Proposition:   CL(H) = H.

Note:  Last time we set C = CL(H) and concluded four things.
1)  If x is in C then xs and xn are in C.
2)  If s is in C, and s is semisimple, then s is in H.
3)  The restriction of k to H is nondegenerate.
4)  C is a nilpotent Lie algebra.

Proof:  Continued from last lecture
            5)  H and [CC] are disjoint.  This follows from the associativity of k.  We know that k(H, [CC]) = k([HC], C) = 0 (as [HC] = 0).  Now if h is in H and [CC] then h is in [CC] so k(H, h) = 0.  As k restricted to H is nondegenerate we know h = 0 (h is in H and Rad k|H).
            6)  C is abelian.  Suppose C is not abelian.  Then [CC] is nonzero.  By part 4 of this proof, C is nilpotent.  Thus by a previous theorem Z(C) intersect [CC] is nonzero (because Z(L) intersects every ideal of a nilpotent Lie algebra L).  Pick x in Z(C) and [CC] and write x = xs + xn.  As x is in [CC], by part 5 of this proof we know x is not in H.  We now know xn is not zero, otherwise x = xs and by part 2, x = xs is in H.  Since (ad x)|C = 0 and ad xn is a polynomial in ad x without constant term, we get (ad xn)|C = (p(ad x)(ad x))|C = 0.  Thus xn is in Z(C).  As xn is in Z(C) it commutes with all y in C.  Thus ad y ad xn = ad [yxn] + ad xn ad y = 0 + ad xn ad y and hence ad y and ad xn commute.  Thus (ad y ad xn)^k = (ad y)^k(ad xn)^k = 0 for sufficiently large k (because ad xn is nilpotent).  This shows (ad y ad xn) is nilpotent and k(y,  xn) = trace(ad y ad xn) = 0.  We have shown that for all y in C, k (y,  xn) = 0.  But xn is in C (by part 1) and in Radk|C.  Thus xn = 0, a contradiction.
            7)  C=H.  If not, then C contains a nonzero nilpotent element n.  This is because we can write x in C - H as x = xs + xn with xs and xn in C, xs in H.  If xn were in H then x would be in H.  Now we can use part 6 on our nilpotent n to say ad x ad n = ad[nx] = ad[xn] = ad n ad x for all x in C.  Thus for all x in C, (ad x ad xn)^k = (ad x)^k(ad xn)^k = 0 and k (x,  xn) = 0.  By the nondegeneracy of k|C, n =0.  This is a contradiction and so we are finished. n
 
Note:  In particular we can use part 3 of the proof of this last theorem to identify H and H*.  Thus for any a in F we get a distinguished element ta in H satisfying k(ta, h) = a(h) for all h in H.  One way to see such a ta exists is to represent a(h) as v^Th for some vector v.  Then we can represent k by a nondegenerate matrix A.  By the nondegeracy of A (in respect to the space C) we can find a vector w so that w^TA = v^T (namely w = v^TA^-1) and then a(h) = v^Th = w^TAh = k(w, h).  Letting ta = w, we have found a ta so k(ta, h) = a(h) for all h in H.

Note:  As CL(H) = H, we now can write our root decomposition as L = H (+)a in F La where La = {x in L : [hx] = a(h)x}.

Lemma:  Suppose W is a subspace of V and (thus) W* does not span V*.  Then there is v in V so that l(v) = 0 for all l in W*.

Proof:  Take a basis of W, w1, w2, w3, ..., wk and extend it to a basis of V to get w1, w2, w3, ..., wk, v1, v2, ..., vj (with j > 0).  Let w1*, w2*, w3*, ..., wk*, v1*, v2*, ..., vj* be the dual basis.  Then vj*(wi) = 0 for all i,j.  Then wi*(vj) = 0 for all i,j by the double dual.  Let v = vj for some j and l(v) = 0 for all l in W*. n

Theorem:  (Properties of the root system F)
1)  F spans H*.
2)  If a is in F then so is -a.
3)  If x is in La and y is in L-a then [xy] = k(x,y)ta.
4)  For any a in F, [La L-a] is one dimensional with basis ta.
5)  a(ta) = k(ta, ta) is nonzero for all a in F.
6)  If xa is in La and nonzero, then we can find ya in L-a and ha in H so that [xa ya] =  ha, [ha xa] =  2xa, [ha ya] = -2ya.  Thus xa ,ya, and ha satisfy the relations of sl(2,C).
7)  ha = -ha and ha = 2ta/k(ta, ta).

Proof:  1)  If F does not span H* then, by the lemma, there would be h in H such that a(h) = 0 for all a in F.  Then [h, La] = a(h)La = 0 for all La.  We know [h, H] = 0 since H is abelian, so h is in Z(L) which is zero.  We have reached a contradiction.
            2)  If a is in F then La is nonempty.  Now if -a is not in F, then La is orthogonal to all Lb, for b in F.  By an earlier proposition, ad x is nilpotent for any x in La.  Thus (ad h ad x)^k = (ad [hx])^k = (ad a(h)x)^k = a(h)(ad x)^k = 0 for large enough k.  Thus ad h ad x is nilpotent and h is orthogonal to La as well.  Then La is in the radical of k which is 0 and we have a contradiction.
            3)  If x is in La and y is in L-a then [xy] is in La-a = L0 = H by a proposition from last class.  Now for any h in H, we can compute k([xy],h) = k(x,[yh]) = k(x,-[hy]) = k(x,-(-a(h)y)) = k(x,a(h)y)) = a(h)k(x, y) = k(ta, h)k(x, y) = k(k(x, y)ta, h) because k(x, y) is a scalar.  Then since k is nondegenerate, k([xy],h) = k(k(x, y)ta, h) implies that [xy] = k(x, y)ta.
            4)  We know by part 3 that we are done if k(x,y) is not always zero, but this is true by the nondegeneracy of k.
            5)  Suppose a(ta) = 0.  We can choose x in La and y in L-a so that k(x, y) = 1 (again by nondegeneracy of k).  Then by 3 we get [xy] = k(x,y)ta = ta.  We also get [ta x] = a(ta)x = 0 and [ta y] = -a(ta)y = 0.  Let S = span {x, y, ta}.  Then [SS] = Cta and hence S is solvable.  Then by Lie's theorem [SS] is nilpotent.  As ta is in [SS] it must be ad-nilpotent, but ta is semisimple.  Thus ta must equal zero, a contradiction.
            6)  Set ha = 2ta / a(ta) and choose ya in L-a such that k(xa,ya) = 2 / a(ta) (we can do this using 5 and 4).  Then by 3 [xaya] = k(xa,ya)ta = (2 / a(ta))ta = ha.  Also [haxa] = a(ha)xa = a(2ta / a(ta))xa = (2 / a(ta))a(ta)xa = 2xa and  [haya] = -a(ha)ya = -a(2ta / a(ta))ya = -(2 / a(ta))a(ta)ya = -2ya.
            7)  By construction ha = 2ta / a(ta) so we have left to show -ha = h-a.  As k(ta, h) = a(h) and k(t-a, h) = -a(h) we know that k(ta, h) = -k(t-a, h) for all h in H.  By the nondegeneracy of k on H, we get that -ta = t-a (In fact ta is linear as a function of a).  Then h-a = 2t-a / -a(t-a) = -(2t-a / k(ta, t-a)) = -(2ta / k(ta, ta)) = -(2ta / a(ta)) = -ha and we have completed the proof. n

3/11/99

Note:  This is a tricky lecture.  I'm abandoning the typical theorem, proof, note style of writing this lecture for one which is more the way both the professor and the book present this material.

            Remember, if L is semisimple we have the decomposition L = H (+)a in F La, F < H*\{0}.  We have also shown that for each a in F we have xa in La, ya in L-a, and ha = 2ta/k(ta, ta) in H which form a sl2(C) triple.  We will now consider the action of Sa = C<xa, ya, ha> on L via the adjoint representation.  Thus we consider the map f: Sa ~> gl(L) which takes any z in  Sa to ad z, an endomorphism in gl(L).  Thus we have a representation of Sa on the vector space L, and we know certain things:
a)  By Weyl's Theorem, L is a direct sum of irreducible modules.
b)  From lecture 2, we know what all irreducible sl2(C) modules look like.  For each n = 0, 1, 2... there is a module Vn of dimension n + 1.  In Vn the eigenvalues of ha (actually the eigenvalues of f(ha) if f is our representation) are n, n-2, n-4, ..., -n.
            Now first we fix a particular a.  Consider the subspace of L of the form W = Sc in C Lca .  Thus W is the direct sum of the root spaces for all roots which are multiples of the particular root a.  Now consider the ad-representation of Sa on vector space W.  We must first show the following lemma.

Lemma:  W = Sc in C Lca is an Sa invariant submodule of L.

Proof:  We must show that given x in Sa, w in W, ad x(w) is in W.  By the linearity of ad (or any representation for that matter) it is enough to check that ad xa(w), ad ya(w), and ad ha(w) are in W for w in W.  Also as we can write any w as a sum of elements in different Lcas, it is safe to check only for w in a particular Lca.  So say wca is in Lca for some c.  Then ad xa(wca) = [xa, wca] which is in La+ca by an earlier proposition ([LaLb] is in La+b).  As La+ca = L(c+1)a which is in W we have shown that ad xa(wca) is in W.  This shows ad xa(w) is in W.  The other two cases are similar.  In the first case, ad ya(Lca) = [ya, Lca] < L(c-1)a (because ya is actually in L-a) in W.  In the next ad ha(Lca) = [ha, Lca] < Lca+0 (as ha is in L0) = Lca < W.  This shows W is an Sa submodule of the Sa module given by ad on L. n

Lemma:  If a is a root and ca is a root, then 2c is an integer.

Proof:  Now notice for xca in Lca, that ad ha(xca) = ca(ha)xca = ca(2ta/k(ta, ta))xca =  ca(2ta/a(ta))xca = (2c/a(ta))a(ta)xca  = 2cxca .  By a result from lecture 2, we know that the eigenvalues of ha (actually ad ha) must be integers.  Thus 2c must be an integer and c is half-integral. n

Note:  As W is equal to Sc in C Lca and each Lca is an eigenspace for ad ha we know that these ca(ha) = 2c are all the eigenvalues of ad ha.

Lemma:  If k = dim H, then Ker a is a k-1 dimensional trivial representation of Sa.

Proof:  Let k = dim H.  For h in H, ad xa(h) = [xa, h] = -[h, xa] = -a(h)xa, ad ya(h) = [ya, h] = -[h, ya] = a(h)ya, and ad ha(h) = [ha, h] = 0.  Thus for h in Ker a we get ad xa(h) = -a(h)xa = 0, ad ya(h) = a(h)ya = 0, and ad ha(h) = 0.  This means Sa(Ker a) < (Ker a) and Ker a is a trivial submodule.  Now as a is in H* and a is not 0, the range of a has dimension 1 and the kernel of a has dimension (dim H)-1 = k-1. n
 
            Now we proceed using Weyl's theorem on W.  This says that we can write W = (+)i Wi with each Wi irreducible.  With this we can use the information we have for irreducible representations of Sa on each Wi.  This tells us that if dim Wi is even then the weights (eigenvalues) of ha (actually ad ha) are n, n-2, ..., 2, 0, -2, ..., -n.  If dim Wi is odd then the weights of ha are n, n-2, ..., 1, -1, ..., -n.  We can write  W = (+)i Ei (+)i Oi where the Ei are the irreducible subrepresentations with all even weights and the Oi have odd weights.

Lemma:  If we write W as above the only Ei are one copy of Sa and the one dimensional trivial representations whose sum is the trivial Ker a.

Proof:  First notice that any Ei has a zero weight space (above and lecture 2) with dimension one.  Thus by looking at the zero weight space of W and calculating its dimension, we can see how many Ei there are.  The zero weight space of W is {x in W : ad ha (x) = 0}.  However if [ha, x] = 0 then x is in H.  We can see this by writing x = (+)c in (1/2)Z xca  and noticing [ha, x] = [ha, (+)c in (1/2)Z xca ] = (+)c in (1/2)Z [ha, xca ] = (+)c in (1/2)Z ca(ha)xca = (+)c in (1/2)Z 2cxca.  If this sum is zero then 2cxca must be zero for each half integral c.  This allows only x0a to be nonzero which means x is in H.
            Thus we know the zero weight space is H, and the number of Ei is equal to the dimension of H (call this k).  Now let us consider some particular Ei we know must exist.  We proved that Ker a is a k-1 dimensional trivial representation of Sa.  This representation, can be written as a direct sum of k-1 one dimensional trivial representations, each having even weights (specifically the weight zero).  Thus there can only be one other even irreducible representation (Every one must have a zero weight space, and there is only room for one more).
            Consider the action of Sa by ad on Sa in W.  As xa, ya, and ha are all in W (xa, ya in La, ha in H)  we can see that C<xa, ya, ha> is an irreducible submodule of W.  This is because ad xa, ad ya, and ad ha all take C<xa, ya, ha> to itself (and ha takes up the final weight of zero).  With this we have found all Ei and are done. n

Lemma:  If a is in F then ca is not in F for any integer c.

Proof:  We still must look at the action of Sa on W.  We have shown that if a is a root and ca is a root, then 2c is an integer.  Fix a c, and say that b = ca is a root.  Then notice for xca in Lb = Lca, ad ha(xca) = ca(ha)xca = ca(2ta/k(ta, ta))xca =  ca(2ta/a(ta))xca = (2c/a(ta))a(ta)xca  = 2cxca.  Thus as c is an integer, 2c is an even weight.  Then the irreducible submodule containing Lb (and there must be one, because W is completely reducible) is an even weight space.  But by the last lemma, there are only two possibilities and Lb is not in either Sa or Ker a < H.  Thus we have reached a contradiction. n

Corollary:  If a is a root then 2a is not a root.

Lemma:  If a is in F then (1/2)a is not in F.

Proof: By corollary.  One might notice we are actually considering the action of S(1/2)a on L in doing this, but the corollary is enough. n

Lemma:  If a is a root then the only multiple of a that is a root, is -a.

Proof:  Assume a is a root and b is a multiple of a.  Then by the lemmas we have recently proved, we know b is a half integral multiple of a.  However, a is also a half integral multiple of b.  Thus the only possibility is one being plus or minus one half of the other.  We can assume then that b = (1/2)a for if b = -(1/2)a then -b is a root and we can replace b with its negative.  But this contradicts our most recent lemma, so we are done. n
 
 

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