by adx = lx − rx. Through straightforward computation it is obvious that lx, rx and adx commute. As a is nilpotent, there exists some nonegative integer q such that (adx)q = 0. Let u ∈ U(a) and fix p. Then

l l x u = lxu l = (rx + adx) u l X l = rl−i(adx)iu i x i=0 q X l = rl−i(adx)iu i x i=0 q X l = (adx)iuxl−i i i=0 ∈ U(a)xl−q.

Choose l such that l ≥ p + q. Then the lemma holds. Lemma 3.10. Let λ, µ ∈ h∗ and α ∈ ∆. Assume

M(σαµ) ⊂ M(µ) ⊂ M(λ) and p = hλ, αi is an integer.

(i) If p ≤ 0, then M(λ) ⊂ M(σαλ).

(ii) If p > 0, then M(σαµ) ⊂ M(σαλ) ⊂ M(λ). Aban Husain 24

Proof. (i) If p ≤ 0, then hσαλ, αi = hλ, σααi = −hλ, αi ≥ 0.

Thus M(λ) ⊂ M(σαλ) by (2.10).

α (ii) Let vµ and vλ be the canonical generators of M(µ) and M(λ) respectively. Choose xα ∈ g α and yα ∈ g such that [xα, yα] = hα where λ(hα) = hλ, αi. Given that M(σαµ) ⊂ M(µ), by m (2.10), M(σα) is generated by yα vµ where m = hµ, αi ∈ Z≥0. Similarly M(σαλ) is a submodule p of M(λ) generated by yαvλ.

Since M(µ) ⊂ M(λ), there exists some u ∈ U(n−) such that vµ = uvλ. As n− is a nilpotent Lie p algebra, there exists some l ∈ Z≥0 such that yαu ∈ U(n−)yα. Hence

l l p yαvµ = yαuvλ ∈ U(n−)yαvλ ⊂ M(σαλ).

m p If l ≤ m then we are done as then yα u ∈ U(n−)yα. Therefore assume that l ≥ m. From (2.9), as vµ is a vector of weight µ − ρ, we have

l l−1 [xα, yα]vµ = lyα (hα − l + 1)vµ l−1 = l(m − l)yα vµ.

Given that vµ is annihilated by n+, we have

l−1 l l l l(m − l)yα vµ = xαyαvµ − yαxαvµ = xαyαvµ ∈ M(σαλ),

m m thus inductively yα vµ ∈ M(σαλ). The module M(σαµ) is generated by yα vµ, hence

M(σαµ) ⊂ M(σαλ) ⊂ M(λ).

∗ Theorem 3.11 ([Hum08]). Let λ ∈ h be an integral weight and α ∈ Φ+. Assume that µ = σαλ ≤ λ. Then there exists an embedding

M(µ) ⊂ M(λ).

Proof. (i) Given that λ is an integral weight, µ must be integral as well. There exists some w ∈ W such that µ0 = w−1µ is a dominant integral weight. If we have a reduced decomposition w = σn . . . σ1, by (3.8) there exists a sequence of embeddings relative to this decomposition such that 0 M(µ ) = M(µ0) ⊃ M(µ1) ⊃ · · · ⊃ M(µn) = M(µ), where µk = σk . . . σ1µ0 for k ∈ {1, . . . , n}. Additionally the weights are ordered by

µ0 ≥ µ1 ≥ · · · ≥ µn.

0 −1 0 (ii) Define a parallel sequence of weights associated to λ, with λ = w λ. Then λ = λ0, and for k ∈ {1, . . . , n} define λk = σk . . . σ1λ0, so λn = λ.

(iii) We may assume without loss of generality that µ < λ, as the theorem holds trivially with equality. This implies µk 6= λk for all k ∈ {0, . . . , n} given that the Weyl group acts transitively on roots. Since µ = σn . . . σ1µ0 and λ = σn . . . σ1λ0, then σn . . . σ1µ0 = σασn . . . σ1λ0. This results in

−1 µk = σk . . . σ1µ0 = σk+1 . . . σnσασn . . . σ1λ0 = (σn . . . σk+1) σα(σn . . . σk+1)λk.

−1 As a conjugate of a reflection, (σn . . . σk+1) σα(σn . . . σk+1) is itself a reflection, hence µk =

σβk λk for some βk ∈ Φ+. This implies that µk − λk = −hλk, βkiβk is an integral multiple of βk, as λk is an integral weight. Aban Husain 25

0 (iv) From (3.8), we have µk ≥ µk+1 for all k ∈ {0, . . . , n − 1}. Additionally µ is the unique dominant integral weight in the Weyl orbit of µ and λ. By assumption µ < λ and µ0 ≥ λ0. Hence there must exists a least k ∈ {0, . . . , n} such that µk > λk but µk+1 < λk+1.

(v) Given such a k, by definition

µk+1 − λk+1 = σk+1 . . . σ1µ0 − σk+1 . . . σ1λ0 = σk+1(µk − λk).

In this case µk+1 − λk+1 is a negative multiple of βk+1 and µk − λk a positive multiple of βk. As σk+1 is a simple reflection, the only positive root mapped to a negative root is αk+1. This forces βk = βk+1 = αk+1. By (2.10), this yields the inclusion

M(µk+1) ⊂ M(λk+1).

(vi) Following (3.8) and the previous step, we have the chain of embeddings

M(µk+2) = M(σk+2µk+1) ⊂ M(µk+1) ⊂ M(λk+1).

This fulfills the initial conditions of (3.10), and either alternative results in the same inclusion M(µk+2) = M(σk+2µk+1) ⊂ M(σk+2λk+1) = M(λk+2).

(vii) Inductively this argument yields

M(µ) = M(σnµn−1) ⊂ M(σnλn−1) = M(λ).

3.4 Existence of Embeddings (General Case) The theorem of existence of embeddings of Verma modules also holds for any arbitrary weight λ ∈ h∗ and may be reformulated as follows.

∗ Theorem 3.12 ([Hum08]). Let λ ∈ h be and α ∈ Φ+. Assume that µ = σαλ ≤ λ. Then there exists an embedding M(µ) ⊂ M(λ). While the previous theorem guarantees the existence of embeddings based on certain criterion, the following states the equivalence between embeddings of Verma modules and the ordering on weights within the Weyl orbit of the highest weight. While I will not be providing the proof, complete proofs can be found in both [Hum08] and [Dix96]. Theorem 3.13 ([Dix96]). Let λ, µ ∈ h∗. The following conditions are equivalent: (i) M(µ) ⊂ M(λ)

(ii) There exists α1, . . . , αn ∈ Φ+ such that

λ ≥ σα1 λ ≥ σα2 σα1 λ ≥ · · · ≥ σαn . . . σα1 λ = µ

3.5 Antidominant Weights and Simple Verma Modules ∗ Definition 3.14. Let λ ∈ h . If, for all α ∈ Φ+:

hλ, αi 6∈ Z>0 then λ is termed an antidominant weight. Theorem 3.15 ([Dix96]). Let λ ∈ h∗. Then the following statements are equivalent: Aban Husain 26

(i) M(λ) = L(λ); (ii) λ is an antidominant weight.

Proof. Not (ii) ⇒ not (i): Assume that λ is not antidominant. Then there exists some α ∈ Φ+ such that hλ, αi ∈ Z>0. Hence

λ − σαλ = λ − (λ − hλ, αiα) = hλ, αiα so σαλ < λ. By (3.12), we have a proper inclusion M(σαλ) ⊂ M(λ). Thus M(λ) is not simple, so M(λ) 6= L(λ).

Not (i) ⇒ not (ii): If M(λ) 6= L(λ) then by (3.6) there exists a smallest proper nonzero submodule of M(λ) isomorphic to M(µ). Following from (3.4), µ < λ and µ ∈ W λ, so there exists at least one α ∈ Φ+ such that σαλ < λ which forces hλ, αi ∈ Z>0. Hence λ is not an antidominant weight.

3.6 The example of M(ρ)

In the case of the sl3, let M(ρ) be the Verma module with highest weight 0. Given that ρ is a dominant integral weight: M(wρ) ⊂ M(ρ) for all w ∈ W . Given the set of positive roots Φ+ = {α, β, ρ} and that the Weyl orbit of ρ corresponds to

σα(ρ) = β, σβ(ρ) = α, σασβ(ρ) = −α, σβσα(ρ) = −β, σασβσα(ρ) = −ρ, there exists the following chains of embeddings of Verma modules within M(ρ).

3.6.1 The chains corresponding to α and β

Beginning with the element σβ = α, we have

ρ − σβ(ρ) = ρ − α = β ∈ Q+

α − σασβ(ρ) = α − −α = 2α ∈ Q+

−α − σβσασβ(ρ) = −α − −ρ = β ∈ Q+ which yields the following order of weights ρ ≥ α ≥ −α ≥ −ρ. Hence

M(ρ) ⊃ M(α) ⊃ M(−α) ⊃ M(−ρ).

Similarly, given σα(ρ) = β ρ − σα(ρ) = ρ − β = α ∈ Q+

β − σβσα(ρ) = β − −β = 2β ∈ Q+

−β − σασβσα(ρ) = −β − −ρ = α ∈ Q+ there exists a chain of embeddings M(ρ) ⊃ M(β) ⊃ M(−β) ⊃ M(−ρ). Aban Husain 27

3.6.2 The chains containing both α and β Given that α 6≤ β and β 6≤ α, there is no embedding between M(α) and M(β). Beginning with the embedding M(ρ) ⊃ M(α) we have σβσασβ(α) = −β and α ≥ −β, hence

M(ρ) ⊃ M(α) ⊃ M(−β) ⊃ M(ρ).

Similarly, σασβσα(β) = −α, which yields the chain of embeddings

M(ρ) ⊃ M(β) ⊃ M(−α) ⊃ M(−ρ).

Given that −α and −β are not connected in the order prescribed on W , the only chains of ordered roots are the following:

ρ ≥ α ≥ −α ≥ −ρ ρ ≥ α ≥ −β ≥ −ρ ρ ≥ β ≥ −β ≥ −ρ ρ ≥ β ≥ −α ≥ −ρ.

Hence there are no further embeddings of Verma modules in M(ρ).

References

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