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 incrediblybeautiful model of a representation theory, since the simple representationsof semisimple Lie algebras are completely classified. Furthermore, such atool 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 applicationand then touch on the general representation theory of semisimple Lie alge-bras.

Note that, in this monograph, a linear Lie algebra is a Lie algebra that isviewed a subalgebra of gl(V ), although Ado’s theorem shows that any Liealgebra can be realized in such a way. Furthermore, a field F will always beof 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 witha characterization of the structure of representations of semisimple Lie alge-bras. This section is mostly a rewriting and expanding of [Hum72] chapter6.

2.1. Theorem (Weyl’s Theorem). Let g be a semisimple Lie algebra. Ifφ : 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 andany submodule W ≤ V , we get V = W ⊕W ′, where W ′ is a complement ofW . The easiest way to do this will be to find some module homomorphism

Date: June 2018.

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f : V → W with ker f ∩W = {0} and W + ker f = V . To construct such ahomomorphism, we will need the following.

2.2. Definition. Let φ : g → gl(V ) be a faithful representation of a Liealgebra g and let β(x, y) := tr(φ(x)φ(y)) be a bilinear form on g. Then, forbasis {x1, . . . , xn} of g and {y1, . . . , yn} of g such that β(xi, yj) = δij , wedefine the Casimir element of φ to be

cφ(β) =∑i

φ(xi)φ(yi) ∈ End(V )

2.3. Example. Let g = sl2 with standard basis {e, h, f}. Then, as shownin [See18], we have dual basis {1

4f,18h,

14e}. Now, if φ is the adjoint repre-

sentation, we get

cφ({e, f, h}) = φ(e)φ(1

4f) + φ(h)φ(

1

8h) + φ(f)φ(

1

4e)

=1

4ade adf +

1

8ad2

h +1

4ad f ade

=1

4(2e1,1 + 2e2,2) +

1

8(4e1,1 + 4e3,3) +

1

4(2e2,2 + 2e3,3)

= e1,1 + e2,2 + e3,3

= Id3

2.4. Lemma. Take a basis {x1, . . . , xn} of g and a basis {y1, . . . , yn} of gsuch that, for nondegenerate symmetric associative bilinear form β on g,β(xi, yj) = δij. For x ∈ g, if

[x, xi] =∑j

aijxj , [x, yi] =∑j

bijyj

then aik = −bki

Proof. We compute

aik =∑j

aijδjk

=∑j

aijβ(xj , yk) by bilinearity

= β([x, xi], yk)

= β(−[xi, x], yk)

= β(xi,−[x, yk]) by associativity

= −∑j

bkjβ(xi, yj)

= −bki�

2

2.5. Example. Continuing the example above, consider basis {e, h, f} ofsl2 and dual basis with respect to the Killing form, {1

4f,18h,

14e}. 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

[x,1

4f ] = −1

2bf +

1

4ah =⇒ b1,1 = −2b, b1,2 = 2a, b1,3 = 0

[x,1

8h] =

1

4cf − 1

4ae =⇒ b2,1 = c, b2,2 = 0, b2,3 = −a

[x,1

4e] = −1

4ch+

1

2be =⇒ b3,1 = 0, b3,2 = −2c, b3,3 = 2b

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 immediatelyabove, we get

[φ(x), cφ(β)] =∑i

[φ(x), φ(xi)φ(yi)] =∑i

[φ(x), φ(xi)]φ(yi)+φ(xi)[φ(x), φ(yi)]

Moreover, we know that [φ(x), φ(xi)] =∑

j aijφ(xj) and [φ(x), φ(yi)] =∑j bijφ(xj), with aik = −bki for any k using the same decomposition as in

the first of the two lemmas. Thus,

[φ(x), cφ(β)] =∑i,j

aijφ(xj)φ(yi) +∑i,j

bi,jφ(xi)φ(yj) = 0

�

This proposition is vitally important for proving Weyl’s theorem. Withthis element, we now have an object that must “behave like a scalar” withrespect to φ(g) in End(V ). Indeed, in the sl2 example, our Casimir elementis 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 inEnd(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 isfaithful, so we will often denote cφ := cφ(β). For more details, see [Hum72, p27].

2.10. Proposition. Given Casimir element cφ of faithful representationφ : g→ gl(V ), we get that

tr(cφ) = dim g

Proof. This follows from simply applying the trace operator.

tr(cφ) =∑i

tr(φ(xi)φ(yi)) =∑i

β(xi, yi) = dim g

�

2.11. Example. The Casimir element of sl2 under the adjoint representationwas precisely Id3 and tr(Id3) = 3 = dim sl2.

2.12. Lemma. Let φ : g → gl(V ) be a representation of a semisimple Liealgebra g. Then, φ(g) ⊆ sl(V ).

Proof. We know that, for a semisimple Lie algebra, [g, g] = g by the factthat 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 Liealgebra and let W ≤ V be irreducible with codimension 1. Then, φ is com-pletely 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) tellsus that cφ is actually a g-module homormophism on V . Thus, cφ(W ) ⊆ Wand ker cφ is a submodule of V . Now, by the lemma above, g must send allof 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 ofactions 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 andV = W ⊕ ker cφ. �

2.14. Lemma. Let φ : g → gl(V ) be a representation of a semisimple Liealgebra 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 notbe irreducible. We will show that we can reduce this lemma to the case of

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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 dimW = 1, W is irreducible and we can appeal to the previouslemma. So, now we proceed by induction and assume the result it true fordimW < n. Let dimW = n and W have proper submodule W ′. Then, wehave short exact sequence

0→W/W ′ → V/W ′ → F → 0

Now, by induction, V/W ′ = W/W ′⊕W ′′/W ′. However, this lifts to a shortexact sequence

0→W ′ →W ′′ → F → 0

However, dimW ′ < dimW =⇒ dimW ′′ = dimW ′+1 < dimW+1 = dimV ,so we can apply induction to get a one dimensional submodule complemen-tary to W ′ in W ′′, say X, so that W ′′ = W ′ ⊕X. Thus, we get

V/W ′ = W/W ′ ⊕W ′′/W ′ = W/W ′ ⊕ (W ′ ⊕X)/W ′ =⇒ V = W ⊕Xsince W ∩ X = 0 and dimW + dimX = dimV . Thus, we can repeat thisprocess 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 haveshort exact sequence

0→W → V → V/W → 0

Let U = {ψ ∈ Hom(V,W ) | ∃c ∈ F,ψ|W (w) = cw,∀w ∈ W} and letT = {ψ ∈ Hom(V,W ) | ψ|W (w) = 0,∀w ∈W} ≤ U . Then, we check that Uand 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 isalso 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 determinedby the scalar c, that is τ ∈ ψ+T if and only if τ |W = ψ|W = c.1W for scalarc. Thus, we have short exact sequence

0→ T → U → F → 0

and thus, by the lemma above (2.14), U = T ⊕ T ′ where T ′ is a one-dimensional g-submodue. If we let T ′ = (ψ), ψ ∈ End(V,W ), then we canscale ψ such that ψ|W = 1W since ψ 6∈ T . Now, since ψ is a g-modulehomomorphism, kerψ ≤ V , but kerψ 6= V since W ∩ kerψ = {0}. However,ψ cannot be injective because W � V , which is finite-dimensional, and thus

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kerψ 6= {0}. Therefore, since ψ is surjective, we have that V = W ⊕ kerψas a non-trivial decomposition of V . Thus, we can iteratively continue thisprocess until V has been completely reduced. �

Weyl’s theorem has many applications in the representation theory ofsemisimple Lie algebras. One immediate consequence is the preservation ofthe Jordan decomposition ([Hum72] section 6.4). That is,

2.15. Theorem. Let g ⊆ gl(V ) be a semisimple linear Lie algebra where Vis finite-dimensional. Then, g contains the semisimple and nilpotent partsin gl(V ) of all its elements. In particular, the abstract and usual Jordandecompositions 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 decom-position of x ∈ g, then φ(x) = φ(s)+φ(n) is the usual Jordan decompositionof φ(x).

2.17. Remark. Note that this theorem is a generalization of the followingfacts.

• 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 decomposinginto simple ideals g = g1 ⊕ · · · ⊕ gt. Then, the semisimple and nilpotentparts of x ∈ g are the sums of the semisimple and nilpotent parts in thevarious 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)sis 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 nilpotentendomorphism 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 Jordandecomposition of x. �

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3. Representations of sl2(F )

Recall that sl2(F ) (where F is an algebraically closed field) is the Liealgebra of 2 by 2 traceless matrices with standard basis

e =

(0 10 0

), f =

(0 01 0

), h =

(1 00 −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 corollaryabove. �

3.2. Proposition. Let V be an arbitrary sl2(F )-module. Then, V decom-poses as a direct sum of eigenspaces.

V =⊕λ∈F

Vλ, Vλ = {v ∈ V | h.v = λv}

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, wecan decompose V into φ(h)-eigenspaces. �

3.3. Definition. Whenever Vλ 6= 0, we call λ a weight of h in V , and wecall 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 λsuch that Vλ 6= 0 but Vλ+2 = 0.

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 amaximal 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 maximalvector of weight λ and set v−1 = 0 and vk = 1

k!fk.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,

f.vk = f.1

k!fk.v0 =

1

k!fk+1.v0 = (k + 1)fk+1.v0 = (k + 1)vk+1

For the final relation, we use induction on k. When k = 0, we have e.v0 = 0since v0 is a highest weight vector, and thus our formula holds since v−1 = 0by 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 thesmallest 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 eigenvaluesfor the h-action. Furthermore, the formulas in the lemma above show usthat span{v0, v1, . . . , vm} is a non-zero submodule of V . However, since V isirreducible, that must mean that V is the span of this linearly independentset. �

3.9. Proposition. Let V be an irreducible sl2(F )-module. Then, given high-est 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, forvm 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 , anirreducible sl2(F )-module, the highest weight of V .

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3.11. Proposition. Let m be the highest weight for irreducible sl2(F )-moduleV . Then, dimV = m+ 1 and dimVµ = 1 if Vµ 6= 0.

Proof. Given the eigenbasis (relative to the h-action) {v0, . . . , vm}, it is im-mediate that dimV = m + 1. Furthermore, for each Vµ 6= 0, there is aunique eigenvector under the h-action spanning it, up to scalar multiple.Thus, dimVµ = 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 = dimV and dimVµ = 1for 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 thelemma above, if the basis is chosen appropriately. In particular, thereexists at most one irreducible sl2(F )-module (up to isomorphism) ofeach possible dimension m+ 1 where m ∈ Z,m ≥ 0.

Proof. Part (a) is 3.11. Since dimV = m + 1, we see by relations that Vhas a highest weight vector of weight m, say vm, and it is unique since itseigenspace, Vµ, is 1-dimensional. Finally, part (c) is just appealing to thelemma 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 itsnegative (an equal number of times). Moreover, in any description of V intoa direct sum of irreducible submodules, the number of summands is preciselydimV0 + dimV1.

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 directsum of irreducible submodules. Now, each of these irreducible summandsmust have integer eigenvalues of h and thus, so must V . For the secondpart of the corollary, we note that an irreducible sl2-module must have anoccurence of a 0-eigenvector or a 1-eigenvector, but it cannot have both sincethe eigenvalues differ by even numbers due to the fact that h.vk = (λ−2k)vkin the lemma above. �

Let us now explicitly describe these representations, which are easy toenumerate due to the theorem above.

3.14. Example. The easiest representation is the trivial one-dimensionalrepresentation C = V0.

9

3.15. Example. Consider V = {e1, e2}, the “standard representation” ofsl2, where h.e1 = e1 and h.e2 = −e2. Then, V = V−1 ⊕ V1 = Ce1 ⊕Ce2. Bythe corollary above, this representation must be irreducible.

3.16. Example. Consider W = Sym2 V for V the standard representationabove. Then, a basis is given by {e2, ef, f2} 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 = −2f2

and so W = Cf2⊕Cef⊕Ce2 = V−2⊕V0⊕V2. Thus, by the corollary above,W is the irreducible representation of dimension 3. Proceeding more gener-ally, one can conclude that the irreducible sl2-representation of dimension nis given by Symn V for V the standard representation by computing

h.(en−kfk) = (n− k) · h.e · en−k−1fk + k · h.f · en−kfk−1 = (n− 2k) · en−kfk

4. Representations of sl3(F )

In this section, we describe the representations of sl3(F ) following theprogram in [FH91]. In sl3(F ), we do not have a single obvious element togive us an eigenspace decomposition, in contrast of sl2(F ) where the singleelement h sufficed. Thus, instead, we will consider a 2-dimensional subspaceof all diagonal matrices in sl3(F ), denoted h ⊆ sl3(F ). We make use of thefact

4.1. Lemma. Commuting diagonalizable matrices are simultaneously diag-onalizable.

Therefore, we will be able to find simultaneous eigenvectors for the ma-trices in h that yield an eigenspace decomposition with eigenvalues α ∈ h∗.

4.2. Proposition. Let V be an arbitrary sl3(F )-module, then V decomposesas a direct sum of eigenspaces

V =⊕α∈h∗

Vα, Vα = {v ∈ V | h.v = α(h)v,∀h ∈ h}

Thus, looking specifically at sl3(F ), we get

4.3. Corollary.

sl3(F ) ∼= h⊕

(⊕α

gα

)where α ranges over a finite subset of h∗ given by {εi−εj | 1 ≤ i, j ≤ 3, i 6= j}where

εi

a1

a2

a3

= ai

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-dimensionalityof sl3(F ). To see the specific decomposition, let D ∈ sl3(F ) be a diagonalmatrix with trace 0 and let (mi,j) = M ∈ sl3(F ). Then, by straightforwardcomputation,

[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 entryequal to 0. Thus, Ei,j generate the eigenspaces of the adjoint action of h onsl3(F ), given by the 6 linear functionals εi − εj . �

Now, given an x ∈ gα, we ask, how does its action affect this decomposi-tion. 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 weare content with this understanding. More generally, we see that

4.5. Proposition. Given a representation V of sl3 with eigenspace decom-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 representa-tion of sl3(F ) differ by integral linear combinations of the vectors {εi−εj} ⊆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 εi − εj ’s generate a lattice of eigenvalues in h∗.

4.7. Definition. The weights of the adjoint representation will be calledroots, denoted Φ (this will be expanded on later) and the lattice generatedby these roots is the root lattice, denoted ΛR. Furthermore, from the decom-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 highestweight vector for a representation of sl3(F ).

4.8. Lemma. Given a representation V of sl3(F ), there exists a v ∈ V suchthat

(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 rootspaces.)

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

`(a1ε1 + a2ε2 + a3ε3) = 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} = {gε1−ε3 , gε2−ε3 , gε1−ε2}each of which corresponds to a basis matrix with a single nonzero entry abovethe 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 amongthe weights with respect to `. �

4.9. Definition. A vector v ∈ V that meets the properties of the lemmaabove is a highest weight vector.

4.10. Proposition. Let V be an irreducible sl3(F ) representation and v ∈ Vbe a highest weight vector. Then, V = 〈E2,1, E3,1, E3,2〉.v, that is, V isgenerated 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 undersuccessive 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, sinceE1,3 = [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}} (rememberthat E3,1 = [E3,2, E2,1]). Then, W =

⋃nWn. 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 isclosed 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 highestweight vector. Then, 〈E2,1, E3,1, E3,2〉.v is an irreducible subrepresentationof V .

Proof. From above, it is clear that W := 〈E2,1, E3,1, E3,2〉.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 inthe proof of 4.8. Assume W is not irreducible, so W = W ′ ⊕W ′′. How-ever, the action of h and projection onto either component commutes, soWα = W ′α ⊕W ′′α . Thus, it must be that either W ′α or W ′′α is zero and so veither 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 high-est weight vector, up to scalar multiple.

Now, similar to sl2(F ), we notice that, given a highest weight vector, sayv ∈ gε1−ε3 , we have a collection of vectors v,E2,1.v, E

22,1.v, . . . until Ek2,1.v = 0

for some k ∈ N. From this, we note that span{E2,1, E1,2, [E1,2, E2,1]} isisomorphic 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

W =⊕k

Vα+k(ε2−ε1)

are integral and symmetric with respect to 0.

Proof. We note that W is a sl2(F )-representation via the isomorphism givenabove. Thus, since all eigenvalues of h ∈ sl2(F ) are integral and symmetricwith respect to 0 for all sl2(F )-representations, that must still be true inthis case. �

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

13

5. Beginning of a Structure Theory: Root SpaceDecomposition

We now seek to understand the structure of a general semisimple Liealgebra.

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 ] = 0for all x ∈ T , which is the same as showing adx = 0 (where ad is takenover T ). Since x is semisimple, so is adx and since we are working over analgebraically closed field, adx is diagonalizable. Thus, we need only showthat 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 uniquelyas a linear combination of eigenvectors of ady, say {−ay, v1, . . . , vn−1}, sincey is semisimple,

x = −c0ay +

n−1∑i=1

civi =⇒ −ay = ady x =

n−1∑i=1

ciλivi

where λi is the eigenvalue of vi. However, the set {−ay, v1, . . . , vn−1} is abasis, so it is linearly independent and −ay cannot be written as a linearcombination of {v1, . . . , vn−1}. Thus, we have arrived at a contradiction. �

5.3. Definition. A toral subalgebra h not properly included in any other iscalled a maximal toral subalgebra of g.

5.4. Example. Consider g = sl2(F ). Then, h = 〈h〉 is a maximal toralsubalgebra.

5.5. Remark. If g is a semisimple Lie algebra, it must have at least oneelement whose semisimple part is nonzero. Otherwise, g would be nilpotentby Engel’s theorem. Thus, g contains nonzero toral subalgebras and thuscontains nonzero maximal toral subalgebras.

5.6. Definition. Let h be a maximal toral subalgebra for semisimple Liealgebra 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 Liealgebra 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

g = Cg(h)⊕⊕α∈Φ

gα, gα = {g ∈ g | [h, g] = α(h)g,∀h ∈ h}

for any maximal toral subalgebra h of g.

5.9. Example. Let g = sl2(F ). The decomposition above corresponds tothe decomposition of sl2(F ) considered as an sl2(F )-modules (the regularrepresentation). Indeed, if h = span{h} for standard basis (e, f, h), thenCg(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 similardecomposition is easy to produce for g = sln(F ) with the standard basisconstruction.

5.10. Remark. In the example above, it is relatively straightforward tocompute Cg(h) = h. In fact, this statement ends up being true for allsemisimple Lie algebras. We wish to show this is true. Furthermore, we endup finding that the roots Φ completely characterize the structure of g. Wewill show that this is true in the remaining parts of this section.

5.11. Proposition. The adh-eigenvector decomposition defines an h∗-gradingon 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.

Proof. For some n ∈ N, gnα = 0, so adnx = 0 using the grading as above. �

5.13. Proposition. Given α, β ∈ h∗ with α + β 6= 0, then, relative to theKilling 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 associativefollows from the definition of the bracket on gl(V ).) �

5.14. Corollary. The restriction of the Killing form to Cg(h) is nondegen-erate.

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 mustbe 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 isnon-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 elementssince x ∈ Cg(h) =⇒ adx(h) = 0.

(b) All semisimple elements of Cg(h) lie in h since x ∈ Cg(h) =⇒ h+Fxis 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 isomor-phism between h and h∗ since the nondegeneracy gives trivial kernel anddimension 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) and0 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 eigenvectorsfor ad[x, y], since they are for all elements of ad h. Thus, span{x, y, [x, y]} isa Lie subalgebra of g.

Let h := [x, y] and let S be our Lie subalgebra. To see that S is isomorphicto 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] wouldbe a nilpotent Lie subalgebra (see [Hum72, Cor 4.1C] or [See18]) and so,in particular, ad([x, y]) is both semisimple and nilpotent, thus implyingad[x, y] = 0 =⇒ [x, y] ∈ Z(S) = 0, which is a contradiction. Thus, itmust 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 rep-resentation theory of semisimple Lie algebras: understanding roots and therepresentation theory of sl2. Luckily, we already understand the latter fromthe section above.

More generally, we have the following orthogonality proposition.

5.18. Proposition. [Hum72, Prop 8.3] Let Φ ⊆ h∗ be as above for semisim-ple 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 dimensionalsimple subalgebra of g isomorphic to sl2(F ) via

xα 7→ e, yα 7→ f, hα 7→ h

(g) hα = 2tακ(tα,tα) and hα = −h−α.

5.19. Remark. This proposition will later show that Φ satisfies the axiomsof a “root system” and is thus justifies the terminology “roots of g relativeto h.”

17

5.20. Example. Let us demonstrate the result for g = sl2(F ) since we havealready 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

[e, f ] = h =⇒ t2 =1

4h∗

Indeed, one easily checks κ(14h, λh) = λ · 2 for all λ ∈ F .

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

4h.

For (e), 2 · 14h∗(h) = 1

16κ(h, h) = 12 .

For (f), the result is trivial. For (g), we note

214h

116κ(h, h)

=12h12

= h

Proof. For (a), assume Φ does not span h∗. Then, there is a 0 6= h ∈ h suchthat 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 techniqueas the proof of the lemma above, assume κ(x, g−α) = 0. Then, κ(x, g) = 0is a contradiction since g is semisimple. Thus, there is a y ∈ g−α such thatκ(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, multiplyingby scalars, we may assume κ(x, y) = 1 =⇒ [x, y] = tα by (c). Thus, usingthe same line of reasoning in the lemma, if we let S = span{x, y, tα}, thenadg 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

κ(xα, yα) =2

κ(tα, tα)

using (e) and the fact κ(xα, g−α) 6= 0. Then, by (c),

hα :=2tα

κ(tα, tα)=⇒[xα, yα] = hα

[hα, xα] =2

α(tα)[tα, xα] =

2α(tα)

α(tα)xα = 2xα

[hα, yα] = − 2

α(tα)[tα, yα] = −2α(tα)

α(tα)yα = −2yα

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) forall h ∈ h. Thus, since hα = 2tα

κ(tα,tα) ,

−h−α = − 2t−ακ(t−α, t−α)

=2tα

κ(tα, tα)= hα

�

As noted above, the correspondence of sl2(F ) inside any semisimple Liealgebra to a root is incredibly important, so much so that we will define anew 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 sub-algebra of the form presented in the lemma or part (f) of the propositionabove, that is, a subalgebra generated by {xα, yα, [xα, yα]} for xα ∈ gα andyα ∈ g−α.

With this terminology, we have the following proposition

5.22. Proposition. Given a semisimple Lie algebra g with root α, g maybe 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 ∈ Φor β = 0. Then, subspace ⊕

k∈Fβ+kα∈Φ

gβ+kα

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 := h⊕⊕kα∈Φ

gkα

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 mustbe integral, so 2c ∈ Z. Now, by direct computation, sl(α) acts trivially onkerα ⊆ h. Next, given h + sl(α), we know that, as sl(α)-modules,

h + sl(α) ∼= kerα⊕ sl(α)

since dim kerα = dim h − dim imα = dim h − 1. Thus, by Weyl’s theoremand the above computation,

h + sl(α) ⊆ h⊕⊕kα∈Φ

gkα ∼= kerα⊕ sl(α)⊕N

where N is some complementary submodule. Now, let V be an irreduciblerepresentation of sl2(F ) with dimension k. Then, if k is even, there is a v ∈ Vwith hα.v = 0 =⇒ α(v) = 0 =⇒ v ∈ kerα by virtue of the classificationof sl2(F ) representations. However, kerα ∩ V = 0 =⇒ v = 0, which is acontradiction. Thus, the only even weights occuring in M are 0 and ±2,which tells us that 2α cannot be a root. However, this also means that 1

2αis not a root, so 1 also cannot be a weight of M . Thus, since the number ofirreducible sl2(F )-module summands is precisely dimM0 + dimM1, it mustbe that N contains no irreducible sl2(F )-submodules, and is thus trivial.Therefore,

M ∼= kerα⊕ sl(α)

but kerα ∼= g0 can only be a direct sum of trivial sl(α)-modules by theclassification of sl2(F ) representations. Thus, gα is one-dimensional sincesl(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 :=⊕k∈Z

β+kα∈Φ

gβ+kα.

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

hα.v = (β(hα) + 2k)v

for v ∈ gβ+kα. Because the weights differ by even integers, it cannot be thatboth 0 and 1 occur as a weight of this form, and either can only appearonce. Thus, M must be an irreducible sl(α)-module and the weights mustbe symmetric around 0. Let the heighest weight be given by β+ qα and thelowest 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 aroot, then hα acts on gα+β by eigenvalue (α+β)hα = β(hα) + 2, which tellsus that β(hα) is not the highest weight, a contradiction. Thus, it must bethat adxα xβ 6= 0 =⇒ [gα, gβ] = gα+β.

Finally, by virtue of (d), g is generated as a Lie algebra by the root spacessince, given generator tα ∈ gα and tβ ∈ gβ, if α+β ∈ Φ, [tα, tβ] is a non-zeroscalar 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 usingthe basis given by the root space decomposition. Namely, the action of his 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 awell-understood strucutre.

5.27. Lemma. For α, β ∈ Φ, we have

κ(tα, tβ) =∑γ∈Φ

γ(tα)γ(tβ)

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

κ(hα, hβ) = tr(adhα adhβ ) =∑γ∈Φ

γ(hα)γ(hβ)

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, wecompute

κ(tα, tβ) = κ

(κ(tα, tα)

2hα,

κ(tβ, tβ)

2hβ

)=κ(tα, tα)κ(tβ, tβ)

4κ(hα, hβ)

However,

1

κ(tα, tα)=

1

4κ

(2tα

κ(tα, tα),

2tακ(tα, tα)

)=

1

4κ(hα, hα) ∈ Q

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-degenerateon h and thus (·, ·) is non-degenerate on h∗. Furthermore, (·, ·) inherits itsbilinearity from the Killing form. �

5.31. Lemma. If β ∈ Φ and {α1, . . . , αn} is a basis of roots for h∗, then βis a rational linear combination of the αi’s.

Proof. Consider β =∑ciαi. Then, we compute

(β, αj) =∑i

(αi, αj)ci

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 andthus 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-linearcombinations of other roots, so EQ is independent of choice of roots for abasis. 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 ofroots for EQ, say {α1, . . . , αn}. Thus, for γ, δ ∈ EQ,

(γ, δ) =(∑

ciαi,∑

djαj

)=∑i,j

cidj · (αi, αj) =∑i,j

cidjκ(tαi , tαj )

but ci, dj ∈ Q by definition of EQ. Finally, given λ ∈ EQ,

(λ, λ) = κ(tλ, tλ)5.27=∑γ∈Φ

γ(tλ)2 =∑γ∈Φ

(γ, λ)2 ≥ 0

If (λ, λ) = 0, then γ(tλ) = 0 for all roots γ, and so λ = 0 by 5.18 (e). Thus,(·, ·) is positive-definite. �

5.34. Lemma. β(hα) = 2(β,α)(α,α)

Proof. Using the fact that hα = 2tακ(tα,tα) , we see

2(β, α)

(α, α)=

2κ(tβ, tα)

κ(tα, tα)= κ

(tβ,

2tακ(tα, tα)

)= κ(tβ, hα) = β(hα)

�

5.35. Definition. Let E := R⊗Q EQ.

5.36. Remark. Using standard linear algebra, the form (·, ·)|EQ extendscanonically 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 struc-ture:

5.37. Proposition. Given a semisimple Lie algebra g, Φ is a root systemin 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.

(c) If α, β ∈ Φ, then β − 2(β,α)(α,α) α ∈ Φ.

(d) If α, β ∈ Φ, then 2(β,α)(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, whichare incredibly symmetric structures that are well-understood. There aremany introductions to root systems, including [Hum72, Ch III], [Car05, Ch5], and [See17]. From this point onwards, a working knowledge of rootsystems is assumed. We see that the structure of irreducible root systemscorresponds to that of simple Lie algebras:

5.38. Proposition. Let g be a simple Lie algebra, with maximal toral sub-algebra 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, sayK, 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 becauseg 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 withmaximal toral subalgebra h and root system Φ. If g = g1 ⊕ · · · ⊕ gt is thedecomposition of g into simple ideals, then hi = h ∩ gi is a maximal toralsubalgebra of gi and the corresponding (irreducible) root system Φi may beregarded 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 sim-ple ideals g = g1 ⊕ · · · ⊕ gt and if h is a maximal toral subalgebra of g, thenh = g1 ∩ h⊕ · · · ⊕ gt ∩ h since, x = x1 + · · ·+ xt ∈ h is semisimple and eachxi ∈ 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 sinceany toral subalgebra of gi larger than hi, say h′i would be toral in g and cen-tralize all hj , j 6= i, and so h1⊕· · ·⊕h′i⊕· · ·⊕ht would be a toral subalgebraof g larger than h.

Let Φi be the root system of gi relative to hi. If α ∈ Φi, we can view α asa linear function on h by defining α(hj) = 0 for j 6= i. So, α is thus a rootof 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 girelative to hi. Note that α cannot be a root of another system since gα ⊆ giand [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, apurely geometric task, we gain understanding of all the possible root systemsa simple Lie algebra might have. Furthermore, one can show, with morework, 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] orby 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 diagonalmatrices generalizes to a maximal toral subalgebra h and the adh-actionalways yields integral eigenvalues on the root spaces.

6.1. Example. The roots Φ = {εi − εj} of sl3(F ) form a root system asdescribed in the proposition concluding the previous chapter. To see thisexplicitly, one checks by straightforward computation that parts (a) and (b)of 5.37 are true. Then, we note that {ε1 − ε2, ε2 − ε3} forms a basis for Φ.We then check{

[Ei,j , Ej,i] = Ei,i − Ej,jκ(Ei,j , Ej,i) = 6

=⇒ tεi−εj =1

6(Ei,i − Ej,j)

and so

(ε1 − ε2, ε1 − ε2) =1

36κ(E1,1 − E2,2, E1,1 − E2,2)

=1

36(ε1 − ε2)(E1,1 − E2,2)

=1

36(1 + 1)

25

=1

18

(ε1 − ε2, ε2 − ε3) =1

36κ(E1,1 − E2,2, E2,2 − E3,3)

=1

36(ε1 − ε2)(E2,2 − E3,3)

=1

36(0− 1)

= − 1

36

(ε2 − ε3, ε2 − ε3) =1

36(ε2 − ε3)(E2,2 − E3,3)

=1

36(1 + 1)

=1

18

and so

(ε1 − ε2)− 2(ε1 − ε2, ε2 − ε3)

(ε2 − ε3, ε2 − ε3)(ε2 − ε3) = ε1 − ε2 − (−1)(ε2 − ε3) = ε1 − ε3

which shows (c) and demonstates (d). Indeed, for (d), one can see that2(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

and thus 2(β,α)(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 imagesvisualize a root string as a line through the lattice. For instance, for α ∈ Φ,

W :=⊕k∈Z

α+k(ε2−ε1)

gα+k(ε2−ε1)

is symmetric about the line {L | 〈E1,1 −E2,2, L〉 = 0} and is thus preservedby reflection across said line.

6.2. Proposition. [FH91, Prop 12.15] All eigenvalues of any irreduciblefinite-dimensional representation of sl3(F ) must lie in the weight latticeΛW ⊆ h∗, that is, the lattice generated by the εi and must be congruentmodulo the lattice ΛR ⊆ h∗ generated by the εi − εj.

6.3. Proposition. [FH91, Prop 12.18] Let V be any irreducible, finite-dimensional representation of sl3(F ). Then, for some α ∈ ΛW ⊆ h∗, theset of eigenvalues occurring in V is exactly the set of linear functionals

26

ε1 − ε2

ε1 − ε3ε2 − ε3

ε2 − ε1

ε3 − ε1ε3 − ε2

ε1ε2

ε3

Figure 1. Weight lattice for sl3. Choice of simple rootsα1, 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 | 〈Ei,i − Ej,j , L〉 = 0}.

Fill in proofs forthese proposi-tions.

From this, we see that, given our choice of `, it must be that any heigh-est weight vector will lie in the region cut out by the inequalities 〈E1,1 −E2,2, L〉 ≥ 0 and 〈E2,2−E3,3, L〉 ≥ 0. In other words, it must be of the form

(a+ b)ε1 + bε2 = aε1 − bε3

Thus, we have reason to believe.

6.4. Theorem. For any pair of natural numbers a, b, there exists a unique ir-reducible, finite-dimensional representation Γa,bof sl3(F ) with highest weightaε1 − bε3.

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

27

6.5. Example. Consider the standard representation of sl3(C) on V ∼= C3

with basis {e1, e2, e3} with associated eigenvalues ε1, ε2, and ε3 since, for

h =

a1

a2

a3

ei = aiei = εi(h)ei

Consider also its dual, V ∗, with dual basis {e∗1, e∗2, e∗3} and associated eigen-

ε1ε2

ε3

Figure 2. Weight diagram for V , the standard representa-tion of sl3(C)

values −ε1,−ε2, and −ε3.

ε1ε2

ε3−ε1 −ε2

−ε3

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

Using weight diagrams , one can see that Sym2 V and Sym2 V ∗, whose Actually inserttheseweights are the pairwise sums of the weights of V and V ∗, respectively, are

both irreducible. A more interesting analysis comes from examining V ⊗V ∗.

28

ε1ε2

ε3

Figure 4. Weight diagram for Sym2 V ∗

One can quickly check that the weights of V ⊗V ∗ are {ε1± ε2, ε1± ε3, ε2±ε3}, each with multiplicity 1, and 0 with multiplicity 3. Since the linear map

v ⊗ u∗ 7→ 〈v, u∗〉 = u∗(v)

has a non-trivial kernel. One can easily check that this map is equivalentto the trace since V ⊗ V ∗ ∼= Hom(V, V ) ∼= M3(C) and thus the kernel isall traceless matrices, ie the adjoint representation of sl3(C), which is anirreducible 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 vectorsv, w with weights α, β, respectively, then v ⊗w ∈ V ⊗W is a highest weightvector 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 ε1.(b) V ∗ has a highest weight vector with weight −ε3.

29

(c) Symn V has a highest weight vector with weight n · ε1.(d) Symn V ∗ has a highest weight vector with weight −n · ε3.

Proof of Theorem 6.4. Let V be the standard representation of sl3(F ). Then,Syma V ⊗ Symb V ∗ contains an irreducible subrepresentation generated bythe highest weight vector with weight aε1 − bε3.

To show uniqueness, let V,W be two sl3(F )-representations with highestweight α and corresponding highest weight vectors v, w. Then. consider(v, w) ∈ V ⊕W will be a highest weight vector with highest weight α. LetU be the irreducible subrepresentation generated by (v, w). Then, we havenon-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, mosteasily using the Weyl Character Formula, which has not been proven yet.However, the final result will be given by first considering the contractionmap

ιa,b : Syma V ⊗ Symb V ∗ → Syma−1 V ⊗ Symb−1 V ∗

(v1 · · · va)⊗ (v∗1 · · · v∗b ) 7→∑〈vi, v∗j 〉(v1 · · · vi · · · va)⊗ (v∗1 · · · v∗j · · · v

∗b )

Such a map is surjective and the codomain cannot have eigenvalue aε1−bε3,so Γa,b ⊆ ker ιa,b. In fact, using the Weyl Character Formula, we immediatelyconclude

6.8. Proposition. ker ιa,b = Γa,b.

Also, as an immediate corollary, we conclude

6.9. Corollary. If b ≤ a, Syma V ⊗ Symb V ∗ ∼=⊕b

i=0 Γa−i,b−i and similarlyif a ≤ b.

7. More Structure Theory of Semisimple Lie Algebras

We state some results here without proof since they can be superseced bySerre’s theorem. However, we wish to establish that maximal toral subal-gebras and Cartan subalgebras coincide for semisimple Lie algebras. Alongthe way, we establish a few other distinguished subalgebras of semisimpleLie algebras which appear throughout the literature on Lie algebras.

7.1. Proposition. [Hum72, p 74] Let g be a semisimple Lie algebra, h ⊆ ga maximal toral subalgebra, and Φ the root system of g relative to h. Fix abase ∆ of Φ. Then, g is generated as a Lie algebra by the root spaces gα, g−αfor α ∈ ∆. Equivalently, g is generated by arbitrary nonzero root vectorsxα ∈ gα, yα ∈ g−α for α ∈ ∆.

30

7.2. Theorem. [Hum72, p 75] Let g, g′ be simple Lie algebra over a field Fwith respective maximal toral subalgebras h, h′ and corresponding root sys-tems Φ,Φ′. Suppose there is an isomorphism of Φ onto Φ′ given by α 7→ α′

inducing π : h→ h′. Fix a base ∆ ⊆ Φ so that Φ′ = {α′ | α ∈ ∆} is a base ofΦ′. For each α ∈ ∆, α′ ∈ ∆′, choose arbitrary (nonzero) xα ∈ gα, x

′α ∈ g′α,

that is, choose an arbitrary Lie algebra isomoprhism πα : gα → g′α. Then,there eixsts a unique isomorphism π : g → g′ extending π : h → h′ and ex-tending all of the πα, α ∈ ∆.

h g gα

h′ g′ g′α

π π πα

7.3. Proposition. [Hum72, p 77] Given g as above, but not necessarilysimple, 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 negativeis in Aut Φ. �

7.4. Definition. Let t ∈ EndV , a be a root of the characteristic polynomialof t with multiplicity m, and Va := ker(t−a·1)m. Then, given an x ∈ g, a Liealgebra over algebraically closed field F , we call g0(adx) for adx ∈ End gan Engel subalgebra.

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

Proof. Compute the expansion

(adx− a− b)m[y, z] =

m∑i=0

(m

i

)[(adx− a)i(y), (adx− b)m−i(z)]

for y ∈ ga(adx), z ∈ gb(adx). For sufficiently large m the expression will be0. �

7.6. Lemma. [Hum72, p 79] Let g′ be a subalgebra of g. Choose z ∈ g′ suchthat g0(ad z) is minimal in the collection of all g0(adx) for x ∈ g′. Supposethat g′ ⊆ g0(ad z). Then, g0(ad z) ⊆ g0(adx) for all x ∈ g′.

7.7. Lemma. [Hum72, p 79] If g′ is a subalgebra of g containing an En-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 nilpotentsubalgebra 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 charF = 0.Then, the CSA’s of g are precisely the maximal toral subalgebras of g.

7.11. Proposition. [Hum72, p 81] Let φ : h→ h′ be a surjective Lie algebrahomomorphism.

(a) If h is a CSA of g, then φ(h) is a CSA of g′.(b) If h′ is a CSA of g′ and K = φ−1(h′), 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 solvablesubalgebra of g.

7.13. Definition. Let g be a finite dimensional semisimple Lie algebra overC and let h be a CSA of g. Then, g has triangular decomposition

g = n− ⊕ h⊕ n

where n− =⊕

α∈Φ− gα and n =⊕

α∈Φ+ gα.

7.14. Proposition. The n− and n in the triangular decomposition are sub-algebras of g.

Proof. The sum of two positive roots is still positive. So, for α, β ∈ Φ+, wehave

[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 standardrelative 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 g′ be any subalgebra of g properly containing b. Since it is a subal-gebra containing h, g′ must be stable under the ad h action, and so it mustinclude some gα with α ≺ 0. However, then g′ contains a copy of sl2 gen-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 auto-morphisms of Aut g.

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

7.18. Corollary. [Hum72, p 84] The Cartan subalgebras of an arbitrary Liealgebra 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 thatthe representation theory of U(g) is the same as the representation theoryof 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...

TnV = V ⊗ · · · ⊗ V︸ ︷︷ ︸n times

and say T (V ) =⊔T kV equipped with the product

TmV × TnV → Tm+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 anassociative unital algebra over F . Then, we have the following universalproperty:

V T (V )

A

ι

φ∃!Φ

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 suchthat Φ ◦ ι = φ.

8.3. Definition. Let g be a Lie algebra. Then, consider the ideal J =〈x⊗ y − y ⊗ x− [x, y] | x, y ∈ g〉 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 withbasis {x1, . . . , xn} and [xi, xj ] = 0. Then, J E T (g) is given by J = 〈x ⊗y − y ⊗ x | x, y ∈ g〉. Therefore, U(g) is commutative and generated by1, 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 U(g)

A

φ

ι

∃!Φ

that is, there is a unique F -algebra homomorphism Φ: U(g) → A such thatΦ ◦ ι = φ.

8.6. Remark. Note that we have not shown that ι is injective, but we willdo so later with the PBW Theorem.

8.7. Proposition. There is a bijective correspondence between representa-tions θ : g → gl(V ) and representations φ : U(g) → EndV . Correspondingrepresentations 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 U(g)

gl(V ) = EndV

θ

ι

∃!φ

where φ : U(g)→ EndV is a representation of U(g).

Conversely, given φ a representation of U(g), we know that the linear mapι : 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 algebrahomomorphism.

g U(g)

EndV

ι

φ =⇒g [U(g)]

gl(V )

ι

φ◦ι φ

Then, φ ◦ ι is a representation �

8.8. Remark. Note, the representations under consideration need not befinite-dimensional. In fact, it will turn out that a certain important class ofrepresentations called “Verma modules” will always be infinite-dimensional.

This construction will lead us to the all-important PBW Theorem, whichtells us how to obtain a basis for the universal enveloping algebra of a Liealgebra.

8.9. Theorem (Poincare-Birkhoff-Witt Theorem). [Car05, Theorem 9.4]Let g be a Lie algebra with basis {xi | i ∈ I}. Let < be a total order onthe index set I. Let ι : g → U(g) be the natural linear map from g to itsenveloping algebra. We define yi := ι(xi). Then, the elements

yr1i1 · · · yrnin

for all n ≥ 0, all ri ≥ 0, and all i1, . . . , in ∈ I with i1 < I2 < · · · < in forma 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 andstates this version as Corollary C). The PBW Theorem has many importantcorollaries 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 inde-pendent. Since the xi ∈ g form a basis for g and ι(xi) = yi, it must be thatker ι = {0}. �

8.11. Corollary. The subspace ι(g) ⊆ [U(g)] is isomorphic to g as a Liesubalgebra.

Proof. From the corollary above, ι : g → ι(g) is bijective and yi, i ∈ I forma 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,

a =∑

λi1,...,in,r1,...,rnyr1i1· · · yrnin

=∑

r1+···+rn=rmaximal degree

λi1,...,in,r1,...,rnyr1i1· · · yrnin +

∑r1+···+rn<r

λi1,...,in,r1,...,rnyr1i1· · · yrnin

= f(yi) + sum of terms of smaller degree

b =∑

µi1,...,in,r1,...,rnyr1i1· · · yrnin

=∑

r1+···+rn=rmaximal degree

µi1,...,in,r1,...,rnyr1i1· · · yrnin +

∑r1+···+rn<r

µi1,...,in,r1,...,rnyr1i1· · · yrnin

= g(yi) + sum of terms of smaller degree

Recall also, since ι(g) is a Lie subalgebra,

yiyj − yjyi = [yi, yj ] =∑k

νkyk

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 thatab 6= 0. �

9. Irreducible Modules of Semisimple Lie Algebras

9.1. Lemma. [Car05, Lemma 10.1] Let g be a finite dimensional Lie algebraover C, and let g′ be a subalgebra of g. Then, there exists a unique algebra

36

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

g′ U(g′)

g U(g)

i

ιg′

θ

ιg

where i is the embedding of g′ into g and ιg′ , ιg are the standard embeddingsof a Lie algebra into its universal enveloping algebra. Furthermore, θ isinjective.

Proof. For x ∈ g′, the definition

θ(ιg′(x)) = ιg(i(x))

is well-defined and, since U(g′) is generated by ι(g′)(g′), such a θ is uniquelydetermined. To see existence, define one notices

g′ T (g′) T (g′)/J ′ ∼= U(g′)

g T (g) T (g)/J ∼= U(g)

i i∗

since i∗(J ′) ⊆ J , where J, J ′ are the ideals used to define the universal en-veloping algebra.

To see θ is injective, let x1, . . . , xr be a basis for g′ and 0 6= u ∈ U(g′).Then, u is a non-zero linear combination of monomials xe11 · · ·xerr . Then, letx1, . . . , xr be part of a basis of g. The PBW theorem tells us that such alinear combination of monomials cannot be zero in U(g). Thus, θ(u) 6= 0and so ker θ = 0. �

9.2. Definition. Let g be a finite-dimensional semisimple Lie algebra overC and let λ ∈ h∗. We define, for h = {h1, . . . , hk},

Kλ :=∑α∈Φ+

U(g)xα +

k∑i=1

U(g)(hi − λ(hi))

for basis {xα | α ∈ Φ} ∪ {hi | i = 1, . . . , k} of g. Since Kλ is a left ideal ofU(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 alsodefine

K ′λ :=∑α∈Φ+

U(b)xα +

k∑i=1

U(b)(hi − λ(hi))

which is a left ideal of U(b).

37

9.3. Proposition. [Car05, Prop 10.4] Let g be a semisimple Lie algebra overC with positive roots Φ+ = {β1, . . . , βN} and basis h1, . . . , hk, xβ1 , . . . , xβN .

(a) dimU(b)/K ′λ = 1(b) The elements

(h1 − λ(h1))s1 · · · (hk − λ(hk))skxt1β1 . . . x

tnβN

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

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

hs′11 · · ·h

s′kk x

t′1β1· · ·xt

′NβN≺ hs11 · · ·h

skk x

t1β1· · ·xtNβN if s′1 ≤ s1, . . . , s

′k ≤ sk, t′1 = t1, . . . , t

′N = tN

Thus,

(h1−λ(h1))s1 · · · (hk−λ(hk))skxt1β1 . . . x

tnβN

= hs11 · · ·hskk x

t1β1· · ·xtNβN+ strictly lower terms

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

(h1 − λ(h1))s1 · · · (hk − λ(hk))skxt1β1 . . . x

tnβN, si ≥ 0, ti ≥ 0

span U(b) and they are linearly independent.

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

Consider the representation λ : h → C. Since b/n ∼= h, one can lift to a1-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 usxα ∈ ker ρ α ∈ Φ+

hi − λ(hi) ∈ ker ρ i = 1, . . . , k

1 6∈ ker ρ

=⇒

{K ′λ ⊆ ker ρ

1 6∈ ker ρ=⇒ 1 6∈ K ′λ

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,

Kλ =∑α∈Φ+

U(g)xα +k∑i=1

U(g)(hi − λ(hi))

=∑α∈Φ+

U(n−)U(b)xα +k∑i=1

U(n−)U(b)(hi − λ(hi))

= U(n−)

∑α∈Φ+

U(b)xα +k∑i=1

U(b)(hi − λ(hi))

= U(n−)K ′λ

Therefore, every element of Kλ is a linear combination of terms of the form

yr1β1 · · · yrNβN︸ ︷︷ ︸

∈U(n−)

(h1 − λ(h1))s1 · · · (hk − λ(hk))skxt1β1 · · ·x

tNβN︸ ︷︷ ︸

∈K′λ

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-zeroelement of U(n−) can be a a linear combination of such terms, since its basismust be of the form f r1β1 · · · f

rNβN

. Therefore, Kλ ∩ U(n−) = 0. �

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 forsome u ∈ U(n−), that is, M(λ) is cyclically generated by mλ as aU(n−)-module.

(b) Moreover, the elements yr1β1 · · · yrnβNmλ for all ri ≥ 0 form a basis for

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 theform

yr1β1 · · · yrNβNhs11 · · ·h

skk x

t1β1· · ·xtNβN r1, si, ti ≥ 0

Now, observe

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

himλ = λ(hi)mλ

39

and so

yr1β1 · · · yrNβNhs11 · · ·h

skk x

t1β1· · ·xtNβNmλ =

{0 if any ti > 0

yr1β1 · · · yrNβNγmλ for some γ ∈ C if all ti = 0

Therefore, since umλ is a linear combination of elements of the form of theproduct above, such a linear combination simplifies to a linear combinationof elements of the form

yr1β1 · · · rrNβNmλ, ri ≥ 0

and so elements of this form span M(λ).

To see linear independence, let∑r1,...,rN

γr1,...,rN yr1β1· · · yrNβNmλ = 0, γr1,...,rN ∈ C

and so, by the proposition above,∑r1,...,rN

γr1,...,rN yr1β1· · · yrNβN ∈ Kλ ∩ U(n−) = 0 =⇒ all γr1,...,rN = 0

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 formumλ 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).(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) dimM(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 underthe action of h.

Since the elements yr1β1 · · · yrNβNmλ with ri ≥ 0 form a basis for M(λ), part

(b) follows by showing that

xyr1β1 · · · yrNβNmλ = (λ− r1β1 − · · · − rNβN )(x)yr1β1 · · · y

rNβNmλ, ∀x ∈ h

by induction on r1 + · · · + rN . Thus, yr1β1 · · · yrNβNmλ ∈ M(λ)µ for µ = λ −

r1β1 − · · · − rNβN and so

M(λ) =∑µ

M(λ)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, foreach λ ∈ Λ, let

Nµ = span{yr1β1 · · · yrNβNmλ | λ− r1β1 − · · · − rNβN = µ} ⊆M(λ)µ ⊆M(λ)

By the PBW theorem, we also get

M(λ) =⊕µ∈Λ

Nµ

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,

dimM(λ)µ = dimNµ = |{(r1, . . . , rN ) ∈ NN0 | λ− r1β1 − · · · − rNβN = 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 iscalled the weight space of M(λ) with weight µ.

9.9. Theorem. [Car05, Theorem 10.9] M(λ) has a unique maximal submod-ule, denoted J(λ).

Proof. Take V to be a proper U(g)-submodule of M(λ). The key idea is toshow that V ⊆

∑µ,µ 6=λM(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 submoduleof M(λ) lying in this codimension 1 subspace.

By the theorem above,

v =n∑i=1

vµi , vµi ∈M(λ)µi

where the µi are distinct. To show each vµi ∈ V , we observe, for x ∈ h andfixed i,

xvµi = µi(x)vµi

=⇒∏j,j 6=i

(x− µj(x))v =∑k

∏j,j 6=i

(x− µj(x))vµk

=∑k

∏j,j 6=i

(µk(x)− µj(x))vµk

=∏j,j 6=i

(µi(x)− µj(x))vµi

41

Moreover, since any vector space cannot be a union of finitely many propersubspaces, we can find an x ∈ h with µi(x) 6= µj(x) for all j 6= i. Thus, ifwe pick such an x,∏

j,j 6=i(x− µj(x))v ∈ V =⇒

∏j,j 6=i

(µi(x)− µj(x))vµi ∈ V

and so, since∏j.j 6=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 dimM(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-tained in

∑µ,µ 6=λM(λ)µ. �

9.10. Definition. [Car05, Definition 10.10] For λ, µ ∈ h∗, we say that λ � µif 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∗ withµ ≺ λ. Thus, λ is the highest weight of M(λ) with respect to this partialorder 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) isa maximal submodule of M(λ).

(b) λ is a weight of L(λ) since J(λ)λ = 0.(c) dimL(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 dimL(λ)µ?

9.12. Theorem. Let V be a finite dimensional irreducible g-module withhighest weight vector vλ. Then, there exists a surjective homomorphismθ : M(λ)→ V of U(g)-modules such that θ(mλ) = vλ.

9.13. Corollary. A finite dimensional irreducible g-module V with highestweight 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 uniquemaximal 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) isfinite dimensional, then λ(hi) is a non-negative integer for {h1, . . . , hn} abasis 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), itmust be that λ(hi) ∈ Z≥0. �

9.15. Theorem. [Car05, Thm 10.20] Suppose λ ∈ h∗ is dominant and inte-gral. Then, the irreducible g-module L(λ) is finite dimensional and its setof weights is permuted by W with

dimL(λ)µ = dimL(λ)wµ ∀w ∈ W

The proof of this theorem is quite long (see [Car05] or [Hum72, §21]), butthe following lemma must be used in either proof.

9.16. Lemma. [Hum72, Lem 21.2] The following identities hold in U(g) fork ≥ 0, 1 ≤ i, j ≤ `.

(a) [xj , yk+1i ] = 0 when i 6= j

(b) [hj , yk+1i ] = −(k + 1)αi(hj)y

k+1i

(c) [xi, yk+1i ] = −(k + 1)yki (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)yiNow, asume the result holds for all values less than k + 1. We get

[hj , yk+1i ] = hjy

k+1i − yk+1

i hj

= (hjyki − yki hj)yi + yki (hjyi − yihj)

= [hj , yki ]yi + yki [hj , yi]

= −kαi(hj)yki yi + yki (−αi(hj)yi)

= −(k + 1)αi(hj)yk+1i

(c) Similar to the above, we check

[xi, yk+1i ] = xiy

k+1i − yk+1

i xi

= [xi, yi]yki + yi[xi, y

ki ]

= hiyki + yi(−kyk−1

i ((k − 1) · 1− hi))

= hiyki − yki hi + (k + 1)yki hi − kyki (k − 1)

= −kαi(hi)yki + (k + 1)yki hi − kyki (k − 1)

43

= −2kyki + (k + 1)yki hi − kyki (k − 1)

= (k + 1)yki hi − k(k + 1)yki

= −(k + 1)yki (k · 1− hi)

�

10. Characters

10.1. Definition. Given g a semisimple Lie algebra and V be a g-modulewith weight space decomposition

V =⊕λ∈h∗

Vλ

where dimVλ < ∞. Then, we define the character of V to be the functionchV : h∗ → Z given by

(chV )(λ) = dimVλ

and we define the formal character of V to be

chV =∑µ

dimVµeµ ∈ Z[ΛW ]

where the sum is over all weights of V and eµ(λ) = δµλ. Explain the ringstructure here;the multiplica-tion is not clear.

10.2. Proposition. Let V, V1, V2 be g-modules that admit characters. Then,

(a) If U ⊆ V ,

chU + ch(V/U) = chV

(b) Therefore,

ch(V1 ⊕ V2) = chV1 + chV2

(c)

ch(V1 ⊗ V2) = chV1 · chV2

10.3. Lemma. Given a Verma module M(λ) with λ ∈ h∗, we have that

(chM(λ))(µ) = Number of ways to write λ−µ as a positive sum of positive roots.

Proof. This follows immediately from the definition of chM(λ) and 9.7(d).�

10.4. Proposition. If P(ν) is the number of ways to write ν as a positivesum of positive roots, then

chM(λ) = eλ

∑ν∈h∗

P(ν)e−ν

44

Proof. This follows since

chM(λ) =∑µ∈h∗

P(λ− µ)eµ

=∑ν∈h∗

P(ν)eλ−ν

=∑ν∈h∗

P(ν)eλe−ν

= eλ∑ν∈h∗

P(ν)e−ν

�

10.5. Lemma. In the ring of all function f : h∗ → Z with finite support,∑ν∈h∗

P(ν)e−ν

−1

=∏α∈Φ+

(1− e−α)

Proof. Let Φ+ = {β1, . . . , βn}. Then,∑ν

P(ν)e−ν =∑

r1,...,rn≥0

e−r1β1−···−rnβn

=∑

r1,...,rn≥0

er1−β1 · · · ern−βn

=n∏i=1

∑r1≥0

eri−βi

but then

(1− e−βi)(1 + e−βi + e2−βi + · · · ) = 1

and so we have our desired inversen∏i=1

(1− e−βi) =∏α∈Φ+

(1− e−α)

�

We state the following theorem without proof.

10.6. Theorem. The character of the Verma module with highest weight λis given by

chM(λ) =eλ+ρ

eρ∏α∈Φ+(1− e−α)

Proof.

chM(λ) = eλ

∑ν∈h∗

P(ν)e−ν

=eλ∏

α∈Φ+(1− e−α)=

eλ+ρ

eρ∏α∈Φ+(1− e−α)

45

�

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

M(λ) = N0 ⊇ N1 ⊇ · · · ⊇ Nr = 0

where each Ni is a submodule of M(λ) and Ni+1 is a maximal submodule ofNi. Furthermore, Ni/Ni+1

∼= L(w · λ) for some w ∈ W .

10.8. Theorem (Weyl Character Formula). [Car05, Theorem 12.17] Givensemisimple Lie algebra g with Cartan subalgebra h. Then, for λ a dominantweight,

chL(λ) =

∑w∈W (−1)`(w)ew(λ+ρ)∑w∈W (−1)`(w)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 characterschM(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, sincew·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

chM(w · λ) = chL(w · λ) +∑

y∈W ,y·λ≺w·λaw,y chL(y · λ)

where aw,y ∈ N>0, so the characters are unitriangularily related, meaningwe can create an inverse, giving

chL(w · λ) = chM(w · λ) +∑

y∈W ,y 6=wbw,y chM(y · λ)

with bw,y ∈ Z (not necessarily non-negative). If we specify w = 1, then weget

chL(λ) =∑y∈W

b1,y chM(y · λ) =∑y∈W

b1,yey(λ+ρ)

eρ∏α∈Φ+(1− e−α)

Now, by 9.15,

dimL(λ)µ = dimL(λ)w(µ) =⇒ w(chL(λ)) = chL(λ)

for all w ∈ W . Also,

si

eρ ∏α∈Φ+

(1− e−α)

= si

(∑w∈W

(−1)`(w)ew(ρ)

)= −

∑w∈W

(−1)`(w)ew(ρ)

46

and so

w

eρ ∏α∈Φ+

(1− e−α)

= (−1)`(w)eρ∏α∈Φ+

(1− e−α)

Therefore,

chL(λ) = w(chL(λ)) = w

∑y∈W

b1,yey(λ+ρ)

eρ∏α∈Φ+(1− e−α)

= (−1)`(w)∑y∈W

b1,yw(ey(λ+ρ))

eρ∏α∈Φ+(1− e−α)

tells us that

w

∑y∈W

b1,yey(λ+ρ)

= (−1)`(w)

∑y∈W

b1,yey(λ+ρ)

and since weλ = ewλ, we get b1,w−1y = (−1)`(w)b1,y from the linear inde-

pendence of the eλ’s. In particular, b1,w−1 = (−1)`(w) since b1,1 = 1 and so

b1,w = (−1)`(w−1) = (−1)`(w). Thus, since this is true for every w ∈ W , we

get

chL(λ) =

∑w∈W (−1)`(w)ew(λ+ρ)

eρ∏α∈Φ+(1− e−α)

=

∑w∈W (−1)`(w)ew(λ+ρ)∑w∈W (−1)`(w)ew(ρ)

�

Note that, for λ = 0,

chL(0) = e0 =⇒∑w∈W

(−1)`(w)ew(ρ) =

eρ ∏α∈Φ+

(1− e−α)

so the Weyl Denominator Formula is a special case of the Weyl CharacterFormula (although we used the Weyl Denominator Formula in our proof ofthe Weyl Character Formula).

10.9. Example. Let g = sl2(C). Then, W = S2 and ρ = 1. So, for n ∈ Z≥0,

chL(n) =(−1)0en+1 + (−1)1e−(n+1)

(−1)0e1 + (−1)1e−1=en+1 − e−n−1

e1 − e−1= en+en−2+· · ·+e−n

recovering our knowledge for finite dimensional irreducible sl2(C)-modules.

10.10. Theorem (Kostant’s Multiplicity Formula). Let λ be a dominantweight and µ a weight. Then,

dimL(λ)µ =∑w∈W

(−1)`(w)P(w(λ+ ρ)− (µ+ ρ))

10.11. Theorem (Weyl’s dimension formula). Let λ be a dominant weight.Then,

dimL(λ) =

∏α∈Φ+〈λ+ ρ, α〉∏α∈Φ+〈ρ, α〉

47

10.12. Example. Let g = sl2(C). Then, since Φ+ = {1} and ρ = 1,

dimL(n) =〈n+ 1, 1〉〈1, 1〉

= n+ 1

once again recovering knowledge from previous sections.

10.13. Example. More generally, for g = sln(C), let λ =∑n−1

k=1 λkεk. Then,

ρ = ω1 +ω2 + · · ·+ωn−1 = ε1 +(ε1 +ε2)+ · · ·+(ε1 + · · ·+εn−1) =

n−1∑k=1

(n−k)εk

and so

dimL(λ) =∏α∈Φ+

〈λ+ ρ, α〉〈ρ, α〉

=∏

1≤i<j≤n

〈∑n−1

k=1(λk + n− k)εk, εi − εj〉〈∑k

n=1(n− k)εk, εi − εj〉

=∏

1≤i<j≤n

(λi + n− i)− (λj + n− j)(n− i)− (n− j)

=∏

1≤i<j≤n

λi − λj + j − ij − i

In fact, the formula above is given by specializing the “Schur function”sλ(x1 . . . , xn) to sλ(1, . . . , 1). Furthermore, since

sλ(x1, . . . , xn) =∑

T∈SSYTn(λ)

xweight(T )

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

dimL(λ) = sλ(1, . . . , 1) = |SSYTn(λ)|

10.14. Remark. In fact, in type A, one can see the Weyl Character Formulagives a “Schur function” using the Jacobi Bialternate Formula characteriza-tion of Schur functions. In type A, W ∼= Sn and ΛW = spanZ{ε1, . . . , εn}.

Then, since eλ = eλ1ε1+···+λnεn , we get

∑w∈Sn

(−1)`(w)ew(λ+ρ) =∑w∈Sn

sgn(w)

n∏i=1

e(λi+n−i)εw(i)= det

e(λ1+n−1)ε1 e(λ1+n−1)ε2 · · · e(λ1+n−1)εn

e(λ2+n−2)ε1 e(λ2+n−2)ε2 · · · e(λ2+n−2)εn...

......

e(λn)ε1 e(λn)ε2 · · · e(λn)εn

Then, the initiated observe that this is the Vandermonde determinant ∆λ+ρ

where e(λi+n−i)εj corresponds to the variable xλi+n−1j . Thus, in our type A

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setting,

chL(λ) =

∑w∈Sn(−1)`(w)ew(λ+ρ)∑w∈Sn(−1)`(w)ew(ρ)

=∆λ+ρ

∆ρ= sλ(eε1 , eε2 , . . . , eεn)

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.

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