5.4 Toral subalgebras and Cartan subalgebras.

The strategy is now to show that any two BSA™s of an arbitrary Lie algebra are

conjugate. Any CSA is nilpotent, hence solvable, hence contained in a BSA.

This reduces the proof of the conjugacy theorem for CSA™s to that of BSA™s

as we know the conjugacy of CSA™s in a solvable algebra. Since the radical is

contained in any BSA, it is enough to prove this theorem for semi-simple Lie

algebras. So for this section the Lie algebra g will be assumed to be semi-simple.

Since g does not consist entirely of ad nilpotent elements, it contains some x

which is not ad nilpotent, and the semi-simple part of x is a non-zero ad semi-

simple element of g. A subalgebra consisting entirely of semi-simple elements

is called toral, for example, the line through xs .

Lemma 7 Any toral subalgebra t is abelian.

Proof. The elements ad x, x ∈ t can be each be diagonalized. We must

show that ad x has no eigenvectors with non-zero eigenvalues in t. Let y be an

eigenvector so [x, y] = ay. Then (ad y)x = ’ay is a zero eigenvector of ad y,

which is impossible unless ay = 0, since ad y annihilates all its zero eigenvectors

and is invertible on the subspace spanned by the eigenvectors corresponding to

non-zero eigenvalues. QED

One of the consequences of the considerations in this section will be:

Theorem 11 A subalgebra h of a semi-simple Lie algebra gis a CSA if and

only if it is a maximal toral subalgebra.

To prove this we want to develop some of the theory of roots. So ¬x a

maximal toral subalgebra h. Decompose g into simultaneous eigenspaces

g = Cg (h) g± (h)

where

Cg (h) := {x ∈ g|[h, x] = 0 ∀ h ∈ h}

is the centralizer of h, where ± ranges over non-zero linear functions on h and

g± (h) := {x ∈ g|[h, x] = ±(h)x ∀ h ∈ h}.

As h will be ¬xed for most of the discussion, we will drop the (h) and write

g = g0 • g±

where g0 = Cg (h). We have

• [g± , gβ ] ‚ g±+β (by Jacobi) so

• ad x is nilpotent if x ∈ g± , ± = 0

• If ± + β = 0 then κ(x, y) = 0 ∀ x ∈ g± , y ∈ gβ .

80 CHAPTER 5. CONJUGACY OF CARTAN SUBALGEBRAS.

The last item follows by choosing an h ∈ h with ±(h) + β(h) = 0. Then

0 = κ([h, x], y) + κ(x, [h, y]) = (±(h) + β(h))κ(x, y) so κ(x, y) = 0. This implies

that g0 is orthogonal to all the g± , ± = 0 and hence the non-degeneracy of κ

implies that

Proposition 14 The restriction of κ to g0 — g0 is non-degenerate.

Our next intermediate step is to prove:

Proposition 15

h = g0 (5.3)

if h is a maximal toral subalgebra.

Proceed according to the following steps:

x ∈ g0 ’ xs ∈ g0 xn ∈ g0 . (5.4)

Indeed, x ∈ g0 ” ad x : h ’ 0, and then ad xs , ad xn also map h ’ 0.

x ∈ g0 , x semisimple ’ x ∈ h. (5.5)

Indeed, such an x commutes with all of h. As the sum of commuting semi-

simple transformations is again semisimple, we conclude that h + Cx is a toral

subalgebra. By maximality it must coincide with h.

We now show that

Lemma 8 The restriction of the Killing form κ to h — h is non-degenerate.

So suppose that κ(h, x) = 0 ∀ x ∈ h. This means that κ(h, x) = 0 ∀ semi-

simple x ∈ g0 . Suppose that n ∈ g0 is nilpotent. Since h commutes with n,

(ad h)(ad n) is again nilpotent. Hence has trace zero. Hence κ(h, n) = 0, and

therefore κ(h, x) = 0 ∀ x ∈ g0 . Hence h = 0. QED

Next observe that

Lemma 9 g0 is a nilpotent Lie algebra.

Indeed, all semi-simple elements of g0 commute with all of g0 since they belong

to h, and a nilpotent element is ad nilpotent on all of g so certainly on g0 .

Finally any x ∈ g0 can be written as a sum xs + xn of commuting elements

which are ad nilpotent on g0 , hence x is. Thus g0 consists entirely of ad nilpotent

elements and hence is nilpotent by Engel™s theorem. QED

Now suppose that h ∈ h, x, y ∈ g0 . Then

κ(h, [x, y]) = κ([h, x], y)

= κ(0, y)

=0

and hence, by the non-degeneracy of κ on h, we conclude that

5.5. ROOTS. 81

Lemma 10

h © [g0 , g0 ] = 0.

We next prove

Lemma 11 g0 is abelian.

Suppose that [g0 , g0 ] = 0. Since g0 is nilpotent, it has a non-zero center con-

tained in [g0 , g0 ]. Choose a non-zero element z ∈ [g0 , g0 ] in this center. It can

not be semi-simple for then it would lie in h. So it has a non-zero nilpotent part,

n, which also must lie in the center of g0 , by the B ‚ A theorem we proved in

our section on linear algebra. But then ad n ad x is nilpotent for any x ∈ g0

since [x, n] = 0. This implies that κ(n, g0 ) = 0 which is impossible. QED

Completion of proof of (5.3). We know that g0 is abelian. But then, if

h = g0 , we would ¬nd a non-zero nilpotent element in g0 which commutes with

all of g0 (proven to be commutative). Hence κ(n, g0 ) = 0 which is impossible.

This completes the proof of (5.3). QED

So we have the decomposition

g =h• g±

±=0

which shows that any maximal toral subalgebra h is a CSA.

Conversely, suppose that h is a CSA. For any x = xs + xn ∈ g, g0 (ad xs ) ‚

g0 (ad x) since xn is an ad nilpotent element commuting with ad xs . If we choose

x ∈ h minimal so that h = g0 (ad x), we see that we may replace x by xs

and write h = g0 (ad xs ). But g0 (ad xs ) contains some maximal toral algebra

containing xs , which is then a Cartan subalgebra contained in h and hence must

coincide with h. This completes the proof of the theorem. QED

5.5 Roots.

We have proved that the restriction of κ to h is non-degenerate. This allows us

to associate to every linear function φ on h the unique element tφ ∈ h given by

φ(h) = κ(tφ , h).

The set of ± ∈ h— , ± = 0 for which g± = 0 is called the set of roots and is

denoted by ¦. We have

• ¦ spans h— for otherwise ∃h = 0 : ±(h) = 0 ∀± ∈ ¦ implying that

[h, g± ] = 0 ∀± so [h, g] = 0.

• ± ∈ ¦ ’ ’± ∈ ¦ for otherwise g± ⊥ g.

• x ∈ g± , y ∈ g’± , ± ∈ ¦ ’ [x, y] = κ(x, y)t± . Indeed,

κ(h, [x, y]) = κ([h, x], y)

= κ(t± , h)κ(x, y)

= κ(κ(x, y)t± , h).

82 CHAPTER 5. CONJUGACY OF CARTAN SUBALGEBRAS.

• [g± , g’± ] is one dimensional with basis t± . This follows from the preceding

and the fact that g± can not be perpendicular to g’± since otherwise it

will be orthogonal to all of g.

• ±(t± ) = κ(t± , t± ) = 0. Otherwise, choosing x ∈ g± , y ∈ g’± with κ(x, y) =

1, we get