LIE ALGEBRAS AND LIE GROUPSMAT6633

ABRAHAM BROER

References

[1] Bourbaki, Nicolas, Elements de mathematique. XXVI. Groupes et algebres de Lie. Chapitre 1: Algbres de Lie.

Actualites Sci. Ind. No. 1285. Hermann, Paris 1960.

[2] Duistermaat, J.J.; Kolk, J.A.C., Lie Groups. Universitext, Springer-Verlag, New York, 1999.

[3] Fulton, W.; Harris, J., Representation Theory. A first Course. Graduate Texts in Mathematics 129, Springer-

Verlag, New York, 1991.

[4] Helgason, S. Differential Geometry and Symmetric Spaces. AMS Chelsea Publishing, AMS, Providence, Rhode

Island, 2000.

[5] Jacobson, Nathan, Lie algebras. Republication of the 1962 original. Dover Publications, Inc., New York, 1979.

[6] Knapp, Anthony W., Lie groups. Beyond an Introduction. Second Edition. Progress in Mathematics 140,

Birkhauser, Boston, 2002.

[7] Warner, Frank W., Foundations of Differential Manifolds and Lie Groups. Graduate Texts in Mathematics 94,

Springer-Verlag, New York, 1983.

1. Lie algebras

Let k be a field. A Lie algebra over k is a k vector-space g together with a k-bilinear product,

g× g→ g : x, y 7→ [x, y],

called the (Lie-)bracket, such that [x, x] = 0 for all x ∈ g and the the Jacobi-identity

[x, [y, z]] + [y, [z, x]] + [z, [x, y]] = 0

holds for all x, y, z ∈ g.We shall always assume that g is finite dimensional as a vector space and that the characteristic

of k is zero.Let V be a (finite dimensional) vector-space over k. Then

Endk(V ) := {φ : V → V ; φ is k-linear}

is of course an associative unitary algebra over k, but it is also a Lie algebra if we define the bracketby

[X,Y ] := XY − Y X,

where X,Y ∈ Endk(V ). The Jacobi-identity is checked by writing it out and using the associativityof the usual composition of endomorphisms.

A morphism of Lie algebras over k is a linear map between Lie algebras over k

φ : g→ h

Date: November 28, 2013.

1

2 ABRAHAM BROER

such that φ([x, y]) = [φ(x), φ(y)]. The collection of automorphisms is a group Aut(g). In particular,any map σ : g → g is in Aut(g) if and only if it is a k-linear isomorphism such that σ([X,Y ]) =[σ(X), σ(Y )], for all X,Y ∈ g.

A Lie sub-algebra (over k) is a sub vector-space h ⊂ g such that if x, y ∈ h then [x, y] ∈ h. A Liesub-algebra a is called an ideal if even [x, y] ∈ a for all x ∈ g and y ∈ a; we write aC g.

The kernel of a morphism φ : g→ h of Lie algebras over k is an ideal in g, and the image is a Liesub-algebra of h. If aC g then the quotient space g/a is a Lie algebra with bracket [x, y] := [x, y]and the natural map ν : g→ g/a : X → X is a surjective Lie algebra morphism.

Like in group theory, ring theory or module theory we have the usual isomorphism theorems.For every morphism of Lie algebras g/Ker ' Im. If a and b are ideals, then a+b and a∩b are alsoideals, and (a + b)/a ' b/(a ∩ b). There is a correspondence between the ideals (Lie sub-algebras)of g/a and the ideals (Lie sub-algebras) of g containing a.

If A and B are k-subspaces of g, define [A,B] to be the k-subspace spanned by the set {[a, b]; a ∈A, b ∈ B}. If a and b are ideals, then [a, b] is also an ideal.

A derivation D of a Lie algebra g is a k-linear map D : g→ g such that

D([x, y]) = [D(x), y] + [x,D(y)].

The collection Der(g) of all derivations of a Lie algebra g is a Lie sub-algebra of Endk(g) is aLie sub-algebra, since if D1 and D2 are derivations then also [D1, D2] := D1 ◦ D2 − D2 ◦ D1 is aderivation, as is checked easily.

Any X ∈ g gives rise to the derivation ad(X) : Y 7→ [X,Y ] (called inner derivation). The mapad : g→ Der(g) is a morphism of Lie algebras over k. Its kernel is the center z(g) of g:

z(g) = {X ∈ g : ∀Y ∈ g : [X,Y ] = 0}.

The image of ad is even an ideal in Der(g), since for any derivation D and X ∈ g we have

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

A derivation D is called nilpotent if there is an n ≥ 1 such that Dn = D ◦D ◦D ◦ . . . ◦D = 0.Define the Lie algebra automorphism exp(D) ∈ Aut(g) by

exp(D) =∞∑i=0

1i!Di

which makes sense since the field is of characteristic zero and Dn = 0 if n is big enough. Thenexp(tD) ◦ exp(sD) = exp((s + t)D), for any s, t ∈ k, and in particular exp(D)−1 = exp(−D). Ifk = R or k = C then the definition of exp(D) makes even sense for any endomorphism, since thelimit exists, and exp(D) ∈ Aut(g) if D ∈ Der(g).

Proof. Let X,Y ∈ g. By induction on r we prove that

Dr([X,Y ]) =r∑i=0

(r

i

)[Di(X), Dr−i(Y )].

For r = 1 it is true by the definition of D being a derivation. Supposing true for r then

LIE ALGEBRAS AND LIE GROUPS MAT6633 3

DDr([X,Y ]) =r∑i=0

(r

i

)D([Di(X), Dr−i(Y )])

=r∑i=0

(r

i

)([Di+1(X), Dr−i(Y )] + [Di(X), Dr+1−i(Y ))

=r+1∑i=0

(r + 1i

)([Di(X), Dr+1−i(Y )]).

And so

[exp(D)(X), exp(D)(Y ) = = [∑i

1i!Di(X),

∑j

1j!Dj(Y )]

=∑r

∑i+j=r

1i!j!

[Di(X), Di(Y )]

=∑r

1r!Dr([X,Y ])]

= exp(D)([X, y]).

�

A subalgebra is an ideal if and only if it is stable under all inner derivations. We shall say thata is a characteristic ideal and write a J g if it is even stable under all derivations. For example,the center of a Lie algebra is characteristic z(g) J z. The derived subalgebra is also g′ := [g, g] J g.

That aC bC g does not imply that aC g. But a J bC g does imply that aC g, and a J b J g

does imply that a J g. If a and b are characteristic ideals, then [a, b] is also a characteristic ideal.So if we define g(n) by g(1) = [g, g] and g(n+1) = [g(n), g(n)], then g(n) J g. So if we define gn by

g1 = [g, g] and gn+1 = [g, gn], then gn J g. We have gn ⊆ g(n).Let a be any linear subspace such that g′ ⊆ a ⊆ g. Then aC g, since we even have [X,Y ] ∈

[g, g] ⊆ a for any X and Y .A Lie algebra over k is called solvable if for some n > 0 g(n) = 0. A Lie algebra over k is called

nilpotent if for some n > 0 gn = 0. So if g is nilpotent it is also solvable. We have gn = 0 if andonly if for all X1, X2, . . . , Xn, Y ∈ g we have

ad(X1) ad(X2) . . . ad(Xn)(Y ) = [X1, [X2, [X3, . . . [Xn, Y ]]] . . .] = 0.

Let aC g. Then g is solvable if and only if both a and g/a are solvable. And if g is nilpotentthen a and g/a both are nilpotent.

If a and b are solvable ideals, then a + b is also a solvable ideal.

Proof. Since a∩ b and b are solvable, also b/(a∩ b) ' (a + b)/a is solvable. Since a is also solvable,it follows that a + b is solvable. �

If follows from the finite dimensionality of g there is a biggest solvable ideal containing all othersolvable ideals, called the (solvable) radical radg of g. (It is in fact a characteristic ideal, but theproof is not direct.)

4 ABRAHAM BROER

A Lie algebra is called semi-simple if its radical is 0. A Lie algebra is called simple if it isnon-abelian and has no non-trivial ideals.

Example 1.1. Let V be an n-dimensional vector-space over k. A (complete) flag is a sequence ofsub vector-spaces

0 = V0 ⊂ V1 ⊂ V2, . . . , Vn−1 ⊂ Vn = V

with dimK Vi = i.Then

{X ∈ Endk V ;∀1 ≤ i ≤ n : XVi ⊆ Vi}is a solvable Lie algebra over k and

{X ∈ Endk V ;∀1 ≤ i ≤ n : XVi ⊆ Vi−1}

is a nilpotent Lie algebra.

Example 1.2. Let

sl2(k) := {

(c a

b −a

); a, b, x ∈ k}

be the Lie algebra of 2× 2-matrices with coefficients in k and trace 0. It has basis

E =

(0 10 0

), F := {

(0 01 0

), H := {

(1 00 −1

)with brackets [E,F ] = H, [H,E] = 2E and [H,F ] = −2F . It is a simple Lie algebra.

Proof. Suppose aC sl2(k) is a non-zero ideal. Suppose A = aE+bF +cH ∈ a and a 6= 0. Then also[E, aE+bF +cH] = bH−2cE ∈ a and [E, bH−2cE] = −2bE ∈ a and so E ∈ a and [E,F ] = H ∈ a

and [H,F ] = 2F ∈ a. So a = sl2(k). Similarly if b 6= 0 or if A = cH. �

1.1. Representations. The notion of representation of a Lie algebra plays the same role as anmodule over a ring. Let k ⊂ K be a field extension and g a Lie-algebra over k. A (finite dimensional)K-representation for g on V is a morphism of Lie algebras over k:

F : g→ EndK(V ),

where V is (finite dimensional) vector space over K. We then use often the short-hand notation

X · v = Xv := F (X)(v),

for v ∈ V and X ∈ g. Then in particular

X · (Y · v)− Y · (X · v) = [X,Y ] · v,

for v ∈ V and X,Y ∈ g. We remark that although XY is not defined in the Lie algebra, the linearoperator F (X)F (Y is defined, and we define XY · v as X · (Y · v).

On the other hand, suppose we have a product

g× V → V : (X, v) 7→ X · v

which is K-linear in v, and k-linear in X such that

X · (Y · v)− Y · (X · v) = [X,Y ] · v,

LIE ALGEBRAS AND LIE GROUPS MAT6633 5

then we get a representation F : g → EndK(V ) if we define F (X)(v) := X · v. We shall then alsosay that V is a K-representation (where the action is defined).

If k = K we just say representation, instead of k-representation.A sub K-representation U ⊂ V is a linear subspace such that for all X ∈ g and u ∈ U we

have X · u ∈ U . If U1 and U2 are sub K-representations, then also U1 + U2 and U1 ∩ U2 are subK-representations.

If U ⊂ V is a sub K-representation then V/U becomes a K-representation when we defineX · v := X · v, where X ∈ g and v = v + U ∈ V/U . Using the natural map ν : V → V/U we geta correspondence between the K-sub-representations of V/U and the K-sub-representations of Vcontaining U .

One special representation is the adjoint representation

ad : g→ Endk(g) : X 7→ ad(X).

A sub-representation of g is then the same notion as an ideal. The Lie algebra Derk(g) also has arepresentation on g, by D ·X := D(X). Then a sub-representation of g for the Lie algebra Derk(g)is the same as a characteristic ideal.

Suppose V = Kv is a one-dimensional K-representation, then there is a function λ : g → K

defined by

X · v = λ(X)v

such that λ is K-linear and λ([X,Y ]) = 0 since

[X,Y ] · v = X · Y · v − Y ·X · v = λ(X)λ(Y )v − λ(Y )λ(X)v = 0.

So every bracket acts trivially on any K-representation. In fact, λ does not depend on the choice ofbasis element of V . Inversely, any such function λ gives rise to one-dimensional K-representation.So one-dimensional K-representations are in bijection with the k-linear maps g/[g, g]→ K.

1.2. Lie’s theorem. One of the important classical theorems of Sophus Lie concerns the repre-sentation theory of solvable Lie algebras.

Theorem 1.1 (Lie). Let g be a solvable Lie algebra over k. Let k ⊂ K a field extension and V bea (finite dimensional) K-representation on V . Suppose that for every X ∈ g all the eigenvalues ofthe linear maps v 7→ X · v are in K (this is of course the case if K is algebraically closed).

Then there is a complete flag

0 = V0 ⊂ V1 ⊂ V2, . . . , Vn−1 ⊂ Vn = V

of sub K-representations.Hence on an appropriate basis all matrices associated to the elements of g are simultaneously

represented by triangular matrices.

Proof. We start by proving that there is a one-dimensional sub K-representation V1. Or in otherwords, there is a simultaneous common eigenvector v1 ∈ V , i.e., v1 6= 0 and for all Z ∈ g thereexists a λ(Z) ∈ K such that Z · v1 = λ(Z)v1. Then V1 = Kv1. We shall use induction on dimk g.

6 ABRAHAM BROER

If g = kX is one-dimensional then the eigenvalues of X acting K-linearly on V are all containedin K, so by linear algebra there is indeed an eigenvector v for X with some eigenvalue λ. So for allcX ∈ g (c ∈ k) we have cX · v = cλv.

Assume that the dimension of g is ≥ 2 and that the result is true for solvable Lie algebras oflower dimension. Since g is solvable, the derived Lie algebra g′ = [g, g] is unequal to g. Take anyk-linear subspace a of codimension one in g such that a ⊇ g′. Then a is an ideal of g, and thereforeis also a solvable Lie algebra. Take any X 6∈ a, then kX ⊕ a = g as vector spaces. By inductionthere is a w ∈ V (w 6= 0) and a function λ : a→ K such that for all Y ∈ a

Y w = λ(Y )w.

Write now Vλ = {u ∈ V ; ∀Y ∈ a : Y u = λ(Z)u}. It is a non-zero linear subspace, since itcontains w, and is a-stable. We want to show that Vλ is even a K-subrepresentation of V for theLie algebra g.

If Z = cX + Y , Y ∈ a and u ∈ Vλ we have

Z · u = cX · u+ Y · u = cX · u+ λ(Y )u.

So it suffices to prove that X · u ∈ Vλ. Now

Y · (X · u) = [Y,X] · u+X · Y · u = λ([Y,X])u+X · (λ(Y )u) = λ([Y,X])u+ λ(Y )(X · u),

hence it suffices to prove that λ([Y,X]) = 0 for all Y ∈ a.For this we use the following argument. Define inductively e0 = w and ei+1 = X · ei for i ≥ 0.

Put Ui (i ≥ 0) for the K-linear subspace of V spanned by e0, . . . , ei, and put U−1 = 0. Let m beminimal such that X · em = em+1 ∈ Um, then necessarily Ur = Um for all r ≥ m. Put U = Um,then dimK U = m+ 1 and U is X-stable.

We claim that Y · ei−λ(Y )ei ∈ Ui−1 for any Y ∈ a. This is certainly true for i = 0 by the choiceof e0 = w. Suppose it is true for i ≥ 0, then

Y · ei+1 − λ(Y )ei+1 = Y · (X · ei)− λ(Y )X · ei= [Y,X] · ei +X · (Y · ei)−X · (λ(Y )ei) ∈ Ui

since [Y,X]ei ∈ Ui by induction and also

X · (Y · ei)−X · (λ(Y )ei) = X · (Y · ei − λ(Y )ei) ∈ X · Ui−1 ⊆ Ui.

So the claim follows and U is K-sub-representation for g and we get a Lie algebra homomorphismF : g→ EndK(U) such that F (X)(u) = X · u (for X ∈ g and u ∈ U).

We calculate the trace of F ([X,Y ]) acting on U in two ways. Since [X,Y ] ∈ a, we can cal-culate this trace using the basis e0, . . . em and the claim, and we get the value (m + 1)λ([X,Y ]).On the other hand the trace of a commutator is always 0 and tr(F ([X,Y ]) = tr(F (X)F (Y )) −tr(F (Y )F (X)) = 0. Hence the trace of F ([X,Y ]) acting on U is 0 = (m + 1)λ([X,Y ]). Weconclude that λ([X,Y ]) = 0.

So indeed Vλ is g-stable. Now let v1 ∈ Vλ be an eigenvector for X, then it is a simultaneouseigenvector for g acting on V . So there is a function λ1 : g→ K such that X · v1 = λ(X)v1.

LIE ALGEBRAS AND LIE GROUPS MAT6633 7

Now we shall prove the full version of Lie’s theorem by induction on the dimension of V . IfdimK V = 1 we have nothing left to do. Suppose dimK V > 1. We just showed there is a one-dimensional K-sub-representation V1 for g.

Now we consider the quotient K-representation V/V1 having lower dimension. So we can useinduction and the correspondence at the end of last sub-section to get we get a flag 0 = V0 ⊂ V1 ⊂V2 ⊂ . . . Vn = V of K-sub-representations of V such that 0 = V1/V1 ⊂ V2/V1 ⊂ . . . Vn/V1 is a flagof K-sub-representations of V/V1. We finished the proof of Lie’s theorem. �

1.3. Engel’s theorem. The following theorem due to Friedrich Engel deals with the situationwhen all the elements of g act nilpotently in some representation V .

Theorem 1.2 (Engel). Let g be a Lie algebra over k. Let k ⊂ K a field extension and V be a(finite dimensional) K-representation. Suppose that every X ∈ g acts nilpotently on V .

Then there is a complete flag

0 = V0 ⊂ V1 ⊂ V2, . . . , Vn−1 ⊂ Vn = V

of sub K-representations such that even for all X ∈ g and 1 ≤ i ≤ n we have

X · Vi ⊆ Vi−1.

So there is a K-basis v1, . . . , vn of V such that for all X ∈ g: X · v1 = 0 and for all i ≥ 2:X · vi ∈ Kv1 ⊕ . . . ⊕Kvi−1. Or on this basis all X ∈ g are simultaneously represented by strictlyupper-triangular matrices.

In particular, the image of g inside EndK(V ) is a nilpotent Lie algebra.

Proof. First a preparation. Replacing g by its image in EndK(V ) we can assume that g ⊂ EndK(V )is a Lie sub algebra over k. Also E = EndK(V ) is a K-vector space. For any M ∈ EndK(V ) letL(M) ∈ EndK(E) be left multiplication on E and let R(M) ∈ EndK(E) be right multiplication,hence for any N ∈ EndK(V ) we have L(M)(N) = MN and R(M)(N) = NM and for any M1,M2

we have that L(M1) and R(M2) commute. So

(L(M)−R(M))r =r∑i=0

(r

i

)L(M)i(−R(M))r−i =

r∑i=0

(r

i

)L(M i)R((−M)r−i).

If M ∈ g then M ∈ EndK(V ) is nilpotent and hence also L(M) and R(M) are nilpotent andtherefore L(M) − R(M)) is nilpotent, by the equation above. If X,Y ∈ g then ad(X)(Y ) =[X,Y ] = L(X)(Y )−R(X)(Y ) = {L(X)−R(X)}(Y ). Hence ad(X) ∈ Endk g is nilpotent for everyX ∈ g.

By induction on dimk g we shall show that that for any non-zero K-representation V for g, whereall X ∈ g act nilpotently on V , there always exist a non-zero vector v ∈ V such that X · v = 0 forall X ∈ v.

If g = kX is one-dimensional and X acts nilpotently on V 6= 0, then 0 is the only eigenvaluefor X acting K-linearly on V , and therefore by linear algebra there is an eigenvector v with zeroeigenvalue 0, i.e., tXv = 0 for all t ∈ k and hence for all tX ∈ g. We assume that the result is truefor Lie algebras of smaller dimension than g.

8 ABRAHAM BROER

Let now a ⊂ g be a maximal proper Lie sub-algebra of g. Then for any Y ∈ a and Z ∈ a we havead(Y )(Z) ∈ a, so we get a representation of a on the k-vector space g/a, and any element Y ∈ a

acts nilpotently on g/a. By induction there is an X ∈ g, X 6∈ a such that ad(Y )(X) = 0 for allY ∈ a, or in other words [Y,X] ∈ a for all Y ∈ a. It follows that kX ⊕ a is a strictly larger Liesub-algebra than a. By the maximality hypothesis, it follows that g = kX ⊕ a and aC g.

Returning to V , let V0 := {v ∈ V ;∀Y ∈ a : Y ·v = 0}. By the induction hypothesis it is non-zero.Let v ∈ V0 and Y ∈ a then

Y Xv = [Y,X]v +XY v = 0 + 0 = 0.

Hence V0 is also stable by X, hence stable by g. Now take any non-zero v ∈ V0 that is an eigenvectorof X with eigenvalue 0, then Zv = 0 for all Z ∈ g. This ends the induction proof.

We finish the proof of Engel’s theorem by induction on the dimension of V . If dimK V = 1,then there exists a non-zero v such that Zv = 0 for all Z ∈ g. Then {v} is the K-basis of Engel’stheorem. We assume Engel’s theorem to be true for any representation that has smaller dimensionthan V . Let v1 6= 0 be as above, such that Zv1 = 0 for all Z ∈ g. Then we get a g representationon the quotient space V/Kv1, such that any Z ∈ g acts nilpotently. So we can use the inductionhypothesis and conclude there are v2, . . . , vn such that for all Z ∈ g we have that Zv2 = 0 and fori ≥ 3 that Zvi ∈ Kx2 ⊕ . . . ⊕ vi−1. Or for all i ≥ 2 that Zvi ∈ Kv1 ⊕ . . . ⊕ Kvi−1. This provesEngel’s theorem. �

We apply Engel’s theorem using the adjoint representation.

Corollary 1.1. Let g be a Lie algebra. Then g is a nilpotent Lie algebra if and only if ad(X) isnilpotent for all X ∈ g.

Proof. (i) Suppose that g is nilpotent and that gm = 0. Then in particular for all Y ∈ g we have

ad(X)m(Y ) = [X, [X, [X, . . . [X, [X,Y ]] . . .] = 0,

with m terms X. Hence ad(X) acts nilpotently on g.(ii) Suppose ad(X) is nilpotent for all X ∈ g. Then Engel’s theorem implies that ad(g) is

contained in a nilpotent Lie algebra, hence is a nilpotent Lie algebra itself. The kernel of ad is thecenter z of g, so g/z ' ad(g). There is an m such that [ad(g), [ad(g), [. . . , [ad(g), ad(g)]] . . .] = 0(with m terms ad(g)). So [g, [g, [. . . , [g, g]] . . .] ⊂ z and bracketing with one more copy of g gives

[g, [g, [g, [. . . , [g, g]] . . .] ⊂ [g, z] = 0.

We conclude that g is nilpotent as well. �

Now we apply Engel’s theorem and its corollary in the situation of a solvable Lie algebra.

Corollary 1.2. Let g be a solvable Lie algebra. The collection

n := {X ∈ g; ad(X) is nilpotent}

is the unique largest nilpotent ideal, called its nilpotent radical, i.e., all other nilpotent ideals arecontained in it.

For any derivation D of g we get D(g) and [g, g] are nilpotent and hence contained in n.

LIE ALGEBRAS AND LIE GROUPS MAT6633 9

Proof. Since g is solvable, its image ad(g) is a solvable Lie sub-algebra of Endk g. Let D ∈ Endk(g)be a derivation of g. Consider the k-linear subspace g := kD+ad(g) ⊂ Endk(g). Since [D, ad(X)] =ad(D(X)), it follows that g is a Lie subalgebra of Endk(g) with [g, g] ⊂ ad(g) and so it is solvableitself.

Let k ⊂ K, with K algebraically closed. Put V = g⊗k K, then

g ⊆ Endk(g) ⊆ EndK V,

and V is a K-representation of g.We can apply Lie’s theorem and get a complete flag

0 = V0 ⊂ V1 ⊂ . . . ⊂ Vn−1 ⊂ Vn = V

of K-representations of g.Define functions λi : g→ K as the common eigenvalue function on Vi/Vi−1: if v ∈ Vi then for all

Z ∈ g we get Z ·v−λ(Z)v ∈ Vi−1. We know that any bracket acts trivially on any one-dimensionalK-representation. So in particular, for any X,Y ∈ g and i it holds that λi(ad([X,Y ])) = 0 =adi(ad(D(X))).

Suppose Z ∈ g is nilpotent and Zm = 0 then necessarily for all i we have λi(Z)m = 0, henceλi(Z) = 0. And conversely.

So in particular, for every X,Y ∈ g and any derivation D we have ad([X,Y ]) and ad(D(X)) arenilpotent. We define µi : g→ K by µi(X) := λi(ad(X)) which is a Lie algebra homomorphism andwhose kernel is hence an ideal. It follows that n as defined in the corollary is in fact an ideal of g

containing [g, g] and D(g) for any derivation. �

This last corollary allows us to define the nilpotent radical of any Lie algebra.

Corollary 1.3. Let g be a Lie algebra. Any nilpotent ideal is contained in a unique largest nilpotentideal nC g, called the nilpotent radical of g. It equals the nilpotent radical of radg.

Proof. Let a be any nilpotent ideal. Since it is solvable too, it is contained in the solvable radicalrad(g) and so in the largest nilpotent ideal n of rad(g). On the other hand if X ∈ g, then therestriction of ad(X) to rad(g) is a derivation, and so by the previous corollary [X, n] ⊆ [X, radg] ⊆ n

and so nC g. �

A Lie algebra whose nilpotent radical is zero is called reductive.

1.4. Bilinear forms. To any representation V of a Lie algebra we can associate a bilinear formon g. So let us discuss bilinear forms on a vector space first.

Let V be a finite-dimensional vector space over a field k and

C(·, ·) : V × V → k

a bilinear form over k on V × V . If C(u, v) = C(v, u) for all u, v ∈ V , the bilinear form is calledsymmetric. If C(u, v) = −C(v, u) it is called anti-symmetric. We shall suppose that the form iseither symmetric or anti-symmetric.

Then at least

a = {X ∈ Endk(V );∀u ∈ V, v ∈ V : C(X(u), v) + C(u,X(v)) = 0}

10 ABRAHAM BROER

is a Lie sub-algebra of Endk(V ).

Proof. The bilinearity of C(·, ·) gives that a is a linear subspace. Let X,Y ∈ a and u, v ∈ V . Then

C((XY − Y X)(u), v) = C(XY (u), v)− C(u, Y X(v)) = −C(Y (u), X(v)) + C(Y (u), X(v)) = 0

so [X,Y ] ∈ a. �

If U ⊂ V is a linear subspace then we define U⊥ by

U⊥ := {v ∈ V ;∀u ∈ U : C(v, u) = 0}.

The radical radC of C is defined by

radC := {v ∈ V ;∀u ∈ V : C(u, v) = 0} = V ⊥.

We can restrict the bilinear form C to U×U , say C|U×U . Then the radical of C|U×U is just U ∩U⊥.The bilinear form C is called non-degenerate if its radical is 0. Let v1, . . . , vn be a basis of V .

Put Cij := C(vi, vj) for 1 ≤ i ≤ n and 1 ≤ i ≤ n. Then C is non-degenerate if and only if thematrix (Cij) (called Gramm-matrix) is invertible.

Let C(v, ·) ∈ V ∗ = Homk(V, k) be defined by C(v, ·)(u) := C(v, u), with u ∈ V . Then we get ak-linear map V → V ∗ : v 7→ C(v, ·) with kernel V ⊥ = radC; so this map is an isomorphism if andonly if the radical of C is zero.

Lemma 1.1. Suppose C(·, ·) is non-degenerate and U ⊂ V is a linear subspace.Then dimV = dimU + dimU⊥. And V = U + U⊥ if and only if V = U ⊕ U⊥ if and only if

U ∩ U⊥ = 0 if and only if the radical of C|U×U is 0 if and only if the restriction of C to U × U isa non-degenerate bilinear form.

Proof. We define the linear map ψ : V → U∗ by [ψ(v)](u) := C(v, u) where v ∈ V and u ∈ U . Itskernel is just U⊥. We want to show that ψ is surjective.

Choose a vector space complement U ′ of U , i.e., V = U ⊕ U ′. For any linear form µ : U → K

define the linear form µ : V → K by µ(u+ u′) := µ(u), where u ∈ U and u′ ∈ U ′.Since C(·, ·) is non-degenerate, there is a unique v ∈ V such that µ = C(v, ·). Then for this v:

[ψ(v)](u) = C(v, u) = µ(u) = µ(u),

for all u ∈ U. So ψ is surjective. So

dimV − dimU⊥ = dim(V/U⊥) = dimU∗ = dimU

so

dimV = dimU + dimU⊥.

Now U + U⊥ = V if and only if dim(U + U⊥) = dimV = dimU + dimU⊥ if and only ifU + U⊥ = U ⊕ U⊥ if and only if U ∩ U⊥ = radC|U×U = 0. �

Lemma 1.2. Suppose C(·, ·) is non-degenerate and v1, . . . , vn a basis for V . Then there is a uniquebasis (called dual basis) v′1, . . . , v

′n with the property that C(vi, v′j) = δij (=Kronecker δ).

If v =∑

i aivi and w =∑

i biv′i then ai = C(v, v′i), bi = C(vi, v) and C(v, w) =

∑i aibi.

LIE ALGEBRAS AND LIE GROUPS MAT6633 11

Proof. Take 1 ≤ i ≤ n and put Vi be the linear subspace of V of co-dimension one spanned byall the basis vectors vj where j 6= i. By the previous lemma dimV ⊥i = 1. On the other handdim(V ⊥i )⊥ = dimV − 1 and Vi ⊆ (V ⊥i )⊥, hence Vi = (V ⊥i )⊥. So there exists a v′i ∈ V ⊥i such thatC(vi, v′i) = 1. So we found elements v′1, . . . , v

′n such that C(vi, v′j) = δij (Kronecker δ). Suppose

there is a linear relation∑

j cjv′j = 0, then for all i we have

0 = C(vi, 0) = C(vi,∑j

cjv′j) = ci,

so the relation is trivial and v′1, . . . , v′n is also a basis. �

1.5. Bilinear forms on g. Let g be a Lie algebra over k and F : g→ Endk(V ) any representationon the finite dimensional k-vector space V . Then it induces a symmetric bilinear form on g:

BV (X,Y ) := tr(F (X)F (Y )).

It has the additional property that for all X,Y, Z ∈ g:

BV ([X,Z], Y ) = BV (X, [Z, Y ]),

or

BV ([X,Z], Y ) +BV (Z, [X,Y ]) = 0.

Since

tr(F ([X,Z])F (Y )) = tr(F (X)F (Z)F (Y ))− tr(F (Z)F (X)F (Y ))

= tr(F (X)F (Z)F (Y ))− tr(F (X)F (Y )F (Z))

= tr(F (X)F ([Z, Y ])).

Lemma 1.3. If U ⊂ V is a subrepresention for g with quotient representation on V/U , then

BV (X,Y ) = BU (X,Y ) +BV/U (X,Y ).

Proof. Let v1, . . . , vn be a basis for V such that v1, . . . , vm is a basis for U and vm+1, · · · , vn a basisfor V/U . If X · Y · vj = cijvi then the n × n matrix with coefficients cij (1 ≤ i ≤ n, 1 ≤ j ≤ n)is the matrix associated to v → X · Y · v acting on V ; the m × m-matrix with coefficients cij(1 ≤ i ≤ m, 1 ≤ j ≤ m is the matrix acting on U , and the (n − m) × (n − m)-matrix withcoefficients cij (m+ 1 ≤ i ≤ n,m+ 1 ≤ j ≤ n) is the matrix on V/U . Now taking traces gives thelemma. �

Lemma 1.4. Let g be a Lie algebra with representation V .(i) Let aC g then a⊥C g (the orthogonal space with respect to BV ). In particular, radBV =

g⊥C g.(ii) Suppose a is an ideal such that all A ∈ a act nilpotently on V . Then a ⊆ radBV .

Proof. (i) At least a⊥ is a linear subspace of g. Let H ∈ a⊥. For all X ∈ g and A ∈ a we have[X,A] ∈ a and so

BV ([H,X], A) = BV (H, [X,A]) = 0

so [H,X] ∈ a⊥ and therefore a⊥C g.

12 ABRAHAM BROER

(ii) We shall use induction on V . There is nothing to do if dimV = 0. Let dimV > 0. LetV0 = {v ∈ V ; ∀A ∈ a : A · v = 0}. By Engel’s theorem V0 6= 0. It is even a sub-representation for g,since if X ∈ g and v ∈ V0, then for all A ∈ a we have AXv = [A,X]v +XAv = 0, since [A,X] ∈ a.So AXv = 0 for all v ∈ V0 and so V0 is a subrepresentation for g and also BV0(A,X) = 0.

We get a quotient representation of g on V/V0 and any A ∈ a still acts nilpotently on V/V0 soby induction we can assume that BV/V0

(A,X) = 0.Since BV (A,X) = BV0(A,X) +BV/V0

(A,X) we can conclude. �

1.6. Killing’s symmetric bilinear form on g. Any finite dimensional Lie algebra g over k hasthe adjoint representation on g itself. Its associated k-bilinear form B(·, ·) = Bg(·, ·) is calledKilling’s bilinear form (or the Killing form):

B(X,Y ) = tr(ad(X) ad(Y )).

Lemma 1.5. If aC g then the restriction of the Killing form of g to a× a is the Killing form of a.

Proof. We have a is a subrepresentation of the (adjoint) representation g and so B(X,Y ) =Ba(X,Y ) +Bg/a(X,Y ) and Bg/a(X,Y ) is 0 if X,Y ∈ a, since a is an ideal. �

Proposition 1.1. The nilpotent radical is contained in the radical of the Killing form.In particular, if g is solvable then [g, g] is nilpotent and therefore contained in the radical of the

Killing form.

Proof. Since an ideal aC g is nilpotent if and only if ad(A) is nilpotent for all A ∈ a, we get thatevery nilpotent ideal is contained in the radical of the Killing form, by Lemma 1.4, in particularthe nilpotent radical is contained in the radical of the Killing form. �

The converse is also true.

Theorem 1.3 (Cartan’s solvability criterion). Let g be a finite dimensional k-algebra. Then g issolvable if and only if [g, g] ⊂ radB.

We shall give a proof in the next subsection.

Corollary 1.4. Let g be a finite dimensional k-algebra. Then radB ⊆ radg (and radB containsthe nilpotent radical).

Proof. It suffices to prove that the ideal a := radB is solvable using Cartan’s criterion. Therestriction of the Killing form to a × a is the Killing form C on a. So for any X,Y ∈ a we haveC(X,Y ) = B(X,Y ) = 0 since X is in the radical of B. So we conclude using Cartan’s criterion. �

Theorem 1.4 (Cartan). Let g be a Lie algebra. The following are equivalent.(i) The Killing form on g is non-degenerate;(ii) g is semi-simple;(iii) g is the direct sum of simple ideals.

Proof. Suppose (i). The Killing form is non-degenerate hence g does not have any non-zero nilpotentideals and in particular the nilpotent radical of g is zero. Since [rad(g), rad(g)] is nilpotent itfollows that the radical of g is abelian, say a. Let X ∈ g and A ∈ a. For any B ∈ a we have

LIE ALGEBRAS AND LIE GROUPS MAT6633 13

ad(X) ad(A)(B) = [X, [A,B]] = 0, since a is abelian. So Ba(X,A) = 0. On the other handBg/a(X,A) = 0 since a is an ideal and A ∈ a. So

B(X,A) = Ba(X,A) +Bg/a(X,A) = 0.

So A ∈ rad(B) = 0. We conclude that rad(g) = 0 and g is semi-simple. Hence (i) implies (ii).Suppose still that the Killing form is non-degenerate. We will show (iii) holds. Let aC g be a

minimal non-zero ideal. If a = g then g is simple and we are done. Then [a, a]C g, since [a, a] J a

and aC g. So by minimality either a′ = 0 or a′ = a. In the first case a is an abelian ideal, hencesolvable. But we just proved that the solvable radical is zero if the Killing form is non-degenerate.So a′ = a.

Then a⊥ and a ∩ a⊥ are also ideals. Again by minimality of a there are two possibilities.Suppose a ∩ a⊥ = a. Let Z ∈ g and A ∈ a be arbitrary. There are Ai, A

′i ∈ a such that

A =∑

i[Ai, A′i], since a′ = a. Then

B(A,Z) =∑i

B([Ai, A′i], Z) =∑i

B(Ai, [A′i, Z]) = 0,

since [A′i, Z] ∈ A and A ⊂ A⊥. So A ∈ radB = 0. Contradiction.So a ∩ a⊥ = 0. If X ∈ a and Y ∈ a⊥ then [X,Y ] ∈ a ∩ a⊥ = 0. So g = a⊕ a⊥ is a direct sum of

ideals. Since B(X,Y ) = 0 if X ∈ a and Y ∈ a⊥, the Killing form on a and a⊥ are non-degenerate.And we can use induction to complete the proof. Hence (i) implies (iii).

For the two converses we shall use Cartan’s criterion.Suppose (iii) holds, i.e. g = g1⊕ . . .⊕gn a direct sum of ideals. We want to show that the Killing

form is non-degenerate. The Killing form on g is non-degenerate if and only if the Killing form oneach gi is non-degenerate. So it suffices to prove that the Killing form of a simple Lie algebra isnon-degenerate. So suppose g is simple. The radical of the Killing form is an ideal, hence either g

or 0. In the first case it follows from Cartan’s criterion that g is solvable. Contradiction. Hence(iii) implies (i).

If the Killing form is degenerate, its radical is a non-trivial solvable ideal, so the radical of g isnon-zero, by Cartan’s criterion, and g is not semisimple. So (ii) implies (i). �

14 ABRAHAM BROER

1.7. Proof of Cartan’s solvability criterion. We still have to show that if rad(g) ⊆ rad(B)then g is solvable.

We shall use some facts from undergraduate algebra.

Theorem 1.5 (Chinese remainder theorem). Let k be a field and f1, . . . , fs ∈ k[T ] be non-constantpolynomials that are pairwise relatively prime. Then for any polynomials g1, . . . , gs there exists apolynomial g such that for all 1 ≤ i ≤ s we have

g ≡ gi mod (fi),

i.e., there exist polynomials hi such that g = g+fihi. This polynomial is unique modulo (f1f2 . . . fs).

The second result is called Lagrange interpolation.

Proposition 1.2. Let k be a field and a0, a1, a2, . . . , an be n+ 1 different elements of k. Then forany sequence b0, b1, . . . , bn of elements in k, there exists a unique polynomial F (T ) ∈ k[T ] of degreen such that

F (ai) = bi,

for 0 ≤ i ≤ n.

Proof. We are looking for a polynomial F (T ) = x0 + x1T + x2T2 + . . . + xnTn with unknown

coefficients in k such that F (ai) = bi. So we have to solve the system of linear equations

1 · x0 + ai · x1 + a2i · x2 + . . . ani xn = bn.

The corresponding matrix M ∈ Mat(n+ 1× n+ 1, k) is the Vandermonde matrix

Mij = aji ,

with 0 ≤ i ≤ n and 0 ≤ j ≤ n. An exercise in linear algebra shows that its determinant is∏i<j(ai − aj) 6= 0. So we can solve uniquely the coefficients x0, x1, . . . , xn of the polynomial.For example with n = 2 we have to solve1 a0 a2

0

1 a1 a21

1 a2 a22

x0

x1

x2

=

b0b1b2

.

and we can if a0, a1, a2 are all different. �

The third result follows from the existence of the Jordan normal form of a matrix.

Theorem 1.6 (Jordan normal form). Suppose V is an n-dimensional vector space over a field Kand η ∈ EndK(V ) a linear endomorphism such that all the eigenvalues lie in K (this is the certainlythe case if K = C).

Let λ1, . . . , λs be the various different eigenvalues of η. Define for each 1 ≤ i ≤ s the generalizedeigenspace

Vλi := {v ∈ V ; (η − λi1V )n(v) = 0.}

If v ∈ Vλ then η(v) ∈ Vλ and

V = ⊕si=1Vλi .

LIE ALGEBRAS AND LIE GROUPS MAT6633 15

Let σ ∈ EndK(V ) be uniquely defined by σ(v) = λiv, if v ∈ Vλi. In particular, there exists a basis ofeigenvectors for the linear map σ. There exists a polynomial P (T ) ∈ K[T ], without constant term,such that P (η) = σ. The difference ν = η − σ is nilpotent and [ν, σ] = 0.

If η = σ′+ ν ′ is another decomposition, such that there exists a basis of eigenvectors for σ′, η′ isnilpotent and [σ′, ν ′] = 0, then σ′ = σ and η′ = η.

The decomposition η = σ + ν is called the Jordan decomposition.

Proof. We do not need the full force of the Jordan normal form to prove this, nor the theory ofinvariant factors et cetera. Therefore we think it is still useful to give a proof here.

For any λ ∈ K and i ≥ 0 define

Vλ,i = {v ∈ V ; (η − λ1)i = 0.}

In particular Vλ,1 is the classical eigenspace of η with eigenvalue λ. We get a sequence of linearsubspaces

0 = Vλ,0 ⊆ Vλ,1 ⊆ Vλ,2 ⊆ . . .

If for some i ≥ 0 we have Vλ,i = Vλ,i+1 then we claim that for all j ≥ i+ 1 we also have Vλ,i = Vλ,j .Indeed, let v ∈ Vλ,j , i.e.

0 = (η − λ1)j(v) = (η − λ1)i+1((η − λ1)j−i−1(v))

and so (η − λ1)j−i−1(v) ∈ Vλ,i+1 = Vλ,i and so

(η − λ1)j−1(v) = (η − λ1)i((η − λ1)j−i−1(v)) = 0

and v ∈ Vλ,j−1. It follows that if Vλ,j−1 6= Vλ,j then dimVλ,j ≥ j. We conclude that since dimV = n

we must have that

Vλ,n = {v ∈ V ; ∃r ≥ 0 : (η − λ1)r(v) = 0}.

We also conclude that Vλ,j = 0 for all j if λ is not an eigenvalue for η.Let v ∈ Vλ,n then also (η − λ1)n+1 = 0 and so η(v)− λv ∈ Vλ,n and so η(v) ∈ Vλ,n.Let F (T ) = det(T1 − η) =

∏si=1(T − λi)mi be the characteristic polynomial of η. Then by

the Cayley-Hamilton theorem F (η) acts trivally on V . By the Euclidian algorithm we can findf(T ), g(T ) ∈ K[T ] such that f(t)

∏si=2(T − λi)mi + g(T )(T − λ1)m1 = 1. Define the linear endo-

morphisms π1 and π2 by

π1(v) = f(η)s∏i=2

(η − λi1)mi(v)

and

π2(v) = g(η)(η − λ11)m1(v).

Then v = π1(v)+π2(v), π1 ◦π2 = π2 ◦π1 = 0, η(π1(v)) = π1(η(v)), η(π1(v)) = π1(η(v)) and π1(v) ∈Vλ1 . If v ∈ Vλ1 then π1(v) = v and π2(v) = 0 so π2

1 = π1 and π22 = π2. Put U1 = π2(V ) = Kerπ1.

Then V = Vλ1 ⊕ U1 and both Vλ1 and U1 are stable under η. The characteristic polynomial of ηacting on U1 is

∏si=2(T −λi)mi and its characteristic polynomial acting on Vλ1 is (T −λ1). We can

then use induction to show that

V = Vλ1 ⊕ Vλ2 ⊕ . . .⊕ Vλs .

16 ABRAHAM BROER

By the Chinese remainder theorem there exists a polynomial P (T ) ∈ K[T ] such that P (T ) ≡ λimod (T − λi)n, for 1 ≤ i ≤ s and additionally P (T ) ≡ 0 mod T , if all eigenvalues are non-zero.In any case, P (T ) has no constant term and for any i there exists a polynomial hi(T ) such thatP (T ) = λi + hi(T )(T − λi)n.

We define σ = P (η). Let v ∈ Vλi then

σ(v) = P (η)(v) = λiv + hi(σ)(σ − λ1)n(v) = λiv.

So any non-zero vector of Vλi is an eigen-vector for σ with eigen-value λi. And σ(η(v)) =P (η)(η(v)) = η(P (η)(v)) = η(σ(v)) hence [σ, η] = 0. Define ν = η − σ then for any v ∈ Vλiwe have

νn(vi) = (η − σ)n(v) = (η − σ)n−1(η(v)− λiv) = (η − σ)n−1(η − λi1)(v) = . . . = (η − λi1)n(v) = 0

(by induction). So νn = 0 and so ν acts nilpotently and [σ, ν] = [σ, η − σ] = 0.Let η = σ′+ η′ be another decomposition as in the theorem. Then σ′ and ν ′ also commutes with

η and hence with P (η) = σ and with ν.Now let (ν ′)m = νm = 0 then (ν ′− ν)2m =

∑2mi=0

(2mi

)(ν ′)i(−ν)2m−i = 0 since ν and ν ′ commute

and for 0 ≤ i ≤ m either (ν ′)i = 0 or (−ν)2m−i = 0. So ν ′ − ν = σ − σ′ is nilpotent.Let v ∈ Vλi then σ(σ′(v)) = σ′(σ(v)) = λi(σ′(v) so σ′(v) ∈ Vλi and so σ′ preserves every Vλi . Let

v ∈ V be any eigenvector for σ′ with eigenvalue µ. We can write uniquely v = v1 + v2 + . . . + vs,where vi ∈ Vλi and since σ′(v) = µv we get

∑si=1 µvi =

∑si=1 σ(vi). Since σ(vi) ∈ Vλi we get

σ′(vi) = µvi for each i. We get (σ − σ′)(vi) = (λi − µ)vi and so if vi 6= 0 then (λi − µ) is aneigenvalue of σ − σ′ = ν ′ − ν which is nilpotent (who has only 0 as eigenvalue). So λi = µ andthere is a unique i such that vi 6= 0. We conclude that any eigenvector for σ′ is necessarily alsoan eigenvector for σ. So σ′ − σ also has a basis of eigenvectors. Since 0 is the only eigenvalue weconclude that σ′ − σ = 0. This finishes the proof. �

Lemma 1.6. Let V be a vector space over a field, η ∈ EndK(V ) such that all its eigenvalues are inK. Let η = σ+ν be its Jordan decomposition. Let ad be the adjoint representation of the Lie algebraEndK(V ). Then ad(η) = ad(σ) + ad(ν) is the Jordan decomposition of ad(η) ∈ EndK(EndK(V )).If λ1, . . . , λs are the eigenvalues of η, then {λi − λj , i 6= j} are the eigenvalues of ad(η).

Proof. We pick bases in each Vλi , for each eigenvalue λi; and the reunion of those basis gives abasis e1, . . . , en of V . There are eigenvalues µi such that σei = µiei. Let x1, . . . , xn ∈ V ∗ be thedual basis. Then a basis for EndK(V ) is Eij defined by Eij(v) = xj(v)ei. Then for v =

∑r crer:

[ad(σ)(Eij)](v) =∑r

cr(σ(Eij(er))− Eij(σ(er)))

=∑r

cr(σ(xj(er)ei)− Eij(µrer))

=∑r

cr(xj(er)µiei − (µrxj(er)ei)

= cj(µi − µj)ei= (µi − µj)xj(v)ei

= (µi − µj)E(ij)(v)

LIE ALGEBRAS AND LIE GROUPS MAT6633 17

so ad(σ)(Eij) = (µi − µj)E(ij) and the Eij form a basis of eigenvectors for ad(σ).We saw earlier that if ν is nilpotent, then ad(ν) is also nilpotent. Finally [ad(σ), ad(ν)] =

ad([σ, ν]) = 0. So indeed ad(η) = ad(σ) + ad(nu) is the Jordan decomposition of ad(η). �

Now we are ready to prove Cartan’s solvability criterion.

Proof of Cartan’s solvability criterion. Assuming that [g, g] ⊂ rad(B) it remains to be shown thatg is solvable. We shall show that ad(Y ) is nilpotent for any Y ∈ [g, g], then from Engel’s theoremit will follow that [g, g] is nilpotent and hence g will be solvable.

We shall only prove the case where k ⊂ C. It can be proved in general too by modifyingthe arguments somewhat. Let V = g ⊗k C be the complex vector space associated to g. Inparticular, any k-basis of g becomes a C-basis of EndC(V ). On EndC V we have the C-bilinearform C(X,Y ) = tr(XY ) ∈ C. By replacing g by ad(g) we may suppose that g ⊂ EndC V is a Liesub-algebra over k, and [g, g] ⊂ radC. The hypothesis becomes then that if Y1, Y2, Y3 ∈ g thenC([Y1, Y2], Y3) = 0

Let Y ∈ [g, g]. We must prove that it is nilpotent. Let λ1, . . . , λs be its eigenvalues acting onV and the generalized eigenspace decomposition V = Vλ1 ⊕ . . .⊕ Vλs . There are S,N ∈ EndC(V )such that Y = S + N , [S,N ] = 0, N is nilpotent and S acts by on Vλi by multiplication by λi.Define S by Sv = λiv if v ∈ Vλi . Since Y and S preserve each Vλi and S acts by a scalar on Vλi itfollows that S also commutes with N and so SN is nilpotent.

We will calculate C(S, Y ) = tr(SY ) = tr(SS) + tr(SN) = tr(SS) in two different ways. Re-stricted to Vλi its trace becomes λiλi dimC(Vλi) ≥ 0 and = 0 if 0 is the only eigenvalue, hence if Yis nilpotent. This is what we want to prove.

The Lie algebra EndC(V ) also has an adjoint representation, call it ad0. And by the previouslemma ad0(Y ) = ad0(S) + ad0(N) is the Jordan decomposition of ad0(Y ). So there is a polynomialwithout constant term P (T ) ∈ C[T ] such that P (ad0(Y )) = ad0(S).

By Lagrange interpolation, there is a polynomial Q(T ) ∈ C[T ] such that Q(λi − λj) = λi − λj .Then Q(ad0(S)) = ad0(S). By composition we get a polynomial R(T ) = r1T + r2T

2 + . . .+mTm,with ri ∈ C.R(T ), such that R(ad0(Y )) = ad(S) and R(0) = 0. Or ad(S) =

∑mj=1 rj ad0(Y )j .

Since Y ∈ [g, g] there exist Xi, Zi ∈ g such that Y =∑

i[Xi, Zi].

C(S, Y ) =∑i

C(S, [Xi, Zi])

=∑i

C([S,Xi], Zi)

=∑i

C(ad0(S)(Xi), Zi)

=∑i

m∑j=1

rjC(ad0(Y )j(Xi), Zi)

=∑i

m∑j=1

rjC([Y, ad0(Y )j−1(Xi)], Zi)

= 0

18 ABRAHAM BROER

since for any j ≥ 1 we have Y, ad0(Y )j−1(Xi), Zi ∈ g and so by hypothesis C([Y, ad0(Y )j−1(Xi)], Zi) =0. (Remark: our hypothesis cannot be used directly to show that C([S,Xi], Zi) = 0, since we haveno reason to believe that S ∈ g.)

So indeed Y is nilpotent. This finishes the proof. �

Having done so much work, it is pleasant to be able to use the same proof for another result.We call a representation ρ : g→ Endk V faithful if ρ is injective; or equivalently if X · v = 0 for allv ∈ V then X = 0.

Proposition 1.3. Let V be a faithful representation for the Lie algebra g. And suppose that [g, g]is contained in the radical of BV (·, ·). Then g is solvable.

Proof. We shall assume k ⊆ C, although the proof can be adjusted for the more general case.Replacing V by its complexification V C, and by defining the form C(X,Y ) = tr(X,Y ) on EndC(V C)we end up in the same situation as we were after the second paragraph of the proof of Cartan’ssolvability criterion. So we can use the same proof to conclude that every element of [g, g] is actingnilpotently on V , so by Engel’s theorem, [g, g] is nilpotent, and g is solvable. �

We get a corollary generalizing the implication (ii)⇒(i) of Cartan’s theorem.

Corollary 1.5. Let g be a semi-simple Lie algebra and V a faithful g-representation. Then theform BV (·, ·) is non-degenerate.

Proof. The radical of BV (·, ·) is a Lie-algebra acting faithfully on V , so the hypothesis of theproposition is satisfied. Hence this radical is a solvable ideal of g but g has no non-zero solvableideals. �

LIE ALGEBRAS AND LIE GROUPS MAT6633 19

2. Representation theory for sl2(k)

We shall now give a classification of the irreducible (finite dimensional) representations for thesimple Lie algebra sl2(k). We will not need to assume that k is algebraically closed (but we stillassume k to be of characteristic 0). (There are also simple infinite dimensional representations.)

Consider the polynomial ring k[X0, X1]. Partial derivation ∂0 := ∂∂X0

and ∂1 := ∂∂X1

are definedas usual. They are k-linear operators on k[X0, X1] satisfying Leibnitz’ rule, for example

∂0(PQ) = ∂0(P )Q+ P∂0(Q),

for polynomials P and Q. These operators commute [∂0, ∂1] = 0.For any polynomials f0, f1 the operator f0∂0 + f1∂1 also acts on k[X0, X1] by

(f0∂0 + f1∂1)(P ) := f0∂0(P ) + f1∂1(P );

it is also k-linear and satisfies Leibnitz’ rule. Two linear operators of this kind do no longercommute in general, but their Lie bracket is of the same form. We calculate this Lie bracket ofD1 = (f0∂0 + f1∂1) and D2 = (g0∂0 + g1∂1):

[D1, D2](P ) = [(f0∂0 + f1∂1), (g0∂0 + g1∂1)](P )

= (f0∂0 + f1∂1) ((g0∂0 + g1∂1)(P ))− (g0∂0 + g1∂1) ((f0∂0 + f1∂1)(P ))

= (f0∂0 + f1∂1) ((g0∂0(P ) + g1∂1(P ))− (g0∂0 + g1∂1) ((f0∂0(P ) + f1∂1(P ))

= f0∂0(g0∂0(P )) + f0∂0(g1∂1(P )) + f1∂1(g0∂0(P )) + f1∂1(g1∂1(P )

− (g0∂0(f0∂0(P )) + g0∂0(f1∂1(P )) + g1∂1(f0∂0(P )) + g1∂1(f1∂1(P )))

= f0∂0(g0)∂0(P ) + f0g0∂0(∂0(P )) + f0∂0(g1)∂1(P ) + f0g1∂0(∂1(P ))

+f1∂1(g0)∂0(P ) + f1g0∂1(∂0(P )) + f1∂1(g1)∂1(P ) + f1g1∂1(∂1(P ))

−g0∂0(f0)∂0(P )− g0f0∂0(∂0(P ))− g0∂0(f1)∂1(P )− g0f1∂0(∂1(P ))

−g1∂1(f0)∂0(P )− g1f0∂1(∂0(P ))− g1∂1(f1)∂1(P )− g1f1∂1(∂1(P ))

= (f0∂0(g0) + f1∂1(g0)− g0∂0(f0)− g1∂1(f0)) ∂0(P )

+ (f0∂0(g1) + f1∂1(g1)− g0∂0(f1)− g1∂1(f1)) ∂1(P )

= ((f0∂0 + f1∂1) (g0)− (g0∂0 + g1∂1) (f0)) ∂0(P )

+ ((f0∂0 + f1∂1) (g1)− (g0∂0 + g1∂1) (f1)) ∂1(P )

= (D1(g0)−D2(f0)) ∂0(P ) + (D1(g1)−D2(f1)) ∂1(P )

We conclude that

[D1, D2] = (D1(g0)−D2(f0)) ∂0 + (D1(g1)−D2(f1)) ∂1

is indeed of the same form. The linear space of these linear operators on k[X0, X1]

L := {f0∂0 + f1∂1; f0, f1 ∈ k[X0, X1]}

20 ABRAHAM BROER

becomes an infinite dimensional Lie algebra over k with infinite dimensional k-representation onk[X0, X1].

It has a simple Lie sub-algebra of dimension three generated by

E := X0∂1, F := X1∂0, H := X0∂0 −X1∂1

with brackets

[E,F ] = H, [H,E] = 2E, [H,F ] = −2F.

It follows already that it equals its derived Lie-algebra.We calculate

E(X0) = 0, E(X1) = X0, F (X0) = X1, F (X1) = 0, H(X0) = X0, ,H(X1) = −X1

so V1 := kX0 ⊕ kX1 is a sub-representation; and on the basis X0, X1 the generators take the form

E =

(0 10 0

), F =

(0 01 0

), H =

(1 00 −1

), aE + bF + cH =

(c a

b −c

),

and so we have an isomorphism

kE + kF + kH = sl(V1) ' sl2(k).

We observe E(Xi0) = 0 and E(Xi

0Xj1) = jXi+1

0 Xj−11 if j ≥ 0, F (Xj

1) = 0 and F (Xi0X

j1) =

iXi−10 Xj+1

1 if j ≥ 0 and H(Xi0X

j1) = (i − j)Xi

0Xj1 . So the collection Vd of all polynomials

of degree d is a sub-representation of k[X0, X1] for our Lie algebra kE + kF + kH, with basisXd

0 , Xd−10 X1, X

d−20 X2

1 , . . . , Xd1 .

In fact, we prefer a slightly different basis. Put v0 = Xd0 , and by induction vi+1 = F (vi).

Explicitly,

v0 = Xd0 , v1 = dXd−1

0 X1, v2 = d(d− 1)Xd−20 X2

1 , . . . , vi =d!

(d− i)!Xd−i

0 Xi1, . . . , vd = d!Xd

1 , vd+1 = 0.

And

E(vi) = X0∂1(d!

(d− i)!Xd−i

0 Xi1) =

d!(d− i)!

(i)Xd−i+10 Xi−1

1 = i(d− i+ 1)vi−1.

And

Ei(vi) =i!d!

(d− i)!v0 and so F iEivi =

i!d!(d− i)!

vi.

Since every polynomial is a unique sum of homogeneous polynomials of various degrees, we get adirect sum k[X0, X1] = ⊕∞i=0Vi of kE + kF + kH-representations.

We calculate the bilinear form associated to V1 using this basis X0, X1:

BV1

(aE + bF + cH, a′E + b′F + c′H

)= tr

((c a

b −c

)·

(c′ a′

b′ −c′

))= ab′ + ba′ + 2cc′.

The dual bases with respect to this form becomes E′ = F, F ′ = E,H ′ = 12H. Form the quadratic

differential operator, called the Casimir operator

∆ = EE′ + FF ′ +HH ′ = EF + FE +12HH.

LIE ALGEBRAS AND LIE GROUPS MAT6633 21

It still acts linearly on k[X0, X1], but the Liebniz’ rule does no longer hold. In fact it acts on anysl2(k)-representation. Let V be any kE + kE + kH representation. Then we define

∆ · v := E · F · v + F · E · v +12H ·H · v

for v ∈ V . It commutes with the action of kE + kE + kH = sl(V1) ' sl2(k). For example

[∆, E] = EFE + FEE +12HHE − EEF − EFE − 1

2EHH

= ([F,E]E + EFE) +12

(H[H,E] +HEH)− (E[E,F ] + EFE)− 12

([E,H]H +HEH)

= −HE +12

(2HE)− EH − 12

(−2EH)

= 0.

How does ∆ act on Vd? We calculate on monomials.

∆(Xi0X

j1) = (EE′ + FF ′ +HH ′)(Xi

0Xj1) = (EF + FE +

12HH)(Xi

0Xj1)

= E(iXi−10 Xj+1

1 ) + F (jXi+10 Xj−1

1 ) +12H((i− j)Xi

0Xj1)

= (i(j + 1) + (i+ 1)j +12

(i− j)2)(Xi0X

j1)

= ((i+ j) +12

(i+ j)2)(Xi0X

j1).

So ∆ acts on Vd by scalar multiplication with d(d+2)2 . The Casimir operator can be defined for any

simple Lie algebra; it acts on any representation, commuting with the Lie algebra action, by

∆ · v = E · F · v + F · E · v +12H ·H · v.

So if V is any sl2(k)-representation, the Casimir operator gives rise to an endomorphism

∆ : V → V

of sl2(k)-representations.The Euler operator η = X0∂0 +X1∂1 acting on k[X0, X1] also commutes with kE+ kF + kH. A

polynomial P ∈ k[X0, X1] is homogeneous of degree d if and only if η(P ) = dP (it suffices to checkthis on monomials, and then it is easy). But the Euler-operator is not as useful in Lie-theory asthe Casimir operator, since it does not act naturally on all sl2(k)-representations and it does notgeneralize to all simple Lie algebras. In fact, kE + kF + kH + kη is isomorphic to gl(V1) with η

corresponding to the matrix

(1 00 1

), when we choose the basis X0, X1 for V1.

Lemma 2.1. Each representation Vd is a simple sl2(k)-module.

Proof. Let U ⊆ Vd be an sl2(k) subrepresentation, and u =∑d

i=0 ciXi0X

d−i1 ∈ U be any non-zero

element. Let r be maximal, such that cr 6= 0. Then Ed−r · u = E ·E ·E · . . . ·E · u = (d− r)!crXd0

is also in U . Now applying F -several times, shows that each Xi0X

d−i1 ∈ U . Hence U = Vd. �

22 ABRAHAM BROER

Theorem 2.1. Let V be a simple sl2(k)-representation of dimension d+ 1.(i) There is an isomorphism Vd ' V of representations.(ii) Let φ : Vd → Vd be an endomorphism as sl2(k)-representations. Then there is a unique c ∈ k

such that φ = c1Vd

Proof. (i) Let λ ∈ K ⊃ k be an eigenvalue for H acting on V , in some extension field of k. Wewant to show that λ ∈ Z, and we can therefore take k = K after all. Hence there is a non-zerov ∈ V ⊗k K such that H · v = λv. Then the following calculation

H · (E · v) = [H,E] · v + E ·H · v = 2E · v + λ = (λ+ 2)(E · v)

shows E · v is also an eigenvector for H of eigenvalue λ + 2, if it is non-zero. Without loss ofgenerality we can assume that E · v = 0. Then

H · (F · v) = [H,F ] · v + F ·H · v = −2F · v + λ = (λ− 2)(F · v)

shows that if F · v 6= 0 then it is also an eigenvector for H with eigenvalue λ− 2. Put v0 = v andinductively , vi+1 = F · vi. Let s be minimal such that vs+1 = 0. Then H · vi = (λ − 2i)vi andso v0, v1, . . . , vs are linearly independent, since they are eigenvectors with different eigenvalues. Byinduction on for i ≥ 1 we prove that

E · vi = i(λ− i+ 1)vi−1.

For i = 1:

E · v1 = E · F · v0 = [E,F ] · v0 + F · E · v0 = H · v0 = λv0;

and supposing true for i− 1 :

E · vi = E · F · vi−1 = [E,F ] · vi−1 + F · E · vi−1

= H · vi−1 + F · (i− 1)(λ− i+ 2)vi−2

= ((λ− 2(i− 1)) + (i− 1)(λ− i+ 2))vi−1

= i(λ− i+ 1)vi−1

But since vs+1 = 0 we get 0 = E · vs+1 = (s+ 1)(λ− s)vs = 0 hence λ = s.So the eigenvalue λ is in fact an integer, and we could have taken K = k and U := kv0 + . . .+kvs

is a non-zero subrepresentation of the simple representation V . Hence we get equality, and inparticular s = d.

Now we send Xd0 to v0 and F i ·Xd

0 = d!(d−i)!X

d−i0 Xi

1 to vi to get the isomorphism, since the actionsof E,F and H coincident.

(ii) Since φ is an endomorphism of sl2(k)-representations and the F i · Xd0 form a basis of Vd,

φ is completely determined by φ(Xd0 ), since φ(F i · Xd

0 ) = F i · φ(Xd0 ). If φ(Xd

0 ) = 0 then φ = 0,otherwise φ(Xd

0 ) is an H-eigenvector with eigenvalue d, hence there exists a unique c ∈ k such thatφ(Xd

0 ) = cXd0 and so φ(F i ·Xd

0 ) = cF i ·Xd0 and φ = c1. �

Example 2.1. Since sl2(k) is a simple Lie algebra, the adjoint representation is simple of dimensionthree and so isomorphic to V2. The representation on V ∗d is also simple with dimension d+1, henceVd ' V ∗d as sl2(k)-representations.

LIE ALGEBRAS AND LIE GROUPS MAT6633 23

Corollary 2.1. Let V be a finite dimensional sl2(k)-representation.(i) All the eigenvalues of ∆ acting on V are non-negative half-integers of the form d(d+2)

2 ford ∈ Z≥0.

(ii) trV (∆) = 0 if and only if 0 is the only eigenvalue for ∆ acting on V if and only if ∆ acts as0 on V if and only if sl2(k) acts trivially on V .

Proof. Proof by induction on dimV . If it is one-dimensional, V ' V0 is the trivial sl2(k)-representation, with ∆ acting by 0 and all statements become trivial.

If V is simple, it is isomorphic to Vd for some d, on which ∆ acts by multiplication by thenon-negative half-integer d(d+2)

2 , which is 0 if and only if d = 0 if and only if sl2(k) acts trivially.If V is not simple, let U ' Vd be a simple sub-representation. Since ∆ commutes with the

action of sl2(k), it acts on both U and V/U . So by induction all eigenvalues of ∆ are non-negativehalf-integers on U and V/U , hence also on V . This proves i.

Since trV (∆) = trU (∆) + trV/U (∆) and all eigenvalues for ∆ are non-negative, it follows thattrV (∆) = 0 if and only if trU (∆) = 0 and trV/U (∆) = 0.

Suppose trV (∆) = 0, hence trU (∆) = 0 and trV/U (∆) = 0. Then by induction ∆ acts triviallyon both U and V/U , so in particular 0 is the only eigenvalue of ∆ acting on V , and both U andV/U are trivial representations for sl2(k).

Let v ∈ V , then E·v = 0, for v ∈ V/U , so F ·E·v = E·v ∈ U and E·E·v = 0. This is because sl2(k)acts trivially on U and V/U . Similarly F ·v and H ·v are all in U , and F ·E ·v = E ·F ·v = H ·H ·v = 0and so ∆ · v = 0. i.e., ∆ acts trivially on V . Also H · v = [E,F ] · v = E · F · v − F · E · v = 0,and similarly E · v = 1

2 [H,E] · v = 0 and F · v = −12 [H,F ] · 0, so indeed sl2(k) acts trivially on the

whole of V .Conversely, if sl2(k) acts trivially on V , then ∆ acts as 0 and with trace 0. This finishes the

proof. �

Lemma 2.2. Let U ⊂ V be a sub-representation for sl2(k) of codimension one. There is a v ∈ Vsuch that X · v = 0 for all X ∈ sl2(k) and such that V = U ⊕ kv as sl2(k)-representations.

Proof. We know that all eigenvalues of ∆ acting on V are non-negative integers. Let λ1, . . . , λs bethe different eigenvalues. For λ ∈ k define

Vλ,i := {v ∈ V ; (∆− λ)i · v = 0}

and Vλ = Vλ,n the generalized eigenspace (here n = dimk V ). Since ∆ commutes with the sl2(k)action, each Vλ is an sl2(k)-representation and we get a direct sum decomposition

V = Vλ1 ⊕ Vλ2 ⊕ . . .⊕ Vλs

of sl2(k)-representations. Put Ui = U ∩ Vλi , then V/U ' ⊕si=1Vλi/Ui, and so there is a unique isuch that Ui 6= Vλi . Without loss of generality we can assume that i = 1

Let j be maximal such that Vλ1,j ⊂ U1, then U1 ⊂ Vλ1,j+1 is a proper inclusion. Sincedim(Vλ1/U1) = 1 we get necessarily that Vλ1 = Vλ1,j+1. Since λ1 is the only eigenvalue for ∆acting on Vλ1 and U1 we get

λ1 dimk Vλ1 = trVλ1(∆) = trU1(∆) + trVλ1

/U1(∆) = λ1 dimk U1 + 0 = λ1(dimk Vλ1 − 1)

24 ABRAHAM BROER

since sl2(k) and ∆ act trivially on any one-dimensional sl2(k)-representation.We conclude that λ1 = 0, and we can apply the corollary before. So ∆ and sl2(k) act trivially

on Vλ1 . Now let v ∈ Vλ1 such that v 6∈ U1. Then Vλ1 = U1 ⊕ kv as (trivial) sl2(k)-representationsand we get a direct sum decomposition as sl2(k) representations

V = (kv ⊕ U1)⊕ Vλ2 ⊕ . . .⊕ Vλs = kv ⊕ U1 ⊕ U2 ⊕ . . .⊕ Us = kv ⊕ U.

This finishes the proof. �

Theorem 2.2 (Reductivity of sl2(k)-representations). Let V be any (finite dimensional) represen-tation for sl2(k).

(i) Let U ⊂ V be a subrepresentation. Then there is a subrepresentation U ′ such that V = U⊕U ′

is a direct sum of representations.(ii) There is a direct sum decomposition as sl2(k)-representations

V = U1 ⊕ U2 ⊕ . . .⊕ Us

where each Ui is a simple sl2(k) representation isomorphic to Vd, when dimUi = d+ 1.The direct sum V(d) of all Ui with dimension d+1 is the same as the ∆-eigenspace with eigenvalue

d(d+1)2 , so does not depend on the decomposition as a direct sum of simple sub-representations.

Proof. (i) The vector space Endk(V ) is also an sl2(k)-representation when we define X · φ (forX ∈ sl2(k) and φ ∈ Endk(V )) by

[X · φ](v) = X · φ(v)− φ(X · v).

It has the subrepresentations

M := {φ ∈ EndV ;φ(V ) ⊂ U ; ∃λ ∈ k : ∀u ∈ U : φ(u) = λu},

andN := {φ ∈ EndV ;φ(V ) ⊂ U ;∀u ∈ U : φ(u) = 0}.

Then N ⊂M and M/N is one-dimensional.Using the previous lemma there exists a φ ∈ M such kφ is the trivial representation and M =

N⊕kφ. By normalizing we can assume that φ restricted to U acts by the scalar 1. Which means forall v ∈ V : φ(v) ∈ U and for all X ∈ sl2(k) we have X ·φ(v) = φ(X · v) and φ(u) = u for all u ∈ U .Or φ is a morphism of representations. Take U ′ = Kerφ, then U ′ is a subrepresentation such thatU ∩U ′ = 0. If v ∈ V then u := φ(v) ∈ U and φ(φ(v)− v) = φ(v)− φ(v) = 0 so u′ := φ(v) ∈ U ′ andv = u+ u′ We proved V = U ⊕ U ′ as representations, and so (i) follows.

(ii) Start with any simple sub-representation U1 with complement U ′1 as in (i), then decomposeU ′1. In this way we end u with a direct sum decomposition where each factor is simple and soisomorphic to some Vd. The remaining statement directly. �

To sum up.

Corollary 2.2. Let V be any (finite dimensional) representation for sl2(k). Both E and F actnilpotently on V . There is a basis for V that consists of simultaneous eigenvectors for both H

and the Casimir operator ∆. All the eigenvalues for E,F,H are integers, and for ∆ non-negativehalf-integers.

LIE ALGEBRAS AND LIE GROUPS MAT6633 25

The space V E=0 := {v ∈ V ; E · v = 0} is H-stable, and all its eigenvalues are non-negativeintegers. Let nd be the dimension of the H-eigenspace on V E=0 of eigenvalue d. Then nd isthe number of copies isomorphic to Vd in any direct sum decomposition of V of simple sl2(k)-representations. And trV (∆) =

∑d≥0

ndd(d+1)(d+2)2 .

Proof. Since V is the driect sum of simple representations, we can assume that V is simple, henceisomorphic to one of the Vn. By the explicit construction, the properties were already checked. �

Corollary 2.3. Let V be a any (finite dimensional) representation for sl2(k). The eigenvalues forH are all integers. Put ni for the dimension of the H-eigenspace with eigenvalue i.

(i) Then ni = n−i.(ii) n0 is the number of irreducible components of V isomorphic to Vd, with d even. And n1 is

the number of irreducible components of V isomorphic to Vd, with d odd. So n0 +n1 is the numberof irreducible components.

Proof. If suffices to prove in the case of simple V . Then it is clear by the explicit knowledge of thesimple represenatations. �

2.1. An infinite-dimensional simple representation. There are also infinite dimensional sim-ple sl2(k)-representations. E := X0∂1, F := X1∂0, H := X0∂0 − X1∂1, still act by derivations onk(X0, X1), the collection of rational functions

k(X0, X1) = {PQ

;P,Q ∈ k[X0, X1], Q 6= 0}.

Let V be the smallest sl2(k)-subrepresentation containing v0 := 1X0

. Define for i ≥ 1 by inductionvi = F · vi−1. Then necessarily vi ∈ V . For example

v1 = X1∂0(1X0

) = −X1

X20

, v2 = X1∂0(−X1

X20

) = 2X2

1

X30

,

and by induction

vi = (−1)ii!Xi

1

Xi+10

.

Then

H · vi = (X0∂0 −X1∂1)(−1)ii!

Xi1

Xi+10

)= (−(i+ 1)− i)

(−1)ii!

Xi1

Xi+10

)= −(2i+ 1)vi

so all vi are H-eigenvectors with eigenvalue −(2i + 1), and so they are all linearly independent.Next E · v0 = 0 and for i ≥ 1

E · vi = X0∂1

(−1)ii!

Xi1

Xi+10

)= i

(−1)ii!

Xi−11

Xi0

)= −i2vi−1.

26 ABRAHAM BROER

Finally

∆ · vi = E · F · vi + F · E · vi +12H ·H · vi

= E · vi+1 − i2F · vi−1 −2i+ 1

2H · vi

= −(i+ 1)2vi − i2vi +(2i+ 1)2

2vi

= −12vi,

shows that ∆ acts on V by multiplication by −12 .

So the space U with basis {vi; i = 0, 1, 2, . . .} is indeed an sl2(k)-representation, and the smallestsub-representation of k(X0, X1) containing 1

X0.

Lemma 2.3. V = 1X0k[X1X0

] is a simple infinite dimensional sl2(k)-representation. All H-eigenspacesare one-dimensional; the H-eigenspace with the highest eigenvalue −1 is the only eigenspace of E(with 0 eigenvalue).

Proof. The proof is the same as in the finite case. Let U ⊂ V be a non-zero subrepresentation.So there is a non-zero u ∈ U . By definition u is a linear combination of finitely many of the basisvectors, say u =

∑mi=0 civi, with cm 6= 0. Now Em(u) ∈ U is of the form cv0 for some non-zero

c ∈ k. Hence v0 ∈ U and therefore all ui ∈ U and so U = V . �

Remark. In a dual fashion it can be shown that V ′ := 1X1k[X0X1

] is a simple infinite dimensionalsl2(k)-module whose H-eigenspace with lowest eigenvalue +1 is the only eigenspace of F (with 0eigenvalue). V is an example of a simple highest weight representation, and V ′ is an example of asimple lowest weight representation.

Put x = X1X0

and consider the subalgebra R = k[x] of k(X0, X1). Our Lie algebra sl2(k) still actson the polynomial ring k[x] by derivations, explicitly:

E =d

dx, F = −x2 d

dx,H = −2x

d

dx,∆ = 0.

In fact, it is the smallest sl2(k)-subrepresentation of k(X0, X1) containing x = X1X0

, with basis

{xi; i = 0, 1, 2, . . .}.

But it is not simple, since it contains the trivial sub-representation of constant functions k ·1. Thereis no complementary subrepresentation U such that R = k · 1 ⊕ U , since by a similar method asin the proof above, any non-zero sl2(k)-submodule U contains 1 (which you can see by repeatedlyusing the operator E = d

dx). So in the infinite case it is no longer true that every subrepresentationhas a complement.

Let us try to exponentiate the sl2(k)-action to obtain an SL2(k)-action. For E there is noproblem.

exp(tE)(f) := 1(f) + td

dx(f) +

t2

2d2

dx2(f) + . . .+

ti

i!di

dxi(f) + . . .

= f(x) + tf ′(x) +t2

2f ′′(x) + . . .+

ti

i!f (i)(x)) + . . .

= f(x+ t)

LIE ALGEBRAS AND LIE GROUPS MAT6633 27

by Taylor’s formula; so exp(t(E)(f(X)) = f(X + t). We take k = C, then for any exponent r ≥ 1

exp(tH)(xr) = (1 + (−2rt) +(−2rt)2

2+ . . .+

(−2rt)i

i!+ . . .)xr

= (exp(−2t)x)r

or exp(tH)(f(x)) = f(exp(−2t)x) makes sense for polynomials f(x) ∈ C[x]. By induction F i(x) =(−1)ii!xi+1 and

exp(tF )(x) =∞∑i=0

(tF )i(x)i!

=∞∑i=0

(−t)ixi+1 = x

∞∑i=0

(−tx)i =x

tx+ 1

which clearly is not a polynomial. So we did not succeed in exponentiating the action. But beingso close we should not give up, but enlarge our ring. One possibility is the following.

Let O be the ring of analytic functions f(z) on the upper half plane {z ∈ C; Im(z) > 0}, thenfor every real number t ∈ R the functions f(z + t), f(exp(−2t)z) and f( z

tz+1) are also analytic onthe upper half plane, i.e., remain in O (since t ∈ R). Or in that case we do get an SL(2,R)-actionon O by algebra automorphisms (

a b

c d

)· f(z) := f(

az + b

cz + d),

for

(a b

c d

)∈ SL2(R), f(z) ∈ O. By differentiating this action we get our sl2(R)-action back again,

but extended to O:

E =d

dz, F = −z2 d

dz,H = −2z

d

dz.

This action is famous in number theory.

2.2. Extension of complete reductivity to semi-simple Lie algebras. Working with sl2(k)certainly has the advantage of being explicit. But most of the results proved above remain true insome form for semi-simple Lie algebras. The most important result being the complete reducibilityof any representation.

Theorem 2.3 (Weyl). Let g be a semi-simple Lie algebra. If U ⊂ V is a sub-representation, thenthere is a sub-representation U ′ ⊂ V such that V = U ⊕ U ′.

Proof. It suffices to prove the special case where U has codimension one, as in the sl2(k)-case.Since g is semi-simple it acts trivially on all one-dimensional representations, in particular it actstrivially of V/U . We will prove this special case of Weyl’s theorem by induction on the dimensionof V . If dimV = 1 then U = 0 and we can take U ′ = V .

Suppose U is not a simple g-representation. Then there is a proper sub-representation W ⊂ U .And U/W ⊂ V/W is a sub-representation of codimension 1, so by induction there is a v0 ∈ V ,not in U , such that kv0 = (W + kv0)/W is a complementary sub-representation, or V/W =U/W ⊕ (W + kv0)/W is a direct sum of representations. In particular (W + kv0) is itself arepresentation with sub-representation W of codimension one. We can use induction again andconclude there is v ∈W + kv0 such that kv is a complementary one dimensional representation or

28 ABRAHAM BROER

W + kv0 = W ⊕ kv. Now v is not in U , since otherwise W + kv0 would be part of U but this is acontradiction with the choice of v0. So we can take kv as the complementary module: V = U ⊕ kv.

We can therefore suppose that U is simple. Suppose first that U is the trivial one-dimensionalrepresentation. Also V/U is the trivial representation. For any v ∈ V and X ∈ g we have X ·v ∈ U ,and so for any Y ∈ g: Y · X · v = 0 and so [Y,X] · v = 0. Since g = [g, g] it follows that g actstrivially on the whole of V . Now let v ∈ V be any element not in U , then kv is also a representationand U ⊕ kv = V and we proved this special case.

Now suppose that U is simple but not trivial. If ρ : g → Endk(V ) is the homomorphismassociated to V . We can replace g by ρ(g), since the kernel of ρ acts trivially anyway. Also ρ(g)is semi-simple too (the kernel of ρ, like any ideal, is the direct sum of some of the simple ideals inthe direct sum of g into simple ideals).

So we can suppose that g acts faithfully on V . By Corollary 1.5 the form BV (·, ·) is non-degenerate. Let X1, . . . , Xm be a basis of g with dual basis Y1, . . . , Ym with respect to BV (·, ·). Weget a Casimir operator ∆ acting on V by ∆ · v =

∑iXi · Yi · v.

We prove that it commutes with the g-action. Let X ∈ g. There are aij , bij ∈ k such that

[X,Xi] =∑j

aijXj ; [X,Yi] =∑j

bijYj ,

soaij = BV ([X,Xi], Yj) = −BV (Xi, [X,Yj) = −bji.

We check

(X ·∆−∆ ·X) · v =∑i

(X ·Xi · Yi −Xi · Yi ·X) · v

=∑i

([X,Xi] · Yi −Xi · [Yi, X]) · v

=∑i,j

(aijXj · Yi + bijXi · Yi) · v

=∑i,j

(aijXj · Yi + bjiXj · Yi) · v

= 0

So the map∆ : V → V : v 7→ ∆ · v

is a homomorphism of representations. Since g acts trivially on V/U , also ∆ acts trivially on V/U ,i.e. ∆ · V ⊂ U . Its trace is

tr(∆) =∑i

trV (ρ(Xi)ρ(Yi)) =∑i

BV (Xi, Yi) = dimV.

Since BV = BU +BV/U and g acts trivially on V/U we get BV = BU . So ∆ does not act trivially onU (since its trace is dim g) so ∆ ·U is a non-zero sub-representation of the simple representation U ,so the image of ∆ is U . It follows that the kernel of ∆, say U ′, is a one-dimensional sub-represntationwith zero-intersection with U . We get the complement we were looking for: V = U ⊕ U ′. Thisfinishes the proof of Weyl’s theorem. �

LIE ALGEBRAS AND LIE GROUPS MAT6633 29

3. Complexification of simple Lie algebras

For any Lie algebra g, the quotient g/rad(g) is semi-simple, i.e., isomorphic to a direct sum ofideals, each of which is simple. We would like to have more information on simple Lie algebras. Infact, there is a lot of information available. And there is a classification : we know them all.

3.1. Complexification. Let us recall the process of complexification of a vector space. Let V bea real vector space. We define a new real vector space V C = V ⊗R C as the set of formal expressions

V C := {v ⊗ 1 + w ⊗ i; v, w ∈ V }

(where i ∈ C is as usual a square root of −1). In particular, v1 ⊗ 1 + 1⊗ i = v2 ⊗ 1 +w2 ⊗ i if andonly if v1 = v2 and w1 = w2. We define the addition

(v1 ⊗ 1 + w1 ⊗ i) + (v2 ⊗ 1 + w2 ⊗ i) := (v1 + v2)⊗ 1 + (w1 + w2)⊗ i

and multiplication by a real number λ ∈ R:

λ(v ⊗ 1 + w ⊗ i) := λv ⊗ 1 + λw ⊗ i.

With this structure V C becomes a real vector space isomorphic to V ⊕ V . Now we define for acomplex number z = a+ bi (a, b ∈ R) and v ∈ V :

v ⊗ z := av ⊗ 1 + bv ⊗ i ∈ V C

Then

v ⊗ λz = λv ⊗ z, (v1 + v2)⊗ z = (v1 ⊗ z) + (v2 ⊗ z), v ⊗ (z1 + z2) = (v ⊗ z1) + (v ⊗ z2),

for z, z1, z2 ∈ C, λ ∈ R, v, v1, v2 ∈ V .Finally we define multiplication by a complex number as

z · (v ⊗ 1 + w ⊗ i) := v ⊗ z + w ⊗ zi

= (av ⊗ 1 + bv ⊗ i) + (aw ⊗ i− bw ⊗ 1)

= (av − bw)⊗ 1 + (bv + aw)⊗ i.

So in particular, if z1, z2 ∈ C and v ∈ V then

z1 · (v ⊗ z2) = v ⊗ z1z2.

With this addition and complex multiplication V C becomes a complex vector space. It comesequipped with a map induced by the complex conjugation map

θ : V C → V C; v ⊗ 1 + w ⊗ i 7→ v ⊗ 1− w ⊗ i,

v, w ∈ V . So with the usual complex conjugation θ(v ⊗ z) = v ⊗ z, with v ∈ V, z ∈ C. So θ is aninvolution (θ2 = 1), an isomorphism as real vector space, but θ(zu) = zθ(u), for u ∈ V C, z ∈ C.The fixed points for θ is the ”real part” i.e. V ⊗ 1.

We remark that an arbitrary complex vector space does not come equipped automatically withsuch a map θ !

But if a complex vector space comes equipped with such a θ, then it comes from complexificationof a specific real vector space. Suppose W is a (finite dimensional) complex vector space togetherwith a map θ : W → W , which is an involution, real linear and θ(zw) = zθ(w), for z ∈ C and

30 ABRAHAM BROER

w ∈ W . Take V = {w ∈ W ; θ(w) = w}. Then V is a real sub vector-space of W , iV is also a realsub vector space and V ∩ iV = 0 (since if v = iw, for v, w ∈ W , then v = θv = θiw = −iw = −vand so v = 0). Let w ∈W , then w + θ(w) and −iw + iθ(w) are in V , since

θ(w + θ(w)) = θ(w) + θ(θ(w)) = w + θ(w); θ(−iw + iθ(w)) = iθ(w)− iw.

So

w =12

(w + θ(w)) + i12

(−iw + iθ(w)) ∈ V + iV

and so W = V ⊕ iV . Then

V C 'W : (v1 ⊗ 1 + v2 ⊗ i) 7→ v1 + iv2

gives an isomorphism as complex vector spaces.If v1, . . . , vn is a real basis of V , then v1⊗1, v2⊗1, . . . , vn⊗1 is a basis for V C as complex vector

space.Sometimes we identify V with V ⊗ 1, then iV = V ⊗ i and V C = V ⊕ iV . And then a real basis

for V is at the same time a complex basis for V C.We will avoid this identification when V is a complex vector space seen as a real vector space.

In that case we have to be careful to distinguish between the old complex multiplication and thenew complex multiplication, say in that case z1v⊗ z2 (using the old complex multiplication) is notthe same as z1 · (v ⊗ z2) = v ⊗ z1z2 (the new complex multiplication), where z1, z2 ∈ C, v ∈ V .

If g is a real Lie algebra, then gC is a complex Lie algebra, with Lie bracket

[X1 ⊗ z1, X2 ⊗ z2] := [X1, X2]⊗ z1z2,

where X1, X2 ∈ g, z1, z2 ∈ C. And so

θ([Z1, Z2]) = [θ(Z1), θ(Z2)],

for Z1, Z2 ∈ gC.On the other hand, let k be a complex Lie algebra together with an involution θ : k→ k which is

an automorphism of order two of real Lie algebras, such that θ(uZ) = uθ(Z) for u ∈ C and Z ∈ k.Then g = {Z ∈ k; θ(Z) = Z} is a real sub-algebra and

gC → k : X ⊗ 1 + Y ⊗ i 7→ X + iY

with X,Y ∈ g is an isomorphism of complex Lie algebras.Similarly, if V is a real representation of the real Lie algebra g, then V C is a complex represen-

tation of the complex Lie algebra gC, with multiplication

(X ⊗ z1) · (v ⊗ z2) := (X · v)⊗ z1z2.

The two natural complex conjugation maps θ are related by

θ(Xu) = θ(X)θ(u),

where X ∈ gC and u ∈ V C.For the associated bilinear forms we get

BV C(X1 ⊗ z1, X2 ⊗ z2) = z1z2BV (X1, X2) ∈ C

LIE ALGEBRAS AND LIE GROUPS MAT6633 31

for X1, X2 ∈ g, z1, z2 ∈ V . A real basis of g can be identified with a complex basis of gC, so theassociated Gramm-matrices are the same. Therefore BV is non-degenerate on g if and only if BV C

is non-degenerate on gC . In particular, g is a real semisimple Lie algebra if and only if gC is acomplex semisimple Lie algebra, by Cartan’s theorem.

3.2. Complexification of real simple Lie algebras. If g is a simple real Lie algebra, then itscomplexification gC is not necessarily a simple complex Lie algebra, but at least it is semi-simple,hence a direct sum of simple ideals.

Example 3.1. sl2(R)C ' sl2(C), suC2 ' sl2(C) but sl2(C)C ' sl2(C)⊕ sl2(C).

Proposition 3.1. Suppose that g is a simple complex Lie algebra, but we consider it as a real Liealgebra.

(i) g is also simple considered as a real Lie algebra.(ii) There is an isomorphism of real Lie algebras

gC → g⊕ g : (A⊗ 1 +B ⊗ i) 7→ (A+ iB,A− iB).

Its inverse is (X,Y ) 7→(

12(X + Y )⊗ 1

)+(

12i(X − Y )⊗ i

).

Complex conjugation θ for gC corresponds to the map

g⊕ g : (X,Y ) 7→ (Y,X).

Multiplication by the new i on gC (on the righthsand side of g⊗R C) corresponds to the map

g⊕ g : (X,Y ) 7→ (iX,−iY ).

Proof. (i) Suppose a is a minimal non-zero real ideal of g. So it is a real sub vector-space suchthat [g, a] ⊆ g. Now a + ia is a complex ideal of g, hence a + ia = g. Also [g, ia] = i[g, a] ⊂ ia

shows that ia is also a real ideal. If ia = a then a is even a complex ideal, hence a = g. If not,by minimality of a we get a ∩ ia = 0 and g = a ⊕ ia, a direct sum of real ideals. Let A,B ∈ a

then [iA,B] = i[A,B] ∈ a ∩ ia = 0 so [A,B] = 0 and hence also g is abelian. But g is simple.Contradiction.

(ii) It is certainly real linear. Since

[A⊗ 1 +B ⊗ i, A′ ⊗ 1 +B′ ⊗ i] = ([A,A′]− [B,B′])⊗ 1 + ([A,B′] + [B,A′])⊗ i

is mapped to

(([A,A′]− [B,B′])− i([A,B′] + [B,A′]), ([A,A′]− [B,B′]) + i([A,B′] + [B,A′]) =

= ([A− iB,A′ − iB′], [A+ iB,A′ + iB′]),

it follows that the bracket is preserved. That the two maps are inverses to each other is checkeddirectly. �

We have a converse too.

Proposition 3.2. Suppose g is a simple real Lie algebra such that gC is not simple. Then g is asimple complex Lie algebra, considered as a real Lie algebra.

32 ABRAHAM BROER

Proof. Let a be a simple complex ideal of gC, then θ(a) is also a simple complex ideal of gC. Puta0 = a∩(g⊗1). Then a0 is an ideal of the simple real Lie algebra g⊗1 ' g. So a0 = 0 or a0 = g⊗1.

In case a0 = g⊗ 1, then also g⊗ i ⊂ a and so gC = a. Which is a contradiction, since gC is notsimple. So a0 = 0.

Let X ⊗ 1 + Y ⊗ i ∈ a ∩ θa then12

(X ⊗ 1 + Y ⊗ i+X ⊗ 1− Y ⊗ i) = X ⊗ 1 ∈ a0

and so X = 0 and −i(Y ⊗ i) = Y ⊗ 1 ∈ a0 and so Y = 0. We conclude that a ∩ θa = 0.Let X ⊗ 1 + Y ⊗ i ∈ a be non-zero, then also X ⊗ i − Y ⊗ 1 ∈ a and both X ⊗ 1 − Y ⊗ i and

−X ⊗ i− Y ⊗ 1 ∈ θa and so 2X ⊗ 1 and 2Y ⊗ 1 ∈ (a⊕ θa) ∩ g⊗ 1, and so (a⊕ θa) ∩ (g⊗ 1) 6= 0.From the simplicity of g it follows that (g ⊗ 1) ⊂ a ⊕ θa. And so we get a direct sum of complexideals

gC = a⊕ θa.

The complex linear map

aC → a⊕ θa : (A⊗ 1 +B ⊗ i) 7→ (A+ iB, θ(A− iB))

has inverse linear map (X,Y ) 7→ (12(X + θ(Y ))⊗ 1) + ( 1

2i(X − θ(Y ))⊗ i). It is an isomorphism ofcomplex Lie algebras.

We also have a homomorphism of real Lie algebras

α : a→ g⊗ 1 : Z 7→ (Z + θZ)

since

α([Z,W ]) = ([Z,W ] + θ[Z,W ]) = [Z + θZ,W + θW ]− [Z, θW ]− [θZ,W ] = [α(Z), α(W )].

Let X ⊗ 1 ∈ g⊗ 1. There are unique X1 ⊗ 1 + Y1 ⊗ i and X2 ⊗ 1 + Y2 ⊗ i ∈ a such that

X ⊗ 1 = (X1 ⊗ 1 + Y1 ⊗ i) + (X2 ⊗ 1− Y2 ⊗ i) = (X1 +X2)⊗ 1 + (Y1 − Y2)⊗ i.

So Y2 = Y1. But then

(X1 ⊗ 1 + Y1 ⊗ i)− (X2 ⊗ 1 + Y2 ⊗ i) = (X1 −X2)⊗ 1 ∈ a0

hence X1 = X2 = 12X. And for any X ∈ g there is a unique Z ∈ a such that X = Z + θZ = α(Z).

Or in other words: α : a → g is an isomorphism of real Lie algebras, and the complexificationsof a and g are isomorphic. �

Suppose we know the classification of complex simple Lie algebras. They are also real simple Liealgebras. If we also know all the possible θ’s on the complex simple Lie algebras, we can deducethe simple real Lie algebras.

Up to only five exceptions any complex simple Lie algebra is a member of one of three families

sl(n,C), so(n,C), sp(2n,C).

The last of the three families is associated to non-degenerate anti-symmetric complex bilinear forms,in the same way as the second family is associated to non-degenerate symmetric complex bilinearforms.

LIE ALGEBRAS AND LIE GROUPS MAT6633 33

3.3. Bilinear forms on a complex simple Lie algebra. Being a simple complex Lie algebrag implies that there is up to a constant only one invariant bilinear form, the Killing form. AC-bilinear form C on g is called g-invariant if for all X,Y, Z ∈ g:

C(ad(X)(Y ), Z) + C(Y, ad(X)(Z)) = 0

or C(Y, [X,Z]) = C([Y,X], Z). If the form comes from a complex representation, it has this invari-ancy.

Proposition 3.3. Let g be a simple complex Lie algebra. For any g-invariant bilinear form C ong there exists a c ∈ C such that for all X,Y ∈ g we have C(X,Y ) = cB(X,Y ), where B is theKilling form.

Proof. For any X ∈ g we get a linear form C(X, ·) ∈ g∗. Since the Killing form is non-degeneratethere is a unique α(X) ∈ g such that C(X, ·) = B(α(X), ·). The map α : g → g is clearly linear,but we have more. Since both forms are g-invariant, we have for all Y,Z ∈ g

B(α(ad(Y )(X)), Z) = C(ad(Y )(X), Z) = −C(X, ad(Y )(Z)) = −B(α(X), ad(Y )(Z))

= B(ad(Y )(α(X)), Z),

so B(α(ad(Y )(X)), ·) = B(ad(Y )(α(X)), ·), and by unicity

α(ad(Y )(X)) = ad(Y )(α(X)).

In other terms the map α : g→ g is not quite a homomorphism of Lie algebras, but it is a morphsimof g-representations. Its kernel is a sub representation of g, hence an ideal. If the kernel is g then αis the zero-map and the form C is zero. As constant we can take c = 0. If not, then α is injective,hence α is an automorphism of g-representations. There is an eigenvalue c ∈ C for α acting on g.Then (α − c1) is also an endomorphism of g-representations, so its kernel is a sub-representationhence either 0 or g, by simplicity of the Lie algebra. This kernel is just the eigenspace of eigenvaluec, which is non-zero, hence the whole space. This means that α(X) = cX for all X ∈ g. OrC(X,Y ) = B(cX, Y ) = cB(X,Y ). �

Remark. It also follows there is up to scalar only one Casimir operator associated to a simplecomplex Lie algebra (for some more information, see proof of Weyl’s theorem).

34 ABRAHAM BROER

Departement de mathematiques et de statistique, Universite de Montreal, C.P. 6128, succursale

Centre-ville, Montreal (Quebec), Canada H3C 3J7

E-mail address: broera@DMS.UMontreal.CA