classification of finite-dimensional simple lie algebras over the
TRANSCRIPT
Classification of Finite-Dimensional Simple Lie
Algebras over the Complex Numbers
by
Shawn Baland
A thesis submitted to the
Faculty of the University of Colorado
in partial fulfillment of the requirements
for graduation with departmental honors
Department of Mathematics
November 12, 2007
Updated December 6, 2007
Defense Committee
Prof. Arlan Ramsay, Mathematics (Thesis Advisor)
Prof. Richard Clelland, Mathematics
Prof. James Meiss, Applied Mathematics
ii
Baland, Shawn
Classification of Finite-Dimensional Simple Lie Algebras over the Complex Numbers
Thesis directed by Prof. Arlan Ramsay
Lie algebras were first introduced by S. Lie in 1880 as algebraic structures used to
study local problems in Lie groups. Specifically, the tangent space of a Lie group at the
identity has the natural structure of a Lie algebra. Aside from their many applications,
Lie algebras have attracted interest in their own right. This paper deals primarily with
the classification of finite-dimensional simple Lie algebras over the complex numbers,
first achieved through the independent work of E. Cartan and W. Killing during the
decade 1890-1900. Contemporary methods of studying Lie algebras and their represen-
tations were later developed by H. Weyl.
The classification theorem begins by studying the decomposition of a semisimple
Lie algebra into a direct sum of subspaces with respect to a carefully chosen subalgebra
called the Cartan subalgebra. One then forms a Euclidean space from the nonzero
linear functionals associated with these subspaces via an inner product called the Killing
form. The geometry of this Euclidean space allows us to encode vital information about
a Lie algebra’s structure into a graph called the Dynkin diagram. It is then shown
that there are only a limited number of possible connected Dynkin diagrams, and that
these diagrams are in one-to-one correspondence with the finite-dimensional simple Lie
algebras over the complex numbers. The paper’s structure closely follows that of [Car05].
It is assumed that the reader has a thorough knowledge of linear algebra, although this
is the only major prerequisite.
Dedication
To my mother, for supporting me when I was less ambitious.
iv
Acknowledgments
I would first like to thank Professor Peter Elliott for encouraging me to go beyond
what is usually expected of an undergraduate student. I would also like to thank Profes-
sor Richard Green for suggesting this project, and for pointing me toward many valuable
resources. I owe the greatest thanks to Professor Arlan Ramsay, my thesis advisor, for
spending a great deal of time answering all of my questions, and for asking many ques-
tions that I had not considered. Thanks also to Mike Noyes, my reader, for improving
this paper’s clarity and accuracy. Finally, I would like to thank my wife, Lyndsie, for
her support and understanding throughout my undergraduate career.
v
Contents
Chapter
1 Basic concepts 1
1.1 Lie algebras, subalgebras, and ideals . . . . . . . . . . . . . . . . . . . . 1
1.2 Lie homomorphisms and quotient algebras . . . . . . . . . . . . . . . . . 3
1.3 Nilpotent and solvable Lie algebras . . . . . . . . . . . . . . . . . . . . . 6
2 Representations of solvable and nilpotent Lie algebras 11
2.1 Basic representation theory . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.2 A transition to finite-dimensional complex Lie algebras . . . . . . . . . . 14
2.3 Representations of solvable Lie algebras . . . . . . . . . . . . . . . . . . 14
2.4 Representations of nilpotent Lie algebras . . . . . . . . . . . . . . . . . . 19
3 Cartan subalgebras 26
3.1 Existence of Cartan subalgebras . . . . . . . . . . . . . . . . . . . . . . 26
3.2 Derivations and automorphisms . . . . . . . . . . . . . . . . . . . . . . . 28
3.3 Concepts from algebraic geometry . . . . . . . . . . . . . . . . . . . . . 30
3.4 Conjugacy of Cartan subalgebras . . . . . . . . . . . . . . . . . . . . . . 35
4 The Cartan decomposition 39
4.1 Root spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
4.2 The Killing form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
4.3 The Cartan decomposition of a semisimple Lie algebra . . . . . . . . . . 48
vi
5 Root systems and the Weyl group 57
5.1 Positive systems and fundamental systems of roots . . . . . . . . . . . . 57
5.2 The Weyl group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
6 The Cartan matrix and the Dynkin diagram 70
6.1 The Cartan matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
6.2 The Dynkin diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
6.3 Classification of Dynkin diagrams . . . . . . . . . . . . . . . . . . . . . . 77
6.4 Classification of Cartan matrices . . . . . . . . . . . . . . . . . . . . . . 83
7 The existence and uniqueness theorems 94
7.1 Properties of structure constants . . . . . . . . . . . . . . . . . . . . . . 94
7.2 The uniqueness theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
7.3 Some generators and relations in a simple Lie algebra . . . . . . . . . . 103
7.4 The Lie algebras g(A) and g(A) . . . . . . . . . . . . . . . . . . . . . . . 105
7.5 The existence theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114
Bibliography 133
vii
Figures
Figure
6.1 Cartan matrices and Dynkin diagrams for l = 1 and 2 . . . . . . . . . . 74
6.2 Dynkin diagrams and quadratic forms for l = 1 and 2 . . . . . . . . . . 76
6.3 Standard list of connected Dynkin diagrams . . . . . . . . . . . . . . . . 84
Chapter 1
Basic concepts
1.1 Lie algebras, subalgebras, and ideals
Let k be a field. A k-algebra is a k-vector space A equipped with a bilinear
operation A × A → A called multiplication. The dimension of A is the dimension of
A as a k-vector space.
A Lie algebra is a k-algebra g with multiplication (x, y) 7→ [x, y] satisfying the
following conditions:
(i) [x, x] = 0 for all x ∈ g.
(ii) [[x, y], z] + [[y, z], x] + [[z, x], y] = 0 for all x, y, z ∈ g.
Condition (ii) is called the Jacobi identity.
Proposition 1.1. [y, x] = −[x, y] for all x, y ∈ g, i.e., multiplication in a Lie algebra
is anticommutative.
Proof. If x, y ∈ g, then
0 = [x+ y, x+ y] = [x, x] + [x, y] + [y, x] + [y, y]
= 0 + [x, y] + [y, x] + 0.
It follows that [y, x] = −[x, y].
2
Notice that anticommutativity implies [x, x] = −[x, x] for all x ∈ g, yielding
2[x, x] = 0. Hence if char k 6= 2, then anticommutativity is equivalent to condition (i)
for a Lie algebra.
Example 1.2. An associative algebra is a k-algebra A satisfying the associative law,
that is,
(xy)z = x(yz) for all x, y, z ∈ A.
Lie algebras are easily obtained from associative algebras in the following way: Given
any associative algebra A, we define the commutator of x, y ∈ A by
[x, y] = xy − yx.
The commutator is clearly bilinear. Moreover, [x, x] = xx− xx = 0 and
[[x, y], z] + [[y, z], x] + [[z, x], y] = (xy − yx)z − z(xy − yx) + (yz − zy)x− x(yz − zy)
+ (zx− xz)y − y(zx− xz) = 0
for all x, y, z ∈ A. Thus A forms a Lie algebra via the commutator product. We call
this algebra the Lie algebra of A and denote it by [A].
If a and b are subspaces of a Lie algebra g, we define [a, b] to be the subspace of
g generated by all elements of the form [x, y] with x ∈ a and y ∈ b.
Proposition 1.3. [a, b] = [b, a] for all subspaces a, b of g.
Proof. Any element in [a, b] is of the form
λ1[x1, y1] + · · ·+ λr[xr, yr]
where xi ∈ a, yi ∈ b, λi ∈ k for i = 1, . . . , r. Now [xi, yi] = −[yi, xi] ∈ [b, a] by
Proposition 1.1. It follows that [a, b] ⊂ [b, a]. By a symmetric argument, we also have
[b, a] ⊂ [a, b], and so [a, b] = [b, a].
A subalgebra of a Lie algebra g is a subspace h ⊂ g such that [h, h] ⊂ h. Hence
h forms a Lie algebra by restricting the multiplication on g to the subspace h.
3
Proposition 1.4. If h and k are subalgebras of g, then h ∩ k is a subalgebra of g.
Proof. We know that h ∩ k is a subspace of g from linear algebra. Now [h ∩ k, h ∩ k] is
generated by elements of the form [x, y] with x, y ∈ h∩ k. We have x, y ∈ h and x, y ∈ k
so that [x, y] ∈ [h, h] ∩ [k, k] ⊂ h ∩ k. It follows that [h ∩ k, h ∩ k] ⊂ h ∩ k, and so h ∩ k is a
subalgebra of g.
An ideal of g is a subspace a ⊂ g such that [a, g] ⊂ a. This condition is equivalent
to [g, a] ⊂ a by Proposition 1.3, and so every ideal of a Lie algebra is two-sided. Note
also that [a, a] ⊂ a since a ⊂ g. Thus ideals are also subalgebras.
Proposition 1.5. If a and b are ideals of g, then a∩ b, a+ b, and [a, b] are ideals of g.
Proof. We know that these sets are subspaces of g from linear algebra. For the first
part, notice that [a ∩ b, g] is generated by elements of the form [x, y] with x ∈ a ∩ b,
y ∈ g. We have x ∈ a and x ∈ b so that [x, y] ∈ [a, g] ∩ [b, g] ⊂ a ∩ b. It follows that
[a ∩ b, g] ⊂ a ∩ b, and so a ∩ b is an ideal of g.
For the second part, notice that [a + b, g] is generated by elements of the form
[x+y, z] with x ∈ a, y ∈ b, z ∈ g. We have [x+y, z] = [x, z]+[y, z] ∈ [a, g]+[b, g] ⊂ a+b.
It follows that [a + b, g] ⊂ a + b, and so a + b is an ideal of g.
For the third part, notice that [[a, b], g] is generated by elements of the form
[[x, y], z] with x ∈ a, y ∈ b, z ∈ g. We have [[x, y], z] = [[x, z], y]− [[y, z], x] by the Jacobi
identity. Now [x, z] ∈ a and [y, z] ∈ b since a and b are ideals. Thus [[x, y], z] ∈ [a, b]. It
follows that [[a, b], g] ⊂ [a, b], and so [a, b] is an ideal of g.
1.2 Lie homomorphisms and quotient algebras
As with Lie subalgebras and ideals, the concepts of Lie homomorphisms and
quotient algebras are analogous to those of associative algebras. Let g, h be Lie algebras
over a field k. A map φ : g → h is a homomorphism of Lie algebras if φ is k-linear
4
and
φ([x, y]) = [φ(x), φ(y)] for all x, y ∈ g.
An isomorphism of Lie algebras is a bijective homomorphism of Lie algebras. If
there exists an isomorphism φ : g → h, then we say g and h are isomorphic and write
g ∼= h.
Proposition 1.6. If g is a Lie algebra and a is an ideal of g, then the quotient space
g/a forms a Lie algebra via the multiplication
[x+ a, y + a] = [x, y] + a for all x, y ∈ g.
Proof. To prove this multiplication is well defined, suppose x+ a = x′ + a and y + a =
y′+a. We must show that [x, y]+a = [x′, y′]+a. Now x+a = x′+a implies x = x′+a1
for some a1 ∈ a. Likewise, y + a = y′ + a implies y = y′ + a2 for some a2 ∈ a. Thus
[x, y] + a = [x′ + a1, y′ + a2] + a = [x′, y′] + [x′, a2] + [a1, y
′] + [a1, a2] + a.
The last three summands are in a since a is an ideal, and so [x, y] + a = [x′, y′] + a.
Next, we have
[x+ a, x+ a] = [x, x] + a = 0 + a = a for all x ∈ a.
Finally, the Jacobi identity follows from the Jacobi identity in g.
Proposition 1.7. Let φ : g → h be a homomorphism of Lie algebras. Then imφ is a
subalgebra of h, kerφ is an ideal of g, and g/ kerφ ∼= imφ.
Proof. We know that imφ is a subspace of h from linear algebra. If x, y ∈ g, then
[φ(x), φ(y)] = φ([x, y]) ∈ imφ.
This shows that [imφ, imφ] ⊂ imφ, and so imφ is a subalgebra of h. We also know that
kerφ is a subspace of g. If x ∈ kerφ and y ∈ g, then
φ([x, y]) = [φ(x), φ(y)] = [0, φ(y)] = 0,
5
showing that [x, y] ∈ kerφ. It follows that [kerφ, g] ⊂ kerφ, and so kerφ is an ideal of
g. To prove the last statement, consider the linear map
g/ kerφ→ imφ
x+ kerφ 7→ φ(x).
This map is well defined since
x+ kerφ = y + kerφ ⇔ x− y ∈ kerφ ⇔ φ(x− y) = 0 ⇔ φ(x) = φ(y).
The leftward implication proves that the map is injective. If φ(x) ∈ imφ, then x+kerφ
maps to φ(x), showing that the map is surjective. Finally, we prove that the map is a
homomorphism of Lie algebras. If x, y ∈ g, then
[x+ kerφ, y + kerφ] = [x, y] + kerφ 7→ φ([x, y]) = [φ(x), φ(y)].
Hence the map respects Lie multiplication, and so it is an isomorphism of Lie algebras.
Lemma 1.8. If h is a subalgebra of g and a is an ideal of g, then h + a is a subalgebra
of g.
Proof. We know that h + a is a subspace of g from linear algebra. Now [h + a, h + a] is
generated by elements of the form [x+ y, x′ + y′] with x, x′ ∈ h and y, y′ ∈ a. We have
[x+ y, x′ + y′] = [x, x′] + [x, y′] + [y, x′] + [y, y′]
∈ [h, h] + ([h, a] + [a, h] + [a, a])
⊂ h + a.
It follows that [h + a, h + a] ⊂ h + a, and so h + a is a subalgebra of g.
Proposition 1.9. Let h be a subalgebra of g and a an ideal of g. Then a is an ideal of
h + a, h ∩ a is an ideal of h, and (h + a)/a ∼= h/(h ∩ a).
6
Proof. We know that h+a and h∩a are subalgebras of g by Lemma 1.8 and Proposition
1.4, respectively. We have [a, h + a] ⊂ [a, g] ⊂ a, and so a is an ideal of h + a. Also,
[h ∩ a, h] ⊂ [h, h] and [h ∩ a, h] ⊂ [a, h] so that [h ∩ a, h] ⊂ [h, h] ∩ [a, h] ⊂ h ∩ a. Thus
h ∩ a is an ideal of h.
Notice that (h + a)/a forms a Lie algebra by Proposition 1.6. Consider the linear
map φ : h → (h + a)/a defined by φ(x) = x+ a. We have
φ([x, y]) = [x, y] + a = [x+ a, y + a] = [φ(x), φ(y)],
and so φ is a homomorphism of Lie algebras. It is surjective since each element of
(h + a)/a can be written as x+ a with x ∈ h. Finally, we have
kerφ = {x ∈ h | φ(x) = 0} = {x ∈ h | x+ a = a} = {x ∈ h | x ∈ a} = h ∩ a.
Thus h/(h ∩ a) ∼= (h + a)/a by Proposition 1.7.
1.3 Nilpotent and solvable Lie algebras
The key to studying a Lie algebra’s structure is knowing about its ideals. The
existence of the right kind of ideals will be useful in the sequel and is introduced in this
section.
Let g be a Lie algebra. We consider the subspaces of g defined recursively by
g1 = g, gn+1 = [gn, g] for n ≥ 1.
Proposition 1.10. gn is an ideal of g and
g = g1 ⊃ g2 ⊃ g3 ⊃ · · · .
Proof. We prove the first statement by induction on n. If n = 1, then g1 = g is an ideal
since [g, g] ⊂ g. Now suppose the statement is true for n = r, that is, gr is an ideal.
Then gr+1 = [gr, g] is an ideal by Proposition 1.5, completing the induction. To prove
7
the second statement, let n ≥ 1. Then
gn+1 = [gn, g] ⊂ gn
since gn is an ideal.
Given Proposition 1.10, we call the chain of ideals g = g1 ⊃ g2 ⊃ g3 ⊃ · · · the
lower central series of g. We say that g is nilpotent if gn = 0 for some n ≥ 1. If
g2 = 0, then we say g is abelian. This is equivalent to the condition [g, g] = 0, i.e.,
[x, y] = 0 for all x, y ∈ g.
We now consider the subspaces of g defined recursively by
g(0) = g, g(n+1) =[g(n), g(n)
]for n ≥ 0.
Proposition 1.11. g(n) is an ideal of g and
g = g(0) ⊃ g(1) ⊃ g(2) ⊃ · · · .
Proof. We prove the first statement by induction on n. If n = 0, then g(0) = g is clearly
an ideal. Now suppose the statement is true for n = r, that is, g(r) is an ideal. Then
g(r+1) =[g(r), g(r)
]is an ideal by Proposition 1.5, completing the induction. To prove
the second statement, let n ≥ 0. Then
g(n+1) =[g(n), g(n)
]⊂[g(n), g
]⊂ g(n)
since g(n) is an ideal.
Given Proposition 1.11, we call the chain of ideals g = g(0) ⊃ g(1) ⊃ g(2) ⊃ · · ·
the derived series of g. We say that g is solvable if g(n) = 0 for some n ≥ 0.
Proposition 1.12.
(i) [gm, gn] ⊂ gm+n for all m,n ≥ 1.
(ii) g(n) ⊂ g2nfor all n ≥ 0.
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(iii) Every nilpotent Lie algebra is solvable.
Proof. For part (i), we fix m and use induction on n. If n = 1, then [gm, g] = gm+1.
Now suppose the statement is true for n = r. By definition,[gm, gr+1
]= [gm, [gr, g]].
This subspace is generated by elements of the form [x, [y, z]] with x ∈ gm, y ∈ gr, z ∈ g.
We have
[x, [y, z]] = [[z, x], y] + [[x, y], z] (by the Jacobi identity)
∈ [[g, gm] , gr] + [[gm, gr] , g]
⊂[gm+1, gr
]+[gm+r, g
]⊂ gm+r+1 (by the inductive hypothesis).
It follows that [gm, gn] ⊂ gm+n for all n.
For part (ii), we again use induction on n. For n = 0, we have g(0) = g = g20.
Now suppose the statement is true for n = r. We have
g(r+1) =[g(r), g(r)
]⊂[g2r
, g2r] ⊂ g2·2r= g2r+1
by part (i), completing the induction.
For part (iii), if g is nilpotent, then gn = 0 for some n ≥ 1. Thus g(n) ⊂ g2n ⊂
gn = 0 by part (ii), and so g is solvable.
Proposition 1.13. If g is solvable, then every subalgebra and every quotient algebra of
g is solvable.
Proof. We first show that if φ : g → h is a surjective homomorphism of Lie algebras,
then φ(g(n)
)= h(n) for all n ≥ 0. We proceed by induction on n. For n = 0, we have
φ(g(0))
= φ(g) = h = h(0). Now suppose the statement is true for n = r. We have
φ(g(r+1)
)= φ
([g(r), g(r)
])=[φ(g(r)), φ(g(r))]
=[h(r), h(r)
]= h(r+1),
completing the proof of the claim.
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To prove the proposition, choose n such that g(n) = 0. If h is a subalgebra of g,
then h(n) ⊂ g(n) = 0, and so h is solvable. Finally, if a is an ideal of g, then (g/a)(n) is
the image of g(n) under the canonical homomorphism g → g/a by the preceding result.
We have g(n) = 0, yielding (g/a)(n) = 0, and so g/a is solvable.
Proposition 1.14. If a is an ideal of g such that a and g/a are solvable, then g is
solvable.
Proof. We first show that g(m+n) =(g(m)
)(n)for all m,n ≥ 0. We fix m and use
induction on n. For n = 0, we have g(m+0) = g(m) =(g(m)
)(0). Now suppose the
statement is true for n = r. We have
g(m+r+1) =[g(m+r), g(m+r)
]=[(
g(m))(r)
,(g(m)
)(r)]
=(g(m)
)(r+1),
completing the proof of the claim.
To prove the proposition, notice that since g/a is solvable, we have (g/a)(m) = 0
for some m ≥ 0, that is, g(m) ⊂ a. And since a is solvable, we have a(n) = 0 for some
n ≥ 0. By the preceding result, this yields
g(m+n) =(g(m)
)(n)⊂ a(n) = 0,
and so g is solvable.
Proposition 1.15. Every finite-dimensional Lie algebra g contains a unique maximal
solvable ideal r.
Proof. We first show that if a and b are solvable ideals of g, then a+b is a solvable ideal
of g. Notice that a + b is an ideal by Proposition 1.5. Because a is solvable, we know
that a/(a∩b) is solvable by Proposition 1.13. Now (a+b)/b ∼= a/(a∩b) by Proposition
1.9. Thus (a + b)/b is solvable. Since b is also solvable, we see that a + b is solvable by
Proposition 1.14.
10
To prove the proposition, we choose a solvable ideal r of g having maximal di-
mension. Let a be any solvable ideal of g. Then a+r is a solvable ideal by the preceding
result. We have a + r = r since dim r is maximal, and so a ⊂ r. Thus r contains every
solvable ideal of g.
Given Proposition 1.15, we call the maximal solvable ideal r the solvable radical
of g. A Lie algebra g is semisimple if r = 0, that is, if g contains no nonzero solvable
ideal. A Lie algebra g is simple if it contains no proper ideals, i.e., no ideals other than
0 and g.
Suppose g is 1-dimensional over k. Then for any nonzero x ∈ g, {x} forms a
basis of g. Since [x, x] = 0, we see that g2 = 0, and so g is abelian. This is true
for any Lie algebra of dimension 1. Thus any two 1-dimensional Lie algebras over k
are isomorphic. Because the dimension of any subspace is either 0 or 1, we see that g
contains no proper subspaces. Hence g contains no proper ideals, and so g is simple.
We call the 1-dimensional Lie algebra the trivial simple Lie algebra.
Proposition 1.16. Every non-trivial simple Lie algebra is semisimple.
Proof. Suppose g is simple but not semisimple. Then r 6= 0, and since r is an ideal, we
must have r = g. Because r is solvable, g must be solvable, and so g(n) = 0 for some
n ≥ 0. Now g(1) is an ideal. Hence g(1) = g or g(1) = 0. But g(1) cannot equal g,
because this would imply g(n) = g for all n ≥ 0, contradicting the solvability of g. Thus
g(1) = 0, i.e., [g, g] = 0. If a is any subspace of g, then [a, g] ⊂ [g, g] = 0 ⊂ a. It follows
that every subspace of g is an ideal, and so g cannot have any proper subspaces. Hence
dim g = 1, and so g is the trivial simple Lie algebra.
Chapter 2
Representations of solvable and nilpotent Lie algebras
2.1 Basic representation theory
Recall from Example 1.2 that any associative algebra A gives rise to a Lie algebra
[A] via the commutator product. We consider the case where A = Mn(k), the ring of
n×n matrices over k. One easily checks that this is an associative k-algebra. We define
the Lie algebra gln(k) by
gln(k) = [Mn(k)].
A representation of a Lie algebra g is a homomorphism of Lie algebras
ρ : g → gln(k)
for some n ≥ 0. In this case, ρ is called a representation of degree n. Two representations
ρ, ρ′ of degree n are equivalent if there exists an invertible matrix A ∈ Mn(k) such
that ρ′(x) = A−1ρ(x)A for all x ∈ g.
A left g-module is a k-vector space V equipped with a left action
g× V → V
(x, v) 7→ xv
satisfying the following conditions:
(i) (x, v) 7→ xv is linear in x and v.
(ii) [x, y]v = x(yv)− y(xv) for all x, y ∈ g, v ∈ V .
12
Note that we can define right g-modules in a similar way, although we will not have
occasion to use them in this paper. Thus from now on, we will assume that all g-modules
are left g-modules.
There is a one-to-one correspondence between representations of degree n and
n-dimensional g-modules. Suppose ρ is a representation of g of degree n. Let V be an
n-dimensional vector space and choose a basis e1, . . . , en of V . For each x ∈ g, we define
the action of x on an element of V by
x
∑j
λjej
=∑ij
λjρij(x)ei.
Thus for each x ∈ g, the matrix ρ(x) acts as a linear transformation on V with respect
to the basis e1, . . . , en. For all x, y ∈ g, v ∈ V , we have
ρ([x, y])v = [ρ(x), ρ(y)]v = (ρ(x)ρ(y)− ρ(y)ρ(x))v = ρ(x)ρ(y)v − ρ(y)ρ(x)v,
showing that this action transforms V into a g-module.
Conversely, suppose V is an n-dimensional g-module and let e1, . . . , en be a basis
of V . For each x ∈ g, we write
xej =∑
i
ρij(x)ei
where ρij(x) ∈ k. This gives rise to the matrix ρ(x) = (ρij(x)). We have
[x, y] ej = x(yej)− y(xej)
= x
(∑k
ρkj(y)ek
)− y
(∑k
ρkj(x)ek
)
=∑
k
ρkj(y)xek −∑
k
ρkj(x)yek
=∑
k
ρkj(y)
(∑i
ρik(x)ei
)−∑
k
ρkj(x)
(∑i
ρik(y)ei
)
=∑
i
(∑k
(ρik(x)ρkj(y)− ρik(y)ρkj(x))
)ei
=∑
i
(ρ(x)ρ(y)− ρ(y)ρ(x))ijei.
13
It follows that ρ([x, y]) = ρ(x)ρ(y)−ρ(y)ρ(x) = [ρ(x), ρ(y)], and so ρ is a representation
of g.
Now let f1, . . . , fn be another basis of V and ρ′ the representation obtained from
this basis. I claim that ρ′ is equivalent to ρ. We know there exists an invertible n × n
matrix A such that fj =∑
iAijei. We have
xfj =∑
k
Akjxek =∑
k
Akj
(∑i
ρik(x)ei
)=∑
i
(∑k
ρik(x)Akj
)ei.
We also have
xfj =∑
k
ρ′kj(x)fk =∑
k
ρ′kj(x)
(∑i
Aikei
)=∑
i
(∑k
Aikρ′kj(x)
)ei.
It follows that ρ(x)A = Aρ′(x) so that ρ′(x) = A−1ρ(x)A for all x ∈ g. Thus ρ′ is
equivalent to ρ as desired.
Let V be a g-module, a a subspace of g, and U a subspace of V . We define aU
to be the subspace of V spanned by all elements of the form xu with x ∈ a and u ∈ U .
A submodule of V is a subspace U of V such that gU ⊂ U . Notice that both 0 and
V are submodules of V . A proper submodule of V is a submodule distinct from 0
and V .
A g-module V is irreducible if it contains no proper submodules. We say that V
is completely reducible if it is a direct sum of irreducible submodules. Finally, V is
said to be indecomposable if it cannot be written as a sum of two proper submodules.
We see that every irreducible g-module is indecomposable, although the converse need
not hold.
Example 2.1. Let V be a g-module and W a submodule of V . Then the quotient
space V/W forms a g-module via the action
(x, v +W ) 7→ xv +W.
To prove this action is well defined, let v +W = v′ +W . We must show that
xv +W = xv′ +W.
14
Now v +W = v′ +W implies v = v′ + w for some w ∈W . Thus
xv +W = x(v′ + w) +W = xv′ + xw +W = xv′ +W
since xw ∈W . The module properties in V/W follow easily from those in V .
Example 2.2. The vector space g can be made into a g-module via the action
(x, y) 7→ [x, y].
This is certainly linear in x and y. By the Jacobi identity, we also have
[[x, y], z] = [x, [y, z]]− [y, [x, z]],
showing that g is indeed a g-module. We call this module the adjoint module and
denote the action of x on the vector y by ad x · y. By the module properties, we have
ad [x, y] = ad x ad y − ad y ad x.
Notice that a submodule a of g is defined by the property [g, a] ⊂ a. Hence the sub-
modules of the adjoint module g are precisely the ideals of g.
2.2 A transition to finite-dimensional complex Lie algebras
With the exception of Propositions 1.15 and 1.16, every definition and result in
this paper has been applicable to Lie algebras with arbitrary base fields and dimensions.
We wish now to draw a line in the sand. From this point forward, we shall assume that
every Lie algebra g is finite-dimensional and that the base field k is the field C of complex
numbers. These types of Lie algebras have particularly nice representations, as we shall
witness in the following sections.
2.3 Representations of solvable Lie algebras
This section will produce some very useful results about solvable Lie algebras.
15
As one might expect, a 1-dimensional representation of g is a homomorphism
of Lie algebras ρ : g → [C].
Lemma 2.3. A linear map ρ : g → C is a 1-dimensional representation of g if and only
if ρ vanishes on g2.
Proof. Suppose ρ is a 1-dimensional representation of g and let x, y ∈ g. Then
ρ([x, y]) = [ρ(x), ρ(y)] = ρ(x)ρ(y)− ρ(y)ρ(x) = 0,
and so ρ vanishes on g2. Conversely, if ρ vanishes on g2, then
ρ([x, y]) = 0 = ρ(x)ρ(y)− ρ(y)ρ(x) = [ρ(x), ρ(y)]
so that ρ is a homomorphism from g into [C].
Theorem 2.4 (Lie’s theorem). Let g be a solvable Lie algebra and V a finite-dimensional
irreducible g-module. Then dimV = 1.
Proof. We proceed by induction on dim g. If dim g = 1, then g = Cx for some x ∈ g.
Since C is algebraically closed, we can find an eigenvector v of x in V . Then xv ∈ Cv,
and so Cv is a g-submodule of V . Because V is irreducible, this forces V = Cv and
dimV = 1.
Now suppose dim g > 1 and that the theorem is true for all solvable Lie algebras
of dimension dim g − 1. Notice that since g is solvable, we have g2 6= g, or else we
would have g(n) = g for all n ≥ 0. Let a be a subspace of g containing g2 such that
dim a = dim g− 1. Then
[a, g] ⊂ [g, g] = g2 ⊂ a,
and so a is an ideal of g. We may regard V as an a-module. Let W be an irreducible
a-submodule of V . By induction, we have dimW = 1. Then yw = λ(y)w for all y ∈ a,
where λ is the 1-dimensional representation induced by W . Define the set
U = {u ∈ V | yu = λ(y)u for all y ∈ a}.
16
We have 0 6= W ⊂ U ⊂ V . I claim that U is a g-submodule of V . To prove this, let
x ∈ g, y ∈ a, u ∈ U . Then
y(xu) = x(yu)− [x, y]u = x(λ(y)u)− λ([x, y])u = λ(y)xu− λ([x, y])u.
We shall show that λ([x, y]) = 0. Having proved this, it follows that xu ∈ U , and so U
is a g-submodule of V . Since V is irreducible, this forces U = V , and so yv = λ(y)v
for all y ∈ a, v ∈ V . Now because a has codimension 1, we have g = a ⊕ Cz for some
z ∈ g\a. Let v be an eigenvector of z in V . Then Cv is invariant under the actions of
both a and z, and so Cv is a g-submodule of V . Because V is irreducible, this forces
V = Cv and dimV = 1.
To prove that λ([x, y]) = 0 for all x ∈ g, y ∈ a, it suffices to consider the element
x = z such that g = a⊕Cz since λ vanishes on a2 by Lemma 2.3. Let u be any nonzero
vector in U . We define the vectors
v0 = u, v1 = zu, v2 = z(zu), . . .
Let V0 = 0 and Vi = 〈v0, v1, . . . , vi−1〉 for i > 0. We have V0 ⊂ V1 ⊂ V2 ⊂ · · · . I claim
that yvi = λ(y)vi (mod Vi) for all y ∈ a, i ≥ 0. We proceed by induction on i. For
i = 0, we have
yv0 = yu = λ(y)u = λ(y)v0.
Now suppose i > 0 and that the statement is true for i− 1. We have
yvi = y(zvi−1) = z(yvi−1)− [z, y]vi−1.
By induction, yvi−1 = λ(y)vi−1 + v′ and [z, y]vi−1 = λ([z, y])vi−1 + v′′ for some v′, v′′ ∈
Vi−1. Hence
yvi = z(λ(y)vi−1 + v′)− (λ([z, y])vi−1 + v′′)
= λ(y)vi + zv′ − λ([z, y])vi−1 − v′′
≡ λ(y)vi (mod Vi),
17
completing the proof of the claim. Now because V is finite-dimensional, there exists
a minimal r > 0 such that Vr = Vr+1 = Vr+2 = · · · . By construction, Vr is invariant
under the action of z, and given the above result, Vr is also invariant under the action
of a. Hence Vr is a g-submodule of V , and since V is irreducible, we have V = Vr.
Then by the above result, any element y ∈ a acts on V via an upper-triangular matrix
with λ(y) along the main diagonal. This is certainly true for the element [z, y], and so
trV ([z, y]) = rλ([z, y]). But we also have
trV ([z, y]) = trV (zy − yz) = trV (zy)− trV (yz) = 0.
Hence rλ([z, y]) = 0, and since C is an integral domain, this yields λ([z, y]) = 0.
Corollary 2.5. Let g be a solvable Lie algebra and V a finite-dimensional g-module.
Then a basis can be chosen for V with respect to which we obtain a matrix representation
ρ of g of the form
ρ(x) =
∗ ∗
·
·
·
0 ∗
for all x ∈ g.
Thus elements of g are represented by upper-triangular matrices.
Proof. It suffices to show there exists a chain of submodules
0 = V0 ⊂ V1 ⊂ · · · ⊂ Vn = V
such that dimVi = i for i = 0, . . . , n. We proceed by induction on dimV . The statement
is trivial for dimV = 1. So suppose dimV = n > 1 and that the statement is true for
all g-modules of dimension n − 1. Choose an irreducible g-submodule W of V . By
Theorem 2.4, we have dimW = 1. Then V/W is a g-module of dimension n − 1, and
18
by the inductive hypothesis, we may assume there exists a chain of submodules
0 = V 0 ⊂ V 1 ⊂ · · · ⊂ V n−1 = V/W
with the desired properties. For i = 1, . . . , n, let Vi denote the preimage of V i−1 in V
under the canonical homomorphism. We claim that the chain
0 = V0 ⊂ V1 ⊂ · · · ⊂ Vn = V
has the desired properties for V . Let v1, . . . , vi−1 be a basis of V i−1 where vj ∈ V j\V j−1
for j = 1, . . . , i − 1. Consider the elements w, v1, . . . , vi−1 ∈ Vi where 0 6= w ∈ W and
vj is in the preimage of vj . We shall prove these elements forms a basis of Vi. Let φ be
the canonical homomorphism V → V/W . If λ0w +∑i−1
j=1 λjvj = 0 where λj ∈ C, then
0 = φ(0) = φ
λ0w +i−1∑j=1
λjvj
= λ0φ(w) +i−1∑j=1
λjφ(vj) =i−1∑j=1
λjvj .
Because v1, . . . , vi−1 are linearly independent, we must have λ1 = · · · = λi−1 = 0. It
follows that λ0w = 0, and since w 6= 0, this yields λ0 = 0. Thus w, v1, . . . , vi−1 are
linearly independent, and because
dimVi = dimVi/W + dimW = (i− 1) + 1 = i,
we see that w, v1, . . . , vi−1 is indeed a basis of Vi. Finally, we show that Vi is a submodule
of V . Let x ∈ g, v ∈ Vi. Then φ(v) ∈ V i−1, and by induction, we have ρ(x)v+W ∈ V i−1.
Hence φ(ρ(x)v) ∈ V i−1, and so ρ(x)v ∈ Vi.
Corollary 2.6. Every solvable Lie algebra g has a chain of ideals
0 = a0 ⊂ a1 ⊂ · · · ⊂ an = g
where dim ai = i.
Proof. We apply the proof of Corollary 2.5 to the adjoint module g. The statement
follows immediately since each submodule of g is an ideal by Example 2.2.
19
2.4 Representations of nilpotent Lie algebras
Recall from Proposition 1.12 that every nilpotent Lie algebra is solvable. In this
section, we will apply our results about solvable Lie algebras to nilpotent Lie algebras
in order to gain some very useful representations.
We first introduce some standard results from linear algebra. Let V be a finite-
dimensional vector space and let T : V → V be a linear transformation with eigenvalues
λ1, . . . , λr. The generalized eigenspace of V with respect to λi is the set Vi of all
v ∈ V annihilated by some power of T−λi1. Let χ(t) = (t−λ1)m1(t−λ2)m2 · · · (t−λr)mr
be the characteristic polynomial of T . It follows from the Jordan canonical form (see
[DF04, §12.3]) that
(i) V = V1 ⊕ · · · ⊕ Vr.
(ii) Each Vi is invariant under the action of T .
(iii) dimVi = mi, and the characteristic polynomial of T |Vi is (t− λi)mi .
The relation between generalized eigenspaces and representations of nilpotent Lie alge-
bras will be exhibited in the next theorem. A useful proposition is needed first.
Proposition 2.7. Let g be a Lie algebra and V a g-module. For each y ∈ g, define the
linear map ρ(y) : V → V by ρ(y)v = yv. Let v ∈ V , x, y ∈ g, α, β ∈ C. Then
(ρ(y)− (α+ β)1)nxv =n∑
i=0
(n
i
)((ad y − β1)ix
) ((ρ(y)− α1)n−iv
).
Proof. We proceed by induction on n, the case where n = 0 being obvious. Suppose
the statement is true for n = r. For each i, let xi = (ad y − β1)ix ∈ g. Then
(ρ(y)− (α+ β)1)r+1xv = (ρ(y)− (α+ β)1)r∑
i=0
(r
i
)ρ(xi)(ρ(y)− α1)r−iv.
20
Now
(ρ(y)− (α+ β)1)ρ(xi) = ρ([y, xi]) + ρ(xi)ρ(y)− (α+ β)ρ(xi)
= ρ((ad y − β1)xi) + ρ(xi)(ρ(y)− α1)
= ρ(xi+i) + ρ(xi)(ρ(y)− α1).
It follows that
(ρ(y)− (α+ β)1)r+1xv
=r∑
i=0
(r
i
)ρ(xi+1)(ρ(y)− α1)r−iv +
r∑i=0
(r
i
)ρ(xi)(ρ(y)− α1)r−i+1v
=r+1∑i=0
(r
i− 1
)ρ(xi)(ρ(y)− α1)r−(i−1)v +
r+1∑i=0
(r
i
)ρ(xi)(ρ(y)− α1)r+1−iv(
since(r
−1
)= 0 and
(r
r + 1
)= 0)
=r+1∑i=0
(r + 1i
)((ad y − β1)ix
) (ρ(y)− α1)r+1−iv
).
(The last line follows from the identity(r+1
i
)=(
ri−1
)+(ri
).) This completes the induc-
tion.
Theorem 2.8. Let g be a nilpotent Lie algebra and V a finite-dimensional g-module.
For y ∈ g, let ρ(y) be the linear map described in Proposition 2.7. Then the generalized
eigenspaces of V associated with ρ(y) are all submodules of V .
Proof. Let Vi be a generalized eigenspace of V . If x ∈ g and v ∈ Vi, then
(ρ(y)− λi1)nxv =n∑
i=0
(n
i
)((ad y)ix
) ((ρ(y)− λi1)n−iv
)by Proposition 2.7 with α = λi, β = 0. Since g is nilpotent, we have (ad y)ix = 0 if i is
sufficiently large. And since v ∈ Vi, we have (ρ(y)− λi1)n−iv = 0 if n− i is sufficiently
large. Hence (ρ(y) − λi1)nxv = 0 if n is sufficiently large. It follows that xv ∈ Vi, and
so Vi is a submodule of V .
Corollary 2.9. Let g be a nilpotent Lie algebra and V a finite-dimensional indecom-
posable g-module. Then a basis can be chosen for V with respect to which we obtain a
21
representation ρ of g of the form
ρ(x) =
λ(x) ∗
·
·
·
0 λ(x)
for all x ∈ g.
Proof. By Corollary 2.5, we may choose a basis of V with respect to which ρ(x) is upper-
triangular. The generalized eigenspaces of V with respect to ρ(x) are all submodules
of V by Theorem 2.8, and V is their direct sum. Since V is indecomposable, only one
of the generalized eigenspaces can be nonzero. Thus ρ(x) has only one eigenvalue. Let
λ(x) be this eigenvalue. Then all of the diagonal entries of the upper-triangular matrix
ρ(x) will be equal to λ(x).
One easily checks that the map x 7→ λ(x) is a 1-dimensional representation of g.
Theorem 2.8 showed us that for any element y ∈ g, we obtain a decomposition of V
into a direct sum of g-submodules. We wish now to obtain a more general decomposition
of V , i.e., one that does not rely on the choice of y.
Let g be a nilpotent Lie algebra and V a finite-dimensional g-module. For any
1-dimensional representation λ of g, we define the set
Vλ = {v ∈ V | (∀x ∈ g)(∃N(x) ≥ 1) (ρ(x)− λ(x)1)N(x)v = 0}.
Theorem 2.10. V =⊕
λ Vλ, and each Vλ is a submodule of V .
Proof. We first decompose V into a direct sum of indecomposable submodules. Then
by Corollary 2.9, each such submodule will induce a 1-dimensional representation λ of
g. Let Wλ be the direct sum of all indecomposable submodules giving rise to λ. We
have
V =⊕
λ
Wλ.
22
I shall prove that Wλ = Vλ, and thus Wλ is independent of the way we obtained our
indecomposable components. Suppose Wλ 6= Vλ. Observe that Wλ ⊂ Vλ by Corollary
2.9. It follows from a standard result in linear algebra thatWλ +⊕µ 6=λ
Wµ
∩ Vλ = Wλ +
Vλ ∩⊕µ 6=λ
Wµ
.
We have Wλ +⊕
µ 6=λWµ = V , and thus Vλ = Wλ +(Vλ ∩
⊕µ 6=λWµ
). Since Vλ 6= Wλ,
there must exist a nonzero v ∈ Vλ∩⊕
µ 6=λWµ. We write v =∑
µ 6=λwµ where wµ ∈Wµ.
Now the vector space g over the infinite field C cannot be expressed as the union
of finitely-many proper subspaces. One easily checks that for each µ 6= λ, the set of
x ∈ g satisfying µ(x) = λ(x) is a proper subspace. Thus there exists an x ∈ g such that
µ(x) 6= λ(x) for all µ 6= λ. Now since v ∈ Vλ, there exists an Nλ ≥ 1 such that
(ρ(x)− λ(x)1)Nλv = 0.
And since wµ ∈Wµ, there exists an Nµ ≥ 1 such that (ρ(x)− µ(x)1)Nµwµ = 0. Hence
∏µ 6=λ
(ρ(x)− µ(x)1)Nµv = 0.
Because µ(x) 6= λ(x) for all µ 6= λ, we see that the polynomials
(t− λ(x))Nλ ,∏µ 6=λ
(t− µ(x))Nµ
are coprime. Thus there exist polynomials a(t), b(t) ∈ C[t] such that
a(t)(t− λ(x))Nλ + b(t)∏µ 6=λ
(t− µ(x))Nµ = 1.
It follows that
a(ρ(x))(ρ(x)− λ(x)1)Nλv + b(ρ(x))∏µ 6=λ
(ρ(x)− µ(x)1)Nµv = v.
The left-hand side of this equation is zero by the above results, and thus v = 0. This is
a contradiction, so we must have Wλ = Vλ. It follows that V =⊕
λ Vλ, and each Vλ is
a submodule of V .
23
If λ is a 1-dimensional representation of g and Vλ 6= 0, then we call λ a weight of
V , and we call Vλ the weight space of λ. Given Theorem 2.10, we call the decompo-
sition⊕
λ Vλ the weight space decomposition of V . Since each Vλ is the direct sum
of the indecomposable components giving rise to λ, it follows from Corollary 2.9 that a
basis can be chosen for Vλ with respect to which a representation ρ of g on Vλ has the
form
ρ(x) =
λ(x) ∗
·
·
·
0 λ(x)
for all x ∈ g.
This fact will prove extremely useful throughout the rest of this paper.
The final theorem of this chapter illustrates the connection between nilpotent Lie
algebras and nilpotent linear transformations.
Theorem 2.11 (Engel’s theorem). A Lie algebra g is nilpotent if and only if ad x :
g → g is nilpotent for all x ∈ g.
Proof. Suppose g is nilpotent. Then gn = 0 for some n. Let y ∈ g. We have
ad x · y ∈ g2, (ad x)2y ∈ g3, . . . , (ad x)i−1y ∈ gi, . . .
Thus (ad x)n−1y ∈ gn = 0. It follows that (ad x)n−1 = 0, and so ad x is a nilpotent
linear map.
Now suppose ad x is nilpotent for all x ∈ g. We wish to show that g is nilpotent.
Suppose g is not nilpotent. Choosing a maximal nilpotent subalgebra h of g, we may
regard g as an h-module. Then h is an h-submodule of g. Let m be an h-submodule
of g properly containing h such that m/h is irreducible. By Theorem 2.4, we have
dim m/h = 1. Then m/h = C(x+ h) for some x ∈ m\h, and m = h⊕ Cx.
24
The 1-dimensional representation induced by m/h must be the zero map. To see
why, suppose there existed a y ∈ h such that ad y · (x+ h) = λx+ h with λ 6= 0. Then
ad y · x = λx+ z for some z ∈ h. Thus (ad y)nx = λnx+ (ad y)nz 6= 0 for all n since the
second term is in h. Hence ad y would fail to be nilpotent, contrary to our assumption.
Because the 1-dimensional representation induced by m/h is the zero map, we
have [h,m] ⊂ h. It follows that
[m,m] = [h⊕ Cx, h⊕ Cx] ⊂ [h, h] + [h,Cx] + [Cx, h] + [Cx,Cx]
= [h, h] + [h,Cx] ⊂ h.
Thus m is a subalgebra of g and h is an ideal of m.
We shall prove that for each i ≥ 1, there exists a positive integer e(i) such that
me(i) ⊂ hi. We proceed by induction on i. The statement is true for i = 1 since m2 ⊂ h.
Now suppose the statement is true for i = r. Then
me(r)+1 =[me(r), h⊕ Cx
]⊂ [hr, h] +
[me(r),Cx
]= hr+1 + ad x ·me(r).
I claim that me(r)+j ⊂ hr+1 + (ad x)jme(r) for all j ≥ 1. We proceed by induction on j,
the case for j = 1 being proved. Assuming the statement is true for j = s, we have
me(r)+s+1 ⊂[hr+1 + (ad x)sme(r),m
]⊂ hr+1 +
[(ad x)sme(r), h⊕ Cx
]⊂ hr+1 + (ad x)s+1me(r)
since hr+1 is an ideal of m and (ad x)sme(r) ⊂ (ad x)shr ⊂ hr. Thus we have shown
me(r)+j ⊂ hr+1 + (ad x)jme(r) for all j. Since ad x is nilpotent, (ad x)j = 0 if j is
sufficiently large. For such j, we have me(r)+j ⊂ hr+1. Setting e(r + 1) = e(r) + j, we
obtain me(r+1) ⊂ hr+1, completing the induction.
Now because h is nilpotent, hi = 0 if i is sufficiently large. For such i, we have
me(i) = 0, showing that m is nilpotent. This contradicts the maximality of h, and so g
must be nilpotent.
25
Note: The traditional statement of Theorem 2.11 applies to arbitrary finite-
dimensional g-modules and linear maps. (See [Ser92, §V.2].) The proof, however,
involves embedding g into gln(C) for some n. This requires a result known as Ado’s
theorem, the proof of which lies beyond the scope of this paper. (For a proof, refer to
[Jac71, §VI.2].)
Corollary 2.12. A Lie algebra g is nilpotent if and only if g has a basis with respect
to which the adjoint representation of g has the form
ρ(x) =
0 ∗
·
·
·
0 0
for all x ∈ g.
Proof. Suppose g is nilpotent. Then g has a chain of ideals
g ⊃ g2 ⊃ g3 ⊃ · · · ⊃ gr = 0
for some r. We make this chain finer by choosing a series of subspaces between con-
secutive ideals, each having codimension 1 in its predecessor. Let a be a subspace such
that gi ⊃ a ⊃ gi+1. Then
[a, g] ⊂[gi, g
]= gi+1 ⊂ a,
showing that a is an ideal. Hence we obtain a chain of ideals
g = an ⊃ an−1 ⊃ · · · ⊃ a1 ⊃ a0 = 0
with dim ak = k and [g, ak] ⊂ ak−1. Choosing a basis of g adapted to this chain of
ideals, the representation ρ(x) will be of strictly upper-triangular form for all x ∈ g.
Conversely, suppose a basis can be chosen for g with respect to which ρ(x) is
strictly upper-triangular for all x ∈ g. Then ρ(x) is clearly nilpotent, whence ad x is
nilpotent for all x ∈ g. Thus g is nilpotent by Theorem 2.11.
Chapter 3
Cartan subalgebras
If g is a Lie algebra and h is a subalgebra of g, then g forms an h-module via the
adjoint action. If g is semisimple and h is chosen carefully, then the action of h induces
a decomposition of g that tells us essentially all we need to know about the structure of
g. In this chapter, we focus on finding the subalgebra h yielding such a decomposition.
3.1 Existence of Cartan subalgebras
Let g be a Lie algebra and h a subalgebra of g. We define the normalizer of h
to be the set
N(h) = {x ∈ g | [h, x] ∈ h for all h ∈ h}.
Proposition 3.1. N(h) is a subalgebra of g, h is an ideal of N(h), and N(h) is the
largest subalgebra of g containing h as an ideal.
Proof. For the first statement, let x, y ∈ N(h), h ∈ h. Then
[h, [x, y]] = [[y, h], x] + [[h, x], y] ∈ h
by the Jacobi identity, and so N(h) is a subalgebra of g. The second statement is obvious
from the definition of N(h). For the third statement, notice that if h is an ideal of m,
then [h,m] ⊂ h so that m ⊂ N(h).
A subalgebra h of g is a Cartan subalgebra if h is nilpotent and N(h) = h. We
wish to show that every Lie algebra contains a Cartan subalgebra.
27
Let x ∈ g and consider the map ad x : g → g. Define g0,x to be the generalized
eigenspace of ad x with eigenvalue 0, that is,
g0,x = {y ∈ g | (ad x)ny = 0 for some n ≥ 1}.
We call g0,x the null component of g with respect to x.
An element x ∈ g is regular if dim g0,x is as small as possible. Clearly, any Lie
algebra will contain regular elements.
Theorem 3.2. If x is a regular element of g, then the null component g0,x is a Cartan
subalgebra of g.
Proof. Let h = g0,x. We first show that h is a subalgebra of g. Let y, z ∈ h. Then
(ad x)n[y, z] =n∑
i=0
[(ad x)iy, (ad x)n−iz
]by Proposition 2.7 with V = g and α = β = 0. Since y ∈ h, we have (ad x)iy = 0 if i is
sufficiently large. And since z ∈ h, we have (ad x)n−iz = 0 if n− i is sufficiently large.
Hence (ad x)n[y, z] = 0 if n is sufficiently large. This shows that [y, z] ∈ h, and so h is
a subalgebra of g.
We next show that h is nilpotent. Set dim h = l and let b1, . . . , bl be a basis of h.
Let y = λ1bi+· · ·+λlbl ∈ h where λ1, . . . , λl ∈ C. Since h is a subalgebra, the linear map
ady : g → g restricts to ady : h → h. This induces the linear map ady : g/h → g/h. Let
χ(t) be the characteristic polynomial of ad y on g, χ1(t) the characteristic polynomial
of ad y on h, and χ2(t) the characteristic polynomial of ad y on g/h. We have
χ(t) = χ1(t)χ2(t).
Now χ(t) = det(t1− ad y) and y depends linearly on λ1, . . . , λl. Hence the coefficients
of χ(t) are polynomial functions of λ1, . . . , λl. The same applies to χ1(t) and χ2(t). Let
χ2(t) = d0 + d1t+ d2t2 + · · ·
28
where d0, d1, d2, . . . are polynomial functions of λ1, . . . , λl. Note that d0 is not the zero
polynomial, for in the special case y = x, the eigenvalues of ad y on g/h are nonzero
and d0 is their product. Let
χ1(t) = tm(c0 + c1t+ c2t
2 + · · ·)
where c0, c1, c2, . . . are polynomial functions of λ1, . . . , λl and c0 is not the zero polyno-
mial. Then m ≤ l = degχ1(t). It follows that
χ(t) = tm(c0d0 + a sum involving positive powers of t).
Now c0d0 is not the zero polynomial, and thus we may choose elements λ1, . . . , λl in
C such that c0d0 is nonzero. For such an element y we have dim g0,y = m. Since x is
a regular element with dim g0,x = l, we have l ≤ m. But we also have m ≤ l, and so
m = l. Thus tl divides χ1(t), and since dimχ1(t) = l, we have χ1(t) = tl. It follows by
the Cayley-Hamilton theorem that for any y ∈ h, (ad y)l : h → h is the zero map. Thus
ad y is nilpotent for all y ∈ h, and so h is nilpotent by Theorem 2.11.
Finally, we show that N(h) = h. Clearly h ⊂ N(h). So let z ∈ N(h). Then
[x, z] ∈ h, and hence (ad x)n[x, z] = 0 for some n ≥ 1. But this means (ad x)n+1z = 0,
and so z ∈ h. It follows that N(h) ⊂ h, and thus N(h) = h.
3.2 Derivations and automorphisms
A derivation of a Lie algebra g is a linear map D : g → g satisfying
D[x, y] = [Dx, y] + [x,Dy] for all x, y ∈ g.
Proposition 3.3. ad x is a derivation for all x ∈ g.
Proof. If y, z ∈ g, then
ad x · [y, z] = [x, [y, z]] = [[x, y], z] + [y, [x, z]] = [ad x · y, z] + [y, ad x · z]
by the Jacobi identity. Thus ad x is a derivation of g.
29
An automorphism of g is an isomorphism φ : g → g. The automorphisms of g
form a group Aut(g) under composition of maps.
Proposition 3.4. If D is a nilpotent derivation of g, then exp(D) is an automorphism
of g.
Proof. Since D is nilpotent, we have Dn = 0 for some n. It is easy to check that the
Leibniz rule holds for the linear map D, that is,
Dr[x, y] =r∑
i=0
(r
i
)[Dix,Dr−iy
]for all x, y ∈ g.
It follows that
[exp(D)x, exp(D)y] =
n−1∑i=0
Dix
i!,
n−1∑j=0
Djy
j!
=
2n−2∑k=0
k∑l=0
[Dlx
l!,Dk−ly
(k − l)!
]
=2n−2∑k=0
1k!
k∑l=0
(k
l
)[Dlx,Dk−ly
]=
2n−2∑k=0
Dk[x, y]k!
(by the Leibniz rule)
= exp(D)[x, y].
Hence exp(D) : g → g is a homomorphism. Likewise, exp(−D) : g → g is a homomor-
phism, and we see that exp(D) · exp(−D) = 1. Thus exp(D) is an automorphism.
An inner automorphism of g is an automorphism of the form exp(ad x) for
x ∈ g with ad x nilpotent. The inner automorphism group Inn(g) is the subgroup
of Aut(g) generated by all inner automorphisms.
Proposition 3.5. Inn(g) is a normal subgroup of Aut(g).
Proof. Let φ ∈ Aut(g). Every element of Inn(g) is of the form
exp(ad x1) · exp(ad x2) · · · · · exp(ad xr)
30
where x1, . . . , xr ∈ g and ad x1, . . . , ad xr are nilpotent. We have
exp(ad x1) · exp(ad x2) · · · · · exp(ad xr) = exp(ad x1 + ad x2 + · · ·+ ad xr)
= exp(ad (x1 + x2 + · · ·+ xr)).
Let x = x1 + x2 + · · ·+ xr. For all y ∈ g, we have
φ(ad x)φ−1y = φ[x, φ−1y] = [φx, y] = ad φx · y.
Thus φ(ad x)φ−1 = ad φx. By linearity, we have φ(exp(ad x))φ−1 = exp(ad φx). Hence
φ(exp(adx1)·exp(adx2)·· · ··exp(adxr))φ−1 = exp(adφx1)·exp(adφx2)·· · ··exp(adφxr).
Since φ is an automorphism, we see that ad φx1, . . . , ad φxr are nilpotent, and so
φ(exp(ad x1) · exp(ad x2) · · · · · exp(ad xr))φ−1 ∈ Inn(g).
Two subalgebras h, k of g are conjugate in g if there exists a φ ∈ Inn(g) such
that φ(h) = k. We shall prove that any two Cartan subalgebras of g are conjugate in g.
We first require some concepts from algebraic geometry.
3.3 Concepts from algebraic geometry
If h is a nilpotent subalgebra of a Lie algebra g, then g forms an h-module via the
restriction of the adjoint action. This action gives rise to the weight space decomposition
g =⊕
λ
gλ
as in Theorem 2.10 where
gλ = {x ∈ g | (∀h ∈ h)(∃n ≥ 1) (ad h− λ1)nx = 0}.
It follows from Corollary 2.12 that h ⊂ g0. Suppose our nilpotent subalgebra h is equal
to g0. Then there exist nonzero 1-dimensional representations λ1, . . . , λr of h such that
g = h⊕ gλ1 ⊕ · · · ⊕ gλr .
Thus if x ∈ g, then x = x0 + x1 + · · ·+ xr where x0 ∈ h and xi ∈ gλifor i = 1, . . . , r.
31
Proposition 3.6. ad xi is nilpotent for all i 6= 0.
Proof. Let µ : h → C be a weight of the h-module g and let y ∈ gµ. We have
(ad h− µ(h)1− λi(h)1)n[xi, y] =n∑
j=0
[(ad h− λi(h)1)jxi, (ad h− µ(h)1)n−jy
]for all h ∈ h by Proposition 2.7. Since xi ∈ gλi
, we have (ad h − λi(h)1)jxi = 0 if j is
sufficiently large. And since y ∈ gµ, we have (adh−µ(h)1)n−jy = 0 if n−j is sufficiently
large. Hence (ad h − µ(h)1 − λi(h)1)n[xi, y] = 0 if n is sufficiently large. This shows
that [xi, y] ∈ gλi+µ, and thus ad xi · gµ ⊂ gλi+µ. Now because λi 6= 0 and there are only
finitely-many µ : h → C such that gµ 6= 0, we see that (ad xi)N = 0 if N is sufficiently
large. Thus ad xi is nilpotent.
It follows immediately that exp(ad xi) ∈ Inn(g) for i 6= 0. We define the map
f : g → g by
f(x) = exp(ad x1) · exp(ad x2) · · · · · exp(ad xr) · x0.
Let {bij} be a basis of g where i = 1, . . . , r and, for a fixed i, the elements bij form
a basis of gλiwith respect to which elements of h are represented by upper-triangular
matrices as in Corollary 2.5. (Here λ0 = 0.)
Proposition 3.7. f : g → g is a polynomial function, that is,
f(∑
λijbij
)=∑
µijbij
where each µij is a polynomial in the λkl.
Proof. Each map ad xi : g → g is clearly linear. We also have
exp(ad xi) =N∑
k=0
(ad xi)k
k!
for some N ≥ 1 since ad xi is nilpotent. Thus exp(ad xi) : g → g is a polynomial
function. The map f is a composition of the linear map x 7→ x0 with the polynomial
functions exp(ad xi) and is thus a polynomial function.
32
We now write µij = fij(λkl). We define the Jacobian matrix
J(f) = (∂fij/∂λkl)
and the Jacobian determinant det J(f) of f . Clearly det J(f) is an element of the
polynomial ring C[λkl].
Proposition 3.8. det J(f) is not the zero polynomial.
Proof. We shall show that det J(f) 6= 0 when evaluated at a carefully chosen element
of h. Let h ∈ h and consider (∂fij/∂λkl)h. Suppose k 6= 0. We have
(∂f/∂λkl)h = limt→∞
f(h+ tbkl)− f(h)t
= limt→∞
exp(ad tbkl)h− h
t
= limt→∞
h+ t[bkl, h] + · · · − h
t
= [bkl, h] = −[h, bkl]
≡ −λk(h)bkl (mod 〈bk1, . . . , bkl−1〉).
Now suppose k = 0. We have
(∂f/∂λ0l)h = limt→∞
f(h+ tb0l)− f(h)t
= limt→∞
h+ tb0l − h
t= b0l.
Thus J(f) is a block matrix of the form
J0 0
J1
·
·
·
0 Jr
33
where J0 is the identity matrix and Ji is of the form
−λi(h) ∗
·
·
·
0 −λi(h)
for i = 1, . . . , r. It follows that (det J(f))h = ±
∏ri=1 λi(h)di where di = dim gλi
. Now
the vector space h over the infinite field C cannot be expressed as the union of finitely-
many proper subspaces. For each i, the set of h ∈ h satisfying λi(h) = 0 is a proper
subspace of h since the linear maps λi are all nonzero. Thus there exists an h ∈ h such
that λi(h) 6= 0 for all i. For such an h, we have (det J(f))h 6= 0, showing that detJ(f)
is not the zero polynomial.
Proposition 3.9. The polynomial functions fij are algebraically independent.
Proof. Suppose there exists a nonzero polynomial F (xij) ∈ C(xij) such that F (fij) = 0.
We choose an F with minimal total degree in the variables xij . For each index k, l, we
have∂
∂λklF (fij) = 0
so that ∑i,j
∂F
∂fij
∂fij
∂λkl= 0.
Let v be the vector (∂F/∂fij). Then vJ(f) = (0, . . . , 0). Since det J(f) = 0, we must
have v = (0, . . . , 0), and hence ∂F/∂fij = 0 for all fij . Notice that ∂F/∂xij is a
polynomial in C[xij ] of smaller total degree than F . By our choice of F , ∂F/∂xij must
be the zero polynomial, that is, F does not involve the indeterminate xij . This is true
for all xij , and so F must be a constant polynomial. Since F (fij) = 0, this constant
must be zero. This contradicts the fact that F is not the zero polynomial, and so the
fij must be algebraically independent.
34
Let B = C[fij ] be the polynomial ring in the fij and A = C[λij ] the polynomial
ring in the λij . Consider the homomorphism θ : B → A uniquely determined by
θ(fij) = fij(λkl) ∈ A.
Proposition 3.10. The homomorphism θ : B → A is injective.
Proof. Let F ∈ ker θ, that is, θ(F ) = 0. Then F (fij) = 0 as a function of the λkl. Since
the fij are algebraically independent, this implies F = 0. Thus ker θ = 0, and so θ is
injective.
It follows that θ is an embedding of B into A. The following is a standard result
from algebraic geometry.
Proposition 3.11. Let A and B be integral domains such that B ⊂ A, A,B have
common identity element 1, and A is finitely-generated over B. If p is a nonzero element
of A, then there exists a nonzero element q of B such that any homomorphism φ : B → C
with φ(q) 6= 0 extends to a homomorphism ψ : A→ C with ψ(p) 6= 0.
Proof. We refer the reader to [Car05, Proposition 3.10].
We wish to apply this result to our earlier situation. Let d = dim g. We have the
polynomial function
f : Cd → Cd
(λij) 7→ (fij(λkl)).
Let V = Cd. For each polynomial p ∈ C[xij ], we define the set
Vp = {v ∈ V | p(v) 6= 0}.
Corollary 3.12. For each nonzero polynomial p ∈ C[xij ], there exists a nonzero poly-
nomial q ∈ C[xij ] such that Vq ⊂ f(Vp).
35
Proof. Consider the integral domains B ⊂ A described earlier, that is, B = C[fij ] and
A = C[λij ]. We may regard p as a polynomial in A. By Proposition 3.11, we know
there exists a nonzero q ∈ B such that any homomorphism φ : B → C with φ(q) 6= 0
extends to a homomorphism ψ : A→ C with ψ(p) 6= 0. Now each v ∈ Vq induces such a
homomorphism φ via the relation φ(fij) = vij . Let ψ be an extension of φ as described
above and define the vector w ∈ Cd by wij = ψ(λij). Clearly, w ∈ Vp. By the way we
embedded B into A, we have
vij = φ(fij) = ψ(fij(λkl)) = fij(ψ(λkl)) = fij(w),
and so v = f(w). Hence v ∈ f(Vp) and Vq ⊂ f(Vp) as desired.
3.4 Conjugacy of Cartan subalgebras
We saw in Theorem 3.2 that the null component of any regular element of g is a
Cartan subalgebra. In this section we prove the converse: that every Cartan subalgebra
is the null component of some regular element of g. Finally, we prove that the null
components of any two regular elements are conjugate, and thus any Cartan subalgebra
of g is unique up to isomorphism.
Proposition 3.13. If h is a Cartan subalgebra of g, then there exists a regular element
x ∈ g such that h = g0,x.
Proof. Since h is nilpotent, we may regard g as an h-module and decompose g into
weight spaces with respect to h as in Theorem 2.10. It is clear from Corollary 2.12 that
h is contained in the zero weight space g0. Suppose h 6= g0. Let m/h be an irreducible h-
submodule of g0/h. By Theorem 2.4, we have dim m/h = 1. As in the proof of Theorem
2.11, the 1-dimensional representation induced by m/h must be the zero map. Hence
[h,m] ⊂ h, and so m ⊂ N(h). This contradicts h = N(h), so we must have h = g0. Let
g = h⊕ gλ1 ⊕ · · · ⊕ gλr λ1, . . . , λr 6= 0
36
be the weight space decomposition of g with respect to h. For x ∈ g, we write x =
x0 + x1 + · · ·+ xr where x0 ∈ h and xi ∈ gλifor i = 1, . . . , r. We define the polynomial
function f : g → g by
f(x) = exp(ad x1) · exp(ad x2) · · · · · exp(ad xr) · x0
as in the previous section. We also define p : g → C by
p(x) = λ1(x0)λ2(x0) · · ·λr(x0).
Then p is a polynomial function on g. Clearly p is not the zero polynomial since we
can find an element x0 ∈ h such that λi(x0) 6= 0 for all i = 1, . . . , r as in the proof of
Proposition 3.8. By Corollary 3.12, there exists a nonzero polynomial function q : g → C
such that gq ⊂ f(gp).
Now consider the set R of regular elements of g. Let y ∈ g and let
χ(y) = det(t1− ad y) = tn + µ1(y)tn−1 + · · ·+ µn(y)
be the characteristic polynomial of ad y on g. Then µ1, . . . , µn are polynomial functions
on g. There exists a unique integer k such that µn−k is not the zero polynomial but
µn−k+1, . . . , µn are identically zero. The generalized eigenspace of ad y with eigenvalue
0 has dimension k if µn−k = 0 and dimension greater than k if µn−k(y) = 0. Hence
y ∈ R if and only if µn−k(y) 6= 0.
Now q and µn−k are nonzero polynomials. Thus qµn−k is a nonzero polynomial,
and so there exists a y ∈ g such that (qµn−k)(y) 6= 0. Then q(y) 6= 0 and µn−k(y) 6= 0
so that y ∈ gq ∩R. Since gq ⊂ f(gp), there exists an x ∈ gp such that y = f(x), that is
y = exp(ad x1) · exp(ad x2) · · · · · exp(ad xr) · x0.
Thus y and x0 are conjugate in g. Since y is regular, x0 must also be regular.
I claim that h = g0,x0 . We know that x0 ∈ h and that h is nilpotent. Hence
h ⊂ g0,x0 by Corollary 2.12. Suppose h 6= g0,x0 and choose z ∈ g0,x0\h. We write
37
z = z0 + z1 + · · ·+ zr where z0 ∈ h and zi ∈ gλifor i = 1, . . . , r. Because z ∈ g0,x0 , we
have (ad x0)nz = 0 for some n ≥ 1, that is,
(ad x0)nz0 + (ad x0)nz1 + · · ·+ (ad x0)nzr = 0.
For any i = 1, . . . , r, we have
(ad x0)nzi = −(ad x0)nz0 − · · · − (ad x0)nzi−1 − (ad x0)nzi+1 − · · · − (ad x0)nzr.
Each weight space is invariant under the action of ad x0, so the left-hand side is in gλi
and right-hand side is in its complement. This implies (ad x0)nzi = 0, and so zi ∈ g0,x0
for all i. We also know that z /∈ h. Thus zi 6= 0 for some i ≥ 1. Since zi ∈ gλi, we have
(ad x0 − λi(x0)1)mzi = 0 for some m ≥ 1. Hence zi is in the generalized eigenspace of
ad x0 with eigenvalue λi(x0). Because zi ∈ g0,x0 , we must have λi(x0) = 0. But x ∈ gp,
and so
λ1(x0)λ2(x0) · · ·λr(x0) 6= 0.
In particular, this implies λi(x0) 6= 0, a contradiction. It follows that h = g0,x0 .
Theorem 3.14. Any two Cartan subalgebras of g are conjugate.
Proof. Let h, h′ be Cartan subalgebras of g. We regard g as an h-module and decompose
g into weight spaces with respect to h. We know from the proof of Proposition 3.13 that
h = g0. Let
g = h⊕ gλ1 ⊕ · · · ⊕ gλr λ1, . . . , λr 6= 0
be the weight space decomposition of g with respect to h. For x ∈ g, we write x =
x0 + x1 + · · ·+ xr where x0 ∈ h and xi ∈ gλifor i = 1, . . . , r. For each x0 ∈ h, we have
h ⊂ g0,x0 . By dimension considerations, we see that x0 is regular if and only if h = g0,x0 .
We know from the proof of Proposition 3.13 that the condition
λ1(x0)λ2(x0) · · ·λr(x0) 6= 0
38
implies h = g0,x0 . Now suppose h = g0,x0 but that λi(x0) = 0 for some i ≥ 1. Choosing
a nonzero y ∈ gλi, we have (ad x0 − λi(x0)1)ny = 0 for some n ≥ 1. Thus y ∈ g0,
contradicting the fact that λi 6= 0. Hence h = g0,x0 if and only if
λ1(x0)λ2(x0) · · ·λr(x0) 6= 0.
It follows that this condition is also equivalent to x0 being regular.
Consider the polynomial function f : g → g defined by
f(x) = exp(ad x1) · exp(ad x2) · · · · · exp(ad xr) · x0.
Define the polynomial function p : g → C by
p(x) = λ1(x0)λ2(x0) · · ·λr(x0).
Note that p is not the zero polynomial since p(x0) 6= 0 when x0 is regular. By Corollary
3.12, there exists a nonzero polynomial q : g → C such that gq ⊂ f(gp).
We now consider the second Cartan subalgebra h′ and define the corresponding
functions f ′ : g → g and p′ : g → C. We know there exists a nonzero polynomial
function q′ : g → C such that gq′ ⊂ f ′(gp′).
Since q and q′ are nonzero polynomials, qq′ is a nonzero polynomial, and so there
exists a z ∈ g such that (qq′) (z) 6= 0. Then q(z) 6= 0 and q′(z) 6= 0 so that z ∈ gq ∩ gq′ .
It follows that z ∈ f(gp) ∩ f ′(gp′). Thus there exists an x ∈ g such that z = f(x) and
p(x) 6= 0. Similarly, there exists an x′ ∈ g such that z = f ′ (x′) and p′ (x′) = 0. We have
z = exp(ad x1) · exp(ad x2) · · · · · exp(ad xr) · x0.
Hence z and x0 are conjugate in g. Since p(x0) 6= 0, we see that x0 is regular. Likewise,
z is conjugate to x′0 and x′0 is regular. Thus we have found regular elements x0 ∈ h and
x′0 ∈ h′ such that x0 and x′0 are conjugate in g. Now h = g0,x0 and h′ = g0,x′0since x0
and x′0 are regular. Any automorphism mapping x0 to x′0 will transform h into h′. It
follows that h and h′ are conjugate in g.
The dimension of the Cartan subalgebras of g is called the rank of g.
Chapter 4
The Cartan decomposition
4.1 Root spaces
We saw in the last chapter that any two Cartan subalgebras of a Lie algebra g
are conjugate in g. Thus any weight space decomposition of g with respect to a Cartan
subalgebra h will be unique up to isomorphism. Let⊕
λ gλ be such a decomposition.
By the proof of Proposition 3.13, we have g0 = h. Thus
g = h⊕ gλ1 ⊕ · · · ⊕ gλr λ1, . . . , λr 6= 0.
A 1-dimensional representation λ of h is called a root of g with respect to h if λ 6= 0
and gλ 6= 0. We denote the set of all roots of g with respect to h by Φ. Thus
g = h⊕⊕α∈Φ
gα.
We call this decomposition the Cartan decomposition of g with respect to h. Each
gα is called the root space of α.
Proposition 4.1. If λ and µ are 1-dimensional representations of h, then
[gλ, gµ] ⊂ gλ+µ.
Proof. We know that [gλ, gµ] is generated by elements of the form [y, z] with y ∈ gλ and
z ∈ gµ. Let x ∈ h. By Proposition 2.7, we have
(ad x− λ(x)1− µ(x)1)n[y, z] =n∑
i=0
(n
i
)[(ad x− λ(x)1)iy, (ad x− µ(x)1)n−iz
].
40
Since y ∈ gλ, we have (ad x − λ(x)1)iy = 0 if i is sufficiently large. And since z ∈ gµ,
we have (ad x− µ(x)1)n−iz = 0 if n− i is sufficiently large. Hence
(ad x− λ(x)1− µ(x)1)n[y, z] = 0
if n is sufficiently large. It follows that [y, z] ∈ gλ+µ, and so [gλ, gµ] ⊂ gλ+µ.
Corollary 4.2. Let α, β ∈ Φ be roots of g with respect to h. Then
[gα, gβ] ⊂ gα+β if α+ β ∈ Φ
[gα, gβ] ⊂ h if β = −α
[gα, gβ] = 0 if α+ β 6= 0 and α+ β /∈ Φ.
Proof. This is immediate from Proposition 4.1 and the fact that h = g0.
Proposition 4.3. Let α ∈ Φ. Given any β ∈ Φ, there exists a number r ∈ Q, depending
on α and β, such that β = rα on the subspace [gα, g−α] of h.
Proof. If −α is not a weight of g with respect to h, then g−α = 0, and the proof is
trivial. So assume −α is a weight. Then since α 6= 0, we must have −α ∈ Φ. For i ∈ Z,
we consider the function iα + β : h → C. Since Φ is finite, there exist integers p and q
with p ≥ 0 and q ≥ 0 such that
−pα+ β, . . . , β, . . . , qα+ β
are all in Φ but −(p+1)α+β and (q+1)α+β are not in Φ. If either −(p+1)α+β = 0
or (q + 1)α + β = 0, then the result is obvious. So assume −(p + 1)α + β 6= 0 and
(q + 1)α+ β 6= 0. Let m be the subspace of g given by
m = g−pα+β ⊕ · · · ⊕ gqα+β.
Now [gα, g−α] is generated by elements of the form x = [y, z] with y ∈ gα and z ∈ g−α.
We have ad y · gqα+β ⊂ g(q+1)α+β by Proposition 4.1. Because (q + 1)α + β 6= 0 and
41
(q + 1)α + β /∈ Φ, we must have g(q+1)α+β = 0. Thus ad y · m ⊂ m. By a similar
argument, we have ad z ·m ⊂ m, and so
ad x ·m = (ad y ad z − ad z ad y)m ⊂ m.
We calculate the trace trm(ad x). Since x ∈ h, each weight space giα+β is invariant
under ad x. Thus
trm(ad x) =q∑
i=−p
trgiα+β(ad x).
Now ad x acts on giα+β via a matrix of the form
(iα+ β)(x) ∗
·
·
·
0 (iα+ β)(x)
.
Thus trgiα+β(ad x) = dim giα+β(iα+ β)(x). It follows that
trm(ad x) =q∑
i=−p
dim giα+β(iα+ β)(x)
=
q∑i=−p
idim giα+β
α(x) +
q∑i=−p
dim giα+β
β(x).
But we also have
trm(ad x) = trm(ad y ad z − ad z ad y) = trm(ad y ad z)− trm(ad z ad y) = 0.
Hence q∑i=−p
idim giα+β
α(x) +
q∑i=−p
dim giα+β
β(x) = 0.
We know that dim giα+β > 0 for all −p ≤ i ≤ q. Thus
β(x) = −
(∑qi=−p idim giα+β
)(∑q
i=−p dim giα+β
) α(x).
Hence β(x) = rα(x) for some r ∈ Q independent of x. It follows that β = rα on
[gα, g−α].
42
4.2 The Killing form
In order to further understand the Cartan decomposition of g, we define a map
g× g → C
(x, y) 7→ 〈x, y〉
by
〈x, y〉 = tr(ad x ad y).
One easily checks that this map is bilinear. We call this bilinear form the Killing form
of g.
Proposition 4.4.
(i) The Killing form is symmetric, i.e., 〈x, y〉 = 〈y, x〉 for all x, y ∈ g.
(ii) The Killing form is invariant, i.e.,
〈[x, y], z〉 = 〈x, [y, z]〉 for all x, y, z ∈ g.
Proof. The first statement follows form the fact that tr(AB) = tr(BA) for all square
matrices A,B. To prove the second statement, we calculate
〈[x, y], z〉 = tr(ad [x, y] ad z) = tr((ad x ad y − ad y ad x) ad z)
= tr(ad x ad y ad z)− tr(ad y ad x ad z)
= tr(ad x ad y ad z)− tr(ad x ad z ad y)
= tr((ad x (ad y ad z − ad z ad y)) = tr(ad x ad [y, z]) = 〈x, [y, z]〉.
Proposition 4.5. Let a be an ideal of g and let x, y ∈ a. Then
〈x, y〉a = 〈x, y〉g.
Hence the killing form of g restricted to a is the Killing form of a.
43
Proof. We choose a basis of a and extend it to a basis of g. With respect to this basis,
ad x : g → g is represented by a matrix of the formA1 A2
0 0
since x ∈ a. Similarly, ad y : g → g is represented by a matrix of the formB1 B2
0 0
.
Thus ad x ad y : g → g is represented by the matrixA1B1 A1B2
0 0
.
Hence tra(ad x ad y) = tr(A1B1) = trg(ad x ad y), and so 〈x, y〉a = 〈x, y〉g.
For any subspace m of g, we define the set
m⊥ = {x ∈ g | 〈x, y〉 = 0 for all y ∈ m}.
One easily checks that m⊥ is a subspace of g.
Proposition 4.6. If a is an ideal of g, then a⊥ is an ideal of g.
Proof. The subspace [a⊥, g] is generated by elements of the form [x, y] with x ∈ a⊥ and
y ∈ g. For all z ∈ a, we have
〈[x, y], z〉 = 〈x, [y, z]〉 = 0
since [y, z] ∈ a. This shows that [x, y] ∈ a⊥. It follows that [a⊥, g] ⊂ a⊥, and so a⊥ is
an ideal of g.
In particular, we see that g⊥ is an ideal of g. The Killing form of g is said to be
nondegenerate if g⊥ = 0. This is equivalent to the condition that if 〈x, y〉 = 0 for all
y ∈ g, then x = 0. The Killing form of g is identically zero if g⊥ = g. This means
that 〈x, y〉 = 0 for all x, y ∈ g.
44
Proposition 4.7. Let g be a Lie algebra such that g 6= 0 and g2 = g. Let h be a Cartan
subalgebra of g. Then there exists an x ∈ h such that 〈x, x〉 6= 0.
Proof. Let g = ⊕gλ be the Cartan decomposition of g. Then
g2 = [g, g] =
[⊕λ
gλ,⊕
λ
gλ
]=∑λ,µ
[gλ, gµ].
We have [gλ, gµ] ⊂ gλ+µ by Proposition 4.1. Thus [gλ, g−λ] ⊂ g0, while [gλ, gµ] is
contained in the complement of g0 in g if µ 6= −λ. Since g = g2, we must have
g0 =∑
λ
[gλ, g−λ]
summed over all weights λ such that −λ is also a weight. We also have g0 = h by the
proof of Proposition 3.13. Thus
h = [h, h] +∑α
[gα, g−α]
summed over all roots α such that −α is also a root. Note that g is not nilpotent since
g2 = g 6= 0. But we know that h is nilpotent, and so h 6= g. Thus there exists at least
one root β ∈ Φ. Now β is a 1-dimensional representation of h, and so β vanishes on [h, h]
by Lemma 2.3. But β does not vanish on h since β 6= 0. Using the above decomposition
of h, we see there exists some root α ∈ Φ such that −α ∈ Φ and β does not vanish on
[gα, g−α]. Choose an x ∈ [gα, g−α] such that β(x) 6= 0. Then
〈x, x〉 = tr(ad x ad x) =∑
λ
dim gλ(λ(x))2
since ad x is represented on gλ by a matrix of the form
λ(x) ∗
·
·
·
0 λ(x)
.
45
For each λ, there exists an rλ,α ∈ Q such that λ(x) = rλ,αα(x) by Proposition 4.3. Thus
〈x, x〉 =
(∑λ
dim gλr2λ,α
)α(x)2.
Now β(x) = rβ,αα(x) and β(x) 6= 0. Thus rβ,α 6= 0 and α(x) 6= 0. It follows that
〈x, x〉 6= 0.
Theorem 4.8. If the Killing form of g is identically zero, then g is solvable.
Proof. We proceed by induction on dim g. If dim g = 1, then g is clearly solvable. So
assume dim g > 1. By the contrapositive of Proposition 4.7, we see that g 6= g2. Now
g2 is an ideal of g, so the Killing form of g2 is the restriction of the Killing form of g by
Proposition 4.5. Hence the Killing form of g2 is identically zero. It follows by induction
that g2 is solvable. We also have(g/g2
)2 = 0, and so g/g2 is solvable. Thus g is solvable
by Proposition 1.14.
Theorem 4.9 (Cartan’s criterion). A Lie algebra g is semisimple if and only if the
Killing form of g is nondegenerate.
Proof. We shall prove the contrapositive of the theorem. Suppose the Killing form of g
is degenerate. Then g⊥ 6= 0. We know that g⊥ is an ideal by Proposition 4.6. Thus the
Killing form of g⊥ is identically zero by Proposition 4.5. This implies g⊥ is solvable by
Theorem 4.8. Thus g has a nonzero solvable ideal, and so g is not semisimple.
Now suppose g is not semisimple. Then the solvable radical r of g is nonzero.
Consider the chain of subspaces
r = r(0) ⊃ r(1) ⊃ r(2) ⊃ · · · ⊃ r(k−1) ⊃ r(k) = 0.
Each subspace r(i) is an ideal of g since the product of two ideals is an ideal. Let
a = r(k−1). Then a is a nonzero ideal such that a2 = 0. We choose a basis of a and
extend it to a basis of g. Let x ∈ a, y ∈ g. With respect to our chosen basis, ad x is
46
represented by a matrix of the form 0 A
0 0
since a2 = 0 and a is an ideal of g, and ad y is represented by a matrix of the formB1 B2
0 B3
.
Thus ad x ad y is represented by the matrix 0 AB3
0 0
.
Hence 〈x, y〉 = tr(ad x ad y) = 0. This holds for all x ∈ a, y ∈ g, and so a ⊂ g⊥. Thus
g⊥ 6= 0, and so the Killing form of g is degenerate.
We now define the direct sum of Lie algebras g1, g2 by taking the vector space
g1 ⊕ g2 and defining the multiplication component-wise, that is,
[(x1, x2), (y1, y2)] = ([x1, y1], [x2, y2]).
We also define the subspaces a1 = {(x1, 0) | x1 ∈ g1} and a2 = {(0, x2) | x2 ∈ g2}.
One easily checks that a1 and a2 are ideals of g1 ⊕ g2, that a1 ∩ a2 = 0, and that
a1 + a2 = g1 ⊕ g2. Furthermore, we have g1∼= a1 and g2
∼= a2 via the projections
(x1, 0) 7→ x1 and (0, x2) 7→ x2, respectively.
Conversely, let g be a Lie algebra containing two ideals a1, a2 such that a1∩a2 = 0
and a1 + a2 = g. We shall show that the Lie algebra a1 ⊕ a2 is isomorphic to g under
the map
φ : a1 ⊕ a2 → g
(x1, x2) 7→ x1 + x2.
Clearly, φ is an isomorphism of vector spaces. We have [a1, a2] ⊂ a1 ∩ a2 = 0 so that
[φ(x1, x2), φ(y1, y2)] = [x2 + x2, y1 + y2] = [x1, y1] + [x2, y2]
= φ([x1, y1], [x2, y2]) = φ[(x1, x2), (y1, y2)].
47
Hence φ preserves Lie multiplication. This shows that if a Lie algebra has complemen-
tary ideals a1, a2, then it is isomorphic to a1 ⊕ a2.
We may generalize these results to include the direct sum g1 ⊕ g2 ⊕ · · · ⊕ gn of
more than two Lie algebras.
Theorem 4.10. A Lie algebra g is semisimple if and only if g is isomorphic to a direct
sum of non-trivial simple Lie algebras.
Proof. Suppose g is semisimple. If g is simple, then g must be non-trivial since the
trivial simple Lie algebra is not semisimple. So suppose g is not simple. Let a be a
maximal ideal of g. Then a 6= 0 and a 6= g. Consider the subspace a⊥ of g. This is
an ideal of g by Proposition 4.6. The Killing form of g is nondegenerate by Theorem
4.9. Thus an element x ∈ g lies in a⊥ if and only if the coordinates of x with respect to
a basis of g satisfy dim a homogeneous linear equations that are linearly independent.
Hence
dim a⊥ = dim g− dim a.
Now consider the subspace a ∩ a⊥. This is an ideal of g by Proposition 1.5. Thus the
Killing form of a∩a⊥ is the restriction of the Killing form of g by Proposition 4.5. Hence
the Killing form of a ∩ a⊥ is identically zero, and so a ∩ a⊥ is solvable by Theorem 4.8.
Since g is semisimple, we have a ∩ a⊥ = 0. Thus
dim(a + a⊥) = dim a + dim a⊥ − dim(a ∩ a⊥)
= dim a + dim a⊥ = dim g.
Hence a + a⊥ = g, and so g is the direct sum of the ideals a and a⊥. It follows that
g ∼= a⊕ a⊥.
We shall show that a is a simple Lie algebra. Let b be an ideal of a. We have
[b, g] ⊂ [b, a] + [b, a⊥] = [b, a] ⊂ b
since [b, a⊥] ⊂ [a, a⊥] ⊂ a ∩ a⊥ = 0. Thus b is an ideal of g contained in a. Since a is a
maximal ideal of g, we must have b = 0 or b = a. Thus a is simple.
48
We now show that a⊥ is semisimple. Let b be a solvable ideal of a⊥. Then
[b, g] ⊂ [b, a] + [b, a⊥] = [b, a⊥] ⊂ b
since [b, a] ⊂ [a⊥, a] ⊂ a ∩ a⊥ = 0. Thus b is an ideal of g. Since g is semisimple and b
is solvable, we must have b = 0. Thus a⊥ is semisimple.
Because a 6= 0, we see that dim a⊥ < dim g. By induction, we may assume a⊥ is
a direct sum of non-trivial simple Lie algebras. Since g = a ⊕ a⊥ and a is simple and
non-trivial, g is also a direct sum of non-trivial simple Lie algebras.
Conversely, suppose
g = g1 ⊕ g2 ⊕ · · · ⊕ gr
where each gi is a non-trivial simple Lie algebra. Then each gi is semisimple by Propo-
sition 1.16, and thus has a nondegenerate Killing form by Theorem 4.9. Each gi is also
an ideal of g. Suppose i 6= j and let xi ∈ gi, xj ∈ gj . Then
ad xi ad xj · y ∈ gi ∩ gj = 0 for all y ∈ g,
and so 〈xi, xj〉 = tr(ad xi ad xj) = 0. Now let x = x1 + x2 + · · ·+ xr ∈ g⊥ with xi ∈ gi.
If yi ∈ gi, then by the above result, we have
0 = 〈x, yi〉 =r∑
j=1
〈xj , yi〉 = 〈xi, yi〉.
Since this holds for all yi ∈ gi, we must have xi = 0. This holds for all i, and so x = 0.
Thus the Killing form of g is nondegenerate, and so g is semisimple.
4.3 The Cartan decomposition of a semisimple Lie algebra
In this section, without further comment, we shall assume g is a semisimple
Lie algebra. The Cartan decomposition of a semisimple Lie algebra gives us more
information than that of the general case. This information will be useful in studying
the simple case, which is the goal of this paper.
49
As usual, we denote the Cartan decomposition of g with respect to h by g =⊕
λ gλ
where g0 = h.
Proposition 4.11. If µ 6= −λ, then gλ and gµ are orthogonal with respect to the Killing
form.
Proof. Let x ∈ gλ, y ∈ gµ. For every weight space gν , we have
ad x ad y · gν ⊂ gλ+µ+ν
by Proposition 4.1. We choose a basis of g adapted to the Cartan decomposition. With
respect to such a basis, ad x ad y is represented by a block matrix of the form
0 ∗
·
·
·
∗ 0
since λ+ µ+ ν 6= ν. It follows that 〈x, y〉 = tr(ad x ad y) = 0, and so gλ is orthogonal
to gµ.
Proposition 4.12. If α is a root of g with respect to h, then −α is also a root.
Proof. Recall that α is a root if α 6= 0 and gα 6= 0. Suppose −α is not a root. Then
since −α 6= 0, we must have g−α = 0. By Proposition 4.11, this implies that gα is
orthogonal to all gλ, and thus gα ⊂ g⊥. But g is semisimple, and so g⊥ = 0 by Theorem
4.9. Thus gα = 0, contradicting the fact that α is a root.
Proposition 4.13. The Killing form of g remains nondegenerate on restriction to h.
Proof. Let x ∈ h and suppose 〈x, y〉 = 0 for all y ∈ h. We also have 〈x, y〉 = 0 for all
y ∈ gα where α ∈ Φ by Proposition 4.11. Thus 〈x, y〉 = 0 for all y ∈ g, and so x ∈ g⊥.
But g⊥ = 0 since g is semisimple, and so x = 0.
50
Note that the Killing form of g restricted to h is not the same as the Killing form
of h, because h is solvable and thus not semisimple.
Theorem 4.14. The Cartan subalgebras of a semisimple Lie algebra are abelian.
Proof. Let h be a Cartan subalgebra of g. For all x ∈ [h, h], y ∈ h, we have
〈x, y〉 = tr(ad x ad y) =∑
λ
dim gλ λ(x)λ(y)
since ad x ad y is represented on gλ by a matrix of the form
λ(x)λ(y) ∗
·
·
·
0 λ(x)λ(y)
.
But λ is a 1-dimensional representation of h, and so λ vanishes on [h, h]. Thus λ(x) = 0.
It follows that 〈x, y〉 = 0 for all y ∈ h. Since the Killing form of g restricted to h is
nondegenerate, this implies x = 0. Hence [h, h] = 0, and so h is abelian.
Let h∗ = Hom(h,C) be the dual space of h. It follows from [Hal74, Theorem 15.2]
that this is a vector space of linear maps from h into C and that dim h∗ = dim h. We
define a map h → h∗ using the Killing form of g. Given h ∈ h, we define h∗ ∈ h∗ by
h∗(x) = 〈h, x〉 for all x ∈ h.
Lemma 4.15. The map h 7→ h∗ is an isomorphism of vector spaces between h and h∗.
Proof. This map is clearly linear. Suppose h lies in the kernel. Then h∗ is the zero
map, that is, 〈h, x〉 = 0 for all x ∈ h. It follows that h = 0 by Proposition 4.13, showing
that the map is injective. By the rank-nullity theorem, this also proves that the map is
surjective.
51
Notice that Φ is a finite subset of h∗. Because the map h 7→ h∗ is bijective, we
know that for each α ∈ Φ, there exists a unique element h′α ∈ h such that h′∗α (x) = α(x)
for all x ∈ h, that is,
α(x) =⟨h′α, x
⟩for all x ∈ h.
(The notation hα would seem more natural, but this will be reserved for the coroot of
α, which will be introduced in Chapter 7.)
Proposition 4.16. The vectors h′α for α ∈ Φ span h.
Proof. Suppose the vectors h′α are contained in a proper subspace of h. Then the
annihilator of this subspace is nonzero by [Hal74, Theorem 17.1]. Thus there exists a
nonzero x ∈ h such that x∗ (h′α) = 0 for all α ∈ Φ, that is,⟨h′α, x
⟩= 0. Hence α(x) = 0
for all α ∈ Φ. Let y ∈ h. Then
〈x, y〉 = tr(ad x ad y) =∑
λ
dim gλ λ(x)λ(y) = 0
since λ(x) = 0 for all weights λ. Thus 〈x, y〉 = 0 for all y ∈ h. Since the Killing form of
g restricted to h is nondegenerate, this implies x = 0, a contradiction.
Proposition 4.17. h′α ∈ [gα, g−α] for all α ∈ Φ.
Proof. We know that gα is an h-module by Theorem 2.10. Since all irreducible h-
modules are 1-dimensional, gα contains a 1-dimensional h-submodule Ceα. We have
[x, eα] = α(x)eα for all x ∈ h. Let y ∈ g−α. Then [eα, y] ∈ [gα, g−α] ⊂ h. I claim that
[eα, y] = 〈eα, y〉h′α. We define the element
z = [eα, y]− 〈eα, y〉h′α ∈ h.
Let x ∈ h. Then
〈x, z〉 = 〈x, [eα, y]〉 − 〈eα, y〉⟨x, h′α
⟩= 〈[x, eα], y〉 − 〈eα, y〉α(x)
= α(x)〈eα, y〉 − 〈eα, y〉α(x) = 0.
52
Thus 〈x, z〉 = 0 for all x ∈ h. Since the Killing form of g restricted to h is nondegenerate,
this implies z = 0. Hence [eα, y] = 〈eα, y〉h′α for all y ∈ g−α.
Now there exists a y ∈ g−α such that 〈eα, y〉 6= 0. For otherwise eα would be
orthogonal to g−α, and thus to the whole of g by Proposition 4.11. This would imply
eα ∈ g⊥. But g⊥ = 0 since g is semisimple, and so eα = 0, a contradiction. Choosing a
y ∈ g−α such that 〈eα, y〉 6= 0, we have
h′α =1
〈eα, y〉[eα, y] ∈ [gα, g−α].
Proposition 4.18.⟨h′α, h
′α
⟩6= 0 for all α ∈ Φ.
Proof. Suppose⟨h′α, h
′α
⟩= 0 for some α ∈ Φ. Let β be any element of Φ. By Proposition
4.3, there exists an rβ,α ∈ Q such that β = rβ,αα on [gα, g−α]. Now h′α ∈ [gα, g−α] by
Proposition 4.17. Thus
β(h′α)
= rβ,αα(h′α),
that is,⟨h′β, h
′α
⟩= rβ,α
⟨h′α, h
′α
⟩= 0. This holds for all β ∈ Φ. But the vectors h′α
for α ∈ Φ span h by Proposition 4.16, and so⟨x, h′α
⟩= 0 for all x ∈ h. Since the
Killing form of g restricted to h is nondegenerate, this implies h′α = 0. Thus α = 0,
contradicting the fact that α ∈ Φ.
Having obtained many useful results, we are now in a position to prove one of the
most important theorems about the Cartan decomposition of a semisimple Lie algebra.
Theorem 4.19. dim gα = 1 for all α ∈ Φ.
Proof. Choose a 1-dimensional h-submodule Ceα of gα as in the proof of Proposition
4.17. By the same proof, we can find an e−α ∈ g−α such that [eα, e−α] = h′α. Consider
the subspace m of g given by
m = Ceα ⊕ Ch′α ⊕ g−α ⊕ g−2α ⊕ · · · .
53
There are only finitely-many summands of m since Φ is finite. Thus there are only
finitely-many non-negative integers r such that g−rα 6= 0. Observe that ad eα · m ⊂ m
because
[eα, eα] = 0,[eα, h
′α
]= −α
(h′α)eα,
[eα, y] = 〈eα, y〉h′α for all y ∈ g−α
by the proof of Proposition 4.17, and
ad eα · g−rα ⊂ g−(r−1)α for all r ≥ 2
by Proposition 4.1. Similarly, ad e−α ·m ⊂ m because
[e−α, eα] = h′α,[e−α, h
′α
]= α
(h′α)e−α,
and ad eα · g−rα ⊂ g−(r+1)α for all r ≥ 1. Now h′α = [eα, e−α], and so
ad h′α = ad eα ad e−α − ad e−α ad eα.
Thus ad h′α ·m ⊂ m. We calculate the trace of ad h′α on m in two different ways. First,
we have
trm
(ad h′α
)= α
(h′α)
+ dim g−α
(−α
(h′α))
+ dim g−2α
(−2α
(h′α))
+ · · ·
= α(h′α)(1− dim g−α − 2 dim g−2α − · · · ).
Second, we have
trm
(h′α)
= trm(ad eα ad e−α)− trm(ad e−α ad eα) = 0.
Thus
α(h′α)(1− dim g−α − 2 dim g−2α − · · · ) = 0.
Now α (h′α) =⟨h′α, h
′α
⟩6= 0 by Proposition 4.18, and so
1− dim g−α − 2 dim g−2α − · · · = 0.
54
This can happen only if dim g−α = 1 and dim g−rα = 0 for all r ≥ 2. Now α ∈ Φ if and
only if −α ∈ Φ by Proposition 4.12. Thus dim gα = 1 for all α ∈ Φ.
Note that while all of the root spaces gα are 1-dimensional, the space g0 = h need
not be 1-dimensional.
Proposition 4.20. If α ∈ Φ and rα ∈ Φ where r ∈ Z, then r = 1 or r = −1.
Proof. By the proof of Theorem 4.19, we have dim g−rα = 0 for all r ≥ 2, that is, −rα
is not a root. Now rα ∈ Φ if and only if −rα ∈ Φ by Proposition 4.12. Thus only α
and −α can be roots.
We are now ready to examine some stronger properties of the set Φ of roots. Let
α, β ∈ Φ be roots such that β 6= α and β 6= −α. Then by Proposition 4.20, β is not
an integer multiple of α. There do, however, exist integers p ≥ 0, q ≥ 0 such that the
elements
−pα+ β, . . . ,−α+ β, β, α+ β, . . . , qα+ β
all lie in Φ but −(p+ 1)α+ β and (q + 1)α+ β do not. The set of roots
−pα+ β, . . . , qα+ β
is called the α-chain of roots through β.
Proposition 4.21. Let α, β be roots such that β 6= α and β 6= −α. Let
−pα+ β, . . . , β, . . . , qα+ β
be the α-chain of roots through β. Then
2
⟨h′α, h
′β
⟩⟨h′α, h
′α
⟩ = p− q.
Proof. Consider the subspace m of g given by
m = g−pα+β ⊕ · · · ⊕ gqα+β.
55
Recall that h′α = [eα, e−α] ∈ [gα, g−α] by the proof of Proposition 4.17. Now β is
not an integer multiple of α, and so −(p + 1)α + β 6= 0 and (q + 1)α + β 6= 0. We
have ad eα · gqα+β ⊂ g(q+1)α+β by Proposition 4.1. Because (q + 1)α + β 6= 0 and
(q + 1)α + β /∈ Φ, we must have g(q+1)α+β = 0. Thus ad eα · m ⊂ m. By a similar
argument, we have ad e−α ·m ⊂ m, and so
ad h′α ·m = (ad eα ad e−α − ad e−α ad eα)m ⊂ m.
We calculate the trace of ad h′α on m in two different ways. Following the reasoning of
Proposition 4.3, we have
trm
(ad h′α
)=
q∑i=−p
(iα+ β)(h′α)
since dim giα+β = 1 by Theorem 4.19. Second, we have
trm
(ad h′α
)= trm(ad eα ad e−α)− trm(ad e−α ad eα) = 0.
Thusq∑
i=−p
(iα+ β)(h′α)
= 0,
that is, (q(q + 1)
2− p(p+ 1)
2
)α(h′α)
+ (p+ q + 1)β(h′α)
= 0.
Since p+ q + 1 6= 0, this yields
(q − p)2
⟨h′α, h
′α
⟩+⟨h′α, h
′β
⟩= 0,
and so
2
⟨h′α, h
′β
⟩⟨h′α, h
′α
⟩ = p− q
since⟨h′α, h
′α
⟩6= 0 by Proposition 4.18.
This proposition has some very useful corollaries. The first is a strengthening of
Proposition 4.20.
56
Corollary 4.22. If α ∈ Φ and ζα ∈ Φ where ζ ∈ C, then ζ = 1 or ζ = −1.
Proof. Suppose ζ 6= ±1 and let β = ζα. Then β (h′α) = ζα (h′α), that is,
⟨h′α, h
′β
⟩= ζ⟨h′α, h
′α
⟩.
Applying Proposition 4.21, this yields
2ζ = 2
⟨h′α, h
′β
⟩⟨h′α, h
′α
⟩ = p− q.
Hence 2ζ ∈ Z. If ζ ∈ Z, then ζ = ±1 by Proposition 4.20. Thus ζ /∈ Z. It follows that
p− q is odd. The α-chain of roots through β is
−(p+ q
2
)α, . . . , β =
(p− q
2
)α, . . . ,
(p+ q
2
)α.
Since p− q is odd and consecutive roots differ by α, we see that all roots in the α-chain
are odd multiples of 12α. Also, p − q 6= 0, and so p and q cannot both be zero. Thus
p+ q 6= 0. Because the first and last roots are negatives of one another, 12α must lie in
the α-chain. Thus 12α ∈ Φ. But α ∈ Φ, and so 2
(12α)∈ Φ, contradicting Proposition
4.20. It follows that ζ must be 1 or −1.
Hence the only roots that are scalar multiples of a root α are α and −α.
Corollary 4.23.⟨h′α, h
′β
⟩∈ Q for all α, β ∈ Φ.
Proof. We already know that⟨h′α, h
′β
⟩∈ C. We also have
2
⟨h′α, h
′β
⟩⟨h′α, h
′α
⟩ ∈ Z by Proposition 4.21.
Thus〈h′α,h′β〉〈h′α,h′α〉
∈ Q. It is therefore sufficient to show that⟨h′α, h
′α
⟩∈ Q. We have
⟨h′α, h
′α
⟩= tr
(ad h′α ad h′α
)=∑β∈Φ
(β(h′α))2 =
∑β∈Φ
⟨h′α, h
′β
⟩2.
Dividing by⟨h′α, h
′α
⟩2, this yields
1⟨h′α, h
′α
⟩ =∑β∈Φ
(⟨h′α, h
′β
⟩⟨h′α, h
′α
⟩)2
∈ Z.
Hence⟨h′α, h
′α
⟩∈ Q, completing the proof.
Chapter 5
Root systems and the Weyl group
5.1 Positive systems and fundamental systems of roots
As before, let g be a semisimple Lie algebra, h a Cartan subalgebra of g, and Φ
the set of roots of g with respect to h. Recall from Proposition 4.16 that the elements
h′α for α ∈ Φ span h. We can thus find a subset that forms a basis of h. Let h′α1, . . . , h′αl
be such a basis.
Proposition 5.1. If α ∈ Φ, then h′α =∑l
i=1 µih′αi
where each µi lies in Q.
Proof. We know that h′α =∑l
i=1 µih′αi
for uniquely determined elements µi ∈ C. Let⟨h′αi
, h′αj
⟩= ξij . Then ξij ∈ Q by Corollary 4.23. Consider the system of equations⟨
h′α, h′α1
⟩= µ1ξ11 + µ2ξ21 + · · ·+ µlξl1⟨
h′α, h′α2
⟩= µ1ξ12 + µ2ξ22 + · · ·+ µlξl2
...⟨h′α, h
′αl
⟩= µ1ξ1l + µ2ξ2l + · · ·+ µlξll.
This is a system of l linear equations in l variables µ1, . . . , µl. Since the Killing form of g
restricted to h is nondegenerate, it follows from [HK71, Corollary 10.2] that det(ξij) 6= 0.
Thus we may solve this system of equations for µ1, . . . , µl in terms of the⟨h′α, h
′αi
⟩and
ξij . Because⟨h′α, h
′αi
⟩∈ Q and ξij ∈ Q, it follows that µi ∈ Q for i = 1, . . . , l.
We define hQ to be the set of all elements of the form∑l
i=1 µih′αi
with µi ∈ Q.
Similarly, we define hR to be the set of all such elements with µi ∈ R. Proposition 5.1
58
shows that hQ and hR are independent of the choice of basis h′α1, . . . , h′αl
. Since the
elements h′α for α ∈ Φ span h, this also shows that hQ is the set of all rational linear
combinations of the h′α, while hR is the set of real linear combinations of such elements.
We now show that the Killing form of g behaves favorably when restricted to hR.
Proposition 5.2. Let x ∈ hR. Then 〈x, x〉 ∈ R and 〈x, x〉 ≥ 0. If 〈x, x〉 = 0, then
x = 0.
Proof. Let x =∑l
i=1 µih′αi
where µi ∈ R. Then
〈x, x〉 =l∑
i=1
l∑j=1
µiµj
⟨h′αi
, h′αj
⟩=∑
i
∑j
µiµjtr(ad h′αi
ad h′αj
)=∑
i
∑j
µiµj
∑λ∈Φ
λ(h′αi
)λ(h′αj
)=∑λ∈Φ
∑i
∑j
µiµjλ(h′αi
)λ(h′αj
)=∑λ∈Φ
(∑i
µiλ(h′αi
))2
.
Now λ(h′αi
)=⟨h′λ, h
′αi
⟩∈ Q by Corollary 4.23. Thus 〈x, x〉 ∈ R, and also 〈x, x〉 ≥ 0.
Suppose now that 〈x, x〉 = 0. Then∑
i µiλ(h′αi
)= 0 for all λ ∈ Φ. In particular,∑
i µiαj
(h′αi
)= 0 for j = 1, . . . , l. This yields
∑i µi
⟨h′αi
, h′αj
⟩= 0, that is,
∑i µiξij = 0.
Since the matrix (ξij) is non-singular, we must have µi = 0 for all i. Hence x = 0.
This proposition shows that the Killing form of g restricted to hR is a map
hR × hR → R
that is a symmetric positive definite bilinear form. The vector space hR endowed with
this positive definite form is thus a Euclidean space containing all vectors h′α for α ∈ Φ.
Recall from Lemma 4.15 that the map h 7→ h∗ given by h∗(x) = 〈h, x〉 is an
isomorphism from h into h∗. We define h∗R to be the image of hR under this isomorphism.
59
Thus h∗R is the real subspace of h∗ spanned by Φ. We may also define a symmetric
bilinear form on h∗R by ⟨h∗1, h
∗2
⟩= 〈h1, h2〉 ∈ R.
Thus h∗R becomes a Euclidean space containing all roots α ∈ Φ. We wish to investigate
the structure of the roots in the Euclidean space h∗R. For convenience, we write V = h∗R.
A total ordering on V is a relation ≺ satisfying the following conditions:
(i) If λ ≺ µ and µ ≺ ν, then λ ≺ ν.
(ii) If λ, µ ∈ V , then exactly one of the following holds: λ ≺ µ, λ = µ, or µ ≺ λ.
(iii) If λ ≺ µ, then λ+ ν ≺ µ+ ν.
(iv) If λ ≺ µ and ξ ∈ R with ξ > 0, then ξλ ≺ ξµ.
Every real vector space has such orderings. For example, let v1, . . . , vl be a basis of
V and let λ =∑λivi, µ =
∑µivi with λ 6= µ. We define λ ≺ µ if the first nonzero
coefficient µi − λi is positive. One easily checks that this is a total ordering on V .
A positive system Φ+ ⊂ Φ is the set of all roots α ∈ Φ satisfying 0 ≺ α for some
total ordering on V . Given a positive system Φ+, we define a fundamental system
Π ⊂ Φ+ by
Π = {α ∈ Φ+ | α cannot be expressed as the sum of two elements of Φ+}.
We define Φ− to be the corresponding set of negative roots.
Proposition 5.3. Every root in Φ+ is a sum of roots in Π.
Proof. Let α ∈ Φ+ and suppose α /∈ Π. Then α = β + γ where β, γ ∈ Φ+ and
β ≺ α, γ ≺ α. Since Φ+ is finite, we may continue decomposing the summands of α into
lesser summands until each summand of α is in Π.
Proposition 5.4. If α, β ∈ Π and α 6= β, then 〈α, β〉 ≤ 0.
60
Proof. We first show that α − β /∈ Φ. Suppose α − β ∈ Φ. Then α − β ∈ Φ+ or
β − α ∈ Φ+. If α − β ∈ Φ+, then α = (α − β) + β, contradicting α ∈ Π. Similarly, if
β − α ∈ Φ+, then β = (β − α) + α, contradicting β ∈ Π. It follows that α− β /∈ Φ.
Now consider the α-chain of roots through β. This is of the form
β, α+ β, . . . , qα+ β
since −α+ β /∈ Φ. By Proposition 4.21, we deduce that
2
⟨h′α, h
′β
⟩⟨h′α, h
′α
⟩ = −q.
Now⟨h′α, h
′α
⟩> 0 by Proposition 5.2, so we must have
⟨h′α, h
′β
⟩≤ 0. It follows that
〈α, β〉 ≤ 0 by the way we defined the bilinear form on V .
Thus any two distinct roots in the fundamental system Π are inclined at an obtuse
angle. The next theorem illustrates the importance of the concept of a fundamental
system.
Theorem 5.5. A fundamental system Π forms a basis of V = h∗R.
Proof. We first show that Π spans V . We know that Φ spans V . Since α ∈ Φ if and
only if −α ∈ Φ, we see that Φ+ spans V . Thus Π spans V by Proposition 5.3.
We now show that the set Π is linearly independent. Suppose it is not. Then
there exists a non-trivial linear combination of roots αi ∈ Π that is equal to zero. Taking
the terms with positive coefficients to one side of the equation, we obtain
µi1αi1 + · · ·+ µirαir = µj1αj1 + · · ·+ µjrαjr
where µi1 , . . . , µir , µj1 , . . . , µjs > 0 and αi1 , . . . , αir , αj1 , . . . , αjs are distinct elements in
Π. Writing
v = µi1αi1 + · · ·+ µirαir = µj1αj1 + · · ·+ µjrαjr ,
we have 〈v, v〉 = 〈µi1αi1 + · · · + µirαir , µj1αj1 + · · · + µjrαjr〉. Expanding this linearly,
we see that 〈v, v〉 ≤ 0 by Proposition 5.4. Since the bilinear form is positive definite,
61
this implies v = 0. But 0 ≺ v since we have 0 ≺ αi and µi > 0. This is a contradiction,
and so Π must be linearly independent.
Now let Π be a fundamental system in Φ. By Theorem 5.5, we have |Π| = dim h∗R.
But h∗R is isomorphic to hR, and dim hR = dim h. Thus |Π| = l = dim h, that is, the
number of roots in a fundamental system is equal to the rank of the Lie algebra g.
We may now write Π = {α1, . . . , αl}. Since the roots α1, . . . , αl span h∗R, the
corresponding vectors h′α1, . . . , h′αl
must span hR. By our construction of hR, we also
know that hR spans h. Thus h′α1, . . . , h′αl
are l vectors that span h, and hence form a
basis of h. This fact will become useful as we enter Chapter 7.
Corollary 5.6. Let Π be a fundamental system of roots. Then each α ∈ Φ can be
expressed in the form α =∑niαi where αi ∈ Π, ni ∈ Z, and ni ≥ 0 for all i or ni ≤ 0
for all i.
Proof. If α ∈ Φ+, then α is a positive integer sum of elements of Π by Proposition 5.3.
If α ∈ Φ−, then α is the negative of an integer sum of elements of Π.
5.2 The Weyl group
Inside the root system Φ, a positive system Φ+ can be chosen in many different
ways. In this section, we will show that any two positive systems in Φ can be transformed
into one another by an element of a certain finite group W that acts on Φ.
For each α ∈ V , we define a linear map sα : V → V by
sα(x) = x− 2〈α, x〉〈α, α〉
α for all x ∈ V
where, as before, V = h∗R. One sees immediately that sα satisfies
sα(α) = −α
sα(x) = x if 〈α, x〉 = 0.
62
There is a unique linear map satisfying these conditions – the reflection in the hyperplane
of V orthogonal to α. Thus sα is this reflection.
The group W of all non-singular maps on V generated by the sα for all α ∈ Φ is
called the Weyl group. This group plays a vital role in the Lie theory.
Proposition 5.7. The Weyl group is a group of isometries on V , that is,
〈w(x), w(y)〉 = 〈x, y〉 for all x, y ∈ V , w ∈W.
Proof. Let α ∈ Φ. We have
〈sα(x), sα(y)〉 =⟨x− 2
〈α, x〉〈α, α〉
α, y − 2〈α, y〉〈α, α〉
α
⟩= 〈x, y〉 −
⟨x, 2
〈α, y〉〈α, α〉
α
⟩−⟨
2〈α, x〉〈α, α〉
α, y
⟩+⟨
2〈α, x〉〈α, α〉
α, 2〈α, y〉〈α, α〉
α
⟩= 〈x, y〉 − 2
〈α, y〉〈α, α〉
〈x, α〉 − 2〈α, x〉〈α, α〉
〈α, y〉+ 4〈α, x〉〈α, y〉〈α, α〉〈α, α〉
〈α, α〉
= 〈x, y〉.
Thus sα is an isometry for all α ∈ Φ. The proof follows immediately sinceW is generated
by the elements sα.
Proposition 5.8. If α ∈ Φ and w ∈W , then w(α) ∈ Φ, i.e., W permutes the roots.
Proof. It suffices to show that sα(β) ∈ Φ for all α, β ∈ Φ since the elements sα generate
W . The statement is clear if β is either α or −α. So suppose β 6= ±α and let
−pα+ β, . . . , β, . . . , qα+ β
be the α-chain of roots through β. Then we have
sα(β) = β − 2〈α, β〉〈α, α〉
α = β − (p− q)α
by Proposition 4.21. Now β − (p− q)α lies in the α-chain, and so sα(β) ∈ Φ.
In fact we observe that sα inverts the above α-chain. In particular, we have
sα(qα+ β) = −pα+ β and sα(−pα+ β) = qα+ β.
63
Proposition 5.9. The Weyl group W is finite.
Proof. Each element of W induces a permutation of Φ. If two elements of W induce
the same permutation of Φ, then they must be equal since Φ spans V . Thus there is a
one-to-one correspondence between the elements of W and the permutations of Φ. Since
Φ is finite, there can be only finitely-many permutations of Φ. Hence W is finite.
Now suppose Φ+ is a positive system in Φ and let Π be the corresponding funda-
mental system.
Lemma 5.10. Let α ∈ Π. If β ∈ Φ+ and β 6= α, then sα(β) ∈ Φ+.
Proof. We can express β in the form
β =∑
i
niαi αi ∈ Π, ni ∈ Z, ni ≥ 0
by Corollary 5.6. Since β 6= α, there must be some ni 6= 0 with αi 6= α. We have
sα(β) = β − 2〈α, β〉〈α, α〉
α.
We express sα(β) as a linear combination of the elements of Π. Since β 6= 0, we have
sα(β) ∈ Φ+ or sα(β) ∈ Φ−. It follows from Corollary 5.6 that if sα(β) ∈ Φ+, then the
integer coefficients of the elements of Π will all be nonnegative, whereas if sα(β) ∈ Φ−,
then the integer coefficients of the elements of Π will all be nonpositive. Since the
coefficient of αi remains ni, we must have sα(β) ∈ Φ+.
Theorem 5.11. If Φ+1 ,Φ
+2 are two positive systems in Φ, then there exists a w ∈ W
such that w(Φ+
1
)= Φ+
2 .
Proof. Let m =∣∣Φ+
1 ∩ Φ−2
∣∣. We proceed by induction on m. If m = 0, then Φ+1 = Φ+
2 ,
and so w = 1 satisfies the theorem. Now assume m > 0, that is, Φ+1 ∩ Φ−
2 6= ∅. We
cannot have Π1 ⊂ Φ+2 , as this would imply Φ+
1 ⊂ Φ+2 , contrary to m > 0. Thus there
64
exists an α ∈ Π1 ∩ Φ−2 . Consider the positive system sα
(Φ+
1
)in Φ. By Lemma 5.10,
sα
(Φ+
1
)contains all roots in Φ+
1 except α, together with −α. It follows that
∣∣sα
(Φ+
1
)∩ Φ−
2
∣∣ = m− 1.
By induction, there exists a w′ ∈ W such that w′sα
(Φ+
1
)= Φ+
2 . Letting w = w′sα, we
have w(Φ+
1
)= Φ+
2 as desired.
Corollary 5.12. If Π1,Π2 are two fundamental systems in Φ, then there exists a w ∈W
such that w(Π1) = Π2.
Proof. Let Φ+1 ,Φ
+2 be positive systems containing Π1,Π2, respectively. By Theorem
5.11, there exists a w ∈ W such that w(Φ+
1
)= Φ+
2 . Then w(Π1) is a fundamental
system contained in Φ+2 , and so w(Π1) = Π2.
Proposition 5.13. Let Π be a fundamental system in Φ. Then for each α ∈ Φ, there
exist αi ∈ Π and w ∈W such that α = w(αi).
Proof. Let Φ+ be the positive system with fundamental system Π. Suppose first that
α ∈ Φ+. Then we have
α =∑
i
niαi αi ∈ Π, ni ∈ Z, ni ≥ 0
by Corollary 5.6. We define the height of α by
htα =∑
i
ni.
We proceed by induction on htα. If htα = 1, then α = αi for some i, and so α ∈ Π.
Thus αi = α and w = 1 satisfy the proposition. Now suppose htα > 1. Then ni > 0
for at least two values of i by Proposition 4.20. We have
〈α, α〉 =∑
i
ni〈α, αi〉.
65
Since 〈α, α〉 > 0 and each ni > 0, there exists an αi ∈ Π such that 〈α, αi〉 > 0. Let
β = sαi(α). Then β ∈ Φ and
β = α− 2〈α, αi〉〈αi, αi〉
αi.
Since 〈α, αi〉 > 0, we see that the coefficient of αi in β is less than the coefficient of αi
in α. Thus htβ < htα. On the other hand, we know that β ∈ Φ+ since there are at
least two positive coefficients in α and only one coefficient is reduced in passing from α
to β. By induction, there exist αj ∈ Π and w′ ∈W such that β = w′(αj). This yields
α = sαi(β) = sαiw′(αj)
as desired.
Now suppose α ∈ Φ−. Then α = sα(−α) and −α ∈ Φ+. Thus there exist w′ ∈W
and αi ∈ Π such that −α = w′(αi). Hence α = sαw′(αi), completing the proof.
Theorem 5.14. Let Π = {α1, . . . , αl} be a fundamental system in Φ. Then the reflec-
tions sα1 , . . . , sαlgenerate W .
Proof. Let W0 be the subgroup of W generated by sα1 , . . . , sαl. Since the elements
sα for α ∈ Φ generate W , it is sufficient to prove that each sα lies in W0. We may
assume α ∈ Φ+ since sα = s−α. Applying Proposition 5.13 to the subgroup W0 (each
time we used the inductive hypothesis, we could have assumed that w was in W0), we
see that there exist αi ∈ Π and w ∈ W0 such that α = w′(αi). Consider the element
wsαiw−1 ∈W0. We have
wsαiw−1(α) = wsαi(αi) = w(−αi) = −α.
We shall show that wsαiw−1(x) = x if 〈α, x〉 = 0. Now 〈α, x〉 = 0 implies
⟨w−1(α), w−1(x)
⟩= 0
since w−1 is an isometry. Hence⟨αi, w
−1(x)⟩
= 0. It follows that w−1(x) is orthogonal
to αi, and so sαiw−1(x) = w−1(x), that is, wsαiw
−1(x) = x. Thus wsαiw−1 is the
66
reflection in the hyperplane orthogonal to α, and so wsαiw−1 = sα. This shows that
sα ∈W0, and thus W0 = W .
We wish to further study the way in which the Weyl group is generated by the
reflections of the fundamental roots. As before, let Π = {α1, . . . , αl} be a fundamental
system of roots. For convenience, we write
s1 = sα1 , s2 = sα2 , . . . , sl = sαl.
Each element of W can be expressed as a product of the si. (We do not need to worry
about inverses since s−1i = si for all i.) For each w ∈W , we define l(w) to be the least
nonnegative integer m such that w can be expressed as a product of m fundamental
reflections. We call l(w) the length of w. We have l(1) = 0 and l(si) = 1 for all i.
An expression of w as a product of l(w) fundamental reflections is called a reduced
expression for w.
Recall that each element of W permutes the roots in Φ by Proposition 5.8. For
each w ∈ W , we define n(w) to be the number of roots α ∈ Φ+ such that w(α) ∈ Φ−.
Thus n(w) is the number of positive roots made negative by w. We wish to show that
l(w) = n(w).
Proposition 5.15. n(w) ≤ l(w) for all w ∈W .
Proof. We fix i and compare n(w) with n(wsi). We know that si transforms αi to −αi
and all other positive roots to positive roots by Lemma 5.10. Hence
n(wsi) = n(w)± 1.
In order to determine the sign, we study the effect of w and wsi on αi. If w(αi) ∈ Φ+,
then w transforms αi into a positive root, and so wsi transforms αi into a negative root.
Hence n(wsi) = n(w) + 1. If, on the other hand, w(αi) ∈ Φ−, then wsi transforms αi
into a positive root so that n(wsi) = n(w)− 1.
67
Now let w = si1si2 · · · sir be a reduced expression for w where r = l(w). Then
n(w) ≤ n(si1 · · · sir−1) + 1 ≤ n(si1 · · · sir−2) + 1 ≤ · · · ≤ r.
Thus n(w) ≤ l(w) as desired.
In order to prove the converse, that l(w) ≤ n(w), we need a very useful theorem
called the deletion condition.
Theorem 5.16. Let w = si1 · · · sir be any expression of w ∈ W as a product of the
fundamental reflections. If n(w) < r, then there exist integers j, k with 1 ≤ j < k ≤ r
such that
w = si1 · · · sij · · · sik · · · sir
where ˆ denotes omission.
Proof. Recall from Proposition 5.15 that for all w ∈ W , we have n(wsi) = n(w) ± 1.
Consider the given expression w = si1 · · · sir . Looking at the chain
n(w) ≤ n(si1 · · · sir−1) + 1 ≤ n(si1 · · · sir−2) + 1 ≤ · · · ≤ r,
we see that since n(w) < r, we must have
n(si1 · · · sik) = n(si1 · · · sik−1)− 1
for some k with 1 < k ≤ r. This implies si1 · · · sik−1(αik) ∈ Φ− as in the proof of
Proposition 5.15. Since αik ∈ Φ+, there exists a j with 1 ≤ j < k such that
sij+1 · · · sik−1(αik) ∈ Φ+
sijsij+1 · · · sik−1(αik) ∈ Φ−.
Now sj transforms only one positive root into a negative root by Lemma 5.10, namely
αij . It follows that
sij+1 · · · sik−1(αik) = αij . (1)
68
Applying siksik−1· · · sij+1 to both sides of (1) yields
−αik = siksik−1· · · sij+1(αij ). (2)
But (1) also gives us the relation
sij+1 · · · sik−1(−αik) = −αij . (3)
Plugging (2) into (3), we get
sij+1 · · · sik−1siksik−1
· · · sij+1(αij ) = −αij .
Since sij+1 · · · sik−1siksik−1
· · · sij+1 transforms αij into −αij , we must have
sij = sij+1 · · · sik−1siksik−1
· · · sij+1 .
It follows that
sijsij+1 · · · sik−1= sij+1 · · · sik−1
sik .
Plugging this relation into our original expression for w yields
si1 · · · sir = si1 · · · sij−1sij+1 · · · sik−1sik+1
· · · sir ,
and so w = si1 · · · sij · · · sik · · · sir as desired.
Corollary 5.17. n(w) = l(w)
Proof. We know that n(w) ≤ l(w) by Proposition 5.15. So suppose n(w) < l(w). Let
w = sii · · · sir be a reduced expression for w, that is, r = l(w). Since n(w) < r, we
may use the deletion condition and obtain an expression for w that is a product of
r − 2 fundamental reflections. This contradicts the definition of l(w), so we must have
n(w) = l(w).
This shows that the length of w is equal to the number of positive roots made
negative by w.
69
Proposition 5.18.
(i) The maximal length of any element of W is |Φ+|.
(ii) There is a unique element w0 ∈W with l(w0) = |Φ+|.
(iii) w0 (Φ+) = Φ−.
(iv) w20 = 1.
Proof. Part (i) follows from the fact that l(w) = n(w) and n(w) is bounded by |Φ+|. To
prove part (ii), notice that for each fundamental system Π, −Π is also a fundamental
system, arising from the opposite total ordering on h∗R. By Corollary 5.12, there exists
a w0 ∈ W such that w0(Π) = −Π. Hence w0 (Φ+) = Φ−, and so n(w0) = |Φ+|. Thus
l(w) = |Φ+|, that is, w0 is an element of W of maximal length. Let w′0 be another
element of W such that l (w′0) = |Φ+|. Then n (w′
0) = |Φ+|, that is, w′0 (Φ+) = Φ−. Let
w = w′−10 w0. Then w (Φ+) = Φ+ so that n(w) = 0. It follows that l(w) = 0, and so
w = 1. Thus w′0 = w0, showing that w0 is unique. This proves part (ii), and part (iii)
follows immediately. Finally, we have w20 (Φ+) = w0 (Φ−) = Φ+, and so n
(w2
0
)= 0.
Hence l(w2
0
)= 0 and w2
0 = 1.
Now consider a group having presentation
⟨s1, . . . , sl | (sisj)
mij = 1⟩
where mii = 1 for all i and mij ≥ 2 for i 6= j. A group having such a presentation is
called a Coxeter group. One can show that every Weyl group is isomorphic to a finite
Coxeter group. For a proof, we refer the reader to [Car05, §5.3].
Chapter 6
The Cartan matrix and the Dynkin diagram
6.1 The Cartan matrix
We recall from Proposition 5.2 that the vector space V = h∗R forms a Euclidean
space under the scalar product 〈, 〉. We also recall that the roots Φ span V but are not
linearly independent. Any fundamental system Π ⊂ Φ does, however, form a basis of V
by Theorem 5.5. The goal of this section is to further study the geometry of the root
system Φ in the Euclidean space V .
We shall first find the possible angles between any two roots α, β ∈ Φ and the
relation between the lengths of the roots α, β. We consider only the angles θ satisfying
0 ≤ θ ≤ π.
Proposition 6.1. Let α, β ∈ Φ such that β 6= ±α. Then
(i) The angle between α, β is one of π/6, π/4, π/3, π/2, 2π/3, 3π/4, 5π/6.
(ii) If α, β are inclined at π/3 or 2π/3, then α, β have the same length.
(iii) If α, β are inclined at π/4 or 3π/4, then the ratio of their lengths is√
2.
(iv) If α, β are inclined at π/6 or 5π/6, then the ratio of their lengths is√
3.
Proof. Let θ be the angle between α, β. Then
〈α, β〉 = |α||β| cos θ
71
where, as usual, |α| =√〈α, α〉 and |β| =
√〈β, β〉. Thus
cos2 θ =〈α, β〉2
〈α, α〉〈β, β〉=〈α, β〉〈α, α〉
· 〈β, α〉〈β, β〉
so that
4 cos2 θ = 2〈α, β〉〈α, α〉
· 2〈β, α〉〈β, β〉
.
Recall from Proposition 4.21 that 2 〈α,β〉〈α,α〉 and 2 〈β,α〉
〈β,β〉 are integers. Hence 4 cos2 θ ∈ Z.
Since 0 ≤ 4 cos2 θ ≤ 4 and β 6= ±α, we must have 4 cos2 θ ∈ {0, 1, 2, 3}. In each case, we
consider the possible factorization of 4 cos2 θ into the product of two integers. Without
loss of generality, we assume that |α| ≤ |β|.
First, suppose 4 cos2 θ = 0. Then cos θ = 0 so that θ = π/2.
Second, suppose 4 cos2 θ = 1. Then cos θ = 1/2 or −1/2 so that θ = π/3 or 2π/3.
The possible factorizations of 4 cos2 θ are
1 = 1 · 1 or 1 = (−1)(−1).
In either case, we have
2〈α, β〉〈α, α〉
= 2〈β, α〉〈β, β〉
so that 〈α, α〉 = 〈β, β〉, that is, α and β have the same length.
Now suppose 4 cos2 θ = 2. Then cos θ = 1/√
2 or −1/√
2 so that θ = π/4 or 3π/4.
The possible factorizations of 4 cos2 θ are
2 = 1 · 2 or 2 = (−1)(−2).
In either case, since |α| ≤ |β|, we must have
2〈α, β〉〈α, α〉
= 2 · 2〈β, α〉〈β, β〉
.
Then 〈β, β〉 = 2〈α, α〉 and |β| =√
2|α|. Hence the ratio of the lengths of α, β is√
2.
Finally, suppose 4 cos2 θ = 3. Then cos θ =√
3/2 or −√
3/2 so that θ = π/6 or
5π/6. The possible factorizations of 4 cos2 θ are
3 = 1 · 3 or 3 = (−1)(−3).
72
In either case, since |α| ≤ |β|, we must have
2〈α, β〉〈α, α〉
= 3 · 2〈β, α〉〈β, β〉
.
Then 〈β, β〉 = 3〈α, α〉 and |β| =√
3|α|. Hence the ratio of the lengths of α, β is√
3.
This completes the proof. Notice that we obtain no information about the ratio
of the lengths of α, β in the case where θ = π/2.
Corollary 6.2. Let Π be a fundamental system of roots in Φ and let α, β ∈ Π with
β 6= α. Then the angle between α, β is one of π/2, 2π/3, 3π/4, 5π/6.
Proof. We saw in Proposition 5.4 that the angle θ between two distinct fundamental
roots satisfies π/2 ≤ θ < π. Looking at the list of possible angles in Proposition 6.1,
π/2, 2π/3, 3π/4, 5π/6 are the only angles satisfying this property.
Now let Π = {α1, . . . , αl} be a fundamental system. We wish to encode the
information about the angles between the αi and the ratios of their lengths into a
matrix. For i, j = 1, . . . , l, we define the elements
Aij = 2〈αi, αj〉〈αi, αi〉
.
We have Aij ∈ Z for all i, j. The l × l matrix A = (Aij) is called the Cartan matrix.
Proposition 6.3. The Cartan matrix A has the following properties:
(i) Aii = 2 for all i.
(ii) Aij ∈ {0,−1,−2,−3} if i 6= j.
(iii) If Aij = −2 or −3, then Aji = −1.
(iv) Aij = 0 if and only if Aji = 0.
Proof. Properties (i) and (iv) follow from the definition of Aij . Properties (ii) and (iii)
follow from the proof of Proposition 6.1.
73
If we permute the indices of the roots in Π, we will not, in general, get the same
Cartan matrix A. But aside from the ambiguity of indexing the roots, the Cartan matrix
is uniquely determined by the semisimple Lie algebra g.
Proposition 6.4. The Cartan matrix of g depends only on the indexing of the funda-
mental roots. It is independent of the choice of Cartan subalgebra h and fundamental
system Π.
Proof. The independence of the choice of Cartan subalgebra follows from the conjugacy
of Cartan subalgebras, which was Theorem 3.14.
Now let Π′ be another fundamental system. By Corollary 5.12, there exists a
w ∈ W such that w(Π) = Π′. For each i, we write w(αi) = α′i. Because w is an
isometry of V , we have
2〈αi, αj〉〈αi, αi〉
= 2
⟨α′i, α
′j
⟩⟨α′i, α
′i
⟩ .It follows that the Cartan matrices defined by Π and Π′ with respect to the chosen
labellings are the same.
We now determine the Cartan matrices for l = 1 and 2 using Proposition 6.3.
The only possible 1× 1 Cartan matrix is (2). We also see that any 2× 2 Cartan matrix
must be one of the following: 2 0
0 2
2 −1
−1 2
2 −1
−2 2
2 −2
−1 2
2 −1
−3 2
2 −3
−1 2
.
The pair 2 −1
−2 2
2 −2
−1 2
are obtained by switching the indices 1 and 2, as are the pair 2 −1
−3 2
2 −3
−1 2
.
74
6.2 The Dynkin diagram
In order to determine the possible Cartan matrices for larger values of l, it is
useful to introduce a combinatorial structure called the Dynkin diagram. The Dynkin
diagram is determined by the Cartan matrix. It is a graph with vertices labeled 1, . . . , l.
For i 6= j, the vertices i, j are joined by nij edges where
nij = AijAji.
By Proposition 6.4, the Dynkin diagram is uniquely determined by the semisimple Lie
algebra g. The Dynkin diagrams of the Cartan matrices for l = 1 and 2 are given in
Figure 6.1.
Cartan matrix Dynkin diagram
(2)
2 −1
−1 2
2 0
0 2
2 −1
−2 2
2 −2
−1 2
2 −1
−3 2
2 −3
−1 2
Figure 6.1: Cartan matrices and Dynkin diagrams for l = 1 and 2.
Proposition 6.5. nij ∈ {0, 1, 2, 3} for all i 6= j.
Proof. This follows immediately from Proposition 6.3 since nij = AijAji.
75
This shows that the number of edges joining any two vertices of the Dynkin
diagram is either 0, 1, 2, or 3. Notice that the Dynkin diagram need not be a connected
graph. If it is connected, however, then it will split into connected components. If
we renumber the vertices so that those in each connected component are numbered
consecutively, then the Cartan matrix will split into blocks of the form
A =
∗ 0 0 0
0 ∗ 0 0
0 0 ∗ 0
0 0 0 ∗
where each block corresponds to a connected component. Each block will, in fact, be the
Cartan matrix for its corresponding connected component. The set Π = {α1, . . . , αl}
will be partitioned into subsets in a corresponding way. One easily checks that roots in
different subsets are mutually orthogonal.
At first glance, it may appear that there are many ways to arrange vertices and
edges to form connected components. But the set of graphs that can occur as connected
Dynkin diagrams turns out to be quite restricted. In order to reduce our list of possi-
bilities, it is useful to introduce a quadratic form Q(x1, . . . , xl) defined in terms of the
Dynkin diagram. We define
Q(x1, . . . , xl) = 2l∑
i=1
x2i −
l∑i,j=1i6=j
√nij xixj .
Figure 6.2 gives the quadratic forms of the Dynkin diagrams for l = 1 and 2.
Proposition 6.6. The quadratic form Q(x1, . . . , xl) is positive definite.
Proof. For i 6= j, we have
nij = AijAji = 2〈αi, αj〉〈αi, αi〉
· 2 〈αj , αi〉〈αj , αj〉
.
76
Dynkin diagram Quadratic form
2x21
2x21 + 2x2
2
2x21 + 2x1x2 + 2x2
2
2x21 + 2
√2 x1x1 + 2x2
2
2x21 + 2
√3 x1x1 + 2x2
2
Figure 6.2: Dynkin diagrams and quadratic forms for l = 1 and 2.
By Proposition 5.4, we know that 〈αi, αj〉 ≤ 0. Hence −√nij = 2 〈αi,αj〉|αi||αj | . For i = j, we
have 2 〈αi,αj〉|αi||αj | = 2. Thus the quadratic form is
Q(x1, . . . , xl) =l∑
i,j=1
2〈αi, αj〉|αi||αj |
xixj = 2
⟨l∑
i=1
xiαi
|αi|,
l∑j=1
xjαj
|αj |
⟩= 2〈y, y〉
where y =∑l
i=1xiαi|αi| . Thus Q(x1, . . . , xl) ≥ 0 since the bilinear form 〈, 〉 is positive
definite. Furthermore, if Q(x1, . . . , xl) = 0, then y = 0. In this case, since α1, . . . , αl are
linearly independent, we must have xi = 0 for all i. Thus the quadratic form is positive
definite.
Thus we have shown that the the connected components of the Dynkin diagram
of a semisimple Lie algebra have the following properties:
(A) The graph is connected.
(B) Any pair of distinct vertices are joined by 0, 1, 2, or 3 edges.
(C) The corresponding quadratic form Q(x1, . . . , xl) is positive definite.
In order to find all possible Dynkin diagrams, we are going to to determine all graphs
satisfying conditions (A), (B), (C). Then, having found all such graphs, we will deter-
mine which ones occur as Dynkin diagrams.
77
6.3 Classification of Dynkin diagrams
In this section, we will focus on proving the following theorem.
Theorem 6.7. The graphs satisfying conditions (A), (B), (C) from the previous section
are precisely those in the following list:
· · ·A1
E6
F4
G2
E7 E8
· · ·
A4
B4
D5
A3
B3
D4
A2
B2
A5
B5
D6
· · ·
Proof. These graphs clearly satisfy conditions (A) and (B). We now concentrate on
proving that they satisfy condition (C). It is a standard result in linear algebra that
a quadratic form∑aijxixj is positive definite if and only if the leading minors of its
symmetric matrix (aij) have positive determinant, that is,
|a11| > 0,
∣∣∣∣∣∣∣a11 a12
a21 a22
∣∣∣∣∣∣∣ > 0, . . . , det(aij) > 0.
Given a graph Γ on the list with l vertices, we shall show that Q(x1, . . . , xl) is positive
definite by induction on l.
If l = 1, then Γ = A1 and Q(x1) = 2x21 is positive definite. If l = 2, then Γ = A2,
78
B2, or G2. In each case, the symmetric matrix corresponding to Q(x1, x2) is
A2 :
2 −1
−1 2
B2 :
2 −√
2
−√
2 2
G2 :
2 −√
3
−√
3 2
.
One sees that the leading minors of these matrices have positive determinant. Now
suppose l ≥ 3. Looking at the above list, we see that Γ contains at least one vertex
that is joined to just one other vertex, and joined to it by a single edge. Label such a
vertex l and label the vertex it is joined to l−1. We write Γ = Γl. We denote the graph
obtained from Γl by removing the vertex l by Γl−1 and the graph obtained from Γl−1
by removing the vertex l − 1 by Γl−2. Observe that Γl−1 and Γl−2 are also on the list.
Let det Γl be the determinant of the symmetric matrix representing the quadratic form
Q(x1, . . . , xl) associated with Γl. We obtain the equality
det Γl =
∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣
0...
0
2 −1
0 . . . 0 −1 2
∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣= 2det Γl−1 − det Γl−2
by expanding the determinant by its last row. This gives an inductive way to calculate
det Γl. We perform this calculation for each Γl on the list.
We have
detA1 = 2, detA2 = 3, . . . , detAl = 2detAl−1 − detAl−2.
Thus detAl = l + 1. We also have
detA1 = 2, detB2 = 2, detB3 = 2, . . . , detBl = 2detBl−1 − detBl−2.
Thus detBl = 2. Next, we have
detA3 = 4, detD4 = 4, detD5 = 4, . . . , detDl = 2detDl−1 − detDl−2.
79
Thus detDl = 4. Finally, we have
detE6 = 2detD5 − detA4 = 3
detE7 = 2detD6 − detA5 = 2
detE8 = 2detE7 − detA6 = 1
detF4 = 2detB3 − detA2 = 1
and detG2 = 1 from the previous list of matrices. Thus we have shown that det Γl > 0
for all Γl. The leading minors of the symmetric matrix associated with Γl are the sym-
metric matrices associated with certain subgraphs of Γl. We may number the vertices
of Γl such that every subgraph is connected. But the list of graphs has the property
that every subgraph of any graph on the list is also on the list. Thus every leading
minor of the symmetric matrix associated with Γl is positive, and so the quadratic form
Q(x1, . . . , xl) associated with Γl is positive definite.
So far, we have shown that the graphs on the list satisfy conditions (A), (B),
(C). The following lemmas will prove the converse, that is, that any graph satisfying
conditions (A), (B), (C) is on the list.
Lemma 6.8. For each of the graphs on the following list, the corresponding quadratic
form Q(x1, . . . , xl) has determinant 0:
· · ·
A4 A5A2 A3
· · ·
B4 B5B3
· · ·C3 C4C2
· · ·
D5D4 D6
80
E8E6 E7
F4 G2
Proof. For the graphs Γ = Al, each row of the symmetric matrix associated with the
given quadratic form contains one entry 2 and two entries -1. The remaining entries are
0. Thus the sum of the columns is zero, and so det Al = 0.
In all of the other graphs Γ on the list, we can find a vertex l that is joined to
just one vertex l − 1. Moreover, we may choose l such that it is connected to l − 1 by
either a single or a double edge. If there is a single edge, then we may use the formula
det Γl = 2det Γl−1 − det Γl−2
as before. If there is a double edge, then we may modify the way we obtained the
previous formula to get the formula
det Γl = 2det Γl−1 − 2 det Γl−2.
As before, we calculate the determinants of the above graphs inductively. We have
det B3 = 2detA3 − 2(detA1)2 = 0
det Bl = 2detDl − 2 detDl−1 = 0 for l ≥ 4
det C2 = 2detB2 − 2 detA1 = 0
det Cl = 2detBl − 2 detBl−1 = 0 for l ≥ 3
det D4 = 2detD4 − (detA1)3 = 0
det Dl = 2detDl − detDl−2 · detA1 = 0 for l ≥ 5
det E6 = 2detE6 − detA5 = 0
det E7 = 2detE7 − detD6 = 0
det E8 = 2detE8 − detE7 = 0
81
det F4 = 2detF4 − detB3 = 0
det G2 = 2detG2 − detA1 = 0.
This exhausts all cases.
Lemma 6.9. Let Γ be a graph satisfying conditions (A), (B), (C). Let Γ′ be a con-
nected subgraph of Γ, that is, a connected graph obtained from Γ by omitting vertices
or decreasing the number of edges between vertices or both. Then Γ′ satisfies conditions
(A), (B), (C), also.
Proof. The subgraph Γ′ clearly satisfies conditions (A) and (B). We need to show that
it satisfies (C). Let Q(x1, . . . , xl) be the quadratic form of Γ and Q′(x1, . . . , xm) the
quadratic form of Γ′ where m ≤ l. We have
Q(x1, . . . , xl) = 2l∑
i=1
x2i −
l∑i,j=1i6=j
√nij xixj
Q′(x1, . . . , xm) = 2m∑
i=1
x2i −
m∑i,j=1i6=j
√n′ij xixj
where n′ij ≤ nij for i, j = 1, . . . ,m. Suppose Q′ is not positive definite. Then there exist
y1, . . . , ym ∈ R, not all zero, such that Q′(y1, . . . , ym) ≤ 0. Now (|y1|, . . . , |ym|, 0, . . . , 0)
is not the zero vector in Rl, but
Q(|y1|, . . . , |ym|, 0, . . . , 0) = 2m∑
i=1
|yi|2 −m∑
i,j=1i6=j
√nij |yi||yj |
≤ 2m∑
i=1
y2i −
m∑i,j=1i6=j
√n′ij |yi||yj |
≤ 2m∑
i=1
y2i −
m∑i,j=1i6=j
√n′ij yiyj
= Q′(y1, . . . , ym) ≤ 0.
This contradicts the fact that Q(x1, . . . , xl) is positive definite, so Q′(x1, . . . , xm) must
be positive definite, also.
82
Having proved Lemmas 6.8 and 6.9, we are now ready to complete the proof of
Theorem 6.7.
Let Γ be a graph satisfying conditions (A), (B), (C). Then by Lemmas 6.8 and
6.9, Γ can have no subgraph of type Al, Bl, Cl, Dl, E6, E7, E8, F4, or G2. We will use
this information to show that Γ must be one of the graphs on the list in Theorem 6.7.
We first notice that Γ contains no cycles, for otherwise Al would be a subgraph
of Γ for some l ≥ 2.
Suppose Γ contains a triple edge. Then Γ must be the graph G2, as otherwise G2
would be a subgraph of Γ.
Thus we may assume Γ contains no triple edge. Suppose Γ contains a double
edge. Then Γ cannot contain more than one double edge, for otherwise Cl would be a
subgraph of Γ for some l ≥ 2. Now Γ cannot contain a branch point in addition to a
double edge, as otherwise Bl would be a subgraph of Γ for some l ≥ 3. Thus Γ is a chain
containing only one double edge. If the double edge lies at the end of the chain, then
Γ = Bl for some l ≥ 2. If the double edge lies elsewhere, then Γ = F4, for otherwise F4
would be a subgraph of Γ.
Now suppose Γ contains no double or triple edges. If Γ contains no branch point,
then Γ = Al for some l ≥ 1. Thus we assume that Γ has at least one branch point.
We see that Γ cannot contain more than one branch point, for otherwise Dl would be
a subgraph of Γ for some l ≥ 5. Thus Γ contains exactly one branch point. There can
be at most three branches stemming from this branch point, since otherwise D4 would
be a subgraph of Γ. Let l1, l2, l3 be the number of vertices on the three branches where
l1 ≥ l2 ≥ l3. Then the total number of vertices of Γ is l1 + l2 + l3 + 1.
We observe that l3 must equal 1, for otherwise we would have li ≥ 2 for i = 1, 2, 3,
and so E6 would be a subgraph of Γ. If l2 = 1, then Γ = Dl for some l ≥ 4. Thus we may
assume l2 ≥ 2. But we cannot have l2 > 2, as otherwise we would have l1 ≥ 3, l2 ≥ 3,
and so E7 would be a subgraph of Γ. Hence l3 = 1 and l2 = 2. We must have l1 ≤ 4,
83
for otherwise E8 would be a subgraph of Γ. Thus Γ is either E6, E7, or E8.
This exhausts all possibilities for Γ, and in each case Γ is a graph on the list in
Theorem 6.7. This completes the proof.
Corollary 6.10. Let ∆ be the Dynkin diagram of a semisimple Lie algebra. Then each
connected component of ∆ must be one of the following graphs:
Al, l ≥ 1; Bl, l ≥ 2; Dl, l ≥ 4; E6; E7; E8; F4; G2.
Proof. This is immediate from Theorem 6.7 since the connected components of the
Dynkin diagram of a semisimple Lie algebra satisfy conditions (A), (B), (C).
We will consider whether all of the graphs listed above actually occur as Dynkin
diagrams in the following chapter.
6.4 Classification of Cartan matrices
Recall that the Dynkin diagram is uniquely determined by the Cartan matrix via
the relation
nij = AijAji i 6= j.
But the Cartan matrix is not necessarily determined by the Dynkin diagram. Given
the integers nij ∈ {0, 1, 2, 3} for all i, j with i 6= j, we now consider the extent to which
the Aij are determined. If nij = 0, then we must have Aij = Aji = 0 since Z is an
integral domain. If nij = 1, then we must have Aij = Aji = −1 since Aij , Aji ∈ Z
and Aij , Aji ≤ 0. However, for nij = 2, there are two ways to factor nij = AijAji,
namely Aij = −1, Aji = −2 and Aij = −2, Aji = −1. Similarly, if nij = 3, then we have
Aij = −1, Aji = −3 or Aij = −3, Aji = −1.
Out of the list in Corollary 6.10, the only graphs giving rise to such an ambiguity
are Bl, l ≥ 2, F4, and G2. In order to remove this ambiguity, we place an arrow on the
double and triple edges. The direction of the arrow is determined as follows: The arrow
84
points from vertex i to vertex j if and only if |αi| > |αj | in the Euclidean space V . This
is equivalent to the condition |Aji| > |Aij |. Thus in the situation
i j
we have |αi| =√
2|αj | and Aij = −1, Aji = −2. In the situation
i j
we have we have |αi| =√
3|αj | and Aij = −1, Aji = −3. Thus we may regard the arrow
as an inequality sign on the lengths of the fundamental roots at the vertices.
· · ·A1
E6
F4
G2
E7 E8
· · ·
A2
B2
· · ·
D4
A3
B3
C3
A4
D5
B4
C4
A5
D6
B5
C5
· · ·
Figure 6.3: Standard list of connected Dynkin diagrams.
The set of possible Dynkin diagrams, including arrows, is shown in Figure 6.3.
85
Note that, since the diagrams B2, F4, G2 are symmetric, it does not matter in which
direction the arrow is drawn in these graphs. By the preceding remarks, the connected
components of the Dynkin diagram of any semisimple Lie algebra must appear on this
list.
From this list, we obtain a standard list of corresponding Cartan matrices. Two
Cartan matrices (Aij),(A′
ij
)are called equivalent if there exists a permutation σ of
1, . . . , l such that
A′ij = Aσ(i)σ(j) for all i, j.
We see that equivalent Cartan matrices arise from different labellings of the vertices of
the same Dynkin diagram. By choosing a suitable labeling of vertices of the Dynkin
diagrams in Figure 6.3, we obtain the following list of corresponding Cartan matrices
determined uniquely up to equivalence:
Al =
2 −1
−1 2 −1
−1 2 −1
−1 · ·
· · ·
· · −1
−1 2 −1
−1 2 −1
−1 2
l ≥ 1
86
Bl =
2 −1
−1 2 −1
−1 2 −1
−1 · ·
· · ·
· · −1
−1 2 −1
−1 2 −1
−2 2
l ≥ 2
Cl =
2 −1
−1 2 −1
−1 2 −1
−1 · ·
· · ·
· · −1
−1 2 −1
−1 2 −2
−1 2
l ≥ 3
Dl =
2 −1
−1 2 −1
−1 2 −1
−1 · ·
· · ·
· · −1
−1 2 −1
−1 2 −1 −1
−1 2 0
−1 0 2
l ≥ 4
87
E6 =
2 −1
−1 2 −1
−1 2 −1 −1
−1 2
−1 2 −1
−1 2
E7 =
2 −1
−1 2 −1
−1 2 −1
−1 2 −1 −1
−1 2
−1 2 −1
−1 2
E8 =
2 −1
−1 2 −1
−1 2 −1
−1 2 −1
−1 2 −1 −1
−1 2
−1 2 −1
−1 2
F4 =
2 −1
−1 2 −1
−2 2 −1
−1 2
G2 =
2 −1
−3 2
A Cartan matrix is indecomposable if its Dynkin diagram is connected. We
know that every Cartan matrix uniquely determines a Dynkin diagram, and that this di-
88
agram decomposes into connected components. Each connected component corresponds
to an indecomposable Cartan matrix, and thus every Cartan matrix can be expressed
as the direct sum of indecomposable Cartan matrices.
If A is the Cartan matrix of any semisimple Lie algebra, then each indecomposable
component of A will be equivalent to some Cartan matrix from the previous list.
Proposition 6.11. If a semisimple Lie algebra g has a connected Dynkin diagram, then
g is simple.
Proof. Let h ⊕∑
α∈Φ gα be a Cartan decomposition of g giving rise to the Dynkin
diagram ∆. Let a be a nonzero ideal of g. We will show that a = g, thus proving that
g is simple.
We first prove that a ∩ h 6= 0. Suppose a ∩ h = 0. For each α ∈ Φ, let eα be a
nonzero element in gα. We choose a nonzero element x ∈ a and write
x = h+∑α∈Φ
µαeα h ∈ h, µα ∈ C
such that the number of nonzero µα is as small as possible. Now since a ∩ h = 0, there
exists some β ∈ Φ with µβ 6= 0. We have
[h′β, x
]=∑α∈Φ
µα
[h′β, eα
]=∑α∈Φ
µαα(h′β)eα
since h is abelian. By Proposition 4.17, there exist elements eβ ∈ gβ and e−β ∈ g−β
such that [eβ, e−β] = h′β. Thus
[[h′β, x
], e−β
]=∑α∈Φ
µαα(h′β)[eα, e−β] = µββ
(h′β)h′β +
∑α∈Φα 6=β
µαα(h′β)Nα,−βeα−β
where [eα, e−β] = Nα,−βeα−β. Now [[h′β, x], e−β] ∈ a since x ∈ a. Also, [[h′β, x], e−β] 6= 0
since µβ 6= 0 and β(h′β)
=⟨h′β, h
′β
⟩6= 0 by Proposition 4.18. Notice that the number of
nonzero components from the root spaces gα is less for [[h′β, x], e−β] than it was for x.
This contradicts our choice of x, so we must have a ∩ h 6= 0.
89
We next show that h ⊂ a. Suppose this is not the case. Then 0 6= a ∩ h 6= h.
Now there exist αi ∈ Π and x ∈ a ∩ h such that⟨h′αi
, x⟩6= 0, for otherwise a ∩ h would
be orthogonal to each h′αi, hence to the whole of h, and would thus be 0. We choose
eαi ∈ gαi and e−αi ∈ g−αi such that [eαi , e−αi ] = h′αi. We have
[x, eαi ] = αi(x)eαi =⟨h′αi
, x⟩eαi ∈ a.
Since⟨h′αi
, x⟩6= 0, we see that eαi ∈ a. It follows that [eαi , e−αi ] = h′αi
∈ a.
We may now divide the αi ∈ Π into two classes: those with h′αi∈ a, and those
with h′αi/∈ a. Since we assumed h was not contained in a, we see that both classes are
nonempty. I claim that if h′αi∈ a and h′αj
/∈ a, then⟨h′αj
, h′αi
⟩= 0. Suppose this is not
true. We choose eαj ∈ gαj and e−αj ∈ g−αj such that [eαj , e−αj ] = h′αj. Then
[h′αi
, ej]
= αj
(h′αi
)ej =
⟨h′αi
, h′αj
⟩ej ∈ a.
Since⟨h′αi
, h′αj
⟩6= 0, we must have eαj ∈ a. But then [eαj , e−αj ] = h′αj
∈ a, a con-
tradiction. This completes the proof of the claim. Thus if h′αi∈ a and h′αj
/∈ a, then⟨h′αj
, h′αi
⟩= 0, that is, 〈αi, αj〉 = 0. This means that the vertices corresponding to the
roots in both classes are not joined in the Dynkin diagram ∆, and so ∆ is disconnected.
This is a contradiction, so we must have h ⊂ a.
We are now ready to prove that a = g. Let α ∈ Φ. We have
[h′α, eα
]= α
(h′α)eα =
⟨h′α, h
′α
⟩eα.
Since h ⊂ a, we have h′α ∈ a, and so [h′α, eα] ∈ a. Because⟨h′α, h
′α
⟩6= 0, this implies
that eα ∈ a. This is true for all α ∈ Φ, and so a = g. Thus g is simple.
Our next goal is to prove the converse of Proposition 6.11. We first define an
action of the Weyl group W on the Cartan subalgebra h. Recall that the Weyl group
is a group of non-singular linear transformations on the real vector space h∗R. We can
extend the action of W on h∗R by linearity to give an action of W on h∗ by C-linear
90
transformations. We now define the action of W on h by h 7→ wh where
λ(wh) =(w−1λ
)h for all h ∈ h, λ ∈ h∗, w ∈W.
To prove this action is well defined, suppose x1, x2 ∈ h satisfy λ(x1) =(w−1λ
)h and
λ(x2) =(w−1λ
)h for all λ ∈ h∗. Then x1 = x2. For otherwise x1−x2 6= 0, and so there
would exist a λ ∈ h∗ such that λ(x1 − x2) 6= 0. But λ(x1) = λ(x2) since x1, x2 satisfy
the above conditions. Thus λ(x1 − x2) = 0, a contradiction.
The action h 7→ wh has some very nice properties. We observe that
w1(w2h) = (w1w2)h for all h ∈ h, w1, w2 ∈W.
This follows from the fact that
λ(w1(w2h)) =(w−1
1 λ)w2h =
(w−1
2
(w−1
1 λ))h =
((w−1
2 w−11
)λ)h
=((w1w2)
−1 λ)h = λ((w1w2)h)
for all λ ∈ h∗. Moreover, the actions of W on h∗ and h are compatible with the
isomorphism h∗ → h given by λ 7→ h′λ where λ(x) =⟨h′λ, x
⟩for all x ∈ h. For suppose
w(λ) = µ for λ, µ ∈ h∗. Then
⟨w(h′λ), x⟩
=⟨h′λ, w
−1(x)⟩
= λ(w−1(x)
)= (wλ)x = µ(x) =
⟨h′µ, x
⟩for all x ∈ h.
Hence w(λ) = µ implies w (h′λ) = h′µ. Having this relation, we deduce that since
sα(λ) = λ− 2〈α, λ〉〈α, α〉
α for α ∈ Φ, λ ∈ h∗,
we must have
sα(x) = x− 2
⟨h′α, x
⟩⟨h′α, h
′α
⟩h′α for x ∈ h.
Proposition 6.12. Let g be a semisimple Lie algebra whose Dynkin diagram ∆ splits
into connected components ∆1, . . . ,∆r. Then we have
g = g1 ⊕ · · · ⊕ gr
where gi is a simple Lie algebra with Dynkin diagram ∆i.
91
Proof. We have ∆ = ∆1 ∪∆2 ∪ · · · ∪∆r. For each i, let Πi be the subset of Π corre-
sponding to the vertices in ∆i. We have
Π = Π1 ∪Π2 ∪ · · · ∪Πr.
We also have 〈α, β〉 = 0 if α ∈ Πi, β ∈ Πj , and i 6= j. Let hi be the subspace of h
spanned by the elements h′α with α ∈ Πi. Then
h = h1 ⊕ h2 ⊕ · · · ⊕ hr
where⟨h, h′
⟩= 0 if h ∈ hi, h′ ∈ hj , and i 6= j.
Now let α ∈ Πi and consider the fundamental reflection sα ∈ W . One easily
checks that sα transforms hi into itself and fixes every vector in hj for all j 6= i. Thus
sα (hj) = hj j = 1, . . . , r.
Because the elements sα generate W , we see that
w (hj) = hj j = 1, . . . , r w ∈W.
Now for all α ∈ Φ, we have h′α = w(h′αi
)for some αi ∈ Π and w ∈ W by Proposition
5.13 and the way we defined the W -action on h. It follows that for each α ∈ Φ, h′α lies
in hi for some i. For each i, let Φi be the set of all α ∈ Φ such that h′α ∈ hi. Then
Φ = Φ1 ∪Φ2 ∪ · · · ∪Φr.
We now define gi to be the subspace of g spanned by hi and the eα for all α ∈ Φi.
Given the Cartan decomposition g = h⊕∑
α∈Φ Ceα and the above decomposition of h,
we see that
g = g1 ⊕ g2 ⊕ · · · ⊕ gr,
a direct sum of subspaces. I claim that each gi is a subalgebra of g. Since h is abelian
and [h′α, eβ] = β (h′α) eβ for all α ∈ Πi, β ∈ Φi, we need only verify that [eα, eβ] ∈ gi if
92
α, β ∈ Φi. Now if α+ β ∈ Φ, then α+ β ∈ Φi since h′α+β = h′α + h′β ∈ hi. If α+ β = 0,
then [eα, eβ] is a scalar multiple of h′α and thus lies in hi, and hence in gi. Finally, if
α+ β 6= 0 but is not a root, then [eα, eβ] = 0. In all cases, we have [eα, eβ] ∈ gi, and so
gi is a subalgebra of g.
Next, we show that [gi, gj ] = 0 if i 6= j. Let α ∈ Φi and β ∈ Φj . We have
[h′α, eβ
]= β
(h′α)eβ =
⟨h′β, h
′α
⟩eβ = 0.
Similarly, we have [eα, h′β] = 0. We also have [eα, eβ] = 0, because α + β /∈ Φ since
h′α + h′β does not lie in any subspace hk of h. It follows that [gi, gj ] = 0.
Now each gi is an ideal of g since
[gi, g] =∑
j
[gi, gj ] = [gi, gi] ⊂ gi.
This implies that
[x1 + · · ·+ xr, y1 + · · ·+ yr] = [x1, y1] + · · ·+ [xr, yr]
where xi, yi ∈ gi. Thus
g = g1 ⊕ g2 ⊕ · · · ⊕ gr
is a direct sum of Lie algebras.
We also observe that each gi is a semisimple Lie algebra. Let a be a solvable
ideal of gi. Since [a, gj ] = 0 for all j 6= i, we see that a is an ideal of g. Because g is
semisimple, we have a = 0. Hence gi is semisimple.
Next, we show that hi is a Cartan subalgebra of gi. We know that hi is abelian,
and hence nilpotent. Let x ∈ N(hi), the normalizer of hi in gi. Then [x, h] ∈ hi for all
h ∈ hi. We also have [x, h] = 0 for all h ∈ hj with j 6= i. It follows that [x, h] ∈ h for all
h ∈ h. But N(h) = h since h is a Cartan subalgebra of g. Thus x ∈ h. It follows that
x ∈ h ∩ gi = hi. This shows that N(hi) = hi, and so hi is a Cartan subalgebra of gi.
93
We now consider the Cartan decomposition
gi = hi ⊕∑α∈Φi
Ceα
of gi with respect to hi. We see that Φi is the root system of gi, that Πi is a fundamental
system in Φi, and that ∆i is the Dynkin diagram of gi. Now ∆i is connected, and so gi
is simple by Proposition 6.11. Thus we have obtained a decomposition of g as a direct
sum of Lie algebras gi whose respective Dynkin diagrams are the connected components
∆i of ∆.
Corollary 6.13. A semisimple Lie algebra g has a connected Dynkin diagram if and
only if g is simple.
Proof. This follows immediately from Propositions 6.11 and 6.12.
Chapter 7
The existence and uniqueness theorems
In the last chapter, we saw that each non-trivial simple Lie algebra g has a Dynkin
diagram ∆ that appears on the standard list of connected Dynkin diagrams in Figure
6.3. But does the converse hold? That is, given a Dynkin diagram ∆ on the standard
list, is there a simple Lie algebra g with Dynkin diagram ∆? If so, does ∆ uniquely
determine g up to isomorphism? In this chapter, we will show that both existence
and uniqueness hold. The proof of the uniqueness property is somewhat easier, so we
will prove this first. In order to do so, we first study some properties of the structure
constants of the Lie algebra g.
7.1 Properties of structure constants
Let g be a simple Lie algebra with Dynkin diagram ∆. Let h be a Cartan subal-
gebra of g and
g = h⊕∑α∈Φ
gα
the Cartan decomposition of g with respect to h. We know form Theorem 4.19 that
dim gα = 1 for all α ∈ Φ. For each α, we choose a nonzero element eα ∈ gα. Let
Π = {α1, . . . , αl} be a fundamental system of roots in Φ. Then the elements h′αiform a
basis of h. We wish now to choose a more convenient basis consisting of scalar multiples
95
of the h′αi. For 1 = 1, . . . , l, we define the elements hi ∈ h by
hi =2h′αi⟨
h′αi, h′αi
⟩ .Notice that αi(hi) = 2 for all i. We see that {hi, i = 1, . . . , l ; eα, α ∈ Φ} is a basis
of g. We know from Proposition 4.17 that h′α ∈ [gα, g−α] for all α ∈ Φ. Thus, having
chosen the elements eα for all α ∈ Φ+, we may choose e−α for all α ∈ Φ+ to satisfy
[eα, e−α] =2h′α⟨h′α, h
′α
⟩ .This relation will then be automatically satisfied for all α ∈ Φ−.
For each α ∈ Φ, we define the element hα ∈ h by
hα =2h′α⟨h′α, h
′α
⟩ .The element hα is called the coroot corresponding to the root α. In particular, we have
hi = hαi . By construction, we have
[eα, e−α] = hα for all α ∈ Φ.
We now consider the product [eα, eβ] when α 6= β. We know that [eα, eβ] = 0 if
α + β 6= 0 and α + β /∈ Φ. If α + β ∈ Φ, then [gα, gβ] ⊂ gα+β. We define the elements
Nα,β ∈ C by the relation
[eα, eβ] = Nα,βeα+β.
The numbers Nα,β for α, β, α + β ∈ Φ are called structure constants of g. Notice
that the structure constants of g depend on our choice of elements eα ∈ gα.
Next, we consider the multiplication of the basis vectors {hi, eα} of g. We have
[hi, hj ] = 0
[hi, eα] = α(hi)eα
[eα, e−α] = hα
[eα, eβ] = Nα,βeα+β if α, β, α+ β ∈ Φ
[eα, eβ] = 0 if α+ β 6= 0 and α+ β /∈ Φ.
96
In order to express [eα, e−α] as a linear combination of the basis elements, we simply
express hα as a linear combination of the hi.
We now derive some relations among the structure constants Nα,β.
Proposition 7.1. The structure constants Nα,β satisfy the following relations:
(i) Nβ,α = −Nα,β.
(ii) If α, β, γ ∈ Φ satisfy α+ β + γ = 0, then
Nα,β
〈γ, γ〉=
Nβ,γ
〈α, α〉=
Nγ,α
〈β, β〉.
(iii) Nα,βN−α,−β = −(p + 1)2 where −pα + β, . . . , β, . . . , qα + β is the α-chain of
roots through β.
(iv) If α, β, γ, δ ∈ Φ satisfy α + β + γ + δ = 0, and no pair are negatives of one
another, then
Nα,βNγ,δ
〈α+ β, α+ β〉+
Nβ,γNα,δ
〈β + γ, β + γ〉+
Nγ,αNβ,δ
〈γ + α, γ + α〉= 0.
Proof. Part (i) follows immediately from the definition. To prove part (ii), suppose
α+ β + γ = 0. We consider the Jacobi identity
[[eα, eβ], eγ ] + [[eβ, eγ ], eα] + [[eγ , eα], eβ] = 0.
This yields
Nα,β[eα+β, e−(α+β)] +Nβ,γ [e−α, eα] +Nγ,α[e−β, eβ] = 0,
that is,
2Nα,β
h′α+β⟨h′α+β, h
′α+β
⟩ = 2Nβ,γh′α⟨
h′α, h′α
⟩ + 2Nγ,α
h′β⟨h′β, h
′β
⟩ .Now the roots α, β are linearly independent, for if they were not, then α + β = −γ
would fail to be a root. Thus h′α, h′β are linearly independent, and h′α+β = h′α + h′β. It
follows thatNα,β⟨
h′α+β, h′α+β
⟩ =Nβ,γ⟨h′α, h
′α
⟩ =Nγ,α⟨h′β, h
′β
⟩ ,
97
that is,Nα,β
〈γ, γ〉=
Nβ,γ
〈α, α〉=
Nγ,α
〈β, β〉.
To prove part (iii), suppose α, β ∈ Φ are linearly independent. Without loss of
generality, we may assume that |α| ≤ |β|. Consider the Jacobi identity
[[eα, e−α], eβ] + [[e−α, eβ], eα] + [[eβ, eα], e−α] = 0.
This yields
2[h′α, eβ]⟨h′α, h
′α
⟩ +N−α,βN−α+β,αeβ +Nβ,αNα+β,−αeβ = 0.
Since [h′α, eβ] = β (h′α) eβ =⟨h′α, h
′β
⟩eβ and eβ is not the zero vector, it follows that
2
⟨h′α, h
′β
⟩⟨h′α, h
′α
⟩ +N−α,βN−α+β,α +Nβ,αNα+β,−α = 0.
Rearranging terms and using part (i), this expression becomes
Nα,βNα+β,−α +N−α,βNα,−α+β = 2〈α, β〉〈α, α〉
. (1)
From part (ii), we can derive the relations
Nα+β,−α
〈β, β〉=
N−α,−β
〈α+ β, α+ β〉,
N−α,β
〈−α+ β,−α+ β〉=Nα−β,−α
〈β, β〉.
Plugging these relations into (1) gives us
Nα,βN−α,−β〈β, β〉
〈α+ β, α+ β〉−Nα,−α+βN−α,α−β
〈−α+ β,−α+ β〉〈β, β〉
= 2〈α, β〉〈α, α〉
. (2)
(Note: If −α + β is not a root, then N−α,β is interpreted as 0, so the second term on
the left-hand side of (1) disappears.) We next consider the α-chain of roots through β,
−pα+ β, . . . , β, . . . , qα+ β.
We plug the pairs (α, β), (α,−α + β), . . . , (α,−pα + β) into (2) to get the system of
equations
Nα,βN−α,−β〈β, β〉
〈α + β, α + β〉 − Nα,−α+βN−α,α−β〈−α + β,−α + β〉
〈β, β〉 = 2〈α, β〉〈α, α〉
Nα,−α+βN−α,α−β〈−α + β,−α + β〉
〈β, β〉 − Nα,−2α+βN−α,2α−β〈−2α + β,−2α + β〉〈−α + β,−α + β〉 = 2
〈α,−α + β〉α, α
...
Nα,−pα+βN−α,pα−β〈−pα + β,−pα + β〉
〈−(p − 1)α + β,−(p − 1)α + β〉 = 2〈α,−pα + β〉
〈α, α〉 .
98
(The last equation has only one term on the left since −(p + 1)α + β is not a root.)
Adding these equations and using the bilinearity of 〈, 〉, we get
Nα,βN−α,−β〈β, β〉
〈α+ β, α+ β〉= 2(p+ 1)
〈α, β〉〈α, α〉
− 2p(p+ 1)
2.
But we know from Proposition 4.21 that 2 〈α,β〉〈α,α〉 = p− q. It follows that
Nα,βN−α,−β〈β, β〉
〈α+ β, α+ β〉= −(p+ 1)q.
In order to obtain the desired result, Nα,βN−α,−β = −(p+ 1)2, we must show that
〈α+ β, α+ β〉〈β, β〉
=p+ 1q
.
Let γ = −pα + β, the initial root in the α-chain through β. Let r, s be non-
negative integers such that −rα + γ, . . . , γ, . . . , sα + γ are all roots but −(r + 1)α + γ
and (s+ 1)α+ γ are not. Recall from the proof of Proposition 6.1 that
2〈α, γ〉〈α, α〉
· 2〈γ, α〉〈γ, γ〉
= 4 cos2 θ
where θ is the angle between α, γ, and hence that 2 〈α,γ〉〈α,α〉 ∈ {0,±1,±2,±3}. We also
have 2 〈α,γ〉〈α,α〉 = r − s by Proposition 4.21. But γ is the initial root in the α-chain, so
r = 0, and hence s = p + q ≤ 3. This shows that any α-chain contains at most four
roots. Thus the possible positions of β in its α-chain are
α + ββ
α + ββ
β
β
β
−2α + β
α + β
α + β
−2α + β
β
α + β
−2α + β
−α + β
−α + β
−α + β
−3α + β
−2α + β
α + β
p = 0
p = 0
p = 1
p = 0
p = 1
p = 2
q = 1
q = 2
q = 1
q = 3
q = 2
q = 1
99
Since |α| ≤ |β|, we can compute the ratio of 〈α+ β, α+ β〉 to 〈β, β〉 in each case using
the proof of Proposition 6.1. The only case that is not straightforward is that in which
p = q = 1. In this case, we have
2〈α, α+ β〉〈α, α〉
= 2.
We see from Proposition 6.1 that this is possible only when 〈α + β, α + β〉 = 2〈α, α〉.
But we also have 〈α + β, α + β〉 = 〈α, α〉 + 〈β, β〉 since 〈α, β〉 = 0. It follows that
〈α, α〉 = 〈β, β〉 and 〈α+ β, α+ β〉 = 2〈β, β〉. The remaining computations give us
〈α+ β, α+ β〉〈β, β〉
= 1, 12 , 2,
13 , 1, 3
in our six cases, respectively. In each case, we see that
〈α+ β, α+ β〉〈β, β〉
=p+ 1q
,
and so Nα,βN−α,−β = −(p+ 1)2.
To prove part (iv), we assume that α, β, γ, δ ∈ Φ satisfy α + β + γ + δ = 0 with
no pair being negatives. Consider the Jacobi identity
[[eα, eβ], eγ ] + [[eβ, eγ ], eα] + [[eγ , eα], eβ] = 0.
This yields
Nα,βNα+β,γ +Nβ,γNβ+γ,α +Nγ,αNγ+α,β = 0. (1)
From part (ii), we have the relations
Nα+β,γ
〈δ, δ〉=
Nγ,δ
〈α+ β, α+ β〉,
Nβ+γ,α
〈δ, δ〉=
Nα,δ
〈β + γ, β + γ〉,
Nγ+α,β
〈δ, δ〉=
Nβ,δ
〈γ + α, γ + α〉.
Plugging these expressions into (1) gives us
Nα,βNγ,δ
〈α+ β, α+ β〉+
Nβ,γNα,δ
〈β + γ, β + γ〉+
Nγ,αNβ,δ
〈γ + α, γ + α〉= 0.
(As usual, we interpret Nθ,φ as 0 if θ + φ is not a root.)
Proposition 7.1(iii) has a very useful corollary.
Corollary 7.2. If α, β, α+ β ∈ Φ, then Nα,β 6= 0, that is, [gα, gβ] = gα+β.
100
7.2 The uniqueness theorem
We now use these relations between the structure constants to prove that the Lie
algebra g is uniquely determined (up to isomorphism) by its Dynkin diagram. Given a
Dynkin diagram on the standard list in Figure 6.3, this diagram uniquely determines
a Cartan matrix A = (Aij) on the standard list in Section 6.4. Each Cartan matrix,
in turn, determines the set Φ of roots as linear combinations of the fundamental roots
Π = {α1, . . . , αl}. To see this, recall from Proposition 5.13 that each root α ∈ Φ is
of the form α = w(αi) for some αi ∈ Π, w ∈ W . Moreover, each element w ∈ W is
a product of the elements s1, . . . , sl by Theorem 5.14. The action of s1, . . . , sl on the
fundamental roots α1, . . . , αl are given in terms of the Cartan matrix by
si(αj) = αj −Aijαj .
Thus by applying the fundamental reflections successively to the fundamental roots, we
obtain all roots as linear combinations of the fundamental roots.
Next, we show that the scalar products⟨h′α, h
′β
⟩for α, β ∈ Φ are uniquely deter-
mined by the Cartan matrix. By Proposition 4.21, the scalars 2〈h′α,h′β〉〈h′α,h′α〉
are determined
by the root system, and hence by the Cartan matrix as illustrated above. Then⟨h′α, h
′α
⟩is determined by the formula
1⟨h′α, h
′α
⟩ =∑β∈Φ
(⟨h′α, h
′β
⟩⟨h′α, h
′α
⟩)2
as in the proof of Corollary 4.23. It follows that⟨h′α, h
′β
⟩is also determined by the
Cartan matrix. Thus we see that if the structure constants Nα,β are known, then the
multiplication of the basis elements
[hi, hj ] = 0
[hi, eα] = α(hi)eα
[eα, e−α] = hα
101
[eα, eβ] = Nα,βeα+β if α, β, α+ β ∈ Φ
[eα, eβ] = 0 if α+ β 6= 0 and α+ β /∈ Φ
will be completely determined by the Cartan matrix.
Our next goal is to show that for certain pairs (α, β) of roots, the structure
constants Nα,β can be arbitrarily chosen, and that the remaining structure constants
are uniquely determined in terms of these by the relations in Proposition 7.1.
We choose a total ordering ≺ on the vector space V = h∗R as in Section 5.1. This
ordering will give rise to a positive system Φ+ and fundamental system Π+ of roots.
An ordered pair (α, β) of roots is called special if α + β ∈ Φ and 0 ≺ α ≺ β. A pair
(α, β) is called extraspecial if (α, β) is special and, in addition, if for all pairs (γ, δ)
such that α+ β = γ + δ, we have α � γ.
Lemma 7.3. The structure constants Nα,β for extraspecial pairs (α, β) can be chosen
as arbitrary nonzero elements of C by appropriate choice of the elements eα.
Proof. We begin by choosing the eα for α ∈ Φ+ in the order given by ≺. Suppose the
pair of roots (α, β) is extraspecial. Then we have
[eα, eβ] = Nα,βeα+β
where eα, eβ have already been chosen. Now there is only one extraspecial pair giving
rise to the sum α+ β. Thus eα+β can be chosen to give any nonzero value of Nα,β.
Proposition 7.4. All of the structure constants Nα,β are determined by the structure
constants for extraspecial pairs.
Proof. We consider the set of all pairs of roots (α, β) such that α + β is a root. Let
(α, β) be such a pair and let γ = −α − β. One easily checks that the following twelve
pairs are of the given type:
(α, β), (β, γ), (γ, α), (β, α), (γ, β), (α, γ)
(−α,−β), (−β,−γ), (−γ,−α), (−β,−α), (−γ,−β), (−α,−γ).
102
Since α+ β + γ = 0, we see that either two or one of α, β, γ are positive. Hence either
two of α, β, γ are positive or two of −α,−β,−γ are positive. By choosing two positive
roots from α, β, γ or from −α,−β,−γ and writing them in an appropriate order, we
obtain a special pair. Thus only one of the above twelve pairs is a special pair.
Now the relations in Proposition 7.1 (i), (ii), (iii) enable us to express Nβ,α, Nβ,γ ,
Nγ,α, and N−α,−β in terms of Nα,β. Thus, regardless of which pair (θ, φ) on the above
list is special, we can express all of the structure constants corresponding to these pairs
in terms of the Nθ,φ.
Next, we show that the Nα,β for all special pairs (α, β) can be written in terms
of the Nα,β for extraspecial pairs. Suppose (α, β) is special but not extraspecial. Then
there exists an extraspecial pair (γ, δ) such that α+ β = γ + δ. Then
α+ β + (−γ) + (−δ) = 0
and no pair of α, β,−γ,−δ are negatives. By Proposition 7.1(iv), we have
Nα,βN−γ,−δ
〈α+ β, α+ β〉+
Nβ,−γNα,−δ
〈β − γ, β − γ〉+
N−γ,αNβ,−δ
〈−γ + α,−γ + α〉= 0.
Now the roots α, β, γ, δ are ordered by
0 ≺ γ ≺ α ≺ β ≺ δ.
Thus we may use relations (i), (ii), and (iii) from Proposition 7.1 to express N−γ,−δ in
terms of Nγ,δ, Nβ,−γ in terms of Nγ,β−γ , Nα,−δ in terms of Nα,δ−α, N−γ,α in terms of
Nγ,α−γ , and Nβ,−δ in terms of Nβ,δ−β. Thus Nα,β is expressed in terms of
Nγ,δ, Nγ,β−γ , Nα,δ−α, Nγ,α−γ , Nβ,δ−β.
Now (γ, δ) is an extraspecial pair and (γ, β − γ), (α, δ − α), (γ, α − γ), and (β, δ − β)
are all pairs of positive roots whose sums are roots less than α+ β = γ + δ in the given
ordering. We may therefore argue by induction on α+ β that Nα,β can be expressed in
terms of Nθ,φ for extraspecial pairs (θ, φ).
103
We are now ready to prove the uniqueness theorem.
Theorem 7.5. If two simple Lie algebras have the same Cartan matrix, then they are
isomorphic.
Proof. By Lemma 7.3, we may choose the basis elements {hi, eα} of such a Lie algebra
g such that Nα,β = 1 for all extraspecial pairs of roots (α, β). The remaining structure
constants Nα,β are then uniquely determined by Proposition 7.4. Thus the formulae
expressing a Lie product of basis elements as a linear combination of basis elements
are completely determined by the Cartan matrix. It follows that the Lie algebra g is
uniquely determined up to isomorphism.
7.3 Some generators and relations in a simple Lie algebra
We now turn to the question of the existence of a Lie algebra having a Cartan
matrix on the standard list in Section 6.4. Let g be a simple Lie algebra with Cartan
matrix A. Let h be a Cartan subalgebra of g and
g = h⊕∑α∈Φ
gα
be the Cartan decomposition of h with respect to h. As before, we consider the coroots
hi ∈ h given by
hi =2h′αi⟨
h′αi, h′αi
⟩where Π = {α1, . . . , αl} is a fundamental system in Φ. For each i, we can choose
elements ei ∈ gαi , fi ∈ g−αi such that [ei, fi] = hi as in Section 7.1.
We will show that the elements e1, . . . , el, f1, . . . , fl, h1, . . . , hl generate g. (This is
equivalent to saying that e1, . . . , el, f1, . . . , fl generate g, but it will be useful to include
h1, . . . , hl in the generating set.)
Lemma 7.6. If α ∈ Φ+ and α /∈ Π, then there exists an αi ∈ Π such that α−αi ∈ Φ+.
Thus every non-fundamental root is the sum of a fundamental root with a positive root.
104
Proof. Suppose the statement is false. Then α−αi is not a root and is nonzero for each
i. (We can use Corollary 5.6 to show that α − αi is not a negative root.) For each i,
consider the αi-chain of roots through α. This is of the form
α, αi + α, . . . , qαi + α.
By Proposition 4.21, we have
2〈αi, α〉〈αi, αi〉
= −q.
It follows that 〈αi, α〉 ≤ 0. Now α ∈ Φ+ has the form α =∑
i niαi where ni ≥ 0 for all
i. Thus
〈α, α〉 =∑
i
ni〈αi, α〉 ≤ 0.
This is a contradiction since we know that 〈α, α〉 > 0.
Proposition 7.7. The elements e1, . . . , el, f1, . . . , fl, h1, . . . , hl generate g.
Proof. Since h1, . . . , hl generate h, it suffices to show that each gα for α ∈ Φ+ is con-
tained in the subalgebra generated by e1, . . . , el, and that each gα for α ∈ Φ− is contained
in the subalgebra generated by f1, . . . , fl.
Let α ∈ φ+. If α = αi for some i, then we have gα = Ceα. If α /∈ Π, then we may
write α = αi + β for some αi ∈ Π, β ∈ Φ+ by Lemma 7.6. We then have [gαi , gβ] = gα
by Corollary 7.2. We may thus choose eα = [ei, eβ] for some nonzero eβ ∈ gβ. By
repeating this process, we obtain
eα = [[ei1 , ei2 ], · · · eik ]
for some sequence i1, . . . , ik. Thus each gα for α ∈ Φ+ is contained in the subalgebra
generated by e1, . . . , el. A similar argument proves that each gα for α ∈ Φ− is contained
in the subalgebra generated by f1, . . . , fl.
Proposition 7.8. The generators e1, . . . , el, f1, . . . , fl, h1, . . . , hl of g satisfy the follow-
ing relations:
105
(a) [hi, hj ] = 0
(b) [hi, ej ] = Aijej
(c) [hi, fj ] = −Aijfj
(d) [ei, fi] = hi
(e) [ei, fj ] = 0 if i 6= j
(f) [ei, [ei, · · · [ei,︸ ︷︷ ︸1−Aij times
ej ] · · · ]] = 0 if i 6= j
(g) [fi, [fi, · · · [fi,︸ ︷︷ ︸1−Aij times
fj ] · · · ]] = 0 if i 6= j
Note that since Aij ≤ 0 for all i 6= j, the number 1−Aij is a positive integer.
Proof. Relation (a) follows from the fact that [h, h] = 0. For relation (b), we have
[hi, ej ] = 2
[h′αi
, ej]⟨
h′αi, h′αi
⟩ = 2αj
(h′αi
)⟨h′αi
, h′αi
⟩ej = 2
⟨h′αi
, h′αj
⟩⟨h′αi
, h′αi
⟩ ej = 2〈αi, αj〉〈αi, αi〉
ej = Aijej .
A similar argument proves relation (c). Relation (d) is just the definition of ei, fi.
Relation (e) holds because [ei, fi] ∈ gαi−αj and αi − αj is not a root when i 6= j, as
follows from Corollary 5.6. To prove relation (f), we consider the αi-chain of roots
through αj . Since −αi + αj is not a root, this chain is of the form
αj , αi + αj , . . . , qαi + αj .
By Proposition 4.21, we have Aij = −q. Thus (1 − Aij)αi + αj is not a root. Since
the element [ei, [ei, · · · [ei, ej ]]] lies in g(1−Aij)αi+αj, this element must be 0. A similar
argument proves relation (g).
7.4 The Lie algebras g(A) and g(A)
Let A be a Cartan matrix on the standard list in Section 6.4. Motivated by
Propositions 7.7 and 7.8, we now seek to construct a Lie algebra g(A) that will be
shown to be a finite-dimensional simple Lie algebra with Cartan matrix A.
106
Suppose A is an l × l matrix. Let F be the free associative algebra over C
on the 3l generators e1, . . . , el, h1, . . . , hl, f1, . . . , fl. The set of all monomials in these
generators form a basis of F. Let [F] be the Lie algebra formed from F via the
commutator product, and let L be the subalgebra of [F] generated by the elements
e1, . . . , el, h1, . . . , hl, f1, . . . , fl. Let v be the ideal of L generated by the elements
[hi, hj ]
[hi, ej ]−Aijej
[hi, fj ] +Aijfj
[ei, fi]− hi
[ei, fj ] for i 6= j
[ei, [ei, · · · [ei, ej ]]] for i 6= j
[fi, [fi, · · · [fi, fj ]]] for i 6= j
where the number of occurrences of ei, fi in the last two elements, respectively, is 1−Aij .
We define g(A) = L/v. We will eventually be able to show that g(A) is the
Lie algebra we require to prove the existence theorem. This construction of g(A) by
generators and relations is due to J.P. Serre.
In order to more easily study the Lie algebra g(A), we define a second, larger Lie
algebra g(A). Let v be the ideal of L generated by the elements
[hi, hj ]
[hi, ej ]−Aijej
[hi, fj ] +Aijfj
[ei, fi]− hi
[ei, fj ] for i 6= j.
Let g(A) = L/v. Since v ⊂ v, we have surjective Lie algebra homomorphisms
L → g(A) → g(A).
107
We will investigate the properties of the Lie algebra g(A). This is generated by the
images of the generators of L under the above homomorphism. For convenience, we
continue to write these images as e1, . . . , el, h1, . . . , hl, f1, . . . , fl. These elements satisfy
the relations
[hi, hj ] = 0
[hi, ej ] = Aijej
[hi, fj ] = −Aijfj
[ei, fi] = hi
[ei, fj ] = 0 for i 6= j.
Proposition 7.9. Let F− be the free associative algebra over C with generators f1, . . . , fl.
Then F− may be made into a g(A)-module inducing a representation ρ : g(A) → [EndF−]
defined by
ρ(fi)fi1 · · · fir = fifi1 · · · fir
ρ(hi)fi1 · · · fir = −
(r∑
k=1
Aiik
)fi1 · · · fir
ρ(ei)fi1 · · · fir = −r∑
k=1
δiik
(r∑
h=k+1
Aiih
)fi1 · · · fik · · · fir
where, as usual, the symbol fik means that fik is omitted from the product.
Proof. Since the monomials fi1 · · · fir form a basis of F−, the endomorphisms ρ(fi),
ρ(hi), ρ(ei) are uniquely determined by the above formulae. Thus there is a unique
homomorphism F → End F− mapping ei, hi, fi to ρ(ei), ρ(hi), ρ(fi), respectively. This
induces a Lie algebra homomorphism [F] → [End F−] and hence, by restriction, a Lie
algebra homomorphism L → [EndF−]. In order to obtain a Lie algebra homomorphism
g(A) → [End F−], we must verify the following relations:
(a) [ρ(hi), ρ(hj)] = 0
(b) [ρ(hi), ρ(ej)] = Aijρ(ej)
108
(c) [ρ(hi), ρ(fj)] = −Aijρ(fj)
(d) [ρ(ei), ρ(fi)] = ρ(hi)
(e) [ρ(ei), ρ(fj)] = 0 if i 6= j.
Relation (a) is automatic since ρ(hi) multiplies each basis element of F− by a scalar.
To prove relation (b), we have
ρ(hi)ρ(ej)fi1 · · · fir = −r∑
k=1
δjik
(r∑
h=k+1
Ajih
)−∑g 6=k
Aiig
fi1 · · · fik · · · fir ,
ρ(ej)ρ(hi)fi1 · · · fir = −r∑
k=1
δjik
(r∑
h=k+1
Ajih
)− r∑g=1
Aiig
fi1 · · · fik · · · fir .
Thus
(ρ(hi)ρ(ej)− ρ(ej)ρ(hi))fi1 · · · fir = −r∑
k=1
δjik
(r∑
h=k+1
Ajih
)Aiikfi1 · · · fik · · · fir
= Aij
(−
r∑k=1
δjik
(r∑
h=k+1
Ajih
))fi1 · · · fik · · · fir
= Aijρ(ej)fi1 · · · fir .
To prove relation (c), we have
ρ(hi)ρ(fj)fi1 · · · fir = −
(Aij +
r∑k=1
Aiik
)fjfi1 · · · fir ,
ρ(fj)ρ(hi)fi1 · · · fir = −
(r∑
k=1
Aiik
)fjfi1 · · · fir .
Thus
(ρ(hi)ρ(fj)− ρ(fj)ρ(hi))fi1 · · · fir = −Aijfjfi1 · · · fir = −Aijρ(fj)fi1 · · · fir .
Next, we prove relation (d). We have
ρ(ei)ρ(fi)fi1 · · · fir =−
(r∑
h=1
Aiih
)fi1 · · · fir
−r∑
k=1
δiik
(r∑
h=k+1
Aiih
)fifi1 · · · fik · · · fir ,
ρ(fi)ρ(ei)fi1 · · · fir =−r∑
k=1
δiik
(r∑
h=k+1
Aiih
)fifi1 · · · fik · · · fir .
109
Thus
(ρ(ei)ρ(fi)− ρ(fi)ρ(ei))fi1 · · · fir = −
(r∑
h=1
Aiih
)fifi1 · · · fir = ρ(hi)fi1 · · · fir .
Finally, we prove relation (e). Suppose i 6= j. Then
ρ(ei)ρ(fj)fi1 · · · fir = −r∑
k=1
δiik
(r∑
h=k+1
Aiih
)fjfi1 · · · fir
= ρ(fj)ρ(ei)fi1 · · · fir .
Thus ρ preserves all relations, and so ρ : g(A) → [End F−] is a homomorphism of Lie
algebras.
This homomorphism will allow us to deduce some useful information about the
Lie algebra g(A).
Proposition 7.10. The elements h1, . . . , hl of g(A) are linearly independent.
Proof. We will show that the elements ρ(h1), . . . , ρ(hl) of End F− are linearly indepen-
dent. We have
ρ(hi)fj = −Aijfj .
Thus if∑λiρ(hi) = 0, then we have
∑λiAij = 0 for all j = 1, . . . , l. Since the
Cartan matrix A = (Aij) is non-singular, This implies that λi = 0 for each i. Hence
ρ(h1), . . . , ρ(hl) are linearly independent, and so h1, . . . , hl must be linearly independent,
also.
Let h be the subspace of g(A) spanned by h1, . . . , hl. Then we have dim h = l.
We also have [h, h] = 0 by our relations, and so h is an abelian subalgebra of g(A).
We consider the weight spaces of g(A) with respect to h. We are no longer dealing
with finite-dimensional h-modules as in Theorem 2.10, but analogous ideas apply in
this situation. For each λ ∈ Hom(h,C), we define the set
g(A)λ = {x ∈ g(A) | [h, x] = λ(h)x for all h ∈ h}.
If g(A)λ 6= 0, then we call λ a weight, and we call g(A)λ the weight space of λ.
110
Proposition 7.11. g(A) =⊕
λ g(A)λ. Thus g(A) is the direct sum of its weight spaces.
Proof. We first show that g(A) =∑
λ g(A)λ. A vector that lies in a weight space will be
called a weight vector. Let x, y ∈ g(A) be weight vectors of weights λ, µ, respectively.
Then since
[h, [x, y]] = [[h, x], y] + [x, [h, y]] = λ(h)[x, y] + µ(h)[x, y]
= (λ+ µ)(h)[x, y] for all h ∈ h,
we see that [x, y] is a weight vector of weight λ+ µ.
Now g(A) is generated by the elements ei, hi, fi. Let αi ∈ Hom(h,C) be the map
defined by
αi(hj) = Aji.
One easily checks that ei is a weight vector of weight αi, fi is a weight vector of weight
−αi, and hi is a weight vector of weight 0. Thus all Lie products of the generators ei,
hi, fi are weight vectors. It follows that
g(A) =∑
λ
g(A)λ.
We next show that this is a direct sum. Suppose this is not the case. Then there exists
a λ and a nonzero x ∈ g(A)λ such that x =∑
µ xµ where xµ ∈ g(A)µ and µ runs over a
finite set of weights, all of which are distinct from λ.
Now the vector space h over the infinite field C cannot be expressed as the union
of finitely-many proper subspaces. One easily checks that for each µ, the set of h ∈ h
satisfying µ(h) = λ(h) is a proper subspace. Thus there exists an h ∈ h such that
µ(h) 6= λ(h) for all µ. Now since x ∈ g(A)λ, we have
(ad h− λ(h)1)x = 0.
And since x =∑
µ xµ with xµ ∈ g(A)µ, we have
∏µ
(ad h− µ(h)1)x = 0.
111
Because µ(h) 6= λ(h) for all µ, we see that the polynomials
t− λ(h),∏µ
(t− µ(h))
are coprime. Thus there exist polynomials a(t), b(t) ∈ C[t] such that
a(t)(t− λ(h)) + b(t)∏µ
(t− µ(h)) = 1.
It follows that
a(ad h)(ad h− λ(h)1)x+ b(ad h)∏µ
(ad h− µ(h)1)x = x.
The left-hand side of this equation is zero by the preceding results, and thus x = 0.
This is a contradiction, so the sum g(A) =∑
λ g(A)λ must be direct.
We next obtain some useful information about the weights of g(A). The weights
α1, . . . , αl ∈ Hom(h,C) are linearly independent since the Cartan matrix A is non-
singular. Because Hom(h,C) has the same dimension as h, we see that α1, . . . , αl form
a basis of Hom(h,C). Thus any weight is of the form n1α1 + · · · + nlαl with ni ∈ C.
Our goal is to show that all weights of g(A) have this form with ni ∈ Z and ni ≥ 0 for
all i or ni ≤ 0 for all i.
Let
Q = {n1α1 + · · ·+ nlαl | ni ∈ Z}
Q+ = {n1α1 + · · ·+ nlαl | ni ≥ 0 for all i}
Q− = {n1α1 + · · ·+ nlαl | ni ≤ 0 for all i}.
Let
g(A)+ =∑
λ∈Q+
g(A)λ
g(A)− =∑
λ∈Q−
g(A)λ.
It follows from Proposition 7.11 that the sum g(A)−+ h+ g(A)+ is direct. We will show
that, in fact,
g(A) = g(A)− ⊕ h⊕ g(A)+.
112
Let n be the subalgebra of g(A) generated by e1, . . . , el and n− the subalgebra
generated by f1, . . . , fl. Since ei has weight αi and fi has weight −αi, we see that
n ⊂ g(A)+ and n− ⊂ g(A)−. Thus the sum n + h + n− is direct.
Proposition 7.12.
(i) g(A) = n− ⊕ h⊕ n.
(ii) n = g(A)+, n− = g(A)−, and h = g(A)0.
(iii) Every nonzero weight of g(A) lies in Q+ or in Q−.
Proof. We first show that [hi, [[ei1 , ei2 ], · · · , eir ]] ∈ n for all products [[ei1 , ei2 ], · · · , eir ]
using induction on r. The relations [hi, ej ] = Aijej prove the statement for r = 1.
Now let r > 1 and write y = [[ei1 , ei2 ], · · · eir ], x = [[ei1 , ei2 ], · · · eir−1 ]. We assume by
induction that [hi, x] ∈ n. Then
[hi, y] = [hi, [x, eir ]] = [[hi, x], eir ] + [x, [hi, eir ]]
by the Jacobi identity. The right-hand side is a sum of elements in [n, n] ⊂ n, and so
[hi, y] ∈ n. This completes the induction. It follows that [h, n] ⊂ n since the elements hi
generate h and [h, h] = 0. Having this result, we see that
[h + n, h + n] = [h, h] + [h, n] + [n, n] ⊂ h + n.
Thus h + n is a subalgebra of g(A). Similarly, n− + h is a subalgebra of g(A).
Next, we show that [ei, [[fi1 , fi2 ], · · · fir ]] ∈ n−+h for all products [[fi1 , fi2 ], · · · fir ].
The relations [ei, fi] = hi and [ei, fj ] = 0 for i 6= j prove the statement for the generators
of n−, and an inductive argument similar to that given above proves that it is true for
all such products in n. It follows that [ei, n−] ⊂ n− + h. Similarly, [fi, n] ⊂ h + n. These
relations give us
[ei, n− + h + n] ⊂ n− + h + n
113
since h + n is a subalgebra of g(A),
[fi, n− + h + n] ⊂ n− + h + n
since n− + h is a subalgebra of g(A), and
[hi, n− + h + n] ⊂ n− + h + n
since n− + h and h + n are both subalgebras. It follows that the subspace consisting of
all x ∈ g(A) such that
[x, n− + h + n] ⊂ n− + h + n
contains the generators ei, hi, fi of g(A). However, the relation
[[x, y], z] = [[x, z], y] + [x, [y, z]]
for x, y in the subspace, z ∈ n− + h + n shows that this subspace is also a subalgebra.
Since it contains the generators of g(A), this subalgebra must be the whole of g(A).
Thus n− + h + n is an ideal of g(A). Since g(A) is generated by the elements ei, hi, fi,
it follows that n− + h + n = g(A). We know this sum is direct, so we have
g(A) = n− ⊕ h⊕ n.
Because n− ⊂ g(A)−, n ⊂ g(A)+, and the sum g(A)−+h+g(A)+ is direct, we deduce that
n− = g(A)− and n = g(A)+. Moreover, since g(A) = g(A)−⊕ h⊕ g(A)+, h ⊂ g(A)0, and
the weights occurring in g(A)− and g(A)+ are all nonzero, we deduce from Proposition
7.11 that h = g(A)0. Thus all parts of the proposition have been proved.
Proposition 7.13. dim g(A)αi = 1 and dim g(A)−αi = 1.
Proof. We know that ei ∈ g(A)αi . Also, the element ei ∈ g(A) is nonzero since it induces
a nonzero endomorphism ρ(ei) on the g(A)-module F− introduced in Proposition 7.9.
Hence dim g(A)αi ≥ 1. On the other hand, we have
g(A)αi ⊂ g(A)+ = n.
114
Now n is generated by e1, . . . , el and is thus spanned by monomials in these elements.
All such monomials are weight vectors. The only monomial having weight αi is ei since
the αi are linearly independent. Thus we have dim g(A)αi = 1. A similar argument
yields dim g(A)−αi = 1.
7.5 The existence theorem
We now turn to a study of the Lie algebra g(A) in order to show that it is a
finite-dimensional Lie algebra with Cartan matrix A. From the definitions of g(A) and
g(A), we see that g(A) is isomorphic to g(A)/u where u is the ideal of g(A) generated
by the elements
[ei, [ei, · · · [ei, ej ]]]
[fi, [fi, · · · [fi, fj ]]]
for all i 6= j. As usual, we have 1−Aij factors of ei and fi.
Proposition 7.14.
(i) If u+ is the ideal of n generated by the elements [ei, [ei, · · · [ei, ej ]]] for all i 6= j,
then u+ is an ideal of g(A).
(ii) If u− is the ideal of n− generated by the elements [fi, [fi, · · · [fi, fj ]]] for all i 6= j,
then u− is an ideal of g(A).
(iii) u = u+ ⊕ u−.
Proof. We write Xij = [ei, [ei, · · · [ei, ej ]]] and Yij = [fi, [fi, · · · [fi, fj ]]]. Then u+ is the
set of all linear combinations of elements
[[Xij , ek1 ], · · · ekr ]
for all i 6= j and ki, . . . , kr ∈ {1, . . . , l} since all such linear combinations lie in u+ and
form an ideal of n.
115
Now Xij is a weight vector, being the Lie product of the weight vectors ei, ej .
Similarly, each [[Xij , ek1 ], · · · ekr ] is a weight vector. It is therefore transformed by each
of the elements h1, . . . , hl into a scalar multiple of itself. In order to show that u+ is an
ideal of g(A), it is thus sufficient to show that
[fk, [[Xij , ek1 ], · · · ekr ]] ∈ u+
for all i, j, k, k1, . . . , kr. We prove this by induction on r, beginning with r = 0. In the
following lemma, we will show that [fk, Xij ] = 0, thus beginning the induction. Now
let r > 0 and write y = [[Xij , ek1 ], · · · ekr ], x = [[Xij , ek1 ], · · · ekr−1 ]. We assume by
induction that [fk, x] ∈ u+. Then
[fk, y] = [fk, [x, ekr ]] = [[fk, x], ekr ] + [x, [fk, ekr ]]
by the Jacobi identity. If kr 6= k, then [fk, y] = [[fk, x], ekr ] ∈ u+. On the other hand, if
kr = k, then we have
[fk, y] = [[fk, x], ekr ] + [hk, x] ∈ u+.
This completes the induction. Thus u+ is an ideal of g(A). Similarly, u− is an ideal of
g(A). It follows that u+ ⊕ u− is an ideal of g(A) containing the elements Xij and Yij .
Moreover, any ideal of g(A) containing the Xij and Yij must contain u+ and u−. Hence
u+ ⊕ u− = u.
To complete the proof of Proposition 7.14, we require the following lemma.
Lemma 7.15. [fk, Xij ] = 0 for all i, j, k with i 6= j.
Proof. Suppose k /∈ {i, j}. Then [fk, ei] = 0 and [fk, ej ] = 0 so that
[fk, [ei, ej ]] = [[fk, ei], ej ] + [ei, [fk, ej ]] = 0.
Using induction, we deduce that [fk, Xij ] = 0 for all i, j.
116
Now suppose k = j. Then
[fj , [ei, ej ]] = [[fj , ei], ej ] + [ei, [fj , ej ]] = [hj , ei] = Ajiei,
[fj , [ei, [ei, ej ]]] = [[fj , ei], [ei, ej ]] + [ei, [fj , [ei, ej ]]] = Aji[ei, ei] = 0,
[fj , [ei, [ei, · · · [ei,︸ ︷︷ ︸r times
ej ]]]] = 0 for r ≥ 2
by induction on r. Hence [fj , Xij ] = 0 if 1−Aij ≥ 2, that is, Aij ≤ −1. If 1−Aij = 1,
then Aji = 0, and so [fk, Xij ] = 0 in this case, also.
Finally, suppose k = i. We have
[fi, [ei, [ei, · · · [ei,︸ ︷︷ ︸r times
ej ]]]] = −[hi, [ei, [ei, · · · [ei,︸ ︷︷ ︸r − 1 times
ej ]]]] + [ei, [fi, [ei, [ei, · · · [ei,︸ ︷︷ ︸r − 1 times
ej ]]]]] (1)
by the Jacobi identity. I claim that
[ei, [fi, [ei, [ei, · · · [ei,︸ ︷︷ ︸s times
ej ]]]]] = −s(Aij + s− 1) [ei, [ei, · · · [ei,︸ ︷︷ ︸s times
ej ]]]
for all s ≥ 1. For s = 1, we have
[ei, [fi, [ei, ej ]]] = [ei, [[fi, ei], ej ] + [ei, [fi, ej ]]] = [ei, [−hi, ej ]] = −Aij [ei, ej ].
Now let s > 1. For convenience, we write
[ei, [ei, · · · [ei,︸ ︷︷ ︸t times
ej ]]] = Xtij
for all t ≥ 1. We assume by induction that[ei,[fi, X
s−1ij
]]= −(s−1)(Aij + s−2)Xs−1
ij .
Then[ei,[fi, X
sij
]]=[ei,[fi,[ei, X
s−1ij
]]]=[ei,[[fi, ei], Xs−1
ij
]+[ei,[fi, X
s−1ij
]]]=[ei,[[fi, ei], Xs−1
ij
]]+[ei,[ei,[fi, X
s−1ij
]]]=[ei,[−hi, X
s−1ij
]]+[ei,−(s− 1)(Aij + s− 2)Xs−1
ij
]= −
[[ei, hi], Xs−1
ij
]−[hi,[ei, X
s−1ij
]]− (s− 1)(Aij + s− 2)
[ei, X
s−1ij
]=[Aiiei, X
s−1ij
]−[hi, X
sij
]− (s− 1)(Aij + s− 2)Xs
ij
= 2Xsij −
[hi, X
sij
]− (s− 1)(Aij + s− 2)Xs
ij . (2)
117
One easily verifies by induction that[hi, X
tij
]= (Aij + 2t)Xt
ij for all t ≥ 1. Plugging
this into (2) for t = s gives us
[ei,[fi, X
sij
]]= −s(Aij + s− 1)Xs
ij
as desired. Inserting this result into (1) for s = r − 1 then yields
[fi, [ei, [ei, · · · [ei,︸ ︷︷ ︸r times
ej ]]]]
= −[hi, [ei, [ei, · · · [ei,︸ ︷︷ ︸r − 1 times
ej ]]]]− (r − 1)(Aij + r − 2) [ei, [ei, · · · [ei,︸ ︷︷ ︸r − 1 times
ej ]]]
= (−(Aij + 2r − 2)− (r − 1)(Aij + r − 2)) [ei, [ei, · · · [ei,︸ ︷︷ ︸r − 1 times
ej ]]]
= −r(Aij + r − 1) [ei, [ei, · · · [ei,︸ ︷︷ ︸r − 1 times
ej ]]].
We now set r = 1−Aij and obtain [fi, Xij ] = 0.
Corollary 7.16. g(A) = n− ⊕ h ⊕ n where h is isomorphic to h, n is isomorphic to
n−/u−, and n is isomorphic to n/u+.
Proof. This follows from the fact that g(A) is isomorphic to g(A)/u, g(A) = n−⊕ h⊕ n,
and u = u+ ⊕ u−.
We will continue to denote the generators of g(A) by ei, hi, fi. These are the
images of the generators of L under the natural homomorphism L → g(A).
Proposition 7.17. The maps ad ei : g(A) → g(A) and ad fi : g(A) → g(A) are locally
nilpotent.
Proof. To show that ad ei is locally nilpotent, we must show that for each x ∈ g(A),
there exists an integer n(x) ≥ 1 such that (ad ei)n(x)x = 0. Now if ad ei acts locally
nilpotently on x and y, then ad ei acts locally nilpotently on [x, y]. This follows from
the relation
(ad ei)n[x, y] =n∑
r=0
(n
r
)[(ad ei)rx, (ad ei)n−ry
].
118
We have (ad ei)rx = 0 if r is sufficiently large and (ad ei)n−ry = 0 if n− r is sufficiently
large. Thus (ad ei)n[x, y] = 0 if n is sufficiently large.
It follows that the set of elements in g(A) on which ad ei acts locally nilpotently
is a subalgebra of g(A). However, we have
ad ei · ei = 0
(ad ei)1−Aijej = 0 if i 6= j
(ad ei)2hj = 0 for all j
(ad ei)3fi = 0
ad ei · fj = 0 if i 6= j.
Thus this subalgebra contains all generators of g(A), so it must be the whole of g(A).
A similar argument shows that ad fi is locally nilpotent on g(A).
Now the proof of Proposition 3.4 shows that if D : g → g is a locally nilpotent
derivation of a Lie algebra g, then exp(D) is an automorphism of g. Thus exp(ad ei)
and exp(ad fi) are automorphisms of g(A). For each i, we define φi ∈ Aut(g(A)) by
φi = exp(ad ei) · exp(ad (−fi)) · exp(ad ei).
Proposition 7.18.
(i) φi(h) = h.
(ii) φi(h) = si(h) where si : h → h is the linear map defined by
si(hj) = hj −Ajihi.
Proof. We have
exp(ad ei) · hj = (1 + ad ei)hj = hj −Ajiei,
119
exp(ad (−fi)) · exp(ad ei) · hj = exp(ad (−fi)) · (hj −Ajiei)
=(
1− ad fi +(ad fi)2
2
)(hj −Ajiei)
= hj −Ajiei −Ajifi −Ajihi +Ajifi
= hj −Ajihi −Ajiei,
exp(ad ei) · exp(ad (−fi)) · exp(ad ei) · hj = exp(ad ei)(hj −Ajihi −Ajiei)
= (1 + ad ei)(hj −Ajihi −Ajiei)
= hj −Ajihi −Ajiei −Ajiei + 2Ajiei
= hj −Ajihi.
Thus φi restricted to the subalgebra h is the map si described in the second statement.
Since φi(h) ⊂ h and φi is an automorphism of g(A), we must have φi(h) = h.
Now the action of si on h is the same as that of the fundamental reflection si = sαi
defined in Section 6.4. We recall that
si(h) = h− 2
⟨h′αi
, h⟩⟨
h′αi, h′αi
⟩h′αifor h ∈ h
= h−⟨h′αi
, h⟩hi.
In particular, we have
si(hj) = hj −⟨h′αi
, hj
⟩hi = hj − 2
⟨h′αi
, h′αj
⟩⟨h′αj
, h′αj
⟩hi = hj −Ajihi.
Thus Proposition 7.18 shows that the automorphism φi induces the fundamental reflec-
tion si on h.
We now consider the decomposition of g(A) into weight spaces with respect to h.
This time, the weights will be the elements λ ∈ Hom(h,C) for which the set
g(A)λ = {x ∈ g(A) | [h, x] = λ(h)x for all h ∈ h}
is nonzero.
Proposition 7.19. g(A) =⊕
λ g(A)λ.
120
Proof. The Lie algebra g(A) is the sum of its weight spaces since its generators ei, hi,
fi are weight vectors. Moreover, the sum of weight spaces is direct, just as in the proof
of Proposition 7.11.
It also follows from Proposition 7.12 that g(A) = n− ⊕ h ⊕ n where all weights
coming from n are in Q+ and all weights coming from n− are in Q−.
Proposition 7.20. dim g(A)αi = 1 and dim g(A)−αi = 1.
Proof. By Proposition 7.13, we know that dim g(A)αi ≤ 1. Now the ideal u+ of n
such that n ∼= n/u+ has the property that u+ is a sum of weight spaces, and all weights
occurring in u+ are sums of α1, . . . , αl involving at least two terms. This can be deduced
from the proof of Proposition 7.14. Thus αi is not a weight of u+. Hence
dim g(A)αi = dim g(A)αi = 1.
A similar argument shows that dim g(A)−αi = 1.
Proposition 7.21. The automorphism φi of g(A) transforms g(A)λ to g(A)siλ. Hence
dim g(A)λ = dim g(A)siλ.
Proof. Let x ∈ g(A)λ. Then [h, x] = λ(h)x for all h ∈ h. We now apply the automor-
phism φi. This fixes h by Proposition 7.18. We thus have
[φih, φix] = λ(h)φix.
Hence
[h, φix] = λ(φ−1
i h)φix = λ
(s−1i h
)φix = (siλ(h))φix,
again by Proposition 7.18. Thus we have φix ∈ g(A)siλ. Hence
φi(g(A)λ) ⊂ g(A)siλ.
Replacing φi by φ−1i , λ by siλ, and recalling that s2i = 1, we also have
φ−1i (g(A)siλ) ⊂ g(A)λ.
121
Hence g(A)siλ ⊂ φi(g(A)λ), giving us φi(g(A)λ) = g(A)siλ.
We now define W to be the group of non-singular transformations on h∗ =
Hom(h,C) generated by s1, . . . , sl and define Φ to be the set of elements w(αi) for
w ∈ W and i = 1, . . . , l. Then Φ is the root system determined by the given Cartan
matrix A and W is the Weyl group.
Proposition 7.22. dim g(A)α = 1 for all α ∈ Φ.
Proof. We have α = w(αi) for some i and some w ∈ W . Since W is generated by
s1, . . . , sl, w is a product of such elements. It follows from Proposition 7.21 that
dim g(A)α = dim g(A)αi = 1.
Our goal is to show that g(A) is finite-dimensional. As a step in this direction,
we will show that the Weyl group W is finite. We know that W is isomorphic to
the group of non-singular linear transformations on hR generated by s1, . . . , sl where
hR = Rh1 + · · · + Rhl. We have dim hR = l. We do not have a scalar product on hR
available from the Killing form, so we define a scalar product directly from the Cartan
matrix A.
Proposition 7.23. The Cartan matrix can be factorized as A = DB where D is di-
agonal and B is symmetric. Moreover, D is the diagonal matrix with entries d1, . . . , dl
defined as follows:
• If the Dynkin diagram has only single edges, then di = 1 for all i.
• If the Dynkin diagram has a double edge, then di = 1 if αi is a short root and
di = 2 if αi is a long root.
• If the Dynkin diagram has a triple edge, then di = 1 if αi is a short root and
di = 3 if αi is a long root.
122
Proof. This can be checked from the standard list of Cartan matrices in Section 6.4.
For example, for the matrix of type G2, we have 2 −1
−3 2
=
1 0
0 3
2 −1
−1 23
.
We now define a bilinear form on hR by 〈hi, hj〉 = didjBij . This form is symmetric
since B is a symmetric matrix.
Proposition 7.24. This scalar product is positive definite.
Proof. We have nij = AijAji = didjB2ij . Thus −
√nij =
√di
√djBij . The matrix of
our scalar product is
DBD =
√d1 0
·
·
·
0√dl
2 −√nij
·
·
·
−√nij 2
√d1 0
·
·
·
0√dl
.
This matrix is congruent to the matrix
2 −√nij
·
·
·
−√nij 2
of the quadratic form Q(x1, . . . , xl) of Proposition 6.6, which is positive definite. Thus
DBD is also positive definite.
Proposition 7.25. Our scalar product on hR is invariant under W .
Proof. We first observe that 〈hi, x〉 = diαi(x) for all h ∈ hR. This follows from the fact
that
〈hi, hj〉 = didjBij = diAji = diαi(hj)
123
for all hi, hj .
To prove the proposition, it suffices to show that 〈six, siy〉 = 〈x, y〉 for all x, y ∈ hR
since the reflections si generate W . We note that si(x) = x − αi(x)hi since si(hj) =
hj −Ajihi. Thus
〈six, siy〉 = 〈x− αi(x)hi, y − αi(y)hi〉
= 〈x, y〉 − αi(y)〈hi, x〉 − αi(x)〈hi, y〉+ αi(x)αi(y)〈hi, hi〉
= 〈x, y〉 − diαi(y)αi(x)− diαi(x)αi(y) + 2diαi(x)αi(y)
= 〈x, y〉.
Thus the Weyl group acts as a group of isometries on the Euclidean space hR.
We now define certain subsets of hR in terms of the scalar product 〈, 〉 as follows:
hi = {x ∈ hR | 〈hi, x〉 = 0}
h+i = {x ∈ hR | 〈hi, x〉 > 0}
h−i = {x ∈ hR | 〈hi, x〉 < 0}
C = h+1 ∩ · · · ∩ h+
l .
The set C is called the fundamental chamber.
Let Wij be the subgroup of W generated by si, sj where i 6= j. I claim that each
element sisj has finite order mij given in terms of the Cartan matrix by 2 cos(π/mij) =
√nij . To prove this, recall that
nij = AijAji = 4 cos2 θ
where θ is the angle between αi, αj . It follows that
si(αj) = αj − 2〈αi, αj〉〈αi, αi〉
αi = αj − 2 cos θ|αj ||αi|
αi = αj −√nij
|αj ||αi|
αi.
Now consider the map s′i defined by
s′i(αj) = αj − 2 cos(π/mij)|αj ||αi|
αi.
124
For i = j, we have mii = 1 so that s′i(αi) = −αi. If αj is orthogonal to αi under our
inner product, then an interesting calculation shows that mij = 2 so that si(αj) = αj .
It follows that s′i is an orthogonal reflection about a hyperplane that intersects the
origin. Thus s′i is a linear transformation. We see, in fact, that s′i is the reflection in
the hyperplane orthogonal to αi. It follows that s′i = si. Hence for all i, j, we have
αj − 2 cos(π/mij)|αj ||αi|
αi = αj −√nij
|αj ||αi|
αi.
Thus 2 cos(π/mij) = √nij as desired. This shows that Wij is a finite dihedral group.
Lemma 7.26. Let w ∈Wij with i 6= j. Then either w(h+
i ∩h+j
)⊂ h+
i , or w(h+
i ∩h+j
)⊂
h−i and l(siw) = l(w)− 1.
Proof. Let U be the 2-dimensional subspace of hR spanned by hi, hj , and let U⊥ be the
orthogonal subspace. Then hR = U ⊕ U⊥ and the elements in Wij act trivially on U⊥.
It is therefore sufficient to prove the result in U . Let Γ = U ∩ h+i ∩ h+
j . Because U is
2-dimensional, we obtain the following configuration of chambers in U :
hi
si(Γ)hj
Γ
sj(Γ)sisj(Γ)
sjsi(Γ)
125
From this diagram, we see that the chambers
Γ, sj(Γ), sjsi(Γ), . . . , sjsi · · ·︸ ︷︷ ︸mij − 1 times
(Γ)
all lie in h+i , whereas
si(Γ), sisj(Γ), . . . , sisj · · ·︸ ︷︷ ︸mij times
(Γ)
all lie in h−i . The elements
si, sisj , . . . , sisj · · ·︸ ︷︷ ︸mij times
all satisfy l(siw) = l(w) − 1 since s2i = 1. Thus for each w ∈ Wij , we have either
w(h+
i ∩ h+j
)⊂ h+
i , or w(h+
i ∩ h+j
)⊂ h−i and l(siw) = l(w)− 1.
Proposition 7.27.
(i) Let w ∈W . Then either w(C) ⊂ h+i , or w(C) ⊂ h−i and l(siw) = l(w)− 1.
(ii) Let w ∈ W and i 6= j. Then there exists a w′ ∈ Wij such that w(C) ⊂
w′(h+i ∩ h+
j
)and l(w) = l (w′) + l
(w′−1w
).
Note. Part (i) is the result we will need. To prove it, we must also prove part (ii) at
the same time.
Proof. We prove both statements together by induction on l(w). If l(w) = 0, then
w = 1, and so (i) and (ii) are true with w′ = 1. So suppose l(w) > 0. Then w = sjw′
with l (w′) = l(w)− 1 for some j and w′ ∈W . We prove part (i).
First suppose j = i. By induction, we have w′(C) ⊂ h+i , or w′(C) ⊂ h−i and
l (siw′) = l (w′)− 1. But l (siw
′) = l (w′) + 1, and so w′(C) ⊂ h+i . Thus w(C) ⊂ h−i and
l(siw) = l(w)− 1.
Now suppose j 6= i. By induction, there exists a w′′ ∈ Wij such that w′(C) ⊂
w′′(h+i ∩h+
j
)and l (w′) = l (w′′)+ l
(w′′−1w′). Thus w(C) ⊂ sjw
′′(h+i ∩h+
j
). By Lemma
126
7.26, we have either sjw′′(h+
i ∩ h+j
)⊂ h+
i , or sjw′′(h+
i ∩ h+j
)⊂ h−i and l (sisjw
′′) =
l (sjw′′) − 1. In the first case, we have w(C) ⊂ h+
i . In the second case, we have
w(C) ⊂ h−i and
l(siw) = l(sisjw
′) = l(sisjw
′′w′′−1w′)≤ l(sisjw
′′)+ l(w′′−1w′)
= l(sjw
′′)− 1 + l(w′′−1w′)
≤ l(w′′)+ l
(w′′−1w′) = l
(w′) = l(w)− 1.
Thus l(siw) = l(w)− 1, and (i) is proved.
We now prove (ii). If w(C) ⊂ h+i ∩h+
j , then (ii) is true with w′ = 1. Without loss of
generality, we now assume that w(C) 6⊂ h+i . Hence by (i), which is now proved under the
assumption of the inductive hypothesis, w(C) ⊂ h−i and l(siw) = l(w)−1. By induction,
there exists a w′ ∈Wij such that siw(C) ⊂ w′(h+i ∩h+
j
)and l(siw) = l (w′)+l
(w′−1siw
).
Thus w(C) = sisiw(C) ⊂ siw′(h+
i ∩ h+j
)and
l(w) = 1 + l(siw) = 1 + l(w′)+ l
(w′−1siw
)≥ l(siw
′)+ l((siw
′)−1w)≥ l(w).
Thus we have equality throughout, and so l(w) = l (siw′) + l
((siw
′)−1w). Hence
siw′ ∈Wij is the required element, and (ii) is proved.
Proposition 7.28. If w ∈W satisfies C ∩ w(C) 6= ∅, then w = 1.
Proof. We prove the contrapositive. Suppose w 6= 1. Then w = siw′ with l(w′) =
l(w) − 1 for some i and w′ ∈ W . By Proposition 7.27(i), we have w′(C) ⊂ h+i . Thus
w(C) ⊂ h−i , and so
C ∩ w(C) ⊂ h+i ∩ h−i = ∅.
Now the Euclidean space hR has an orthonormal basis, and the isometries of hR
are represented by orthogonal matrices with respect to this basis. Thus W ⊂ Ol where
Ol is the group of l × l orthogonal matrices. We have Ol ⊂ Ml, the set of all l × l
matrices over R.
127
For any matrix M = (mij) ∈Ml, we define ‖M‖ =√∑
i,j m2ij . Similarly, for any
column vector v = (λ1, . . . , λl)t ∈ Rl, we define ‖v‖ =√∑
i λ2i .
Lemma 7.29.
(i) If M ∈ Ol, v ∈ Rl, then ‖Mv‖ = ‖v‖.
(ii) If M ∈ Ol, N ∈Ml, then ‖MN‖ = ‖N‖.
(iii) If M ∈Ml, v ∈ Rl, then ‖Mv‖ ≤ ‖M‖ ‖v‖.
Proof. Parts (i) and (ii) are straightforward calculations using the fact that the columns
of M all have length 1. Part (iii) follows from a similarly easy calculation.
Proposition 7.30.
(i) W is finite.
(ii) Φ is finite.
Proof. Since Φ = W (Π) where Π = {α1, . . . , αl}, it is clear that (i) implies (ii). Thus we
show that W is finite. We consider the W -action on the Euclidean space hR. We assign
coordinates to elements of hR relative to our orthonormal basis. Let v = (λ1, . . . , λl)t ∈
C. By the definition of C, there exists an r > 0 such that Br(v) ⊂ C where
Br(v) = {x ∈ Rl | ‖x− v‖ < r}.
Let w ∈W with w 6= 1. Then C ∩w(C) = ∅ by Proposition 7.28. Thus w(v) /∈ C, and
so
‖w(v)− v‖ ≥ r.
Hence
‖w − 1‖ ‖v‖ ≥ ‖w(v)− v‖ ≥ r
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by Lemma 7.29, yielding ‖w − 1‖ ≥ r‖v‖ . Let ε = r
‖v‖ . Then ‖w − 1‖ ≥ ε for all w 6= 1
in W . Now let w,w′ ∈W such that w 6= w′. Then
∥∥w − w′∥∥ =∥∥w′ (w′−1w − 1
)∥∥ =∥∥w′−1w − 1
∥∥ ≥ ε
since w′ ∈ Ol. Thus distinct elements of W are separated by a distance of at least ε.
Since Ol, and hence W , is bounded, it follows that W is finite.
We now return to our Lie algebra g(A). We know that dim g(A)0 = l and
dim g(A)α = 1 for all α ∈ Φ. If we can prove that every weight of g(A) lies in Φ ∪ {0},
then we will be able to deduce that g(A) is finite-dimensional.
Proposition 7.31. Every nonzero weight of g(A) lies in Φ.
Proof. Let λ be a nonzero weight of g(A). Since dim g(A)λ ≤ dim g(A)λ, we see from
Proposition 7.12(iii) that λ ∈ Q+ or λ ∈ Q−. In particular, λ lies in the vector space
h∗R of real linear combinations of α1, . . . , αl.
Suppose first that λ is a multiple of a root α ∈ Φ. Then λ = nα or λ = −nα
for some n > 0 and α ∈ Φ+. We know that α = w(αi) for some w ∈ W and αi ∈ Π,
and we have dim g(A)nα = dim g(A)nαi by Proposition 7.21. Hence g(A)nαi 6= 0. Now
n is generated by the elements e1, . . . , el, and no nonzero Lie product of these can have
weight nαi unless n = 1. Thus λ = α or λ = −α, that is, λ ∈ Φ.
Now suppose λ is not a multiple of a root. Let
hλ = {h ∈ hR | λ(h) = 0}
hα = {h ∈ hR | α(h) = 0}.
Then hλ is distinct from all the hα, α ∈ Φ. Thus hλ\⋃
α∈Φ(hλ ∩ hα) 6= ∅ since Φ is
finite and the vector space hλ over R cannot be expressed as the union of finitely-many
proper subspaces. Hence there exists an h ∈ hλ such that h /∈ hα for all α ∈ Φ. Then
w(h) /∈ hα for all α ∈ Φ since W permutes the hα.
129
I claim there exists a w ∈W such that αi(w(h)) > 0 for all i = 1, . . . , l. To prove
this, we define the height of an element of hR by
ht(∑
nihi
)=∑
ni.
We choose an element w ∈W such that htw(h) is maximum. This is possible since W
is finite. Then
si(w(h)) = w(h)− αi(w(h))hi.
Since ht si(w(h)) ≤ htw(h), we have αi(w(h)) ≥ 0. But αi(w(h)) = 0 would imply
w(h) ∈ hαi , a contradiction. Thus αi(w(h)) > 0 for all i, and so w(h) ∈ C.
Now we have (w(λ))(w(h)) = λ(h) = 0 by the way we defined the W -action on h
in Section 6.4. We write w(λ) =∑l
i=1miαi. Then
l∑i=1
miαi(w(h)) = 0.
Since αi(w(h)) > 0 for all i, we must have some mi > 0 and some mj < 0 in this
sum. Thus w(λ) /∈ Q+ and w(λ) /∈ Q−. Hence g(A)w(λ) = 0. By Proposition 7.21, this
implies that g(A)λ = 0. Thus g(A)λ = 0, a contradiction.
Corollary 7.32.
(i) g(A) = h⊕∑
α∈Φ g(A)α.
(ii) dim g(A) = l + |Φ|.
Proof. Part (i) is clear since g(A) is the direct sum of its weight spaces. The 0-weight
space is h, and this has dimension l. The only nonzero weights are those belonging to
Φ, and the corresponding weight spaces are 1-dimensional by Proposition 7.22. Thus
(ii) is the correct formula for the dimension of g(A).
We have thus shown that g(A) is a finite-dimensional Lie algebra – indeed, it has
the dimension required for a simple Lie algebra with Cartan matrix A. We next show
that g(A) has the other required properties.
130
Proposition 7.33. The Lie algebra g(A) is semisimple.
Proof. Let r be the solvable radical of g(A) and consider its derived series
r = r(0) ⊃ r(1) ⊃ · · · ⊃ r(n−1) ⊃ r(n) = 0
where, as usual, r(i+1) =[r(i), r(i)
]. Let a = r(n−1). Suppose r 6= 0. Then a is a nonzero
abelian ideal of g(A). We know that r is unique, and hence it is invariant under the
automorphisms of g(A). Thus [r, r] is invariant under the automorphisms of g(A), and
an inductive argument shows that r(i) is invariant under these automorphisms for all i.
Thus a is invariant under all automorphisms of g(A).
Since [h, a] ⊂ a, we may regard a as an h-submodule. We decompose a into its
weight spaces with respect to h. This weight space decomposition is
a = (h ∩ a)⊕∑α∈Φ
(g(A)α ∩ a).
For let x ∈ a with x = x0 +∑
α∈Φ xα where x0 ∈ h and xα ∈ g(A)α. We show that
x0 ∈ a and each xα ∈ a. Now the vector space h over the infinite field C cannot be
expressed as the union of finitely-many proper subspaces. Fixing α, one easily checks
that for each β ∈ Φ with β 6= α, the set of h ∈ h satisfying β(h) = α(h) is a proper
subspace. We also know that kerα is a proper subspace of h. Thus there exists an h ∈ h
such that α(h) 6= 0 and β(h) 6= α(h) for all β ∈ Φ with β 6= α. Then
ad h∏β∈Φβ 6=α
(ad h− β(h)1)x = α(h)∏β∈Φβ 6=α
(α(h)− β(h))xα.
The left-hand side is in a since a is invariant under ad h. Hence xα ∈ a. Since this is
true for all α ∈ Φ, we also have x0 ∈ a. It follows that
a = (h ∩ a)⊕∑α∈Φ
(g(A)α ∩ a).
I claim that g(A)α ∩ a = 0 for each α ∈ Φ. For otherwise we would have g(A)α ⊂ a.
Now α = w(αi) for some w ∈ W and i ∈ {1, . . . , l}. By Proposition 7.21, we can find
131
an automorphism of g(A) that transforms g(A)α into g(A)αi . Since a is invariant under
all automorphisms, we would have g(A)αi ⊂ a. Hence ei ∈ a. But then [ei, fi] = hi ∈ a,
and we would have [hi, ei] = 2ei 6= 0, contradicting the fact that a is abelian. Thus
g(A)α ∩ a = 0 for all α ∈ Φ, and so a ⊂ h. Now let x ∈ a. Then for i = 1, . . . , l, we have
[x, ei] = αi(x)ei ∈ a, yielding αi(x) = 0. Since α1, . . . , αl span h∗, we have λ(x) = 0 for
all λ ∈ h∗. Thus x = 0. It follows that a = 0, a contradiction.
Proposition 7.34. h is a Cartan subalgebra of g(A).
Proof. Since we already know that h is abelian, it suffices to show that h = N(h). Let
x ∈ N(h). Then x = h′ +∑
α∈Φ λαeα for h′ ∈ h, eα ∈ g(A)α. Then for all h ∈ h, we
have
[h, x] =∑α∈Φ
λαα(h)eα ∈ h.
But we can find an h ∈ h such that α(h) 6= 0 for all α ∈ Φ. We deduce that λα = 0 for
all α ∈ Φ, and hence x ∈ h.
Proposition 7.35. g(A) is a simple Lie algebra with Cartan matrix A.
Proof. The Cartan decomposition of g(A) with respect to h is
g(A) = h⊕∑α∈Φ
g(A)α.
Thus Φ is the root system of g(A). The Cartan matrix A′ =(A′
ij
)of g(A) is determined
by the relations
si(αj) = αj −A′ijαi.
But we also have
si(hj) = hj −Ajihi by Proposition 7.18.
Thus for any k = 1, . . . , l, we have
αj(sihk) = αj(hk)−Akiαj(hi) = αj(hk)−Aijαi(hk) = (αj −Aijαi)hk.
132
Since αj(sihk) = (siαj)hk, we deduce that
si(αj) = αj −Aijαi.
Hence A′ = A, and so the Cartan matrix of g(A) is A.
Since the Dynkin diagram that determined A was assumed to be connected, g(A)
must be a simple Lie algebra by Proposition 6.11.
Thus for each Cartan matrix on the standard list in Section 6.4, we have con-
structed a finite-dimensional simple Lie algebra g(A) with Cartan matrix A. We sum-
marize our results in the final theorem.
Main Theorem. The finite-dimensional non-trivial simple Lie algebras over C are
Al l ≥ 1
Bl l ≥ 2
Cl l ≥ 3
Dl l ≥ 4
E6, E7, E8
F4
G3
These Lie algebras are pairwise non-isomorphic.
Proof. For each Cartan matrix on the standard list in Section 6.4, there exists a finite-
dimensional simple Lie algebra which, by Theorem 7.5, is determined up to isomor-
phism. Simple Lie algebras with different Cartan matrices cannot be isomorphic since,
by Proposition 6.4, the Cartan matrix on the standard list is uniquely determined by
the Lie algebra.
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