3 Universal enveloping algebras

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1 3 UNIVERSAL ENVELOPING ALGEBRAS 26 3 Universal enveloping algebras 3.1 Definition and first properties In this section, k will be an arbitrary field, not necessarily of characteristic zero. The idea is to introduce an associative unital k-algebra U(L) associated to a Lie algebra L, such that L and U(L) have the same representation theory. One could then use tools from representation theory of associative algebras to understand the representation theory of L. Recall that any associative unital k-algebra A can be seen as a Lie algebra via the bracket [, ] :A A A, (a, a ) aa a a. We will denote by Lie(A) the algebra A viewed as a Lie algebra. Definition 3.1. Let L be an arbitrary Lie algebra over the field k. A universal enveloping algebra of L (or simply enveloping algebra of L) is a pair (U(L),ϵ) where U(L) is an associative, unital k-algebra and ϵ : L Lie(U(L)) ahomomorphismofliealgebrassuchthatthefollowinguniversal property holds: for any associative unital k-algebra A and any homomorphism of Lie algebras ϕ : L Lie(A), thereexistsauniquehomomorphism ϕ : U(L) A of associative algebras such that ϕ = ϕ ϵ. L ϵ U(L) ϕ ϕ A Exercise. Show using the definition above that If L =0,thenU = k with the unique possible map ϵ is a universal enveloping algebra of L, If L = k, thenu = k[x], withthemapϵ : α αx is a universal enveloping algebra of L. Notice that the associative algebra k[x] is infinite dimensional! In fact, except in case L =0,auniversalenvelopingalgebrawillalwaysbeinfinite dimensional. Lemma 3.2 (Unicity of the universal enveloping algebra). Assume that (U 1,ϵ 1 ) and (U 2,ϵ 2 ) are two enveloping algebras of L. There is a canonical isomorphism ψ : U 1 U 2.

2 3 UNIVERSAL ENVELOPING ALGEBRAS 27 We therefore say that a universal enveloping algebra is unique "up to unique isomorphism" (but there may be more than one isomorphism of associative algebras U 1 U 2 ;the"canonical"inlemma3.2isheretoemphasize that such an isomorphism should be compatible with the maps ϵ 1, ϵ 2 ). Hence "unique up to unique isomorphism" means "unique up to unique isomorphism satisfying some compatibility conditions". We may therefore simply call a universal enveloping algebra "the" universal enveloping algebra. Proof. Since ϵ 2 : L Lie(U 2 ) is a morphism of Lie algebras from L into an associative k-algebra viewed as Lie algebra, the universal property of the universal enveloping algebra applied with U 1 as universal algebra yields a homomorphism of associative algebras ϵ 2 : U 1 U 2 such that ϵ 2 ϵ 1 = ϵ 2. Similarly by reversing the roles of U 1 and U 2,onegetsahomomorphismof associative algebras ϵ 1 : U 2 U 1 such that ϵ 1 ϵ 2 = ϵ 1.Wethenhavethat ϵ 1 ϵ 2 ϵ 1 = ϵ 1 ϵ 2 = ϵ 1, Hence f := ϵ 1 ϵ 2 : U 1 U 1 is a homomorphism of associative unital k- algebras such that f ϵ 1 = ϵ 1. But by the universal property of (U 1,ϵ 1 ), there should be a unique homomorphism with such a property, and f = id U1 satisfies it. Hence f = id U1. Similarly one shows that ϵ 2 ϵ 1 = id U2,hence ψ = ϵ 2 is the looked for isomorphism. Lemma 3.3. Let V be an abelian group, L aliealgebraoverk and ϵ : L U(L) auniversalenvelopingalgebraforl. Thereisabijection Structures of U(L) module on V Structures of representation of L over k on V, Where an U(L)-module V is made into a k-linear representation of L via α v =(α1 U(L) ) v, x v = ϵ(x) v, α k, x L, v V. Proof. AstructureofV as U(L)-module is by definition a homomorphism of rings ϕ : U(L) End(V ). The restriction of ϕ to k U(L) endows V with a structure of k-vector space. Hence ϕ can be seen as a morphism of associative unital k-algebras U(L) End k (V ), andthereforealsoasa homomorphism of Lie algebras Lie(U(L)) Lie(End k (V )) = gl(v ); we abuse notation and write this homomorphism again ϕ. The composition ϕ ϵ then yields a morphism of Lie algebras ϕ ϵ : L gl(v ). This is by definition a k-linear representation of L.

3 3 UNIVERSAL ENVELOPING ALGEBRAS 28 To show that our assigment is bijective, we give the inverse map. A k- linear representation of V is a pair (V,ρ) where V is a k-vector space and ρ : L gl(v ) = Lie(End k (V )) ahomomorphismofliealgebras. Bythe universal property of (U(L),ϵ), there exists a unique way of extending ρ to ahomomorphism ρ : U(L) End k (V ) of associative unital k-algebras. One then shows that the two maps are inverse of each other. We proved that if the universal enveloping algebra U(L) of L exists, then it is unique. We postpone the proof of existence to Subsection 3.3. Lemma 3.4. Let L be a Lie algebra with basis {X i } i I,whereI is totally ordered, and let U(L) be the universal enveloping algebra of L. Then U(L) is generated as k-vector space by the monomials ϵ(x i1 )ϵ(x i2 ) ϵ(x im ),where m Z 0, i 1 i 2 i m. Proof. Let U := ϵ(x i1 )ϵ(x i2 ) ϵ(x im ) m Z 0, (i 1,...,i m ) I m k vector space. It is clear that U is a k-subalgebra of U(L). Moreover, Im(ϵ) U. Hence consider ϕ : L Lie(U ), x ϵ(x). BytheuniversalpropertyofU(L), there is a unique homomorphism of associative unital k-algebras ϕ : U(L) U with ϕ = ϕ ϵ. Let ψ : U U be the inclusion. Then it is clear that ψ ϕ = ϵ. Hencewehavethat ψ ϕ ϵ = ψ ϕ = ϵ, hence by the universal property one must have ψ ϕ = id U(L). Hence ψ is surjective, but since it is also injective, one has that U = U. Now the products of the form ϵ(x i1 )ϵ(x i2 ) ϵ(x im ), m>1, withi j > i j+1 for some 1 j < m can be written as linear combinations of order monomials using inductively the relation ϵ(x ij )ϵ(x ij+1 )=ϵ([x ij,x ij+1 ]) + ϵ(x ij+1 )ϵ(x ij ). 3.2 Poincaré-Birkhoff-Witt Theorem Theorem 3.5 (Poincaré-Birkhoff-Witt). Let L be a Lie algebra over a field k with universal enveloping algebra (U(L),ϵ), {X i } i I abasisofl, withi

4 3 UNIVERSAL ENVELOPING ALGEBRAS 29 totally ordered. The set A := {ϵ(x i1 )ϵ(x i2 ) ϵ(x im ) i 1 i 2 i m, (i j ) I m } m Z 0 is a k-basis of U(L). Inparticular,ϵ : L U(L) is injective. Proof. Thanks to Lemma 3.3, it suffices to show that the elements of A are linearly independent. For M =(i 1,...,i m ) I m,setl(m) =m. Let V be the vector space with basis m Z 0 {z M M =(i 1 i 2 i m ) I m }. In the sequel we write M for the set of ordered monomials in the elements of I. For i I, wewritei M if i i 1 and we set im := (i, i 1,...,i m ). We claim that V can be equipped with a structure of L-module, such that X i z M = z im whenever i M. Thatis,thereisak-linear map L V V, (a, v) a v such that X i z M = z im whenever i M, andsuchthatthe induced map L gl(v ), a {v a v} is a morphism of Lie algebras. Since {X i } i I is a k-basis of L and {z M } a k-basis of V,itsufficestogive aformulaforx i z M and check that it defines an L-module structure on V. Such a formula is defined inductively on the lexicographical order on M I, where M is ordered by l( ). That is, we fix (M,i) M I and assume that X j z N is already defined if l(n) <l(m), orifl(n) =l(m) and j<i. We furthermore assume that in that in case (N,j) < (M,i), then X j z N z P l(p ) l(n)+1 k vector space. (4) To start the induction, we just define X i z = z (i) for all i I. Wethen set { zim if i M X i z M := X j (X i z N )+[X i,x j ] z N if M = jn,i > j. Notice that all the operations occurring in the second row are alreadydefined since l(n) =l(m) 1,andX i z N z P l(p ) l(n)+1=l(m) k vector space and j<i. First step. We show that condition 4 is still satisfied. This is clear for the first case since l(im) =l(m)+1and for the second by induction. Second step. We show that such an inductive formula defines an operation of L on V,thatis,weneedtoshowthatforanyi, j I and N M, the following equality holds: [X i,x j ] z N = X i (X j z N ) X j (X i z N ).

5 3 UNIVERSAL ENVELOPING ALGEBRAS 30 The proof is by induction on l(n) and min(i, j). Ifi = j, thenbothsides are equal to zero. If j i, thenbyantisymmetrywecanassumethati>j. If j N, thenx j z N = z jn,hencewehave X i (X j z N )=X i z jn = by definition X j (X i z N )+[X i,x j ] z N, hence the formula holds. Now assume that j N. Write N = kk, with k<j<i.hencewehavez N = X k z K.Wehavetoshowthat X i X j X k z K X j X i X k z K =[X i,x j ]X k z K, and we denote this equality simply by (ijk). Weknowthatforanyx, y L, the equality xy z K = yx z K +[x, y] z K holds since l(k) =l(n) 1. Therefore,therighthandsideof(ijk) can be rewritten as [X i,x j ]X k z K = X k [X i,x j ] z K + [[X i,x j ],X k ] z K = X k X i X j z K X k X j X i z K + [[X i,x j ],X k ] z K. Notice that the equalities (jki) and (kij) hold by induction since k<j<i. By adding the right hand sides of (ijk), (jki) and (kij) we get X i X j X k z K X j X i X k z K +X j X k X i z K X k X j X i z K +X k X i X j z K X i X k X j z K, While by adding the right hand sides as rewritten above we get X k X i X j z K X k X j X i z K + [[X i,x j ],X k ] z K + X i X j X k z K X i X k X j z K + [[X j,x k ],X i ] z K + X j X k X i z K X j X i X k z K + [[X k,x i ],X j ] z K. The sum of the three terms with brackets vanishes thanks to the Jacobiidentity. Hence the sum of the right hand sides of (ijk), (jki) and (kij) and the sum of the left hand sides are the same. Since (jki) and (kij) hold, it implies that (ijk) must also hold. Hence V is an L-module. Third step. For M =(i 1,i 2,...,i m ) M,weset X M := ϵ(x i1 )ϵ(x i2 ) ϵ(x im ).

6 3 UNIVERSAL ENVELOPING ALGEBRAS 31 We want to show that X M z = z M. The proof is by induction on l(m). If l(m) =0,thenX M = X =1,and 1 z = z.nowassumethatl(m) > 0, writem = in, i N. Onehasthat X M z = ϵ(x i )X N z = ϵ(x i ) z N = X i z N = z in = z M. Fourth step. We conclude the proof by showing that the X M, M M are linearly independent. Assume that 0= M M c MX M,wheretheC M are scalars, only finitely many of which are nonzero. Then one has that ( ) 0= c M z M z = c M X M z = c M z M V, M M M M M M hence c M =0for any M Msince {z M } M M is a k-basis of V. 3.3 Tensor algebra and existence of universal enveloping algebras Definition 3.6. Let V be a vector space over a field k. Thetensor algebra T (V ) of V is the associative, unital k-algebra defined as vector space by T (V )= r Z 0 V r = k V (V V ) (V V V )..., endowed with a product defined by (v 1 v 2 v n ) (w 1 w m )=v 1 v n w 1 w m, and extended k-bilinearly. Lemma 3.7 (Universal property of the tensor algebra). Let V be a k-vector space and c : V T (V ) the inclusion (as k-linear map). If A is an associative unital k-algebra and ϕ : V A a k-linear map, then there exists exactly one morphism of associative, unital k-algebras ˆϕ : T (V ) A such that ϕ =ˆϕ c. c V T (V ) ϕ ˆϕ A

7 3 UNIVERSAL ENVELOPING ALGEBRAS 32 Proof. Since the set V generates T (V ) as associative unital k-algebra, there can be at most one such morphism. One the other hand, ˆϕ can be defined explicitly by setting ˆϕ(v 1 v n )=ϕ(v 1 )ϕ(v 2 ) ϕ(v n ). Proposition 3.8. Let L be a Lie algebra over k. ConsidertheidealI(L) of T (L) generated by the elements of the form (x y y x [x, y]), x, y L. Then the associative unital algebra U(L) := T (L)/I(L) together with the map ϵ : L c T (L) π U(L), whereπ is the canonical quotient map, is a universal enveloping algebra of L. Proof. One first has to show that ϵ = π c is a morphism of Lie algebras, since c is only a k-linear map. We have ϵ([x, y]) = π([x, y]) = π(x y y x) =π(x)π(y) π(y)π(x) = ϵ(x)ϵ(y) ϵ(y)ϵ(x) =[ϵ(x),ϵ(y)], Hence ϵ is a morphism of Lie algebras. It remains to show that U(L) satisfies the universal property of the universal enveloping algebra. Hence let ϕ : L Lie(A) be a morphism of Lie algebras, where A is an associative, unital k-algebra. In particular, ϕ is a k-linear map. By the universal property of the tensor algebra T (L), there exists a unique morphism of associative unital k-algebras ˆϕ : T (L) A such that ˆϕ c = ϕ. Butϕ is a morphism of Lie algebras, hence ˆϕ(x y y x [x, y]) = ϕ(x)ϕ(y) ϕ(y)ϕ(x) ϕ([x, y]) = 0, which implies that ˆϕ(I) = 0. Henceˆϕ factors through U(L) via a map ϕ : U(L) A, thatis, ˆϕ = ϕ π. T (L) π U(L) ˆϕ ϕ A It remains to show that ϕ is unique. Assume that we have another morphism of associative unital k-algebras ψ : U(L) A such that ψ π c = ϕ. Since π is surjective, it suffices to show that ψ π =ˆϕ. Butwehave ψ π c = ϕ π c =ˆϕ c, and since L generates T (L) as k-algebra, it follows that ψ π and ˆϕ agree on the set c(l) of generators of T (L), hencetheymustagreeeverywhere. We therefore have ψ π =ˆϕ, whichconcludes.

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