LIE ALGEBRAS 1 Definition of a Lie algebra k is a fixed field. Let L be a k-vector space (or vector space). We say that L is a k-algebra (or algebra) if we are given a bilinear map L×L −! L denoted (x; y) 7! xy. If L; L0 are algebras, an algebra homomorphism f : L −! L0 is a linear map such that f(xy) = f(x)f(y) for all x; y. We say that f is an algebra isomorphism if it is an algebra homomorphism and a vector space isomorphism. In this case f −1 : L0 −! L is an algebra isomorphism. If L is an algebra, a subset L0 ⊂ L is a subalgebra if it is a vector subspace and x 2 L0; y 2 L0 =) xy 2 L0. Then L0 is itself an algebra in an obvious way. If L is an algebra, a subset I ⊂ L is an ideal if it is a vector subspace; x 2 L; y 2 I =) xy 2 I; x 2 I; y 2 L =) xy 2 I. Then I is a subalgebra. Moreover, the quotient vector space V=I is an algebra with multiplication (x + I)(y + I = xy + I. (Check that this is well defined.) The canonical map L −! L=I is an algebra homomorphism with kernel I. Conversely, if f : L −! L0 is any algebra homomorphism then ker(f) = f −1(0) is an ideal of L. We say that the algebra L is a Lie algebra if (a) xx = 0 for all x; (b) x(yz) + y(zx) + z(xy) = 0 for all x; y; z. (Jacobi identity). Note that (a) implies (a') xy = −yx for all x; y. Indeed xy + yx = (x + y)(x + y) − xx − yy = 0. Traditionally in a Lie algebra one writes [x; y] instead of xy and one calls [x; y] the bracket. Example (a). Let A be an algebra. Assume that (xy)z = x(yz), that is, the algebra A is associative. Define a new algebra structure on A by [x; y] = xy − yx. This makes A into a Lie algebra. We check the Jacobi identity: [x; [y; z]] + [y; [z; x]] + [z; [x; y]] = x(yz − zy) − (yz − zy)x + y(zx − xz) − (zx − xz)y + z(xy − yx) − (xy − yx)z = 0: Example (b). Let V be a vector space. Let End(V ) be the vector space of endomorphisms of V (linear maps V −! V ). This is an associative algebra where the product xy is the composition of endomorphisms (xy)(v) = x(y(v)) for v 2 V . By example (a), End(V ) is a Lie algebra with bracket [x; y] = xy − yx. This Lie algebra is also denoted by gl(V ). Typeset by AMS-TEX 1 2 LIE ALGEBRAS 1 Example (c). Let V be as in (b). Assume that dim(V ) < 1. Let L = fx 2 End(V )jtr(x) = 0g. This is a (Lie) subalgebra of End(V ). Indeed, if x; y 2 End(V ) = gl(V ) then tr[x; y] = tr(xy − yx) = 0 hence [x; y] 2 L. Thus L is even an ideal of gl(V ). The Lie algebra L is also denoted by sl(V ). Example (d). Let V be as in (b). Let (; ) : V × V −! k be a bilinear map. Let L = fx 2 End(V )j(x(v); v0) + (v; x(v0)) = 08v; v0 2 V g. We show that L is a Lie subalgebra of gl(V ). Indeed, if x; y 2 L and v; v0 2 V we have ([x; y](v); v0) + (v; [x; y](v0)) = (x(y(v)); v0) − (y(x(v); v0) + (v; x(y(v0)) − (v; y(x(v0)) = 0: Example (e). Let V be as in (c). Assume that we are given a sequence of vector subspaces 0 = V0 ⊂ V1 ⊂ V2 ⊂ : : : ⊂ Vn = V , dim Vi = i. Let t = fx 2 End(V )jx(Vi) ⊂ Vi for i 2 [0; n]g n = fx 2 End(V )jx(Vi) ⊂ Vi−1 for i 2 [1; n]g. Then t is a Lie subalgebra of End(V ) and n is a Lie subalgebra of t. Example (f). Let A be an algebra. A derivation of A is a linear map d : A −! A such that d(xy) = d(x)y + xd(y) = 0 for all x; y. (Leibniz rule). Let Der(A) be the set of all derivations of A. This is a vector subspace of End(A). We show that Der(A) is a Lie subalgebra of End(A). Indeed, let d; d0 2 Der(A) and x; y 2 A. We have [d; d0](xy) = d(d0(xy)) − d0(d(xy)) = d(d0(x)y + xd0(y)) − d0(d(x)y + xd(y)) = d(d0(x))y + d0(x)d(y) + d(x)d0(y) + xd(d0(y)) − d0(d(x))y − d(x)d0(y) − d0(x)d(y) − xd0(d(y)) = d(d0(x))y − d0(d(x))y + xd(d0(y)) − xd0(d(y)) = [d; d0](x)y + x[d; d0](y) hence [d; d0] 2 Der(A). Example (g). Let L be a Lie algebra. By (f), Der(L) is again a Lie algebra. If x 2 L then ad(x) : L −! L, y 7! [x; y] is a derivation of L, that is, ad(x)[y; y0] = [ad(x)(y); y0] + [y; ad(x)(y0)] for all y; y0. (Equivalently, [x; [y; y0]] = [[x; y]; y0] + [y; [x; y0]] which follows from Jacobi's identity.) The map x 7! ad(x) is a Lie algebra homomorphism L −! Der(L), that is, it is linear (obviously) and, for x; x0; y 2 L we have ad([x; x0])(y) = ad(x)(ad(x0)y) − ad(x0)(ad(x)y). (Equivalently, [[x; x0]; y] = [x; [x0; y]] − [x0; [x; y]], which again follows from Jacobi's identity.) Example (h). Let V be as in (a). Define an algebra structure on A on V by xy = 0 for all x; y. This is a Lie algebra. (An abelian Lie algebra.) Example (i). Let L be a Lie algebra. Let [L; L] be the vector subspace of L spanned by f[x; y]jx 2 L; y 2 Lg. This is an ideal of L. One calls [L; L] the derived algebra of L. Example (j). Let L be a Lie algebra. Let ZL = fx 2 Lj[x; y] = 08y 2 Lg. This is an (abelian) subalgebra of L called the centre of L. LIE ALGEBRAS 1 3 Example (k). Let L be a Lie algebra. Let K be a vector subspace of L. Then 0 NL(K) = fx 2 Lj[x; y] 2 K8y 2 Kg is a Lie subalgebra of L. Indeed, if x; x 2 0 0 0 0 NL(K) and y 2 K we have [[x; x ]; y] = [[x; y]; x ] + [x; [x ; y]] 2 [x ; K] + [x; K] ⊂ K + K ⊂ K. One calls NL(K) the normalizer of K in L. Example (l). Let L = sl(V ) where V is a 2-dimensional vector space. Identify L with the set of 2 × 2 matrices with trace 0. A basis is given by 0 1 0 0 1 0 e = 0 0 ; f = 1 0 ; g = 0 −1 : We have [e; f] = h; [h; e] = 2e; [h; f] = −2f. A Lie algebra L is said to be simple if it is not abelian and L has no ideals other than 0; L. We show that L in Ex.(l) is simple. (We assume 2−1 2 k.) Let I be an ideal other than 0. We show that I = L. Let x = ae + bf + ch 2 I − f0g. Now [e; [e; x]] = [e; bh − 2e] = −2be 2 I; [f; [f; x]] = [f; ah + 2cf] = 2af 2 I. If a 6= 0 we deduce f 2 I hence h = [e; f] 2 I and e = −2−1[e; h] 2 I; thus I = L. Similarly if b 6= 0 then I = L. Thus we may assume that a = b = 0. Then h 2 I and e = −2−1[e; h] 2 I; f = 2−1[f; h] 2 I so that I = L. ∼ Let A be an algebra. An automorphism of A is an algebra isomorphism L −! L. Lemma. Assume that k has characteristic 0. If d : A −! A is a derivation such n that n for some then d d is an automorphism of d = 0 n > 0 e = Pn2N n! : A −! A A. From the definition of a derivation we get ds dp dq s! (xy) = Pp+q=s p! (x) q! (y) for all s ≥ 0. (Use induction on s.) For x; y 2 A we have dp dq dp dq ds ed(x)ed(y) = (x) (y) = (x) (y) = (xy) = ed(xy): X p! X q! X X p! q! X s! p q s p+q=s s Thus ed is an algebra homomorphism. We have ede−d = e−ded = 1. Thus ed is a vector space isomorphism. The lemma is proved. Solvable, nilpotent Lie algebras Let L be a Lie algebra. If X; X0 are subsets of L, let [X; X0] be the subspace spanned by f[x; x0]jx 2 X; x0 2 X0g. If I; I0 are ideals of L then [I; I0] is an ideal of L. Hence L(0) = L; L(1) = [L; L]; L(2) = [L(1); L(1)]; L(3) = [L(2); L(2)]; : : : are ideals of L and L(0) ⊃ L(1) ⊃ L(2) ⊃ : : : . L is said to be solvable if L(n) = 0 for some n ≥ 1.