Lemma:  Suppose L is semisimple, f : L ~> gl(V) is an irreducible nontrivial representation.  Then the Casmir element cf acts on V by the nonzero scalar dim L / dim V.

Proof:  Recall that the trace form b(x,y) = tr (f(x)f(y)) is nondegenerate and cf = Si f(xi)f(yi) where x1, ..., xdimL and y1,...,ydimL are dual bases for b.  Then because cf commutes, it acts by some scalar, l, on V due to Schur's Lemma.  Thus traceV(cf) = (dim V)l.  But trace(cf) =  trace (Si f(xi)f(yi)) = Si  trace (f(xi)f(yi)) = Si b(xi, yi) = dim L.  Thus (dim V)l=dim L and l=(dim L)/(dim V). n

Theorem:  Suppose L is a semisimple Lie algebra and f : L ~> gl(V) is a faithful representation.  Let x in L have abstract Jordan decomposition x = xs + xn.  Then f(x) =  f(xs) + f(xn) is the Jordan decomposition of f(x) in gl(V).

Proof:  Let f(x) = S + N be the Jordan decomposition of f(x) in gl(V).  It suffices to show that S and N belong to f(L).  If this is so then there are s, n in L so that  f(x) =  f(s) + f(n) = S + N and ad x = ad xs + ad xn .  However adL x is the same transformation as adf(L) f(x) as f(L) is isomorphic to L.  Then the decompositions of adf(L) f(x) and ad x must agree by uniqueness.  (Fill in details!)
            Now by a previous proposition(lecture 7), if A < B < V are subspaces and x maps B into A, then xs and xn also map B into A.  Also by a previous lemma (Not done in class but from page 18 of Humphreys, lemma A) ad f(x) = ad S + ad L is the Jordan decomposition of ad f(x).  As ad f(x) maps f(L) to f(L), ad(S)(f(L)) = [S, f(L)] is in f(L) and ad(N)(f(L)) = [N, f(L)] is in f(L).  Thus S and N are in the normalizer of  f(L) in gl(V).
            If the normalizer of f(L) was equal to f(L) then we would be done.  However this is not true (consider the scalars for instance).  Now, we can write V = W1 (+) ... (+) Wk with Wi each irreducible due to Weyl's Theorem.  Then L ~> gl(W1) (+) ... (+) gl(Wk) in gl(V).  Since L = [LL] by previous theorem, we get that f maps L to sl(W1) (+) ... (+) sl(Wk) (Because [gl(V)gl(V)]=sl(V)).
            Define N = {Z in the normalizer of f(L) : Z(W) is a subset of W and trace|W (Z) = 0 for each Wi}.  We must prove three short things about N.  First f(L) < N as f(L) acts as a subset of sl(W1) (+) ... (+) sl(Wk).  Second S and N are in N.  Being polynomials of f(x) they map each Wi back to itself.  Also trace|W f(x) = 0 and trace|W N = 0 because N is nilpotent.  Now trace|W S = trace|W f(x) - trace|W N = 0.  Third we must prove N is a Lie algebra. If A,B are in N then A(W) and B(W) are subsets of W so [AB](W) = AB(W) - BA(W) is a subset of W.  Also trace|W [AB] = 0 and soN is a Lie algebra.
            Now we will be done if we can show f(L) =N.  Consider the ad representation of f(L) on N.  As [f(L)f(L)] < f(L), we know f(L) is a submodule of N.  By Weyl's theorem we can write N = f(L) (+) I  for some ideal I.  Now as I is in the normalizer of f(L) (because I < N < the normalizer of f(L)) we know [I, f(L)] <  f(L).  However as I is also a submodule in the ad representation of f(L), [f(L), I] < I.  Thus [f(L), I] is in I intersect f(L) = 0.  If z is in I, then [z, f(x)] = 0 for all x in L.  Thus by Schur's Lemma z acts by a scalar on each irreducible Wi.  But trace z|Wi = 0 (by definition of N) for all i, so z acts by zero on each Wi.  Finally zV = zW1 (+) ... (+) zWk = 0, so z is 0.  This means I is zero and N = f(L) so we are done. n

Note:  Recall sl(2,C) = Cx + Cy + Ch with [hx] = 2x, [hy] = -2y, [xy] = h.  Here h is semisimple and x and y are nilpotent.  ad h is semisimple so x,y,h is an eigenbasis for ad h.  We used this to characterize representations of sl(2,C) in the second lecture.  We will begin to generalize this method.

Def:  A subalgebra H of a semisimple Lie algebra L is called toral if H consists entirely of semisimple elements.

Note:  Such a subalgebra always exists.  For x in L look at xn and xs in the abstract decomposition.  If xs = 0 for all x then L is ad nilpotent, hence nilpotent and not semisimple.  Thus we can take some xs and look at the 1 dimensional subspace spanned by it.  This subalgebra is toral.

Lemma:  Every toral subalgebra is abelian.

Proof:  Let H be a toral subalgebra of L.  Suppose x is in H and we will show ad x |H is zero.  To do this we will show ad x |H is both semisimple and nilpotent.  Since x is semisimple, ad x |H is semisimple.  Suppose ad x |H is not nilpotent, then it has a nonzero eigenvalue l.  Then ad x (y) = [xy] = ly for some nonzero y in H.  But then ad y (x) = [yx] = -ly and (ad y)² (x) = 0.  We can write ad y as a diagonal matrix A because it is a semisimple endomorphism.  However for diagonal A, if A²x = (a1²x1, a2²x2, ... , ak²xk) = 0 then Ax = (a1x1, a2x2, ... , akxk) = 0.  Thus -ly = 0 and l=0.  Hence, ad x cannot have nonzero eigenvalues and ad x is both nilpotent and solvable, thus ad x = 0. n

3/4/99

Def:  A maximal toral subalgebra is called a Cartan subalgebra.

Ex:  Consider sl(2, C) with the usual basis x, y, h.  Here sl(2, C) is made up of three toral subalgebras as sl(2, C) < H1 (+) H2 (+) H3 where H1 = Ch, H2 = C(x + y), and H3 = C(x - y).

Exercise:  Show that H1, H2, and H3 are conjugate under SL(2, C).  Thus there exists gij in SL(2, C) so that gijHigij^-1 = Hj for i,j in {1,2,3}.

Note:  As H consists only of semisimple elements, ad(H) consists of semisimple elements in gl(L).  As H is abelian ad(H) is a family of commuting semeisimple elements in gl(L) hence we can simultaneously diagonalize ad(H) of obtain the decomposition L = (+)a in H* La where La = {x in L : [hx] = ad h (x) = a(h)x for all h in H}.
            Notice that a is in H* because ad is linear.  If ad h (x) = a(h)x  for all h in H then a(h1 + h2)x = ad (h1 + h2) (x) = ad h1 (x) + ad h2 (x) = a(h1)x + a(h2)x.  It is easy to see that La = 0 for all but finitely many a in H*.  Also observe for a = 0, L0 = {x in L : [hx] = 0 for all h in H} = CL(H).  As H is abelian H < CL(H) so L0 is nonempty.
            We call any nonzero a in H* so that La is non-zero a root of L.  As L0 is always nonempty we choose not to call zero a root.  The set of all roots of L is denoted by F < H* and called the root space.  We get the following decomposition which we call the root space decomposition, L = CL(H) (+) (+)a in F La.  This sum is a direct sum of vector spaces and not a direct sum of ideals as discussed before.

Proposition:  F satisfies the three following properties
1)  [La Lb] < La+b if a + b is in F union {0}.  [La Lb] = 0 otherwise.
2)  If a is in F and x is in La then ad x is nilpotent.
3)  If a and b are in F union {0} and a is not equal to -b then k(La ,Lb) = 0.

Proof:  (of 1)  Suppose x is in La, y is in Lb, and h is in H.  Then ad h ([xy]) = [ad h(x), y] + [x, ad h(y)] (because ad is a derivation) = [a(h)x, y] + [x, b(h)y] = a(h)[xy] + b(h)[xy] = a(h) + b(h)[xy] = (a + b)(h)[xy].  Thus [xy] is in La+b.  If a + b is neither 0 nor a root, then [xy] must equal 0.
            (of 2)  If y is in Lb for some b in F union {0} then by 1, (ad x)^k(y) is in Lb+ka.  As F is finite Lb+ka must equal zero for large enough values of k.  Then for any z in L we can write z = z1 + z2 + ... + zn where each zk is in Lb for some b in F union {0}.  Then for each zi there is a ki so that (ad x)^ki(zi) = 0 so for k = maxi ki we get (ad x)^k(z) =  (ad x)^k(z1 + z2 + ... + zn) = (ad x)^k(z1) + (ad x)^k(z2) + ... + (ad x)^k(zn) = 0.
            (of 3)  Let x be in La, and y be in Lb.  Choose any h in H.  Now -a(h)k(x, y) = k(-a(h)x, y) = k(-[hx], y) = k([xh], y) = k(x, [hy]) = k(x, b(h)y) = b(h)k(x, y).  Thus we get the equation (b-a)(h)k(x, y) = 0.  Either (b-a)(h) = 0 for all h, in which case a is equal to -b, or k(x, y) = 0 for all x in La, y in Lb in which case k(La ,Lb) = 0. n

Note:  Proposition 2 requires that a is in F, and not just in F union {0}, as in proposition 1.  If a = 0 then for y in Lb, (ad x)^k(y) is in Lb+ka = Lb.  In fact we will soon show that if x is in La for a = 0, then x is not just in CL(H), x is in H itself.  Thus if ad x is nilpotent then x is zero.

Note:  Notice that by proposition 3, L0 is orthogonal to La for all a in F.

Proposition: (4) k|C_L(H) is nondegenerate.

Proof:  Suppose x in CL(H) is such that k(x, y) = 0 for all y in CL(H).  We know that k(x, z) = 0 for all z in La, a in F.  Hence x is in Rad(k), but Rad(k) =0 because L is semisimple. n

Proposition:   CL(H) = H.

Proof:  (The beginning at least, we will finish next lecture)  Our strategy is to show that CL(H) does not contain nonzero nilpotent elements.  We will complete this proof in several small steps which we will number.  We also let C = CL(H) to simplify notation.
            1)  If x = xs + xn is the Jordan decomposition of x in C, then both xs and xn are in C.  ad xs is a polynomial in ad x with no constant term, so we can write ad xs = p(ad x)ad x for some polynomial p.  Then ad xs (H) = p(ad x)ad x (H) = p(ad x)([xH]) = 0 as [xH] = 0.  Thus xs is in CL(H).  We can show xn is in CL(H) similarly.
            2)  If s in C is semisimple then s is in H.  Any semisimple element of C commutes with all of H.  Then H + Cs is a toral subalgebra containing H, thus contradicting the maximality of H.
            3)  The restriction of k to H is nondegenerate.  Notice that we already know the restriction of k to C is nondegenerate.  Suppose x is in Rad k|H, thus k(x, y) = 0 for all y in H.  Let z in C be arbitrary.  Then by part 1 of this proof, zn and zs are in C.  By part 2 of this proof zs is in H.  Thus k(x, z) = k(x, zn + zs) = k(x, zn) + k(x, zs) = (because x and zs are in H) k(x, zn) = trace(ad x ad zn).  Now ad zn is nilpotent and commutes with ad x (as [ad x, ad zn] = ad [xzn] = 0) so (ad x ad zn)^k = (ad x)^k(ad zn)^k = 0 for large enough k, hence (ad x ad zn) is nilpotent and trace(ad x ad zn) = 0.  Thus  k(x, z) = trace(ad x ad zn) = 0 for all z in C, hence x is in Rad k|C which is equal to 0 by proposition 4.  We have shown Rad k|H = 0 hence the restriction of k to H is nondegenerate.
            4)  C is nilpotent as a Lie algebra.  This statement may seem like a strange statement at first, but we are considering C as a subalgebra here and the killing form of C is not the same as the killing form of L restricted to C.  Notice this will be one step closer to proving C is abelian, which will help us finish this proof.  By Engels theorem we need to show that for all x in C, adc x is nilpotent.  Write x in C as xs + xn.  By part 1 of our proof xs is in C, and by part 2 xs is in H.  Thus adc xs = 0.  We get adc x = adc xs + xn = adc xs + adc xn = adc xn.  Because ad xn is nilpotent on L, it is nilpotent on any invariant subspace thus it is nilpotent on C.  This shows adc x is nilpotent. n
 

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