DIGITAL TEXT BOOK

1. DIGITAL TEXT BOOK
SUBMITTED BY
VEENA VALSAN
ROLL NO : 15
MOUNT TABOR TRAINING COLLEGE

2. INTRODUCTION
In mathematics, a lie algebra is a vector space together with a non- associative
multiplication called “Lie bracket” when an algebraic product is defined on the
space, the lie bracket is the commutator.
Lie algebra were introduced to study the concept of infinitesimal
transformations. Herman wegll introduced the term “Lie Algebra” (after sophus
Lie) in the 1930s, In order texts, the name “infinitesimal group” is used.
Lie algebras are closely related to lie groups which are groups that are
also smooth manifolds, with the property that the group operations of
multiplication and inversion are smooth maps.
Any Lie group gives rise to a Lie algebra. Conversely to any finite
dimensional Lie algebra over real or complex numbers, there is a
corresponding connected Lie group unique up to covering. This correspondence
between Lie groups and Lie algebras allow one to study Lie groups in terms of
Lie algebras.

3. We begin by defining Lie algebras and giving a collection of typical examples
to which we shall refer throughout the project and we explore some of the
constructions in which ideals are involved. We know that in the theory of Lie
algebras ideals play a role similar to that played by normal subgroup in the
theory of groups.
We would like to know how many essentially different Lie algebras there
are what approaches we can use to classify them. To get some feeling for these
questions, we shall look at Lie algebras of dimension 1,2 and 3. Another reason
for looking at there low-dimensional Lie algebras is that they often occur as sub
algebras of the larger Lie algebras.
Abelian Lie algebras are easily understood. Many of the Lie algebras
occur naturally as sub algebras of the Lie algebras of linear transformation of
vector space. Even more easily seen to be isomorphic to such sub algebras.

4. CHAPTER – I
PRELIMINARIES
In this chapter we recall some elementary facts of linear algebra, which are
needed throughout the chapters.
Definition 1.1
Let F be a field. An F – vector space 𝑣 is a set with two operations.
𝑣 × 𝑣 ⟼ 𝑣
( 𝑣1 𝑤) → 𝑣 + 𝑤 and
𝐹 × 𝑣 → 𝑣
(𝜆1 𝑤) ⟼ 𝜆. 𝑤
Called addition and multiplication by scalars with the usual axioms.
Example 1.1
The set of row vectors of length n containing complex numbers is denoted by
ℂ1 × 𝑛 ∶= {[ 𝛼1 … 𝛼 𝑛] 𝛼1 … 𝛼 𝑛 ∈ ℂ}
If is a ℂ -vector space, we add vectors and multiply them by scalar as exhibited
in the following examples
[11 − 2,3] + [4,5,6] = (1 + 4,−2 + 5,3 + 6] = [5,3,9]

5. And
(−3):[4,1/2,−2] = [−3,4,(−3)(1/2),(−2),(−3)]
= [−12,−
3
2
, 6] = (−1)[12,3/2,6]
Definition 1.2
(Linear combination, span, spanning set)
If 𝑣 is a ℂ − vector space 𝑢1 … 𝑢 𝑘 ∈ 𝑣 and ∈ 𝜆1 … 𝜆 𝑘 ∈ ℂ then
kkk
k
i
ii vvvvw  
...: 211
1
is called a linear combination defined
The set of linear combinations
Span ( 𝑣1,… 𝑣 𝑘 ) ≔ {𝑤 ∈ 𝑣/𝑤 is a linear combination of the 𝑣𝑖} is called span of
the vectors 𝑣1 … 𝑣 𝑘
Note : Linear combinations are always finite sum
Definition 1.3
Let 𝑣 be a ℂ − vector space. A non empty subset 𝑤 ⊆ 𝑣 is called subspace if
𝑢 + 𝑣 ∈ 𝑤 and 𝜆𝑢 ∈ 𝑤 for all 𝑢, 𝑣 ∈ 𝑤 and all 𝜆 ∈ ℂ

6. Example : The following is a subspace of ℂ1x3
Span ([1,0,−1],[0,2,1],[1,2,0]) = {[𝑥, 𝑦1𝑧] ∈ ℂ1×3
}2/ xyz 
Definition 1.4
A tuple (𝑣1 … 𝑣 𝑘 ) of vectors is 𝐴 ℂ- vector space 𝑣 are called linearly
independent, if one of the following statement are true
(i) for arbitrary numbers 𝜆1 … 𝜆 𝑘 ∈ ℂ the following implication holds.
(ii) Every vector is span (𝑣1 … 𝑣 𝑘 ) can be expressed a linear combination
of the vectors 𝑣1 … 𝑣 𝑘 is a unique way
(iii) No vector n is contained in the span of the others
𝑣𝑖𝜀 span (
𝑣𝑗
1
≤ 𝑗, 𝑗 ≠ 𝑖) for all i.
Other wise the touple is linearly independent
Example 1.1.2
The tuple of vectors ( [5,0,2],[2,3,0],[−1,0,0]) is linearly independent
Definition 1.5
Let v be a ℂ −vector space. A taple (𝑣1 … 𝑣 𝑛) of vectors in v is called a basis of
v. if
V = span (𝑣1 … 𝑣 𝑛) and
(𝑣1 … 𝑣 𝑛) is linearly independents

7. Theorem 1.6
In a ℂ- vector space v any two bases base the same on of elements. The number
of elements in an arbitrary basis of v is called the dimension of v,
Example 1.1.3
The ℂ- Vector space ℂ1×𝑛
is n-dimensional because
([1,0…0],[0,1,0…]…[0… 0,1])
Is a basis of length n, it is called the standard basis
Definition 1.6
Let v and w be ℂ-vector space. A map 𝜑 ∶ 𝑣 → 𝑤 is called ℂ- linear, if
( 𝑢 + 𝑣) 𝜑 = 𝑢𝜑 + 𝑣𝜑 and ( 𝜆𝑣) 𝜑 = 𝜆(𝑢𝜑) for all 𝑢, 𝑣 ∈ 𝑣 and 𝜆 ∈ ℂ
Example 1.1.4
The map
ℂ1×3
→ ℂ1×3
, [ 𝑥, 𝑦, 𝑧] ⟼ (2𝑥 − 𝑦 + 3𝑧, 𝑥 + 𝑧, −𝑥 + 7𝑧,6𝑦]
is a ℂ - linear. It in uniquely defined by doing
[1,0,0] ⟼ [2,1,−1] and [0,1,0] ⟼ [−1,0,6] and
[0,0,1] ⟼ [3,1,7]
Definition 1.7

8. The ℂ- vector space v is said to be the direct sum 𝑢⨁𝑤 of two subspace 𝑢 and
𝑤 of 𝑣, if one and thus both of the following equivalent conditions hold,
𝑣 = 𝑢 + 𝑤 = {𝑢 + 𝑤|𝑢 ∈ 𝑢, 𝑤 ∈ 𝑊} and 𝑢 ∩ 𝑤 = {0}
Every vector 𝑣 ∈ 𝑉 can be written as a sum 𝑢 + 𝑤 of a vector 𝑢 ∈ 𝑈 and vector
𝑤 ∈ 𝑊 is a unique way.
Theorem .18
If 𝑣 = 𝑢⨁𝑤 and (𝑢1 … 𝑢 𝑚) is a basis of 𝑢 and (𝑤1 … 𝑤 𝑛)is a basis of W, then
(𝑢1 … 𝑢 𝑚, 𝑤1 … 𝑤 𝑛)is a basis of 𝑣 and we have
dim( 𝑣) = dim( 𝑢) + dim(𝑤)
Example: We have
ℂ1×3
= span [1,2,3])⨁{[ 𝑥, 𝑦, 𝑧] ∈ ℂ1×3
/𝑧 = 𝑥 − 𝑦}
Remark : Note that for every subspace 𝑢 of a ℂ − vector space 𝑣 there is a
subspace w of 𝑣 such that 𝑣 = 𝑢⨁𝑊

9. CHAPTER – II
Introduction to Lie Algibra
In this chapter we define Lie algebras and giving a collection of typical
examples to which we shall prove the definition. The remaining sections in this
chapter introduce the basic vocabulary of Lie algebras.
Definition 2.1
Let F be a field. A Lie algebra ouer F is an F vector space 𝐿1 together with a
bilinear map, the lie bracket
𝐿 × 𝐿 → 𝐿1( 𝑥, 𝑦) → [𝑥, 𝑦]
Satisfying the following properties.
[ 𝑥, 𝑥] = for all 𝑥 ∈ 𝐿, (𝐿1)
[𝑥, [ 𝑦, 𝑧]] + [𝑦,[ 𝑧, 𝑥]] + [𝑧, [ 𝑥, 𝑦]] = 0 for all 𝑥, 𝑦, 𝑧 ∈ 𝐿 (𝐿2)
The lie bracket [𝑥, 𝑦] is often afforded to as the commentator of 𝑥 and 𝑦.
Condition (𝐿2) is known as the Jacobi identity. As the lie bracket [−, −] is
bilinear , we have
0 = [ 𝑥 + 𝑦, 𝑥 + 𝑦] = [ 𝑥, 𝑥] + [ 𝑥, 𝑦] + [ 𝑦, 𝑥] + [𝑦, 𝑦]
= [ 𝑥, 𝑦] + [𝑦, 𝑥]

10. Hence condition (𝐿1) implies
[ 𝑥, 𝑦] = −[𝑦, 𝑥] for all 𝑥, 𝑦 ∈ 𝐿 (𝐿1
′
)
If the field F does not have characteristic 2, then putting 𝑥 = 𝑦 in (𝐿1
′
) implies
(𝐿1)shows that (𝐿1
′
) implies (𝐿1
′
)
Unless specifically stated otherwise, all lie algebras should taken to be finite
dimensional.
Examples 2.1.1
(1) Let F = R The vector product (𝑥, 𝑦) → 𝑥 ∧ 𝑦 defines the structure of a lie
algebra a on R3. We denote this Lie algebra by R3.
Explicitly, if 𝑥 = (𝑥1, 𝑥2, 𝑥3) and 𝑦 = ( 𝑦1, 𝑦2, 𝑦3), the
𝑥 ∧ 𝑦 = (𝑥2 𝑦3 − 𝑥3 𝑦2, 𝑥3 𝑦1 − 𝑥1 𝑦3, 𝑥2 𝑦2 − 𝑥2 𝑦1
Examples 2.1.2
(2) Any vector space v has a Lie bracket define by [ 𝑥, 𝑦] = 0 for all 𝑥, 𝑦 ∈ 𝑣
This is the abelian Lie algebra structure on v. In particular, the field F may be
regarded as a 1-dimensional abelian Lie algebra.
Examples 2.1.3

11. (3) Suppose that v is finite – dimensional vector space over F. Write 𝑔𝑙(𝑣) for
the set of all linear maps from v to V. This is again a vector space our R, and it
becomes a Lie algebra, known as the general linear algebra, if we define the Lie
bracket [-,-] by
[ 𝑥, 𝑦] = 𝑥 ∘ 𝑦 − 𝑦 ∘ 𝑥 for 𝑥, 𝑦 ∈ 𝑔𝑙(𝑣)
Where ∘ denote the composition of maps
Examples 2.1.4
(4) let 𝑏(𝑛, 𝐹) be the upper triangular matrices in 𝑔𝑙(𝑛, 𝐹) ( A matrix x is said
to be upper triangular if 𝑥𝑖𝑗 = 0 when our 𝑖 > 𝑗. )
This is a Lie algebra with the same Lie bracket as 𝑔𝑙( 𝑛, 𝐹).
Similarly, Let 𝑛(𝑛, 𝐹) be the strictly upper triangular matrix in 𝑔𝑙(𝑛, 𝐹)
(A matrix is said to be strictly upper triangular if 𝑥𝑖𝑗 = 0 whenever 𝑖 ≥ 𝑗)
Again this is Lie algebra with the same Lie bracket as 𝑔𝑙(𝑛, 𝐹)
Definition 2.2
The last examples suggest that, a given Lie algebra L, we might define a Lie
sub algebra of L to be a vector subspace 𝑘 ⊆ 𝐿 such that
[𝑥, 𝑦] ∈ 𝑘 for all 𝑥, 𝑦 ∈ 𝑘

12. Lie sub algebra are easily seen to be Lie algebras in their own right. In the
above example we saw there Lie sub algebras of 𝑔𝑙(𝑛, 𝐹)
Definition 2.3
We also define an ideal of a Lie algebra L to be a subspace I of L such that
[𝑥, 𝑦] ∈ 𝐼 for all 𝑥 ∈ 𝐿, 𝑔 ∈ 𝐼
By (𝐿1
′
), [ 𝑥, 𝑦] = −[𝑦, 𝑥], so we do not need to distinguish between left and
right ideals
Examples 2.3.1
(1) 𝑠𝑙(𝑛, 𝐹) is an ideal of 𝑔𝑙(𝑛, 𝐹) and 𝑛(𝑛, 𝐹) is an ideal of 𝑏( 𝑛, 𝐹).
Note 2.4
An ideal is always a sub sub algebra on the other hand, a sub algebra need not
be an ideal
Example 2.4.1
Let 𝑏(𝑛, 𝐹) is a sub algebra of 𝑔𝑙( 𝑛, 𝐹), but provided 𝑛 ≥ 2, if is not an ideal.
To see this, note that 𝑒11 ∈ 𝑏 (𝑛, 𝐹) and 𝑒21 ∈ 𝑔𝑙(𝑛, 𝐹).
However
[ 𝑒21 𝑒11] = 𝑒21 ∉ 𝑏(𝑛, 𝐹)

13. Remark 2.5
The lie algebra L is itself an ideal of L. At the other extreme, {0} is an ideal of
L. we call there the trivals ideals of L.
Examples 2.5.1
An important example of an ideal which frequently is non-trivial is the center of
L, defined by
𝑧( 𝐿): = {𝑥 ∈ 𝐿 ∶ [ 𝑥, 𝑦] = 0 for all 𝑦 ∈ 𝐿}
We know that 𝐿 = 𝑧(𝐿) as this is the case iff L is abelian on the other hand, it
might take some work to decide whether or not 𝑧( 𝐿) = {0}
Definition 2.6
If 𝐿1 and 𝐿2 are lie algebras our a field F , then we say that a map 𝜑is a
homomorphism if 𝜑 is a linear map and
𝜑([ 𝑥1 𝑦]) = [𝜑( 𝑥), 𝜑( 𝑦)] for all 𝑥, 𝑦 ∈ 𝐿1
In this equation the first lie bracket is taken in 𝐿1 and the second lie bracket is
taken 𝐿2
Note 2.7

14. We say that 𝜑 is an isomorphism if 𝜑 is also bijective. An exteremely important
homomorphism is the adjoin homomorphism
𝑎𝑑 ∶ 𝐿 → 𝑔𝑙(𝐿)
by ( 𝑎𝑑𝑥)( 𝑦):(𝑥, 𝑦) for 𝑥, 𝑦 ∈ 𝐿. If follows from the bilinearity of the bracket
that the map 𝑎𝑑𝑥 is linear for each 𝑥 ∈ 𝐿. for the some reason, the map 𝑥 ⟼
𝑎𝑑𝑥 is itself linear. So to show that ad is a homomorphism
ie
𝑎𝑑([ 𝑥, 𝑦)] = 𝑎𝑑 𝑥 ∘ 𝑎𝑑𝑦 for all 𝑥, 𝑦 ∈ 𝐿. This turn out to be equivalent to the
jacabi identity. The kernel of ad is the centre of L.
Definition 2.8
An algebra over a field F is a vector space A our F together with a bilinear map
𝐴 × 𝐴 ⟶ 𝐴, (𝑥, 𝑦) → 𝑥𝑦
We say that 𝑥𝑦 is the product of 𝑥 and 𝑦. Usually algebras has some further
properties. Lie algebras are the algebras satisfying identies (𝐿1) and (𝐿2).
(1) The algebra A is said to be associative if
( 𝑥𝑦) 𝑧 = 𝑥(𝑦𝑧) for all 𝑥, 𝑦, 𝑧 ∈ 𝐴

15. (2) unitial if there is an elements 𝐼𝐴 in A such that 𝐼𝐴 𝑥 = 𝑥 = 𝑥𝐼𝐴 for all non
zero elements of A.
Example 2.8.1
(1) Let 𝑔𝑙( 𝑣), the vector space of a linear transformation of the vector space 𝑣,
has the structure of a unital associative algebra where the product is given
by the composition of maps. The identity transformation is the identity
element in this algebra, Likewise 𝑔𝑙( 𝑛, 𝐹), the set of 𝑛 × 𝑛 matrices our F,
is a unital associative algebra with respect to noutrix multiplication
Example 2.8.2
(2) Let L be a lie algebra show that the Lie bracket is associated, that is
[𝑥, [𝑦, 𝑧]] = [[ 𝑥, 𝑦], 𝑧] for all 𝑥, 𝑔, 𝑧 ∈ 𝐿 iff for all 𝑎, 𝑏, ∈ 𝐿 the commutator
[𝑎, 𝑏] lies in 𝑧(𝐿).
If A is an associative algebra over F, the we define a new bilinear operation
[−,−] on A by
[ 𝑎, 𝑏] = 𝑎𝑏 − 𝑏𝑎 for all 𝑎𝑏 ∈ 𝐴
Then A together with [−,−] is a Lie algebra. The Lie algebra 𝑔𝑙(𝑣)n and
𝑔𝑙(𝑛, 𝐹) are special cases of this construction
Definition 2. 9

16. Let A be an algebra over a field F. A derivation of A is an f- linear map
𝐷 ∶ 𝐴 → 𝐴 such that
𝐷( 𝑎𝑏) = 𝑎𝐷( 𝑏) + 𝐷( 𝑎) 𝑏 for all 𝑎, 𝑏 ∈ 𝐴
Example 2.9.1
(1)Let A = 𝑐∞
𝑅 be the vector space of all infinitely differentiable functions
𝑅 → 𝑅 for f,g ∈ 𝐴, we define the product fg by pointwise multiplication
( 𝑓𝑔)( 𝑥) = 𝑓( 𝑥) 𝑔 ( 𝑥). With this definition, A is an associative algebra. The
usual derivative 'fDf  is a derivation of A since by the product rule
𝐷( 𝑓𝑔) = (𝑓𝑔)′
= 𝑓′
𝑔 + 𝑓𝑔′
= ( 𝐷𝑓) 𝑔 + (𝐷𝑔)
(2) Let L be a Lie algebra and let 𝑥 ∈ 𝐿. The map → is a derivation of L since L
of the Jacobi identity.
( 𝑎𝑑𝑥)[ 𝑔, 𝑧] = [𝑥, [ 𝑔, 𝑧]] = [[ 𝑥, 𝑔], 𝑧] + [𝑔1[ 𝑥, 𝑧]]
= [ 𝑎𝑑( 𝑥) 𝑦, 𝑧], [𝑔1 ( 𝑎𝑑𝑥) 𝑧]
for all 𝑦, 𝑧 ∈ 𝐿
Definition 2.10

17. If L is a lie algebra over a field F with basis ( 𝑥1 … 𝑥 𝑛), then [-,-] is completely
determined by the products [𝑥𝑖, 𝑥𝑗]. We define scalars 𝑎𝑖𝑗
𝑘
∈ 𝐹 such that
k
n
k
k
ijji xaxx 

1
],[
The 𝑎𝑖𝑗
𝑘
are the structure constants of L with respect to this basis.
Example 2.10.1
Let 𝐿1 and 𝐿2 be Lie algebras. Show that 𝐿1 is isomorphic to 𝐿2 iff there is a
basis 𝐵1 of 𝐿1 and a basis 𝐵2 of 𝐿2 such that the structural constants of 𝐿1 with
respect to 𝐵1 are equal structure constants of 𝐿2 with respect of 𝐵2.
Definition 2.11
Suppose that I and J are ideals of a Lie algebra L. There are several ways we
can constract new ideal from I and J. First we shall show that I n J is an ideals of
L.
We know that InJ is a subspace of L, so all we need check is that if 𝑥 ∈ 𝐿 and
𝑦 ∈ InJ, then [𝑥, 𝑦] ∈ I nJ. This shows that at one I and J are ideals.
Example 2.11.1
Show that I + J is an ideal of L,

18. where I + J = { 𝑥 + 𝑦 ∶ 𝑥 ∈ 𝐼, 𝑦 ∈ 𝐽}
We can also define a product of ideals
Let 𝐼, 𝐽] = 𝑠𝑝𝑎𝑛 {[ 𝑥, 𝑦]: 𝑥 ∈ 𝐼, 𝑦 ∈ 𝐽}
We claim that [I, J] is an ideal of L
Firstly, it is by definition a subspace
Secondly, if 𝑥 ∈ 𝐼, 𝑦𝑒J and 𝑢 ∈ 𝐿, then the Jacobi identity gives
[𝑎1[ 𝑥, 𝑦]] = [𝑥, [ 𝑢, 𝑦]] + [[ 𝑢, 𝑥], 𝑦]
Here [𝑢, 𝑦] ∈ 𝑔 as J is an ideal , so [x, [u,y]] ∈ [I, J]
Similarly [[ 𝑢, 𝑥], 𝑦 ] ∈ [ I, J ]
∴ Their sum belongs to [ I, J ]
A general element 𝑡 of [I, J ] is a linear combination of brackets [𝑥, 𝑦] with 𝑥 ∈
I, 𝑦 ∈ J say 𝑡 = ∑ 𝑐𝑖 [𝑥𝑖, 𝑦𝑖] where 𝑐𝑖 are scalars and 𝑥𝑖 ∈ J and 𝑦𝑖 ∈J.
Then, for any 𝑢 ∈ 𝐿, we have
[ 𝑎, 𝑡] = [ 𝑎1 ∑ 𝑐𝑖 [𝑐𝑖 𝑦𝑖]] = ∑ 𝑐𝑖[ 𝑎1[𝑥𝑖, 𝑦𝑖]]
Where [𝑎1[ 𝑥𝑖 𝑦𝑖]] ∈ [I, J] as shown above
Hence [ 𝑢, 𝑡] ∈ [I, J] and so [ I, J ] is an ideal of L

19. Remark 2.12
It is necessary to define [ I, J] to be the spanof the commutators of elements of I
and J rather than just the set of such commutators.
Definition 2.13
If I is an ideal of the Lie algebra L, then I is in particular a subspace of L, and so
we may consider the cosets.
z + I = {z + x ∶ x ∈ I}for z ∈ I and the quotient vector space
𝐿/𝐼 = {𝑧 + 𝐼: 𝑧 ∈ 𝐿}
We claim that a Lie bracket on L/I may be defined by
[ w + I, z + I] = [w,z] + I fir w,z ∈ I
As we know Lie bracket on L/ I is well defines, we most check that [ 𝑤 + 𝑧] + I
depends only on the cosets containing 𝑤 and 𝑧 and not on the particular coset
representations 𝑤 and z. Suppose 𝜔 + 𝐼 = 𝜔′
+ 𝐼 and 𝑧 + 𝐼 = 𝑧′
+ 𝐼
Then 𝜔 − 𝜔′
∈ 𝐼 and 𝑧 − 𝑧′
∈ 𝐼. By bilinearity of the lie bracket in L.
[ 𝜔′
, 𝑧′] = [𝜔′
+ ( 𝜔 − 𝜔′), 𝑧′
+ ( 𝑧 − 𝑧′)]
= [ 𝜔, 𝑧] + [ 𝜔 − 𝜔′
, 𝑧′] + [ 𝜔′
, 𝑧 − 𝑧′] + [𝜔 − 𝜔′
, 𝑧 − 𝑧′
]

20. Where the final there sammands all belong to I.
∴ [ω′
+ I, z′
+ I] = [ω,z] + I
So L/ I is a lie algebra. It is called the quotient or factor algebra of L by I.
Theorem 2.14
First Isomorphism Theorem
Let 𝜑 ∶ 𝐿 → 𝐻 a homomorphism of a Lie algebra over a field F and k : ker (𝜑).
Then 𝜑 ∶ 𝐿/𝐾 → im (𝜑)𝑥 + 𝑘 ⟼ 𝑥𝜑 is an isomorphism of Lie algebras.
Pf : The map 𝜑 is well defined since 𝑥 + 𝑘 = 𝑦 + 𝑘 is equivalent to 𝑥 − 𝑔 ∈
𝐾 = ker(𝜑)and thus 𝑥 𝜑 = 𝑦𝜑. This also prove that 𝜑 is injective, and
subjectivity to the image of 𝜑 is obvious.
The map 𝜑 is clearly F – linear and a homomorphism of Lie algebra as because
𝜑
Second Isomorphism Theorem 2.15
Let L be a Lie algebra, K an ideal H a subalgebra. Then Hnk is an ideal of H
and the map
𝜑 ∶ 𝐻 /(𝐻𝑛𝑘) → (𝐻 + 𝑘)/𝑘
ℎ + ( 𝐻𝑛𝑘) ⟼ ℎ + 𝑘

21. is an isomorphism of Lie algebra
Pf: The ideal k of L is automatically an ideal of the subalgebra h+k
Define 0 map 𝜑 ∶ 𝐻 → (𝐻 + 𝑘)/𝑘 by setting ℎ′
𝜑:ℎ + 𝑘
This is clearly linear and homomorphism if lie algebras.
Its image is all of (H+k) /k since wery coset in there has a representative in H.
The kernel of 𝜑 is exactly Hnk and it follows that this is an ideal in H.
Example 2.15.1
(1) The trace of an 𝑛 × 𝑛 matrix is the sum of its diagonal entries. Fix a field F
and consider the linear map 𝑡𝑟 ∶ 𝑔}(𝑛, 𝐹) → 𝐹 which sends a matrix to its trace.
This is a Lie algebra homomorphism, for if 𝑥, 𝑦 ∈ 𝑔|(𝑛, 𝐹) then
𝑡𝑟 [ 𝑥, 𝑦] = 𝑡𝑟 ( 𝑥𝑦 − 𝑦𝑥) = 𝑡𝑟 𝑥𝑦 − 𝑡𝑟𝑦𝑥 = 0
So 𝑡𝑟 [ 𝑥, 𝑦] = [ 𝑡𝑟, 𝑡𝑟𝑦] = 0
Here the first Lie bracket is taken in 𝑔|𝑛, 𝐹) and second in the abelian Lie
algebra F.
As we know 𝑡𝑟is surjective. Its kernel is 𝑠|( 𝑛, 𝐹), the Lie algebra of matrix with
trace 0.
So by the first isomorphism theorem
𝑔′
(𝑛, 𝐹)/𝑆(𝑛, 𝐹) ≅ 𝐹

22. Example 2.15.2
(2) Suppose that L is a Lie algebra and I is an ideal in L such that L/I is abelian.
In this case, the ideals of L/I are just the subspaces of L/I. By the ideal
correspondence, the ideals of L which contain I are exactly the subspace of L
which contain L.

23. CHAPTER – III
LOW DIMENSIONAL LIE ALGEBRA AND SUBALGEBRAS
Definition 3.1
Any 1 dimensional Lie algebra is abelian
Suppose L is an non abelian Lie algebra of dimension 2 over a field F. The
derived algebra of L cannot be more than 1 – dimensional since if [𝑘, 𝑦} is a
basis of L, then 𝐿′
is spanned by [ 𝑥, 𝑦].
Therefore 𝐿′
must be 1-dimensional. Take a non zero element 𝑥 ∈ 𝐿′
and
extend it in any way to a vector space basis {𝑥, 𝑦̃} of L. Then ( 𝑥, 𝑦̃ ] ∈ 𝐿′
: This
element must be non zero, as otherwise L would be abelian.
Theorem 3.2
Let F be any field up to isomorphism there is a unique two dimensional non
abelian Lie algebra over F. This lie algebra has a basis [𝑥, 𝑦} such that its Lie
bracket is described by [ 𝑥, 𝑦] = 𝑥 Then centre of this Lie algebra is 0.
Dimension 3.3

24. If L is a non – abelian 3-dimensional Lie algebra over a field F, then we know
only that the derived algebra 𝐿′
is non zero. It might have dimension 10r 20r
even 3.
We also know that the centre 𝑧(𝐿) is a proper ideal of L.
Definition 3.4
Assume first that 𝐿′
is 1-dimensional and 𝐿′
is contained in 𝑧( 𝐿). We shall
show that there is a unique such lie algebra, and that it has a basis 𝑓, 𝑔, 𝑧 where
[ 𝑓, 𝑓] = 𝑧 and 𝑧 lies in 𝑧( 𝐿). This algebra is known as the Heisenberg algebra.
Theorem 3.5
Let F be any field. There is a unique 3-dimension Lie algebra over F such that 𝐿′
is 1-dimensional and 𝐿′
is not contained in 𝑧( 𝐿). This Lie algebra is the direct
sum of the 2-dimensional non – alberian Lie algebra with the 1-dimensional
Lie algebra.
Note 3.6
(1) Suppose that dim L = 3 and dim 𝐿′
= 2. We shall see that, over ℂ at least
there are infinitely many non isomorphic such lie algebras.
3.6.2 (2) suppose that L is a complex Lie algebra of dim 3.
Such that L = 𝐿′

25. Definition 3.7
The lie algebra L is said to be solvable if for 𝑚 ≥ 1 we have 𝐿(𝑚)
= 0
Example 3.7.1
1. The Heisenberg algebra is solvable
2. The algebra of upper triangular matrices is solvable
3. The classification of 2-dimensional Lie algebra with any 2-dimensional
Lie algebra is solvable
Note 3.8
If L = | (2c) with L = 𝐿′
and therefore 𝐿(𝑚)
= 𝐿 for all 𝑚 ≥ 1, so 𝑠|(2, 𝑐) is not
solvable
If L solvable, then the derived series of L provides with an approximation of L
by a finite series of ideal with abelian quotients.
Corollary : let L be a finite – dimensional Lie algebra. There is a unique
solvable ideal of L containing very solvable ideal of L.
Proof
Let R be a solvable ideal of largest dimension
Suppose that I is any solvable ideal
So R + I is solvable ideal

26. Now 𝑅 ⊆ 𝑅 + 𝐼 and hence dim 𝑘 ≤ dim( 𝑅 + 𝐼). We choose R of maximal
possible dimension and there we must have dim 𝑅 = (𝑅 + 𝐼) and hence R = R
+ I, so I is contained in R.
Definition 3.10
Let L be a Lie algebra and 𝐻1 𝐻2 soluble ideal of L. Then 𝐻1 + 𝐻2 is soluble
ideal of L, too. Further more if L is a finite – dimensional, there is soluble ideal
rod (L) of L that contain every soluble ideal of L. It is caked radical or L.
Definition 3.11
A non – zero lie algebra L is said to be semi simple if it has no non – zero
solvable ideals or equivalentely if rad L = 0
Definition 3.12
A Lie algebra L is called simple, if it is non abelian and has no ideals than 0 and
L.
Definition 3.13
A one – dimensional Lie algebra is automatically abelian and is called the trivial
algebra
Definition 3.14

27. The Lie algebra L is said to be nilpotent if for 𝑠𝑜𝑚 𝑚 ≥ 1 we have 𝐿 𝑚
= 0
Lemma 3.15
For every finite – dimensional Lie algebra 𝐿1 the quotient Lie algebra
𝐿 | 𝑟𝑎𝑑 (𝐿) is semi simple
Proof
The pre image of any soluble ideal of < ? 𝑟𝑎𝑑 (𝑐) ander canonical map 𝐿 →
𝐿|𝑟𝑎𝑑(𝐿) would be a soluble ideal of L that properly contain 𝑟𝑎𝑑 ( 𝐿).
Example 3.15.1
Every simple lie algebra L is semi simple, since it contain no ideals other than L
and 0 and L is not soluble. The direct sum 𝐿1 ⊕ 𝐿2 ⊕ …⊕ 𝐿 𝑘 of simple Lie
algebra 𝐿1 … 𝐿 𝑘 is semi simple
By the direct sum we mean the direct sum of vector space with component-wise
Lie product. It is a routine that this make the direct sum into Lie algebra, such
that every summand 𝐿1 is an ideal, since [𝐿 𝑖, 𝐿𝑗] = 0 for 𝑖 ≠ 𝑗 in this lie
algebra
Assume that k is any ideal of the sum 𝐿1 ⨁…⨁𝐿 𝑘. We claim tbat for every
summand 𝐿 𝑖. We either have 𝐿 𝑖 ⊆ 𝑘 or 𝐿 𝑖 𝑛𝑘 = {0}. This is true because 𝐿 𝑖 𝑛𝑘 is
an ideal in 𝐿 𝑖 and 𝐿𝑗 is simple. Thus, k is the sum of some of the 𝐿 𝑖. However if
𝑘 ≠ {0}, then k is not soluble,

28. Since ( 𝐿 𝑖 𝐿 𝑖) = 𝐿 𝑖 for all 𝑖 and at …one 𝐿 𝑖 fully contatned in k in this care.

29. CHAPTER – IV
REPRESENTATION OF LIE ALGEBRA
Definition 4.1
Let L be a lie algebra over the field. A representation of L is a lie algebra
homomorphism.
𝜌 ∶ 𝐿 → 𝐿𝑖𝑒 ⊂ 𝑒𝑛𝑑 (𝑣))
For some F-vector space v of dimension 𝑛 ∈ ℕ, which is called the degree of 𝜌.
This means nothing but 𝜌 is a linear map and
[ 𝑥, 𝑦] 𝜌 = [ 𝑥𝜌, 𝑦𝜌] = ( 𝑥𝜌).( 𝑦𝜌) − ( 𝑦𝜌)(𝑥𝜌)
for all 𝑥, 𝑦 ⊂ 𝐿
Definition 4.2
Two representation 𝜌: 𝐿 → 𝐿𝑖𝑒 (𝑒𝑛𝑑 ( 𝑣)) and 𝜌: 𝐿 ⟶ 𝐿𝑖𝑒 (𝑒𝑛𝑑 ( 𝑗)) of degree
𝑛 are called equivalent, if there is an invertible linear map 𝑇: 𝑉 ⟶ 𝑣′
such that
( 𝑥𝜌). 𝑇 = 𝑇( 𝑥, 𝜌) for all 𝑥 ∈ 𝐿.

30. Definition 4.3
Let L be a Lie algebra over a field 𝔽. An L-module is a finite dimensional P-
vector space v together with an action.
𝑉 × 𝐿 → 𝑣, (𝑣, 𝑡) ⟼ 𝑣𝑙
Such that
 ( 𝑣 + 𝑤) 𝑥 = 𝑣𝑥 + 𝑤𝑥 and
(𝜆𝑣)𝑥 = 𝜆(𝑣𝑥)
 𝑣( 𝑥 + 𝑔) = 𝑣𝑥 + 𝑣𝑔 and
 𝑣[ 𝑥, 𝑦] = ( 𝑣𝑥) 𝑦 − ( 𝑣𝑦) 𝑥
For all 𝑣, 𝑤 ∈ 𝑉 and 𝑥, 𝑦 ∈ 𝐿 and all 𝜆 ∈ 𝐹 respectively.
Lemma 4.4
Let L be a lie algebra over a field F. A representation 𝜌: 𝐿 → 𝐿𝑖𝑒 𝑒𝑛𝑑 (𝐹1×𝑛
))
makes the row space 𝔽 𝑒×𝑛
into an L – module by setting 𝑣𝑥 = 𝑣(𝑥𝜌)
Coversily if 𝑣 is an L-module then expressing the linear action as
endomorphism defines a representation of L of degree 𝑛. Thus the two concepts
two aspects of the same thing.
Example 4.4.1 (A presentation)
Let < be the Lie subalgebra of Lie (⊄ 𝑛×𝑛
) of lower triangular matrices

31. The map
)1( 2
2

 endL 







2,21,2
2,11,1
,1],[
aa
aa
njijai 
is a Lie algebra homomorphism and thus a representation. This make ⊄1×2
into
an L – module
Example 4.4.2
The adjoint representation
Let L be any Lie algebra over field 𝔽. The adjoint representation of L is its
action on itself.
ad : 𝐿 → 𝐿𝑖𝑒 (𝑒𝑛𝑑 ( 𝐿))
𝑋 ⟼ 𝑥 𝑎𝑑
: (𝑦 ⟼ [𝑦, 𝑥]
The map and is in fact a Lie algebra homomorphism and thus a representation.
To verify this, we first check that 𝑥 𝑛𝑑
is a linear map from L to L for every 𝑥 ∈
𝐿.
( 𝑦 + 𝜆𝑧) 𝑥 𝑎𝑑
= [𝑦 + 𝜆 𝑧, 𝑥]
= [ 𝑦, 𝑥] + 𝜆 [𝑧, 𝑥]
= 𝑦𝑥 𝑎𝑑
+ 𝜆(𝑧𝑥 𝑎𝑑
)

32. For 𝑦, 𝑧 ∈ 𝐿 and 𝜆 ∈ 𝐹. The map ad itself is linear, since
𝑧 (𝑥 + 𝜆𝑦) 𝑎𝑑
= [𝑧, 𝑥 + 𝜆𝑦]
= [ 𝑧, 𝑥] + 𝜆[𝑧, 𝑥]
= 𝑧𝑥 𝑎𝑑
+ 𝑧(𝜆𝑦 𝑎𝑑
)
for all 𝑥𝑥, 𝑦, 𝑧 ∈ 𝐿 and 𝜆 ∈ 𝔽.
Finally, the Jacobi identity shows that ad is homomorphism of Lie algebras:
𝑧 [𝑥, 𝑦] 𝑎𝑑
= [𝑧1[ 𝑦, 𝑧] − [𝑦1[ 𝑧, 𝑥]]
= [[ 𝑧, 𝑥], 𝑦] − [[ 𝑧, 𝑦], 𝑥]
= ( 𝑧𝑥 𝑎𝑑) 𝑧 𝑎𝑑
− (𝑧𝑦 𝑎𝑑
)𝑥 𝑎𝑑
for all 𝑥, 𝑦, 𝑧 ∈ 𝐿
Example 4.4.3
One – dimensional representation.
A one – dimensional representation of a Lie – algebra Lower 𝔽 is simply a
linear map 𝜌 ∶ 𝐿 → 𝐹 with
[ 𝑥, 𝑦] 𝜌 = ( 𝑦𝜌).( 𝑦𝜌) − ( 𝑥𝜌) = 0

33. For all 𝑥, 𝑦 ∈ 𝐿. since 𝔽 is commutative. So the one – dimensional
representation of L are precisely the 𝔽 − linear maps to 𝔽 that vvanish on the
subspace 𝐿′
= 𝐿(1)
= ( 𝐿1 𝐿).
This shows for examples that the simple Lie algebra 𝑠𝑙 𝑎 has only one –
dimensional representation which is the zero map.
𝑠𝑙2 → 𝑐, 𝑥 ⟼ 0
Anyway, the Kernel of such a representation is an ideal so it can only be 0 or
𝑠𝑙2 because 𝑠𝑙2 is simple.
Definition 4.5
Let B be a lie algebra ouer a field F and v be an L-module. A subspace W of v
is called a submodule, it if is invariant under the action of L:
𝑤𝑥 ∈ 𝑊 for all 𝑤 ∈ 𝑊 and 𝑥 ∈ 𝐿
Definition 4.6
A module v is called irredcible, if it has no sabmodules other than 0 and v
itself. A module v is the direect sum
𝑤1⨁…..⨁𝑤 𝑘 of submodules 𝑤1, 𝑤2 … 𝑤 𝑘 if it is direct vector space direct sum
of the Wi. A module v is called indecomposable if it not the diarect sum of two
non –trival submodules.

34. Remark 4.7.1
Irreducible implies indecomposable
An urreducible L – module is clearly indecomposable, However, the reverse
implication does not hold in general. There are Lie algebra with modules v that
have a proper submodule 0 < 𝑤 < 𝑣, for which there is no other submodule 𝑣
with 𝑣 = 𝑊⨁𝑢.
Remark 4.7.2 : Irreducible adjoint representation
Let 𝑣 ∶= 𝐿 be the L – module given by the adjoint representation. A submodule
of v is the same as an ideal of L. The module v is irreducible if and only if L is a
simple Lie algebra.
Definition 4.8
Let L be a Lie algebra ouer a field F. A homomorphism of L-modules in F-
linear map
𝑇 ∶ 𝑣 ⟶ 𝑣′

35. Between two L –modules v and 𝑣′
, such that ( 𝑣𝑇) 𝑥 = ( 𝑥𝑣) 𝑇 for all 𝑣 ∈ 𝑉 and
𝑥 ∈ 𝐿. It is called an isomorphism if there is a homomorphism 𝑠 ∶ 𝑣′
→ 𝑣 of L
module with 𝑇𝑠 = 𝑖𝑑 𝑣 and 𝑆𝑡 = 𝑖𝑑 𝑣′
Definition 4.9
Let v be an 𝔽 - vector space and 𝑇 ∶ 𝑣 → 𝑉 a linear map. Then an eigenvalue is
an element 𝜆 ∈ 𝔽, for which a vector 𝑣 ∈ 𝑉{0} exists with
𝑣𝑇 = 𝜆. 𝑣
Every such v is called an eigen vector for the eigenvalue 𝜆. The set of eigen
vectors for the eigen value 𝜆 together with the zero vector called eigen space
for eigen value?
For 𝔽 − ℂ, every endomorphism T has an eigen value, since the characteristic
polynomial of T has a root. Since ℂ is algebraically closed.
Lemma 4.10
Let v and 𝑣′
be irreducible L-modules for a Lie algebra l ouer 𝔽 and let 𝑇 ∶
𝑣 → 𝑣′
be an L – module homomorphism. Then either T maps every elements
of v to zero or it is an isomorphism.
Proof

36. The image in T and the kernel Ker T of T are submodules of 𝑣′
and v
respectively, since both v and 𝑣′
are irreducible, either in 𝑇 = 0 and ker 𝑇 =
𝑉, 𝑜𝑟 𝑖𝑚 𝑇 = 𝑣′
and ker 𝑇 = 0
Corollary 4.11
Let v be an irreducible L – module for a Lie algtebra Louer ℂ and 𝑇 ∶ 𝑣 → 𝑉 be
an L-module homomorphsm then T is a scalar multiple of the identity map.
Proof
Let 𝑇 ∶ 𝑣 → 𝑉 be any L-endomorphism. The T is in particular a linear map from
v to V so it has an eigen value 𝜆 with corresponding eigen vector 𝑣 ∈ 𝑉. Thus
the linear map − 𝜆. 𝑖𝑑 𝑣 must be equal to zero and thus T = 𝜆. 𝑖𝑑 𝑦. Note that 𝜆
can be equal to 0.
Theorem 4.12
Let L be a semisimple Lie algebra ouer ℂ and v a finite-dimensional L-module.
Then V has irreducible sub modules 𝑤1, 𝑤2 … 𝑤 𝑘 such that 𝑉 = 𝑤1⨁…⨁𝑤 𝑘
for some 𝑘 ∈ ℕ. That is v is the direct sum of irreducible sub modules.

37. BIBLIOGRAPHY
(1)J.S Milne, Lie algebra, Algebraic groups and Lie groups
(2)Karin erdmann and Mark J. Wildon, Indroduction to Lie – algebra,
springer international edition
(3)Max neunhoffer, Lie algebras.