On the Conjugacy Theorems of Cartan and Borel Subalgebras
by
Mary Clair Thompson
A thesis submitted to the Graduate Faculty of
Auburn University
in partial ful?llment of the
requirements for the Degree of
Master of Science
Auburn, Alabama
May 14, 2010
Keywords: Cartan subalgebras, Borel subalgebras, conjugacy
Copyright 2010 by Mary Clair Thompson
Approved by:
Tin-Yau Tam, Chair, Professor of Mathematics
Randall R. Holmes, Professor of Mathematics
Huajun Huang, Assistant Professor of Mathematics
Abstract
We study the conjugacy theorems of Cartan subalgebras and Borel subalgebras of gen-
eral Lie algebras. We present a history of the problem, along with two proofs of the the-
orems which stay completely within the realm of Lie algebras. The ?rst is a reworking by
Humphreys of an earlier proof, relying upon the ideas of Borel subalgebras and using dou-
ble induction. The second proof is a newer proof presented by Michael which substantially
simpli?es the theory.
ii
Acknowledgments
The author would like to thank Dr. Tin-Yau Tam for his skillful teaching and guidance
throughout the course of this research, and her parents for their continual support, love, and
encouragement.
iii
Table of Contents
Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii
Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii
1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
2 De?nitions and Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
3 Properties of Cartan Subalgebras . . . . . . . . . . . . . . . . . . . . . . . . . . 14
4 Humphreys? Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
5 Michael?s Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
6 Some remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
iv
Chapter 1
Introduction
A Cartan subalgebra of a Lie algebra L over the ?eld F, often abbreviated CSA, is a
subalgebra H of L that is (1) nilpotent and (2) self-normalizing. For example, any nilpotent
Lie algebra is its own Cartan subalgebra.
Any nilpotent subalgebra is also solvable, thus contained in a maximal solvable subal-
gebra of L. Such subalgebras are of su?cient importance to merit their own name, Borel
subalgebras, and are often denoted BSAs.
In a Lie algebra over an algebraically closed ?eld of characteristic 0, it is a truly re-
markable fact that any pair of CSAs are conjugate, in the sense that the subgroup E(L) of
IntL acts transitively on the set of CSAs. Indeed the BSAs are also conjugate under the
same set of automorphisms. This paper will examine two di?erent elementary proofs of the
conjugacy theorems.
CSAs overCwere ?rst introduced by ?Elie Cartan in his 1894 doctoral dissertation [8] in
order to better study complex semisimple Lie algebras. His work was a major contribution to
Lie algebras in that Cartan completed the classi?cation of the complex semisimple algebras
which Wilhelm Killing had begun. CSAs play an important role in the structure theory of
semisimple Lie algebras. Due to their importance and to Cartan?s contribution to the theory,
Chevalley [9] proposed naming them Cartan subalgebras.
If the ?eld is algebraically closed and of characteristic 0 and the algebra is ?nite di-
mensional, Chevalley [9] proved that two CSAs of L are conjugate via IntL. Also see the
comments of Borel [4, p.148] on the history of the development. The special case where L is
semisimple had already been proved previously by Weyl [23] (also see Hunt [12] for a metric
proof) using analytic methods (F = C) and by Weil [22] by topological methods (also see
1
Hopf and Samelson [10]). However, Chevalley assumed that more generally the ?eld is alge-
braically closed and has characteristic zero so that he did not have a ready-made analogue
of the adjoint group IntL to perform the conjugacy [4, p.148]. Chevalley?s proof uses the
methods of algebraic geometry. Of particular importance in the proof is the use of Pl?ucker
coordinates.
Winter [25], based on the techniques developed by Mostow, gave an elementary algebraic
(non-geometric) proof of the conjugacy theorem. The proof presented in Humphreys? book
[11] follows the approach of Winter [25].
Michael [16] gave a new elementary proof for the conjugacy theorem of CSAs of a ?nite-
dimensional Lie algebra over an algebraically closed ?eld of characteristic zero. The approach
?ts into the theme of the presentation in Bourbaki [7].
Humphreys begins his proof of the conjugacy of CSAs by proving the theorem directly
for solvable Lie algebras. However, his focus soon shifts to BSAs, and after establishing
several properties of BSAs and providing connections between the mechanics of BSAs and
CSAs, he proceeds to prove that BSAs of a semisimple Lie algebra are conjugate. The proof
is highly technical and employs double induction.
Humphreys then shows that the BSAs of a general Lie algebra L are in 1-1 correspon-
dence with the BSAs of the semisimple Lie algebra L/R. As the conjugacy of BSAs of
semisimple algebras has already been established, the conjugacy in a general algebra follows
readily. Humphreys ?nally returns his attention to CSAs; since any CSA is contained in a
BSA, the conjugacy of CSA is almost immediate.
The proof contributed by Michael begins, as with that of Humphreys, with a proof of
the conjugacy of CSAs of solvable Lie algebras; this particular proof is a reworking of a proof
given in Bourbaki [5].
Similar to Humphreys approach, Michael relies heavily upon BSAs. However, he uses
properties of BSAs that are readily established and involve relatively simple facts from linear
2
algebra. In particular, he uses dimension arguments and orthogonal complements to simplify
the proofs.
Michael extends the vector space properties of BSAs in semisimple Lie algebras in such
a way that the conjugacy of CSAs of semisimple Lie algebras is relatively easily proven.
He then uses a ?connecting lemma? to pull the argument back to the general case, thus
establishing the conjugacy of CSAs of any Lie algebra.
Finally, in order to establish the conjugacy of BSAs, Michael proves that any BSA must
contain a CSA. The proof of the conjugacy of BSAs then follows by a standard argument.
3
Chapter 2
De?nitions and Preliminaries
All Lie algebras L are  nite dimensional, and the  eld F is algebraically closed with
characteristic 0.
De nition 2.1. ([11, p.1]) Let L be a vector space over a ?eld F. Then L is a Lie algebra
if it is endowed with a bilinear operation [?,?] such that for all x, y, and z ?L,
1. [x,x] = 0.
2. [x,[y,z]] + [y,[z,x]] + [z,[x,y]] = 0 (the Jacobi Identity).
A subalgebra of C of L is a vector subspace of L that is closed under [?,?].
Example 2.2. The following are Lie algebras:
1. The general linear algebra gl(V) (V a vector space overF) is the set End(V) endowed
with the operation [x,y] = xy?yx, x,y ? gl(V). Here End(V) denotes the set of all
endomorphisms of V.
2. The algebra of n?n matrices gl(n,F) over F, with [?,?] given by [A,B] = AB?BA,
A,B ? gl(n,F).
3. L in which [?,?] is trivially de?ned, i.e. [x,y] = 0 for every pair of elements x and y of
L; in such a case we call L abelian.
De nition 2.3. ([11, p.7]) The normalizer of a subalgebra C of L is
NL(C) := {x?L| [x,C] ?C}.
4
We call C self-normalizing if NL(C) = C. The centralizer of C is
ZL(C) := {x?L| [x,C] = 0}.
We write Z(L) = ZL(L) and call it the center of L. So L is abelian if and only if
Z(L) = L.
De nition 2.4. ([11, p.6]) An ideal I of L is a subalgebra such that [x,y] ?I for all x?L
and y ?I.
We note that all ideals are two-sided, for bilinearity combined with condition (1) of
De?nition 2.1 imply that [x,y] = ?[y,x] for any x,y ? L. If L has no ideals except itself
and 0, then we call L simple.
Given an idealI ofL, the quotient spaceL/I is endowed with a bracket: [x+I,y+I] :=
[x,y] +I, x,y ? L. The operation is unambiguous so that L/I is a Lie algebra, called the
quotient algebra.
In general, the vector space sum I +J = {x+y | x ? I,y ? J} of two subalgebras I
and J of L need not be a subalgebra; however, if I is an ideal and J is a subalgebra, then
I +J is indeed a subalgebra. If both I and J are ideals, then I +J is an ideal as well.
De nition 2.5. ([11, p.7]) A vector space homomorphism ? : L ? L? is a Lie algebra
homomorphism if ?([x,y]) = [?(x),?(y)] for all x, y ? L. If ? : L ? L is an isomorphism,
we call ? an automorphism. The group of all automorphisms of L is denoted by AutL.
The image Im? is a subalgebra of L? and the kernel Ker? is an ideal of L. The inverse
image ??1(C?) of a subalgebra C? of L? is a subalgebra of L.
5
Theorem 2.6. [11, p.7] (First Isomorphism Theorem) If ? : L ? L? is a Lie algebra
homomomorphism, Ker? = K, then L/K ?= Im?, and the following diagram commutes:
L
? !!CCC
CCCC
CC
? //L/K
?
  
L?
Proof. Let ? be the canonical projection of L onto L/K, and de?ne ? : L/K ? L? by
?(x+K) = ?(x). Then ?(?(x)) = ?(x+K) = ?(x), and the diagram commutes.
Givena = ?(x) ? Im?, we have?(x+K) = a, and? is onto. If?(x+K) = ?(y+K) we
have 0 = ?((x?y)+K) = ?(x?y), and x?y ? Ker?. So ?(x?y) ?K, i.e. x+K = y+K,
and ? is one to one, thus an isomorphism.
Given a Lie algebra L, denote by adx? EndL the endomorphism de?ned by
adx(y) = [x,y], y ?L.
If the subspaceC ofLis adxstable, we denote the restriction of adxtoC by adCx : C ?C.
De nition 2.7. An endomorphismtis nilpotent iftk = 0 for somek> 0. An endomorphism
is semisimple if the roots of its minimal polynomial over F are distinct.
An element x ? L is called ad-nilpotent if the endomorphism adx is a nilpotent one.
Similarly, we call x ad-semisimple if adx is a semisimple endomorphism.
De nition 2.8. ([11, p.8]) The adjoint map ad : L? gl(L) is a representation of L, i.e., a
homomorphism of L with a Lie algebra of endomorphisms. Clearly the kernel of ad, denoted
Kerad, is the center Z(L) of L.
As we shall see, the adjoint representation has many useful properties which make it of
fundamental importance in the study of Lie algebras.
We now introduce two important classes of Lie algebras, namely, nilpotent and solvable
algebras.
6
De nition 2.9. ([11, p.11]) The Lie algebra L is nilpotent if the descending central (or
lower central) series de?ned by L0 := L, L1 := [L,L],...,Li := [L,Li?1] terminates.
Example 2.10. The algebra of all strictly upper triangular matrices n(n,F) overFis nilpo-
tent.
It is not hard to see that adx ? gl(L) is nilpotent for all x ? L (i.e. x is ad-nilpotent)
if L is nilpotent. The converse is true and is known as Engel?s theorem ([11, p.12-13] or [5,
p.39-40]).
Theorem 2.11. (Engel)
1. Let L? gl(V) be a subalgebra of gl(V), V ?nite dimensional. If L consists of nilpotent
endomorphsims and V ?= 0, then there exists nonzero v ? V such that ?v = 0 for all
??L.
2. A Lie algebra L is nilpotent if and only if adx ? EndL is a nilpotent endomorphism
for all x?L.
De nition 2.12. ([11, p.10]) The Lie algebra L is solvable if the derived series de?ned by
L(0) := L, L(1) := [L,L],...,L(i) := [L(i?1),L(i?1)] terminates, i.e., if there is n ? N such
that L(n) = 0.
Every nilpotent algebra is solvable: clearlyL(i) ?Li for alli, so if the descending central
series terminates, the derived series does as well.
The following is Lie?s theorem ([11, p.15-16] or [5, p.46]) on the characterization of
solvable algebras.
Theorem 2.13. (Lie) The Lie algebra L is solvable if and only if the derived algebra [L,L]
is nilpotent.
An alternate characterization of solvable Lie algebras is given by Cartan?s Criterion([11,
p.20] or [5, p.48]):
7
Theorem 2.14. (Cartan?sCriterion)The LiealgebraLissolvableif and onlyif tr(adxady) =
0 for all x? [L,L], y ?L.
Every Lie algebra L has a unique maximal solvable ideal R [11, p.11]: for if I is any
other solvable ideal of L, then R +I is a solvable ideal as well. By the maximality of R,
R+I = R, or I ?R.
De nition 2.15. ([11, p.11]) The radical ofL, denoted byR = RadL, is the unique maximal
solvable ideal. If R = 0, then L is called semisimple.
Lemma 2.16. ([11, p.11]) Let L be a Lie algebra with radical R. Then the Lie algebra L/R
is semisimple, i.e., the radical of L/R is 0.
Proof. Let R? be the radical of L/R and ? : L?L/R be the canonical projection. Since ?
is a homomorphism, ??1(R?) is an ideal of L. As R? is solvable, we know that the derived
series of R? terminates, that is (R?)(n) = R for some n; then ?((??1(R?))(n)) ? (R?)(n) means
that (??1(R?))(n) ?R, that is ??1(R?) is a solvable ideal, and by maximality is contained in
R.
Indeed the radical R of L is the smallest ideal of L such that L/R is semisimple.
Given x?L, adx is a vector space endomorphism of L. According to Jordan-Chevalley
decomposition [11, p.17-18], L is the direct sum of generalized eigenspaces
La(adx) := Ker(adx?a?1)m
where m is the multiplicity of the eigenvalue a of adx. Each La(adx) is invariant under
adx and the restriction of adx to La(adx) is the sum of a scalar multiple (namely a) of the
identity and a nilpotent endomorphism. For each nonzero x ? L, 0 is an eigenvalue of adx
since adx(x) = [x,x] = 0, so that we have L0(adx) ?= 0. We set La(adx) = 0 if a is not an
eigenvalue of adx. Thus we have the following
8
Lemma 2.17. ([11, p.78]) Given x?L, the Lie algebra L may be decomposed as
L =
?
a?F
La(adx) = L0(adx)?L?(adx)
where L?(adx) denotes the sum of those La(adx) such that a ?= 0. In addition, any subal-
gebra K of L that is stable under adx can be written as K = K0(adx)?K?(adx), where
Ki(adx) = K?Li(adx).
Lemma 2.18. [11, p.78] If a,b ? F, then [La(adx),Lb(ady)] ? La+b(adx). In particular
L0(adx) is a subalgebra of L. When a?= 0, each element of La(adx) is ad-nilpotent.
Proof. Binomial expansion [11, p.79] yields
(adx?(a+b))m[y,z] = (adx?a?b)m[y,z]
=
m?
i=0
(m
i
)
[(adx?a)iy,(adx?b)m?iz].
For su?ciently large m, all terms on the right side are 0 for y ?La(adx) and z ?Lb(adx).
So [La(adx),Lb(ady)] ?La+b(adx). Then L0(adx) is a subalgebra of L.
Each z ?L can be written z = z0 +za1 +...+zan with zai ?Lai(adx) by Lemma 2.17.
When a?= 0 and y ?La(adx), we have (ady)ri(zai) ?Lria+ai(adx) = 0 for su?ciently large
ri, since there are ?nitely many eigenvalues for adx (recall that L is ?nite dimensional). We
need merely chooser =?ri to force (ady)r(z) = 0. Thus ady is a nilpotent endomorphism,
i.e., elements of La(adx) are ad-nilpotent.
We note that y ?L0(adx) need not be ad-nilpotent.
De nition 2.19. [11, p.82] We call x ? L strongly ad-nilpotent if x ? La(ady) for some
a?= 0. The set of all strongly ad-nilpotent elements of L will be denoted N(L).
By Lemma 2.18 strongly ad-nilpotent elements are ad-nilpotent.
9
De nition 2.20. [7, p.1] Given a nilpotent subalgebra H of L, ??H?, and h?H, we set
L?,h := L?(h)(adh) = Ker(adh??(h)?1)m
where m is the algebraic multiplicity of the eigenvalue ?(h), and set
L?(H) := ?h?HL?,h.
When H is understood, we will often denote L?(H) by L?.
Theorem 2.21. ([7, p.8]) If H is a nilpotent subalgebra of L, then L may be decomposed
as
L =
?
??H?
L?(H).
The decomposition is called the root space decomposition.
Lemma 2.22. [11, p.14] A nilpotent Lie algebra L contains no proper self-normalizing
subalgebras.
Proof. Indeed, a Lie algebra L containing a proper self-normalizing subalgebra K must also
contain a (nonzero) x1 ?L\K, and (as K is self-normalizing) a corresponding k1 ?K such
that x2 = [k1,x1] ?? K. Then there is a k2 ? K with [k2,x2] = x3 ?? K; thus we have a
nonzero sequence x1,...,xn,... with xi ? Li?1. Then the descending central series of L is
non-terminating, i.e. L is not nilpotent.
Lemma 2.23. Let L and L? be isomorphic as Lie algebras via ?. Then the subalgebra C of
L is self-normalizing if and only if ?(C) is self-normalizing.
Proof: It su?ces to show one implication. By virtue of the isomorphism ?, elements of
L and L? are in 1 to 1 correspondence; we are thus justi?ed in denoting ??1(x?) = x for any
x? ? L?. In addition, preimages of subalgebras of L? are isomorphic subalgebras in L; thus
for any subalgebra C? of L, we have C? ?= C ?L, where ??1(C?) = C.
10
Suppose that the subalgebra C of L is self-normalizing, and let x?(= ?(x)) ? NL?(C?),
that is [x?,C?] ? C?. So ?([x,C]) = [?(x),?(C)] ? ?(C). But ? is an isomorphism, so
[x,C] ? C. As C is self-normalizing, we have x ? C so that ?(x) ? ?(C) and ?(C) is
self-normalizing.
Lemma 2.24. Let ?? EndV, where V is a vector space. If K ?V is an invariant subspace
under ? and contains the eigenspace corresponding to the eigenvalue 0, then the induced
endomorphism?? : V/K ?V/K de?ned by??(v+K) = ?(v)+K has no nonzero eigenvalues.
Proof. Choosea basisBK = {k1,...,kt}ofK andextendBK toa basisBV ={k1,...,kt,v1,...,vn?t}
of V. With respect BV, the matrix (denoted by M?) of ? is in block form
M? =
?
??A B
0 C
?
??
ClearlyAis the matrix of?|K : K ?K with respect toBK. The eigenvalues of?are those of
AandC so thatC has only non-zero eigenvalues . The map?? is a well-de?ned endomorphism
on the quotient (vector) space V/K. With respect to the basis {v1 +K,...,vn?t +K} of
V/K, the matrix of ?? is C: for each i = 1,...,n?t,
??(vi +K) = ?(vi) +K
= (b1ik1 +???+btikt +c1iv1 +...+cn?t,ivn?t) +K
= (c1iv1 +...+cn?t,ivn?t) +K
= c1i(v1 +K) +...+cn?t,i(vn?t +K).
Lemma 2.25. Let F ? EndV be a commuting diagonalizable family, where V is a vector
space over F. Suppose W ?V is a subspace stable under F. Then FW := {T|W ? EndW :
T ? F} ? EndW is a commuting diagonalizable family.
11
Proof. If S,T ? F, then S|W ?T|W = (S ?T)|W = (T ?S)|W = T|W ?S|W. Hence FW is
a commuting family. It su?ces to show that each T|W ? EndW is diagonalizable, i.e., each
T|W has k linearly independent eigenvectors (k := dimW), or the geometric multiplicity
m of each eigenvalue ? of T|W is 1. Now w ? Ker(T|W ??? 1W)m implies that w ? W
and 0 = (T|W ???1w)mw = (T ???1)mw. But T is diagonalizable so that the geometric
multiplicity of the eigenvalue ? of T is 1, i.e., (T ???1)w = 0, i.e., m = 1.
Recall from De?nition 2.19 that N(L) denotes the set of strongly ad-nilpotent elements
of L.
Lemma 2.26. ([11, p.82]) For any epimorphism ? : L?L?,
?(La(ady)) = L?a(ad(?(y))), y ?L. (2.1)
So ?(N(L)) = N(L?).
Proof. Given a Lie algebra homomorphism ? : L ? L?, that ?(La(ady)) ? L?a(ad(?(y))) is
readily seen: for x?La(ady) means that (ady?a?1)k(x) = 0 for some k ?N. So
((ad?(y))?a?1)k(?(x)) = ?((ady?a?1)k(x)) = 0,
i.e., ?(x) ?L?a(ad?(y)).
On the other hand, any x? ?L?a(ad?(y)) has a preimage x in L by the surjectivity of ?.
We employ Jordan-Chevalley decomposition [11, p.17] to write
L = La1(ady)?...?Lan(ady),
where a1,...,an are the eigenvalues of ady. Then we may decompose x accordingly:
x = xa1 +...+xan, xt ?Lat(ady).
12
By the above, ?(xt) ? L?at(ad?(y)). In addition, the sum L?a1(ad?(y))?...?L?an(ad?(y))
is direct. Suppose a = ai. Then
L?ai(ad?(y)) ? x? ??(xai) = ?(x)??(xai)
= ?(xa1) +???+?(xai?1) +?(xai+1) +???+?(xan)
? L?a1(ad?(y))????? \L?ai(ad?(y))?????L?an(ad?(y))
where the hat indicates the term L?ai(ad?(y)) is deleted. But x??(xai) ?L?ai(ad?(y)). As
the sum is direct, we have x???(xai) = 0, i.e. x? = ?(xai) and we have found xai ?Lai(ady)
such that ?(xai) = x?. So ?(La(ady)) = L?a(ad(?(y))).
Recall that N(L) is de?ned as the set of all x?L such that x?La(ady), where a is a
nonzero eigenvalue of the endomorphism ady. From the above, it is clear that ?(N(L)) =
N(L?).
13
Chapter 3
Properties of Cartan Subalgebras
We introduce Cartan subalgebras and discuss properties that will be useful in both of
the conjugacy proofs.
De nition 3.1. [11, p.80] Let L be a Lie algebra over F. A Cartan subalgebra H (abbrevi-
ated CSA) of L is a self-normalizing nilpotent subalgebra of L.
Unfortunately, this de?nition does not imply the existance of CSAs of a given Lie alge-
bra.
Both conjugacy proofs will employ the following de?nition:
De nition 3.2. ([11, p.83]) A Borel subalgebra (abbreviated BSA) B of L is a maximal
solvable subalgebra of L.
Note that BSAs are subalgebras, while RadL is required to be an maximal solvable
ideal.
A helpful propety of Borel subalgebras is the following:
Lemma 3.3. ([11, p.83]) Every BSA B of L is self-normalizing.
Proof. If x ? L normalizes B, we may create B +Fx, which is certainly a subalgebra of L
since
[B +Fx,B +Fx] ? [B,B] + [B,Fx] + [Fx,Fx].
The last term is zero, and the term [B,Fx] is inside of B since x is a normalizer. Clearly
[B +Fx,B +Fx] ?B, so B +Fx is solvable. Now B is Borel (maximal solvable) so that x
must be an element of B.
14
While the proof of the following lemma is beyond the scope of this paper, we shall use
the result to establish several properties of CSAs:
Lemma 3.4. [7, p.8] If H is a nilpotent subalgebra of L, then there is an x ? L such that
L0(adx) = L0(H).
We note that H ?L0(H) as H is nilpotent.
Lemma 3.5. [7, p.14] Let H be a nilpotent subalgebra of L. Then H is a CSA of L if and
only if L0(H) = H.
Proof. We ?rst note that any nilpotent subalgebra H is contained in L0(H). If L0(H) =
H, H is self-normalizing (see [7, p.10]), thus a CSA. On the other hand, if H ( L0(H),
considering the nilpotent subalgebra adH of gl(L0(H)/H), we may apply Engel?s theorem
to the (nontrivial) quotient algebra L0(H)/H to ?nd an x?L0(H)\H such that [x,H] ?H,
that is x? NL(H), and H is not self-normalizing, i.e. not a CSA.
Theorem 3.6. [7, p.14] Every CSA H of L may be written in the form H = L0(adx) for
some x?L.
Proof. Lemmas 3.4 and 3.5 allow us to ?nd an x?L such that L0(adx) = L0(H) = H.
De nition 3.7. [16, p.156] The rank of L, rankL, is de?ned as min{dimL0(adx) |x?L}.
Any x with rankL = dimL0(adx) is called regular.
Lemma 3.8. [7, p.17] LetH be a subalgebra ofL. Then every regular element ofLcontained
in H is also regular in H.
Theorem 3.9. [7, p.18] Let x be a regular element of L. Then L0(adx) is a CSA of L.
Proof. If x is regular in L, set H := L0(adx). By de?nition, H0(adx) = H. By the
previous lemma, x is a regular element of H, so rankH = dimH. Then for all h ? H,
15
dimH0(adh) = dimH, so adHh is a nilpotent endomorphism of H. By Lie?s Theorem,
then, H is nilpotent; so we have H ?L0(H) ?L0(adx) = H, and by 3.5, H is a CSA of L.
As a consequence of the previous theorem, we now know that CSAs of a ?nite dimen-
sional Lie algebra over an algebraically closed, characteristic 0 ?eld F always exist.
Lemma 3.10. [7, p.13] A CSA H of L is a maximal nilpotent subalgebra.
Proof. If H is a nilpotent subalgebra of L containing a CSA H?, we note that H? is self-
normalizing not only in L, but also in H; by Lemma 2.22, H = H?.
Compare the following with [11, p.79].
Lemma 3.11. If K ? L is a subalgebra of L containing a CSA H of L, then K is self-
normalizing.
Proof. By Theorem 3.6, we may write H = L0(adx) ? K for some x ? L, and make the
following observations:
(1) [x,x] = 0 so x?H ?K; thus [NL(K),x] ?K.
(2) In particular adx(K) ? K since K ? NL(K) , i.e., K is an invariant subspace of
NL(K) under the endomorphism adx. In addition, K contains the eigenspace of adx
corresponding to the eigenvalue 0 since L0(adx) ?K.
By Lemma 2.24, the endomorphism adx acts on NL(K)/K with no zero eigenvalues. By
(1), every coset of NL(K)/K is mapped by adx into K, i.e.,
adx(m+K) = adx(m) +K = K
where m?NL(K). In other words, adx acts trivially on NL(K)/K. On the other hand, by
(2), adx has no zero eigenvalues on NL(K)/K. So K is the only possible coset of NL(K)/K
and we conclude that NL(K) = K.
16
Lemma 3.12. ([11, p.81]) If ? : L ? L? is an epimorphism of Lie algebras, then ?(H) is a
CSA of L? for every CSA H of L. If H? is a CSA of L?, there is a CSA H of L such that
?(H) = H?.
Proof. Obviously ?(H) is nilpotent so we need only show that it is self-normalizing. Let
A = Ker?. Then by the First Isomorphism Theorem (Theorem 2.6), L/A ?= Im? = L? and
we have the induced (Lie algebra) isomorphism ? : L/A?L?, de?ned by ?(x+A) = ?(x).
The function ? allows us to isomorphically identify the subalgebra H/A of L/A with ?(H),
for by the above ?(H/A) = ?(H). By Lemma 2.23 ?(H) self-normalizing is equivalent to
H/A being self-normalizing. This will be simpler to prove, so we focus our attention upon
L/A, in particular the subalgebra H/A of L/A. In L itself, H is a subalgebra and A is an
ideal, so H+A?L is a subalgebra of L. This subalgebra contains the CSA H. By Lemma
3.11, H +A is self-normalizing as a subalgebra of L.
Now suppose that H/A ? L/A is normalized by the coset x + A, i.e., [x,H]/A =
[x+A,H/A] ?H/A. We must have [x,H] ?H +A so that
[x,H +A] = [x,H] + [x,A] ?H +A?L
(note that [x,A] ?A since A is an ideal). So x normalizes H+A. By the above, x?H+A
so that that x+A?H/A. So H/A is self-normalizing as a subalgebra of L/A, and ?(H) is,
as well; thus ?(H) is a CSA of L?.
Finally, if H? is a CSA of L?, Set K := ??1(H?). We may choose a CSA H of K; then
?(H) is a CSA of H? by the above, that is ?(H) is a self-normalizing, nilpotent subalgebra of
H?. But we may view H? as a Lie algebra in its own right, and note that H? is nilpotent. As
CSAs are maximal nilpotent subalgebras (Lemma 3.10), ?(H) = H?. We must show that H
is a CSA of L. If x?L normalizes H, then ?(x) normalizes ?(H) = H? so that ?(x) ??(H),
i.e. x?H + Ker?. But Ker??K so that x?H +K ?K. Now x?NK(H) = H since H
is a CSA of K. See [7, Corollary 2, p.18].
17
Chapter 4
Humphreys? Proof
We ?rst prove the conjugacy of CSAs following Humphreys? approach. Indeed as
Humphreys [11, p.88] points out the approach is from Winter [25, Section 3.8, p.92-99]
and was inspired by Mostow [23, p.vii]. Since some of Humphreys? arguments are very brief,
elaboration is needed.
De nition 4.1. ([11, p.9,82]) IntL is the subgroup of AutL generated by all
exp(adx) := 1 + adx+ (adx)2/2! + (adx)3/3! +???
where x?L is ad-nilpotent. Note that the sum has a ?nite number of terms, for (adx)n = 0
for some n. We de?ne E(L) as the subgroup of IntL generated by all exp(adx) such that
x? N(L) (see De?nition 2.19).
Remark 4.2. ([11, p.82]) Given a subalgebra K of L, we note that N(K) ? N(L), and
de?ne E(L;K) as the subgroup of E(L) generated by expadx, where x ? N(K). Thus
E(K) is precisely the restriction of the automorphisms of E(L;K) to K. In particular, given
?? ? E(K), we may extend ?? to ? ? E(L), where ?|K = ??.
It turns out that if L is semisimple, E(L) = IntL.
Lemma 4.3. ([11, p.82]) Let ? : L ? L? be an epimorphism of Lie algebras. For any
?? ? E(L?), there exists ? ? E(L) such that the following diagram commutes:
L
?
  
? //L?
??  
L ? //L?
18
i.e., ?? ?? = ???.
Proof. As E(L?) is generated by ?? = expadx?, where x? ? N(L?), it su?ces to show the
theorem true for such ??. By Lemma 2.26, we may choose x ? N(L) such that ?(x) = x?,
and we set ? := expadx? E(L). For any z ?L,
(???)(z) = ?(expadx(z)) = ?(z+ [x,z] + (1/2)[x,[x,z]] +???)
= ?(z) + [?(x),?(z)] + (1/2)[?(x),[?(x),?(z)]] +???
= ?(z) + [x?,?(z)] + (1/2)[x?,[x?,?(z)]] +???
= expadx?(?(z))
= (?? ??)(z).
One of our main goals is to show that CSAs of a Lie algebra L are conjugate via E(L);
we ?rst handle the special case when L is solvable.
Theorem 4.4. ([11, p.82]) CSAs of a solvable L are conjugate via E(L).
Proof: We proceed by induction. The theorem is obvious when dimL = 1.
If L is nilpotent, there is nothing to prove, for L itself will be its only self-normalizing
subalgebra by Lemma 2.22.
Thus we may assume that L is solvable but not nilpotent. As the last term of the
derived series of L must be abelian, we are guaranteed the existence of non-zero abelian
ideals of L; choose one, A, of minimum dimension. Set L? := L/A, the homomorphic image
of L under the canonical projection ? : L?L/A. Then for CSAs H1, H2 of L, ?(H1) = H?1
and ?(H2) = H?2 are themselves CSAs of L? by Lemma 3.12. By the induction hypothesis,
since dimL/A = dimL?dimA< dimL, these are conjugate via ?? ? EL?. By Lemma 4.3,
? ? E(L) exists such that ??? = ??.
19
Setting Ki := ??1(H?i), i = 1,2, we note that the Ki are subalgebras of L. We have
??(K1) = ???(K1) = ??(H?1) = H?2. But ??1(H?2) = K2, so ?(K1) ? K2. Similarly,
?(K2) ?K1. As ? is an automorphism, we must have ?(K1) = K2.
L ????? L?
?
H1 ?K1 ???? H?1?
?y? ??y??
H2 ?K2 ???? H?2
We consider two cases.
Case 1. K2 ( L. Then again by the induction hypothesis, we have ?? ? E(K2) with
???(H1) = H2, since by Lemma 3.12 ?(H1) as well as H2 are CSAs of K2. Extend ?? to
? ? E(L) to complete the proof (see remark 4.2).
Case 2. K2 = L. As before, we have ?(K1) = K2 = L, so K1 = K2 = L. Now L/A =
?(L) = ?(K1) = H?1 ? L/A; so L/A = H?1 = ?(H1). For any y ? L, ?(y) = y +A ? H1/A,
which means y ?H1 +A?L. A similar argument applies for H2, so we have
L = H1 +A = H2 +A.
Now by Lemma 3.6, we write
H2 = L0(adx)
for some x?L. Since A is an ideal, it is stable under adx (i.e., adAx : A?A) and Lemma
2.17 allows us to write
A = A0(adx)?A?(adx).
Wewill showthatA0(adx)is an idealofL: onone hand[H2,A0(adx)] = [L0(adx),A0(adx)] ?
L0(adx) since A0(adx) ?L0(adx); on the other hand [H2,A0(adx)] ? [H2,A] ?A since A
is an ideal ([A0,A] = 0 since A is abelian). Thus [H2,A0(adx)] ?L0(adx)?A = A0(adx).
20
As an ideal of L, A0(adx) must be trivial, otherwise by the minimality of A we have
A = A0(adx). But this is impossible, for it would force A ? L0(adx) = H2, that is
L = H2 +A = H2, a nilpotent algebra. Note that the argument is symmetric, thus applies
to H1 as well.
Since A0(adx) = 0, we have A = A?(adx) ? L?(adx), and since H2 = L0(adx), the
sum L = H2 +A is direct and we have
A = A?(adx) = L?(adx).
SinceL = H1+A, we may writex = y+z, withy ?H1 andz ?L?(adx). The endomorphism
adx is invertible on A = L?(adx), for L?(adx) clearly has no zero eigenvalues under adx.
Thus a z? ?L?(adx) exists such that z = [x,z?]. As A is abelian, (adAz?)2 = 0; in addition,
A is an ideal and H1?A = 0, so adz?(H1) = 0. Thus (adz?)2 = 0. Thus expadz? = 1+adz?
and in particular,
expadz?(x) = x+ [z?,x] = x?[x,z?] = x?z = y.
By Lemma 2.26 with ? := expadz? ? AutL, we have
expadz?(L0(adx)) = L0(ad(expadz?(x))) = L0(ady).
By Lemma 3.12 H := L0(ady), as the isomorphic image of the CSA H2 = L0(adx) of L,
is also a CSA. Now y ? H1, a nilpotent subalgebra, implies that ady acts as a nilpotent
endomorphism on H1 (by Engel?s theorem). Then by H1 ?L0(ady) = H, we have H = H1,
for both are maximum nilpotent subalgebras (Lemma 3.10).
Now expadz? ? AutL sends H2 to H1; it remains to show that expadz? ? E(L).
By Lemma 2.17 we write z? ? L?(adx) as a sum of strongly ad-nilpotent elements, say
21
z? =
?
a
z?a, with z?a ?La(adx), a a non-zero eigenvalue for adx. Then
expadz? = 1 + adz? = 1 + ad(
?
a
z?a) = 1 +
?
a
adz?a.
But each z?a is an element of the abelian ideal A, so (adz?a)(adz?b) = 0 ? EndL. Thus
1 +
?
a
adz?a =
?
a
(1 + adz?a) =
?
a
expadz?a ? E(L)
since (adz?a)2 = 0 (indeed (adz?)2 = 0). Thus H1 and H2 are conjugate via an element of
E(L).
In order to prove the general case, we will employ several useful properties of Borel
subalgebras.
Lemma 4.5. ([11, p.83]) The BSAs of L are in natural 1-1 correspondence (with respect to
the canonical projection) with the BSAs of the semisimple L/R, where R denotes the radical
of L.
Proof. The radical R of L is a maximal solvable ideal, thus B +R is a solvable subalgebra
of L for any BSA B of L since B is solvable. Thus we have R?B by the maximality of B.
Now B/R is solvable in L/R. Any subalgebra K? ? L/R properly containing B/R is
not solvable, otherwise the subalgebra K := ??1(K?) of L containing B would be solvable,
forcing B(K, a contradiction. Thus B/R is indeed a BSA of L/R.
Conversely if B? ? L/R is a BSA of L/R, then the subalgebra B := ??1(B?) of L is
solvable . Any subalgebra ?B properly containing B is not solvable, otherwise the solvable
?B/R properly contains the BSA B? = B/R of L/R.
When L is semisimple, the abstract Jordan decomposition [11, p.24] asserts that any
element x of L may be decomposed as x = xs +xn, where the ad-semisimple part xs of x
22
and the ad-nilpotent part xn of x are also elements of L. The same is true for any BSA of a
semisimple L:
Lemma 4.6. [11, p.85] Any Borel subalgebra B of a semisimple L contains the semisimple
and nilpotent parts of all of its elements.
Proof. Jordan-Chevalley decomposition implies that if the endomorphism? maps a subspace
B of L into A ? B, then both the nilpotent and semisimple parts of ? map B into A [11,
p.17]. For each x ? B, view adx ? EndL. Then by Lemma 3.3, x ? B if and only if adx
maps B into itself. Thus adxs and adxn map B into itself, i.e. xs and xn are normalizers of
B, thus elements of B. We conclude that any Borel subalgebra contains the semisimple and
nilpotent parts of all of its elements.
A subalgebra T of L is said to be toral if T consists of ad-semisimple elements. Toral
subalgebras certainly exist in a semisimple Lie algebra. Any Lie algebra consisting entirely
of ad-nilpotent elements is a nilpotent algebra by Engel?s Theorem (Theorem 2.11). A
semisimple Lie algebra L has no solvable ideals and cannot be nilpotent; we conclude that
L has an ad-semisimple element, thus a nonzero toral subalgebra T.
Lemma 4.7. ([11, p.80]) The CSAs of a semisimple L are precisely the maximal toral
subalgebras of L. In particular, CSAs of semisimple Lie algebras are abelian.
If H is a CSA of the semisimple L, then adH is an abelian subalgebra of semisimple
endomorphisms in EndL and thus adH is simultaneously diagonalizable according to linear
algebra. So given h?H, we have L?,h = L?(h)(adh) = Ker(adh??(h)?1), i.e., m = 1, and
thus L?(H) takes the special form
L?(H) = {x?L| [h,x] = ?(h)x for all h?H},
i.e., each nonzero x ? L?(H) is a common eigenvector for all endomorphisms adh, h ? H.
Note that L0(H) is simply the centralizer of H. The ? such that L?(H) ?= 0 are called
23
roots of L with respect to H, a subcollection of H? which we will denote by ?(H). We will
frequently make use of the root space decomposition of a semisimple L.
Lemma 4.8. ([11, p.35]) Given a CSA H of the semisimple L, we have the root space
decomposition
L = H ?
?
??  (H)
L?(H).
We may ?x a base ? of ?(H) and note that
B(?) := H +
?
??0
L?(H)
is a Borel subalgebra of L, which we call a standard Borel with respect to H. In addition,
N(?) := [B(?),B(?)] =
?
??0
L?(H)
is nilpotent [11, p.84].
The following result on semisimple algebras is similar to Lemma 2.18:
Lemma 4.9. [11, p.36] Let H be a CSA of the semisimple Lie algebra L, and let ?, ? ?H?.
Then [L?(H),L?(H)] ?L?+?(H).
Lemma 4.10. ([11, p.84]) Let H be a CSA of the semisimple L. All standard Borel subal-
gebras of L relative to H are conjugate via E(L).
The second main goal in this chapter is the following conjugacy theorem for BSAs.
Theorem 4.11. ([11, p.84]) The BSAs of a semisimple Lie algebraLare conjugate via E(L).
Proof. We will employ induction on dimL. The proof is trivial if dimL = 1, for L is itself
its only BSA.
Let L be semisimple and H a CSA, and ?x a base ? and thus a standard BSA B with
respect to H. We will show that any other BSA B? of L is conjugate to B via an element of
24
E(L). If B?B? = B, then B = B? by the maximality of B, and there is nothing to prove.
Thus a downward induction on the dim(B?B?) is appropriate: we may assume that a BSA
whose intersection with B has dimension greater than dim(B?B?) is already a conjugate of
B.
We divide the proof into cases, and the ?rst case into two subcases.
Case 1: B?B? ?= 0.
Let N? ? B ?B? be the set of ad-nilpotent elements of B ?B?, i.e., x ? B ?B? and
adx? EndL is nilpotent.
Subcase (i): Suppose N? ?= 0. By Lemma 4.8, B may be decomposed as
B = H +
?
??0
L?(H), N := [B,B] =
?
??0
L?(H)
whereH is the ?xed CSA ofL. So by Lie?s theorem, the derived algebraN = [B,B] =
?
??0
L?
is nilpotent (N? = N ?B?). Now as [B ?B?,B ?B?] is properly contained in both [B,B]
and [B?,B?], we see that [B?B?,B?B?] ? B?B? is nilpotent. Thus, by Engel?s theorem
again, [B?B?,B?B?] ? N?, so N? is an ideal of B?B?. As L is semisimple, it is allowed
no proper solvable (thus no nilpotent) ideals, so N? is not an ideal of L and
K := NL(N?)(L.
We are going to show that B ?B? is properly contained in both B ?K and B? ?K.
Clearly from the above B?B? ?B?K and B?B? ?B??K. For each x?N? ?B, adx is
a nilpotent endomorphism of B (as well as of B?) since B (respectively B?) is stable under
adx. But B?B? is also stable under adx, so the induced action of adx on the vector space
B/(B ?B?) is nilpotent (see the proof of Lemma 2.24). This is true for all x ? N?, so by
Engel?s Theorem, there is a non-zero y+B?B? ?B/(B?B?) killed by all elements of N?.
We have found a y ? B \B ?B? such that [N?,y] ? B ?B?. In addition, [N?,y] ? [B,B]
(since y ? B and N? ? B), which is nilpotent. So every [x,y] ? [N?,y] is nilpotent, forcing
25
[x,y] ?N? which is the set of nilpotent elements of B?B?. Thus y ?NB(N?) = B?K, but
y ??B?B?, i.e., B?B? (B?K. The argument is symmetric for B and B? so
B?B? (B?K, B?B? (B? ?K.
As BSAs, B and B? are solvable in L, so B?K and B? ?K are solvable subalgebras of
K. We choose BSAs of K, C and C?, containing B?K, B??K respectively (see the ?gure).
L
|
K
  
C C?
| |
B?K B? ?K
  
B?B?
|
N?
Since K ( L, by the ?rst induction hypothesis (on dimL), C and C? are conjugate via
?? ? E(K). So there is ? ? E(L;K) ? E(L) such that ?(C?) = C [remark 4.2]. Now B?B?
is a proper subalgebra of both B ?K ? C and B? ?K ? C?. Note that C may not be a
BSA of L, but as it is solvable, it is contained in a BSA M of L. We have
B?B? (B?K ?B?C ?B?M,
26
so dim(B?M) > dim(B?B?). By the second induction hypothesis (downward, on dim(B?
B?)), there is a ? ? E(L) with ?(M) = B. We have ?(C) ?B, i.e.,
??(C?) ?B.
Since B?B? (B? ?K, we have ??(B?B?)(??(B? ?K). Clearly B? ?K = B? ?C?, so
??(B?B?)(??(B? ?K) = ??(B? ?C?) ???(B?)???(C?) ???(B?)?B.
Then dim(B???(B?)) > dim??(B?B?) = dim(B?B?). Clearly B and ??(B?) are BSAs
of L. Then we use the second induction hypothesis (downward, on dim(B?B?)) to see that
B is conjugate to ??(B?) via an element of E(L).
Subcase (ii): Suppose N? = 0. Hence B?B? contains no nonzero nilpotent elements.
By Lemma 4.6, any BSA contains both the semisimple and nilpotent parts of any of its
elements, so all elements of B?B? are semisimple, i.e.,
T := B?B? ?= 0
is a toral subalgebra of the semisimple L. Recall that B = H +N is a standard Borel and
the subalgebra N = [B,B] of L consists entirely of nilpotent elements. Since 0 is the only
element which is both nilpotent and semisimple, T ?N = 0. For any x ? B, [x,T] ? T
means that [x,T] ?T ?N = 0. Thus
NB(T) = CB(T).
As T is toral, T ? CB(T), and we may choose a maximal toral subalgebra C of CB(T)
containing T. By de?nition, C is nilpotent and self-normalizing (in CB(T)). So we have
T ?C ?CB(T) = NB(T) ?NB(C).
27
If n ? NB(C) ? B, then for each t ? T ? C, (adt)kn = 0 for some k ? N since C
is nilpotent. However, adt ? EndL is a semisimple endomorphism since T is toral. So
L0(adt) = Keradt = Ker(adt)k. Then we may choose k = 1, i.e., n ? NB(C) ? CB(T).
But C is its own normalizer in CB(T). Then
C = NCB(T)(C) = NB(C).
As a nilpotent self-normalizing subalgebra of B, C is a CSA of the solvable algebra B.
Clearly H is a CSA of both B and L. By Theorem 4.4, C is conjugate, via an element of
E(B) (hence via E(L)), to H.
Thus without loss of generality we may assume that T ?H.
Now we consider two subcases.
(A) T = H. Now H = B ? B? ( B? (otherwise B = B? and there is nothing to
prove). Recall that B = H +
?
??0
L?. Notice that [H,B?] = [B ?B?,B?] ? B?, i.e., B? is
stable under adLH. Since adLH is simultaneously diagonalizable (as adH ? EndL is a
commuting family of semisimple endomorphisms), adB?H ? EndB? is also simultaneously
diagonalizable by Lemma 2.25. This allows us to decompose B? into root spaces under H:
B? = H +??B?? where
B?? = {x?B? | [h,x] = ?(h)x for all h?H}.
Clearly each ? ? ? and B?? = L? since dimL? = 1 [11, p.39]. Since H = B?B? (B?, we
must have B? = H+???0L?. So there is at least one ?? 0 relative to ? with 0 ?= L? ?B?.
The re?ection ?? [11, p.42] sends ? into ??, H is preserved, and ?? = ??(?) is a base.
Then B? is conjugate to the standard Borel B?? := B(??) whose intersection with B includes
H +L??. So the second induction hypothesis (downward, on dimB?B?) applies and B?? is
conjugate to B.
28
Remark: Instead of using ??, one can use the longest element ?0 ?W [17, p.88] of the
Weyl group W to get the conclusion immediately, i.e., B and B? are conjugate via ?0, since
?0(?+) = ??+ where ?+ denotes the positive roots.
(B) T (H. Then we have two subcases:
(B1) B? ?CL(T). The semisimple L has zero center and we are assuming T = B?B? ?= 0.
So CL(T) ( L as T itself is abelian, and we have dimCL(T) < dimL. We are thus
justi?ed in using the ?rst induction hypothesis on CL(T). Now H = CL(H) [11, p.36]
and T ?H so that H = CL(H) ?CL(T). Since H is abelian and thus solvable, there
is a BSA B?? of CL(T) containing H. By the ?rst induction hypothesis, there exists
? ? E(L;CL(T)) ? E(L) with ?(B?) = B??, since B? and B?? are BSAs of CL(T). By
virtue of the fact that B?? is the isomorphic image of the BSA B? of L, B?? is a BSA of
L as well. Moreover B?? contains H, so
dim(B?B??) ? dimH > dimT = dim(B?B?).
By the second induction hypothesis (downward, on dim(B?B?)), B and B?? are con-
jugate via E(L), thus B and B? are, as well.
(B2) B? ??CL(T). As in (A), since we are assuming T (H, adB?T is simultaneously diago-
nalizable. Since B? ??CL(T), there is t? ?T and a common eigenvector x?B?\CL(T)
of the endomorphisms of adB?T such that [t?,x] = c?x, where c? ?Fis nonzero. Setting
t := t?/c?, we have [t,x] = x. Let ?t := {?? ? : ?(t) > 0 and rational} ? ?. Given ?,
? ? ?t, ?+? ? ?t. Then
S := H +
?
?? t
L?
is subalgebra of L. Notice that x?S since x is of the form
x = xh +
?
?? 
x?, x? ?L?,xh ?H
29
and
xh +
?
?? 
x? = x = [t,x] =
?
?? 
[t,x?] =
?
?? 
?(t)x?
so that xh = 0 and x? = 0 if ??? ?t. By Lemma 4.9 S is solvable and so lies in a BSA
B??. Since T ?B?? and x?B??
B?B? = T (T +Fx?B?? ?B?.
we see that dim(B?? ?B?) > dim(B?B?); by the second induction hypothesis, B?? and
B? are conjugate. In addition, dim(B???B) > dim(B?B?) since H ??B?B?; thus B??
and B are conjugate, so B? and B are as well.
Case 2: B ? B? = 0. As a standard BSA of L, B = H +
?
??0
L?(H). But L itself is
semisimple, thus allowing us to decompose L = H ?
?
??  (H)
L?(H). We know that ? is a
root if and only if ?? is a root [11, p.37]; so dimB > (dimL)/2. However, since B?B? = 0,
we employ a standard argument regarding dimension in vector spaces to see that
dimB + dimB? = dim(B +B?) ? dimL.
Thus B? must be a ?small? Borel, i.e., dimB? < (dimL)/2. Choose a maximal toral subalge-
bra T of B?. If T = 0, then B? consists solely of ad-nilpotent elements. By Engel?s theorem
B? is nilpotent, and as it is Borel, it is also self-normalizing by Lemma 3.3. In other words,
B? is a CSA of L; but this is impossible, for on one hand each CSA of the semisimple L is
toral by Lemma 4.7, while on the other hand T = 0. So we must have T ?= 0. Once again
applying Lemma 4.7, choose a CSA H0 of L containing T, and B?? a standard Borel with
respect to H0. Then B? ?B?? ?= 0, and by Case 1 of the proof, B? and B?? are conjugate so
that dimB?? = dimB? < (dimL)/2. However B?? is standard so that dimB?? > (dimL)/2, a
contradiction.
Theorem 4.12. ([11, p.84]) The BSAs of a Lie algebra L are conjugate under E(L).
30
Proof. We have previously established the theorem for semisimple Lie algebras; if L is not
semisimple, then we construct the semisimple Lie algebra L/R (?= L). Now the BSAs of L
are in 1-1 correspondence with the BSAs of L/R by Lemma 4.5. Given BSAs B1 and B2 of
L, we identify them with the BSAs B?1 and B?2 of L/R. Now B?1 and B?2 are conjugate via
an element ?? of E(L/R), and by Lemma 4.3, there is a ? ? E(L) such that ??? = ?? ??,
where ? is the canonical projection. So
??(B1) = ???(B1) = ??(B?1) = B?2,
thus ?(B1) = B2.
Theorem 4.13. ([11, p.84]) The CSAs of L are conjugate under E(L).
Proof. Any nilpotent subalgebra is solvable, thus contained in a maximal solvable subalgebra.
So given a pair H, H? of CSAs of L, there is a pair B, B? of BSAs containing H, H?
respectively. By Theorem 4.12, there is ? ? E(L) such that ?(B) = B?. We have two CSAs
of the solvable B?, namely ?(H) and H?, which are conjugate via ?? ? E(B?) thanks to
Theorem 4.4. We have ???(H) = H?; once again, we extend ?? to ? ? E(L). Then we have
an endomorphism ?? ? E(L) such that ??(H) = H?.
Corollary 4.14. Every BSA of a Lie algebra L contains a CSA of L.
Proof. We have previously established the existence of CSAs in any Lie algebra (see remark
following Theorem 3.9). Given a CSA H of L, H is contained in some BSA B of L. For any
BSA B? of L, B and B? are conjugate via ?? E(L), thus B? contains the CSA ?(H).
31
Chapter 5
Michael?s Proof
Bourbaki [7, p.22-23] provides a proof for the conjugacy of CSAs using algebraic geom-
etry; the approach is very di?erent from that of Humphreys.
Michael?s approach does not use algebraic geometry and is along the presentation of
Bourbaki [5, 7]. Similar to Humphreys, he begins by proving the conjugacy of CSAs of
solvable Lie algebras. The proof is similar to that in [7, p.25] and will be omitted here.
His next task is to provide a few lemmas to ease the way to a full proof. We will utilize
Lemma 3.12, as well as various technical lemmas. We will also need De?nition 2.20 in the
following discussion:
L?,h := L?(h)(adh) = Ker(adh??(h)?1)m,
where m is the algebraic multiplicity of the eigenvalue ?(h), and
L?(H) := ?h?HL?,h
where H is a nilpotent subalgebra of L. Clearly L?(H) ? N(L), indeed each L?,h ? N(L)
(De?nition 2.19).
De nition 5.1. ([7, p.22], [16, p.156]) Given a Lie algebra L and CSA H, we denote by
EL(H) the subgroup of IntL generated by expadx, where x?L?(H) for some ??= 0.
Michael?s proof relies upon several results which can be viewed as analogues of Lemma
2.26 and Lemma 4.3, which we will state in the following lemmas.
32
Recall Lemma 2.26: an epimorphism ? : L ? L? has the property that ?(La(ady)) =
La(ad?(y)). The following lemma for L?(H) corresponds to Lemma 2.26:
Lemma 5.2. Let ? : L?L? be an epimorphism, and H a nilpotent subalgebra of L. Then
?(L?(H)) = L??(?(H)), where ?? ? (?(H))? is given by ??(?(h)) = ?(h).
Proof. De?ne ?? as above; it is clear that ?(L?(h)(adh)) = L???(?(h))(ad?(h)) for all h ? H,
i.e. ?(L?(H)) = L??(?(H)).
As analogue to Lemma 4.3, we have
Lemma 5.3. Given an epimorphism ? : L?L? and ?? ? EL?(H?), H? a CSA of L?, there is
a CSA H of L with ?(H) = H? and ? ? EL(H) such that ?? ?? = ???.
Proof. Lemma 3.12 allows us to choose a CSA H of L such that ?(H) = H?. Let ? :=
expadx? be a generator of EL?(H?), i.e. x? ?L???(H?); by Lemma 5.2 there is x?L?(H) such
that ?(x) = x?. Then ? := expadx? EL(H) is the promised automorphism.
Similar to Remark 4.2, we note the following:
Lemma 5.4. Given a Lie algebra L and subalgebra B containing a CSA H of L, any
? ? EB(H) may be extended to a ?? ? EL(H) such that ??|B = ?.
Proof. A generator expadBx of EB(H) may be viewed as a generator of EL(H): for x ?
B?(H) means that x?L?(H), as well. Thus expadL x? EL(H).
Lemma 5.5. Given a Lie algebra L, CSA H, and u ? EL(H), we have EL(u(H)) =
u(EL(H))u?1.
Proof. By Lemma 3.12, u(H) is a CSA, thus EL(u(H)) is de?ned.
Given a generator expadx of EL(u(H)), we note that
1. (adu(h)??(u(h))I)k(x) = 0 for all h?H
2. adu(h) = u?(adh)?u?1
33
De?ne ?? ?H? by ??(h) := ?(u(h)). Then for all h?H,
0 = (adu(h)??(u(h))I)k(x) = u?(adh???(h)I)k ?u?1(x).
So u?1(x) ?L??(H), and expad(u?1(x)) ? EL(H). Then we readily see that
expadx = u(expad(u?1(x)))u?1 ?u(EL(H))u?1.
Thus EL(u(H)) ?u(EL(H))u?1.
On the other hand, given y ?L?(H), we wish to show that u(expady)u?1 ? EL(u(H)).
Clearly u(expady)u?1 = expadu(y). De?ning ?? ? (u(H))? by ??(u(h)) = ?(h), we see by
Lemma 5.3 that u(y) ?L??(u(H)). The lemma follows.
The following is a re?nement of [7, p.25-26] and the proof is almost identical.
Theorem 5.6. If L is solvable and H1,H2 are CSAs of L, there exist ui ? EL(Hi), i = 1,2,
such that u1(H1) = u2(H2).
Corollary 5.7. ([16, p.158]) If L is solvable and H1,H2 are CSAs of L, EL(H1) = EL(H2).
Proof. Using Lemma 5.5 and Theorem 5.6, we have
EL(H1) = u1EL(H1)u?11 = EL(u1(H1)) = EL(u2(H2)) = u2EL(H1)u?12 = EL(H2).
So if L is solvable, the choice of H is inconsequential and we may denote EL(H) = EL.
Corollary 5.8. ([16, p.158]) Assume that L is solvable.
1. EL acts transitively on the set of CSAs of L.
2. Any CSA of L has dimension rankL.
3. L0(adx) is a CSA of L if and only if x is regular.
34
Proof. (1) is clear from Lemma 3.12 and Corollary 5.7. For (2), we note that EL is a group
of automorphisms; so by (1), all CSAs have the same dimension. By Lemma 3.9, this must
be the rank of L. Lemma 3.6 combined with Lemma 3.9 immediately give us (3).
De nition 5.9. ([11, p.21]) Given a Lie algebraL, the Killing form?is a symmetric bilinear
form on L given by
?(x,y) = tr(adx)(ady).
Elements x and y of L are orthogonal via ? if ?(x,y) = 0. Two subspaces H1 and H2 are
orthogonal, H1 ?H2, if their elements are mutually orthogonal via ?.
Remark 5.10. We will have occasion to employ a useful identity regarding the Killing form.
Note that, since [adx,ady] ? gl(L), we know the action of [?,?]:
ad[x,y] = [adx,ady] = adx ady?ady adx.
Then
?([x,y],z) = tr([adx,ady]adz) = tr(adx ady adz)?tr(ady adx adz).
But we may commute matrices without changing the trace, so the above is equal to
tr(adx ady adz)?tr(adx adz ady) = tr(adx[ady,adz]) = ?(x,[y,z]).
Thus ?([x,y],z) = ?(x,[y,z]).
Theorem 5.11. ([11, p.22]) Let L be a Lie algebra. Then L is semisimple if and only if its
Killing form is nondegenerate, i.e., the radical S of the Killing form, de?ned by
S := {x?L|?(x,y) = 0 for all y ?L}
is 0.
35
In general, ifV is a ?nite dimensional vector space overFwith nondegenerate symmetric
form [15, Chapter XV] ? and if W ?V is a subspace, we denote by
W? := {v ?V|?(v,w) = 0 for all w ?W}
the orthogonal complement of W with respect ?. See [15, Chapter XV], [24] for the general
theory of bilinear form. An orthogonal basis [15, p.575] always exists for V with dimV ? 1
(but not necessarily an orthonormal basis, for example, ? on F2 de?ned by the matrix?
??0 1
1 0
?
??). Moreover ? induces an isomorphism between V? and V: f 7? (x,?), f ? V? and
x?V. The restriction of ? on a subspace W is nondegenerate if and only if W ?W? = 0.
Lemma 5.12. Let V be a ?nite dimensional vector space with nondegenerate symmetric
bilinear form ?. Let W,V1,V2 be subspaces of V. Then
1. dimW + dimW? = dimV (but W +W? = V is not necessarily true).
2. (W?)? = W.
3. V1 ?V2 if and only if V?2 ?V?1 ; V1 (V2 if and only if V?2 (V?1 .
4. (V1 +V2)? = V?1 ?V?2 .
5. V?1 +V?2 = (V1 ?V2)?.
Proof. 1. Denote by f : V ?W? the map de?ned by f(v)(w) = ?(v,w), w ?W, v ?V.
Clearly Kerf = W? and Imf = W? so that dimImf = dimW? = dimW. Then
apply the dimension theorem
dimKerf + dimImf = dimV.
2. ClearlyW ? (W?)?. Then from (1) dimW+dimW? = dimV = dimW?+dim(W?)?
so that dimW = (W?)?. Hence (W?)? = W.
36
3. The ?rst part is clear and the second part is from (1) and (2) since dimV?1 = dimV ?
dimV1 > dimV ?dimV2 = dimV?2 .
4. From (3) (V1 +V2)? ?V?i , i = 1,2, since Vi ?V1 +V2. Thus (V1 +V2)? ?V?1 ?V?2 .
For any w ? V?1 ?V?2 and for all v = v1 +v2 ? V1 +V2, where vi ? Vi, i = 1,2, we
have ?(w,v1 +v2) = ?(w,v1) +?(w,v2) = 0. We may choose either of v1, v2 to be 0,
so w ? (V1 +V2)?.
5. By (2) each subspace W of V is the orthogonal complement of some subspace, namely,
W?. So it su?ces to show V1 +V2 = (V?1 ?V?2 )? and again by (2), it is simply (4).
We know focus our attention on semisimple Lie algebras.
Lemma 5.13. ([16, p.159]) Let L be a semisimple Lie algebra, H a CSA of L, and B a
solvable subalgebra of L containing H. Then
1. B = H ?[B,B]
2. The set [B,B] coincides with the set of ad-nilpotent elements in L contained in B.
Proof. By virtue of the fact that H is a CSA of L, H is in turn a CSA of B. Since
[H,B] ? B, adBH ? EndB is a simultaneously diagonalizable family by Theorem 2.25.
Thus B has root space decomposition B = H ?
?
?? (H)
B?(H). By Lemma 3.6, we write
H = L0(ads) for some s ? H. Note that the restriction of ads to B+(H) :=
?
??=0
B?(H) is
bijective, for B+(H) contains no eigenvectors of ads with corresponding eigenvalue 0. We
conclude that any x ? B+(H) can be written x = [s,m] for some m ? B+(H); so we have
B+(H) ? [B,B]. Now B is solvable, thus [B,B] is nilpotent; in addition, H is semisimple.
Thus any y ? H ? [B,B] is both ad-nilpotent and ad-semisimple, that is y = 0. We have
B = H ?
?
??=0
B?(H) ?H ?[B,B] ?B, and (1) follows.
37
The second statement follows from
[B,B] = [H ?
?
?? (H)
B?(H),H ?
?
?? (H)
B?(H)] =
?
?? (H)
B?(H).
On one hand, we know that the only endomorphism of L that is simultaneously nilpotent
and semisimple is 0. Since elements of H are ad-semisimple, the ad-nilpotent elements of
L contained in B must be in [B,B]. On the other hand, by Lemma 4.9 and the fact that
B?(H) ?L?(H), adLx is a nilpotent endomorphism for any x? [B,B].
We may now prove
Lemma 5.14. ([16, p.159]) A subalgebra B of a semisimple Lie algebra L is a BSA if and
only if [B,B] = B?.
Proof. SinceLis semisimple, the subalgebraH ?Lis solvable if and only if adH is solvable.
For the forward implication, by Cartan?s criterion [11, p.20] B is solvable if and only if
?(B,[B,B]) = 0. Thus [B,B] ?B?. Suppose that the inclusion is proper; by Lemma 5.12,
B([B,B]?. We setP := [B,B]?/B, a (nontrivial) vector space quotient. The the matrices
of adB form a solvable subalgebra of gl(P). By Lie?s Theorem, the matrices of adB are
upper triangular (with respect to the proper basis of P); thus there is an x ? [B,B]? \B
such that [B,x] ?B +Fx. Set B1 := B +Fx. Then B1 is a subalgebra of L.
By Lemma 5.12, B?1 = (B + Fx)? = B? ?Fx?. Now [B1,B1] ? [B,B] + [B,Fx];
by Cartan?s Criterion, since B is solvable, [B,B] ? B?. But x was chosen from [B,B]?,
so we also have [B,B] ? (Fx)?; thus [B,B] ? B? ? Fx? = B?1 . For any a,b ? B,
?([a,x],b) = ??(x,[a,b]) = 0, since x ? [B,B]?. Thus [B,Fx] ? B?. Finally, given
a ? B, we note that ?([a,x],x) = ?(a,[x,x]) = 0, allowing us to write [B,Fx] ? (Fx)?. So
[B,Fx] ?B?1 as well; we conclude that [B1,B1] ?B?1 . Once again using Cartan?s Criterion,
B1 is solvable; but B ( B1 was chosen as Borel, a contradiction. We are forced to have
[B,B] = B?.
38
We now show the reverse implication. Suppose [B,B] = B?, and choose a solvable
subalgebra B1 of L containing B. Then
[B1,B1] ?B?1 ?B? = [B,B] ? [B1,B1],
and all inclusions are forced to be equalities. So B?1 = B? implies B1 = B by Lemma 5.12,
and B is a BSA.
Theorem 5.15. ([16, p.160]) Let L be semisimple with CSAs H1 and H2, and let B1, B2
be BSAs containing H1 and H2, respectively. Then
1. B1 ?B2 contains a CSA of L.
2. There exist ui ? EL(Hi) (i = 1, 2) such that u1(H1) = u2(H2).
Proof. 1. De?ne Ni := [Bi,Bi]; then by Lemma 5.13 and Lemma 5.14, we have Bi = Hi?Ni
and Bi = N?i . Since Ni is the set of all ad-nilpotent elements of L contained in Bi, we know
that B1 ?N2 = N1 ?N2 = B2 ?N1 is the set of all ad-nilpotent elements of B1 ?B2. By
Lemma 5.12
B1 = N?1 ? (N1 ?N2)? = (B1 ?N2)? = B?1 +N?2 = N1 +B2.
We have shown that B1 ?N1 +B2, thus
B1 = N1 + (B1 ?B2).
Let ri = dimHi. By symmetry, we may assume r1 ?r2. By Corollary 5.8, ri = rankBi.
By Lemma 3.6 the CSA H1 may be written as H1 = L0(adz) for some z ? H1. Since
B1 = N1 + (B1 ?B2) = H1 +N1, we may choose n?N1 such that
w := z+n?B1 ?B2.
39
By Lie?s theorem, there is a basis of L so that adw, adz and adn, in matrix form, are upper
triangular. In addition, because adn is nilpotent, it is also strictly upper triangularizable;
thus the diagonal entries of adz are precisely those of adw, i.e. the pair of endomorphisms
share eigenvalues. Now dimL0(adz) and dimL0(adw) are the algebraic multiplicity of 0 as
an eigenvalue of adz and adw, respectively; thus dimL0(adz) = dimL0(adw). So
r1 = dimH1 = dimL0(adz) = dimL0(adw) ? dim(Bi)0(adw) ? rankBi = ri ?r1.
Thus each step above is an equality, and we have rankB1 = r1 = r2 = rankB2. So on one
hand, every CSA of L has dimension rankL = dimL0(adz), while on the other hand, by
Lemma 3.9, if H = L0(adx) has dimension rankL, then H is a CSA. So a subalgebra is a
CSA if and only if H = L0(adx) for some regular x. In particular, L0(adw) ?B1 ?B2 is a
CSA, and (1) follows.
Considering H = L0(adw) and Hi as CSAs of the (solvable) Bi, we may choose ui ?
EBi(Hi) such that u1(H1) = H = u2(H2) by Theorem 5.6. We employ Lemma 5.4 to extend
each ui to an element u?i ? EL(Hi), and (2) follows.
The following theorem will ?ll in the gap between the solvable and semisimple cases.
Theorem 5.16. ([16, p.158]) Let L be a Lie algebra and ? : L ? L/R the canonical
homomorphism of L, where R := RadL. Then the following statements for CSAs H1, H2 of
L are equivalent:
1. There exist ui ? EL(Hi) such that u1(H1) = u2(H2).
2. There exist vi ? EL/R(?(Hi)) such that v1(?(H1)) = v2(?(H2)).
Proof. That (1) implies (2) is straightforward: given ui ? EL(Hi), ui = ?expadx, de?ne
vi ? End(L/R) by vi = ?expad(x + R); then vi(y + R) = ui(y) + R, so v1(?(H1)) =
v2(?(H2)). All that remains to show is that the vi are indeed elements of EL/R(?(Hi)). Now
ui = ?exp(adx) where x ? L?(Hi), and by Lemma 5.2, ?(x) ? ?(L?(Hi)) means that
40
x + R ? L/R??(?(Hi)). Thus vi ? EL/R(?(Hi)). By construction, we have vi ? EL/R(Hi)
satisfying v1(?(H1)) = v2(?(H2)).
Thereverseimplicationrequires morework: byLemma5.3, givenv1(?(H1)) = v2(?(H2)),
we have ui ? EL(Hi) with ?(u1(H1)) = ?(u2(H2)), that is u1(H1)/R = u2(H2)/R. Consider
the set u1(H1) +R = u2(H2) +R as a subalgebra of L, called T; viewed as a Lie algebra,
we see that each ui(Hi) is a CSA of T. Thus the ui(Hi) are solvable subalgebras of T;
R, the radical of L, is also solvable, so T itself is solvable. We already know that CSAs
of solvable Lie algebras are conjugate (5.6), so we may choose u?i ? ET(ui(Hi)) such that
u?1u1(H1) = u?2u2(H2). By Lemma 5.4 the u?i may be extended to ui?? in EL(ui(Hi)).
Then u?iui ? ET(ui(Hi))ui can be written as u??iui ? EL(ui(Hi))ui. By Lemma 5.5
EL(ui(Hi))ui = uiEL(Hi) = EL(Hi), and we conclude that u??iui ? EL(Hi).
Theorem 5.17. ([16, p.160]) Given CSAs H1 and H2 of the Lie algebra L, there exist
ui ? EL(Hi) such that u1(H1) = u2(H2).
Proof. L/R is semisimple when R = RadL. By Theorem 5.15, with ? : L ? L/R the
canonical projection, ?(H1) and ?(H2) are conjugate, and by Theorem 5.16 H1 and H2 are
as well.
Theorem 5.18. ([16, p.161]) For any Lie algebra L,
1. EL(C) does not depend on the choice of the CSA C, and may be denoted EL.
2. EL acts transitively on the set of CSAs of L.
3. Any CSA has dimension rankL.
4. The element x?L is regular if and only if L0(adx) is a CSA, and any CSA of L may
be written in this form.
Proof. The proof of (1) is identical to the proof of Corollary 5.7; (2), (3), and (4) are similar
to Corollary 5.8.
In order to establish the conjugacy of BSAs, we shall need several preliminary results.
41
Lemma 5.19. ([16, p.161]) Any BSA of a semisimple Lie algebra L contains a CSA of L.
Proof. Given a BSA B of L, choose a CSA H of B. We will show that H is actually a CSA
of L as well; to do so we simply need NL(H) = H. Since H is nilpotent, by Theorem 2.21
L =
?
??H?
L?(H).
Givenh?H, we use abstract Jordan decomposition ([11, p.24]) to decomposeh(inL) as
h = hs+hn, wherehs andhn are the ad-semisimple and ad-nilpotent parts, respectively, ofh.
We need both of the pieceshs andhn to be elements ofH. Now adhn may be put into strictly
upper triangular form and adhs may be diagonalized simultaneously, so the action of adhs
on an element x?L?(H) is simply multiplication by ?(h). Of particular importance is the
fact that [hs,L0(H)] = 0, so hs ?Z(L0(H)). Now adhs is a polynomial in adh ([11, p.17]),
so adhs(H) ?H, thus hs ?NL(H). Also, h?B implies that adh(B) ?B, so hs ?NL(B),
indeedhs ?NL(B)?NL(H). By Lemma 3.3,B = NL(B), sohs ?B?NL(H) = NB(H) = H;
we have hs ?H, so hn ?H as well. Thus H contains the semisimple and nilpotent parts of
all of its elements.
Now L0(H) is reductive by [7, p.10], so we may use [5, p.56] to write
L0(H) = Z(L0(H))?[L0(H),L0(H)].
As H ?L0(H), [Z(L0(H)),H] = 0, so Z(L0(H)) ?NL(H). Given x?NL(H), adh(x) ?H
implies that (adh)n(adh(x)) = 0 for su?ciently large n, since H is nilpotent. Thus we know
thatNL(H) ?L0(H). We haveZ(L0(H)) ?NL(H) ?L0(H), so utilizing the decomposition
of L0(H), we write
NL(H) = Z(L0(H))?(NL(H)?[L0(H),L0(H)]).
Thus we need to show
42
1. Z(L0(H)) ?H
2. NL(H)?[L0(H),L0(H)] ?H
The decomposition of L yields a decomposition of B:
B = B?L = B?
?
??H?
L?(H) =
?
??H?
B?(H),
where B? := B ? L?. Set B+(H) :=
?
??=0
B?(H). As H is a CSA of B, we know that
B0(H) = H by Lemma 3.5. So the decomposition of B is given by
B = H ?B+(H).
Since we have ?nitely many nonzero ? ? H? in the above decomposition such that
B?(H) ?= 0, we can ?nd a nonzero ? ? H? such that ? is not contained in the union of the
orthogonal complements of??s. Transporting back to H, that means that there exists h?H
such that ?(h) ?= 0 for any nonzero ??H? such that B?(H) ?= 0. Thus adh acts bijectively
on B+(H), so B+(H) ? [B,B]. Then
[B,B] = B?[B,B] = (H ?[B,B])?B+(H).
By [11, p.36], ?(L?(H),L?(H)) ?= 0 when ?+? ?= 0, and in particular since Z(L0(H)) ?
L0(H), we have ?(Z(L0(H)),B+(H)) = 0. This forces B+(H) ? (Z(L0(H)))?. In addition,
[Z(L0(H)),H ?[B,B]] = 0 since H ?[B,B] ?H ?L0(H).
As B is solvable, [B,B] is nilpotent. Then for each h?H?[B,B], we know that adBh
is a nilpotent endomorphism. Given x ? Z(L0(H)) and h ? H ? [B,B], adx ady is also
a nilpotent endomorphism of B. We conclude that tr(adx ady) = 0. Thus H ? [B,B] ?
(Z(L0(H)))?. We have previously shown that B+(H) ? (Z(L0(H)))?; as [B,B] = (H ?
43
[B,B])?B+(H), we have shown
[B,B] ?Z(L0(H))?.
As L is semisimple, we know by Lemma 5.14 that [B,B] = B?. Then B? = [B,B] ?
Z(L0(H))? implies
Z(L0(H)) ?B.
Then Z(L0(H)) ?B?NL(H) = H, i.e. we have shown (1): Z(L0(H)) ?H.
We still need to show (2), that
NL(H)?[L0(H),L0(H)] ?H.
Suppose that z ?NL(H)?[L0(H),L0(H)]. Then
[H +Fz,H +Fz] ? [H,H] + [H,Fz] ?H.
Thus H +Fz must be a solvable subalgebra of L, and by Cartan?s Criterion,
?(H +Fz,[H +Fz,H +Fz]) = 0.
In addition, since L is semisimple and H +Fz is solvable, [H +Fz,H +Fz] is precisely
the set of ad-nilpotent elements of L contained in H +Fz by Lemma 5.13. Given h ? H,
adLhn is nilpotent and hn ? H; so hn ? [H +Fz,H +Fz]. Thus ?(z,hn) = 0. Recall that
[hs,L0(H)] = 0; so hs ?Z(L0(H)), and
?(z,hs) ??([L0(H),L0(H)],Z(L0(H))) = ?(L0(H),[L0(H),Z(L0(H))]),
44
where the last equality follows by Remark 5.10. But [L0(H),Z(L0(H))] = 0, so
?(L0(H),[L0(H),Z(L0(H))]) = 0.
We have ?(z,hs) = 0, giving us ?(z,h) = 0 for any h?H. Now z was an arbitrary element
of NL(H)?[L0(H),L0(H)], so
NL(H)?[L0(H),L0(H)] ?H?.
Clearly NL(H) ? [L0(H),L0(H)] ? L0(H), and since B+(H) ?
?
??H?
L?(H), we know
that L0(H) ? (B+(H))?. Then B = H ?B+(H) implies that
NL(H)?[L0(H),L0(H)] ?H? ?(B+(H))? = B?.
Recall that B? = [B,B]. Then
NL(H)?[L0(H),L0(H)] ?B? ?NL(H) ?B?NL(H) = H,
and we have shown (2).
To conclude, we have shown that
NL(H) = Z(L0(H))?(NL(H)?[L0(H),L0(H)])
and as each piece of the summand is a subset of H, it follows that NL(H) = H, i.e. H is a
CSA of L.
Lemma 5.20. The intersection of two BSAs of a semisimple L contains a CSA.
Proof. Follow immediately by Theorem 5.15 and Lemma 5.19.
45
Lemma 5.21. ([16, p.163]) The BSAs of a semisimple Lie algebra L are conjugate under
EL.
Proof. Given a pair B1, B2 of BSAs, by Lemma 5.20 their intersection contains a CSA H.
The Bi are stable under H, indeed the matrices of adBiH are simultaneously diagonalizable.
So we may decompose Bi as
Bi = H +
?
?? i
Bi,?,
where ?i ? ?. But if ? ? ?i then Bi,? = L?. We may then use a standard argument
involving the Weyl group to permute B1 onto B2. (See [11, p.75])
Theorem 5.22. ([16, p.161]) Each BSA B of a Lie algebra L contains a CSA of L.
Proof. Consider the image of B under the canonical homomorphism ? : L?L/R, where R
is the radical ofL. By Lemma 4.5?(B) = B? is a BSA of the semisimpleL/Rand by Lemma
5.19 B? contains a CSA H? of L/R. We may ?nd a CSA H of L such that ?(H) = H? by
Lemma 3.12. H is contained in a BSA ?B, and the BSA ?( ?B) = ?B? of L/R may be permuted
via an element ?? of EL/R onto B? by Lemma 5.21. By Lemma 5.3 there is a ? ? EL with
??? = ??, i.e.
???( ?B) = ??( ?B?) = B? = ??( ?B).
But the BSAs of L are in 1-1 correspondence with the BSAs of L/R by Lemma 4.5, so
?( ?B) = B, and in particular ?(H) is a CSA of L contained in B.
Theorem 5.23. ([16, p.163]) The following statements hold in any Lie algebra L:
1. The intersection of two BSAs contains a CSA.
2. The BSAs of L are conjugate under EL.
Proof. Both statements have already been established for semisimple L; using a process
similar to that in the proof of Theorem 5.22, the theorem follows.
46
Chapter 6
Some remarks
Cartan subalgebras exist for ?nite dimensional Lie algebras whenever the base ?eld is
in?nite. Indeed Barnes [1] (see also [2]) shows that if L is a Lie algebra of dimension n over
a ?eld F of at least n?1 elements, then there exists a Cartan subalgebra of L. Barnes also
showed that every ?nite-dimensional solvable Lie algebra has a Cartan subalgebra over any
?eld.
WhenFhas no more than dimFLelements, the existence of Cartan subalgebras of ?nite
dimensional Lie algebras is still an open problem [11, p.80] [19, p.509].
In [3] Billig and Pianzola give an example of a Lie algebra L of countable dimension
which has no Cartan subalgebras (the de?nition of CSA given in [3] reduces to the classical
case if L is ?nite dimensional).
Real case: In general, Cartan subalgebras in a real Lie algebra g are not necessarily
conjugate. Example: In sl2(R) there are two essentially di?erent Cartan subalgebras
H1 =
?
??
??
?
??? 0
0 ??
?
??: ??R
?
??
??, H2 =
?
??
??
?
?? 0 ?
?? 0
?
??: ??R
?
??
??.
Notice that in sl2(C) these subalgebras are conjugate.
Kostant [14] and Sugiura [21] gave the theory of conjugacy classes of Cartan subalgebras
for a real simple (and hence semisimple) noncompact Lie algebra L.
47
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