MATH 223A NOTES 2011
LIE ALGEBRAS
5
2.
Ideals and homomorphisms
2.1.
Ideals.
Defnition 2.1.1.
An
ideal
in a Lie algebra
L
is a vector subspace
I
so that [
LI
]
⊆
I
.In
other words, [
ax
]
∈
I
for all
a
∈
L, x
∈
I
.
Example 2.1.2.
(1) 0 and
L
are always ideals in
L
.
(2) If
L
is abelian then every vector subspace is an ideal.
(3) [
LL
] is an ideal in
L
called the
derived algebra
of
L
.
Proposition 2.1.3.
Every 2dimensional Lie algebra contains a 1dimensional ideal.
Proof.
As we saw last time, there are only two examples of a 2dimensional Lie algebra:
Either the basis elements commute, in which case
L
is abelian, or they don’t commute,
in which case [
LL
] is 1dimensional. In both cases,
L
has a 1dimensional ideal.
°
Proposition 2.1.4.
The kernel of an homomorphism of Lie algebras
ϕ
:
L
→
L
°
is an
ideal in
L
. (The image is a subalgebra of
L
°
.) Conversely, for any ideal
I
⊆
L
,
L/I
is a
Lie algebra and
I
is the kernel of the quotient map
L
→
L/I
.
Proof.
If
x
∈
ker
ϕ
and
a
∈
L
then
ϕ
[
ax
]=[
ϕ
(
a
)
ϕ
(
x
)] = [
ϕ
(
a
)0] = 0. So, [
ax
]
∈
ker
ϕ
.
Conversely, if
I
⊂
L
is an ideal,
a
∈
L, x
∈
I
then
[(
a
+
I
)(
x
+
I
)]
⊆
[
ax
]+[
aI
Ix
II
]
⊆
[
ax
]+
I
So, the bracket is welldeFned in
L/I
,[
ϕ
(
a
)
ϕ
(
x
)] =
ϕ
[
ax
] and
I
=ker
ϕ
.
°
Proposition 2.1.5.
Any epimorphism (surjective homomorphism) of Lie algebras
ϕ
:
L
→
L
°
gives a 11 correspondence between ideals
I
°
in
L
°
and ideals
I
of
L
containing
ker
ϕ
.
Proof.
The correspondence is given by
I
°
=
ϕ
(
I
).
This is an ideal since [
L
°
,I
°
]=
[
ϕ
(
L
)
,ϕ
(
I
)] =
ϕ
[
LI
]
⊆
ϕ
(
I
)=
I
°
and
I
=
ϕ
−
1
(
I
°
) which is an ideal since it is the
kernel of
L
→
L
°
→
L
°
/I
°
.
°
Example 2.1.6.
The
center
of a Lie algebra
L
is deFned to be
Z
(
L
{
x
∈
L

[
xL
]=0
}
.
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 Fall '11
 K.Igusa
 Algebra, Vector Space, Ring, Lie group, Lie algebra, λx, ker φ

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