Theorem 4 Engel Let ρ : g ’ End(V ) be a representation such that ρ(x) is

nilpotent for each x ∈ g. Then there exists a basis in terms of which ρ(g) ‚

n+ (gl(d)), i.e. becomes strictly upper triangular. Here d =dim V .

Given a single nilpotent operator, we can always ¬nd a non-zero vector, v

which it sends into zero. Then on V /{v} a non-zero vector which the induced

61

62 CHAPTER 4. ENGEL-LIE-CARTAN-WEYL

map sends into zero etc. So in terms of such a ¬‚ag, the corresponding matrix

is strictly upper triangular. The theorem asserts that we are can ¬nd a single

¬‚ag which works for all ρ(x). In view of the above proof for a single operator,

Engel™s theorem follows from the following simpler looking statement:

Theorem 5 Under the hypotheses of Engel™s theorem, if V = 0, there exists a

non-zero vector v ∈ V such that ρ(x)v = 0 ∀x ∈ g.

Proof of Theorem 5 in seven easy steps.

• Replace g by its image, i.e. assume that g ‚ End V .

• Then (ad x)y = Lx y ’ Rx y where Lx is the linear map of End V into itself

given by left multiplication by x, and Rx is given by right multiplication

by x. Both Lx and Rx are nilpotent as operators since x is nilpotent.

Also they commute. Hence by the binomial formula (ad x)n = (Lx ’ Rx )n

vanishes for su¬ciently large n.

• We may assume (by induction) that for any Lie algebra, m, of smaller

dimension than that of g (and any representation) there exists a v ∈ V

such that xv = 0 ∀x ∈ m.

• Let k ‚ g be a subalgebra, k = g, and let

N = N (k) := {x ∈ g|(ad x)k ‚ k}

be its normalizer. The claim is that

3 N (k) is strictly larger than k.

To see this, observe that each x ∈ k acts on k and on g/k by nilpotent

maps, and hence there is an 0 = y ∈ g/k killed by all x ∈ k. But then

ˆ

y ∈ k, and [y, x] = ’[x, y] ∈ k for all x ∈ k. So y ∈ N (k), y ∈ k.

• If g = 0, there is an ideal i ‚ g such that dimg/i = 1. Indeed, let i be a

maximal proper subalgebra of g. Its normalizer is strictly larger, hence all

of g, so i is an ideal. The inverse image in g of a line in g/i is a subalgebra,

and is strictly larger than i. Hence it must be all of g.

• Choose such an ideal, i. The subspace

W ‚ V, W = {v|xv = 0, ∀x ∈ i}

is invariant under g. Indeed, if y ∈ g, w ∈ W then xyw = yxw + [x, y]w =

0.

• W = 0 by induction. Take y ∈ g, y ∈ i. It preserves W and is nilpotent.

Hence there is a non-zero v ∈ W with yv = 0. Since y and i span g, we

have xv = 0 ∀x ∈ g. QED

No assumptions about the ground ¬eld went into this.

4.2. SOLVABLE LIE ALGEBRAS. 63

4.2 Solvable Lie algebras.

Let g be a Lie algebra. Dn g is de¬ned inductively by

D0 g := g, D1 (g) := [g, g], . . . , Dn+1 g := [Dn g, Dn g].

If we take b to consist of all upper triangular n — n matrices, then D1 b = n+

consists of all strictly triangular matrices and then successive brackets eventually

lead to zero. We claim that the following conditions are equivalent and any Lie

algebra satisfying them is called solvable.

1. ∃n |Dn g = 0.

2. ∃n such that for every family of 2n elements of g the successive brackets

of brackets vanish; e.g for n = 4 this says

[[[[x1 , x2 ], [x3 , x4 ]], [[x5 , x6 ], [x7 , x8 ]]], [[[x9 , x10 ], [x11 , x12 ]], [[x13 .x14 ], [x15 , x16 ]]]] = 0.

3. There exists a sequence of subspaces g := i1 ⊃ i2 ⊃ · · · ⊃ in = 0 such each

is an ideal in the preceding and such that the quotient ij /ij+1 is abelian,

i.e. [ij , ij ] ‚ ij+1 .

Proof of the equivalence of these conditions. [g, g] is always an ideal in

g so the Dj g form a sequence of ideals demanded by 3), and hence 1) ’ 3). We

also have the obvious implications 3) ’ 2) and 2) ’ 1). So all these de¬nitions

are equivalent.

Theorem 6 [Lie.] Let g be a solvable Lie algebra over an algebraically closed

¬eld k of characteristic zero, and (ρ, V ) a ¬nite dimensional representation of g.

Then we can ¬nd a basis of V so that ρ(g) consists of upper triangular matrices.

By induction on dim V this reduces to

Theorem 7 [Lie.] Under the same hypotheses, there exists a (non-zero) com-

mon eigenvector v for all the ρ(y), i.e. there is a vector v ∈ V and a function

χ : g ’ k such that

ρ(y)v = χ(y)v ∀ y ∈ g. (4.1)

Lemma 2 Suppose that i is an ideal of g and (4.1) holds for all y ∈ i. Then

χ([x, h]) = 0, ∀ x ∈ g h ∈ i.

Proof of lemma. For x ∈ g let Vi be the subspace spanned by v, xv, . . . , xi’1 v

and let n > 0 be minimal such that Vn = Vn+1 . So Vn is ¬nite dimensional and

xVn ‚ Vn . Also Vn = Vn+k ∀k.

64 CHAPTER 4. ENGEL-LIE-CARTAN-WEYL

Also, for h ∈ i, (dropping the ρ) we have:

hv =

χ(h)v

xhv ’ [x, h]v

hxv =

≡

χ(h)xv mod V1

hx2 v =

xhxv + [h, x]xv

χ(h)x2 v + uxv, mod V1 u ∈ I

≡

χ(h)x2 v + χ(u)xv mod V1

≡

χ(h)x2 v mod V2

=

. .

. .

. .

hxi v ≡ χ(h)xi v mod Vi .

Thus Vn is invariant under i and for each h ∈ i, tr|Vn h = nχ(h). In particular

both x and h leave Vn invariant and tr|Vn [x, h] = 0 since the trace of any

commutator is zero. This proves the lemma.

Proof of theorem by induction on dim g, which we may assume to be positive.

Let m be any subspace of g with g ⊃ m ⊃ [g, g]. Then [g, m] ‚ [g, g] ‚ m so m

is an ideal in g. In particular, we may choose m to be a subspace of codimension

1 containing [g, g]. By induction we can ¬nd a v ∈ V and a χ : m ’ k such

that (4.1) holds for all elements of m. Let

W := {w ∈ V |hw = χ(h)w ∀ h ∈ m}.

If x ∈ g, then

hxw = xhw ’ [x, h]w = χ(h)xw ’ χ([x, h])w = χ(h)xw

since χ([x, h]) = 0 by the lemma. Thus W is stable under all of g. Pick

x ∈ g, x ∈ m, and let v ∈ W be an eigenvector of x with eigenvalue », say.

Then v is a simultaneous eigenvector for all of g with χ extended as

χ(h + rx) = χ(h) + r». QED

We had to divide by n in the above argument. In fact, the theorem is not

true over a ¬eld of characteristic 2, with sl(2) as a counterexample.

Applied to the adjoint representation, Lie™s theorem says that there is a ¬‚ag

of ideals with commutative quotients, and hence [g, g] is nilpotent.

4.3 Linear algebra

Let V be a ¬nite dimensional vector space over an algebraically closed ¬eld of