A bilinear form ( , ) :g×g→kis calledinvariant if
([x, y], z) + (y,[x, z]) = 0 ∀x, y, z∈g. (4.2) Notice that if ( , ) is an invariant form, and iis an ideal, theni⊥ is again an ideal.
One way of producing invariant forms is from representations: if(ρ, V) is a representation ofg, then
(x, y)ρ:= trρ(x)ρ(y) is invariant. Indeed,
([x, y], z)ρ+ (y,[x, z])ρ
= tr{(ρ(x)ρ(y)−ρ(y)ρ(x))ρ(z)) +ρ(y)(ρ(x)ρ(z)−ρ(z)ρ(x))}
= tr{ρ(x)ρ(y)ρ(z)−ρ(y)ρ(z)ρ(x)}
= 0.
In particular, if we take ρ= ad, V =g the corresponding bilinear form is called theKilling formand will be denoted by (, )κ. We will also sometimes writeκ(x, y) instead of (x, y)κ.
Theorem 8 gis semi-simple if and only if its Killing form is non-degenerate.
Proof. Suppose g is not semi-simple and so has a non-zero abelian ideal, a.
We will show that (x, y)κ = 0 ∀x∈a, y∈ g. Indeed, let σ= adxady. Then σmaps g→aanda→0. Hence in terms of a basis starting with elements of a and extending, it (is upper triangular and) has 0 along the diagonal. Hence trσ= 0. Hence if gisnotsemisimple then its Killing form is degenerate.
Conversely, suppose thatgis semi-simple. We wish to show that the Killing form is non-degenerate. So let u := g⊥ = {x|tr adxady = 0 ∀y ∈ g}. If x∈u, z∈gthen
tr{ad[x, z] ady} = tr{adxadzady−adzadxady)}
= tr{adx(adzady−adyadz)}
= tr adxad[z, y]
= 0,
souis an ideal. In particular, tru(adxuadyu) = trg(adgxadgy) forx, y∈u, as can be seen from a block decomposition starting with a basis ofuand extending tog.
If we take y ∈ Du, we see that tr aduDadu = 0, so adu is solvable by Cartan’s criterion. But the kernel of the mapu→aduis the center ofu. So if aduis solvable, so isu. QED
Proposition 8 Let g be a semisimple algebra, i any ideal of g, and i⊥ its orthocomplement with respect to its Killing form. Theni∩i⊥= 0.
Indeed,i∩i⊥ is an ideal on whichtradxady≡0 hence is solvable by Cartan’s criterion. Sincegis semi-simple, there are no non-trivial solvable ideals. QED
Therefore
Proposition 9 Every semi-simple Lie algebra is the direct sum of simple Lie algebras.
4.7. COMPLETE REDUCIBILITY. 69 Proposition 10 Dg=g for a semi-simple Lie algebra.
(Since this is true for each simple component.)
Proposition 11 Letφ:g→sbe a surjective homomorphism of a semi-simple Lie algebra onto a simple Lie algebra. Then if g=L
gi is a decomposition of ginto simple ideals, the restriction,φiofφto each summand is zero, except for one summand where it is an isomorphism.
Proof. Sincesis simple, the image of everyφi is 0 or all ofs. Ifφiis surjective for somei then it is an isomorphism since gi is simple. There is at least one i for which it is surjective sinceφis surjective. On the other hand, it can not be surjective for for two ideals,gi,gji6=jfor thenφ[gi,gj] = 06= [s,s] =s. QED
4.7 Complete reducibility.
The basic theorem is
Theorem 9 [Weyl.] Every finite dimensional representation of a semi-simple Lie algebra is completely reducible.
Proof.
1. If ρ : g → EndV is injective, then the form ( , )ρ is non-degenerate.
Indeed, the ideal consisting of allxsuch that (x, y)ρ= 0∀y∈gis solvable by Cartan’s criterion, hence 0.
2. TheCasimir operator. Let (ei) and (fi) be bases of gwhich are dual with respect to some non-degenerate invariant bilinear form, (,). So (ei, fj) = δij. As the form is non-degenerate and invariant, it defines a map of
g⊗g7→Endg; x⊗y(w) = (y, w)x.
This map is an isomorphism and is agmorphism. Under this map, Xei⊗fi(w) =X
(w, fi)ei =w
by the definition of dual bases. Hence under the inverse map Endg7→g⊗g
the identity element, id, corresponds toP
ei⊗fi (and so this expression is independent of the choice of dual bases). Since id is annihilated by commutator by any element of End(g), we conclude that P
iei ⊗fi is annihilated by the action of all (adx)2 = adx⊗1 + 1⊗adx, x ∈ g.
Indeed, forx, e, f, y∈gwe have
((adx)2(e⊗f))y = (adxe⊗f+e⊗adxf)y
= (f, y)[x, e] + ([x, f], y)e
= (f, y)[x, e]−(f,[x, y])e by (4.2)
= ((adx)(e⊗f)−(e⊗f)(adx))y.
Set
C:=X
i
ei·fi∈U(L). (4.3)
ThusCis the image of the elementP
iei⊗fiunder the multiplication map g⊗g7→U(g), and is independent of the choice of dual bases. Furthermore, Cis annihilated by adxacting onU(g). In other words, it commutes with all elements ofg, and hence with all ofU(g); it is in the center ofU(g).
TheCcorresponding to the Killing form is called theCasimir element, its image in any representation is called theCasimir operator.
3. Suppose that ρ : g → EndV is injective. The (image of the) central element corresponding to ( , )ρ defines an element of EndV denoted by Cρ and
trCρ = trρ(X eifi)
= trX
ρ(ei)ρ(fi)
= X
i
(ei, fi)
= dimg
With these preliminaries, we can state the main proposition:
Proposition 12 Let0→V →W →k→0be an exact sequence ofgmodules, wheregis semi-simple, and the action ofgonkis trivial (as it must be). Then this sequence splits, i.e. there is a line in W supplementary to V on which g acts trivially.
The proof of the proposition and of the theorem is almost identical to the proof we gave above for the special case of sl(2). We will need only one or two additional arguments. As in the case ofsl(2), the proposition is a special case of the theorem we want to prove. But we shall see that it is sufficient to prove the theorem.
Proof of proposition. It is enough to prove the proposition for the case thatV is an irreducible module. Indeed, ifV1is a submodule, then by induction on dimV we may assume the theorem is known for 0→V /V1→W/V1→k→0 so that there is a one dimensional invariant subspaceM inW/V1supplementary toV /V1on which the action is trivial. LetN be the inverse image ofM inW. By another application of the proposition, this time to the sequence
0→V1→N→M →0
we find an invariant line,P, inN complementary toV1. SoN=V1⊕P. Since (W/V1) = (V /V1)⊕M we must have P∩V ={0}. But since dim W = dim V + 1, we must have W = V ⊕P. In other words P is a one dimensional subspace ofW which is complementary toV.
4.7. COMPLETE REDUCIBILITY. 71 Next we can reduce to proving the proposition for the case that g acts faithfully on V. Indeed, leti= the kernel of the action onV. For allx∈gwe have, by hypothesis, xW ⊂V, and forx∈iwe have xV = 0. HenceDiacts trivially onW. Buti=Disincei is semi-simple. Henceiacts trivially on W and we may pass tog/i. This quotient is again semi-simple, sinceiis a sum of some of the simple ideals of g.
So we are reduced to the case that V is irreducible and the action, ρ, ofg onV is injective. Then we have an invariant elementCρwhose image in EndW must mapW →V since every element ofgdoes. (We may assume thatg6= 0.) On the other hand, Cρ 6= 0, indeed its trace is dimg. The restriction of Cρ to V can not have a non-trivial kernel, since this would be an invariant subspace.
Hence the restriction of Cρ toV is an isomorphism. Hence ker Cρ:W →V is an invariant line supplementary toV. We have proved the proposition.
Proof of theorem from proposition. Let 0 → E0 → E be an exact sequence ofg modules, and we may assume that E0 6= 0. We want to find an invariant complement toE0 inE. DefineW to be the subspace of Homk(E, E0) whose restriction to E0 is a scalar times the identity, and let V ⊂ W be the subspace consisting of those linear transformations whose restrictions to E0 is zero. Each of these is a submodule of End(E). We get a sequence
0→V →W →k→0
and hence a complementary line of invariant elements in W. In particular, we can find an element,T which is invariant, mapsE→E0, and whose restriction to E0 is non-zero. Then kerT is an invariant complementary subspace. QED As an illustration of construction of the Casimir operator considerg=sl(2) with
This coincides with theC that we used in Chapter II.
Chapter 5
Conjugacy of Cartan subalgebras.
It is a standard theorem in linear algebra that any unitary matrix can be di-agonalized (by conjugation by unitary matrices). On the other hand, it is easy to check that the subgroup T ⊂ U(n) consisting of all unitary matrices is a maximal commutative subgroup: any matrix which commutes with all diagonal unitary matrices must itself be diagonal; indeed if Ais a diagonal matrix with distinct entries along the diagonal, any matrix which commutes withAmust be diagonal. Notice thatT is a product of circles, i.e. a torus.
This theorem has an immediate generalization to compact Lie groups: Let Gbe a compact Lie group, and letT andT0 be two maximal tori. (SoT andT0 are connected commutative subgroups (hence necessarily tori) and each is not strictly contained in a larger connected commutative subgroup). Then there exists an element a ∈ G such that aT0a−1 = T. To prove this, choose one parameter subgroups of T and T0 which are dense in each. That is, choose x and x0 in the Lie algebra g ofG such that the curve t7→exptxis dense in T and the curvet7→exptx0 is dense inT0. If we could finda∈Gsuch that the
a(exptx0)a−1= exptAdax0
commute with all the expsx, then a(exptx0)a−1 would commute with all ele-ments ofT, hence belong toT, and by continuity,aT0a−1⊂T and hence =T. So we would like to find anda∈Gsuch that
[Adax0, x] = 0.
Put a positive definite scalar product ( , ) ong, the Lie algebra ofGwhich is invariant under the adjoint action ofG. This is always possible by choosing any positive definite scalar product and then averaging it overG.
Choosea∈Gsuch that (Adax0, x) is a maximum. Let y:= Adax0.
73
We wish to show that
[y, x] = 0.
For anyz∈gwe have
([z, y], x) = d
dt(Adexptzy, x)|t=0= 0 by the maximality. But
([z, y], x) = (z,[y, x])
by the invariance of (, ), hence [y, x] is orthogonal to allghence 0. QED We want to give an algebraic proof of the analogue of this theorem for Lie algebras over the complex numbers. In contrast to the elementary proof given above for compact groups, the proof in the general Lie algebra case will be quite involved, and the flavor of the proof will by quite different for the solvable and semi-simple cases. Nevertheless, some of the ingredients of the above proof (choosing “generic elements” analogous to the choice ofxand x0 for example) will make their appearance. The proofs in this chapter follow Humphreys.
5.1 Derivations.
Letδbe a derivation of the Lie algebrag. this means that δ([y, z]) = [δ(y), z] + [y, δ(z)] ∀y, z∈g.
Then, fora, b∈C
(δ−a−b)[y, z] = [(δ−a)y, z] + [y,(δ−b)z]
(δ−a−b)2[y, z] = [(δ−a)2y, z] + 2[(δ−a)y,(δ−b)z] + [y,(δ−b)2z]
(δ−a−b)3[y, z] = [(δ−a)3y, z] + 3[(δ−a)2y,(δ−b)z)] + 3[(δ−a)y,(δ−b)2z] + [y,(δ−b)3z]
... ... (δ−a−b)n[y, z] = X
n k
[(δ−a)ky,(δ−b)n−kz].
Consequences:
• Let ga = ga(δ) denote the generalized eigenspace corresponding to the eigenvalue a, so (δ−a)k = 0 onga for large enoughk. Then
[ga,gb]⊂g[a+b]. (5.1)
• Let s = s(δ) denote the diagonizable (semi-simple) part of δ, so that s(δ) =aonga. Then, fory∈ga, z∈gb
s(δ)[y, z] = (a+b)[y, z] = [s(δ)y, z] + [y, s(δ)z]
sosand hence alson=n(δ), the nilpotent part ofδare both derivations.
5.1. DERIVATIONS. 75
• [δ,adx] = ad(δx)]. Indeed, [δ,adx](u) = δ([x, u])−[x, δ(u)] = [δ(x), u].
In particular, the space of inner derivations, Inngis an ideal in Derg.
• Ifgis semisimple then Inng= Derg. Indeed, split off an invariant comple-ment to Inngin Derg(possible by Weyl’s theorem on complete reducibil-ity). For anyδ in this invariant complement, we must have [δ,adx] = 0 since [δ,adx] = adδx. This says that δx is in the center of g. Hence δx= 0∀xhenceδ= 0.
• Hence any x∈ g can be uniquely written as x= s+n, s ∈ g, n ∈ g where ads is semisimple and adn is nilpotent. This is known as the decomposition into semi-simple and nilpotent parts for a semi-simple Lie algebra.
• (Back to generalg.) Let kbe a subalgebra containing g0(adx) for some x∈ g. Thenxbelongsg0(adx) hence to k, hence adx preservesNg(k) (by Jacobi’s identity). We have
x∈g0(adx)⊂k⊂Ng(k)⊂g
all of these subspaces being invariant under adx. Therefore, the character-istic polynomial of adxrestricted toNg(k) is a factor of the charactristic polynomial of adx acting on g. But all the zeros of this characteristic polynomial are accounted for by the generalized zero eigenspaceg0(adx) which is a subspace ofk. This means that adxacts onNg(k)/kwithout zero eigenvalue.
On the other hand, adxacts trivially on this quotient space since x∈k and hence [Ngk, x]⊂kby the definition of the normalizer. Hence
Ng(k) =k. (5.2)
We now come to the key lemma.
Lemma 3 Let k⊂gbe a subalgebra. Let z∈k be such thatg0(adz)does not strictly contain anyg0(adx), x∈k. Suppose that
k⊂g0(adz).
Then
g0(adz)⊂g0(ady) ∀y∈k.
Proof. Choose z as in the lemma, and let x be an arbitrary element of k.
By hypothesis, x∈g0(adz) and we know that [g0(adz),g0(adz)]⊂g0(adz).
Therefore [x,g0(adz)]⊂g0(adz) and hence
ad(z+cx)g0(adz)⊂g0(adz)
for all constantsc. Thus ad(z+cx) acts on the quotient spaceg/g0(adz). We can factor the characteristic polynomial of ad(z+cx) acting ongas
Pad(z+cx)(T) =f(T, c)g(T, c)
wheref is the characteristic polynomial of ad(z+cx) ong0(adz) andg is the characteristic polynomial of ad(z+cx) ong/g0(adz). Write
f(T, c) = Tr+f1(c)Tr−1+· · ·fr(c) r= dimg0(adz) g(T, c) = Tn−r+g1(c)Tn−r−1+· · ·+gn−r(c) n= dimg.
Thefi and thegi are polynomials of degree at most i in c. Since 0 is not an eigenvalue of adz ong/g0(adz), we see thatgn−r(0)6= 0. So we can findr+ 1 values ofcfor whichgn−r(c)6= 0, and hence for these values,
g0(ad(z+cx))⊂g0(adz).
By the minimality, this forces
g0(ad(z+cx)) =g0(adz)
for these values ofc. This means thatf(T, c) =Tr for these values ofc, so each of the polynomials f1, . . . , fr has r+ 1 distinct roots, and hence is identically zero. Hence
g0(ad(z+cx))⊃g0(adz)
for allc. Takec= 1, x=y−z to conclude the truth of the lemma.