Introduction to the Structure of Semisimple Lie Algebras and Their Representation Theory

Home , Cartan subalgebra, Dual representation, Semisimple Lie algebra, Simple Lie algebra, Toral subalgebra

INTRODUCTION TO THE STRUCTURE OF SEMISIMPLE LIE ALGEBRAS AND THEIR REPRESENTATION THEORY

GEORGE H. SEELINGER

1. Introduction The representation theory of semisimple Lie algebras is an incredibly beautiful model of a representation theory, since the simple representations of semisimple Lie algebras are completely classiﬁed. Furthermore, such a tool is very powerful, since it can be used to • completely classify all simple Lie algebras • completely understand the representation theory of simply connected Lie groups • serve as a guide for understanding the reprsentation theory of other algebraic objects In this monograph, we will be more concerned with the former application and then touch on the general representation theory of semisimple Lie algebras.

Note that, in this monograph, a linear Lie algebra is a Lie algebra that is viewed a subalgebra of gl(V ), although Ado’s theorem shows that any Lie algebra can be realized in such a way. Furthermore, a ﬁeld F will always be of characteristic 0 and algebraically closed unless otherwise stated.

2. Weyl’s Theorem In this section, we seek to prove Weyl’s theorem, which provides us with a characterization of the structure of representations of semisimple Lie algebras. This section is mostly a rewriting and expanding of [Hum72] chapter 6. 2.1. Theorem (Weyl’s Theorem). Let g be a semisimple Lie algebra. If φ: g → gl(V ) is a representation of g, then φ is completely reducible.

To prove this theorem, we will need to show that, for any g-module V and any submodule W ≤ V , we get V = W ⊕ W 0, where W 0 is a complement of W . The easiest way to do this will be to ﬁnd some module homomorphism

Date: June 2018.

1 f : V → W with ker f ∩ W = {0} and W + ker f = V . To construct such a homomorphism, we will need the following. 2.2. Deﬁnition. Let φ: g → gl(V ) be a faithful representation of a Lie algebra g and let β(x, y) := tr(φ(x)φ(y)) be a bilinear form on g. Then, for basis {x1, . . . , xn} of g and {y1, . . . , yn} of g such that β(xi, yj) = δij, we deﬁne the Casimir element of φ to be X cφ(β) = φ(xi)φ(yi) ∈ End(V ) i

2.3. Example. Let g = sl2 with standard basis {e, h, f}. Then, as shown 1 1 1 in [See18], we have dual basis { 4 f, 8 h, 4 e}. Now, if φ is the adjoint representation, we get 1 1 1 c ({e, f, h}) = φ(e)φ( f) + φ(h)φ( h) + φ(f)φ( e) φ 4 8 4 1 1 1 = ad ad + ad2 + ad f ad 4 e f 8 h 4 e 1 1 1 = (2e + 2e ) + (4e + 4e ) + (2e + 2e ) 4 1,1 2,2 8 1,1 3,3 4 2,2 3,3 = e1,1 + e2,2 + e3,3

= Id3

2.4. Lemma. Take a basis {x1, . . . , xn} of g and a basis {y1, . . . , yn} of g such that, for nondegenerate symmetric associative bilinear form β on g, β(xi, yj) = δij. For x ∈ g, if X X [x, xi] = aijxj, [x, yi] = bijyj j j then aik = −bki

Proof. We compute X aik = aijδjk j X = aijβ(xj, yk) by bilinearity j

= β([x, xi], yk)

= β(−[xi, x], yk)

= β(xi, −[x, yk]) by associativity X = − bkjβ(xi, yj) j

= −bki

2 2.5. Example. Continuing the example above, consider basis {e, h, f} of 1 1 1 sl2 and dual basis with respect to the Killing form, { 4 f, 8 h, 4 e}. Then, for x = ae + bh + cf, we check that

[x, e] = 2be − ch =⇒ a1,1 = 2b, a1,2 = −c, a1,3 = 0

[x, h] = −2ae + 2cf =⇒ a2,1 = −2a, a2,2 = 0, a2,3 = 2c

[x, f] = ah − 2bf =⇒ a3,1 = 0, a3,2 = a, a3,3 = −2b 1 1 1 [x, f] = − bf + ah =⇒ b = −2b, b = 2a, b = 0 4 2 4 1,1 1,2 1,3 1 1 1 [x, h] = cf − ae =⇒ b = c, b = 0, b = −a 8 4 4 2,1 2,2 2,3 1 1 1 [x, e] = − ch + be =⇒ b = 0, b = −2c, b = 2b 4 4 2 3,1 3,2 3,3 2.6. Lemma. For x, y, z ∈ End(V ), [x, yz] = [x, y]z + y[x, z] Proof. [x, y]z + y[x, z] = (xy − yx)z + y(xz − zx) = xyz − yxz + yxz − yzx = xyz − yzx = [x, yz]

2.7. Proposition. cφ(β) commutes with φ(g) in End(V ).

Proof. For x ∈ g, consider [φ(x), cφ(β)]. Using the lemma immediately above, we get X X [φ(x), cφ(β)] = [φ(x), φ(xi)φ(yi)] = [φ(x), φ(xi)]φ(yi)+φ(xi)[φ(x), φ(yi)] i i P Moreover, we know that [φ(x), φ(xi)] = j aijφ(xj) and [φ(x), φ(yi)] = P j bijφ(xj), with aik = −bki for any k using the same decomposition as in the ﬁrst of the two lemmas. Thus, X X [φ(x), cφ(β)] = aijφ(xj)φ(yi) + bi,jφ(xi)φ(yj) = 0 i,j i,j This proposition is vitally important for proving Weyl’s theorem. With this element, we now have an object that must “behave like a scalar” with respect to φ(g) in End(V ). Indeed, in the sl2 example, our Casimir element is exactly the identity, although this need not always be true:

2.8. Proposition. If φ is faithful and irreducible, then cφ(β) is a scalar in End(V ).

3 Proof. Since φ is irreducible, we can invoke Schur’s Lemma to get that cφ(β) must be a scalar multiple of the identity map.

2.9. Remark. Note that cφ(β) is independent of choice of basis when φ is faithful, so we will often denote cφ := cφ(β). For more details, see [Hum72, p 27].

2.10. Proposition. Given Casimir element cφ of faithful representation φ: g → gl(V ), we get that tr(cφ) = dim g Proof. This follows from simply applying the trace operator. X X tr(cφ) = tr(φ(xi)φ(yi)) = β(xi, yi) = dim g i i

2.11. Example. The Casimir element of sl2 under the adjoint representation was precisely Id3 and tr(Id3) = 3 = dim sl2. 2.12. Lemma. Let φ: g → gl(V ) be a representation of a semisimple Lie algebra g. Then, φ(g) ⊆ sl(V ).

Proof. We know that, for a semisimple Lie algebra, [g, g] = g by the fact that g is a direct sum of simple Lie algebras. Furthermore, [gl(V ), gl(V )] = sl(V ) by simple computation. Thus, φ(g) = [φ(g), φ(g)] ⊆ [gl(V ), gl(V )] = sl(V ). 2.13. Lemma. Let φ: g → gl(V ) be a representation of a semisimple Lie algebra and let W ≤ V be irreducible with codimension 1. Then, φ is completely reducible.

Proof. Without loss of generality, we may assume φ is faithful . Now, con- why? sider cφ commutes with φ(g), so cφ(g.v) = (cφ ◦φ(g))(v) = (φ(g)◦cφ)(v) tells us that cφ is actually a g-module homormophism on V . Thus, cφ(W ) ⊆ W and ker cφ is a submodule of V . Now, by the lemma above, g must send all of V into W and thus act trivially on the quotient module V/W . There- fore, cφ must also act trivially since it is simply a linear combination of actions in g. So, it must be that the trace of cφ on V/W is 0. However, we know trV (c) = dim g 6= 0. Now, since cφ acts as a (must be non-zero) scalar on W , it must be that ker cφ ∩ W = {0} and ker cφ + W = V . Thus, since ker cφ is a submodule of V , it must be that ker cφ is 1-dimensional and V = W ⊕ ker cφ. 2.14. Lemma. Let φ: g → gl(V ) be a representation of a semisimple Lie algebra and let W ≤ V with codimension 1. Then, φ is completely reducible.

Proof. Note that this is the same as the lemma above, but now W need not be irreducible. We will show that we can reduce this lemma to the case of

4 the previous lemma 2.13. By 2.12, we know g must act trivially on V/W . Thus, V/W =∼ F and we have short exact sequence 0 → W → V → F → 0 Now, if dim W = 1, W is irreducible and we can appeal to the previous lemma. So, now we proceed by induction and assume the result it true for dim W < n. Let dim W = n and W have proper submodule W 0. Then, we have short exact sequence 0 → W/W 0 → V/W 0 → F → 0 Now, by induction, V/W 0 = W/W 0 ⊕ W 00/W 0. However, this lifts to a short exact sequence 0 → W 0 → W 00 → F → 0 However, dim W 0 < dim W =⇒ dim W 00 = dim W 0+1 < dim W +1 = dim V , so we can apply induction to get a one dimensional submodule complementary to W 0 in W 00, say X, so that W 00 = W 0 ⊕ X. Thus, we get V/W 0 = W/W 0 ⊕ W 00/W 0 = W/W 0 ⊕ (W 0 ⊕ X)/W 0 =⇒ V = W ⊕ X since W ∩ X = 0 and dim W + dim X = dim V . Thus, we can repeat this process until we get an irreducible W and appeal to the lemma above (2.13) to get the result. Proof of Weyl’s Theorem 2.1. Let 0 6= W V be g-modules. Thus, we have short exact sequence 0 → W → V → V/W → 0

Let U = {ψ ∈ Hom(V,W ) | ∃c ∈ F, ψ|W (w) = cw, ∀w ∈ W } and let T = {ψ ∈ Hom(V,W ) | ψ|W (w) = 0, ∀w ∈ W } ≤ U. Then, we check that U and T are g-submodules of Hom(V,W ). For g ∈ g, ψ ∈ U, and w ∈ W (g.ψ)(w) = g.ψ(w) − ψ(g.w) by the Jacobi identity = g.(cw) − c(g.w) ψ ∈ U = c(g.w) − c(g.w) = 0

Thus, g.ψ|W = 0 for all g ∈ g. A similar computation shows that T is also a g-sucmodule, specifying c = 0. This also shows that g.U ≤ T . Now, consider the quotient U/T . Every element in U/T is uniquely determined by the scalar c, that is τ ∈ ψ + T if and only if τ|W = ψ|W = c.1W for scalar c. Thus, we have short exact sequence 0 → T → U → F → 0 and thus, by the lemma above (2.14), U = T ⊕ T 0 where T 0 is a one- dimensional g-submodue. If we let T 0 = (ψ), ψ ∈ End(V,W ), then we can scale ψ such that ψ|W = 1W since ψ 6∈ T . Now, since ψ is a g-module homomorphism, ker ψ ≤ V , but ker ψ 6= V since W ∩ ker ψ = {0}. However, ψ cannot be injective because W V , which is ﬁnite-dimensional, and thus

5 ker ψ 6= {0}. Therefore, since ψ is surjective, we have that V = W ⊕ ker ψ as a non-trivial decomposition of V . Thus, we can iteratively continue this process until V has been completely reduced. Weyl’s theorem has many applications in the representation theory of semisimple Lie algebras. One immediate consequence is the preservation of the Jordan decomposition ([Hum72] section 6.4). That is, 2.15. Theorem. Let g ⊆ gl(V ) be a semisimple linear Lie algebra where V is ﬁnite-dimensional. Then, g contains the semisimple and nilpotent parts in gl(V ) of all its elements. In particular, the abstract and usual Jordan decompositions in g coincide. 2.16. Corollary. Let g be a semisimple Lie algebra, φ: g → gl(V ) a (ﬁnite- dimensional) representation of g. If x = s + n is the abstract Jordan decomposition of x ∈ g, then φ(x) = φ(s)+φ(n) is the usual Jordan decomposition of φ(x). 2.17. Remark. Note that this theorem is a generalization of the following facts. • A nilpotent x ∈ g is also ad-nilpotent. • If x ∈ g is diagonalizable, then adx is also diagonailzable. We also prove the following, which does not require the theorems above, but is related to the Jordan decomposition. 2.18. Proposition. Let g be a semisimple Lie algebra, thus decomposing into simple ideals g = g1 ⊕ · · · ⊕ gt. Then, the semisimple and nilpotent parts of x ∈ g are the sums of the semisimple and nilpotent parts in the various gi of the components of x.

Proof. Take x ∈ g, so x = x1 + ··· + xt with xi ∈ gi and xi = (xi)s + (xi)n.

Then, (adg(xi)s)|gi = adgi (xi)s is a semisimple endomorphism of gi and

(adg(xi)s)|gj = 0 for i 6= j by the direct sum decomposition. Thus, adg(xi)s is a semisimple endomorphism of g. Now, let u = (x1)s + ··· + (xt)s. Then, adg u is a semisimple endomorphism of g since [(xi)s, (xj)s] = 0 for all i 6= j.

By a similar argument, if v = (x1)n + ··· + (xt)n, then adg v is a nilpotent endomorphism of g. Therefore, because

[u, v] = [(x1)s, (x1)n] + [(x1)s, (x2)n] + [(x2)n, (x1)s] + [(x2)s, (x2)n] + ···

= [(x1)s, (x1)n] + ··· + [(xt)s, (xt)n] = 0 Then u, v are commuting endomorphisms with u semisimple and v nilpotent, and so by the uniqueness of Jordan decomposition, x = u + v is the Jordan decomposition of x.

6 3. Representations of sl2(F )

Recall that sl2(F ) (where F is an algebraically closed ﬁeld) is the Lie algebra of 2 by 2 traceless matrices with standard basis 0 1 0 0 1 0 e = , f = , h = 0 0 1 0 0 −1 and relations [h, e] = 2e, [h, f] = −2f, [e, f] = h

3.1. Proposition. Let V be an arbitrary sl2(F )-module. Then, h acts diag- onally on V , that is, φ(h) is diagonalizable.

Proof. Since h is semisimple, we get that φ(h) is semisimple by the corollary above.

3.2. Proposition. Let V be an arbitrary sl2(F )-module. Then, V decomposes as a direct sum of eigenspaces. M V = Vλ,Vλ = {v ∈ V | h.v = λv} λ∈F Proof. Since h is semisimple, φ(h) is semisimple and thus diagonalizable, meaning that the set of eigenvectors of φ(h) forms a basis for V . Thus, we can decompose V into φ(h)-eigenspaces.

3.3. Deﬁnition. Whenever Vλ 6= 0, we call λ a weight of h in V , and we call Vλ a weight space.

3.4. Lemma. Let V be a ﬁnite-dimensional g-module. Then, for v ∈ Vλ, e.v ∈ Vλ+2 and f.v ∈ Vλ−2. Proof. This follows from the following computations: h.(e.v) = [h, e].v + e.h.v by the Jacobi identity

= 2e.v + e.(λv) by sl2 relations = (λ + 2)e.v Similarly, we get h.(f.v) = [h, f].v + f.h.v = −2f.v + f.(λv) = (λ − 2)f.v

3.5. Proposition. For ﬁnite-dimensional sl2-module V , there exists a λ such that Vλ 6= 0 but Vλ+2 = 0. F Proof. Since V is ﬁnite-dimensional and V = λ∈F Vλ is direct, such a λ must exist.

3.6. Deﬁnition. A nonzero vector in Vλ annihilated by e ∈ sl2 is called a maximal vector of weight λ or the highest weight vector of weight λ.

7 3.7. Lemma. Let V be an irreducible sl2(F )-module. Let v0 be a maximal 1 k vector of weight λ and set v−1 = 0 and vk = k! f .v0 for k ≥ 0. Then, (a) h.vk = (λ − 2k)vk (b) f.vk = (k + 1)vk+1 (c) e.vk = (λ − k + 1)vk−1

Proof. By the previous lemma, vk ∈ Vλ−2k, so h.vk = (λ − 2k)vk.

By deﬁnition of vk, 1 1 f.v = f. f k.v = f k+1.v = (k + 1)f k+1.v = (k + 1)v k k! 0 k! 0 0 k+1 For the ﬁnal relation, we use induction on k. When k = 0, we have e.v0 = 0 since v0 is a highest weight vector, and thus our formula holds since v−1 = 0 by assumption. Now, observe ke.vk = e.f.vk−1 by part (b), f.vj = (j + 1)vj+1

= [e, f].vk−1 + f.e.vk−1 by the Jacobi identity

= h.vk−1 + f.e.vk−1 by sl2 relations

= (λ − 2(k − 1))vk−1 + (λ − k + 2)f.vk−2 by part (a) and induction

= (λ − 2k + 2)vk−1 + (k − 1)(λ − k + 2)vk−1 by part (b)

= k(λ − k + 1)vk−1 Thus, dividing both sides by k, we get the desired result.

3.8. Proposition. Let V be an irreducible sl2(F )-module. Then, for m the smallest integer such that vm 6= 0, vm+1 = 0, we get basis {v0, v1, . . . , vm} for V .

Proof. From formula (a) in the lemma above, it is clear that {v0, . . . , vm} are linearly independent since they are eigenvectors of diﬀerent eigenvalues for the h-action. Furthermore, the formulas in the lemma above show us that span{v0, v1, . . . , vm} is a non-zero submodule of V . However, since V is irreducible, that must mean that V is the span of this linearly independent set.

3.9. Proposition. Let V be an irreducible sl2(F )-module. Then, given highest weight vector v of weight λ, λ must be an integer.

Proof. In the lemma above, take k = m + 1 for formula (c). We get, for vm 6= 0 but vm+1 = 0,

0 = e.vm+1 = (λ − (m + 1) + 1)vm = (λ − m)vm Thus, it must be that λ = m ∈ Z.

3.10. Deﬁnition. We call the weight of a maximal vector vm ∈ V , an irreducible sl2(F )-module, the highest weight of V .

8 3.11. Proposition. Let m be the highest weight for irreducible sl2(F )-module V . Then, dim V = m + 1 and dim Vµ = 1 if Vµ 6= 0.

Proof. Given the eigenbasis (relative to the h-action) {v0, . . . , vm}, it is immediate that dim V = m + 1. Furthermore, for each Vµ 6= 0, there is a unique eigenvector under the h-action spanning it, up to scalar multiple. Thus, dim Vµ = 1.

To summarize, we have the following theorem.

3.12. Theorem. Let V be an irreduclbe sl2(F )-module.

(a) Relative to h, V is a direct sum of weight spaces Vµ where µ = m, m − 2,..., −(m − 2), −m where m + 1 = dim V and dim Vµ = 1 for each µ. (b) V has (up to nonzero scalar multiples) a unique maximal vector, whose weight is m. (c) The action of sl2(F ) on V is given explicitly by the formulas in the lemma above, if the basis is chosen appropriately. In particular, there exists at most one irreducible sl2(F )-module (up to isomorphism) of each possible dimension m + 1 where m ∈ Z, m ≥ 0.

Proof. Part (a) is 3.11. Since dim V = m + 1, we see by relations that V has a highest weight vector of weight m, say vm, and it is unique since its eigenspace, Vµ, is 1-dimensional. Finally, part (c) is just appealing to the lemma above.

3.13. Corollary. Let V be any (ﬁnite-dimensional) sl2(F )-module. Then, the eigenvalues of h on V are all integers, and each occurs along with its negative (an equal number of times). Moreover, in any description of V into a direct sum of irreducible submodules, the number of summands is precisely dim V0 + dim V1.

Proof. If V = 0, then the result is immediate. So, given a non-trivial V , Weyl’s theorem says V is completely reducible, so we write V as a direct sum of irreducible submodules. Now, each of these irreducible summands must have integer eigenvalues of h and thus, so must V . For the second part of the corollary, we note that an irreducible sl2-module must have an occurence of a 0-eigenvector or a 1-eigenvector, but it cannot have both since the eigenvalues diﬀer by even numbers due to the fact that h.vk = (λ−2k)vk in the lemma above.

Let us now explicitly describe these representations, which are easy to enumerate due to the theorem above. 3.14. Example. The easiest representation is the trivial one-dimensional representation C = V0.

9 3.15. Example. Consider V = {e1, e2}, the “standard representation” of sl2, where h.e1 = e1 and h.e2 = −e2. Then, V = V−1 ⊕ V1 = Ce1 ⊕ Ce2. By the corollary above, this representation must be irreducible. 3.16. Example. Consider W = Sym2 V for V the standard representation above. Then, a basis is given by {e2, ef, f 2} and we have relations h.(e · e) = e · h.e + h.e · e = 2e2 h.(e · f) = e · h.f + h.e · f = 0 h.(f · f) = f · h.f + h.f · f = −2f 2 2 2 and so W = Cf ⊕Cef ⊕Ce = V−2 ⊕V0 ⊕V2. Thus, by the corollary above, W is the irreducible representation of dimension 3. Proceeding more generally, one can conclude that the irreducible sl2-representation of dimension n is given by Symn V for V the standard representation by computing h.(en−kf k) = (n − k) · h.e · en−k−1f k + k · h.f · en−kf k−1 = (n − 2k) · en−kf k

4. Representations of sl3(F )

In this section, we describe the representations of sl3(F ) following the program in [FH91]. In sl3(F ), we do not have a single obvious element to give us an eigenspace decomposition, in contrast of sl2(F ) where the single element h suﬃced. Thus, instead, we will consider a 2-dimensional subspace of all diagonal matrices in sl3(F ), denoted h ⊆ sl3(F ). We make use of the fact 4.1. Lemma. Commuting diagonalizable matrices are simultaneously diagonalizable. Therefore, we will be able to ﬁnd simultaneous eigenvectors for the matrices in h that yield an eigenspace decomposition with eigenvalues α ∈ h∗.

4.2. Proposition. Let V be an arbitrary sl3(F )-module, then V decomposes as a direct sum of eigenspaces M V = Vα,Vα = {v ∈ V | h.v = α(h)v, ∀h ∈ h} α∈h∗

Thus, looking speciﬁcally at sl3(F ), we get 4.3. Corollary. ! ∼ M sl3(F ) = h ⊕ gα α ∗ where α ranges over a ﬁnite subset of h given by {i −j | 1 ≤ i, j ≤ 3, i 6= j} where   a1 i  a2  = ai a3 and, for any h ∈ h, y ∈ gα, we get [h, y] = α(h)y.

10 Proof. The decomposition follows from the proposition and the ﬁnite-dimensionality of sl3(F ). To see the speciﬁc decomposition, let D ∈ sl3(F ) be a diagonal matrix with trace 0 and let (mi,j) = M ∈ sl3(F ). Then, by straightforward computation, [D,M] = ((ai − aj)mi,j) and so [D,M] will be a multiple of M if and only if M has all but one entry equal to 0. Thus, Ei,j generate the eigenspaces of the adjoint action of h on sl3(F ), given by the 6 linear functionals i − j.

Now, given an x ∈ gα, we ask, how does its action aﬀect this decomposition. We get the following.

4.4. Proposition. Given x ∈ gα, y ∈ gβ, [x, y] ∈ gα+β. In other words, ad(gα) sends vectors in gβ to gα+β. Proof. Consider the action of h on [x, y]. [h, [x, y]] = [x, [h, y]] + [[h, x], y] by the Jacobi identity

= [x, β(h)y] + [α(h)x, y] since y ∈ gβ, x ∈ gα = (α(h) + β(h))[x, y] Later, we will discuss how to visualize this information, but for now we are content with this understanding. More generally, we see that

4.5. Proposition. Given a representation V of sl3 with eigenspace decom- L position V = α Vα, then for any x ∈ gα and v ∈ Vβ, x.v ∈ Vα+β. In other words, gα takes Vβ to Vα+β. From the above situation, we also notice the following 4.6. Proposition. The eigenvalues occurring in an irreducible representation of sl3(F ) diﬀer by integral linear combinations of the vectors {i −j} ⊆ h∗. Proof. Consider h.x.v = x.h.v + [h, x].v by the Jacobi identity

= x.β(h)v + α(h)x.v since v ∈ Vβ, x ∈ gα = (α(h) + β(h))x.v ∗ Thus, the i − j’s generate a lattice of eigenvalues in h . 4.7. Deﬁnition. The weights of the adjoint representation will be called roots, denoted Φ (this will be expanded on later) and the lattice generated by these roots is the root lattice, denoted ΛR. Furthermore, from the decom- L position g = α∈h∗ gα in 4.3, the non-zero gα’s will be called root spaces.

11 Now, just as with sl2(F ), we wish to ﬁnd an equivalent notion to a highest weight vector for a representation of sl3(F ).

4.8. Lemma. Given a representation V of sl3(F ), there exists a v ∈ V such that (a) v is an eigenvector for h (b) v is in the kernel of the action by E1,2,E1,3, and E2,3 in sl3(F ). (More generally, v is in the kernel of the action of half of the root spaces.)

Proof. Let us choose linear functional `: h∗ → F given by

`(a11 + a22 + a33) = aa1 + ba2 + ca3 with a + b + c = 0 and a > b > c, such that `|ΛR has trivial kernel. Thus, we get the equality

{gα ⊆ g | `(α) > 0} = {g1−3 , g2−3 , g1−2 } each of which corresponds to a basis matrix with a single nonzero entry above the diagonal. Now, pick a v ∈ Vα for `(α) maximal among the weights of V . Then, each of E1,2,E1,3, and E2,3 must kill v by the maximality of α among the weights with respect to `. 4.9. Deﬁnition. A vector v ∈ V that meets the properties of the lemma above is a highest weight vector.

4.10. Proposition. Let V be an irreducible sl3(F ) representation and v ∈ V be a highest weight vector. Then, V = hE2,1,E3,1,E3,2i.v, that is, V is generated by v under successive application of E2,1,E3,1, and E3,2. Proof. We show that the subspace W ⊆ V spanned by images of v under successive application of E2,1,E3,1, and E3,2 is preserved by all of sl3(F ). Thus, we check

E1,2.v = E2,3.v = E1,3.v = 0 by the lemma above

E1,2.E2,1.v = E2,1.E1,2.v + [E1,2,E2,1].v

= 0 + α([E1,2,E2,1])v since [E1,2,E2,1] is a diagonal matrix(∈ h)

E2,3.E2,1.v = E2,1.E2,3.v + [E2,3,E2,1].v

= 0 + 0 since [E2,3,E2,1] = 0

Similar computations hold when applied to E3,2.v. Furthermore, since E1,3 = [E1,2,E3,2], we do not need to check it.

Now, let Wn = span{w.v | w = wn ··· w1, wi ∈ {E2,1,E2,3}} (remember S that E3,1 = [E3,2,E2,1]). Then, W = n Wn. Now, consider that

E1,2.E2,1.(wn−1 ··· w1).v = E2,1.E1,2.(wn−1 ··· w1).v + [E1,2,E2,1].(wn−1 ··· w1).v

∈ E2,1.Wn−2 + β([E1,2,E2,1]).(wn−1 ··· w1).v ([E1,2,E2,1] ∈ h)

12 ⊆ Wn−1

E2,3.E2,1.(wn−1 ··· w1).v = E2,1.E2,3.(wn−1 ··· w1).v + [E2,3,E2,1].(wn−1 ··· w1).v

∈ E2,1.Wn−2 ([E2,3,E2,1] = 0)

⊆ Wn−1 and similarly for applying E1,2 and E2,3 to E3,2.(wn−1 ··· w1).v. Thus, W is closed under the action sl3(F ), but V is irreducible, so W = V .

In fact, we have slightly more.

4.11. Proposition. Let V be a sl3(F ) representation and v ∈ V be a highest weight vector. Then, hE2,1,E3,1,E3,2i.v is an irreducible subrepresentation of V .

Proof. From above, it is clear that W := hE2,1,E3,1,E3,2i.v is a subrepre- ∗ sentation. Let v have weight α ∈ h . Then, it must be that Wα is one- dimensional since the action will reduce the weight relative to ` deﬁned in the proof of 4.8. Assume W is not irreducible, so W = W 0 ⊕ W 00. How- ever, the action of h and projection onto either component commutes, so 0 00 0 00 Wα = Wα ⊕ Wα . Thus, it must be that either Wα or Wα is zero and so v 0 00 0 00 either lies in W or in W , and thus either W = W or W = W .

4.12. Corollary. Any irreducible representation of sl3(F ) has a unique highest weight vector, up to scalar multiple.

Now, similar to sl2(F ), we notice that, given a highest weight vector, say 2 k v ∈ g1−3 , we have a collection of vectors v, E2,1.v, E2,1.v, . . . until E2,1.v = 0 for some k ∈ N. From this, we note that span{E2,1,E1,2, [E1,2,E2,1]} is isomorphic to sl2(F ) via the isomorphism

e 7→ E1,2, f 7→ E2,1, h 7→ [E1,2,E2,1] This leads us to the proposition

4.13. Proposition. The eigenvalues of H1,2 := [E1,2,E2,1] on the subspace M W = Vα+k(2−1) k are integral and symmetric with respect to 0.

Proof. We note that W is a sl2(F )-representation via the isomorphism given above. Thus, since all eigenvalues of h ∈ sl2(F ) are integral and symmetric with respect to 0 for all sl2(F )-representations, that must still be true in this case.

Of course, this process generalizes, and to each root we can associate a copy of sl2(F ).

13 5. Beginning of a Structure Theory: Root Space Decomposition We now seek to understand the structure of a general semisimple Lie algebra. 5.1. Deﬁnition. A subalgebra of a Lie algebra g consisting of (nonzero) semisimple elements is called a toral subalgebra 5.2. Lemma. A toral subalgebra of a Lie algebra g is abelian.

Proof. Let T be a toral subalgebra of g. We wish to show that [x, T ] = 0 for all x ∈ T , which is the same as showing adx = 0 (where ad is taken over T ). Since x is semisimple, so is adx and since we are working over an algebraically closed ﬁeld, adx is diagonalizable. Thus, we need only show that all the eigenvalues of adx are 0.

Assume that [x, y] = ay for some 0 6= a ∈ F and y ∈ T . Then, ady x = −ay is an eigenvector of ady with eigenvalue 0. We may also write x uniquely as a linear combination of eigenvectors of ady, say {−ay, v1, . . . , vn−1}, since y is semisimple,

n−1 n−1 X X x = −c0ay + civi =⇒ −ay = ady x = ciλivi i=1 i=1 where λi is the eigenvalue of vi. However, the set {−ay, v1, . . . , vn−1} is a basis, so it is linearly independent and −ay cannot be written as a linear combination of {v1, . . . , vn−1}. Thus, we have arrived at a contradiction. 5.3. Deﬁnition. A toral subalgebra h not properly included in any other is called a maximal toral subalgebra of g.

5.4. Example. Consider g = sl2(F ). Then, h = hhi is a maximal toral subalgebra. 5.5. Remark. If g is a semisimple Lie algebra, it must have at least one element whose semisimple part is nonzero. Otherwise, g would be nilpotent by Engel’s theorem. Thus, g contains nonzero toral subalgebras and thus contains nonzero maximal toral subalgebras. 5.6. Deﬁnition. Let h be a maximal toral subalgebra for semisimple Lie algebra g. Then, for α ∈ h∗, we deﬁne

gα := {g ∈ g | [h, g] = α(h)g, ∀h ∈ h} 5.7. Deﬁnition. Let h be a maximal toral subalgebra for semisimple Lie algebra g. Then, we deﬁne the roots of g relative to h to be ∗ Φ := {α ∈ h | gα 6= 0}

14 5.8. Theorem. A semisimple Lie algebra g admits a decomposition M g = Cg(h) ⊕ gα, gα = {g ∈ g | [h, g] = α(h)g, ∀h ∈ h} α∈Φ for any maximal toral subalgebra h of g.

5.9. Example. Let g = sl2(F ). The decomposition above corresponds to the decomposition of sl2(F ) considered as an sl2(F )-modules (the regular representation). Indeed, if h = span{h} for standard basis (e, f, h), then Cg(h) = h and ∼ sl2(F ) = h ⊕ g−2 ⊕ g2 where g2 = span{e} since [h, e] = 2e and g−2 = span{f}. In fact, a similar decomposition is easy to produce for g = sln(F ) with the standard basis construction. 5.10. Remark. In the example above, it is relatively straightforward to compute Cg(h) = h. In fact, this statement ends up being true for all semisimple Lie algebras. We wish to show this is true. Furthermore, we end up ﬁnding that the roots Φ completely characterize the structure of g. We will show that this is true in the remaining parts of this section. ∗ 5.11. Proposition. The adh-eigenvector decomposition deﬁnes an h -grading on g.

∗ Proof. Let x ∈ gα and y ∈ gβ for α, β ∈ h . Then, for h ∈ h, adh([x, y]) = [[h, x], y] + [x, [h, y]] = α(h)[x, y] + β(h)[x, y] = (α + β)(h)[x, y]

Thus, [x, y] ∈ gα+β. ∗ 5.12. Corollary. If x ∈ gα for 0 6= α ∈ h , then adx is nilpotent.

n Proof. For some n ∈ N, gnα = 0, so adx = 0 using the grading as above. 5.13. Proposition. Given α, β ∈ h∗ with α + β 6= 0, then, relative to the Killing form κ of g, gα is orthogonal to gβ.

Proof. Let h ∈ h such that (α + β)(h) 6= 0. Then, we get α(h)κ(x, y) = κ([h, x], y) = −κ([x, h], y) = −κ(x, [h, y]) by the associativity of the Killing form. = −β(h)κ(x, y) Thus, (α + β)(h)κ(x, y) = 0. (Recall that the Killing form is associative follows from the deﬁnition of the bracket on gl(V ).)

5.14. Corollary. The restriction of the Killing form to Cg(h) is nondegenerate.

15 Proof. Since g is semisimple, then the Killing form on g is nondegenerate (see [Hum72, §5.1] or [See18, §6]). However, by the proposition above, Cg(h) = g0 is orthogonal to all other gα’s with respect to the Killing form. Thus, it must be that, for z ∈ g0,

κ(z, g0) = 0 =⇒ κ(z, g) = 0 =⇒ z = 0 by the non-degeneracy of the Killing form on g. Thus, the Killing form is non-degenerate on g0 = Cg(h). 5.15. Proposition. Let h be a maximal toral subalgebra of g. Then, h = Cg(h). Proof. We present only an outline of the proof given in [Hum72, §8.2].

(a) Cg(h) contains the semisimple and nilpotent parts of its elements since x ∈ Cg(h) =⇒ adx(h) = 0. (b) All semisimple elements of Cg(h) lie in h since x ∈ Cg(h) =⇒ h + F x is toral and h is maximal. (c) The restriction of κ to h is nondegenerate by the above items since [x, h] = 0 and adx is nilpotent =⇒ κ(x, h) = 0. (d) Cg(h) is nilpotent by Engel’s theorem since all its elements are ad- nilpotent. (e) h∩[Cg(h),Cg(h)] = 0 since [h,Cg(h)] = 0 =⇒ κ(h, [Cg(h),Cg(h)]) = 0. (f) Cg(h) is abelian. (g) Cg(h) = h, otherwise Cg(h) contains a nonzero nilpotent element by above. 5.16. Corollary. The Killing form restricted to h is non-degenerate.

Proof. Deﬁne map θh : h → h via

θh(k) = κ(h, k).

Since the Killing for is non-degenerate on h, the map h 7→ θh is an isomorphism between h and h∗ since the nondegeneracy gives trivial kernel and dimension count gives surjectivity. Thus, a maximal toral subalgebra (or a Cartan subalgebra) is self-centralizing. Now, we seek to understand the structure of the roots.

5.17. Lemma. Given α ∈ Φ of semisimple Lie algebra g (over C) and 0 6= x ∈ gα, then −α ∈ Φ and there is a y ∈ g−α such that {x, y, [x, y]} spans a Lie subalgebra of g isomorphic to sl2(C). Proof. If −α 6∈ Φ, then, for all β ∈ Φ,

κ(gα, gβ) = 0 =⇒ κ(gα, g) = 0 by the orthogonality in 5.13, contradicting the non-degeneracy of κ.

16 Next, we note that α 6= 0, so there is a t ∈ h such that α(t) 6= 0. Thus, for any y ∈ g−α, κ(t, [x, y]) = κ([t, x], y) = α(t)κ(x, y) 6= 0 =⇒ [x, y] 6= 0

Now, we observe that [x, y] ∈ h = g0 and x, y are simultaneous eigenvectors for ad[x, y], since they are for all elements of ad h. Thus, span{x, y, [x, y]} is a Lie subalgebra of g.

Let h := [x, y] and let S be our Lie subalgebra. To see that S is isomorphic to sl2, we ﬁrst observe that α(h) 6= 0. Assume otherwise. [h, x] = α(h)x = 0 = −α(h)y = [h, y] thus showing that S is a solvable Lie subalgebra. Therefore, [S, S] would be a nilpotent Lie subalgebra (see [Hum72, Cor 4.1C] or [See18]) and so, in particular, ad([x, y]) is both semisimple and nilpotent, thus implying ad[x, y] = 0 =⇒ [x, y] ∈ Z(S) = 0, which is a contradiction. Thus, it must be that S is a 3-dimensional complex Lie algebra with [S, S] = S, ∼ which immediately implies that S = sl2(C). In fact, this will hold over more general ﬁelds, which we will show below, but this method provides quick insight into the guiding idea behind the representation theory of semisimple Lie algebras: understanding roots and the representation theory of sl2. Luckily, we already understand the latter from the section above.

More generally, we have the following orthogonality proposition. 5.18. Proposition. [Hum72, Prop 8.3] Let Φ ⊆ h∗ be as above for semisimple Lie algebra g over . Then, (a) Φ spans h∗. (b) If α ∈ Φ, then −α ∈ Φ. ∗ (c) If α ∈ Φ, x ∈ gα, y ∈ g−α, then [x, y] = κ(x, y)tα where tα ∈ h is the unique element such that α(h) = κ(tα, h) for all h ∈ h. (d) Let α ∈ Φ. Then, [gα, g−α] is one dimensional, with basis tα. (e) α(tα) = κ(tα, tα) 6= 0 for α ∈ Φ. (f) If α ∈ Φ and xα is any nonzero element of gα, then there exists yα ∈ L−α such that xα, yα, hα = [xα, yα] span a three dimensional simple subalgebra of g isomorphic to sl2(F ) via

xα 7→ e, yα 7→ f, hα 7→ h (g) h = 2tα and h = −h . α κ(tα,tα) α −α 5.19. Remark. This proposition will later show that Φ satisﬁes the axioms of a “root system” and is thus justiﬁes the terminology “roots of g relative to h.”

17 5.20. Example. Let us demonstrate the result for g = sl2(F ) since we have already shown what needs to be shown in the previous section. Then, h = span{h} =∼ F and so h∗ = span{h∗} =∼ F where h∗(h) = 1, h∗(e) = 0 = h∗(f) and Φ = {±2h∗} since

sl2 = h ⊕ g2 ⊕ g−2 In this case, (a) and (b) above are immediate. For (c), we check

κ(e, f) = tr(ade adf ) = 4 and 1 [e, f] = h =⇒ t = h∗ 2 4 1 Indeed, one easily checks κ( 4 h, λh) = λ · 2 for all λ ∈ F .

For (d), [g2, g−2] = span{[e, f]} = span{h} is one dimensional with basis 1 given by 4 h.

1 ∗ 1 1 For (e), 2 · 4 h (h) = 16 κ(h, h) = 2 .

For (f), the result is trivial. For (g), we note 1 1 2 4 h 2 h 1 = 1 = h 16 κ(h, h) 2 Proof. For (a), assume Φ does not span h∗. Then, there is a 0 6= h ∈ h such that α(h) = 0 for all α ∈ Φ. However, this gives, for all α ∈ Φ,

[h, gα] = 0 =⇒ [h, g] = 0 ⇐⇒ h ∈ Z(g) = 0 which is a contradiction.

(b) is already proven in the lemma above.

For (c), take α ∈ Φ, x ∈ gα, y ∈ g−α. Then,

κ(h, [x, y]) = κ([h, x], y) = α(h)κ(x, y) = κ(tα, h)κ(x, y) = κ(κ(x, y)tα, h) = κ(h, κ(x, y)tα) and so, subtracting the two sides, we have [x, y] − κ(x, y)tα is perpendic- ular to all h ∈ h. However, since κ is non-degenerate, this means that [x, y] − κ(x, y)tα = 0.

For (d), we see that tα spans [gα, g−α] assuming [gα, g−α] 6= 0 by part (c). Thus, we need only show that [gα, g−α] 6= 0. Following the same technique as the proof of the lemma above, assume κ(x, g−α) = 0. Then, κ(x, g) = 0 is a contradiction since g is semisimple. Thus, there is a y ∈ g−α such that κ(x, y) 6= 0 =⇒ [x, y] 6= 0.

18 For (e), assuming α(tα) = 0 =⇒ [tα, x] = 0 = [tα, y] for all x ∈ gα, y ∈ g−α. However, we can ﬁnd an x, y such that κ(x, y) 6= 0 and, multiplying by scalars, we may assume κ(x, y) = 1 =⇒ [x, y] = tα by (c). Thus, using the same line of reasoning in the lemma, if we let S = span{x, y, tα}, then adg tα = 0 =⇒ tα ∈ Z(S) = 0, which is a contradiction.

For (f), let 0 6= xα ∈ gα and ﬁnd a yα ∈ g−α such that 2 κ(xα, yα) = κ(tα, tα) using (e) and the fact κ(xα, g−α) 6= 0. Then, by (c),

2tα hα := =⇒[xα, yα] = hα κ(tα, tα) 2 2α(tα) [hα, xα] = [tα, xα] = xα = 2xα α(tα) α(tα) 2 2α(tα) [hα, yα] = − [tα, yα] = − yα = −2yα α(tα) α(tα)

Thus, {xα, yα, hα} span a three dimensional subalgebra of g with the sl2(F ) relations.

For (g), we note that tα = −t−α since κ(−t−α, h) = α(h) = κ(tα, h) for all h ∈ h. Thus, since h = 2tα , α κ(tα,tα)

2t−α 2tα −h−α = − = = hα κ(t−α, t−α) κ(tα, tα)

As noted above, the correspondence of sl2(F ) inside any semisimple Lie algebra to a root is incredibly important, so much so that we will deﬁne a new notation to refer to such a copy of sl2(F ). ∼ 5.21. Deﬁnition. For a pair of roots α, −α, let sl(α) = sl2(F ) be a subalgebra of the form presented in the lemma or part (f) of the proposition above, that is, a subalgebra generated by {xα, yα, [xα, yα]} for xα ∈ gα and yα ∈ g−α.

With this terminology, we have the following proposition 5.22. Proposition. Given a semisimple Lie algebra g with root α, g may be considered as an sl(α)-module via the adjoint action, that is, for a ∈ sl(α), g ∈ g

a.g = ada g = [a, g]

Proof. This module is simply a restriction of the adjoint representation.

19 5.23. Definition. Let g be a semisimple Lie algebra with roots Φ. Let β ∈ Φ or β = 0. Then, subspace M gβ+kα k∈F β+kα∈Φ is an α-root string through β. Note that such a root string is an sl(α)-module. Now, using this notion, we prove the following. 5.24. Lemma. Let Φ ⊆ h∗ be a set of roots for a semisimple Lie algebra g. Then, for α ∈ Φ, the only multiples of α which lie in Φ are ±α. Furthermore, gα is one-dimensional for every root α ∈ Φ. Proof. Let M M := h ⊕ gkα kα∈Φ be the α-root string through 0. If cα is a root, then hα.xcα = 2cα(hα)xcα = 2cxcα, so hα has 2c as an eigenvalue. However, the eigenvalues of hα must be integral, so 2c ∈ Z. Now, by direct computation, sl(α) acts trivially on ker α ⊆ h. Next, given h + sl(α), we know that, as sl(α)-modules, h + sl(α) =∼ ker α ⊕ sl(α) since dim ker α = dim h − dim im α = dim h − 1. Thus, by Weyl’s theorem and the above computation, M ∼ h + sl(α) ⊆ h ⊕ gkα = ker α ⊕ sl(α) ⊕ N kα∈Φ where N is some complementary submodule. Now, let V be an irreducible representation of sl2(F ) with dimension k. Then, if k is even, there is a v ∈ V with hα.v = 0 =⇒ α(v) = 0 =⇒ v ∈ ker α by virtue of the classification of sl2(F ) representations. However, ker α ∩ V = 0 =⇒ v = 0, which is a contradiction. Thus, the only even weights occuring in M are 0 and ±2, 1 which tells us that 2α cannot be a root. However, this also means that 2 α is not a root, so 1 also cannot be a weight of M. Thus, since the number of irreducible sl2(F )-module summands is precisely dim M0 + dim M1, it must be that N contains no irreducible sl2(F )-submodules, and is thus trivial. Therefore, M =∼ ker α ⊕ sl(α) ∼ but ker α = g0 can only be a direct sum of trivial sl(α)-modules by the classification of sl2(F ) representations. Thus, gα is one-dimensional since sl(α) = gα + g−α + [gα, g−α]. 5.25. Proposition. [Hum72, Prop 8.4] Let α, β ∈ Φ with β 6= ±α. Then,

(a) β(hα) ∈ Z (b) There exist integers r, q ≥ 0 such that β +kα ∈ Φ if and only if k ∈ Z and −r ≤ k ≤ q. Furthermore, β(hα) = r − q.

20 (c) β − β(hα)α ∈ Φ (d) If α + β ∈ Φ, then [gα, gβ] = gα+β. (e) g is generated (as a Lie algebra) by the root spaces gα.

Proof. We will now look at the action of sl(α) on M M := gβ+kα. k∈Z β+kα∈Φ

Since hα.w = β(hα)w for w ∈ gβ, it must be that β(hα) ∈ Z. From the lemma above, there can be no k such that β + kα = 0 since, otherwise, β would be a multiple of α. Thus, the action of hα on each one-dimensional root space is given by

hα.v = (β(hα) + 2k)v for v ∈ gβ+kα. Because the weights diﬀer by even integers, it cannot be that both 0 and 1 occur as a weight of this form, and either can only appear once. Thus, M must be an irreducible sl(α)-module and the weights must be symmetric around 0. Let the heighest weight be given by β + qα and the lowest weight by β − rα. Thus,

−(β(hα) + 2q) = β(hα) − 2r =⇒ β(hα) = r − q which also tells us that

β − β(hα)α = β − (r − q)α ∈ Φ since −r ≤ q − r ≤ q.

Now, for (d), let v ∈ gβ and adxα xβ = 0. Then, xβ is a highest weight vector of M with weight β(hα). However, we also know that if α + β is a root, then hα acts on gα+β by eigenvalue (α + β)hα = β(hα) + 2, which tells us that β(hα) is not the highest weight, a contradiction. Thus, it must be that adxα xβ 6= 0 =⇒ [gα, gβ] = gα+β. Finally, by virtue of (d), g is generated as a Lie algebra by the root spaces since, given generator tα ∈ gα and tβ ∈ gβ, if α+β ∈ Φ, [tα, tβ] is a non-zero scalar multiple of tα+β.

5.26. Deﬁnition. The integers β(hα) are called the Cartan integers. At this point, we now understand a lot of the structure constants of g using the basis given by the root space decomposition. Namely, the action of h is determined by the roots, [xα, xβ] is determined by the roots for β 6= ±α, and [xα, x−α] is a scalar multiple of hα. Finally, we wish to prove Φ has a well-understood strucutre. 5.27. Lemma. For α, β ∈ Φ, we have X κ(tα, tβ) = γ(tα)γ(tβ) γ∈Φ

21 Proof. By deﬁnition,

κ(tα, tβ) = tr(adtα adtβ )

However, as a map, adtα adtβ : g → g has adtα adtβ (h) = 0 for all h ∈ h and, for x ∈ gγ, we have

x 7→ [tα, [tβ, x]] = [tα, γ(tβ)x] = γ(tα)γ(tβ)x So, the trace of the map will be the sum in the proposition.

5.28. Lemma. If α, β ∈ Φ, then κ(hα, hβ) ∈ Z and κ(tα, tβ) ∈ Q. Proof. We first realize that X κ(hα, hβ) = tr(adhα adhβ ) = γ(hα)γ(hβ) γ∈Φ since the trace is the sum of the eigenvalues with multiplicity as in 5.27. However, all these eigenvalues are integers, so κ(hα, hβ) ∈ Z. Next, we compute κ(t , t ) κ(t , t ) κ(t , t )κ(t , t ) κ(t , t ) = κ α α h , β β h = α α β β κ(h , h ) α β 2 α 2 β 4 α β However, 1 1 2tα 2tα 1 = κ , = κ(hα, hα) ∈ Q κ(tα, tα) 4 κ(tα, tα) κ(tα, tα) 4 and thus κ(tα, tβ) ∈ Q since it is a product of rational numbers. 5.29. Definition. Let us define form (·, ·): h∗ × h∗ → F via

(γ, δ) := κ(tγ, tδ) for γ, δ ∈ h∗ 5.30. Proposition. Such a form is non-degenerate and bilinear.

Proof. Since the Killing form is non-degenerate on g, it is non-degenerate on h and thus (·, ·) is non-degenerate on h∗. Furthermore, (·, ·) inherits its bilinearity from the Killing form. ∗ 5.31. Lemma. If β ∈ Φ and {α1, . . . , αn} is a basis of roots for h , then β is a rational linear combination of the αi’s. P Proof. Consider β = ciαi. Then, we compute X (β, αj) = (αi, αj)ci i which gives us n equations with n unknowns (the ci’s) with rational coeﬃ- cients.

(β, α1) = (α1, α1)c1 + (α2, α1)c2 + ··· + (αn, α1)cn

(β, α2) = (α1, α2)c1 + (α2, α2)c2 + ··· + (αn, α2)cn

22 . .

(β, αn) = (α1, αn)c1 + (α2, αn)c2 + ··· + (αn, αn)cn Since (·, ·) is non-degenerate, the associated matrix is non-degenerate and thus invertible. Furthermore, since all the coefficients are rational, its solu- tions will be rational. Thus, ci ∈ Q. 5.32. Definition. Let g be a semisimple Lie algebra with roots Φ. Then, ∗ we define EQ ⊆ h to be the Q-subspace spanned by all the roots Φ. By virtue of the proposition above, we know that all roots are Q-linear combinations of other roots, so EQ is independent of choice of roots for a basis. We now wish to show (·, ·) has more structure on EQ.

5.33. Proposition. The restriction of (·, ·) to EQ is rational and positive- deﬁnite

Proof. From lemma 5.28, we know that κ(tα, tβ) ∈ Q for any roots α, β ∈ Φ, but Φ Q-spans EQ by the proposition above, and so we can pick a basis of roots for EQ, say {α1, . . . , αn}. Thus, for γ, δ ∈ EQ, X X X X (γ, δ) = ciαi, djαj = cidj · (αi, αj) = cidjκ(tαi , tαj ) i,j i,j but ci, dj ∈ Q by deﬁnition of EQ. Finally, given λ ∈ EQ, 5.27 X 2 X 2 (λ, λ) = κ(tλ, tλ) = γ(tλ) = (γ, λ) ≥ 0 γ∈Φ γ∈Φ

If (λ, λ) = 0, then γ(tλ) = 0 for all roots γ, and so λ = 0 by 5.18 (e). Thus, (·, ·) is positive-deﬁnite. 2(β,α) 5.34. Lemma. β(hα) = (α,α)

Proof. Using the fact that h = 2tα , we see α κ(tα,tα) 2(β, α) 2κ(tβ, tα) 2tα = = κ tβ, = κ(tβ, hα) = β(hα) (α, α) κ(tα, tα) κ(tα, tα)

5.35. Deﬁnition. Let E := R ⊗Q EQ. 5.36. Remark. Using standard linear algebra, the form (·, ·)| extends EQ canonically to E and is still positive-deﬁnite. Thus, E is a Euclidean space. Thus, we see that we have already shown Φ has a incredibly useful structure: 5.37. Proposition. Given a semisimple Lie algebra g, Φ is a root system in E, that is (a) Φ spans E and 0 6∈ Φ.

23 (b) If α ∈ Φ, then the only multiples of α in Φ are ±α. 2(β,α) (c) If α, β ∈ Φ, then β − (α,α) α ∈ Φ. 2(β,α) (d) If α, β ∈ Φ, then (α,α) ∈ Z. Proof. (a) is 5.18 (a) and the deﬁnition of Φ. (b) is 5.24. (c) is a rephrasing of 5.25 (c) in light of lemma 5.34. (d) is a rephrasing of 5.25 (a) in light of lemma 5.34. Thus, the structure of Φ is given by the geometry of root systems, which are incredibly symmetric structures that are well-understood. There are many introductions to root systems, including [Hum72, Ch III], [Car05, Ch 5], and [See17]. From this point onwards, a working knowledge of root systems is assumed. We see that the structure of irreducible root systems corresponds to that of simple Lie algebras: 5.38. Proposition. Let g be a simple Lie algebra, with maximal toral subalgebra h and Φ ⊆ h∗. Then, Φ is an irreducible root system. Proof. Assume the root system is reducible into orthogonal root systems: Φ = Φ1 ∪ Φ2. Then, take α ∈ Φ1, β ∈ Φ2. We see that (α + β, α) 6= 0 6= (α + β, β) =⇒ α + β 6∈ Φ since α + β does not lie strictly in Φ1 or Φ2. Therefore, it must be that

[gα, gβ] = 0 by the root space decomposition. Observe

(a) For all β ∈ Φ2, gβ centralizes the subalgebra generated by gα, say K, which must be proper since Z(g) = 0 by the simplicity of g. (b) For all α ∈ Φ1, gα normalizes K since K is generated by gα

Therefore, K is normalized by all gγ for γ ∈ Φ and thus by all of g because g is generated by its root spaces (5.25). Thus, K is a proper ideal of g, violating the simplicity of g. 5.39. Corollary. [Hum72, p 74] Let g be a semisimple Lie algebra with maximal toral subalgebra h and root system Φ. If g = g1 ⊕ · · · ⊕ gt is the decomposition of g into simple ideals, then hi = h ∩ gi is a maximal toral subalgebra of gi and the corresponding (irreducible) root system Φi may be regarded canonically as a subsystem of Φ in such a way that Φ = Φ1 ∪· · ·∪Φt is the decomposition of Φ into its irreducible components. Proof. Since g is semisimple, we know that it has a decomposition into simple ideals g = g1 ⊕ · · · ⊕ gt and if h is a maximal toral subalgebra of g, then h = g1 ∩ h ⊕ · · · ⊕ gt ∩ h since, x = x1 + ··· + xt ∈ h is semisimple and each xi ∈ gi is semisimple by 2.18, so xi ∈ gi ∩ h.

24 Now, let hi := gi ∩ h. Then, hi is a maximal toral subalgebra in gi since 0 any toral subalgebra of gi larger than hi, say hi would be toral in g and cen- 0 tralize all hj, j 6= i, and so h1 ⊕· · ·⊕hi ⊕· · ·⊕ht would be a toral subalgebra of g larger than h.

Let Φi be the root system of gi relative to hi. If α ∈ Φi, we can view α as a linear function on h by deﬁning α(hj) = 0 for j 6= i. So, α is thus a root of g relative to h with gα ⊆ gi.

Conversely, let α ∈ Φ. Then, [hi, gα] 6= 0 for some i and so gα ⊆ gi, otherwise h would centralize gα. From this, we get that α|hi is a root of gi relative to hi. Note that α cannot be a root of another system since gα ⊆ gi and [gi, gj] = 0 for all j 6= i.

Thus, we can decompose Φ = Φ1∪· · ·∪Φt into irreducible components. From the above, by understanding all the irreducible root systems, a purely geometric task, we gain understanding of all the possible root systems a simple Lie algebra might have. Furthermore, one can show, with more work, that any irreducible root system corresponds to a simple Lie algebra. This is a consequence of results such as Serre’s Theorem [Hum72, §18.3] or by the existence of the Chevalley basis [Hum72, §25].

6. A Return to Representations of sl3(F )

As seen above, we have successfully generalized our results about sl3(F ) and a little bit more. In particular, we see that the subspace of all diagonal matrices generalizes to a maximal toral subalgebra h and the adh-action always yields integral eigenvalues on the root spaces.

6.1. Example. The roots Φ = {i − j} of sl3(F ) form a root system as described in the proposition concluding the previous chapter. To see this explicitly, one checks by straightforward computation that parts (a) and (b) of 5.37 are true. Then, we note that {1 − 2, 2 − 3} forms a basis for Φ. We then check ( [Ei,j,Ej,i] = Ei,i − Ej,j 1 =⇒ ti−j = (Ei,i − Ej,j) κ(Ei,j,Ej,i) = 6 6 and so 1 ( − , − ) = κ(E − E ,E − E ) 1 2 1 2 36 1,1 2,2 1,1 2,2 1 = ( − )(E − E ) 36 1 2 1,1 2,2 1 = (1 + 1) 36

25 1 = 18 1 ( − , − ) = κ(E − E ,E − E ) 1 2 2 3 36 1,1 2,2 2,2 3,3 1 = ( − )(E − E ) 36 1 2 2,2 3,3 1 = (0 − 1) 36 1 = − 36 1 ( − , − ) = ( − )(E − E ) 2 3 2 3 36 2 3 2,2 3,3 1 = (1 + 1) 36 1 = 18 and so

2(1 − 2, 2 − 3) (1 − 2) − (2 − 3) = 1 − 2 − (−1)(2 − 3) = 1 − 3 (2 − 3, 2 − 3) which shows (c) and demonstates (d). Indeed, for (d), one can see that 2(β,α) (α,α) = −1 or 2 for α, β in the basis.

(Note, for the general sln(F ) case, one can prove that κ(Ei,j,Ej,i) = 2n 2(β,α) and thus (α,α) = 0, ±1 for all β 6= ±α.)

In fact, the root system above can be visualized as the configuration A2 and, with that insight, we can visualize the weight spaces of any sl3(F ) representation as a lattice in a hexagonal configuration. Thus, we can Add images visualize a root string as a line through the lattice. For instance, for α ∈ Φ, M W := gα+k(2−1) k∈Z α+k(2−1) is symmetric about the line {L | hE1,1 − E2,2,Li = 0} and is thus preserved by reflection across said line. 6.2. Proposition. [FH91, Prop 12.15] All eigenvalues of any irreducible finite-dimensional representation of sl3(F ) must lie in the weight lattice ∗ ΛW ⊆ h , that is, the lattice generated by the i and must be congruent ∗ modulo the lattice ΛR ⊆ h generated by the i − j. 6.3. Proposition. [FH91, Prop 12.18] Let V be any irreducible, finite- ∗ dimensional representation of sl3(F ). Then, for some α ∈ ΛW ⊆ h , the set of eigenvalues occurring in V is exactly the set of linear functionals

26 2 − 3 1 − 3

2 1 2 − 1 1 − 2

3 3 − 2 3 − 1

Figure 1. Weight lattice for sl3. Choice of simple roots α1, α2 are given by black arrows.

congruent to α modulo the lattice ΛR and lying in the hexagon with ver- tices the images of α under the group generated by reﬂections in the lines {L | hEi,i − Ej,j,Li = 0}. Fill in proofs for From this, we see that, given our choice of `, it must be that any heigh- these proposi- est weight vector will lie in the region cut out by the inequalities hE1,1 − tions. E2,2,Li ≥ 0 and hE2,2 − E3,3,Li ≥ 0. In other words, it must be of the form

(a + b)1 + b2 = a1 − b3 Thus, we have reason to believe. 6.4. Theorem. For any pair of natural numbers a, b, there exists a unique irreducible, ﬁnite-dimensional representation Γa,bof sl3(F ) with highest weight a1 − b3.

To prove this theorem, we will construct such representations explicitly. Thus, let us look at some examples.

27 ∼ 3 6.5. Example. Consider the standard representation of sl3(C) on V = C with basis {e1, e2, e3} with associated eigenvalues 1, 2, and 3 since, for   a1 h =  a2  ei = aiei = i(h)ei a3 ∗ ∗ ∗ ∗ Consider also its dual, V , with dual basis {e1, e2, e3} and associated eigen-

2 1

Figure 2. Weight diagram for V , the standard representation of sl3(C) values −1, −2, and −3.

−3 2 1

−1 −2 3

Figure 3. Weight diagram for V ∗, the dual representation of the standard representation of sl3(C)

Using weight diagrams , one can see that Sym2 V and Sym2 V ∗, whose Actually insert weights are the pairwise sums of the weights of V and V ∗, respectively, are these both irreducible. A more interesting analysis comes from examining V ⊗V ∗.

28 2 1

Figure 4. Weight diagram for Sym2 V ∗

∗ One can quickly check that the weights of V ⊗V are {1 ±2, 1 ±3, 2 ± 3}, each with multiplicity 1, and 0 with multiplicity 3. Since the linear map v ⊗ u∗ 7→ hv, u∗i = u∗(v) has a non-trivial kernel. One can easily check that this map is equivalent ∗ ∼ ∼ to the trace since V ⊗ V = Hom(V,V ) = M3(C) and thus the kernel is all traceless matrices, ie the adjoint representation of sl3(C), which is an irreducible sl3(C)-representation. Thus, ∗ ∼ V ⊗ V = sl3(C) ⊕ C We will also need the following lemma 6.6. Lemma. Given representations V and W with highest weight vectors v, w with weights α, β, respectively, then v ⊗ w ∈ V ⊗ W is a highest weight vector of weight α + β.

Proof. Consider that, for Ei,j with i ≤ j, we get

Ei,j.(v ⊗ w) = Ei,j.v ⊗ w + v ⊗ Ei,j.w = 0 + 0 = 0 and, for H ∈ h, we get H.(v ⊗ w) = H.v ⊗ w + v ⊗ H.w = α(H)v ⊗ w + v ⊗ β(H)w = (α(H) + β(H))(v ⊗ w) Thus, from the example and the lemma, we get 6.7. Proposition. Given standard representation V , we have the following

(a) V has a highest weight vector with weight 1. ∗ (b) V has a highest weight vector with weight −3.

29 n (c) Sym V has a highest weight vector with weight n · 1. n ∗ (d) Sym V has a highest weight vector with weight −n · 3.

Proof of Theorem 6.4. Let V be the standard representation of sl3(F ). Then, Syma V ⊗ Symb V ∗ contains an irreducible subrepresentation generated by the highest weight vector with weight a1 − b3. To show uniqueness, let V,W be two sl3(F )-representations with highest weight α and corresponding highest weight vectors v, w. Then. consider (v, w) ∈ V ⊕ W will be a highest weight vector with highest weight α. Let U be the irreducible subrepresentation generated by (v, w). Then, we have non-zero projections between irreducibles,

π1 : U → V, π2 : U → W ∼ ∼ which, by Schur’s lemma, must be isomorphisms. Thus, V = U = W .

In fact, we can more explicitly construct the representations Γa,b, most easily using the Weyl Character Formula, which has not been proven yet. However, the final result will be given by first considering the contraction map a b ∗ a−1 b−1 ∗ ιa,b : Sym V ⊗ Sym V → Sym V ⊗ Sym V ∗ ∗ X ∗ ∗ ˆ∗ ∗ (v1 ··· va) ⊗ (v1 ··· vb ) 7→ hvi, vj i(v1 ··· vî ··· va) ⊗ (v1 ··· vj ··· vb )

Such a map is surjective and the codomain cannot have eigenvalue a1 −b3, so Γa,b ⊆ ker ιa,b. In fact, using the Weyl Character Formula, we immediately conclude

6.8. Proposition. ker ιa,b = Γa,b. Also, as an immediate corollary, we conclude a b ∗ ∼ Lb 6.9. Corollary. If b ≤ a, Sym V ⊗ Sym V = i=0 Γa−i,b−i and similarly if a ≤ b.

7. More Structure Theory of Semisimple Lie Algebras We state some results here without proof since they can be superseced by Serre’s theorem. However, we wish to establish that maximal toral subalgebras and Cartan subalgebras coincide for semisimple Lie algebras. Along the way, we establish a few other distinguished subalgebras of semisimple Lie algebras which appear throughout the literature on Lie algebras. 7.1. Proposition. [Hum72, p 74] Let g be a semisimple Lie algebra, h ⊆ g a maximal toral subalgebra, and Φ the root system of g relative to h. Fix a base ∆ of Φ. Then, g is generated as a Lie algebra by the root spaces gα, g−α for α ∈ ∆. Equivalently, g is generated by arbitrary nonzero root vectors xα ∈ gα, yα ∈ g−α for α ∈ ∆.

30 7.2. Theorem. [Hum72, p 75] Let g, g0 be simple Lie algebra over a ﬁeld F with respective maximal toral subalgebras h, h0 and corresponding root systems Φ, Φ0. Suppose there is an isomorphism of Φ onto Φ0 given by α 7→ α0 inducing π : h → h0. Fix a base ∆ ⊆ Φ so that Φ0 = {α0 | α ∈ ∆} is a base of 0 0 0 0 0 Φ . For each α ∈ ∆, α ∈ ∆ , choose arbitrary (nonzero) xα ∈ gα, xα ∈ gα, 0 that is, choose an arbitrary Lie algebra isomoprhism πα : gα → gα. Then, there eixsts a unique isomorphism π : g → g0 extending π : h → h0 and extending all of the πα, α ∈ ∆.

h g gα

π π πα

0 0 0 h g gα 7.3. Proposition. [Hum72, p 77] Given g as above, but not necessarily simple, ﬁx 0 6= xα ∈ gα for α ∈ ∆ and let yα ∈ g−α satisfy [xα, yα] = hα. Then, there exists an automorphism σ of g of order 2, satisfying, for α ∈ ∆, h ∈ h, σ(xα) = −yα, σ(yα) = −xα, σ(h) = −h.

Proof. This follows from the fact that any automorphism of Φ can be ex- tended to Aut Φ, so in particular, the map sending each root to its negative is in Aut Φ. 7.4. Deﬁnition. Let t ∈ End V , a be a root of the characteristic polynomial m of t with multiplicity m, and Va := ker(t−a·1) . Then, given an x ∈ g, a Lie algebra over algebraically closed ﬁeld F , we call g0(ad x) for ad x ∈ End g an Engel subalgebra.

7.5. Lemma. [Hum72, p 78] If a, b ∈ F , then [ga(ad x), gb(ad x)] ⊆ ga+b(ad x). In particular, g0(ad x) is a subalgebra of g, and when char F = 0, a 6= 0, each element of ga(ad x) is ad-nilpotent. Thus, an Engel subalgebra is actually an algebra.

Proof. Compute the expansion m X m (ad x − a − b)m[y, z] = [(ad x − a)i(y), (ad x − b)m−i(z)] i i=0 for y ∈ ga(ad x), z ∈ gb(ad x). For suﬃciently large m the expression will be 0. 7.6. Lemma. [Hum72, p 79] Let g0 be a subalgebra of g. Choose z ∈ g0 such 0 that g0(ad z) is minimal in the collection of all g0(ad x) for x ∈ g . Suppose 0 0 that g ⊆ g0(ad z). Then, g0(ad z) ⊆ g0(ad x) for all x ∈ g . 7.7. Lemma. [Hum72, p 79] If g0 is a subalgebra of g containing an En- 0 0 gel subalgebra, then Ng(g ) = g . In particular, Engel subalgebras are self- normalizing.

31 7.8. Definition. A Cartan subalgebra (CSA) of a Lie algebra g is a nilpotent subalgebra which equals its normalizer in g. 7.9. Theorem. [Hum72, p 80] Let h be a subalgebra of Lie algebra g. Then, h is a CSA of g if and only if h is a minimal Engel subalgebra. 7.10. Corollary. [Hum72, p 80] Let g be semisimple over F with char F = 0. Then, the CSA’s of g are precisely the maximal toral subalgebras of g. 7.11. Proposition. [Hum72, p 81] Let φ: h → h0 be a surjective Lie algebra homomorphism. (a) If h is a CSA of g, then φ(h) is a CSA of g0. (b) If h0 is a CSA of g0 and K = φ−1(h0), then any CSA h of K is also a CSA of g. 7.12. Definition. A Borel subalgebra of a Lie algebra g is a maximal solvable subalgebra of g. 7.13. Definition. Let g be a finite dimensional semisimple Lie algebra over C and let h be a CSA of g. Then, g has triangular decomposition g = n− ⊕ h ⊕ n − L L where n = α∈Φ− gα and n = α∈Φ+ gα. 7.14. Proposition. The n− and n in the triangular decomposition are subalgebras of g.

Proof. The sum of two positive roots is still positive. So, for α, β ∈ Φ+, we have [gα, gβ] ⊆ gα+β ⊆ n − and so n is a subalgebra. Similarly for n . 7.15. Proposition. (a) b = h ⊕ n is a Borel subalgebra, called standard relative to h. (b) n is an ideal of b. (c) b/n =∼ h.

Proof. For (a), we ﬁrst observe that [b, b] = [h + n, h + n] ⊆ h + n = b because h, n are subalgebras and [h, n] ⊆ n. Thus, b is a subalgebra.

Let g0 be any subalgebra of g properly containing b. Since it is a subalgebra containing h, g0 must be stable under the ad h action, and so it must 0 include some gα with α ≺ 0. However, then g contains a copy of sl2 gen- 0 erated by {xα, yα, [xα, yα]} for xα ∈ g−α, yα ∈ gα. Since sl2 is simple, g cannot be solvable.

32 For (b), we observe [n, b] = [n, n + h] ⊆ n For (c), we compute

b/n = (h + n)/n =∼ h/(h ∩ n) =∼ h because h ∩ n = 0. 7.16. Proposition. [Hum72, pp 83–84]

(a) If b is a Borel subalgebra of g, then b = Ng(b). (b) If rad g 6= g, then the Borel subalgebras of g are in natural one-to-one correspondence with those of the semisimple Lie algebra g/ rad g. (c) Let g be semisimple with CSA h and root system Φ. All standard Borel subalgebras of g relative to h are conugate under inner automorphisms of Aut g.

7.17. Theorem. [Hum72, p 84] The Borel subalgebras of an arbitrary Lie algebra g are all conjugate under inner automorphisms of Aut g.

7.18. Corollary. [Hum72, p 84] The Cartan subalgebras of an arbitrary Lie algebra g are conjugate under inner automorphisms of Aut g.

8. Universal Enveloping Algebra In this section, we will construct an associative algebra U(g) such that the representation theory of U(g) is the same as the representation theory of g.

8.1. Definition. Let V be a finite dimensional vector space over a field F . Then, we define

T 0V = F T 1V = V T 2V = V ⊗ V . . T nV = V ⊗ · · · ⊗ V | {z } n times and say T (V ) = F T kV equipped with the product

T mV × T nV → T m+nV

(v1 ⊗ · · · ⊗ vm) · (w1 ⊗ · · · ⊗ wn) 7→ v1 ⊗ · · · ⊗ vm ⊗ w1 ⊗ · · · ⊗ wn is called the tensor algebra on V .

33 8.2. Proposition. Let V be a finite dimensional vector space and A an associative unital algebra over F . Then, we have the following universal property: V ι T (V ) φ ∃!Φ A In words, for any F -linear map φ: V → A and ι: V → T (V ) sending v 7→ v ∈ T 1V , there is a unique homomorphism of F -algebra Φ: T (V ) → A such that Φ ◦ ι = φ. 8.3. Definition. Let g be a Lie algebra. Then, consider the ideal J = hx ⊗ y − y ⊗ x − [x, y] | x, y ∈ gi in T (g). We define U(g) = T (g)/J and call such a quotient the universal enveloping algebra of g.

8.4. Example. Let g be an n-dimensional abelian Lie algebra over C with basis {x1, . . . , xn} and [xi, xj] = 0. Then, J E T (g) is given by J = hx ⊗ y − y ⊗ x | x, y ∈ gi. Therefore, U(g) is commutative and generated by ∼ 1, x1, . . . , xn. Thus, U(g) = C[x1, . . . , xn]. 8.5. Proposition. Let g be an arbitrary Lie algebra. Then, if ι: g → U(g) is a linear map satisyﬁng ι([x, y]) = ι(x)ι(y) − ι(y)ι(x), x, y ∈ g and φ: g → A satisﬁes the same property for A an associative unital F - algebra, then we have the following universal property

g ι U(g) φ ∃!Φ A that is, there is a unique F -algebra homomorphism Φ: U(g) → A such that Φ ◦ ι = φ. 8.6. Remark. Note that we have not shown that ι is injective, but we will do so later with the PBW Theorem. 8.7. Proposition. There is a bijective correspondence between representations θ : g → gl(V ) and representations φ: U(g) → End V . Corresponding representations are related by the condition φ(ι(x)) = θ(x) for all x ∈ g where ι: g → U(g) is the standard inclusion.

34 Proof. Given θ a representation of g, we have

g ι U(g) θ ∃!φ gl(V ) = End V where φ: U(g) → End V is a representation of U(g).

Conversely, given φ a representation of U(g), we know that the linear map ι: g → U(g) has ι(x)ι(y) − ι(y)ι(x) = ι([x, y]) =⇒ [ι(x), ι(y)] = ι([x, y]) if we deﬁne [·, ·]: U(g) × U(g) → U(g) by [x, y] = xy − yx. Thus, ι: g → [U(g)], that is, U(g) with the extra Lie algebra structure, is a Lie algebra homomorphism.

ι g ι U(g) g [U(g)]

=⇒ φ φ φ◦ι End V gl(V )

Then, φ ◦ ι is a representation 8.8. Remark. Note, the representations under consideration need not be ﬁnite-dimensional. In fact, it will turn out that a certain important class of representations called “Verma modules” will always be inﬁnite-dimensional.

This construction will lead us to the all-important PBW Theorem, which tells us how to obtain a basis for the universal enveloping algebra of a Lie algebra. 8.9. Theorem (Poincaré-Birkhoff-WittTheorem). [Car05, Theorem 9.4] Let g be a Lie algebra with basis {xi | i ∈ I}. Let < be a total order on the index set I. Let ι: g → U(g) be the natural linear map from g to its enveloping algebra. We define yi := ι(xi). Then, the elements yr1 ··· yrn i1 in for all n ≥ 0, all ri ≥ 0, and all i1, . . . , in ∈ I with i1 < I2 < ··· < in form a basis for U(g).

For the proof of this theorem, see [Car05, pp 155–159] or [Hum72, pp 93– 94] (Note, [Hum72] gives an equivalent statement of the PBW Theorem and states this version as Corollary C). The PBW Theorem has many important corollaries such as 8.10. Corollary. The map ι: g → U(g) is injective.

35 Proof. By the PBW Theorem, the elements yi for i ∈ I are linearly independent. Since the xi ∈ g form a basis for g and ι(xi) = yi, it must be that ker ι = {0}. 8.11. Corollary. The subspace ι(g) ⊆ [U(g)] is isomorphic to g as a Lie subalgebra.

Proof. From the corollary above, ι: g → ι(g) is bijective and yi, i ∈ I form a basis of ι(g). Furthermore,

[yi, yj] = [ι(xi), ι(xj)] = ι([xi, xj]) ∈ ι(g)

Thus, ι(g) is a Lie subalgebra of [U(g)]. 8.12. Corollary. U(g) has no zero-divisors.

Proof. Take 0 6= a, b ∈ U(g). Then, X a = λ yr1 ··· yrn i1,...,in,r1,...,rn i1 in X X = λ yr1 ··· yrn + λ yr1 ··· yrn i1,...,in,r1,...,rn i1 in i1,...,in,r1,...,rn i1 in r1+···+rn=r r1+···+rn

= f(yi) + sum of terms of smaller degree X b = µ yr1 ··· yrn i1,...,in,r1,...,rn i1 in X X = µ yr1 ··· yrn + µ yr1 ··· yrn i1,...,in,r1,...,rn i1 in i1,...,in,r1,...,rn i1 in r1+···+rn=r r1+···+rn

= g(yi) + sum of terms of smaller degree Recall also, since ι(g) is a Lie subalgebra, X yiyj − yjyi = [yi, yj] = νkyk k and so why?

f(yi)g(yi) = (fg)(yi) + a sum of terms of smaller degree. Therefore,

ab = (fg)(yi) + a sum of terms of smaller degree. and since f 6= 0 6= g, we get fg 6= 0. Thus, the PBW Theorem tells us that ab 6= 0.

9. Irreducible Modules of Semisimple Lie Algebras 9.1. Lemma. [Car05, Lemma 10.1] Let g be a ﬁnite dimensional Lie algebra 0 over C, and let g be a subalgebra of g. Then, there exists a unique algebra

36 homomorphism θ : U(g0) → U(g) such that the following diagram commutes

ι 0 g0 g U(g0)

i θ ι g g U(g)

0 where i is the embedding of g into g and ιg0 , ιg are the standard embeddings of a Lie algebra into its universal enveloping algebra. Furthermore, θ is injective.

Proof. For x ∈ g0, the deﬁnition

θ(ιg0 (x)) = ιg(i(x)) is well-deﬁned and, since U(g0) is generated by ι(g0)(g0), such a θ is uniquely determined. To see existence, deﬁne one notices g0 T (g0) T (g0)/J 0 =∼ U(g0)

i i∗ g T (g) T (g)/J =∼ U(g) since i∗(J 0) ⊆ J, where J, J 0 are the ideals used to deﬁne the universal enveloping algebra.

0 0 To see θ is injective, let x1, . . . , xr be a basis for g and 0 6= u ∈ U(g ). e1 er Then, u is a non-zero linear combination of monomials x1 ··· xr . Then, let x1, . . . , xr be part of a basis of g. The PBW theorem tells us that such a linear combination of monomials cannot be zero in U(g). Thus, θ(u) 6= 0 and so ker θ = 0. 9.2. Definition. Let g be a finite-dimensional semisimple Lie algebra over ∗ C and let λ ∈ h . We define, for h = {h1, . . . , hk}, k X X Kλ := U(g)xα + U(g)(hi − λ(hi)) α∈Φ+ i=1 for basis {xα | α ∈ Φ} ∪ {hi | i = 1, . . . , k} of g. Since Kλ is a left ideal of U(g), we also define M(λ) := U(g)/Kλ + and call M(λ) the Verma module determined by λ. Since xα for α ∈ Φ and hi − λ(hi) for i = 1, . . . , k all lie in U(b) for b = h ⊕ n, we may also define k 0 X X Kλ := U(b)xα + U(b)(hi − λ(hi)) α∈Φ+ i=1 which is a left ideal of U(b).

37 9.3. Proposition. [Car05, Prop 10.4] Let g be a semisimple Lie algebra over + C with positive roots Φ = {β1, . . . , βN } and basis h1, . . . , hk, xβ1 , . . . , xβN . 0 (a) dim U(b)/Kλ = 1 (b) The elements

(h − λ(h ))s1 ··· (h − λ(h ))sk xt1 . . . xtn 1 1 k k β1 βN

with si ≥ 0, ti ≥ 0, excluding the element with si = ti = 0 for all i, 0 form a basis for Kλ.

Proof. We ﬁrst wish to show the elements listed above form a basis for U(b). To see this, one can deﬁne a partial ordering on the basis elements by saying

s0 s0 t0 t0 h 1 ··· h k x 1 ··· x N ≺ hs1 ··· hsk xt1 ··· xtN if s0 ≤ s , . . . , s0 ≤ s , t0 = t , . . . , t0 = t 1 k β1 βN 1 k β1 βN 1 1 k k 1 1 N N Thus,

(h −λ(h ))s1 ··· (h −λ(h ))sk xt1 . . . xtn = hs1 ··· hsk xt1 ··· xtN + strictly lower terms 1 1 k k β1 βN 1 k β1 βN

Thus, by induction on the exponents si, ti and from the triangularity property above, elements of the form

(h − λ(h ))s1 ··· (h − λ(h ))sk xt1 . . . xtn , s ≥ 0, t ≥ 0 1 1 k k β1 βN i i span U(b) and they are linearly independent.

By deﬁnition, it follows immedaitely that all of these elements except for 0 the element with si = 0, ti = 0 for all i are in Kλ, but it is not clear if this 0 exceptional element, which is equal to the unit element 1, is in Kλ or not. 0 We will show that 1 is not in Kλ.

∼ Consider the representation λ: h → C. Since b/n = h, one can lift to a 1-dimensional representation of b with kernel n agreeing with λ on b/n =∼ h. From this, we get a 1-dimensional representation ρ of U(b) which does + xα 7→ 0 α ∈ Φ

hi 7→ λ(hi) i = 1, . . . , k 1 7→ 1 This gives us  x ∈ ker ρ α ∈ Φ+ (  α K0 ⊆ ker ρ h − λ(h ) ∈ ker ρ i = 1, . . . , k =⇒ λ =⇒ 1 6∈ K0 i i 1 6∈ ker ρ λ 1 6∈ ker ρ

Therefore, both (a) and (b) follow. − 9.4. Proposition. [Car05, Prop 10.5] Kλ ∩ U(n ) = 0.

38 Proof. g = n− ⊕ b =⇒ U(g) = U(n−)U(b) by the PBW theorem. Then, k X X Kλ = U(g)xα + U(g)(hi − λ(hi)) α∈Φ+ i=1 k X − X − = U(n )U(b)xα + U(n )U(b)(hi − λ(hi)) α∈Φ+ i=1  k  − X X = U(n )  U(b)xα + U(b)(hi − λ(hi)) α∈Φ+ i=1 − 0 = U(n )Kλ

Therefore, every element of Kλ is a linear combination of terms of the form yr1 ··· yrN (h − λ(h ))s1 ··· (h − λ(h ))sk xt1 ··· xtN β1 βN 1 1 k k β1 βN | {z } | {z } − 0 ∈U(n ) ∈Kλ + where yα = x−α for α ∈ Φ and ri, si, ti ≥ 0 with (s1, . . . , sk, t1, . . . , tN ) 6= (0, 0,..., 0). However, the PBW Theorem also tells us that no non-zero element of U(n−) can be a a linear combination of such terms, since its basis must be of the form f r1 ··· f rN . Therefore, K ∩ U(n−) = 0. β1 βN λ ∼ 9.5. Theorem. [Car05, Theorem 10.6]Let mλ := 1+Kλ ∈ M(λ) = U(g)/Kλ. Then,

(a) Each element of M(λ) is uniquely expressible in the form umλ for − some u ∈ U(n ), that is, M(λ) is cyclically generated by mλ as a U(n−)-module. (b) Moreover, the elements yr1 ··· yrn m for all r ≥ 0 form a basis for β1 βN λ i M(λ). ∼ Proof. Since 1 7→ mλ under the quotient map U(g) → U(g)/Kλ = M(λ), every element of M(λ) has the form umλ for some u ∈ U(g).

Now, we have basis for g, + {xα, yα | α ∈ Φ } ∪ {hi | i = 1, . . . , k} thus, by the PBW theorem, giving us a basis for U(g) of elements of the form yr1 ··· yrN hs1 ··· hsk xt1 ··· xtN r , s , t ≥ 0 β1 βN 1 k β1 βN 1 i i Now, observe

xβi mλ = xβi (1 + Kλ) = xβi + Kλ = 0 + Kλ = 0 ∈ M(λ) since xβi ∈ Kλ

himλ = λ(hi)mλ

39 and so ( r1 rN s1 sk t1 tN 0 if any ti > 0 y ··· y h ··· h x ··· x mλ = β1 βN 1 k β1 βN yr1 ··· yrN γm for some γ ∈ if all t = 0 β1 βN λ C i

Therefore, since umλ is a linear combination of elements of the form of the product above, such a linear combination simpliﬁes to a linear combination of elements of the form yr1 ··· rrN m , r ≥ 0 β1 βN λ i and so elements of this form span M(λ).

To see linear independence, let X γ yr1 ··· yrN m = 0, γ ∈ r1,...,rN β1 βN λ r1,...,rN C r1,...,rN and so, by the proposition above, X γ yr1 ··· yrN ∈ K ∩ U(n−) = 0 =⇒ all γ = 0 r1,...,rN β1 βN λ r1,...,rN r1,...,rN by the PBW Theorem on U(n−). Thus, we have shown that we, in fact, have a basis of M(λ) and so each element is uniquely expressible in the form − umλ for u ∈ U(n ). 9.6. Deﬁnition. For each 1-dimensional representation µ: h → C, we deﬁne

M(λ)µ := {m ∈ M(λ) | xm = µ(x)m, ∀x ∈ h} 9.7. Proposition. [Car05, Theorem 10.7] When considered as an h-module, M(λ) has the following properties.

(a) M(λ)µ is a subspace of M(λ). L (b) M(λ) = µ∈h∗ M(λ)µ. (c) M(λ)µ 6= 0 if and only if λ − µ is a sum of positive roots. (d) dim M(λ)µ is equal to the number of ways of expressing λ − µ as a sum of positive roots.

Proof. Part (a) follows from the deﬁnition of M(λ)µ since it is closed under the action of h.

Since the elements yr1 ··· yrN m with r ≥ 0 form a basis for M(λ), part β1 βN λ i (b) follows by showing that xyr1 ··· yrN m = (λ − r β − · · · − r β )(x)yr1 ··· yrN m , ∀x ∈ h β1 βN λ 1 1 N N β1 βN λ by induction on r + ··· + r . Thus, yr1 ··· yrN m ∈ M(λ) for µ = λ − 1 N β1 βN λ µ r1β1 − · · · − rN βN and so X M(λ) = M(λ)µ µ

40 In fact, this sum is direct by direct linear algebra computation by showing

M(λ)µ ∩ (M(λ)µ1 + ··· + M(λ)µk ) = 0 ∗ for µ, µ1, . . . , µk ∈ h distinct. Finally, take Λ = {µ ∈ h∗ | λ − µ is a sum of positive roots}. Then, for each λ ∈ Λ, let N = span{yr1 ··· yrN m | λ − r β − · · · − r β = µ} ⊆ M(λ) ⊆ M(λ) µ β1 βN λ 1 1 N N µ By the PBW theorem, we also get M M(λ) = Nµ µ∈Λ L However, since M(λ) = µ∈h∗ M(λ)µ from above, we get that M(λ)µ = Nµ ∗ for all µ ∈ Λ and M(λ)µ = 0 for µ ∈ h when µ 6∈ Λ. Thus, N dim M(λ)µ = dim Nµ = |{(r1, . . . , rN ) ∈ N0 | λ − r1β1 − · · · − rN βN = µ}| that is, the number of ways to write λ − µ as a sum of positive roots. ∗ 9.8. Deﬁnition. µ ∈ h is a weight of M(λ) if M(λ)µ 6= 0 and M(λ)µ is called the weight space of M(λ) with weight µ. 9.9. Theorem. [Car05, Theorem 10.9] M(λ) has a unique maximal submodule, denoted J(λ).

Proof. Take V to be a proper U(g)-submodule of M(λ). The key idea is to P show that V ⊆ µ,µ6=λ M(λ)µ $ M(λ) and take J(λ) to be the sum of all the proper submodules of M(λ), thus yielding a unique maximal submodule of M(λ) lying in this codimension 1 subspace.

By the theorem above, n X v = vµi , vµi ∈ M(λ)µi i=1 where the µi are distinct. To show each vµi ∈ V , we observe, for x ∈ h and ﬁxed i,

xvµi = µi(x)vµi Y X Y =⇒ (x − µj(x))v = (x − µj(x))vµk j,j6=i k j,j6=i X Y = (µk(x) − µj(x))vµk k j,j6=i Y = (µi(x) − µj(x))vµi j,j6=i

41 Moreover, since any vector space cannot be a union of ﬁnitely many proper subspaces, we can ﬁnd an x ∈ h with µi(x) 6= µj(x) for all j 6= i. Thus, if we pick such an x, Y Y (x − µj(x))v ∈ V =⇒ (µi(x) − µj(x))vµi ∈ V j,j6=i j,j6=i Q and so, since j.j6=i(µi(x) − µj(x)) 6= 0, we get that vµi ∈ V .

If we deﬁne Vµ := V ∩M(λ)µ, we need only show that Vλ = V ∩M(λ)λ = 0. However, if this were not the case, Vλ = M(λ)λ since dim M(λ)λ = 1 so

mλ ∈ V =⇒ M(λ) = U(g)mλ ⊆ V =⇒ V = M(λ) which is a contradiction. Thus, every proper submodule V of M(λ) is con- P tained in µ,µ6=λ M(λ)µ. 9.10. Definition. [Car05, Definition 10.10] For λ, µ ∈ h∗, we say that λ µ if λ − µ is a sum of positive roots. This defines a partial order on h∗. By , we also know that the weights of M(λ) are precisely the µ ∈ h∗ with µ ≺ λ. Thus, λ is the highest weight of M(λ) with respect to this partial order and so M(λ) is called the Verma module with highest weight λ. Finally, we define L(λ) := M(λ)/J(λ). 9.11. Proposition. (a) L(λ) is an irreducible U(g)-module since J(λ) is a maximal submodule of M(λ). (b) λ is a weight of L(λ) since J(λ)λ = 0. (c) dim L(λ)λ = 1 since J(λ)λ = 0 and λ is the highest weight of L(λ). Now that we have L(λ), we still wish to figure out (a) For which λ is L(λ) finite dimensional? (b) Which weights µ occur in L(λ) and what are their multiplicities? In otherwords, what is dim L(λ)µ? 9.12. Theorem. Let V be a finite dimensional irreducible g-module with highest weight vector vλ. Then, there exists a surjective homomorphism θ : M(λ) → V of U(g)-modules such that θ(mλ) = vλ. 9.13. Corollary. A finite dimensional irreducible g-module V with highest weight λ is isomorphic to L(λ).

Proof of Corollary. Consider the surjective homomorphism θ : M(λ) V Since V is irreducible, ker θ must be maximal in M(λ), but M(λ) has unique maximal submodule J(λ). Therefore, ker θ = J(λ) and so L(λ) =∼ M(λ)/J(λ) = M(λ)/ ker θ =∼ V

42 Note, however, the converse statement is not true. The partial converse is 9.14. Proposition. [Car05, Prop 10.15] or [Hum72, Thm 21.1] If L(λ) is ﬁnite dimensional, then λ(hi) is a non-negative integer for {h1, . . . , hn} a basis of h.

Proof. Take simple root αi corresponding to hi and let sl(αi) = span{xαi , yαi , hi} be the corresponding copy of sl2(C) in g. Since L(λ) is thus also a ﬁnite- dimensional sl(αi)-module, there is a highest weight vector of weight λ, which is the weight for hi. Thus, by virtue of our work on sl2 (see 3.12), it must be that λ(hi) ∈ Z≥0. 9.15. Theorem. [Car05, Thm 10.20] Suppose λ ∈ h∗ is dominant and integral. Then, the irreducible g-module L(λ) is ﬁnite dimensional and its set of weights is permuted by W with

dim L(λ)µ = dim L(λ)wµ ∀w ∈ W

The proof of this theorem is quite long (see [Car05] or [Hum72, §21]), but the following lemma must be used in either proof. 9.16. Lemma. [Hum72, Lem 21.2] The following identities hold in U(g) for k ≥ 0, 1 ≤ i, j ≤ `. k+1 (a) [xj, yi ] = 0 when i 6= j k+1 k+1 (b) [hj, yi ] = −(k + 1)αi(hj)yi k+1 k (c) [xi, yi ] = −(k + 1)yi (k · 1 − hi)

Proof of Lemma. (a) αj − αi is not a root, so the result follows. why? (b) When k = 0, [hj, yi] = −αi(hj)yi Now, asume the result holds for all values less than k + 1. We get k+1 k+1 k+1 [hj, yi ] = hjyi − yi hj k k k = (hjyi − yi hj)yi + yi (hjyi − yihj) k k = [hj, yi ]yi + yi [hj, yi] k k = −kαi(hj)yi yi + yi (−αi(hj)yi) k+1 = −(k + 1)αi(hj)yi (c) Similar to the above, we check k+1 k+1 k+1 [xi, yi ] = xiyi − yi xi k k = [xi, yi]yi + yi[xi, yi ] k k−1 = hiyi + yi(−kyi ((k − 1) · 1 − hi)) k k k k = hiyi − yi hi + (k + 1)yi hi − kyi (k − 1) k k k = −kαi(hi)yi + (k + 1)yi hi − kyi (k − 1)

43 k k k = −2kyi + (k + 1)yi hi − kyi (k − 1) k k = (k + 1)yi hi − k(k + 1)yi k = −(k + 1)yi (k · 1 − hi)

10. Characters 10.1. Definition. Given g a semisimple Lie algebra and V be a g-module with weight space decomposition M V = Vλ λ∈h∗ where dim Vλ < ∞. Then, we define the character of V to be the function ∗ ch V : h → Z given by (ch V )(λ) = dim Vλ and we define the formal character of V to be X ch V = dim Vµeµ ∈ Z[ΛW ] µ where the sum is over all weights of V and eµ(λ) = δµλ. Explain the ring structure here; 10.2. Proposition. Let V,V1,V2 be g-modules that admit characters. Then, the multiplica- (a) If U ⊆ V , tion is not clear. ch U + ch(V/U) = ch V (b) Therefore,

ch(V1 ⊕ V2) = ch V1 + ch V2 (c)

ch(V1 ⊗ V2) = ch V1 · ch V2 10.3. Lemma. Given a Verma module M(λ) with λ ∈ h∗, we have that (ch M(λ))(µ) = Number of ways to write λ−µ as a positive sum of positive roots.

Proof. This follows immediately from the deﬁnition of ch M(λ) and 9.7(d). 10.4. Proposition. If P(ν) is the number of ways to write ν as a positive sum of positive roots, then   X ch M(λ) = eλ  P(ν)e−ν ν∈h∗

44 Proof. This follows since X ch M(λ) = P(λ − µ)eµ µ∈h∗ X = P(ν)eλ−ν ν∈h∗ X = P(ν)eλe−ν ν∈h∗ X = eλ P(ν)e−ν ν∈h∗ ∗ 10.5. Lemma. In the ring of all function f : h → Z with ﬁnite support,  −1 X Y  P(ν)e−ν = (1 − e−α) ν∈h∗ α∈Φ+

+ Proof. Let Φ = {β1, . . . , βn}. Then, X X P(ν)e−ν = e−r1β1−···−rnβn ν r1,...,rn≥0 X = er1 ··· ern −β1 −βn r1,...,rn≥0 n   Y X = eri  −βi  i=1 r1≥0 but then (1 − e )(1 + e + e2 + ··· ) = 1 −βi −βi −βi and so we have our desired inverse n Y Y (1 − e−βi ) = (1 − e−α) i=1 α∈Φ+ We state the following theorem without proof. 10.6. Theorem. The character of the Verma module with highest weight λ is given by eλ+ρ ch M(λ) = Q eρ α∈Φ+ (1 − e−α) Proof.   X eλ eλ+ρ ch M(λ) = eλ  P(ν)e−ν = Q = Q + (1 − e−α) eρ + (1 − e−α) ν∈h∗ α∈Φ α∈Φ

45 10.7. Theorem. [Car05, Theorem 12.16] The Verma module M(λ) has a ﬁnite composition series

M(λ) = N0 ⊇ N1 ⊇ · · · ⊇ Nr = 0 where each Ni is a submodule of M(λ) and Ni+1 is a maximal submodule of ∼ Ni. Furthermore, Ni/Ni+1 = L(w · λ) for some w ∈ W . 10.8. Theorem (Weyl Character Formula). [Car05, Theorem 12.17] Given semisimple Lie algebra g with Cartan subalgebra h. Then, for λ a dominant weight, P (−1)`(w)e ch L(λ) = w∈W w(λ+ρ) P `(w) w∈W (−1) ew(ρ)

Proof. Since λ is dominant and integral, λ(hi) ≥ 0 for all hi. Therefore,

(λ+ρ)(hi) = λ(hi)+1 > 0 =⇒ λ+ρ is in the fundamental Weyl chamber, C Thus, w(λ + ρ) ∈ w(C) and w(λ + ρ) for each w ∈ W . The highest weight for M(w·λ) is w·λ = w(λ+ρ)−ρ and so the characters ch M(w · λ) are linearly independent for w ∈ W and similarly for L(w · λ). Let M(w·λ) have composition factors L(y·λ) for some y ∈ W . Then, since w·λ is the highest weight, it must be that y·λ ≺ w·λ and dim(M(w·λ)w·λ) = 1 =⇒ dim(L(w · λ)w·λ) = 1. Thus, we have X ch M(w · λ) = ch L(w · λ) + aw,y ch L(y · λ) y∈W ,y·λ≺w·λ where aw,y ∈ N>0, so the characters are unitriangularily related, meaning we can create an inverse, giving X ch L(w · λ) = ch M(w · λ) + bw,y ch M(y · λ) y∈W ,y6=w with bw,y ∈ Z (not necessarily non-negative). If we specify w = 1, then we get X X ey(λ+ρ) ch L(λ) = b1,y ch M(y · λ) = b1,y Q eρ + (1 − e−α) y∈W y∈W α∈Φ Now, by 9.15,

dim L(λ)µ = dim L(λ)w(µ) =⇒ w(ch L(λ)) = ch L(λ) for all w ∈ W . Also,   ! Y X `(w) X `(w) si eρ (1 − e−α) = si (−1) ew(ρ) = − (−1) ew(ρ) α∈Φ+ w∈W w∈W

46 and so   Y `(w) Y w eρ (1 − e−α) = (−1) eρ (1 − e−α) α∈Φ+ α∈Φ+ Therefore,   X ey(λ+ρ) `(w) X w(ey(λ+ρ)) ch L(λ) = w(ch L(λ)) = w  b1,y Q  = (−1) b1,y Q eρ + (1 − e−α) eρ + (1 − e−α) y∈W α∈Φ y∈W α∈Φ tells us that     X `(w) X w  b1,yey(λ+ρ) = (−1)  b1,yey(λ+ρ) y∈W y∈W

`(w) and since weλ = ewλ, we get b1,w−1y = (−1) b1,y from the linear inde- `(w) pendence of the eλ’s. In particular, b1,w−1 = (−1) since b1,1 = 1 and so `(w−1) `(w) b1,w = (−1) = (−1) . Thus, since this is true for every w ∈ W , we get P (−1)`(w)e P (−1)`(w)e ch L(λ) = w∈W w(λ+ρ) = w∈W w(λ+ρ) Q P `(w) eρ α∈Φ+ (1 − e−α) w∈W (−1) ew(ρ) Note that, for λ = 0,   X `(w) Y ch L(0) = e0 =⇒ (−1) ew(ρ) = eρ (1 − e−α) w∈W α∈Φ+ so the Weyl Denominator Formula is a special case of the Weyl Character Formula (although we used the Weyl Denominator Formula in our proof of the Weyl Character Formula).

10.9. Example. Let g = sl2(C). Then, W = S2 and ρ = 1. So, for n ∈ Z≥0, 0 1 (−1) en+1 + (−1) e−(n+1) en+1 − e−n−1 ch L(n) = 0 1 = = en+en−2+···+e−n (−1) e1 + (−1) e−1 e1 − e−1 recovering our knowledge for ﬁnite dimensional irreducible sl2(C)-modules. 10.10. Theorem (Kostant’s Multiplicity Formula). Let λ be a dominant weight and µ a weight. Then, X `(w) dim L(λ)µ = (−1) P(w(λ + ρ) − (µ + ρ)) w∈W 10.11. Theorem (Weyl’s dimension formula). Let λ be a dominant weight. Then, Q α∈Φ+ hλ + ρ, αi dim L(λ) = Q α∈Φ+ hρ, αi

47 + 10.12. Example. Let g = sl2(C). Then, since Φ = {1} and ρ = 1, hn + 1, 1i dim L(n) = = n + 1 h1, 1i once again recovering knowledge from previous sections. Pn−1 10.13. Example. More generally, for g = sln(C), let λ = k=1 λkk. Then, n−1 X ρ = ω1 +ω2 +···+ωn−1 = 1 +(1 +2)+···+(1 +···+n−1) = (n−k)k k=1 and so Y hλ + ρ, αi dim L(λ) = hρ, αi α∈Φ+ Pn−1 Y h (λk + n − k)k, i − ji = k=1 Pk 1≤i

Y (λi + n − i) − (λj + n − j) = (n − i) − (n − j) 1≤i

X weight(T ) sλ(x1, . . . , xn) = x

T ∈SSYTn(λ)

where SSYTn(λ) is the set of all semistandard Young tableaux with letters in {1, . . . , n}, we get

dim L(λ) = sλ(1,..., 1) = | SSYTn(λ)| 10.14. Remark. In fact, in type A, one can see the Weyl Character Formula gives a “Schur function” using the Jacobi Bialternate Formula characteriza- ∼ tion of Schur functions. In type A, W = Sn and ΛW = spanZ{1, . . . , n}.

Then, since eλ = eλ11+···+λnn , we get   e(λ1+n−1)1 e(λ1+n−1)2 ··· e(λ1+n−1)n n e e ··· e X `(w) X Y  (λ2+n−2)1 (λ2+n−2)2 (λ2+n−2)n  (−1) ew(λ+ρ) = sgn(w) e(λ +n−i) = det  . . .  i w(i)  . . .  w∈Sn w∈Sn i=1   e(λn)1 e(λn)2 ··· e(λn)n

Then, the initiated observe that this is the Vandermonde determinant ∆λ+ρ λi+n−1 where e(λi+n−i)j corresponds to the variable xj . Thus, in our type A

48 setting, P (−1)`(w)e ∆ ch L(λ) = w∈Sn w(λ+ρ) = λ+ρ = s (e , e , . . . , e ) P (−1)`(w)e ∆ λ 1 2 n w∈Sn w(ρ) ρ References [Car05] R. Carter, Lie Algebras of Finite and Aﬃne Type, 2005. [EW06] K. Erdmann and M. Wildon, Introduction to Lie Algebras, 2006. [FH91] W. Fulton and J. Harris, Representation Theory, 1991. [Hum72] J. E. Humphreys, Introduction to Lie Algebras and Representaton Theory, 1972. Third printing, revised. [Hum08] , Representations of Semisimple Lie Algebras in the BGG Category O, 2008. [See18] G. H. Seelinger, Nilpotent, Solvable, and Semisimple Lie Algebras (2018). https://ghseeli.github.io/grad-school-writings/Monographs/ nilpotent-and-solvable-lie-algebras.pdf. [See17] , Root Systems (2017). https://ghseeli.github.io/ grad-school-writings/Monographs/RootSystems.pdf.