Introduction to Lie Algebras

8/14/2019 Introduction to Lie Algebras

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Workshop on Group Theory: Classification of Reductive Algebraic GroupsIndian Statistical Institute, Bangalore Centre

December 18, 2006 - January 5, 2007

Introduction to Lie Algebras

Amber HabibMathematical Sciences Foundation, Delhi

Abstract

An introduction to Lie algebras. This material sets the stage for

the study of semisimple and reductive Lie algebras via their root sys-

tems. In turn, those Lie algebras are the means of understanding

reductive groups.

Contents

1 Lecture 1 Basic Definitions and Examples 1

2 Lecture 2 Ideals, Quotients & Homomorphisms 6

3 Lecture 3 Solvable and Nilpotent Lie Algebras 9

4 Lecture 4 Theorems of Engel and Lie 13

5 Lecture 5 Jordan Decomposition 19

1 Lecture 1 Basic Definitions and Examples

For our basic example we consider the vector space L(V) of linear operatorson a vector space V (over a field F). Besides the vector space operationsof addition and scaling, this has another natural operation: composition oflinear operators. This operation is not commutative: in general, fg = g f.We can try to capture the amount of non-commutativity by defining

[f, g] = f g g f.

1


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1 LECTURE 1 BASIC DEFINITIONS AND EXAMPLES 2

Now we have a new operation, [ , ] : L(V) L(V) L(V). First, we easily

see it is bilinear:

[f + g,f + g] = [f, f] + [f, g] + [g, f] + [g, g],

where , , , F. Next, it is not commutative. In fact, we have

[f, g] = [g, f] and [f, f] = 0.

Finally, let us consider associativity:

[f, [g, h]] [[f, g], h] = f[g, h] [g, h]f [f, g]h + h[f, g]

= f gh f hg ghf + hgf f gh + gf h + hf g hgf

= g[f, h] [f, h]g = [g, [f, h]].

So the bracket is not associative either. However, the last calculation can berewritten in a form which is quite useful:

[f, [g, h]] + [g, [h, f]] + [h, [f, g]] = 0.

Note the cyclic pattern.

The properties listed above lead to the following abstract notion:

Definition 1.1 A Lie algebra g is a vector space (over a field F) with abilinear operation g g g called the bracket or commutator, and denoted(X, Y) [X, Y], such that:

1. [X, X] = 0 X, Y g.

2. [X, [Y, Z]] + [Z, [X, Y]] + [Y, [Z, X]] = 0 X , Y , Z g.

The first property of the bracket is called anti-commutativitywhile the second

is the Jacobi identity.

Exercise 1.2 Show that in a Lie algebra, [X, Y] = [Y, X].

Lie algebras can be studied for their own sake, but ourinterest in them arisesout of their applications to the study of certain groups. Roughly, to each suchgroup we will assign a Lie algebra which will contain local information aboutthis group. Its job will be to convert problems about group structure toproblems in linear algebra.


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Exercise 1.3 Let A be an associative algebra overF. Define [a, b] = ab ba

for a, b A. Show that this bracket makes A a Lie algebra.

Example 1.4 Consider L(V) with the bracket [f, g] = f g g f. Wehave seen that it becomes a Lie algebra, and we shall call this Lie algebrathe general Lie algebra and denote it by gl(V).

Definition 1.5 Let g be a Lie algebra. We have the following definitions.

1. The Lie algebra g is abelian if the bracket is trivial: [X, Y] 0.

2. A subset h g is a Lie subalgebra ofg if it is a vector subspace and is

closed under the bracket operation.

3. A subset h g is an idealofg if it is a vector subspace and H h, X gimplies [H, X] h.

4. Ifh is another Lie algebra, then : g h is a Lie algebra homomor-phismif it is linear and preserves the bracket:

[X, Y] = [X,Y] X, Y g.

5. A Lie algebra h is isomorphic to g if there is a bijective Lie algebra

homomorphism : g h. Then is called an isomorphism. (Notethat 1 is then an isomorphism from h to g.)

6. Let V be a vector space. A Lie algebra homomorphism g gl(V) iscalled a representation ofg in V.

Exercise 1.6 Classify the one and two dimensional Lie algebras up to iso-morphism.

Exercise 1.7 Let : g h be a Lie algebra homomorphism. Show thatim is a Lie subalgebra ofh and ker is an ideal in g.

Example 1.8 If the vector space V has a basis of size n, it becomes identifiedwith Fn and L(V) with M(n,F) - the nn matrices with entries in F. Underthis identification, composition becomes matrix multiplication and so thebracket is now defined by

[A, B] = AB BA.

M(n,F) with this bracket is denoted by gl(n,F). Clearly gl(V) and gl(n,F)are isomorphic.


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Example 1.9 With the Lie algebra gl(n,F) in hand, we obtain others by

considering various familiar subspaces:

1. sl(n,F) = {X gl(n,F) : Trace(X) = 0.}. (Special Linear Algebra)

2. skew(n,F) = {X gl(n,F) : X+ Xt = 0.}.

3. t(n,F) = {X gl(n,F) : X is upper triangular}.

4. n(n,F) = {X gl(n,F) : X is strictly upper triangular}.

5. d(n,F) = {X gl(n,F) : X is diagonal}.

Exercise 1.10 Which of the above Lie algebras depend on the choice of basis,and to what extent?

Since sl(n,F) is independent of the choice of basis, we can denote it by sl(V).

Definition 1.11 A Lie algebra is called linear if it is a Lie subalgebra ofgl(n,F).

Example 1.12 We shall describe a machine for generating many linear Liealgebras. Let V = Fn and J M(n,F). Then define

gJ := {X gl(n,F) : JX+ XtJ = 0}.

It is easily verified that gJ is a vector subspace and also closed under bracket,hence it is a Lie subalgebra ofgl(n,F). For example, J = I gives gI = o(n,F).

Exercise 1.13 Show that if J and K are orthogonally similar, thengJ and

gK are isomorphic.

Example 1.14 Let us consider various choices of J. (Note: The explicitdescriptions of the Lie algebras below involve the assumption that char(F) =2.)

1. Let n = 2p and consider

J =

0 I

I 0

,


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where I is the p p identity matrix. Then

gJ =

X YZ Xt

: X , Y , Z M(p,F), Y = Yt, Z = Zt

is called the symplectic algebra and denoted by sp(n,F).

2. Let n = 2p and consider

K =

0 II 0

,


gK =

X YZ Xt

: X , Y , Z M(p,F), Y + Yt = Z + Zt = 0

is called the orthogonal algebra and denoted by o(n,F) = o(2p,F).

3. Let n = 2p + 1 and consider

L =

1 0 00 0 I0 I 0

,


gL =

0 b

t ct

c X Yb Z Xt

: b, c Fp, X , Y , Z M(p,F),

Y + Yt = Z+ Zt = 0

is also called the orthogonal algebraand denoted o(n,F) = o(2p + 1,F).

Exercise 1.15 Show thato(n,F) is isomorphic to skew(n,F), provided thatF is algebraically closed.

Exercise 1.16 Consider g = R3 with the vector cross-product

[X, Y] := X Y.

Verify g is a Lie algebra. Show it is isomorphic to o(3,R).


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2 LECTURE 2 IDEALS, QUOTIENTS & HOMOMORPHISMS 6

Exercise 1.17 Let Eij M(n,F) be defined as having all entries equal 0,

except that the (i, j) one equals 1. Show that

[Eij, Ekl] = jkEil liEkj .

Exercise 1.18 Show thatdim(sp(n,F)) = 12

n(n+1), dim(o(n,F)) = 12

n(n1).

Exercise 1.19 Prove the isomorphisms sl(2,F) = o(3,F) = sp(2,F), as-suming char(F) = 2.

2 Lecture 2 Ideals, Quotients & Homomor-phisms

Let g be a Lie algebra. Given A, B g we define their bracket by

[A, B] = span{[X, Y] : X A, y B}.

Thus, a subspace h g is a Lie subalgebra if [h, h] h. It is an ideal if[h, g] h.

Exercise 2.1 Let h1, h2 be ideals of g. Then so are h1 h2, h1 + h2 and[h1, h2].

Definition 2.2 The derived algebra of g is the ideal [g, g]. It is also calledthe commutator ideal.

Let h be an ideal in g. Then the quotient vector space g/h becomes a Liealgebra under the bracket

[X+ h, Y + h] := [X, Y] + h.

Exercise 2.3 Let h be an ideal ing. The quotient map : g g/h, X X+ h is a Lie algebra homomorphism.

One has the usual results about ideals and quotients:

Exercise 2.4 Let : g h be a Lie algebra homomorphism. Then g/ ker is isomorphic to im .


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Exercise 2.5 Let : g h be a Lie algebra homomorphism, and j an ideal

of g contained in ker . Then there is a unique homomorphism : g/j hsuch that the following diagram commutes:

g

g/j

? - h

-

Exercise 2.6 If j, k are ideals of g such that j k, then k/j is an ideal in

g/f j andg/j

k/jis naturally isomorphic to g/k.

Exercise 2.7 If j, k are ideals ofg, then

j + k

jis naturally isomorphic to

k

j k.

Example 2.8 As the Workshop proceeds, we shall see that the Lie algebrag = sl(2,F) has a special role to play. Let us look at its structure in detail.

A natural choice of basis for g is:

X =

0 10 0

, Y =

0 01 0

, H =

1 00 1

.

Since the bracket is bilinear, we only have to understand the bracket relationsbetween these basis elements. They turn out to have a simple form:

[H, X] = 2X, [H, Y] = 2Y, [X, Y] = H.

In fact, consider the linear g g map defined by Z [H, Z]. The map

is diagonalizable and the basis elements are its eigenvectors! This suggeststhat it would be useful to study the bracket via the linear maps it induces.

An implication of these calculations is that [g, g] = g if char(F) = 2.

Definition 2.9 Let g be a Lie algebra. For any X g, define a linear mapad(X) : g g by

ad(X)Y = [X, Y].

This map is called the adjoint of X.


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Exercise 2.10 Show that ad(X) = ad(Y) implies [X, Y] = 0. Is the con-

verse true?

Exercise 2.11 Show that ad(X)[Y, Z] = [ad(X)Y, Z] + [Y, ad(X)Z].

Definition 2.12 A linear map D : g g is a derivation if

D[X, Y] = [DX,Y] + [X,DY], X, Y g.

The collection of all derivations of g is denoted by Der(g).

Exercise 2.13 Show that any derivation D of g satisfies the Leibniz rule:

Dn[X, Y] =ni=0

nCi [DiX, DniY].

Remark The binomial coefficients are defined for arbitrary F recursively by

nC0 =nCn = 1 and

nCi =n1Ci1 +

n1Ci for 0 < i < n.

Exercise 2.14 Show that Der(g) is a Lie subalgebra ofgl(g).

Exercise 2.15 Show that the adjoint map ad : g gl(g) is a representationof g in g.

Note that the image of the adjoint representation lies in Der(g). Members ofthis image are called inner derivations. Derivations which are not inner arecalled outer.

Exercise 2.16 g/[g, g] is an abelian Lie algebra.

Exercise 2.17 Show that t(n,F) = d(n,F) + n(n,F) (vector space directsum). Also:

[d(n,F), n(n,F)] = n(n,F)

[t(n,F), t(n,F)] = n(n,F)

Exercise 2.18 LetX gl(n,F) be diagonalizable with eigenvalues a1, . . . , an.Then ad(X) is diagonalizable with eigenvalues ai aj (1 i, j n).


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3 LECTURE 3 SOLVABLE AND NILPOTENT LIE ALGEBRAS 9

Exercise 2.19 Show that the center ofgl(n,F) equals s(n,F), which consists

of the scalar matrices. In addition,

char(F) n = s(n,F) + sl(n,F) = gl(n,F)

(vector space direct sum),

char(F) = 2 = [sl(n,F),sl(n,F)] = sl(n,F),

2 = char(F) n = [gl(n,F), gl(n,F)] = sl(n,F).

Exercise 2.20 If char(F) n then the center of sl(n,F) is 0. Else, it iss(n,F).

Exercise 2.21 If char(F) = 2 then the only non-trivial ideals in gl(2,F) aresl(2,F) and s(2,F).

Exercise 2.22 Let X g and D Der(g). Then [D, ad(X)] = ad(DX).(Hence the inner derivations form an ideal in Der(g).)

Exercise 2.23 Show thatsl(2,F) has no non-trivial ideals if char(F) = 2.

3 Lecture 3 Solvable and Nilpotent Lie Al-

gebras

Definition 3.1 The center ofg is defined by

Z(g) = {X g : [X, Y] = 0 Y g} = ker(ad).

Z(g) is an ideal ofg. If Z(g) = g, then g is abelian. If Z(g) = {0}, then adis one-one and so g is isomorphic to the linear Lie algebra ad(g).

Definition 3.2 A Lie algebra g is called simple if it has no ideals except

itself and 0, and [g, g] = 0. (The last requirement exactly excludes the one-dimensional Lie algebra!)

For example, sl(2,F) is simple if char(F) = 2.

A simple Lie algebra must have zero center and hence the adjoint represen-tation makes it isomorphic to a linear Lie algebra.

Definition 3.3 An automorphism of g is an isomorphism with itself. Thecollection of all automorphisms of g is denoted Aut(g).


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Example 3.4 Let g = gl(V) or sl(V). Let A L(V) be invertible. Then

the map X AXA1 is an automorphism ofg.

Until further notice, assume char(F) = 0.

Suppose X L(V) is nilpotent: XM = 0. Then we can define its exponential:

exp(X) =N1n=0

1

n!Xn.

Note that exp(X) L(V).

Exercise 3.5 Suppose X, Y L(V) are nilpotent and commute. Then

exp(X+ Y) = exp(X) exp(Y).

In particular, exp(X) has inverse exp(X).

Suppose X gl(V) is nilpotent: XM = 0. Let l(X) denote left multiplicationby X and r(X) denote right multiplication by X. Then l(X) and r(X) arecommuting nilpotent maps. In fact l(X)M = r(X)M = 0. Also, ad(X) =l(X) r(X). Therefore

ad(X)2M = (l(X) r(X))2M =2Mn=0

2MCn l(X)nr(X)2Mn = 0.

Thus ad(X) is nilpotent. So we can define exp(ad(X)) : g g and it is alinear isomorphism. In fact,

exp(ad(X))Y = exp(l(X) r(X))Y = exp(l(X)) exp(r(X))Y

= exp(l(X))Y exp(X) = exp(X)Y exp(X)

It is easy to see from this that exp(ad(X)) Aut(gl(V)).


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More generally, suppose D is a nilpotent derivation of g. Then exp(D) is

defined and is a linear isomorphism of g. If DM = 0, we calculate:

[exp(D)X, exp(D)Y] = Mk=0

Dk

k!X,

Ml=0

Dl

l!Y

=

Mk=0

Ml=0

[Dk

k!X,

Dl

l!Y

=2Mn=0

ni=0

[Di

i!X,

Dni

(n i)!Y

=2Mn=0

1

n!

ni=0

nCi[DiX, DniY

=2Mn=0

1

n!Dn[X, Y] (Leibniz Rule)

= exp(D)[X, Y]

Thus, exp(D) Aut(g). In particular, ifX g such that ad(X) is nilpotent,then exp(ad(X)) Aut(g).

Definition 3.6 Int(g) denotes the subgroup of Aut(g) generated by auto-morphisms of the form exp(ad(X)), where X g and ad(X) is nilpotent.Members of Int(g) are called inner automorphisms.

Exercise 3.7 Int(g) is a normal subgroup of Aut(g): If Aut(g) andX g then exp(ad(X))1 = exp(ad(X))

We now remove the assumption that char(F) = 0.

So far, we have developed the basic theory of Lie algebras without any restric-tion on their dimension (though our examples have been finite dimensional).

From here on, we shall assume the Lie algebras to be finite dimensional(though certain infinite dimensional ones will temporarily appear later inthis Workshop). And if g gl(V) is linear, we assume V is finite dimen-sional.

We shall now start exploring the structure of a Lie algebra via its ideals. Onthe one extreme, we have simple Lie algebras such as sl(2,F) which have nonon-trivial ideals. On the other, are the abelian ones in which every subspaceis an ideal. In between are the algebras in the following example.


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Example 3.8 Consider t= t(n,F). Its commutator ideal is [t, t] = n(n,F),

which is non-trivial. Every superspace of n(n,F) is clearly an ideal in t.However, d(n,F) is not an ideal.

Now let us consider n = n(n,F). Its commutator ideal is

n1 := [n, n] =

0 0 . . .

. . .

. . . 00 0

.

Moreover, n1

is also an ideal of t: [n1

, t] = n1

. We can repeat these calcula-tions using matrices where the non-zero entries keep shifting more and moretowards the top-right corner.

To bring order to these observations, we set up two series of nested ideals:

1. Derived Series: Define g(0) = g, g(1) = [g, g], g(2) = [g(1), g(1)], . . . , andin general g(i+1) = [g(i), g(i)].

2. Lower (or Descending) Central Series: Define g0 = g, g1 = [g, g],

g2

= [g, g1

], ..., and in general gi+1

= [g, gi

].

Exercise 3.9 Prove that each g(i), gi is an ideal ing. (Hence each series isdescending.)

Exercise 3.10 Leth be an ideal ing. Show that eachh(i), hi is an ideal ing.

Definition 3.11 A Lie algebra g is solvable if g(i) = 0 for some i. It isnilpotentifgi = 0 for some i.

Clearly gi g(i) and so nilpotent Lie algebras are also solvable.

Exercise 3.12 Showt(n,F) is solvable but not nilpotent. On the other hand,n(n,F) is nilpotent.

Exercise 3.13 If g is solvable or nilpotent, then so is every subalgebra orhomomorphic image ofg.


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4 LECTURE 4 THEOREMS OF ENGEL AND LIE 14

However, there is the following result.

Exercise 4.5 Ifg/Z(g) is nilpotent, so is g. (Hint: (g/Z(g))i = gi/Z(g).)

Exercise 4.6 A Lie algebrag is solvable or nilpotent iff ad(g) is so.

Exercise 4.7 Every Lie algebrag has a unique maximal solvable ideal.

Definition 4.8 The unique maximal solvable ideal of the Lie algebra g iscalled its radical and is denoted Rad(g).

Definition 4.9 If Rad(g) = 0, g is called semisimple.

A semisimple Lie algebra must have zero center, hence its adjoint represen-tation is injective.

Simple Lie algebras are semisimple. So is the 0 algebra.

Exercise 4.10 For any Lie algebrag, g/Rad(g) is semisimple.

Let us turn to nilpotent Lie algebras. The condition gn = 0 means that forany X1, . . . , X n+1 g we have

[X1, [X2, . . . [Xn, Xn+1] . . . ]] = 0, or ad(X1)ad(X2) ad(Xn) = 0

In particular: ad(X)n = 0 X g.

Definition 4.11 If ad(X) is nilpotent, we say X is ad-nilpotent.

We have just observed that if g is nilpotent, then each element of g is ad-nilpotent. Amazingly, the converse is also true.

Theorem 4.12 Letg gl(V) be a linear Lie algebra, V = 0. If eachX g

is nilpotent, then v V such that v = 0 and Xv = 0 X g

.

Proof. We proceed by induction on dim(g). When dim(g) = 0, the result iseasy. Now consider an arbitrary g satisfying the hypotheses of the theorem.

We first show g has an ideal h of co-dimension one. Choose h to be anymaximal proper subalgebra of g. For any H h, ad(H) preserves h, hencewe can define

ad : h gl(g/h), ad(H)(X+ h) = ad(H)X+ h.


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Since each ad(H) is nilpotent, so is each ad(H). Applying the induction

hypothesis to ad(h), we find a non-zero Y + h such that ad(H)(Y + h) = hfor every H h. Then Y / h and ad(H)Y h for each H h.

This shows h + FY is a subalgebra, properly containing h. Hence we musthave g = h + FY. So h has co dimension one, and is also an ideal.

Now apply the induction hypothesis to h: v V, v = 0, such that Hv = 0for every H h. Let

W = {v V : Hv = 0 H h}.

Also, fix Y / h: then g = h + FY. We have to find a non-zero v W suchthat Y v = 0. For any w W and H h,

HY w = [H, Y]w + Y Hw = 0,

and so Y w W. Therefore Y acts on W. Since the action must be nilpotent,there is a non-zero v W such that Y v = 0.

Corollary 4.13 Let g gl(V) be a linear Lie algebra. If each X g isnilpotent, then there is a basis ofV such that each member ofg is representedby a strictly upper triangular matrix.

Corollary 4.14 (Engels Theorem) Letg be a Lie algebra such that eachelement is ad-nilpotent. Theng is nilpotent.

In fact, the entire family of results 4.12-4.14 is generally grouped under thename of Engels Theorem.

Exercise 4.15 Letg be a nilpotent Lie algebra and h a non-zero ideal ing.Thenh Z(g) = 0.

Now we shall start exploring the structure of Lie algebras via eigenvectorsand eigenvalues, hence:

We assume that the underlying field F is algebraically closed.

We have seen that prime characteristic creates various exceptions to generalpatterns, and so we also assume that char(F) = 0.

Example 4.16 Let g be the non-abelian two dimensional Lie algebra, withbasis {X, Y} such that [X, Y] = X. Then it is solvable: g(1) = [g, g] = FX,


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g(2) = [g(1), g(1)] = 0. Consider its adjoint representation:

ad(X) =

0 10 0

ad(Y) =

1 0

0 0

Hence for a general element T = aX+ bY g,

ad(aX+ bY) =

b a

0 0

.

Clearly, there is no non-zero vector Z such that ad(T)Z = 0 for every T g.However X is at least a common eigenvector for each ad(T):

ad(aX+ bY)X = bX.

Theorem 4.17 Let g gl(V) be a solvable Lie algebra. If V = 0 then itcontains a common eigenvector for all the elements of g.

Proof. We proceed by induction on dim(g), broadly following the scheme used

for Engels Theorem. If dim(g) = 0, the claim is trivial.

For a general g we first locate an ideal of co dimension one. Since g is solvable,[g, g] = g. Hence we can choose some vector subspace k g such that it hasco dimension one and contains [g, g]. But then,

[k, g] [g, g] k,

hence k is an ideal.

By the induction hypothesis, there is a common eigenvector v V for each

element ofk: Kv = (K)v, K k.

It is easy to see that : k F is linear. Now define:

W = {w V : Kw = (K)w, K k}.

Since v W, W is a non-zero subspace of V. We shall show g preserves W.Let X g and w W. Then for any K k,

KXw = [K, X]w + XKw = ([K, X])w + (K)Xw.


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To show Xw W we have to prove that ([K, X]) = 0. Let w W be

non-zero and consider the sequence {w,Xw,X2w . . . }. Let n be largest suchthat

Wn = {w , X w , . . . , X nw}

is linearly independent. We have:

KXiw = KXXi1w

= [K, X]Xi1w + XKXi1w

= ([K, X])Xi1w + (K)Xiw.

Hence the action of K on Wn is given by an upper triangular matrix whosediagonal entries are all (K). Applying this to the action of [K, X], wefind that its trace is Tr [K, X]|Wn = (n + 1)([K, X]). By our choice ofn, we also have XWn Wn, and so [K, X]|Wn = [K|Wn, X|Wn]. But thenTr [K, X]|Wn = 0, and so ([K, X]) = 0 (Since char(F) = 0).

So we have shown that g preserves W. Let g = k+FZ. Since F is algebraicallyclosed, Z has an eigenvector v W. Then v is clearly a common eigenvectorfor g.

Corollary 4.18 (Lies Theorem) Letg gl(V) be a solvable Lie algebra.Then V has a basis such that the matrix of each X g is upper triangular.

Proof. Proceed by induction on n = dim(V). The n = 0 case is trivial.Assume n > 0. We have a common eigenvector v for the g action on V. LetV1 = V /Fv. Define : g gl(V1) by

(X)(w + Fv) = Xw + Fv.

Since (g) is solvable, by the induction hypothesis, there is a basis {v1 +Fv , . . . , vn + Fv} of V1 such that each (g) is upper triangular. Then

{v, v1, . . . , vn}

is a basis ofV which makes g upper triangular.

Exercise 4.19 Let g be solvable and : g gl(V) a representation of g.Then V has a basis in which each (X) is upper triangular.

Exercise 4.20 Letg be solvable. Then there are ideals hi of g such that

0 = h0 h1 h2 hn = g, and dim(hi) = i.


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Exercise 4.21 Let g be solvable. Then ad(X) is nilpotent for each X

[g, g].

Exercise 4.22 A Lie algebrag is solvable if and only if [g, g] is nilpotent.

Exercise 4.23 The sum of two nilpotent ideals is nilpotent. Hence each ghas a maximal nilpotent ideal.

Exercise 4.24 LetX, Y L(V) commute. Let their Jordan decompositionsbe Xs+Xn andYs+Yn respectively. Then the Jordan decomposition of X+Yis (Xs + Ys) + (Xn + Yn).

Exercise 4.25 Show the previous result can fail if X, Y do not commute.

Exercise 4.26 Let g gl(V) be solvable. Show that Tr (XY) = 0 for allX [g, g] and Y g.

Exercise 4.27 Letchar(F) = 2 and letg gl(2,F) be the span of

X =

0 11 0

, Y =

0 00 1

.

Showg is a solvable Lie algebra but its elements have no common eigenvector

inF2.

Exercise 4.28 Let g be as in the previous exercise. Consider the vectorspace direct sumh = g F2 and define a bracket on it by

[X x, Y y] = [X, Y] (Xy Y x).

Show thath is a solvable Lie algebra but [h, h] is not nilpotent.

Exercise 4.29 Ifg is a real Lie algebra, its complexification is the complexvector space gC = gRC, with bracket defined by [Xw, Yz] = [X, Y]wz.

Verify gC is a Lie algebra overC.

Exercise 4.30 Letg be a real Lie algebra. Show it is solvable if and only ifgC is.

Exercise 4.31 If g is a solvable real Lie algebra, then [g, g] is nilpotent.


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5 LECTURE 5 JORDAN DECOMPOSITION 19

5 Lecture 5 Jordan Decomposition

Jordan Decomposition

We keep in place the requirement that F is algebraically closed,but allow arbitrary characteristic.

Diagonalizable linear maps over an algebraically closed field are also calledsemisimple.

We have the following standard result from linear algebra:

Theorem 5.1 (Jordan Decomposition) LetV be a finite dimensional vec-tor space over an algebraically closed field F. Then any X L(V) has aunique decompositionX = S+ N where S is semisimple, N is nilpotent and[S, N] = 0. Moreover,

1. There exist polynomials p, q without constant term such that p(X) = Sand q(X) = N.

2. For any eigenvalue of X, define its generalized eigenspace by

V = {v V : (X I)k

v = 0 for some k}.

Then V is the direct sum of the Vs and S acts on V by .

S and N are called (respectively) the semisimple and nilpotent parts of X.

Exercise 5.2 Let X = S + N be the Jordan decomposition of X L(V).Then

1. IfM L(V) commutes with X, it commutes with S and N.

2. IfA B V are subspaces such that X(B) A, then S(B), N(B) A.

Exercise 5.3 Let X g = gl(V) have Jordan decomposition X = S + N.Then ad(X) gl(g) has Jordan decomposition

ad(X) = ad(S) + ad(N).

Theorem 5.4 Letg be a Lie algebra. Then Der(g) contains the semisimpleand nilpotent parts of all its elements.


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Proof. Let D Der(g) with Jordan decomposition D = S+ N. We have to

show S Der(g). Consider the generalized eigenspaces of D:

g = {X g : (D I)kX = 0 for some k}.

To see that S is a derivation, it is enough to apply it to brackets of the form[X, Y] where X g and Y g. We first show that [g, g] g+. Notethat

(D )[X, Y] = [DX,Y] + [Y,DX] [X, Y] [X, Y]

= [(D )X, Y] + [X, (D )Y].

This is easily generalized by induction to

(D )n[X, Y] =ni=0

nCi[(D )iX, (D )niY].

Let (D I)kX = (D I)kY = 0. Then it follows that

(D ( + )I)2k[X, Y] = 0.

Therefore, for X g and Y g, we have

S[X, Y] = ( + )[X, Y] = [X,Y] + [X,Y] = [SX,Y] + [X,SY].

The direct sum decomposition g = g now implies that S Der(g).

Cartans Criterion

We assume that F is algebraically closed and that char(F) = 0.

For a linear map X L(V) the following criterion for solvability is quiteeasy to obtain: Let A B V be subspaces, and define

M = {T L(V) : T(B) A}.

Suppose X M satisfies Tr (XT) = 0 T M. Then X is nilpotent.

We have a version of this for the adjoint action.

Lemma 5.5 LetA B be subspaces of L(V). Define

M = {X gl(V) : [X, B] A}.

Suppose X M satisfies Tr (XY) = 0 for all Y M. Then X is nilpotent.


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Proof. Let X have Jordan decomposition S+ N. Fix a basis of V in which

S has a diagonal matrix:

[S] =

a1. . .

an

.

Since char(F) = 0 we have the rationals Q F. Treat F as a vector spaceover Q and let E be the span (over Q) of a1, . . . , an. We have to show E = 0(as that will give S = 0).

We will show E = 0. Let f : E Q be linear. Define Y gl(V) by

[Y] =

f(a1). . .

f(an)

.

Then ad(S) has eigenvalues aiaj and ad(Y) has eigenvalues f(ai)f(aj), inboth cases corresponding to the eigenvectors Eij . By Lagrange interpolationthere is a polynomial r(t) F[t] such that

r(ai aj) = f(ai) f(aj), i,j.

It follows that r(ad(S)) = ad(Y). Note that r has zero constant term.

We know ad(S) is a polynomial in ad(X) without constant term, hence ad(Y)is itself a polynomial in ad(X) without constant term.

Since ad(X) maps B into A, so does ad(Y). Hence ad(Y) M. ThereforeTr (XY) = 0, which gives

i aif(ai) = 0. Applying f, we get

i f(ai)

2 = 0,and hence f(ai) = 0 i.

Exercise 5.6 If X , Y , Z L(V) then Tr ([X, Y]Z) = Tr (X[Y, Z]).

Recall the earlier exercise that g solvable implies Tr (ad(X)ad(Y)) = 0 forevery X [g, g] and Y g. We shall establish the converse.

Theorem 5.7 (Cartans Criterion for Solvability) Letg gl(V) be aLie algebra such that Tr (XY) = 0 for every X [g, g] and Y g. Thengis solvable.

Proof. Choose A = [g, g], B = g, and define

M = {T gl(V) : [T, B] A}.


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Clearly, [g, g] g M. Now let X, Y g and T M. Then

Tr ([X, Y]T) = Tr (X[Y, T]) = Tr ([Y, T]X).

By definition of M, [T, Y] B, hence by the hypothesis of the theoremTr ([Y, T]X) = 0. So we have obtained that Tr ([X, Y]T) = 0 for everyT M. It follows that Tr (ZT) = 0 for every Z [g, g] and T M. Theearlier Lemma therefore implies that Z is nilpotent for each Z [g, g].

Therefore [g, g] is nilpotent and g is solvable.

Exercise 5.8 Letg be a Lie algebra such that Tr (ad(X)ad(Y)) = 0 for every

X [g, g] and Y g. Theng is solvable.

Killing Form

Definition 5.9 Let g be a Lie algebra. Its Killing form : g g F isdefined by (X, Y) = Tr (ad(X)ad(Y).

Exercise 5.10 The Killing form ofg is bilinear, symmetric and associative:([X, Y], Z) = (X, [Y, Z]).

Definition 5.11 Given a bilinear symmetric form : V V F, its radicalis defined to be

S = {x V : (x, y) = 0 y V}.

If S = 0 we call the form non-degenerate.

Exercise 5.12 The radical of the Killing form of g is an ideal ofg.

Exercise 5.13 Leth be an ideal ofg. Let be the Killing form ofg and hthe Killing form ofh. Then h = |hh.

For the remaining results we again restrict to a field F which isalgebraically closed with char(F) = 0.

First, note that the Cartan Criterion corollary and its converse can now beexpressed as: g is solvable iff the radical of its Killing form contains [g, g].

Exercise 5.14 The radical of the Killing form of g is solvable. (Hence theradical of the Killing Form is in Rad(g).)


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Theorem 5.15 (Criterion for Semisimplicity) A Lie algebra is semisim-

ple if and only if its Killing form is non-degenerate.

Proof. Let S be the radical of the Killing form ofg.

Ifg is semisimple, S Rad(g) = 0.

Now, suppose Rad(g) = 0. Then the last non-zero term h in its derived seriesis an abelian ideal ofg. If X h and Y g then ad(X)ad(Y) maps g hand (ad(X)ad(Y))2 maps g [h, h] = 0. Therefore ad(X)ad(Y) is nilpotentand so (X, Y) = 0. This shows h S = 0.

Remark The first half of the proof uses Cartans Criterion and so needs Falgebraically closed and char(F) = 0. The second half works for any F andshows that for any g every abelian ideal is in S.

Exercise 5.16 Letg = F. Give representations ofg in which:

1. Every element ofg acts semisimply.

2. Every element ofg acts nilpotently.

3. No element ofg acts semisimply or nilpotently. Nor are the semisimpleor nilpotent parts of the images in the image of the representation.

Exercise 5.17 If g is nilpotent, its Killing form is identically 0.

Exercise 5.18 The radical of a Lie algebra need not equal the radical of itsKilling form.

Exercise 5.19 Compute the Killing form of sl(2,F).

Introduction to Lie Algebras

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