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Home , Adjoint representation, Algebra homomorphism, Dual representation, Structure constants, Trivial representation

REPRESENTATIONS OF sl2(C)

JAY TAYLOR

Basic Definitions and Introduction Definition. A Lie g is a over a field k with an associated [·, ·]: g × g → g, such that the following hold: • [x, x] = 0 for all x ∈ g, • [x, [y, z]] + [y, [z, x]] + [z, [x, y]] = 0 for all x, y, z ∈ g. Note. We call the latter axiom of the above definition the . The idea of this axiom is to be a replacement for associativity, as we do not have that a is an . We refer to the bilinear map [·, ·] as the Lie bracket of g. Example. (a) Let g be any vector space over any field k. Then we can endow g with the trivial bracket operation [x, y] = 0 for all x, y ∈ g. We refer to this as an abelian Lie algebra. (b) Let k = R and let g = R3. We define a product structure on g using the standard vector product x ∧ y for all x, y ∈ g. In other words if x, y ∈ g such that x = (x1, x2, x3) and y = (y1, y2, y3) then

[x, y] = (x2y3 − x3y2, x3y1 − x1y3, x1y2 − x2y1). (c) Let V be any finite-dimensional vector space over a field k. We define the general linear Lie algebra gl(V ) to be the vector space of all linear maps from V to V , endowed with the bracket

[x, y] = x ◦ y − y ◦ x for all x, y ∈ gl(V ). (d) We now define a matrix analogue for the Lie algebra in example (c). Let k be any field and let gl(n, k) be the vector space of all n × n matrices defined over k. Then gl(n, k) is a Lie algebra with Lie bracket given by

[x, y] = xy − yx for all x, y ∈ gl(n, k), i.e. the commutator bracket. Note that a for gl(n, k) as a vector space is given by the n × n matrices eij which have entry 1 in the ijth position and zeros elsewhere. We then see that the commutator bracket is given by

[eij, ek`] = δjkei` − δi`ekj,

where δij is the Kronecker delta. 1 2 JAY TAYLOR

(e) Let k be any field and sl(2, k) = {x ∈ gl(2, k) | tr(x) = 0} ⊂ gl(2, k) be the vector subspace of gl(2, k) whose elements have 0. Now if x, y ∈ sl(2, k) then we will have [x, y] = xy − yx ∈ sl(2, k) hence the commutator brackets gives sl(2, k) a Lie algebra structure. As a vector space it can be shown that sl(2, k) has a basis given by

0 1 0 0 1 0  e = f = h = . 0 0 1 0 0 −1 These elements have Lie bracket relations [e, f] = h,[h, f] = −2f,[h, e] = 2e. (f) Let A be an associative algebra over a field k. Clearly A is a vector space over k and we can give it the structure of a Lie algebra by endowing it with the commutator bracket [x, y] = xy − yx for all x, y ∈ A. Definition. Let g be a Lie algebra over a field k then a derivation D : g → g is a which satisfies the Leibniz rule

D([x, y]) = [D(x), y] + [x, D(y)] for all x, y ∈ g. Let g be a Lie algebra over a field k then Der(g) the vector space of all derivations of g is a Lie algebra whose Lie bracket is given by the commutator bracket [D1,D2] = D1 ◦ D2 − D2 ◦ D1 for all D1,D2 ∈ Der(g). We define a very important derivation known as the adjoint operator. Let x ∈ g then we define a map adx : g → g by adx(y) = [x, y] for all y ∈ g.

Claim. For any Lie algebra g we have adx ∈ Der(g) for all x ∈ g.

Proof. First of all we must show that adx is linear. For any α, β ∈ k and y, z ∈ g we have

adx(αy + βz) = [x, αy + βz] = α[x, y] + β[x, z] = α adx(y) + β adx(z). Hence the map is linear. We now show that this map satisfies the Liebniz rule. For all y, z ∈ g we have

adx([y, z]) = [x, [y, z]] = −[y, [z, x]] − [z, [x, y]], = [y, [x, z]] + [[x, y], z],

= [adx(y), z] + [y, adx(z)].  Definition. For any Lie algebra g we call a derivation D ∈ Der(g) an inner derivation if there exists an element x ∈ g such that D = adx. Any derivation of g which is not an inner derivation is called an outer derivation.

Note that the derivation adx is not to be confused with the adjoint . We define the adjoint homomorphism to be the map ad : g → gl(g) given by x 7→ adx for all x ∈ g. However, for this to make sense we must define what we mean by a Lie algebra homomorphism. REPRESENTATIONS OF sl2(C) 3

Definition. Let g1, be Lie defined over a common field k. Then a homomor- phism of Lie algebras ϕ : g1 → g2 is a linear map of vector spaces such that ϕ([x, y]) = [ϕ(x), ϕ(y)], i.e. it preserves the Lie bracket. Claim. The map ad : g → gl(g) is a homomorphism of Lie algebras. Proof. Clearly this map is linear by the linearity properties of the Lie bracket. Hence to show this is a homomorphism we must show that ad[x,y] = [adx, ady] = adx ◦ ady − ady ◦ adx for all x, y ∈ g. We do this by showing equivalence for all z ∈ g

ad[x,y](z) = [[x, y], z] = −[z, [x, y]], = [x, [y, z]] + [y, [z, x]],

= adx([y, z]) − ady([x, z]), = (adx ◦ ady − ady ◦ adx)(z).  Definition. A representation of a Lie algebra g is a pair (V, ρ) where V is a vector space over k and ρ : g → gl(V ) is a Lie algebra homomorphism. Example. (a) Take V to be any vector space over k and ρ = 0 to be the zero map. We call this the trivial representation of g. (b) The adjoint homomorphism of g is a representation of g with V = g and ρ = ad. We call this the of g. Alternatively instead of thinking of representations we can also consider modules for a Lie algebra g. Definition. Let g be a Lie algebra over a field k.A g-module is a pair (V, ·) where V is a vector space and · : g × V → V is a map satisfying the following conditions for all x, y ∈ g, v, w ∈ V and λ, µ ∈ k. • (λx + µy) · v = λ(x · v) + µ(y · v), • x · (λv + µw) = λ(x · v) + µ(x · w), • [x, y] · v = x · (y · v) − y · (x · v).

The Universal Enveloping Algebra In the beginning one of the main stumbling blocks in the of Lie algebras was that the Lie algebra is not an associative algebra over k. This proves quite irritating as we already know a lot about the representation theory of associative algebras and we would like to apply this to Lie algebras. Enter the Universal Enveloping Algebra.

Definition. Let g be a Lie algebra over k with basis xi and Lie bracket [·, ·] defined P k by [xi, xj] = k cijxk. The universal enveloping algebra U(g) is the associative algebra P k generated by the xi’s with the defining relations xixj − xjxi = k cijxk. We call the k elements cij the of U(g). 4 JAY TAYLOR

The Universal Enveloping Algebra of g is an associative algebra generated as “freely” as possible by g, (to quote [Hum78]), with respect to the bracket relations. This algebra is always infinite dimensional unless the Lie algebra is abelian. Example.

(a) Let g = span{x1} be a 1-dimensional Lie algebra over a field k. The only Lie bracket relation on g comes from [x1, x1] = 0. Therefore we only have one structure 1 constant c11 = 0 and hence U(g) is the polynomial algebra k[x1] in one variable. (b) Consider sl(2, k) for any field k. Now U(sl(2, k)) is a free algebra generated by e, f, g, which is subject to the relations

ef − fe = h hf − fh = −2f he − eh = 2e. We know that sl(2, k) = span{f} ⊕ span{h} ⊕ span{e}. Therefore U(sl(2, k)) will contain U(span{f}), i.e. all polynomials in f. Similarly we will get that U(sl(2, k)) wil contain all polynomials in h and all polynomials in e. As well as this we will get U(sl(2, k)) contains all products of these elements. Theorem (Poincar´e-Birkoff-Witt or PBW Theorem). Let g be a finite dimensional Lie algebra over a field k and let x1, . . . , xn be an ordered basis for g. Then the universal enveloping algebra U(g) will have a basis given by

a1 an {x1 . . . xn | a1, . . . , an > 0}. Proof. See section 17.4 of [Hum78].  We note that we have only stated the PBW Theorem for finite dimensional Lie algebras, (for ease of notation), but this works perfectly well for infinite dimensional Lie algebras as well. See section 17.3 of [Hum78] for more details.

Corollary. The elements x1, . . . , xn in U(g) are linearly independent and hence g is a vector subspace of U(g).

Proof. Clear from the PBW Theorem.  Corollary. Let h ⊆ g be a Lie subalgebra of g then U(h) is a subalgebra of U(g). Proof. Take a basis for h and extend it to a basis for g then this is clear from the PBW Theorem.  Proposition. Let g be a finite dimensional Lie algebra and let U(g) be its universal en- veloping algebra. Then there is a bijective correspondence between g-modules and U(g)- modules. Furthermore, under this correspondence, a g-module is irreducible if and only if its corresponding U(g) module is irreducible.

Proof. See Lemma 15.10 of [EW06].  REPRESENTATIONS OF sl2(C) 5 The above Proposition makes our life simpler as we only have to find irreducible U(g)- modules, which we are much more adept at doing. We make a few remarks about the universal enveloping algebra before continuing. Let g be a Lie algebra over a field k and let A be an associative algebra over k. We have that A has the structure of a Lie algebra using the commutator bracket. Let ϕ : g → A be a homomorphism of Lie algebras then we must have that ϕ factors uniquely through the universal enveloping algebra. This can be most easily seen in the following commutative diagram.

ϕ g > A ∧ ∃! > U(g) Hence the universal enveloping algebra is “universal” in the sense that it satisfies this . The second remark we make is that the definition we have given here depends on the choice of a basis for g. This isn’t very convenient and it can be shown that the following definition is equivalent to the definition given above. Definition. Let g be a Lie algebra over a field k and let T g be the of g. Consider the ideal I = hx ⊗ y − y ⊗ x − [x, y] | x, y ∈ gi of the tensor algebra. Then the universal enveloping algebra of g is defined to be U(g) = T g/I. We finish this section with two more definitions.

Definition. Let g be a Lie algebra over a field k and let (V, ρV ) and (W, ρW ) be two representations of g. Then the of these representations is the pair (V ⊗ W, ρV ⊗W ) where we define ρV ⊗W to be ρV ⊗W (x) = ρV (x) ⊗ Id + Id ⊗ ρW (x) for all x ∈ g.

Definition. Let g be a Lie algebra over a field k and let (V, ρV ) be a representation of g. ∗ ∗ Then the of g is the pair (V , ρV ∗ ) where V is the dual vector space ∗ to V and ρV ∗ is given by ρV ∗ (x) = −ρV (x) for all x ∈ g.

Irreducible Representations of sl(2, C)

Let (V, ρV ) be a finite dimensional representation, (note not necessarily irreducible), of sl(2, C) = span{f, h, e}, we consider V as a g module by setting x · v = ρV (x)v for all x ∈ g. Let λ ∈ k then we define a vector subspace of V by

Vλ = {v ∈ V | h · v = λv}.

If λ is not an eigenvalue of ρV (h) then Vλ = {0}. Whenever Vλ 6= {0} we call λ a weight of h in V and Vλ its associated weight space.

Proposition. We have that for any v ∈ Vλ (a) e · v = 0 or e · v is an eigenvector of h with eigenvalue λ + 2, 6 JAY TAYLOR

(b) f · v = 0 or f · v is an eigenvector of h with eigenvalue λ − 2.

Proof. Recall the relation [h, e] = he − eh = 2e in sl(2, C). Hence for any v ∈ Vλ we have

h · (e · v) = (eh + 2e) · v = λ(e · v) + 2(e · v) = (2 + λ)e · v. Therefore either e · v = 0 or e · v is an eigenvector of h. Similarly recalling the relation [h, f] = hf − fh = −2f we obtain for any v ∈ Vλ

h · (f · v) = (fh − 2f) · v = λ(f · v) − 2(f · v) = (λ − 2)f · v.  Given this proposition we can describe the action of the basis e, f, h of sl(2, C) on the eigenspaces of h in V in a fairly simple pictorial fashion.

e e e e e

{0} V−λ V−λ+2 ··· Vλ−2 Vλ {0}

f f f f f h h h h

Note that in the following discussion it will become clear why we have only included the eigenspaces from −λ to λ for V on this diagram.

Proposition. Let V be a finite dimensional sl(2, C) module then there exists an eigenvector w ∈ V for h such that e · w = 0.

Proof. We are working over an algebraically closed field, which means ρV (h) has at least one eigenvalue and hence at least one eigenvector, say v. The vectors v, e · v, e2 · v, . . . , if non-zero, form an infinite sequence of linearly independent eigenvectors for ρV (h) in V by the above Lemma. However V is finite dimensional so these cannot all be non-zero, k k+1 k hence there exists a k > 0 such that e · v 6= 0 and e · v = 0. Setting w = e we have h · w = (λ + 2k)w and e · w = 0. 

We refer to an eigenvector w ∈ V with the property that Vλ 6= {0} and Vλ+2 = {0} as a maximal vector of weight λ.

Lemma. Let V be a finite dimensional irreducible module for sl(2, C) and let w be a ` `+1 maximal vector of weight λ. Let ` > 0 be such that f · w 6= 0 and f · w = 0 then the ` vectors w, f · w, . . . , f · w form a basis for V for some ` > 0. Proof. We aim to show that span{w, f · w, . . . , f ` · w} = W ⊆ V is a non-zero submodule of V . The vectors w, f ·w, . . . , f ` ·w are linearly independent because they are eigenvectors for ρV (h) with distinct eigenvalues. Clearly these vectors are invariant under the action of f and they are invariant under the action of h as they are eigenvectors for ρV (h). Hence we must check that they are invariant under the action of e. REPRESENTATIONS OF sl2(C) 7 Now e · w = 0 ∈ W as w was chosen this way. Recall the relation [e, f] = ef − fe = h in sl(2, C) then we have

e · (f · w) = (h + fe) · w = h · w + f · (e · w) = λw + f · 0 = λw, e · (f 2 · w) = (h + fe) · f · w = (λ − 2)f · w + λf · w = 2(λ − 2 + 1)f · w, . . e · (f i · w) = i(λ − i + 1)f i−1 · w. So we can see that e will take every vector f i · w to some multiple of f i−1 · w. We now prove the final statement by induction on i, note that we know e · w = 0 and hence the statement is true for i = 0. For the case i + 1 we have

e · (f i+1 · w) = (h + fe) · f i · w = (λ − 2i)f i · w + i(λ − i + 1)f i · w, = (λ + iλ − i2 − i)f i · w, = [(i + 1)λ − (i + 1)i]f i · w, = (i + 1)(λ − i)f i · w. Therefore W is a non-zero submodule of V and as V is irreducible this means V = W .  Corollary. We have λ = ` in the above Lemma. Hence dim V = ` + 1 and every weight of h is a non-negative integer which is dim V − 1.

`+1 ` Proof. Recall that f · w = 0 and f · w 6= 0 for some ` > 0 then we have

`+1 ` 0 = e · (f · w) = (` + 1)(λ − `)f · w ⇒ (` + 1)(λ − `) = 0 ⇒ ` = λ.  Note that we call the weight of a maximal vector the highest weight of V , this is the weight such that dim V = λ + 1.

Corollary. Let µ be a weight with respect to h and Vµ the corresponding weight space. Then as a vector space we have Vµ is one dimensional. i i Proof. The action of h on the basis of V is h · (f · w) = (λ − 2i)f · w.  In the Lemma we wrote down an explicit basis for V and said exactly how e, f and h act on this basis. Therefore the irreducible module V is determined explicitly by its highest weight λ. Therefore we have shown the following theorem.

Theorem. Let V be a finite dimensional irreducible module for sl(2, C). (a) Relative to the action of h, V is the direct sum of weight spaces Vµ, where µ = λ, λ − 2,..., −(λ − 2), −λ where dim V = λ + 1 and dim µ = 1 for each µ. (b) V has, up to non-zero scalar multiples, a unique maximal vector whose weight, called the highest weight of V , is λ. 8 JAY TAYLOR

(c) Let w, f · w, . . . , f λ · w be the basis of V prescribed in the Lemma then sl(2, C) acts on this basis, for 0 6 i 6 λ in the following way

h · (f i · w) = (λ − 2i)f i · w, f · (f i · w) = f i+1 · w, e · (f i · w) = i(λ − i + 1)f i−1 · w. In particular there exists at most one irreducible sl(2, C)-module, up to , (λ) of each possible dimension λ + 1 for λ > 0. We refer to this module as V . We comment that there is a concrete way to construct the irreducible sl(2, C) modules (λ) V . Let V = C[X,Y ] be the vector space of polynomials in two variables. For each λ > 0 we let Vλ be the vector subspace of all homogeneous polynomials of degree λ. This has a basis given by the monomials Xλ,Xλ−1Y,...,XY λ−1,Y λ. We turn this vector subspace into a module for sl(2, C) by defining a Lie algebra homomorphism ϕ : sl(2, C) → gl(Vλ) in the following way

∂ ∂ ∂ ∂ ϕ(e) = X ϕ(f) = Y ϕ(h) = X − Y . ∂Y ∂X ∂X ∂Y It is left as an exercise to verify that this is indeed a representation of sl(2, C) and to construct an explicit isomorphism between this description of V (λ) and the description given above.

Low Dimensional Irreducible Representations (0) (0) Let λ = 0 then V has dimension 1 and V = V0 = {v ∈ V | h · v = 0}. We know there is only one irreducible representation of dimension 1, hence this clearly must be the trivial representation. Let λ = 1 then V (1) has dimension 2. Consider the natural representation of sl(2, C) acting on C2. We have the action of h on the natural basis x, y of C2 is

1 0  1 1 1 0  0  0  = = . 0 −1 0 0 0 −1 1 −1 Therefore h has eigenvalues ±1 on the natural basis of C2 and we have C2 = Cy ⊕ Cx = (1) V−1 ⊕ V1, which means V is isomorphic to the natural representation of sl(2, C). Note that the highest weight vector is clearly given by x or some scalar multiple of x. Let λ = 2 then V (2) has dimension 3. Recall the adjoint representation of sl(2, C) given by x 7→ adx for all x ∈ sl(2, C). Now the action of adh on the basis of sl(2, C) is given by

adh(f) = −2f adh(h) = 0 adh(e) = 2e.

Clearly as a vector space we have sl(2, C) = V−2 ⊕V0 ⊕V2 = span{f}⊕span{h}⊕span{e}, which correspond to the generalised eigenspaces of h. Also e is a highest weight vector for REPRESENTATIONS OF sl2(C) 9

(2) (2) V as ade(e) = 0 and a basis for V = sl(2, C) is given by e, f · e = [f, e] = −h and f 2 · e = −[f, h] = 2f.

Complete Reducibility of Finite Dimensional Representations We would like to prove the following theorem, which is attributed to Weyl.

Theorem (Weyl’s Theorem). let g be a complex . Every finite- dimensional representation of g is completely reducible.

Proof. See section 6.3 of [Hum78]. 

We will not prove this in general as the proof is quite long but will instead give an easier proof for representations of sl(2, C). We first introduce the so called Casimir operator of sl(2, C), which we define to be

h2 c = ef + fe + . 2 The Casimir operator was first discovered by embedding the Lie algebra R3 with the vector into sl(2, C), (see Exercise 8.7 in [EW06]). This operator has relationships with angular momentum in Quantum mechanics. See the wikipedia entry on for more details.

Lemma. Let M be a finite dimensional sl(2, C) module. Then c : M → M is a , in other words c(x · m) = x · c(m) for all x ∈ g and m ∈ M.

Proof. It is clearly sufficient to do this just for the basis of sl(2, C). So we have for e that

 h2  eh2 e · ef + fe + = e2f + efe + , 2 2 eh2 = eh + 2efe + , 2

using the identity e2f = eh + efe

heh = 2efe + , 2 using the identity 2eh = heh − eh2. 10 JAY TAYLOR

Now considering multiplication on the other side we have  h2  h2e ef + fe + · e = efe + fe2 + , 2 2 h2e = 2efe − he + , 2 heh = 2efe + , 2 using similar identities as before. Now considering the action of f we have

 h2  fh2 f · ef + fe + = fef + f 2e + , 2 2 fh2 = 2fef − fh + , 2 using the identity f 2e = fef − fh hfh = 2fef + , 2 using the identity 2fh = fh2 − hfh.

Now considering multiplication on the other side we have  h2  h2f ef + fe + · f = ef 2 + fef + , 2 2 h2f = hf + 2fef + , 2 hfh = 2efe + , 2 using similar identities as before. To prove this for h we recall the identities he − eh = 2e  hef − ehf = 2ef  ⇒ ⇒ hef = efh. hf − fh = −2f ehf − efh = −2ef Similarly we obtain he − eh = 2e  fhe − feh = 2fe  ⇒ ⇒ hfe = feh. hf − fh = −2f hfe − fhe = −2fe This gives us hef + hfe = efh + feh and so it’s clear to see that

 h2  h3 h3  h2  h · ef + fe + = hef + hfe + = efh + feh + = ef + fe + · h. 2 2 2 2  REPRESENTATIONS OF sl2(C) 11 We comment briefly that the Casimir operator is an element in the centre of the universal enveloping algebra of sl(2, C). By Schur’s Lemma we must have that c acts as a scalar on the irreducible modules V (λ).

(λ) Proposition. Let V be an irreducible sl(2, C) module with highest weight λ > 0 then we (λ) 1 have c(v) = cλv for all v ∈ V , where cλ = 2 λ(λ + 2). Proof. Let w ∈ V (λ) be the highest weight vector then we have

 h2  h2 · w c(w) = ef + fe + · w = ef · w + fe · w + , 2 2 λ2w = λw + 0 + , 2 λ(λ + 2) = w. 2 1 Hence cλ = 2 λ(λ + 2). 

Let M be any finite dimensional sl(2, C) module and let µ1, . . . , µr be the distinct eigen- values of c acting on M. Now, as c is a module homomorphism, we will have c − µi1M is a module homomorphism for each 1 6 i 6 r. Clearly ker(c − µi1M ) is a submodule of M for each 1 6 i 6 r, hence for some multiplicities mi we will have a decomposition

r M mi M = ker(c − λi1M ) . i=1 In other words to express M as a direct sum of irreducible modules we can assume that c has only one eigenvalue on M.

Proposition. Let M be a finite dimensional sl(2, C) module then any irreducible submodule of M is isomorphic to V (λ) for some specific highest weight λ. Proof. By our above comment we can assume that c has only one eigenvalue on the module M. Let U be an irreducible submodule of M such that U is isomorphic to V (λ) for some (λ) 1 highest weight λ. We know c acts on V as multiplication by cλ = 2 λ(λ + 2) and hence 1 we can assume that the eigenvalue of c on M is µ = 2 λ(λ + 2). Assume W ∼= V (λ0) is another irreducible submodule of M then we must have c acts on 1 0 0 W as multiplication by 2 λ (λ + 2). However as cλ is the unique eigenvalue of c on M we must have

1 1 λ(λ + 2) = λ0(λ0 + 2) ⇒ λ = λ0 ⇒ U ∼= W ∼= V (λ). 2 2  Proposition. Let M be a finite dimensional sl(2, C) module and N a submodule of M, then any irreducible submodule of M/N is isomorphic to V (λ). 12 JAY TAYLOR

m Proof. As before we can assume M = ker(c − µ1M ) for some m > 0 and µ the unique eigenvalue of c on M. Now we can consider the action of c on the quotient space M/N and it is clear that

m (c − µ1M ) (v + N) = 0 for all v ∈ M. Therefore c has just one eigenvalue on the quotient space M/N, namely µ, and so as in the (λ) previous proposition we have any irreducible submodule of M/N is isomorphic to V .  We now put all these ingredients together to prove our main result. Theorem. Any finite dimensional sl(2, C) module is completely reducible. Proof. Let M be a finite dimensional module of sl(2, C) and assume that c has only one eigenvalue on M. Suppose U is a maximal proper completely reducible submodule of M then the quotient M/U is non-zero. So M/U must have an irreducible submodule which is isomorphic to V (λ) for some highest weight λ. Looking at the largest eigenvalue of h appearing in this submodule tells us that there exists a eigenvector for h with eigenvalue λ. We will now use the fact, (proved in Theorem 9.16 of [EW06]), that representations of Lie algebras preserve Jordan decompositions. So as h is a diagonal matrix we will have that h will act diagonalisably on M and so there exists a h eigenvector v ∈ M/U such that h · v = λv. If e · v 6= 0 then e · v would be a h eigenvector with eigenvalue λ + 2 but this contradicts the maximality of λ. Hence e · v = 0 and v is a highest weight vector. Let W be the submodule of M generated by v then W is irreducible. As w 6∈ U the irreduciblity of W implies that U ∩ W = {0} and so U ⊕ W is a larger completely reducible submodule of M, a contradiction.  References [EW06] Karin Erdmann and Mark J. Wildon. Introduction to Lie Algebras. Springer Undegraduate Math- ematics Series. Springer-Verlag, 2006. [Hum78] James E. Humphreys. Introduction To Lie algebras and Representation Theory. Number 9 in Graduate Texts in . Springer-Verlag, 1978.