§ 2.
Ideals and homomorphisms
7
§2. Ideals and homomorphisms
§2.1. Ideals
Definition 2.1 (Ideal). A subspace I of a Lie algebra L is called an ideal of L if x ∈ L, y ∈ I
together imply [x, y] ∈ I.
Since [x, y] = −[y, x], the condition could just as well be written: [y, x] ∈ I. Ideals play the
role in Lie algebra theory which is played by normal subgroups in group theory and by two sided
ideals in ring theory: they arise as kernels of homomorphisms §2.2.
Examples 2.2. Obviously 0 (the subspace consisting only of the zero vector) and L itself are
ideals of L.
(1). (center) A less trivial example is the center
Z(L) = {z ∈ L | [x, z] = 0 for all x ∈ L}.
(2.1)
Notice that L is abelian if and only if Z(L) = L.
(2). (derived algebra) Another important example is the derived algebra of L, denoted [L, L],
which is analogous to the commutator subgroup of a group. It consists of all linear combinations of commutators [x, y], and is clearly an ideal.
(3). (sum) If I, J are two ideals of a Lie algebra L, then I + J = {x + y | x ∈ I, y ∈ I} is also
an ideal.
(4). (bracket) Similarly, [I, J] = {
m
P
[xi , yi ] | xi ∈ I, yi ∈ J} is an ideal; the derived algebra
i=1
[L, L] is just a special case of this construction.
Evidently L is abelian if and only if [L, L] = 0. At the other extreme, a study of the
multiplication table for L = sln (F) in (§1.2) (n 6= 2 if char F = 2) shows that L = [L, L] in this
case, and similarly for other classical linear Lie algebras (Exercise 1.9).
Definition 2.3 (Perfect Lie algebra). If L = [L, L], then we call L a perfect Lie algebra.
It is natural to analyze the structure of a Lie algebra by looking at its ideals.
Definition 2.4 (Simple Lie algebra). A Lie algebra L is called simple if L has no ideals
except itself and 0, and if moreover [L, L] 6= 0.
The condition [L, L] 6= 0 (i.e., L nonabelian) is imposed in order to avoid giving undue
prominence to the one dimensional algebra. Clearly, L simple implies Z(L) = 0 and L = [L, L].
Examples 2.5. Let L = sl2 (F), char F 6= 2. Take as standard basis for L the three matrices
(cf. §1.2):
0 1
0 0
1 0
x=
, y=
, h=
.
(2.2)
0 0
1 0
0 −1
The multiplication table is then completely determined by the equations:
[x, y] = h, [h, x] = 2x, [h, y] = −2y.
(2.3)
8
Lie algebras
(Notice that x, y, h are eigenvectors for ad h, corresponding to the eigenvalues 2, −2, 0. Since
char F 6= 2, these eigenvalues are distinct.) If I 6= 0 is an ideal of L, let ax + by + ch be an
arbitrary nonzero element of I. Applying ad x twice, we get −2bx ∈ I, and applying ad y twice,
we get −2ay ∈ I. Therefore, if a or b is nonzero, I contains either y or x (char F 6= 2), and then,
clearly, I = L follows. On the other hand, if a = b = 0, then 0 = ch ∈ I, so h ∈ I, and again
I = L follows. We conclude that L is simple. (If I 6= 0 is an ideal of L, let ax + by + ch be an
arbitrary nonzero element of I. Applying ad h, then we get 2ax − 2by ∈ I and 4ax + 4by ∈ I.
Since char F 6= 2, then ax, by, ch ∈ I. Hence we get x ∈ I, y ∈ I or h ∈ I.)
In case a Lie algebra L is not simple (and not one dimensional) it is possible to “factor out”
a nonzero proper ideal I and thereby obtain a Lie algebra of smaller dimension.
Definition 2.6 (Quotient algebra). The construction of a quotient algebra L/I for an ideal
I of L is formally the same as the construction of a quotient ring: as vector space L/I is just
the quotient space, while its Lie multiplication is defined by [x + I, y + I] , [x, y] + I.
This is unambiguous, since if x + I = x0 + I, y + I = y 0 + I, then we have x0 = x + u
for u ∈ I, y 0 = y + v for v ∈ I, whence [x0 , y 0 ] = [x, y] + ([u, y] + [x, v] + [u, v], and therefore
[x0 , y 0 ] + I = [x, y] + I, since the terms in parentheses all lie in I.
Definition 2.7 (Direct product). If L and L0 are Lie algebras over F, then we can define a
Lie algebra structure on the direct sum L ⊕ L0 as
[x + x0 , y + y 0 ] = [x, y] + [x0 , y 0 ] for x, y ∈ L, x0 , y 0 ∈ L0 .
The Lie algebra L ⊕ L0 is called the direct product of L and L0 .
♣. For L = gl2 (F), it is easy to know that FI2 L and sl2 (F) L. Moreover, L/FI2 ∼
= sl2 (F)
∼
and L/sl2 (F) = FI2 . Indeed we have gl2 (F) = FI2 ⊕ sl2 (F).
For later use we mention a couple of related notions, analogous to those which arise in group
theory.
Definition 2.8 (Normalizer, self-normalizing, centralizer).
(1). The normalizer of a subalgebra (or just subspace) K of L is defined by NL (K) = {x ∈
L | [x, K] ⊆ K}. By the Jacobi identity, NL (K) is a subalgebra of L; it may be described
verbally as the largest subalgebra of L which includes K as an ideal (in case K is a
subalgebra to begin with). If K = NL (K), we call K self-normalizing; some important
examples of this behavior will emerge later.
(2). The centralizer of a subset X of L is CL (X) = {x ∈ L | [x, X] = 0}. Again by the Jacobi
identity, CL (X) is a subalgebra of L. For example, CL (L) = Z(L).
♣. For L = sl2 (F) = Fx ⊕ Fy ⊕ Fh,
(1).
(2).
(3).
(4).
if
if
if
if
K
K
K
K
= Fx, then NL (K) = Fx ⊕ Fh, CL (K) = K = Fx;
= Fx ⊕ Fh, then NL (K) = Fx ⊕ Fh, CL (K) = 0;
= Fx ⊕ Fy, then NL (K) = Fh, CL (K) = 0;
= Fh, then NL (K) = CL (K) = K = Fh.
♣. For L = Fx ⊕ Fy with [x, y] = y,
(1). if K = Fx, then NL (K) = Fx, CL (K) = K = Fx;
(2). if K = Fy, then NL (K) = L, CL (K) = Fy = K.
§ 2.
Ideals and homomorphisms
9
§2.2. Homomorphisms and representations
Lecture 3. March 9, 2016
The definition should come as no surprise.
Definition 2.9 (Homomorphism). A linear transformation ϕ : L → L0 (L, L0 Lie algebras
over F) is called a homomorphism if ϕ([x, y]) = [ϕ(x), ϕ(y)], for all x, y ∈ L. ϕ is called a
monomorphism if Ker(ϕ) = 0, an epimorphism if Im(ϕ) = L0 , an isomorphism (as in §1.1)
if it is both monomorphism and epimorphism.
Remark 2.10. ♣. The first interesting observation to make is that Kerϕ is an ideal of L:
indeed, if ϕ(x) = 0, and any y ∈ L, then ϕ([x, y]) = [ϕ(x), ϕ(y)] = 0. It is also apparent
that Im(ϕ) is a subalgebra of L0 .
♣. As in other algebraic theories, there is a natural 1-1 correspondence between homomorphisms and ideals: to ϕ is associated Ker(ϕ), and to an ideal I is associated the canonical
map x 7→ x + I of L onto L/I.
It is left as an easy exercise to verify the standard homomorphism theorems:
Proposition 2.11. (a). If ϕ : L → L0 is a homomorphism of Lie algebras, then L/Ker(ϕ) ∼
=
Im(ϕ). If I is any ideal of L included in Ker(ϕ), there exists a unique homomorphism
ψ : L/I → L0 making the following diagram commute (π = the canonical map):
ϕ
L
π
/ L0
O
ψ
L/I
(b). If I and J are ideals of L such that I ⊆ J, then J/I is an ideal of L/I and (L/I)/(J/I) is
naturally isomorphic to L/J.
(c). If I, J are ideals of L, there is a natural isomorphism between (I + J)/J and I/(I ∩ J).
Definition 2.12 (representation). A representation of a Lie algebra L is a homomorphism
ϕ : L → gl(V ) (V = vector space over F). Sometimes we say (V, ϕ) is a representation of L.
Although we require L to be finite dimensional, it is useful to allow V to be of arbitrary
dimension: gl(V ) makes sense in any case.
Examples 2.13. However, for the time being the only important example to keep in mind is
the adjoint representation ad : L → gl(L) introduced in §1.3, which sends x to ad x, where
ad x(y) = [x, y]. (The image of ad is in Der(L) ⊆ gl(L), but this does not concern us at the
moment.) It is clear that ad is a linear transformation. To see that it preserves the bracket, we
calculate:
[ad x, ad y](z) =ad x ad y(z) − ad y ad x(z)
=ad x([y, z]) − ad y([x, z])
=[x, [y, z]] + [y, [z, x]]
= − [z, [x, y]] = ad [x, y](z),
i.e., [ad x, ad y] = ad [x, y].
10
Lie algebras
♣. What is the kernel of ad ? It consists of all x ∈ L for which ad x = 0, i.e., for which
[x, y] = 0 all y ∈ L. So Ker(ad ) = Z(L).
This already has an interesting consequence:
Proposition 2.14. Any simple Lie algebra is isomorphic to a linear Lie algebra.
Proof. If L is simple, then Z(L) = 0, so that ad : L → gl(L) is a monomorphism. Hence L is
isomorphic to a subalgebra of gl(L).
Remark 2.15. Denote ad (L) the set of all inner derivations of L, then ad (L) is a subspace of
Der(L). Moreover, ad (L) is an ideal of Der(L), whence a Lie subalgebra. Indeed, let D ∈ Der(L)
be a derivation of L and x, y ∈ L, then we have
[D, ad (x)](y) = D([x, y]) − [x, D(y)] = [D(x), y] = ad (D(x))(y),
i.e., [D, ad (x)] = ad (D(x)) ∈ ad (L).
§2.3. Automorphisms
Definition 2.16 (Automorphism). An automorphism of L is an isomorphism of L onto
itself. Aut(L) denotes the group of all automorphisms of L.
Examples 2.17. Important examples occur when L is a linear Lie algebra L ⊆ gl(V ).
If g ∈ GL(V ) is any invertible endomorphism of V , and if moreover gLg −1 = L, then it is
immediate that the map x 7→ gxg −1 is an automorphism of L. For instance, if L = gl(V ) or
even sl(V ), the second condition is automatic, so we obtain in this way a large collection of
automorphisms. (Cf. Exercise 12.)
Now specialize to the case: char F = 0.
Examples 2.18. Suppose x ∈ L is an element for which ad x is nilpotent, i.e., (ad x)k = 0 for
some k > 0. Then the usual exponential power series for a linear transformation over C makes
sense over F, because it has only finitely many terms:
exp(ad x) = 1 + ad x +
(ad x)2 (ad x)3
(ad x)k−1
+
+ ··· +
.
2!
3!
(k − 1)!
We claim that exp(ad x) ∈ Aut(L).
More generally, this is true if ad x is replaced by an arbitrary nilpotent derivation δ of L.
For this, use the familiar Leibniz rule:
n n h i
i
X
X
δn
n i
δ
δ n−i
δ ([x, y]) =
δ (x), δ n−i (y) ⇐⇒
([x, y]) =
(x),
(y)
i
n!
i!
(n − i)!
n
i=0
i=0
Proposition 2.19. If δ ∈ Der(L) and δ k = 0 for some k > 0, then exp(δ) ∈ Aut(L).
Proof. First, we will show that exp(δ) is a homomorphism from L to L, i.e.,
exp(δ)([x, y]) = [exp(δ)(x), exp(δ)(y)].
§ 2.
Ideals and homomorphisms
11
Indeed, we have
X
k−1
k−1 i
δ (x) X δ j (y)
,
[exp(δ)(x), exp(δ)(y)] =
i!
j!
j=0
i=0
=
=
2k−2
X
n i
X
δ
n=0
2k−2
X
i=0
n=0
δ n−i
(x),
(y)
i!
(n − i)!
!
k−1
X δn
δn
([x, y]) =
([x, y])
n!
n!
n=0
=exp(δ)([x, y])
The fact that exp(δ) is invertible follows (in the usual way) by exhibiting the explicit inverse
1 − η + η 2 − η 3 + · · · + (−η)k−1 , if exp(δ) = 1 + η.
Remark 2.20. ♣. Indeed, exp(δ)−1 = exp(−δ). Usually, if δ, η ∈ gl(V ), δ k = η k = 0 and
δη = ηδ, then we have exp(δ)exp(η) = exp(δ + η). If δη 6= ηδ, then exp(δ)exp(η) could
not equal to exp(δ + η). For example, let V = F2 be the two dimensional vector space and
take δ = x ∈ sl2 (F), η = y ∈ sl2 (F). Since (δ + η)(1, 1)t = (1, 1)t , then we have
e
1
=
exp(δ + η)
e
1
and
exp(δ)exp(η)
1
1
=
1 1
0 1
1 0
1
3
1
2 1
.
=
·
=
2
1
1 1
1 1
1
♣. For V = L = sl2 (F), δ = ad (x) and η = ad (y), we have (δ + η)(x + y) = 0 and this implies
exp(δ + η)(x + y) = x + y. But we know
1
exp(η)(x + y) = y + x + ad (y)(x) + (ad (y))2 (x) = x − h
2
and
exp(δ)exp(η)(x + y) = exp(δ)(x − h) = x − h + 2x = 3x − h.
♣. Let V = F3 be the three dimensional vector space and take δ = e23 ∈ gl3 (F), η = e12 ∈
gl3 (F). Now we have
1
1
exp(δ + η) = 1 + (δ + η) + (δ + η)2 = 1 + e12 + e23 + e13 ,
2
2
but
exp(δ)exp(η) = (1 + e23 )(1 + e12 ) = 1 + e12 + e23 6= exp(δ + η).
Proposition 2.21. An automorphism of the form exp(ad x), ad x nilpotent, is called inner;
more generally, the subgroup of Aut(L) generated by these is denoted Int(L) and its elements
called inner automorphisms. It is a normal subgroup, i.e., Int(L) Aut(L).
Proof. If ϕ ∈ Aut(L), x, y ∈ L, then
ϕ(ad x)ϕ−1 (y) = ϕ[x, ϕ−1 (y)] = [ϕ(x), y] = ad (ϕ(x))(y)
i.e., ϕ(ad x)ϕ−1 = ad ϕ(x). Hence ϕ exp(ad x) ϕ−1 = exp(ad ϕ(x)).
(2.4)
12
Lie algebras
Examples 2.22. Let L = sl2 (F), with standard basis x, y, h. Define
σ = exp(ad x) exp(ad (−y)) exp(ad x) : L → L
(so σ ∈ Int(L)). It is easy to compute the effect of σ on the basis (Exercise 10):
σ(x) = −y, σ(y) = −x, σ(h) = −h.
In particular, σ has order 2, i.e., σ 2 = idL . Notice that exp(x), exp(−y) are well defined
elements of SL2 (F), the group of 2 × 2 matrices of det 1, conjugation by which leaves L invariant
(as noted at the start of this subsection), so the product s = (exp(x))(exp(−y))(exp(x))
induces
0
1
an automorphism θ : z 7→ szs−1 of L. A quick calculation shows that s =
and that
−1 0
conjugating by s has precisely the same effect on L as applying σ.
The phenomenon just observed is not accidental:
Proposition 2.23. If L ⊆ gl(V ) is an arbitrary linear Lie algebra (char F = 0), and x ∈ L is a
nilpotent matrix, then we claim that
(exp(x))y(exp(x))−1 = (exp(ad x))(y) for all y ∈ L.
(2.5)
Proof. To prove this, notice that ad x = λx +ρ−x where λx , ρx denote left and right multiplication
by x in the ring End(V ) (these commute, of course, and are nilpotent). Then the usual rules of
exponentiation show that
exp(ad x) = exp(λx + ρ−x ) = exp(λx )exp(ρ−x ) = λexp(x) ρexp(−x) ,
which implies (2.5).
By the Proposition, we have
−1
szs−1 =(exp(x))(exp(−y))(exp(x))z (exp(x))(exp(−y))(exp(x))
=(exp(x)) (exp(−y)) (exp(x))z(exp(−x)) (exp(y)) (exp(−x))
=(exp(ad x))(exp(ad (−y)))(exp(ad x))(z)
=σ(z),
i.e., θ = σ.