One way of producing invariant forms is from representations: if(ρ, V ) is a

representation of g, then

(x, y)ρ := tr ρ(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 the Killing form and will be denoted by ( , )κ . We will also sometimes

write κ(x, y) instead of (x, y)κ .

Theorem 8 g is 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 = ad x ad y. Then

σ maps g ’ a and a ’ 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

tr σ = 0. Hence if g is not semisimple then its Killing form is degenerate.

Conversely, suppose that g is semi-simple. We wish to show that the Killing

form is non-degenerate. So let u := g⊥ = {x| tr ad x ad y = 0 ∀y ∈ g}. If

x ∈ u, z ∈ g then

tr{ad[x, z] ad y} = tr{ad x ad z ad y ’ ad z ad x ad y)}

= tr{ad x(ad z ad y ’ ad y ad z)}

= tr ad x ad[z, y]

= 0,

so u is an ideal. In particular, tru (ad xu ad yu ) = trg (adg x adg y) for x, y ∈ u, as

can be seen from a block decomposition starting with a basis of u and extending

to g.

If we take y ∈ Du, we see that tr ad uD ad u = 0, so ad u is solvable by

Cartan™s criterion. But the kernel of the map u ’ ad u is the center of u. So if

ad u is solvable, so is u. QED

Proposition 8 Let g be a semisimple algebra, i any ideal of g, and i⊥ its

orthocomplement with respect to its Killing form. Then i © i⊥ = 0.

Indeed, i © i⊥ is an ideal on which tr ad x ad y ≡ 0 hence is solvable by Cartan™s

criterion. Since g is 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 Let φ : g ’ s be a surjective homomorphism of a semi-simple

Lie algebra onto a simple Lie algebra. Then if g = gi is a decomposition of

g into simple ideals, the restriction, φi of φ to each summand is zero, except for

one summand where it is an isomorphism.

Proof. Since s is simple, the image of every φi is 0 or all of s. If φi is surjective

for some i then it is an isomorphism since gi is simple. There is at least one i

for which it is surjective since φ is surjective. On the other hand, it can not be

surjective for for two ideals, gi , gj i = j for then φ[gi , gj ] = 0 = [s, s] = s. QED

4.7 Complete reducibility.

The basic theorem is

Theorem 9 [Weyl.] Every ¬nite dimensional representation of a semi-simple

Lie algebra is completely reducible.

Proof.

1. If ρ : g ’ End V is injective, then the form ( , )ρ is non-degenerate.

Indeed, the ideal consisting of all x such that (x, y)ρ = 0 ∀y ∈ g is solvable

by Cartan™s criterion, hence 0.

2. The Casimir operator. Let (ei ) and (fi ) be bases of g which are dual

with respect to some non-degenerate invariant bilinear form, (, ). So

(ei , fj ) = δij . As the form is non-degenerate and invariant, it de¬nes

a map of

g — g ’ End g; x — y(w) = (y, w)x.

This map is an isomorphism and is a g morphism. Under this map,

ei — fi (w) = (w, fi )ei = w

by the de¬nition of dual bases. Hence under the inverse map

End g ’ g — g

ei — fi (and so this expression

the identity element, id, corresponds to

is independent of the choice of dual bases). Since id is annihilated by

i ei — fi is

commutator by any element of End(g), we conclude that

annihilated by the action of all (ad x)2 = ad x — 1 + 1 — ad x, x ∈ g.

Indeed, for x, e, f, y ∈ g we have

((ad x)2 (e — f )) y = (ad xe — f + e — ad xf ) y

= (f, y)[x, e] + ([x, f ], y)e

= (f, y)[x, e] ’ (f, [x, y])e by (4.2)

= ((ad x)(e — f ) ’ (e — f )(ad x)) y.

70 CHAPTER 4. ENGEL-LIE-CARTAN-WEYL

Set

ei · fi ∈ U (L).

C := (4.3)

i

Thus C is the image of the element i ei —fi under the multiplication map

g—g ’ U (g), and is independent of the choice of dual bases. Furthermore,

C is annihilated by ad x acting on U (g). In other words, it commutes with

all elements of g, and hence with all of U (g); it is in the center of U (g).

The C corresponding to the Killing form is called the Casimir element,

its image in any representation is called the Casimir operator.

3. Suppose that ρ : g ’ End V is injective. The (image of the) central

element corresponding to ( , )ρ de¬nes an element of End V denoted by

Cρ and

tr Cρ = tr ρ( ei fi )

= tr ρ(ei )ρ(fi )

= (ei , fi )

i

= dim g

With these preliminaries, we can state the main proposition:

Proposition 12 Let 0 ’ V ’ W ’ k ’ 0 be an exact sequence of g modules,

where g is semi-simple, and the action of g on k is 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 of sl(2), the proposition is a special case

of the theorem we want to prove. But we shall see that it is su¬cient to prove

the theorem.

Proof of proposition. It is enough to prove the proposition for the case

that V is an irreducible module. Indeed, if V1 is a submodule, then by induction

on dim V we may assume the theorem is known for 0 ’ V /V1 ’ W/V1 ’ k ’ 0

so that there is a one dimensional invariant subspace M in W/V1 supplementary

to V /V1 on which the action is trivial. Let N be the inverse image of M in W .

By another application of the proposition, this time to the sequence

0 ’ V1 ’ N ’ M ’ 0

we ¬nd an invariant line, P , in N complementary to V1 . So N = 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 of W which is complementary to V .

4.7. COMPLETE REDUCIBILITY. 71

Next we can reduce to proving the proposition for the case that g acts

faithfully on V . Indeed, let i = the kernel of the action on V . For all x ∈ g we

have, by hypothesis, xW ‚ V , and for x ∈ i we have xV = 0. Hence Di acts

trivially on W . But i = Di since i is semi-simple. Hence i acts trivially on W

and we may pass to g/i. This quotient is again semi-simple, since i is 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, of g

on V is injective. Then we have an invariant element Cρ whose image in End W

must map W ’ V since every element of g does. (We may assume that g = 0.)

On the other hand, Cρ = 0, indeed its trace is dim g. 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 to V is an isomorphism. Hence ker Cρ : W ’ V is