REPRESENTATIONS OF sl2(C) JAY TAYLOR Basic Definitions and Introduction Definition. A Lie algebra g is a vector space over a field k with an associated bilinear map [·; ·]: g × g ! g, such that the following hold: • [x; x] = 0 for all x 2 g, • [x; [y; z]] + [y; [z; x]] + [z; [x; y]] = 0 for all x; y; z 2 g. Note. We call the latter axiom of the above definition the Jacobi Identity. The idea of this axiom is to be a replacement for associativity, as we do not have that a Lie algebra is an associative algebra. 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 2 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 2 g. In other words if x; y 2 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 commutator bracket [x; y] = x ◦ y − y ◦ x for all x; y 2 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 2 gl(n; k); i.e. the commutator bracket. Note that a basis for gl(n; k) as a vector space is given by the n × n unit 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) = fx 2 gl(2; k) j tr(x) = 0g ⊂ gl(2; k) be the vector subspace of gl(2; k) whose elements have trace 0. Now if x; y 2 sl(2; k) then we will have [x; y] = xy − yx 2 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 2 A. Definition. Let g be a Lie algebra over a field k then a derivation D : g ! g is a linear map which satisfies the Leibniz rule D([x; y]) = [D(x); y] + [x; D(y)] for all x; y 2 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 2 Der(g). We define a very important derivation known as the adjoint operator. Let x 2 g then we define a map adx : g ! g by adx(y) = [x; y] for all y 2 g. Claim. For any Lie algebra g we have adx 2 Der(g) for all x 2 g. Proof. First of all we must show that adx is linear. For any α; β 2 k and y; z 2 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 2 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 2 Der(g) an inner derivation if there exists an element x 2 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 homomorphism. We define the adjoint homomorphism to be the map ad : g ! gl(g) given by x 7! adx for all x 2 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; g2 be Lie algebras defined over a common field k. Then a homomor- phism of Lie algebras ' : g1 ! g2 is a linear map of vector spaces such that '([x; y]) = ['(x);'(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 2 g. We do this by showing equivalence for all z 2 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 adjoint representation 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 2 g, v; w 2 V and λ, µ 2 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 representation theory 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 structure constants 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 = spanfx1g 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) = spanffg ⊕ spanfhg ⊕ spanfeg. Therefore U(sl(2; k)) will contain U(spanffg), 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 fx1 : : : xn j a1; : : : ; an > 0g: 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).