Def:  If L is a Lie Algebra, the killing form is defined by k(x,y)=tr(ad x ad y).

Def:  Rad k = {x : k(x,y)=0 for all y in L}

Lemma:  If L is a Lie algebra then for all k, [LL]^k is an ideal of L.

Proof: [x[LL]^k]=[x[L[LL]^k-1]=[L[x[LL]^k-1] + [[LL]^k-1[Lx]] < (by induction) [L[LL]^k-1] = [LL]^k. n

Theorem:  L is solvable iff [LL] is a subset of Rad k.
 
 Proof:  First assume [LL] is a subset of Rad k.  Then for all x in [LL], y in L, tr(ad x ad y)=0.  By Cartan's criterion L is solvable.  Next assume L is solvable.  Then by a corollary of Lie's theorem [LL] is nilpotent.  Take x in [LL], y in L.  Notice if z is in L, (ad x ad y)(z) is in ([LL],[LL]) = [LL]². If z is in [LL]^k then (ad x ad y)(z) = [x[yz]].  Now [yz] is in [LL]^k as [LL]^k is an ideal and x is in [LL], so [x[yz]] is in [LL]^k+1.  Thus for any x in [LL], y,z in L there is some k, so that (ad x ad y)^k is in [LL]^k = 0.  Thus (ad x ad y) is nilpotent as an endomorphism (and has trace zero) for any x in [LL].  Thus [LL] is a subset of Rad k and we are done. n

Lemma:  If L is a Lie algebra, every term in the derived series of L is an ideal.

Proof:  Take x in L.  [xL^(k)] = [x[L^(k-1)L^(k-1)]] = [L^(k-1)[xL^(k-1)]] + [[xL^(k-1)]L^(k-1)] < (by induction) [L^(k-1)L^(k-1)] = L^(k). n

Theorem:  L is semisimple iff Rad k=0.

Proof:  We assume L is semisimple first.  Let S=Rad k.  We now will show S is an ideal of L.  We use the "associative" property of the killing form, that k([xy],z) = k(x,[yz]).  Thus if x is in S, and y,z in L then k(x,[yz])=0=k([xy],z).  Thus for any y in L [xy] is in S and S is an ideal.
            Next we use the lemma that says if I is an ideal of L and x, y are in I, kI(x,y)=kL(x,y).  S=Rad k implies kL|SxS=0 so kS=0.  Then by Cartan S is solvable.  As L is semisimple, S must be zero.
            Now we assume Rad k=0 and want to show Rad L=0 so L will be semisimple.  First we claim every abelian ideal of L is contained in Rad k.  Suppose I is abelian and x is in I.  Then for any y in L Im(ad x ad y) is in I (as I is an ideal).  Then (ad x ad y)² maps L ~> [x[yI]] < [I, I]=0.  Thus (ad x ad y) is nilpotent for all y and x.  Hence trace(ad x ad y) =0 for all x in I and all y.  Thus I is contained in Rad k.
            Now assume J is a solvable ideal of L, and we will show it contains a nonzero abelian ideal (reaching a contradiction as L has no abelian ideals).  As J is solvable, every term in the derived series of J is an ideal.  For some k, J^k+1=0 whereas J^k is nonzero.  Thus [J^kJ^k]=0 and J^k is abelian, a contradiction. n

Def:  A Lie algebra L is called simple if L has no proper ideals and dim L > 1.

Def:  If L1 and L2 are Lie algebras then we define their direct sum L = L1 (+) L2 = {(x,y) : x in L1 y in L2} with operation [(x,y)(x',y')] = ([x,x'],[y,y']).
 
Note:  We want to spot Lie algebras which can be written in the form (x,0) and (0,y).  Well if we have something of this sort L1 = {(x,0) x in L1} then this is an ideal, as [(x,0),(x'y')] = ([x,x'],[0,y']) = ([xx'],0).  Similarly L2 will also be an ideal.  However in this situation notice that both L1 and L2 are disjoint.  Notice that if we have an arbitrary L with two ideals I1 and I2 whose intersection is 0, then if L = I1 + I2 we can write L in the form  I1 (+) I2.

Note:  In general if L is a Lie algebra and  I1, I2, ..., Ik are ideals such that L = I1 (+) I2 (+) ... (+) Ik as vector spaces, then L =  I1 (+) I2 (+) ... (+) Ik as a Lie algebra.

Theorem:  If L is semisimple then there exist simple ideals L1, L2, ..., Lk of L so that L =  L1 (+) L2 (+) ... (+) Lk, and moreover these ideals are unique up to permutation (thus every simple ideal of L is equal to one of the Lk).

Proof:  Assume L is semisimple and suppose I is an ideal of L.  Let J = {x in L : k(x,y)=0 for all y in I}.  Then J is an ideal in L by associativity of the killing form.  Say z is in J, x is in L, then for any y in I, k([zx],y) =  k(z,[xy]) = 0 (as [xy] is in I).  Hence [zx] is in J and J is an ideal.
            Since k is nondegenerate and J is orthogonal to I, dim I + dim J = dim L.  Also the intersection of I and J is an ideal, and we can consider k(I intersect J) = kL restricted to I intersect J by a previous lemma.  This is zero by the definition of J.  Thus by Cartan's criterion, I intersect J is solvable.  Since L is semisimple, this intersection must be zero.  Hence L = I (+) J as a Lie algebra.
            Moreover, I and J are semisimple.  If x is in Rad kI then  kI (x, x') = 0 for all x' in I so kL (x, x') = 0 for all x' in I.  But we know that kL (x, y) = 0 for all y in J so x is in Rad kL and x = 0.  Similarly, if x is in Rad kJ then x=0.  This shows I and J are semisimple.  Now if I or J is not simple we can repeat the process until we get L = L1 (+) L2 (+) ... (+) Lk.
            Suppose I is a simple ideal of L.  Consider [I, L] = {[xy] : x in I, y in L}.  As [I, L] is an ideal of I either [I, L] = 0 or [I, L] = I.  If [I, L] = 0 then I is in Z(L).  As L is semisimple Z(L) = 0 and hence I is 0.  Thus if I is a nontrivial simple ideal then [I, L] = I.  Now I = [I, L] = [I, L1 (+) L2 (+) ... (+) Lk] = [I, L1] (+) [I, L2] (+) ... (+) [I, Lk].  Each [I, Li] is an ideal of Li and hence [I, Li] = 0 or [I, Li] = Li .  Exactly one must be nonzero (if two were nonzero then I would not be simple and if none were nonzero then I is trivial).  If [I, Li] = Li  then also [I, Li] is a subset of I and must equal I thus I = [I, Li] = Li  and I = Li and with this we are done. n

2/18/99

Theorem:  If L is semisimple and x in L, then there exist xs, xn in L such that ad(x) = ad(xs) + ad(xn) is the Jordan composition in gl(L) (Then x = xs + xn is the abstract Jordan decomposition of x in L).

Note:  This will follow from the following lemmas.

Def:  If L is a Lie algebra then Der(L) = {d in gl(L) : d([xy]) = [dx,y] + [x,dy]}

Lemma:  Der(L) is a Lie algebra.

Proof:  Let d and g be two derivations and we must show [d, g] is a derivation or that [dg]([xy]) = [[dg]x,y] + [x,[dg]y].  However [dg]([xy]) = dg([xy]) - gd([xy]) = d([gx,y] + [x,gy]) - g([dx,y] + [x,dy]) =  d([gx,y]) + d([x,gy]) - g([dx,y]) - g([x,dy]) = [dgx,y] + [gx,dy] + [dx,gy] + [x,dgy] - [gdx,y] - [dx,gy] - [gx,dy] - [x,gdy] = [dgx,y] + [x,dgy] - [gdx,y]- [x,gdy] =   [dgx - gdx,y] + [x,dgy - gdy] = [[dg]x,y] + [x,[dg]y]. n

Lemma:  ad(L) is an ideal of Der(L).

Proof:  First we show that ad x even is a derivation for any x in L.  We know ad x([yz]) = x([yz]) = [x[yz]] = [[xy]z] + [y[xz]]= [ad x(y), z] + [y, ad x(z)] and we are done.  Next we must show the commutator of an element of ad(L) and an element of Der(L) is in ad(L) thus showing ad(L) to be an ideal.  [d, ad x](y) = d ad x (y) - ad x d (y) = [d[xy]] - [x[dy]] = [dx,y] = ad (dx) (y). n
 
Lemma:  If L is semisimple then ad L is isomorphic to Der (L).

Proof:  Let I = ad(L) an ideal of D = Der(L).  Now let kD be the killing form on D and let J = {d in D : kD (d, ad x) = 0 for all x in L} the orthogonal complement to I.  First we claim I and J are disjoint.  If ad y is in I and J then for all x in L, 0 = kD (ad y, ad x) = kI (ad y, ad x) (because I is an ideal) =  kL (y, x) (because ad is an isomorphism of L for semisimple L).  This implies y = 0 as kL is nondegenerate (Rad k = 0).  Thus I and J are disjoint.
            Now as dim I + dim J = dim D and I and J are disjoint we know I + J = D.  Also if d is in J, and x is in L then ad (dx) = [d , ad x] which is in the intersection of I and J, hence equals zero.  However, if ad (dx) = 0 for all x, then dx = 0 for all x so d = 0.  We chose d arbitrarily in J, so J = 0 and D = I.  Thus L is isomorphic to ad L which equals Der (L), so we are done. n

Lemma:  Suppose A is an algebra (not necessarily associative) and d in Der (A) < gl(A) has the Jordan decomposition d = s + n.  Then s and n are in Der (A).

Proof:  It is enough to show s is a derivation for then n will be a difference of derivations, hence a derivation.  Let l1, l2, ..., ln be the eigenvalues of d on A and let Al1, Al2, ..., Alk be the corresponding generalized eigenspaces ((d -liI) acts nilpotently on Ali).  Then s acts on Ali by multiplication by li.
            We claim if a is in Al and b is in Am then ab is in Al+m.  Well (d - (l+m)I)ab = d(ab) - (l+m)ab = (da)b + adb - lab - mab = (da)b - lab + adb - amb = ((d-l)a)b + a(d - m)b.  Inductively we can show that (d - (l+m)I)ⁿab = Skc(n,k)((d-l)^ka)(d-m)^n-kb.  For large enough n, all these terms are zero.  Thus AlAm is a subset of Al+m.
            Now for a in Al, b in Am s(ab) = (l+m)ab = (sa)b + a(sb) = (la)b + a(mb) = s(ab).  We can get linearity from writing A = (+)i Ali. By linearity, s is in Der(A) and we are done. n

Proof: (of if L is semisimple and x in L, then there exist xs, xn in L such that ad(x) = ad(xs) + ad(xn) is the Jordan composition in gl(L))  Given x in L, we can write ad(x) = s + n.  By the most recent lemma, s and n are in Der(L).  But Der(L) is isomorphic to ad(L) so s and n are in ad(L).  Thus there exist xs, xn in L such that ad(xs) = s and ad(xn) = n. n

Def:  An L-module is a Lie algebra together with a vector space V and an operation (L,V) ~> V (denoted (x, v) |~> x.v or xv) which satifies the following conditions:
(M1) (ax + by).v = a(x.v) + b(y.v)
(M2) x.(av+bw) = a(x.v) + b(x.w)
(M3) [xy].v = x.y.v - y.x.v (x,y in L; v,w in V; a,b in F)

Recall:  If L is a Lie algebra and V is a vector space, a representation for L on V is a Lie algebra homomorphism f: L ~> gl(V).

Lemma:  Given a Lie algebra L, there is a 1-1 correspondence between L-modules and representations of L.

Proof:  Given any L-module LV~>V consider the map f: L ~> gl(V) given by f(x)v = x.v for all v in gl(V).  Given any representation f: L ~> gl(V) we can consider the L-module we get from x.v = f(x)v. n

Recall:  V is reducible if there is a subspace W < V such that 0 < dim W < dim V and f(x)w is in W for all w in W, x in L.  Then x |~> f(x)|W gives a subrepresentation and W is called a submodule.

Note:  If (f1,V1), (f2,V2) are L-modules then V1 (+) V2 is an L-module via f(x)(v1,v2) = (f(x)v1,f(x)v2).

Theorem: (Herman Weyl) If L is semisimple then every finite dimensional L-module is the direct sum of irreducible L-modules.

Ex:  This example shows that L must be semisimple.  Consider the Lie algebra of 2 by 2 matrices with all zeros except for the upper right hand entry.  Here [L,L] = 0.  L has a 2 dimensional representation but L is not the direct sum of two one dimensional representations.
 

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