(T ’ »i )mi

det(T I ’ u) =

4.3. LINEAR ALGEBRA 65

be the factorization of its characteristic polynomial where the »i are distinct.

Let S(T ) be any polynomial satisfying

S(T ) ≡ »i mod (T ’ »i )mi , S(T ) ≡ 0 mod T,

which is possible by the Chinese remainder theorem. For each i let Vi := the

kernel of (u ’ »i )mi . Then V = Vi and on Vi , the operator S(u) is just the

scalar operator »i I. In particular s = S(u) is semi-simple (its eigenvectors span

V ) and, since s is a polynomial in u it commutes with u. So

u=s+n

where

n = N (u), N (T ) = T ’ S(T )

is nilpotent. Also

ns = sn.

We claim that these two elements are uniquely determined by

u = s + n, sn = ns,

with s semisimple and n nilpotent. Indeed, since sn = ns, su = us so s(u ’

»i )k = (u’»i )k s so sVi ‚ Vi . Since s’u is nilpotent, s has the same eigenvalues

on Vi as u does, i.e. »i . So s and hence n is uniquely determined.

If P (T ) is any polynomial with vanishing constant term, then if A ‚ B

are subspaces with uB ‚ A then P (u)B ‚ A. So, in particular, sB ‚ A and

nB ‚ A.

De¬ne

Vp,q := V — V — · · · — V — V — — · · · — V —

with p copies of V and q copies of V — . Let u ∈ End(V ) act on V — by ’u— and

on Vpq by derivation, so , for example,

u12 = u — 1 — 1 ’ 1 — u— — 1 ’ 1 — 1 — u— .

Similarly, u11 acts on V1,1 = V — V — by

u11 (x — ) = ux — ’ x — u— .

Under the identi¬cation of V —V — with End(V ), the element x— acts on y ∈ V

by sending it into

(y)x.

So the element u11 (x — ) sends y to

(y)u(x) ’ (u— )(y)x = (y)u(x) ’ (u(y))x.

This is the same as the commutator of the operator u with the operator (cor-

responding to) x — acting on y. In other words, under the identi¬cation of

V — V — with End(V ), the linear transformation u11 gets identi¬ed with ad u.

66 CHAPTER 4. ENGEL-LIE-CARTAN-WEYL

Proposition 2 If u = s + n is the decomposition of u then upq = spq + npq is

the decomposition of upq .

Proof. [spq , npq ] = 0 and the tensor products of an eigenbasis for s is an

eigenbasis for spq . Also npq is a sum of commuting nilpotents hence nilpotent.

The map u ’ upq is linear hence upq = spq + npq . QED

If φ : k ’ k is a map, we de¬ne φ(s) by φ(s)|Vi = φ(»i ). If we choose a

polynomial such that P (0) = 0, P (»i ) = φ(»i ) then P (u) = φ(s).

Proposition 3 Suppose that φ is additive. Then

(φ(s))pq = φ(spq ).

Proof. Decompose Vpq into a sum of tensor products of the Vi or Vj— . On each

such space we have

φ(sp,q ) = φ(»i1 + · · · ’ · · · )

= φ(»i1 ) + φ(..)..

= (φ(s))p,q

where the middle equation is just the additivity. QED

As an immediate consequence we obtain

Proposition 4 Notation as above. If A ‚ B ‚ Vp,q with upq B ‚ A then for

any additive map, φ(s)pq B ‚ A

Proposition 5 (over C) Let u = s + n as above. If tr(uφ(s)) = 0 for φ(s) = s

then u is nilpotent.

mi |»i |2 . So the condition implies that all the

Proof. tr uφ(s) = mi »i »i =

»i = 0. QED

4.4 Cartan™s criterion.

Let g ‚ End(V ) be a Lie subalgebra where V is ¬nite dimensional vector space

over C. Then

g is solvable ” tr(xy) = 0 ∀x ∈ g, y ∈ [g, g].

Proof. Suppose g is solvable. Choose a basis for which g is upper triangular.

Then every y ∈ [g, g] has zeros on the diagonal, Hence tr(xy) = 0. For the

reverse implication, it is enough to show that [g, g] is nilpotent, and, by Engel,

that each u ∈ [g, g] is nilpotent. So it is enough to show that tr us = 0, where

s is the semisimple part of u, by Proposition 5 above. If it were true that s ∈ g

we would be done, but this need not be so. Write

u= [xi , yi ].

4.5. RADICAL. 67

Now for a, b, c ∈ End(V )

= tr(abc ’ bac)

tr([a, b]c)

= tr(bca ’ bac)

= tr(b[c, a]) so

tr(us) = tr([xi , yi ]s)

= tr(yi [s, xi ]).

So it is enough to show that ad s : g ’ [g, g]. We know that ad u : g ’ [g, g],

and we can, by Lagrange interpolation, ¬nd a polynomial P such that P (u) = s.

The result now follows from Prop. 4:

Since End(V ) ∼ V1,1 , take A = [g, g] and B = g. Then ad u = u1,1 so

u1,1 g ‚ [g, g[ and hence s1,1 g ‚ [g, g] or [s, x] ∈ [g, g] ∀x ∈ g. QED

4.5 Radical.

If i is an ideal of g and g/i is solvable, then D(n) (g/i) = 0 implies that D(n) g ‚ i.

If i itself is solvable with D(m) i = 0, then D(m+n) g = 0. So we have proved:

Proposition 6 If i ‚ g is an ideal, and both i and g/i are solvable, so is g.

If i and j are solvable ideals, then (i + j)/j ∼ i/(i © j) is solvable, being the

homomorphic image of a solvable algebra. So, by the previous proposition:

Proposition 7 If i and j are solvable ideals in g so is i + j. In particular, every

Lie algebra g has a largest solvable ideal which contains all other solvable ideals.

It is denoted by rad g or simply by r when g is ¬xed.

An algebra g is called semi-simple if rad g = 0. Since Di is an ideal

whenever i is (by Jacobi™s identity), if r = 0 then the last non-zero D(n) r is

an abelian ideal. So an equivalent de¬nition is: g is semi-simple if it has no

non-zero abelian ideals.

We shall call a Lie algebra simple if it is not abelian and if it has no proper

ideals. We shall show in the next section that every semi-simple Lie algebra is

the direct sum of simple Lie algebras in a unique way.

4.6 The Killing form.

A bilinear form ( , ) : g — g ’ k is called invariant if

([x, y], z) + (y, [x, z]) = 0 ∀x, y, z ∈ g. (4.2)

Notice that if ( , ) is an invariant form, and i is an ideal, then i⊥ is again an

ideal.