Lemma:  Suppose L is a Lie algebra and I, J are solvable ideals of L.  Then I+J is a solvable ideal of L.

Proof:  First we claim I+J is an ideal of L.  If x is in L, and i + j is in I+J then [x, i+j]=[xi]+[xj] which is in I+J as [xi] is in I and [xj] is in J.
            Let K={I intersect J}.  K is an ideal.  If we take any x in L [xk] is in I as k is in I and [xk] is in J as k is in J.  K is solvable as K is a subideal of I so K⁽ⁿ⁾ is a subset of I⁽ⁿ⁾ for all n.  As I⁽ⁿ⁾ is zero for some n, K⁽ⁿ⁾ must be zero for the same n and K is solvable. Next we notice that if K is an ideal of I and both K and I/K are solvable then I is solvable.
            We now use the homomorphism theorems to note I+J/J is isomorphic to I/(I int J).  As I and I int J are solvable, I/ I int J is solvable and thus I+J/J is solvable.  As J is solvable, then I+J is solvable and we are done. n

Lemma:  Every Lie algebra has a unique maximal solvable ideal.

Proof:  We are considering finite dimensional Lie algebras.  For any two distinct solvable ideals I and J, we can take I+J to get a solvable ideal of greater dimension.  Repeating this process a finite number of times gives us our result. n

Def:  The maximal solvable ideal of a Lie algebra L is written rad(L) and called the radical of L.

Def:  A Lie algebra is called semisimple if rad(L) = 0.

Lemma:  If L is any Lie algebra then L/rad(L) is semisimple.

Proof:  Suppose I/rad(L) is a solvable ideal of L/rad(L).  Then I is a solvable ideal of L containing rad(L).  As rad(L) is the largest such ideal, I=rad(L).  Then I/rad(L) = 0.  Thus L/rad(L) only has one solvable ideal, namely 0 and rad(L/rad(L))=0. n

Note:  We can also consider the following short exact sequence 0 ~> Rad L ~> L ~> L/Rad L ~> 0.  Thus every semisimple Lie algebra is in some sense "glued together" from a solvable Lie algebra and a semisimple Lie algebra.

Def:  Suppose L is a Lie algebra.  Then we can define a symmetric bilinear form on L by k(x,y)=(tr(ad x) tr(ad y)).  This is called the Killing form.

Note:  We can define the killing form for any representation, and will do so later.

Ex:  Given structure constants [xi,xj] = S cij^k xk, find the trace form.

Theorem: (Cartan's Solvability Criterion)  L is solvable iff kL(x,y) = 0 for all x in [L,L], y in L.

Proof:  We will only worry about one direction now, the other will be done in a later lecture (Lecture 8).  We also will also only prove this for linear Lie algebras for now, so assume L < gl(V).  We assume kL(x,y) = 0 for all x in [L,L], y in L and will show [LL] is nilpotent.  Then as [LL] is nilpotent and L/[LL] is abelian, both [LL] and L/[LL] are solvable and thus L is solvable.
            We will use a lemma from Linear Algebra here that will be proved in the next lecture.  The lemma says suppose A and B are subspaces of gl(V) with A a subspace of B.  Define C = {x in gl(V) : [xB] < A}.  Then if x in C satisfies trace(xy)=0 for all y in C then x is nilpotent.
            We want to show that if x is in [LL] and kL(x,y) = 0 for all y in L, then x is nilpotent.  Thus [LL] will be nilpotent, hence nilpotent.  Apply the lemma with A = [LL], B = L.  Clearly A is a subspace of B and x is in C.  In order to deduce the nilpotence of x we must show that trace(xy) = 0 for all y in M = {y in gl(V) : [yz] is in [LL] for all z in L}.  Now x = [x1x2] for x1,x2 in L.  Then trace(xy) = trace ([x1x2]y) = trace (x1[x2y]) and by definition of M, [x2y] is in [LL].  As x1 is in L, we use our hypothesis to guarantee trace (x1[x2y]) = 0.  With this we are done. n

2/11/99

Note:  We still need to prove a lemma which we used last lecture.  Suppose A and B are subspaces of gl(V) and C = {x in gl(V) : [xB] is in A}.  Then if x in C is such that trace(xy) = 0 for all y in C then x is nilpotent. We will need the following theorem first.

Theorem:  (Chevally-Jordan decomposition) Suppose x is in End(V).  Then there exists a unique decomposition x = xs + xn which satisfies these three properties.
1)  xs is semisimple (diagonalizable, minimum polynomial has distinct linear factors) and xn is nilpotent.
2)  xs and xn commute.
3)  There exist polynomials p, q in one variable t, without constant term, such that xs = p(x) and xn = q(x).
 
Proof:  Suppose the characteristic polynomial of x is (t - a1)^m1(t - a2)^m2...(t - ak)^mk.  Define Vi = Ker((x - aiI)^mi).  Then by a well known decomposition theorem (p.220 Hoffmann/Kuntz theorem 12) we can write V = V1 (+) V2 (+) ... (+)Vk with each Vi stable under x.  On each Vi, the characteristic polynomial of x is (x - aiI)^mi.  Thus for all i, (x - aiI)^mi = 0 on the space Vi.  It is easy to check that each of the polynomials (T - aiI)^mi are pairwise prime in the ring C[T].
            We now can use the Chinese Remainder Theorem on the ring C[T] to find a polynomial p(T) which satisfies the congruences p(T) = ai (mod (T - ai)^mi).  If zero is not an eigenvalue for x we add one more congrence p(T) = 0 (mod T) to assure that p(T) has no constant term.  Remember (x - aiI)^mi = 0 on the space Vi.  Now p(x) = ai (mod (x - ai)^mi) so on the space Vi, p(x) = aiI (mod 0) = aiI (I'm still a little unsure of this argument myself so possibly a correction or more explanation is needed).  Thus p(x) restricted to Vi equals aiI.  Hence p(x) acts diagonally on each Vi with single eigenvalue ai.  This means p(x) is semisimple on V and we let xs = p(x).
            Let q(T) = T - p(T) and xn = q(x).  Notice that q(T) has no constant term because p(T) and T have no constant terms.  Next q(x) restricted to Vi equals x - aiI thus the restriction to each Vi is nilpotent (again this probably needs some more explaining).  Thus q(x) is nilpotent over V = V1 (+) V2 (+) ... (+) Vk.
            Next p(x) and q(x) commute, because q(x)p(x) = (x-p(x))q(x) = xp(x) - p(x)p(x) = p(x)x - p(x)p(x) = p(x)(x-p(x)) = p(x)q(x).  Thus we have only uniqueness left to check.  Suppose both xs, xn and xs', xn' satisfy the three criterion.  Thus all four commute with x.  As xs + xn = x = xs' + xn' we get xs - xs' = xn' - xn.  Now  xn' - xn is nilpotent, being a difference of nilpotent endomorphisms.  As xs and xs' commute (why?) they can be diagonalized simultaneously, thus xs - xs' is semisimple.  However the only semisimple and nilpotent endomorphism is zero and so xs - xs' = 0 = xn' - xn which means xs = xs' and xn' = xn. Thus we have shown uniqueness and are done. n
 
Lemma:  Suppose x is in End(V) and x = xs + xn is its Jordan decomposition in End(V).  Then ad x = ad xs + ad xn is the Jordan decomposition of ad x in End(End(V)).

Proof:  As ad is additive, we know ad xs + ad xn = ad (xs + xn) = ad x.  Thus we have a decomposition.  We must check that it satisfies certain properties and by uniqueness we will be done.  Specifically we need to show that ad xs is semisimple in End(End(V)), ad xn is nilpotent in End(End(V)), and that ad xs and ad xn commute.  First if n is nilpotent in End(V) then n is ad nilpotent, by a previous theorem.  Thus ad xn is nilpotent in End(End(V)).
            Next if s is semisimple then we can choose a basis of V, v1, v2, ..., vn so that s is diagonal under that basis.  Let a1, a2, ..., an be the ordered entries of s along the diagonal.  Let {Eij} be the standard basis of End(V) with respect to this basis.  Then Eijvk = djkvi and hence ad s (Eij) = sEij - Eijs = (ai - aj)Eij.  Thus ad s acts on each basis element by a scalar, so ad s is diagonalizable in End(End(V)) and ad s is semisimple.
            Finally we are left to show that ad xs and ad xn commute.  This is easy as [ad xs ad xn] = ad[xs xn] = 0.  With this we are done. n

Lemma:  Suppose A and B are subspaces of gl(V) and C = {x in gl(V) : [xB] is in A}.  Then if x in C is such that trace(xy) = 0 for all y in C then x is nilpotent.

Proof:  Let x be in C with that property and let x = s + n be the Chevally-Jordan decomposition of x.  We want to show s = 0, so x is nilpotent.  Suppose the eigenvalues of s (or x) are a1, a2, ..., ak in C.  Consider the vector space over Q spanned by these eigenvalues and call it E.  We want to show E = 0.
            To show E = 0 it suffices to show the dual space E* is zero.  Thus it suffices to show that if f: E ~> Q is Q-linear, then f = 0.  Fix an eigenbasis for s (so s = diag(a1, a2, ..., ak)) and let y in End(V) be such that its matrix with respect to this basis is diag(f(a1), f(a2), ..., f(ak)).  Now y is semisimple in End(V), so by the preceeding lemma ad y is semisimple in End(End(V)).  By looking at the proof of the preceeding lemma, we can also see that the eigenvalues of ad y are of form f(ai) - f(aj) for i, j in {1, ..., k}.  As f is linear we get that the eigenvalues of ad y are all the f(ai - aj).
            Now by interpolation we can construct a polynomial without constant term such that r(ai - aj) = f(ai - aj) for all i, j in {1, ..., k}.  There is no ambiguity of the values, because if ai - aj = ak - al, then f(ai - aj) = f(ak - al).  Now ad s(Eij) = (ai - aj)Eij and  ad y (Eij) = (f(ai) - f(aj))Eij = (f(ai - aj))Eij = r(ai - aj)Eij = (I think) r((ai - aj)Eij) = r(ad s(Eij)).  Thus ad y = r(ad s).  By the previous lemma, ad s is a polynomial in ad x without constant term.  As r has no constant term, ad y = r(ad s) is also a polynomial in ad x with no constant term.  We know (ad x)(B) is in A, as x is in C, but this means s is also in C by the fact that ad s is a polynomial in ad x without constant term (so (ad s)(B) = p(ad x)(ad x)(B) a subset of A as A<B).  Then since r has no constant term and ad y = r(ad s), y is in C as well.
            Now by our hypothesis and the fact that y is in C, trace(xy) = 0.  But x and y are diagonal matrices, so the trace of xy is simply Siaif(ai).  Thus we can use the fact that f is linear repeatedly to get 0 = f(0) = f(Siaif(ai)) = Si f(aif(ai)) = Si f(ai)f(ai) (as f is Q-linear and f(ai) is in Q) =Si (f(ai))².  This means for each ai, f(ai) = 0 and thus f equals zero (as f is a linear functional on E* and thus determined by the ai).  However f was chosen to be an arbitrary linear functional in E*, so every linear functional of E is zero, thus E = 0.  However, E is the span of the eigenvalues of s over Q, so the eigenvalues of s must all be zero.  This means s is nilpotent, and as s is semisimple it must be zero.  Thus x = s + n = n and x is nilpotent. n
 

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