KÄHLER STRUCTURES ON

GENERALIZED FLAG MANIFOLDS

DAG NILSSON

Master’s thesis2013:E17

Faculty of ScienceCentre for Mathematical SciencesMathematics

CE

NT

RU

MSC

IEN

TIA

RU

MM

AT

HE

MA

TIC

AR

UM

Abstract

In this Master’s thesis we study complex structures and Kahler forms on homogeneous spacesM = G/K, where G is a Lie group and K a closed subgroup of G. We show some general resultson these manifolds regarding G-invariant complex structures and G-invariant metrics.

In particular we study the case when G is a compact connected semisimple Lie group and K isthe centralizer of a torus in G. In this case M is called a generalized flag manifold. We investigatethese manifolds by looking at the corresponding root space decompositions, with respect to someCartan subalgebra.

Our main goal is to show that for each G-invariant complex structure on a generalized flagmanifold M there exists a family of G-invariant Kahler metrics on M . For this result we follow [4]by Bordemann, Forger and Romer.

For a more detailed discussion regarding complex structures on homogeneous spaces, we rec-ommend [5] by Borel and Hirzebruch and [13] by Wang.

Throughout this work it has been my firm intention to give reference to the stated results andcredit to the work of others. All theorems, propositions, lemmas and examples left unmarked areassumed to be too well known

Sammanfattning

Ett homogent rum ar en mangfald M tillsammans med en Lie grupp G med egenskapen attgivet tva punkter x, y ∈M , sa existerar det g ∈ G sa att g ·x = y. I Proposition 2.6 sa visar vi attett homogent rum ocksa kan skrivas som G/Gx, dar Gx = {g ∈ G | g · x = y }. Vidare sa noterarvi att varje mangfald pa formen G/K, dar K ar en sluten delgrupp av G, ar ett homogent rum.

En generaliserad flagmangfald ar ett homogent rum pa formen G/K, dar G ar en kompakt Liegrupp och

K = C(T ) = { g ∈ G | ghg−1 = h for alla h ∈ T }

dar T ar en torus i G.Lat M vara en reell mangfald. En nastan komplex struktur J pa M ar ett tensorfalt sadant

att J2x = −I for alla x ∈ M . Med andra ord ar Jx en komplex struktur pa TxM for alla x ∈ M .

Om den nastan komplexa strukturen kommer fran en holomorf struktur pa mangfalden sa kallasden for en komplex struktur.

Lat M vara en mangfald med nastan komplex struktur J . En Kahler metrik g ar en metriksadan att g(JX, JY ) = g(X,Y ) och den bilineara formen ω(X,Y ) = g(JX, Y ) ar sluten.

Lat g vara en komplex semienkel Lie algebra och lat h vara en Cartan subalgebra. For ettelement α ∈ h∗ sa definierar vi

gα = {X ∈ g | (ad(H)(X)− α(X)I)n = 0 for alla H ∈ h}

och dar n ar nagot heltal. Lat R vara mangden av alla element α ∈ h∗ sadana att gα 6= {0}.Mangden R kallas for rotsystemet horande till g med avseende pa h. Vi har da att

g = h⊕∑α∈R

gα.

For en generaliserad flagmangfald M = G/K sa har vi att

gC = hC ⊕∑α∈R

gα

dar hC ar en Cartan subalgebra av gC sadan att hC ⊆ kC. Pa grund av detta sa existerar detRK ⊆ R sadant att

kC = hC ⊕∑α∈RK

gα.

Man kan sedan visa att det existerar en bijektion mellan G-invarianta komplexa strukturer pa Moch speciella delrum av gC som kallas Weylkammare.

Slutligen sa kan man visa att varje generaliserad flagmangfald med G-invariant komplex struk-tur J har en Kahler metrik. For att bevisa detta sa definierar vi helt enkelt en metrik pa M ochvisar att denna ar Kahler. Konstruktionen av denna metrik kommer att bero pa den Weylkammaresom motsvarar J.

Acknowledgments

I would like to thank my supervisor Sigmundur Gudmundsson for his advice when writing thisthesis and also for suggesting this interesting topic.

Dag Nilsson

Contents

1 Basic Lie Theory 11.1 Root systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Derivations of Lie Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2 Homogeneous Spaces 72.1 Homogeneous Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.2 Reductive Homogeneous Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

3 Kahler Manifolds 133.1 Complex Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133.2 Complex Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153.3 Complex Homogeneous Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163.4 Kahler Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

4 Generalized Flag Manifolds 234.1 Complex Structures on Generalized Flag Manifolds . . . . . . . . . . . . . . . . . . 234.2 Weyl Chambers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314.3 The Construction of the Kahler Metrics . . . . . . . . . . . . . . . . . . . . . . . . 34

Appendix:

A Differentiable Manifolds 47

Bibliography 49

Chapter 1

Basic Lie Theory

1.1 Root systems

We will in this chapter state some results regarding Lie algebras which will be needed whenconstructing complex structures on generalized flag manifolds. The presentation given here follows[10]. We begin to state some preliminary definitions.

Definition 1.1. A real (complex) vector space V together with an operation [, ] : V × V → V issaid to be a real (complex) Lie algebra if

a) [λX + µY,Z] = λ[X,Z] + µ[Y,Z]

b) [X,Y ] = −[Y,X]

c) [X, [Y, Z]] + [Z, [X,Y ]] + [Y, [Z,X]] = 0

for all X,Y, Z ∈ V and λ, µ ∈ R (C). The equality c) is called the Jacobi identity.

We will denote Lie algebras by Gothic letters g, h,m . . . and will only consider finite-dimensionalLie algebras.

Let g be a Lie algebra. For j ≥ 0, we define

g0 = g, g1 = [g0, g0], . . . , gj+1 = [g, gj ].

A Lie algebra g is said to be nilpotent if gj = 0 for some j. In a similar way, for j ≥ 0, define

g0 = g, g1 = [g0, g0], . . . , gj+1 = [gj , gj ].

A Lie algebra g is said to be solvable if gj = 0 for some j.A subalgebra h of g is a subspace of g, such that [h, h] ⊆ h. An ideal h in g is a subspace of

g satisfying [h, g] ⊆ h.A Lie algebra is said to be semisimple if it has no non-zero solvable ideals. Next we define

ad : g→ EndC(g) byad(X)(Y ) = [X,Y ].

This is a linear map since the bracket is linear in both its arguments. We then define the symmetricbilinear form B:

B(X,Y ) = Tr(ad(X) ◦ ad(Y )).

B is called the Killing form of g. There is a way of defining semisimplicity in terms of the Killingform, which is equivalent to the definition given above. We state this as a theorem.

Theorem 1.2. [10] A Lie algebra g is semisimple if and only if the Killing form of g is nonde-generate.

1

Now let g be a complex Lie algebra with a nilpotent subalgebra h. Let α ∈ h∗ be an elementof the dual of h. We then define

gα = {X ∈ g | (ad(H)− α(H)I)nX = 0 for all H ∈ h},

where I denotes the identity mapping of g and n is some positive integer depending on both Hand X. If gα 6= 0, then α is called a root of g with respect to h, and gα is called the correspondingroot space. The elements of gα are called root vectors. The following theorem is a key result.

Theorem 1.3. [10] Let g be a complex Lie algebra with a nilpotent subalgebra h. Then the followingholds:

a) g = ⊕αgα

b) h ⊆ g0 = {X ∈ g | ad(H)nX = 0 for all H ∈ h}

c) [gα, gβ ] ⊆ gα+β, where [gα, gβ ] = 0 if α+ β is not a root.

From c) we see that g0 is a subalgebra, since [g0, g0] ⊆ g0+0 = g0.

Definition 1.4. A nilpotent Lie subalgebra h of g is called a Cartan subalgebra if h = g0.

It can be shown that any complex Lie algebra has a Cartan subalgebra and that Cartan sub-algbras are conjugate to each other by an automorphism of g. In particular this means that everyCartan subalgebra of g has the same dimension.

We will be particularly interested in the case when g is semisimple. If h is a Cartan subalgebraof a complex semisimple Lie algebra, then one can show that h is Abelian.

Let g be a complex Lie algebra with Cartan subalgebra h. Let R be the set of nonzero roots ofg with respect to h. The set R is called the root system of g with respect to h. From Theorem1.3 we then have the following

g = h⊕⊕α∈R

gα.

We call this the root space decomposition of g with respect to h. We will usually write thisdecomposition as

g = h⊕∑α∈R

gα

where the sum is understood to be direct. We now state some additional results regarding roots.

Proposition 1.5. [10] Let g be a complex semisimple Lie algebra with Cartan subalgebra h andlet R be the corresponding root system. Then

a) If α, β ∈ R ∪ {0} and α+ β 6= 0, then B(gα, gβ) = 0.

b) If α ∈ R ∪ {0}, then B(X, g−α) 6= 0 for all nonzero X in gα.

c) If α ∈ R, then −α ∈ R.

d) B is non-degenerate on h× h

e) For each α ∈ R there exist an element Eα ∈ gα such that ad(H)(Eα) = α(H)Eα for H ∈ h.

f) If α ∈ R, then dim gα = 1.

g) If α, β ∈ R ∪ {0}, such that α+ β 6= 0, then [gα, gβ ] = gα+β.

h) Let α ∈ R. The only roots proportional to α are −α, 0 and α.

2

From e) and f) we see that

gα = {X ∈ g | (ad(H)− α(H)I)X = 0 for all H ∈ h},

so we need no longer worry about the integer n. From d) we get that for all ψ ∈ h∗, there exists aunique Hψ ∈ h such that ψ(H) = B(H,Hψ). We let

h0 = spanR{Hα}α∈R.

One can show that (h0,B) is a real Euclidean vector space and that

h = h0 ⊕ ih0.

Since B(X,Y ) ∈ R for all X,Y ∈ h0, we have in particular that α(H) = B(Hα, H) ∈ R, for allH ∈ h0. So the roots are real-valued on h0.

Since the root spaces are 1-dimensional and [gα, gβ ] = gα+β for α+β 6= 0 we define the numberNα,β such that

[gα, gβ ] = Nα,βEα+β

andNα+β = 0 if α+ β /∈ R.

The following proposition will be very useful later on.

Proposition 1.6. [10] For each α ∈ R the basis vectors Eα ∈ gα can be chosen such that

B(Eα, E−α) = 1[Eα, E−α] = Hα.

Next we introduce the notion of positivity on R. The idea is to choose a subset R+ of R, andsaying that α is positive if α ∈ R+. The set R+ must satisfy the following criterions:

i) given α ∈ R, exactly one of α, −α is positive.

ii) Any sum of positive roots, which belongs to R, is positive.

A root α is said to be simple, if α > 0 and there does not exist β, γ > 0 such that α = β + γ. LetΠ be the set of simple roots of R. We then have the following result:

Proposition 1.7. [10] Let g be a complex semisimple Lie algebra with the set of nonzero roots R,corresponding to some Cartan subalgebra of g. The set of simple roots Π = {α1, . . . , αl} ⊂ R, isa basis of R, such that every α ∈ R can be written uniquely as α =

∑li=1 niαi, where the ni are

integers which are either all non-negative or all non-positive. Moreover, the set of simple roots arelinearly independent over R.

Definition 1.8. A Lie algebra is compact if it is the Lie algebra of a compact Lie group.

We have the following result regarding compact semisimple Lie algebras over R.

Proposition 1.9. [9] Let g be a compact semisimple Lie algebra over R. Then g is compact if andonly if the Killing form of g is strictly negative definite.

1.2 Derivations of Lie Algebras

A derivation of a Lie algebra is a special kind of endomorphism. The goal of this section is to showthat every derivation of a semisimple Lie algebra can be written as ad(X), where X is an elementin the Lie algebra.

Throughout this section g will be a (real or complex) finite dimensional Lie algebra.

Definition 1.10. An endomorphism D of g, which satisfies

D([X,Y ]) = [D(X), Y ] + [X,D(Y )] for all X,Y ∈ g

is called a derivation of g.

3

Let δ(g) be the set of all derivations of g. Recall the Jacobi identity [X, [Y,Z]] + [Z, [X,Y ]] +[Y, [Z,X]] = 0 for X,Y, Z ∈ g. We may rewrite this as

[Z, [X,Y ]] = [X, [Z, Y ]] + [[Z,X], Y ]

which is equivalent to

ad(Z)[X,Y ] = [X, ad(Z)(Y )] + [ad(Z)(X), Y ]

which means that ad(Z) is a derivation for all Z ∈ g. We therefore have that ad(g) ⊆ δ(g). Wewould like to show that ad(g) = δ(g) when g is semisimple. We will begin by proving the followinglemmas.

Lemma 1.11. [10] δ(g) is a Lie algebra and ad: g→ δ(g) is a Lie algebra homomorphism.

Proof. [10] It is easy to see that δ(g) is a vector space. So what we need to show is that [D,E] ∈ δ(g)for all D,E ∈ δ(g). Let D,E ∈ δ(g) and X,Y ∈ g. Then

[D,E]([X,Y ]) = (DE − ED)([X,Y ])= DE([X,Y ])− ED([X,Y ])= D([EX,Y ] + [X,EY ])− E([DX,Y ] + [X,DY ])= [DEX,Y ] + [EX,DY ] + [DX,EY ] + [X,DEY ]− [EDX,Y ]− [DX,EY ]− [EX,DY ]− [X,EDY ]

= [DEX,Y ]− [EDX,Y ] + [X,DEY ]− [X,EDY ]= [(DE − ED)X,Y ] + [X, (DE − ED)Y ]= [[D,E]X,Y ] + [X, [D,E]Y ]

which implies that [D,E] ∈ δ(g).Let X,Y, Z ∈ g. Then, using the Jacobi identity, we see that

ad([X,Y ])(Z) = [[X,Y ], Z]= −[[Y,Z], X]− [[Z,X], Y ]= [X, [Y, Z]]− [Y, [X,Z]]= (ad(X) ◦ ad(Y ))(Z)− (ad(Y ) ◦ ad(X))(Z)= [ad(X), ad(Y )](Z)

which shows that ad is a Lie algebra homomorphism.

Lemma 1.12. [9] Let B be the Killing form of g and let σ be an automorphism of g, then

B(σ(X), σ(Y )) = B(X,Y )B([X,Z], Y ) = B(X, [Z, Y ])

for all X,Y, Z ∈ g.

Proof. [2],[9] Since σ is an automorphism of g, we have that σ([X,Y ]) = [σ(X), σ(Y )] for allX,Y ∈ g. Using this we see that

ad(σ(X))(Y ) = [σ(X), Y ]

= σ([X,σ−1(Y )])

= (σ ◦ ad(X) ◦ σ−1)(Y ).

This holds for all Y ∈ g, hence ad(σ(X)) = σ ◦ ad(X) ◦ σ−1. Using this and the definition of theKilling form, we get

B(σ(X), σ(Y )) = Tr(ad(σ(X)) ◦ ad(σ(Y )))

4

= Tr(σ ◦ ad(X) ◦ σ−1 ◦ σ ◦ ad(Y ) ◦ σ−1)

= Tr(σ ◦ ad(X) ◦ ad(Y ) ◦ σ−1)= Tr(ad(X) ◦ ad(Y ))= B(X,Y ).

To show the second equality we will use the Jacobi identity.

[Z, [X, [Y,W ]]] = [X, [Z, [Y,W ]]] + [[Z,X], [Y,W ]]= [X, [Y, [Z,W ]] + [[Z, Y ],W ]] + [[Z,X], [Y,W ]]= [X, [Y, [Z,W ]]] + [X, [[Z, Y ],W ]] + [[Z,X], [Y,W ]]

from which it follows that

ad(Z) ◦ ad(X) ◦ ad(Y ) = ad(X) ◦ ad(Y ) ◦ ad(Z)+ ad(X) ◦ ad(ad(Z)(Y )) + ad(ad(Z)(X)) ◦ ad(Y )⇔ ad(Z) ◦ ad(X) ◦ ad(Y )− ad(X) ◦ ad(Y ) ◦ ad(Z) = ad(X) ◦ ad(ad(Z)(Y ))

+ ad(ad(Z)(X)) ◦ ad(Y )⇔ [ad(Z), ad(X) ◦ ad(Y )] = ad(X) ◦ ad(ad(Z)(Y )) + ad(ad(Z)(X)) ◦ ad(Y ).

This implies that

B(ad(Z)(X), Y ) + B(X, ad(Z)(Y ))= Tr(ad(ad(Z)(X)) ◦ ad(Y )) + Tr(ad(X) ◦ ad(ad(Z)(Y )))= Tr([ad(Z), ad(X) ◦ ad(Y )])= 0.

So we get that−B([Z,X], Y ) = B(X, [Z, Y ])

or equivalentlyB([X,Z], Y ) = B(X, [Z, Y ]),

which is what we wanted to show.

Lemma 1.13. [10] Let V be a finite dimensional vector space and C a nondegenerate bilinear formon V × V . Then the map ϕ : V → V ∗ given by ϕ(u)(v) = C(u, v) is bijective.

Proof. [10] By assumption C is nondegenerate. This implies that

ker(ϕ) = {v ∈ V | ϕ(v)(u) = 0 for all u ∈ V }= {v ∈ V | C(v, u) = 0 for all u ∈ V }= {0}.

So ϕ is injective and hence bijective, since dim(V )=dim(V ∗).

We are now ready to prove the main result of this section.

Proposition 1.14. [10] If g is semisimple, then ad(g) = δ(g).

Proof. [10] Let D ∈ δ(g). Then

ad(DX)(Y ) = [DX,Y ]= D[X,Y ]− [X,DY ]= (D ◦ ad(X))(Y )− (ad(X) ◦D)(Y )= [D, ad(X)](Y )

which implies that ad(DX) = [D, ad(X)].

5

From Lemma 1.13 we know that the map ϕ : g→ g∗ given by ϕ(X)(Y ) = B(X,Y ) is bijective,since B is nondegenerate. Given D ∈ δ(g), we let l(Y ) = Tr(D ◦ ad(Y )) for Y ∈ g. It is clear thatl ∈ g∗. We know that there exists a unique X ∈ g such that l = ϕ(X). We therefore have that

Tr(D ◦ ad(Y )) = l(Y ) = ϕ(X)(Y ) = B(X,Y ).

Using this we get that for Y,Z ∈ g,

B(DY,Z) = Tr(ad(DY ) ◦ ad(Y ))= Tr([D, ad(Y )] ◦ ad(Y ))= Tr(D ◦ ad(Y ) ◦ ad(Z)− ad(Y ) ◦D ◦ ad(Z))= Tr(D ◦ ad(Y ) ◦ ad(Z))− Tr(ad(Y ) ◦D ◦ ad(Z))= Tr(D ◦ ad(Y ) ◦ ad(Z))− Tr(D ◦ ad(Z) ◦ ad(Y ))= Tr(D(ad(Y ) ◦ ad(Z)− ad(Z) ◦ ad(Y )))= Tr(D([ad(Y ), ad(Z)]))= Tr(D ◦ ad([X,Y ]))= B(X, [Y,Z])= B([X,Y ], Z).

We have shown that B(DY,Z) = B(ad(X)(Y ), Z) for all Y,Z ∈ g. Since B is nondegenerate thisimplies that DY = ad(X)(Y ) for all Y ∈ g or equivalently, D = ad(X) ∈ ad(g). It is then clearthat δ(g) = ad(g), since we already knew that ad(g) ⊆ δ(g).

6

Chapter 2

Homogeneous Spaces

In this chapter, we introduce a special class of manifolds called homogeneous spaces. We thenstate and prove several important results regarding these manifolds. In general we use the samenotation as in [8] throughout this text. We assume that the manifolds are Hausdorff and secondcountable.

2.1 Homogeneous Spaces

We begin by introducing the coset manifolds and the action of a Lie group on a manifold. Weshall see that the two are essentially the same. Let G be a Lie group and K be a closed subgroupof G.

Definition 2.1. The setG/K = {gK : g ∈ G}

is called a coset space. The map

π : G→ G/K, π(g) = gK.

is called the natural projection.

We will use the following topology on the coset spaces.

Definition 2.2. The topology on G/K which is uniquely determined by the condition that πshould be open and continuous, is called the natural topology.

It is possible to equip G/K with a manifold structure, in fact we have the following result:

Theorem 2.3. [14] Let G be a Lie group and K be a closed subgroup of G. Then G/K has aunique manifold structure such that the projection π is a smooth submersion.

For the moment we call G/K, together with the above differentiable structure, a coset mani-fold.

Definition 2.4. Let M be a differentiable manifold and G be a Lie group. A smooth map α :G×M → M is a group action of G on M if

α(e, x) = x, where e is the identity element in Gα(g1g2, x) = α(g1, α(g2, x))

for all g1, g2 ∈ G and x ∈ M .

Definition 2.5. A group action α of G on M is said to be transitive, if for any x, y ∈ M thereexist a g ∈ G such that α(g, x) = y.

7

The set Gx = {g ∈ G : α(g, x) = x} is called the isotropy subgroup at x ∈M .The orbit of a point x ∈ M is the set α(G, x) = {α(g, x) : g ∈ G}.

On a coset manifold G/K we can define an action α of G in the following natural way:

α : G×G/K → G/K with α(g, g1K) = gg1K

This is a transitive action, since given g1K, g2K ∈ G/K we may take g = g2g−11 , and so

α(g2g−11 , g1K) = g2K.

Before we write the definition of a homogeneous space, we state the following result.

Theorem 2.6. [14] Let α be a transitive action of a Lie group G on a differentiable manifold Mand let x ∈M . Then the isotropy group Gx is a closed subgroup of G and the map β defined by

β : G/Gx →M with β(gGx) = α(g, x)

is a diffeomorphism.

Proof. [14] Since M is Hausdorff, the singleton sets are closed in M . Let αx(g) = α(g, x). It iseasy to see that

α−1x ({x}) = Gx.

It then follows from the inverse function theorem that Gx is a submanifold of G. In particular αxis a continuous map from G to M , so α−1

x ({x}) = Gx is a closed subset of G.Let g, h ∈ Gx. Then

α(gh, x) = α(g, α(h, x)) = α(g, x) = x

so gh ∈ Gx. Clearly e ∈ Gx and for any g ∈ Gx

x = α(e, x)

= α(g−1g, x)

= α(g−1, α(g, x))

= α(g−1, x)

which implies that g−1 ∈ Gx. Hence Gx is a closed subgroup of G.We need to show that the map β is well defined. Let gGx = hGx. Then h = gg1 for some

g1 ∈ Gx. So

β(hGx) = α(h, x)= α(gg1, x)= α(g, α(g1, x))= α(g, x)

= β(gGx)

We would now like to show that β is a diffeomorphism. By Lemma A.1 it is enough to show thatβ is smooth, bijective and everywhere non-singular. Let us begin by showing that β is bijective.

Given any y ∈ M there exists g ∈ G such that α(g, x) = y since the action α is transitive.Then β(gGx) = α(g, x) = y which implies that β is surjective.

Suppose β(gGx) = β(hGx) for some g, h ∈ G. This is equivalent to α(g, x) = α(h, x). Wetherefore get that

β(h−1gGx) = α(h−1g, x)

= α(h−1, α(g, x))

= α(h−1, α(h, x))

8

= α(e, x)= x

which means that h−1g ∈ Gx or equivalently gGx = hGx. Hence β is injective.Since β is smooth if and only if β ◦ π is smooth we let β = β ◦ π and consider β instead of β.For any g ∈ G we have

β(g) = β(π(g)) = α(g, x) = (α ◦ ix)(g),

where ix is the smooth map from G into G×M defined by ix(g) = (g, x). Since α and ix are bothsmooth it follows that their composition is smooth, hence β is smooth.

What is left to show is that β is nonsingular. To do this we show that dβgGx has kernel {0} for allgGx ∈ G/Gx. We again consider β. Suppose ker(dβg) = ker(dπg). Then ker(dβgGx) = {0}. If thiswas not the case there would exist a nonzero element XgGx ∈ TgGxG such that dβgGx(XgGx) = 0.Since π is a smooth submersion, there exists Xg ∈ TgG such that dπg(Xg) = XgGx . ClearlyXg /∈ ker(dπg) since XgGx 6= 0. But

(dβ)g(Xg) = (dβ)gGx(XgGx) = 0

and this contradicts the fact that Xg /∈ ker(dπg), since we assumed ker(dβg) = ker(dπg). Wemust therefore have that ker(dβgGx) = {0}. From this we see that it is sufficient to show thatker(dβg) = ker(dπg) for all g ∈ G. Later on in this section we will see that the kernel of dπe isTeGx and from this it is easy to see that ker(dπg) = TggGx. So to show that β is nonsingular wewill show ker(dβg) = TggGx for all g ∈ G.

Let αg(y) = α(g, y). We then see that

(αg ◦ β ◦ Lg−1)(g) = αg(β(e))

= αg(β(Gx))= α(g, α(e, x))= α(g, x)= β(g),

where Lg denotes the left translation. This holds for all g ∈ G, hence β = αg ◦β ◦Lg−1 . From thiswe get that

dβg = (dαg)x ◦ dβe ◦ (dLg−1)g. (2.1)

Suppose we know that ker(dβe) = TeGx. Then it follows easily from (2.1) that the kernel of dβg isTggGx. It is therefore sufficient to show that ker(dβe) = TeGx.

It is clear that TeGx ⊂ ker(dβg) since dβ = dβ ◦ dπ. Let Xe ∈ ker(dβe) ⊆ TeG. Let X ∈ g bethe left-invariant vector field which corresponds to Xe. Our problem then translates into showingthat X ∈ gx, where gx is the Lie algebra of Gx. This is equivalent to showing that exp(tX) ∈ Gxfor all t ∈ R, where exp(tX) is the one parameter subgroup generated by X. If we can show thatthe curve

t :→ β(exp(tX))

is constant then we would be done, because β(exp(0)) = β(e) = x so we would have β(exp(tX)) = xfor all t ∈ R. From this we then get that exp(tX) ∈ Gx for all t ∈ R. To show that the curveis constant we consider the tangent vector at t. Using (2.1), the left-invariance of X and the factthat Xe ∈ ker(dβe) we obtain

dβexp(tX)(Xexp(tX)) = ((dαexp(tX))x ◦ dβe ◦ dLexp(−tX))(Xexp(tX))= ((dαexp(tX))x ◦ dβe(Xe))= 0.

Since this holds for all t ∈ R it follows that the curve β(exp(tX)) is constant. By what we haveshown this implies that β is non-singular.

9

Definition 2.7. A homogeneous space is a differentiable manifold M together with a transitiveaction of a Lie group G.

From the discussion before Proposition 2.6 we saw that every coset manifold can be equippedwith a transitive action. This means that every coset manifold can be seen as a homogeneousspace. Moreover, from Proposition 2.6, we see that a homogeneous space is diffeomorphic to acoset manifold. So equivalently we can define a homogeneous space to be a coset manifold i.e. amanifold of the form G/K, where G is a Lie group and K a closed subgroup of G.

Let g and k be the Lie algebras of G and K, respectively. We now consider the tangent spaceat o = π(e), where e is the identity element of G. For X ∈ g, we have that

dπe(X) =d

dt(π(exp(tX)))|t=0 =

d

dt(exp(tX)K)|t=0,

where exp(tX) is the one-parameter subgroup generated by X. If X ∈ k then exp(tX) ∈ K, forall t ∈ R. From the above calculation it follows that k ⊆ ker(dπe). In fact k = ker(dπe), sinceexp(tX) ∈ K implies X ∈ k.

From Theorem 2.3 we know that π is a submersion, which means that dπg is onto for all g ∈ G.So dπe(g) = To(G/K). This gives us the following isomorphism,

g/k ∼= To(G/K).

We may choose m ⊂ g such thatg = k⊕m.

Here m is a vector subspace of g and not necessarily a subalgebra. Since dπe : g→ ToM has kernelk, it follows that dπe : m→ ToM is an isomorphism.

For a fixed g ∈ G we define Ig by,

Ig : G→ G Ig(x) = gxg−1.

It is easily seen that Ig is a homomorphism, moreover if we let Lg, Rg denote left and righttranslation, we see that Ig = R−g ◦ Lg. This implies that Ig is a diffeomorphism, since left andright translation are diffeomorphisms.

Definition 2.8. Let G be a Lie group with Lie algebra g. Then the map

Ad : G→ Aut(g) with Ad(g) = (dIg)e

is called the adjoint representation of G.

Since Igh = Ig ◦ Ih, we have that Ad(gh) = (dIgh)e = (dIg)e ◦ (dIh)e = Ad(g) ◦ Ad(h). So Adreally is a representation of G.

In Chapter 1 we defined ad : g → End(g), by ad(X)(Y ) = [X,Y ]. One can in fact show thatad(X) = (dAd)e(X).

2.2 Reductive Homogeneous Spaces

In this section we show that there is a one to one correspondence between certain metrics on areductive homogeneous space G/K and certain scalar products on m.

Let M = G/K be a homogeneous space. On M we define τg : G/K → G/K

τg(hK) = ghK, hK ∈ G/K, g ∈ G.

Recall the left translation Lg : G→ G, Lg(h) = gh. We note that

(π ◦ Lg)(h) = π(gh) = ghK = τg(hK) = (τg ◦ π)(h),

10

which implies that for all g, h ∈ G, and all X ∈ ThG,

((dπ)gh ◦ (dLg)h)(X) = ((dτg)hK ◦ (dπ)h)(X).

We will have use of the following lemma.

Lemma 2.9. Let M = G/K and let m be a subspace of g such that g = k⊕m. Then

dπe(Ad(k)(X)) = (dτk)o(dπe(X))

for all k ∈ K and X ∈ m.

Proof. We know that Ig = Lg ◦Rg−1 . For k ∈ K and g ∈ G, we have

(π ◦Rk−1)(g) = π(gk−1) = gk−1K = gK = π(g)

which implies that for X ∈ g,

(dπk−1 ◦ (dRk−1)e)(X) = d(π ◦Rk−1)e(X) = dπe(X).

So for X ∈ m,

dπe(Ad(k)(X)) = dπe((dIk)e(X))= dπe(((dLk)k−1 ◦ (dRk−1)e)(X))= (dτk)o((dπk−1 ◦ (dRk−1)e)(X))= (dτk)o(dπe(X)).

Definition 2.10. A homogeneous space G/K is said to be reductive, if there exists a subspacem of g such that

g = k⊕m (2.2)

and Ad(k)(m) ⊂ m for all k ∈ K.

Definition 2.11. Let M = G/K be a homogeneous space. A metric g on M is said to beG-invariant if

gτg(hK)((dτg)hKXhK , (dτg)hKYhK) = ghK(XhK , YhK)

for all hK ∈M , g ∈ G and XhK , YhK ∈ ThKM .

Proposition 2.12. [15] Let M = G/K be a reductive homogeneous space with a reductive decom-position

g = k⊕m.

Then there is a bijection between G-invariant metrics on M and Ad(K)-invariant inner productson m.

Proof. [15] Let g be a G-invariant metric on M . We define an inner product 〈, 〉 on m by

〈X,Y 〉 = go(dπe(X), dπe(Y ))

for X,Y ∈ m. We show that 〈, 〉 is Ad(K)-invariant, i.e.

〈Ad(k)(X),Ad(k)(Y )〉 = 〈X,Y 〉 for all k ∈ K and X,Y ∈ m.

We have that

〈Ad(k)(X),Ad(k)(Y )〉 = go(dπe(Ad(k)(X)), dπe(Ad(k)(Y )))= go((dτk)o(dπe(X)), (dτk)o(dπe(Y )))= go(dπe(X), dπe(Y ))= 〈X,Y 〉

11

where we have used Lemma 2.9 and the G-invariance of g. This shows that 〈, 〉 is Ad(K)-invariant.Suppose now instead that we are given an Ad(K)-invariant inner product 〈, 〉 on m. Since

τg : M → M is a diffeomorphism, it follows that (dτg)o : ToM → TgKM is an isomorphism. Sogiven XgK , YgK ∈ TgKM there exist unique X0, Y0 ∈ ToM such that

(dτg)o(X0) = XgK , (dτg)o(Y0) = YgK .

Moreover, we know that dπe : m → ToM is an isomorphism, so there exist unique X,Y ∈ m suchthat

dπe(X) = X0, dπe(Y ) = Y0.

We defineggK(XgK , YgK) = 〈X,Y 〉.

We need to show that g is well-defined. We may choose another representative h of gK such thathK = gK. This implies that h = gk for some k ∈ K. For g1K ∈M ,

τh(g1K) = τgk(g1K)= gkg1K

= τg(kg1K)= (τg ◦ τk)(g1K),

from which it follows that τh = τg ◦ τk. As before we know that there exist unique X ′0, Y′0 ∈ ToM

such that

(dτh)o(X ′0) = XhK = XgK = (dτg)o(X0)(dτh)o(Y ′0) = YhK = YgK = (dτg)o(Y0).

Using this and the fact that τh = τg◦τk, we get that (dτg)o(X0) = (dτh)o(X ′0) = (dτg)o((dτk)o(X ′0))and since (dτg)o is an isomorphism, we must have that

(dτk)o(X ′0) = X0

and in the same way(dτk)o(Y ′0) = Y0.

Also in the same way as before, there exist unique X ′, Y ′ ∈ m such that

dπe(X ′) = X ′0, dπe(Y′) = Y ′0 .

Using the above equalities, we see that

X ′ = (dπe|m)−1(X ′0)

= (dπe|m)−1((dτk−1)o(X0))

= (dπe|m)−1((dτk−1)o(dπe(X)))

= Ad(k−1)(X)

where we used Lemma 2.9 in the last equality. In the same way we have that Y ′ = Ad(k−1)(Y ).So

ghK(XhK , YhK) = 〈X ′, Y ′〉= 〈Ad(k−1)X,Ad(k−1)Y 〉= 〈X,Y 〉= ggK(XgK , YgK).

which shows that g is well-defined. By construction g is G-invariant.

12

Chapter 3

Kahler Manifolds

In this chapter we introduce the notion of complex structures, first on vector spaces and thenon manifolds. In particular we consider the case when the manifold is a homogeneous space.

3.1 Complex Vector Spaces

We will here go through some of the theory for complex structures on vector spaces.

Let V be a finite-dimensional real vector space. A linear map J : V → V such that

J2 = −I,

where I is the identity mapping of V , is called a complex structure on V . The space V togetherwith J can be turned into a complex vector space, by defining a multiplication on V , by complexnumbers, in the following way:

(a+ bi)X = aX + bJX for all a, b ∈ R and X ∈ V.

Let {e1, e2, . . . , en} be a complex basis of V , with the multiplication defined as above. Then

(a1 + ib1)e1 + (a2 + ib2)e2 + . . .+ (an + ibn)en = 0

if and only if ai = bi = 0 for all i = 1, 2, . . . , n. But by definition

n∑i=1

(ai + ibi)ei =n∑i=1

aiei + biJ(ei)

which implies that the vectors

{e1, e2, . . . , en, J(e1), J(e2), . . . , J(en)}

are linearly independent over R. It is clear that these vectors also span V , since we assumed that{e1, e2, . . . , en} was a complex basis of V . This means that

{e1, e2, . . . , en, J(e1), J(e2), . . . , J(en)}

is a real basis of V . So if it is possible to define a complex structure on a real vector space V , thenV must necessarily be of even dimension.

Suppose now instead that V is a complex vector space of dimension n. Then we define JX = iXfor X ∈ V . The space V can be considered as a real vector space of dimension 2n. If we considerV in this way it is clear that J is a complex structure on V .

For a real vector space V we let the complexification V C of V be given by

V C = {X + iY | X,Y ∈ V }.

13

If J is a complex structure on V we may extend J to V C by

J(X + iY ) = JX + iJY.

It is clear that J2 = −I on V C as well.

Definition 3.1. Let V be a real vector space with a complex structure J . We then define

V 1,0 = {Z ∈ V C | JZ = iZ}V 0,1 = {Z ∈ V C | JZ = −iZ}.

Proposition 3.2. [16] Let V be a real vector space with complex structure J . Then

i) V 1,0 = {X − iJX | X ∈ V } and V 0,1 = {X + iJX | X ∈ V }.

ii) V C = V 1,0 ⊕ V 0,1

iii) V 1,0 = V 0,1

Proof. [16] Let Z = X + iY ∈ V C. Consider the eigenvalue equation,

JZ = iZ

⇔ JX + iJY = iX − Y⇔ JX + Y + i(JY −X) = 0⇔ Y = −JX.

From the above calculation we see that JZ = iZ if and only if Z = X − iJX. In a similar wayway see that JZ = −iZ if and only if Z = X + iJX. It then follows that

V 1,0 = {X − iJX | X ∈ V }V 0,1 = {X + iJX | X ∈ V }.

It is easily seen that V 1,0 = V 0,1.Now let Z = X + iY be an arbitrary element in V C. Let

Z1 =12

(X − iJX + i(Y − iJY ))

Z2 =12

(i(Y + iJY ) +X + iJX).

We see that Z1 ∈ V 1,0 and Z2 ∈ V 0,1, and

Z1 + Z2 =12

(2X + 2iY + JY − JY − iJX + iJX) = X + iY = Z.

So every element in V C can be written as a sum of two elements, one from V 1,0 and one fromV 0,1. Suppose that Z ∈ V 1,0 ∩ V 0,1, so Z = X − iJX, and Z = Y + iJY for some X,Y ∈ V . Thisimplies

X − iJX = Y − iJY⇔ Y −X + iJ(Y +X) = 0⇔ Y = X, Y = −X⇔ Y = X = 0.

Hence V 1,0 ∩ V 0,1 = {0}.

We have the natural correspondence between Cn and R2n,

(x1 + iy1, . . . , xn + iyn)→ (x1, . . . , xn, y1, . . . , yn).

Multiplication by i in Cn, i(x1 + iy1, . . . , xn + iyn) = (ix1 − y1, . . . , ixn − yn), corresponds to thelinear map Jn : R2n → R2n,

Jn =(

0 −InIn 0

)where In is the identity mapping on Rn. The map Jn is called the canonical complex structure onR2n. Multiplication by i considered as an operator on Cn, will also be denoted by Jn.

14

3.2 Complex Manifolds

We would now like to transfer the notion of complex structures to the manifold setting. We dothis in the following way.

Definition 3.3. Let M be a real differentiable manifold. An almost complex structure J onM is a tensor field

J : C∞(TM)→ C∞(TM)

such that J2x = −I, for all x ∈M .

As before we see that a manifold M together with an almost complex structure must be of evendimension. The pair (M,J) is called an almost complex manifold. We also make the followingdefinition.

Definition 3.4. Let M be a topological manifold with atlas {(Uα, zα) | α ∈ I}, where I is someindex set, and zα(Uα) = Vα is an open set in Cn. M is called a complex manifold if

zα ◦ (zβ)−1 : zβ(Uα ∩ Uβ)→ Cn,

are holomorphic functions, for all α, β ∈ I. The collection of all charts (zα, Uα), is called aholomorphic structure.

Suppose that we are given a complex manifold M of dimension n. Let zα = (zα1 , . . . , zαn ) be

a chart. Then, for each 1 ≤ k ≤ n, we have zαk = xαk + iyαk , where xαk , yαk : Uα → R. The pair

(φα = (xα1 , . . . , xαn, y

α1 , . . . y

αn), Uα) is a real chart. This means that any complex manifold with

dimension n can be seen as a real manifold with dimension 2n.We now show that a complex manifold carries a natural almost complex structure. Let x ∈M ,

and choose a neighborhood Uα such that x ∈ Uα. For X ∈ TxM , we define

Jαx (X) = d((zα)−1 ◦ Jn ◦ zα)x(X).

One can show that this definition is independent of the choice of neighborhood Uα. It is clear thatJ defined in this way is an almost complex structure.

Now let (M,J) be an almost complex manifold. We say that J is a complex structure if Jcomes from a holomorphic structure as described above. For an almost complex manifold (M,J),we define the Nijenhuis tensor N of J by,

N(X,Y ) = 2([JX, JY ]− [X,Y ]− J [X,JY ]− J [JX, Y ]), X, Y ∈ C∞(TM).

The following theorem, which we state without proof, will be useful when determining whether ornot J is a complex structure on M .

Theorem 3.5. An almost complex structure J on a manifold M , is a complex structure on M , ifand only if N ≡ 0.

A proof of this theorem, for the special case when M is real analytic, can be found in [11].We continue to let (M,J) be an almost complex manifold. Let x ∈M . Since TxM is a vector

space and Jx is a complex structure on TxM , we have

TCxM = T 1,0

x M ⊕ T 0,1x M,

where

T 1,0x M = {Z ∈ TC

xM | JxZ = iZ}T 0,1x M = {Z ∈ TC

xM | JxZ = −iZ}.

A complex vector field on M is a vector field of the form Z = X + iY , where X,Y ∈ C∞(TM).A complex vector field Z is said to be of type (1, 0) if Zx ∈ T 1,0

x M , for all x ∈ M . From Section3.1 we see that Z is of type (1, 0) if and only if Z = X − iJX for some vector field X ∈ C∞(TM).Similarly, we say that a vector field Z is of type (0, 1), if Zx ∈ T 0,1

x M , for all x ∈M . In the sameway as before we see that Z is of type (0, 1) if and only if Z = X + iJX, for some X ∈ C∞(TM).It is clear that conjugation takes vector fields of type (1, 0) into vector fields of type (0, 1) and viceversa.

15

Proposition 3.6. [16] Let (M,J) be an almost complex manifold. Then the following are equiva-lent:

a) If Z,W is of type (1, 0), then so is [Z,W ].

b) If Z,W is of type (0, 1), then so is [Z,W ].

c) J is a complex structure on M .

Proof. [16] We first show that a) is equivalent to b). Let Z = X1 + iY1 and W = X2 + iY2, whereX1, X2, Y1, Y2 ∈ C∞(TM). It is then clear that

[Z,W ] = ([X1, X2]− [Y1, Y2]) + i([X1, Y2] + [Y1, X2]),

so

[Z,W ] = ([X1, X2]− [Y1, Y2])− i([X1, Y2] + [Y1, X2])= [X1 − iY1, X2 − iY2]

= [Z,W ].

Now assume that a) holds, and that Z,W is of type (1, 0). By assumption [Z,W ] is of type (1, 0),which implies that [Z,W ] is of type (0, 1). We just saw that [Z,W ] = [Z,W ], which then impliesthat [Z,W ] is of type (0, 1) for all Z,W of type (1, 0).

Given any two complex vector fields Z∗,W ∗ of type (0, 1), we know that Z = Z∗,W = W ∗ isof type (1, 0). By our previous argument [Z,W ] is of type (0, 1). But Z = Z∗,W = W ∗, whichmeans that [Z∗,W ∗] is of type (0, 1) for all vector fields Z∗,W ∗ of type (0, 1), which is exactly b).By symmetry we get that a) is equivalent to b).

Recall that every vector field of type (1, 0) is of the form X − iJX, for some X ∈ C∞(TM).Let

Z = [X − iJX, Y − iJY ].

We see that a) holds if and only if Z is of type (1, 0), for all X,Y ∈ C∞(TM). That Z is of type(1, 0) means in particular that JZ = iZ, which is equivalent to Z + iJZ = 0, and

Z + iJZ = [X − iJX, Y − iJY ] + iJ([X − iJX, Y − iJY ])= [X,Y ]− i[X, JY ]− i[JX, Y ]− [JX, JY ]

+ iJ [X,Y ] + J [X, JY ] + J [JX, Y ]− iJ [JX, JY ]= −([JX, JY ]− J [X, JY ]− J [JX, Y ]− [X,Y ])− iJ([JX, JY ]− J [X, JY ]− J [JX, Y ]− [X,Y ])

= −N(X,Y )2

− iJN(X,Y )2

.

From this calculation we see that Z is of type (1, 0) if and only if N(X,Y ) = 0. From ourprevious discussion this means that a) holds if and only if N ≡ 0. From Theorem 3.5 we knowthat J is a complex structure on M if and only if N ≡ 0, which then implies the equivalence of a)and c).

3.3 Complex Homogeneous Spaces

In this section we will specialize to the case when M is a homogeneous space. Let M = G/K,where G is a Lie group, and K is a closed subgroup of G. We choose m ⊂ g such that

g = k⊕m.

Let J be an almost complex structure on M . We say that J is G-invariant if

(Jτg(hK) ◦ (dτg)hK)(XhK) = ((dτg)hK ◦ JhK)(XhK)

for all g, h ∈ G and all XhK ∈ ThKM . We have the following result:

16

Proposition 3.7. [11] There is a bijection between the set of G-invariant almost complex structureson M and the set of linear endomorphisms I0 on m satisfying

(1) I20 = −I

(2) I0((Ad(k)(X))m) = (Ad(k)(I0X))m for all k ∈ K and X ∈ m.

The subscript m denotes projection on the subspace m.

Proof. [11], [15] Recall that dπe : m → ToM is a vector space isomorphism. We let ψ = dπe|m.Suppose first that we are given a G-invariant almost complex structure J on M . We then knowthat Jo : ToM → ToM is a linear endomorphism satisfying J2

o = −I. We define, for X ∈ m,

I0(X) = (ψ−1 ◦ Jo ◦ dπe)(X).

We see directly thatI20 (X) = (ψ−1 ◦ J2

o ◦ dπe)(X) = −X,hence I0 satisfies (1). Using Lemma 2.9 and the G-invariance of J , we see that

I0((Ad(k)X)m) = (ψ−1 ◦ Jo ◦ dπe)((Ad(k)(X))m)

= (ψ−1 ◦ Jo)(dπe(Ad(k)(X)))

= (ψ−1 ◦ Jo ◦ (dτk)o ◦ dπe)(X)

= ((ψ−1 ◦ (dτk)o ◦ Jo ◦ dπe)(X)

= (ψ−1 ◦ (dτk)o ◦ dπe ◦ I0)(X))

= (ψ−1 ◦ dπe ◦Ad(k) ◦ I0)(X)= (Ad(k)(I0X))m.

So the existence of J on M implies the existence of I0 on m satisfying (1) and (2).To prove the converse, let I0 be as in the statement of the proposition. For gK ∈M we define

JgK : TgKM → TgKM

JgK(XgK) = ((dτg)o ◦ dπe ◦ I0 ◦ ψ−1)(X0),

where X0 ∈ ToM satisfies XgK = (dτg)o(X0). Since τg is a diffeomorphism, (dτg)o is a vectorspace isomorphism and X0 is therefore uniquely determined.

We need to show that JgK is independent of the choice of representative of the coset gK.We therefore suppose gK = hK, for some h ∈ G. This implies that h = gk for some k ∈ K.Let Y0 be the tangent vector such that (dτh)o(Y0) = XhK . Since τh = τg ◦ τk, we get that(dτh)o = (dτg)o ◦ (dτk)o, so

XgK = ((dτg)o ◦ (dτk)o)(Y0).

Since XgK = (dτg)o(X0) and (dτg)o is an isomorphism, we must have that X0 = (dτk)o(Y0). UsingLemma 2.9, (2) and the above calculation, we see that

JhK(XhK) = ((dτh)o ◦ dπe ◦ I0 ◦ ψ−1)(Y0)

= ((dτg)o ◦ (dτk)o ◦ dπe ◦ I0 ◦ ψ−1)(Y0)

= ((dτg)o ◦ dπe ◦Ad(k) ◦ I0 ◦ ψ−1)(Y0)

= ((dτg)o ◦ dπe)((Ad(k)(I0(ψ−1(Y0)))m)

= ((dτg)o ◦ dπe ◦ I0)(Ad(k)(ψ−1(Y0))m)

= ((dτg)o ◦ dπe ◦ I0 ◦ ψ−1)((dτk)o(Y0))

= ((dτg)o ◦ dπe ◦ I0 ◦ ψ−1)(X0)= JgK(XgK).

It then follows that J is well defined. We see directly that

J2gK(XgK) = ((dτg)o ◦ Jo)(Jo(X0)) = −(dτg)o(X0) = −XgK .

So J is an almost complex structure on M , and it is clear from the construction that J is G-invariant.

17

When g = k ⊕ m is a reductive decomposition we immediately get the following result fromProposition 3.7.

Proposition 3.8. [11] If g = k⊕m is a reductive decomposition there is a one-to-one correspon-dence between G-invariant almost complex structures on M and linear endomorphisms I0 on msatisfying

(1) I20 = −I

(2) I0(Ad(k)(X)) = (Ad(k)(I0X)) for all k ∈ K and X ∈ m.

Proof. [11] We may remove the projection on the subspace m, since Ad(k)(X) ∈ m for all k ∈ Kand X ∈ m in a reductive space.

Let I0 : m→ m satisfy (1) and (2). We may extend I0 to g by defining

I0 : g→ g

I0(X + Y ) = I0(Y ) for all Z = X + Y ∈ g, X ∈ k, Y ∈ m

For Z = X + Y ∈ g we have

I20 (Z) = I2

0 (Y ) = −Y = −Z +X = −Z mod k.

For k ∈ K and X ∈ k, Ad(k)(X) ∈ k, so

I0(Ad(k)(Z)) = I0(Ad(k)(X)) + I0(Ad(k)(Y ))

= I0(Ad(k)(Y ))

= I0((Ad(k)(Y ))k + (Ad(k)(Y ))m)= I0((Ad(k)(Y ))m)= (Ad(k)(I0(Y )))m

= (Ad(k)(I0(Z)))m

= (Ad(k)(I0(Z)))k + (Ad(k)(I0(Z)))m − (Ad(k)(I0(Z)))k

= (Ad(k)(I0(Z))) mod k.

Hence I0 is a linear endomorphism of g which satisfies

(a) I0(X) = 0,

(b) I20 (Z) = −Z mod k,

(c) I0(Ad(k)(Z)) = Ad(k)(I0(Z)) mod k

for all X ∈ k, k ∈ K and all Z ∈ g. If I0 and I∗0 are two linear endomorphisms of g satisfying(a),(b) and (c), then we say that I0 = I∗0 if I0 = I∗0 mod k. We may now state and prove thefollowing result:

Proposition 3.9. [11] There is a one to one correspondence between the G-invariant complexstructures on M , and the linear endomorphisms of g satisfying (a),(b) and (c).

Proof. [11] Given an almost complex structure J on M , we know from Proposition 3.7 that thereexists a unique linear endomorphism I0 of m satisfying (1) and (2). In the discussion above weconstructed a linear endomorphism I0 of g, which satisfied (a),(b) and (c). It is clear that thiscorrespondence is injective.

Conversely, suppose that we are given a linear endomorphism I0 of g satisfying (a),(b) and (c).We then define, for X ∈ m, I0(X) = (I0(X))m. It is easy to see that I0 satisfies (1) and (2). ByProposition 3.7 I0 gives rise to a unique G-invariant complex structure on M .

18

Let J be a G-invariant almost complex structure on M . We would like to determine whetheror not J is a complex structure by considering the corresponding linear endomorphism I0. Thefollowing theorem will allow us to do this.

Theorem 3.10. [11] A G-invariant almost complex structure J on M is a complex structure ifand only if

[I0X, I0Y ]− [X,Y ]− I0[X, I0Y ]− I0[I0X,Y ] ∈ k,

for all X,Y ∈ g.

Proof. [11] Let U ∈ C∞(TG). We define the tensor field J in the following way:

Jg(Ug) = ((dLg)e ◦ I0 ◦ (dLg)−1e )(Ug).

We see that for h ∈ G

Jhg((dLh)g(Ug)) = ((dLhg)e ◦ I0 ◦ (dLhg)−1e )((dLh)g(Ug))

= ((dLhg)e ◦ I0 ◦ (dLg−1h−1)hg ◦ (dLh)g)(Ug)

= ((dLhg)e ◦ I0 ◦ (d(Lg−1h−1 ◦ Lh))e)(Ug)

= ((dLhg)e ◦ I0 ◦ (dLg−1)g)(Ug)

= (dLh)g((dLg)e ◦ I0 ◦ (dLg)−1e )(Ug)

= (dLh)(Jg(Ug)).

This means that J is left invariant i.e.

(dLh) ◦ J = J ◦ (dLh)

for all h ∈ G.We say that a vector field U ∈ C∞(TG) is projectable if there exists U ′ ∈ C∞(TM) such

thatdπg(Ug) = U ′π(g) for all g ∈ G,

we write this as dπ(U) = U ′. Let P (G) be the set of all projectable vector fields. We shall showthat P (G) is a subalgebra of C∞(TG). Let U, V ∈ P (G). Then

dπ(U + V ) = dπ(U) + dπ(V )dπ(cU) = cdπ(U) c ∈ R,

so U + V, cU ∈ P (G) for all U, V ∈ P (G) and c ∈ R. This means that P (G) is a vector subspaceof C∞(TG). Moreover,

dπ([U, V ]) = [dπ(U), dπ(V )]

so [U, V ] ∈ P (G) for all U, V ∈ P (G), which shows that P (G) is a subalgebra of C∞(TG).We now want to show that if U ∈ P (G) then J(U) ∈ P (G). Let Ug ∈ TgG and let We =

(dLg)−1e (Ug). Then

dπg(Jg(Ug)) = dπg(Jg((dLg)e(We)))

= (dπg ◦ (dLg)e ◦ Je)(We)

= ((dτg)o ◦ dπe ◦ I0)(We)= ((dτg)o ◦ Jo ◦ dπe)(We)= (JgK ◦ (dτg)o ◦ dπe)(We)= (JgK ◦ dπg ◦ (dLg)e)(We)= JgK(dπg(Ug)),

where we used the definition of I0 and the fact that Je = I0. So if U ∈ P (G), then dπ(J(U)) =J(dπ(U)), which implies that J(U) ∈ P (G).

19

We now define the tensor field

N(U, V ) = [J(U), J(V )] + J2([U, V ])− J([U, J(V )])− J([J(U), V ])

for U, V ∈ C∞(TG). What we have shown so far implies that N(U, V ) ∈ P (G) if U, V ∈ P (G),and also that N is a left invariant tensor field. Applying dπ to N(U, V ), where U, V ∈ P (G) givesus the following:

dπ(N(U, V )) = dπ([J(U), J(V )] + J2([U, V ])− J([U, J(V )])− J([J(U), V ]))

= [J(dπ(U)), J(dπ(V ))] + J2([dπ(U), dπ(V )])− J([dπ(U), J(dπ(V ))])− J([J(dπ(U)), dπ(V )])

=12N(dπ(U), dπ(V ))

Note that the map dπ : P (G)→ C∞(TM) is surjective, so

J complex structure on M ⇔N(U ′, V ′) = 0 for all U ′, V ′ ∈ C∞(TM)⇔

N(dπ(U), dπ(V )) = 0 for all U, V ∈ P (G)⇔

dπ(N(U, V )) = 0 for all U, V ∈ P (G).

Since N is left-invariant the last statement is equivalent to

dπe(N(X,Y )) = 0

for all X,Y ∈ g which is equivalent to N(X,Y ) ∈ k, for all X,Y ∈ g. For left invariant vector fieldsX,Y , we have that

N(X,Y ) = [I0X, I0Y ]− [X,Y ]− I0[X, I0Y ]− I0[I0X,Y ],

which concludes the proof of this theorem.

3.4 Kahler Manifolds

So far we have not considered Riemannian metrics on complex manifolds. We will in this sectionintroduce Riemannian metrics with some special properties with respect to the complex structure.In particular we will consider the case when the manifold is a homogeneous space.

Definition 3.11. Let M be an almost complex manifold with an almost complex structure J . ARiemannian metric g on M is said to be Hermitian if

g(JX, JY ) = g(X,Y )

for all X,Y ∈ C∞(TM). An almost complex manifold with a Hermitian metric is called an almostHermitian manifold, and a complex manifold with a Hermitian metric is called a Hermitianmanifold.

Definition 3.12. Let M be an almost Hermitian manifold with almost complex structure J andHermitian metric g. The alternating 2-form ω defined by

ω(X,Y ) = g(JX, Y )

is called the Kahler form of g. If ω is closed, that is if dω = 0, where d denotes the exteriorderivative, then g is called a Kahler metric and a complex manifold equipped with a Kahlermetric is called a Kahler manifold.

Let M = G/K be a reductive homogeneous space with a G-invariant almost complex structureJ and let g be a G-invariant metric on M . Let I0 be the Ad(K)-invariant complex structure on mwhich corresponds to J , and let 〈, 〉 be the Ad(K)-invariant metric in m which corresponds to g.We then have the following proposition.

20

Proposition 3.13. The metric g is Hermitian with respect to J if and only if 〈I0X, I0Y 〉 = 〈X,Y 〉for all X,Y ∈ m.

Proof. Suppose first that g is Hermitian with respect to J . So g(JU, JV ) = g(U, V ) for all U, V ∈C∞(TM). From the definition of 〈, 〉 we have that

〈I0X, I0Y 〉 = go(dπe(I0X), dπe(I0Y ))

and I0 is defined byI0 = ψ−1 ◦ Jo ◦ dπe,

where ψ = dπe|m. Using this we get that

〈I0X, I0Y 〉 = go(dπe(I0X), dπe(I0Y ))= go(Jo(dπe(X)), Jo(dπe(Y )))= go(dπe(X), dπe(Y ))= 〈X,Y 〉.

Suppose now instead that 〈I0X, I0Y 〉 = 〈X,Y 〉 for all X,Y ∈ m. Recall from the proof ofProposition 2.12 that ggK(XgK , YgK) = 〈X,Y 〉, where X,Y ∈ m such that

XgK = (dτg)o(dπe(X)), YgK = (dτg)o(dπe(Y )).

Using this and the G-invariance of J , we see that

JgK(XgK) = (JgK ◦ (dτg)o ◦ dπe)(X)= ((dτg)o ◦ Jo)(dπe(X))= (dτg)o(dπe(I0(X)))

and in the same way, we have that

JgK(YgK) = (dτg)o(dπe(I0(Y ))).

SoggK(JgK(XgK), JgK(YgK)) = 〈I0(X), I0(Y )〉 = 〈X,Y 〉 = ggK(XgK , YgK),

which shows that g is Hermitian with respect to J .

Definition 3.14. On a reductive homogeneous space with G-invariant almost complex structureJ and Hermitian G-invariant metric g we define the bilinear form ω0 on m by

ω0(X,Y ) = 〈I0(X), Y 〉.

The following Lemma will be of importance later on.

Lemma 3.15. An Ad(K) invariant scalar product on m satisfies

〈[Z,X], Y 〉 = 〈X, [Y,Z]〉

for all X,Y ∈ m and Z ∈ k.

Proof.

〈[Z,X], Y 〉 = 〈ad(Z)(X), Y 〉

= 〈 ddt

(Ad(exp(tZ))(X))|t=0, Y 〉

=d

dt(〈Ad(exp(tZ))(X), Y 〉)|t=0

=d

dt(〈X,Ad(− exp(tZ))(Y )〉)|t=0

21

= 〈X, ddt

(Ad(− exp(tZ))(Y ))|t=0〉

= 〈X,−ad(Z)(Y )〉= 〈X, [Y, Z]〉

Corollary 3.16. The bilinear form ω0 satisfies

ω0(X, [Y,Z]) = −ω0(Y, [Z,X])

for all X,Y ∈ m and all Z ∈ k.

Proof. It is easy to see from the Ad(K)-invariance of I0, that I0(ad(Z)(X)) = ad(Z)(I0(X)) forall Z ∈ k and X ∈ m. We therefore have that

ω0(X, [Y, Z]) = 〈I0(X), [Y,Z]〉= 〈[Z, I0(X)], Y 〉= 〈I0([Z,X]), Y 〉= −〈I0(Y ), [Z,X]〉= −ω0(Y, [Z,X]).

Finally we have the following result, which we will state without proof.

Proposition 3.17. [4] The 2-form ω is closed if and only if

ω0([X,Y ]m, Z) + ω0([Y,Z]m, X) + ω0([Z,X]m, Y ) = 0

for all X,Y, Z ∈ m.

22

Chapter 4

Generalized Flag Manifolds

We now introduce a special kind of homogeneous spaces, namely the generalized flag manifolds.The additional structure of these manifolds allows us to use the theory gathered in Chapter 1 aboutsemisimple Lie algebras. The root space decomposition in particular will be of great importancethroughout this chapter.

4.1 Complex Structures on Generalized Flag Manifolds

In this section we show that there is a one-to-one correspondence between G-invariant complexstructures on a generalized flag manifold and certain subsets of the corresponding root system R.

A Lie group G is said to be semisimple if its Lie algebra is semisimple. Recall that a torus ina Lie group G is a subgroup T such that T ∼= S1 × S1 × . . .× S1. A torus T is said to be maximalin G if for any torus T in G such that T ⊂ T ⊂ G we have T = T .

Definition 4.1. A flag manifold is a homogeneous space of the form G/T , where G is a compactLie group and T is a maximal torus in G.

Definition 4.2. A generalized flag manifold is a homogeneous space of the form G/K, whereG is a compact Lie group, and K is the centralizer of a torus T in G i.e.

K = C(T ) = {g ∈ G | ghg−1 = h for all h ∈ T}.

We state the following result.

Proposition 4.3. [6] Let G be compact Lie group and T a torus in G. Then C(T ) is a closedsubgroup of G. If G is connected then C(T ) is also connected.

By Proposition 4.3, C(T ) is a closed subgroup of G, so G/C(T ) really is a homogeneous space.Note that C(T ) = T when T is a maximal torus in G, so a flag manifold is a special case of ageneralized flag manifold.

Throughout this text we will make two more assumptions on G, namely that G should beconnected and semisimple.

Let G/K be a generalized flag manifold and g, k be the Lie algebras of G and K, respectively.We will now use the theory gathered in Chapter 1 to obtain a root space decomposition of gC andkC.

Let T be a maximal torus in G which contains T . Since T is an abelian subgroup of G,T ⊂ C(T ). Let h be the Lie algebra of T . Then hC is a Cartan subalgebra of gC. We thereforeobtain a root space decomposition of gC with respect to hC:

gC = hC ⊕∑α∈R

gα,

where R is the root system of gC with respect to hC. Since T ⊂ K we have that hC ⊂ kC. Becauseof this, we obtain a root space decomposition of kC with respect to hC:

kC = hC ⊕∑α∈RK

gα,

23

where RK is the root system of kC with respect to hC.We start by considering a more general case. We follow the outline given in [12]. Let M = G/K

be a homogeneous space, so G is a Lie group and K a closed subgroup of G. As usual we let g, kdenote the Lie algebras of G,K respectively. We also let m be a subspace of g which satisfiesg = k ⊕ m. Furthermore we suppose that there exists a G-invariant complex structure J on M.From Chapter 3 we know that J gives rise to the decomposition

TCo M = T 1,0

o M ⊕ T 0,1o M,

whereT 1,0o M = {X ∈ TC

o M | JX = iX}, T 0,1o M = {X ∈ TC

o M | JX = −iX}.

We extend the map dπe to gC in the following way: For Z = X + iY ∈ gC,

dπe(Z) = dπe(X) + idπe(Y ).

Clearly dπe(gC) = TCo M and it is easy to see that the kernel of this extension is kC. We define

a+ = (dπe)−1(T 1,0o M), a− = (dπe)−1(T 0,1

o M).

Let θ : gC → gC be the conjugation with respect to g. Since TCo M = T 1,0

o M ⊕ T 0,1o M ,

T 1,0o M = T 0,1

o M and dπe(gC) = TCo M , it follows that θ(a+) = a− and gC = a+ + θ(a+). Moreover,

since ker(dπe) = kC, it also follows that

a+ ∩ θ(a+) = {Z ∈ gC | dπe(Z) ∈ T 1,0o M ∩ T 0,1

o M} = ker(dπe) = kC.

Next we define subspaces m+ and m− of gC in the following way:

m+ = mC ∩ a+, m− = mC ∩ a−.

It is then easy to see that

a+ = kC ⊕m+, a− = kC ⊕m−, mC = m+ ⊕m−.

We know that dπe : mC → TCo M is an isomorphism, since the kernel of dπe is kC. From this

and the definition of a±, it follows that m+ ' T 1,0o M , m− ' T 0,1

o M in the same way.Recall that J gives rise to a unique linear map I0 : m→ m such that

I20 (X) = −X for all X ∈ m,

I0((Ad(k)(X))m) = (Ad(k)(I0(X)))m for all X ∈ m and all k ∈ K.

Lemma 4.4. The subspaces m± satisfies

m± = {Z ∈ mC | I0(Z) = ±iZ }.

Proof. We know thata+ = kC ⊕m+.

By definitionI0(Z) = ((dπe|m)−1 ◦ Jo ◦ dπe)(Z)

so for Z ∈ m+ we have that

I0(Z) = ((dπe|m)−1 ◦ Jo ◦ dπe)(Z) = (dπe|m)−1(idπe(Z)) = iZ,

where we used that dπe(m+) = T 1,0o M . So

m+ ⊆ {Z ∈ mC | I0(Z) = iZ }.

Conversely, suppose Z ∈ mC satisfies I0(Z) = iZ. Then

iZ = I0(Z) = ((dπe|m)−1 ◦ Jo ◦ dπe)(Z)

24

which impliesidπe(Z) = Jo(dπe(Z))

and this is equivalent to dπe(Z) ∈ T 1,0o M . Clearly this means that Z ∈ a+. By assumption

Z ∈ mC, soZ ∈ a+ ∩mC = m+.

Hence{Z ∈ mC | I0(Z) = iZ } ⊆ m+

which together with our previous result implies

m+ = {Z ∈ mC |I0(Z) = iZ}.

Performing similar calculations for a−,m− gives us

m− = {Z ∈ mC |I0(Z) = −iZ}

We may now present the following lemma:

Lemma 4.5. [12] The subspace a+ is a subalgebra of gC. Moreover, the correspondence betweenthe set of G-invariant complex structures on M and the set of subspaces of gC, given by a+ =(dπe)−1(T 1,0

o M) is injective.

Proof. Suppose J ′ is another G-invariant complex structure on M with corresponding decomposi-tion

TCo M = T 1,0

o M ′ ⊕ T 0,1o M ′,

such that a+ = (dπe)−1(T 1,0o M ′). Then

(dπe)−1(T 1,0o M) = (dπe)−1(T 1,0

o M ′).

Let Xo ∈ T 1,0o M ′. Since dπe is onto there exists X ∈ gC such that dπe(X) = Xo. By definition X ∈

(dπe)−1(T 1,0o M ′). But we knew that (dπe)−1(T 1,0

o M) = (dπe)−1(T 1,0o M ′), so X ∈ (dπe)−1(T 1,0

o M)which implies that Xo = dπe(X) ∈ T 1,0

o M . This means that T 1,0o M ′ ⊆ T 1,0

o M . The oppositeinclusion is shown in the same way, hence T 1,0

o M ′ = T 1,0o M . It follows from this that Jo = J ′o and

since both J and J ′ are G-invariant, we must in fact have that J = J ′.Let X,Y ∈ a+. We define vector fields X ′, Y ′ ∈ C∞(TM) in the following way:

X ′gK = ((dτg)o ◦ dπe)(X), Y ′gK = ((dτg)o ◦ dπe)(Y ).

Since X,Y ∈ a+ it follows that dπe(X), dπe(Y ) ∈ T 1,0o M . Using this and the G-invariance of J ,

we see that

JgK(X ′gK) = (JgK ◦ (dτg)o)(dπe(X))

= ((dτg)o ◦ Jo)(dπe(X))= (dτg)o(idπe(X))= iX ′gK .

Similarly we see thatJgK(Y ′gK) = iY ′gK .

This holds for all g ∈ G, which means that X ′ and Y ′ are of type (1,0). We also see that

dπg(Xg) = (dπg ◦ (dLg)e)(Xe) = ((dτg)o ◦ dπe)(X) = X ′gK = X ′π(g)

or equivalently that dπ(X) = X ′. In the same way we see that dπ(Y ) = Y ′. This implies that

dπ([X,Y ]) = [dπ(X), dπ(Y )] = [X ′, Y ′].

In particular we have thatdπe([X,Y ]) = [X ′, Y ′]o ∈ T 1,0

o M,

where we used that J is a complex structure together with Proposition 3.6. This shows that[X,Y ] ∈ a+ and this holds for all X,Y ∈ a+, which implies that a+ is a subalgebra of gC.

25

We have now shown that the existence of a complex G-invariant structure on M implies theexistence of a unique subalgebra a+ satisfying

gC = a+ + θ(a+), a+ ∩ θ(a+) = kC. (4.1)

Lemma 4.6. [12] Let a+ ⊂ gC be a subalgebra of gC which satisfy (4.1). Then there exists aG-invariant complex structure J on M such that a+ = (dπe)−1(T 1,0

o M).

Proof. Let

T 1,0o M = dπe(a+)

T 0,1o M = dπe(a−) = dπe(θ(a+)) = dπe(a+) = T 1,0

o M.

We see that

TCo M = dπe(gC)

= dπe(a+ + a−)

= dπe(a+) + dπe(a−)

= T 1,0o M + T 0,1

o M.

Suppose that Z ∈ ToM+ ∩ ToM−. Then

Z = dπe(X) = dπe(Y ),

where X ∈ a+ and Y ∈ a−. This implies that dπe(X − Y ) = 0 or equivalently, that X − Y ∈ker(dπe) = kC. So there exists some Y ′ ∈ kC such that X = Y +Y ′. Since Y ∈ a− and Y ′ ∈ kC ⊂ a−

it follows that Y + Y ′ ∈ a−. This means that X ∈ a+ ∩ a− = kC. Hence, Z = dπe(X) = 0. Wethen get that

TCo M = T 1,0

o M ⊕ T 0,1o M.

This allows us to define Jo : TCo M → TC

o M , by

Jo(Z) = iZ for Z ∈ T 1,0o M, Jo(Z) = −iZ for Z ∈ T 0,1

o M.

It is clear that Jo is a complex structure on the vector space TCo M . We extend Jo to the whole of

C∞(TM) by definingJgK(XgK) = ((dτg)o ◦ Jo ◦ (dτg)−1

o )(XgK)

for g ∈ G and XgK ∈ TCgKM . We need to show that JgK does not depend on the choice of

representative of the coset gK. To do this we first note that for X ∈ a+ and k ∈ K, Ad(k)(X) ∈ a+.This is true since kC ⊆ a+ and a+ is a Lie algebra. If we combine this with Lemma 2.9, we get that

(dτk)o(dπe(X)) = dπe(Ad(k)(X)) ∈ T 1,0o M

for all X ∈ a+ and all k ∈ K. Hence it follows that (dτk)o(T 1,0o M) = T 1,0

o M for all k ∈ K. Inthe same way we have that (dτk)o(T 0,1

o M) = T 0,1o M for all k ∈ K. It is then easy to see that

Jo ◦ (dτk)o = (dτk)o ◦Jo for all k ∈ K. Suppose now that hK = gK. Then h = gk for some k ∈ K,which implies that (dτh)o = (dτg)o ◦ (dτk)o. We then see that

JhK(XhK) = ((dτh)o ◦ Jo ◦ (dτh)−1o )(XhK)

= ((dτg)o ◦ (dτk)o ◦ Jo ◦ (dτk−1g−1)gK)(XgK)= ((dτg)o ◦ (dτk)o ◦ Jo ◦ (dτk−1)o ◦ (dτg−1)gK)(XgK)

= ((dτg)o ◦ Jo ◦ (dτg)−1o )(XgK)

= JgK(XgK).

So J is well-defined. It is clear that J2gK = −I for all g ∈ G and that J is G-invariant follows directly

from the definition. Moreover, it is easy to see from the definition of T 1,0o M that (dπe)−1(T 1,0

o M) =a+.

26

We would now like to show that J is a complex structure. To do this we will use Theorem 3.10.We let m± be defined as before. We define, for Z1, Z2 ∈ gC,

N(Z1, Z2) = [I0(Z1), I0(Z2)]− [Z1, Z2]− I0([Z1, I0(Z2)])− I0([I0(Z1), Z2]).

We have that Zi = Xi + Yi, Xi ∈ kC,Yi ∈ mC, i = 1, 2. Since mC = m+ ⊕ m− we may writeYi = Y +

i + Y −i , where Y +i ∈ m+ and Y −i ∈ m−. We want to show that N(Z1, Z2) ∈ kC, for all

Z1, Z2 ∈ gC. We begin by looking at each term in N(Z1, Z2) individually. Using Lemma 4.4 andthe properties of I0, we have that

I0(Xi) = 0, I0(Y +i ) = iY +

i , I0(Y −i ) = −iY −i

for i = 1, 2. We use this to obtain the following identities:

[I0(Z1), I0(Z2)] = [iY +1 − iY

−1 , iY +

2 − iY−2 ]

= −[Y +1 , Y +

2 ] + [Y +1 , Y −2 ] + [Y −1 , Y +

2 ]− [Y −1 , Y −2 ],

[Z1, Z2] = [X1 + Y1, X2 + Y2]

= [X1, X2] + [X1, Y+2 ] + [X1, Y

−2 ] + [Y +

1 , X2] + [Y −1 , X2]

+ [Y +1 , Y +

2 ] + [Y +1 , Y −2 ] + [Y −1 , Y +

2 ] + [Y −1 , Y +2 ],

I0([Z1, I0(Z2)]) = I0([X1 + Y +1 + Y −1 , iY +

2 − iY−2 ])

= iI0([X1, Y+2 ])− iI0([X1, Y

−2 ]) + iI0([Y +

1 , Y +2 ])− iI0([Y +

1 , Y −2 ])

+ iI0([Y −1 , Y +2 ])− iI0([Y −1 , Y −2 ]),

I0([I0(Z1), Z2]) = I0([iY +1 − iY

−1 , X2 + Y +

2 + Y −2 ]

= iI0([Y +1 , X2])− iI0([Y −1 , X2]) + iI0[Y +

1 , Y +2 ])− iI0([Y −1 , Y +

2 ])

+ iI0([Y +1 , Y −2 ])− iI0([Y −1 , Y −2 ]).

From these identities we get that:

N(Z1, Z2) = −2(

[Y +1 , Y +

2 ] + iI0([Y +1 , Y +

2 ]))− 2(

[Y −1 , Y −2 ]− iI0([Y −1 , Y −2 ]))

−(

[X1, Y+2 ] + iI0([X1, Y

+2 ]))−(

[X1, Y−2 ]− iI0([X1, Y

−2 ]))

−(

[Y +1 , X2] + iI0([Y +

1 , X2]))−(

[Y −1 , X2]− iI0([Y −1 , X2]))− [X1, X2].

Let us start by considering the first parenthesis in the above expression. Both Y +1 and Y +

2 belongto m+ which is a subspace of a+. By assumption a+ is a subalgebra of gC. This means that[Y +

1 , Y +2 ] ∈ a+. So we may write

[Y +1 , Y +

2 ] = [Y +1 , Y +

2 ]k + [Y +1 , Y +

2 ]m+

where [Y +1 , Y +

2 ]k ∈ kC and [Y +1 , Y +

2 ]m+ ∈ m+. This implies that

[Y +1 , Y +

2 ] + iI0([Y +1 , Y +

2 ]) = [Y +1 , Y +

2 ]k + [Y +1 , Y +

2 ]m+ + i(i[Y +1 , Y +

2 ]m+)

= [Y +1 , Y +

2 ]k ∈ kC

By using the same method it is easy to see that the other parenthesis belong to kC as well. Also,since kC is a subalgebra of gC, it is clear that [X1, X2] ∈ kC. It now follows that N(Z1, Z2) ∈ kC forall Z1, Z2 ∈ gC and by Theorem 3.10 this is equivalent to J being a complex structure.

27

The following theorem is now obvious.

Theorem 4.7. [12] The map

F : the set of G-invariant complex structures on M → the set of subalgebras of gC satisfying (4.1),

given by F (J) = (dπe)−1(T 1,0o M), is a bijection.

Proof. By Lemma 4.5, F is injective and by Lemma 4.6, F is also surjective.

Let us now consider the case when M = G/K is a generalized flag manifold. As described inthe beginning of this section we have that

gC = hC ⊕∑α∈R

gα

kC = hC ⊕∑α∈RK

gα.

Before moving on, we will need some properties regarding the values of the roots on h and how theconjugation with respect to gC acts on the root spaces. These results will also be of importancelater on.

We have already mentioned the conjugation θ : gC → gC with respect to g defined by

θ(X + iY ) = X − iY for all X,Y ∈ g.

The following properties are easy to verify.

θ(θ(Z)) = Z, θ(Z1 + Z2) = θ(Z1) + θ(Z2),θ(aZ) = aθ(Z), θ([Z1, Z2]) = [θ(Z1), θ(Z2)]

for all Z,Z1, Z2 ∈ gC and all a ∈ C. The set of elements in gC that are fixed by θ are exactly theelements in g, i.e.

g = {X ∈ gC | θ(X) = X}

Proposition 4.8. [4] α(H) is purely imaginary for all α ∈ R and all H ∈ h.

Proof. The quotient M = G/K is a generalized flag manifold, so g is compact and semisimple.That g is compact and semisimple implies that B is negative definite on g, by Proposition 1.9. Itis then clear that −B is an inner product on g. Using this, we define an inner product (, ) on gC by

(X,Y ) = −B(X, θ(Y )).

By Lemma 1.12, B(ad(X)(Y ), Z) = B(Y,−ad(X)(Z)) for all X,Y, Z ∈ gC. Note also that forX ∈ g and Y ∈ gC,

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

Let α ∈ R and Eα ∈ gα such that Eα 6= 0. For any H ∈ h we have that ad(H)(Eα) = α(H)Eα.This implies that

(ad(H)(Eα), Eα) = α(H)(Eα, Eα).

But we also have that

(ad(H)(Eα), Eα) = −B(ad(H)(Eα), θ(Eα))= B(Eα, ad(H)(θ(Eα)))= B(Eα, θ(ad(H)(Eα)))= B(Eα, θ(α(H)Eα))

= α(H)B(Eα, θ(Eα))

= −α(H)(Eα, Eα),

28

Hence we have thatα(H)(Eα, Eα) = −α(H)(Eα, Eα)

and since (Eα, Eα) > 0, we must have that α(H) = −α(H), which implies that α(H) is purelyimaginary. This proves the proposition since α ∈ R and H ∈ h were chosen arbitrarily.

Lemma 4.9. [4] The conjugation θ maps gα into g−α, for all α ∈ R ∪ {0}.

Proof. We begin by noting that this is obvious for α = 0, since g0 = hC and clearly θ(H1 + iH2) =H1 − iH2 ∈ hC for all H1, H2 ∈ h. For each α ∈ R we choose basis elements Eα ∈ gα. Let H beany element in hC. Then there exist H1, H2 ∈ h, such that

H = H1 + iH2.

Since θ maps hC into itself, we have that

[θ(H), Eα] = α(θ(H))Eα.

Applying θ to both sides of the above the equation gives us

θ([θ(H), Eα]) = θ(α(θ(H))Eα)

or equivalently[H, θ(Eα)] = α(θ(H))θ(Eα).

Since θ(H) = θ(H1 + iH2) = H1 − iH2, we get that

α(θ(H)) = α(H1 − iH2) = α(H1)− iα(H2).

From Proposition 4.8 we know that α takes purely imaginary values on h, so −iα(H2) ∈ R andα(H1) ∈ iR. We therefore get that

α(θ(H)) = −iα(H2)− α(H1) = −(α(H1) + iα(H2)) = −α(H).

This implies that[H, θ(Eα)] = α(θ(H))θ(Eα) = −α(H)θ(Eα).

Since we can do this for any H ∈ hC, it follows that θ(Eα) ∈ g−α.

By a theorem in [13], every generalized flag manifold admits a G-invariant complex structure.For a proof of this theorem see [5] page 501-502. We may therefore assume that there exists aG-invariant complex structure J on M . Then we obtain a+ as before which satisfies (4.1). SincehC ⊂ kC ⊂ a+ and a+ is a subalgebra of gC, there exists a root system RA+ with respect to theCartan subalgebra hC, such that

a+ = hC ⊕∑

α∈RA+

gα.

It is clear that RK ⊂ RA+ , since kC ⊂ a+. We define ∆+ = RA+\RK . Then

a+ = kC ⊕∑α∈∆+

gα.

Next let ∆− = {α ∈ R| − α ∈ ∆+} and define

a− = kC ⊕∑α∈∆−

gα.

It is easy to see from Lemma 4.9 that θ(a+) = a−, which implies gC = a+ + a− and a+ ∩ a− = kC.We have the following result:

29

Lemma 4.10. [12] Let RK ,∆+ and ∆− be as above. Then the following holds:

R = RK ∪∆+ ∪∆− disjoint union (4.2)

If α ∈ RK ∪∆+, β ∈ ∆+ and α+ β ∈ R, then α+ β ∈ ∆+ (4.3)

Moreover, the subset ∆+ is uniquely determined by a+.

Proof. Let us start by proving (4.2). By definition ∆+ ∩ RK = ∅. Suppose α ∈ ∆− ∩ RK . Then−α ∈ ∆+∩RK , where we used the fact that RK is a root system. Clearly this contradicts the factthat ∆+ and RK are disjoint. Hence, we must have that ∆− ∩RK = ∅.

Suppose now that α ∈ ∆+ ∩∆−. Then gα ⊂ a+ ∩ a− = kC, which is false since α /∈ RK . Thesets RK , ∆+ and ∆− must therefore be disjoint. We note that

hC ⊕∑α∈R

gα = gC

= a+ + a−

= (kC ⊕∑α∈∆+

gα) + (kC ⊕∑α∈∆−

gα)

= hC ⊕∑α∈RK

gα ⊕∑α∈∆+

gα ⊕∑α∈∆−

gα.

From these calculations we see that the number of roots in RK ∪∆+ ∪∆− must be the same asthe number of roots in R. This, together with the fact that RK ∪ ∆+ ∪ ∆− ⊆ R, implies thatR = RK ∪∆+ ∪∆−. This proves that (4.2) holds.

Assume next that α ∈ RK ∪∆+, β ∈ ∆+ and that α+ β ∈ R. We know from Chapter 1 that

[gα, gβ ] = gα+β ,

and since gα, gβ ⊂ a+ and a+ is a subalgebra, we have that gα+β ∈ a+. This implies that α+β ∈ RKor α+ β ∈ ∆+. Assume first that α ∈ RK , and assume for a contradiction that α+ β = γ ∈ RK .Then β = γ − α. We know that α ∈ RK implies −α ∈ RK , and since γ − α = β ∈ ∆+ we see inparticular that γ − α is a root. Since −α, γ ∈ RK , and −α + γ ∈ R, we must therefore have thatβ = −α + γ ∈ RK . But by (4.2), RK ∩ ∆+ = ∅. So the assumption that α + β ∈ RK must befalse. It is clear that we can apply the same reasoning to ∆−, i.e.

α+ β ∈ ∆− for α ∈ RK , β ∈ ∆−. (4.4)

Now let α ∈ ∆+. We still have that α+β ∈ RK or α+β ∈ ∆+. We again assume for a contradictionthat α + β = γ ∈ RK . Then γ − α ∈ ∆−, by (4.4), since −α ∈ ∆−. But this contradicts the factthat γ − α = β ∈ ∆+, since ∆+ ∩ ∆− = ∅. So the assumption that α + β ∈ RK must be false.This proves (4.3).

Suppose that Γ+ is another subset of R which satisfies

a+ = kC ⊕∑β∈Γ+

gβ .

The sets ∆+ and Γ+ must have the same number of elements, if not then the two decompositionsof a+ would have different dimensions, which is impossible. Let β ∈ Γ+ and assume β /∈ ∆+. Thenβ ∈ ∆− by (4.2). But this means that

gβ ⊂ a+ ∩ a− = kC

which is impossible since β /∈ RK by (4.2). So we must have that β ∈ ∆+. But this means thatΓ+ ⊆ ∆+ and since the two sets have the same number of elements they must in fact be equal.

Now assume that we are given a subset ∆+ of R that satisfies (4.2) and (4.3). Then we define

a+ = kC ⊕∑α∈∆+

gα.

We have the following result.

30

Lemma 4.11. [12] The subspace a+ ⊂ gC is a subalgebra of gC, which satisfies (4.1), that is

gC = a+ + θ(a+), a+ ∩ θ(a+) = kC.

Proof. We leta− = kC ⊕

∑α∈∆−

gα.

It then follows directly from (4.2) and Lemma 4.9 that

gC = a+ + a−, a+ ∩ a− = kC and θ(a+) = a−.

We now show that a+ is a subalgebra of gC. Let Z1 = X1 + Y1, Z2 = X2 + Y2 ∈ a+, withX1, X2 ∈ kC and Y1, Y2 ∈

∑α∈∆+ gα. Then

[Z1, Z2] = [X1, X2] + [X1, Y2] + [Y1, X2] + [Y1, Y2].

Since kC is a subalgebra, [X1, X2] ∈ kC. Next we consider [Y1, X2]. We have that

Y1 =∑α∈∆+

aαEα

X2 = H +∑β∈RK

bβEβ ,

where H ∈ hC, Eα ∈ gα. From Chapter 1, we know that [Eα, H] ∈ gα, and [Eα, Eβ ] ∈ gα+β . Ifβ + α ∈ R, then we know from (4.3) that β + α ∈ ∆+. Using this it is clear that

[Y1, X2] =∑α∈∆+

aα[Eα, H] +∑α∈∆+

∑β∈RK

aαbβ [Eα, Eβ ] ∈∑α∈∆+

gα.

Using (4.3) in the same way as above, it is easy to see that the other two terms in [Z1, Z2]belongs to

∑α∈∆+ gα as well. So [Z1, Z2] ∈ a+, and since Z1, Z2 was arbitrarily chosen, it follows

that a+ is a subalgebra of gC.

We may now prove the following theorem.

Theorem 4.12. [12] There is a bijection between the set of G-invariant complex structures J onM and the set of subsets ∆+ of R satisfying (4.2) and (4.3).

Proof. [12] By Lemma 4.10 we have a well-defined correspondence between subalgebras whichsatisfies (4.1) and subsets which satisfies (4.2) and (4.3). It is clear that this correspondence isinjective. From Lemma 4.11 we get that this correspondence is surjective as well. The theoremthen follows from Theorem 4.7.

4.2 Weyl Chambers

In this section we introduce the Weyl chambers. These are certain subsets of the Lie algebrag and we shall see that there is a one-to-one correspondence between Weyl chambers and subsets∆+ of R.

As before let M = G/K be a generalized flag manifold with root space decomposition

gC = hC ⊕∑α∈R

gα

andkC = hC ⊕

∑α∈RK

gα.

We definet = Z(kC) ∩ h,

where Z(kC) = {X ∈ gC | [X,Y ] = 0 for all Y ∈ kC}. We then have the following lemma:

31

Lemma 4.13. [3] The subspace t satisfies

t = {X ∈ h | α(X) = 0 for all α ∈ RK}.

Proof. [3] We see that

t = Z(kC) ∩ h

= {X ∈ gC | [X,Y ] = 0 for all Y ∈ kC} ∩ h

= {X ∈ gC | [X, hC] = 0, [X,Eα] = 0 for all α ∈ RK} ∩ h

= {X ∈ hC | α(X)Eα = 0 for all α ∈ RK} ∩ h

= {X ∈ h | α(X) = 0 for all α ∈ RK}

where we have used that if [X,H] = 0 for all H ∈ hC, then X ∈ g0 by definition and g0 = hC sincehC is a Cartan subalgebra of gC.

Let ∆ = R\RK and let

Lα = {X ∈ t | α(X) = 0} for α ∈ ∆,

andt′ = t\ ∪α∈∆ Lα.

The connected components in t′ are called Weyl chambers.Let C be a Weyl chamber. Let Z ∈ C and suppose that iα(Z) > 0 for some α ∈ ∆. Assume that

there exists X ∈ C such that iα(X) < 0. Since C is connected there exists a path γ : [0, 1] → Csuch that γ(0) = Z and γ(1) = X. The root α is in the dual of hC, which implies that α iscontinuous. By the intermediate value theorem there must exist ε ∈ [0, 1] such that α(γ(ε)) = 0.However, this implies that γ(ε) /∈ C which is a contradiction. So if iα(Z) > 0 for some Z ∈ C,then iα(Z) > 0 for all Z ∈ C.

Theorem 4.14. [4] There is a bijection between the set of Weyl chambers in t′ and the set ofsubsets of R satisfying (4.2) and (4.3).

Proof. Let C be a Weyl chamber and let Z ∈ C. We define

∆+ = {α ∈ ∆ | iα(Z) > 0}.

We show that ∆+ satisfies (4.2) and (4.3). By definition, ∆ = R\RK , so ∆ ∩RK = ∅. Moreover,

∆− = {α ∈ R | − α ∈ ∆+} = {α ∈ ∆ | iα(Z) < 0}.

It follows that ∆ = ∆+ ∪∆−, since iα(Z) is either greater than 0 or less than 0 for all α ∈ ∆. Itis also clear that ∆+ ∩∆− = ∅. This implies that

R = RK ∪∆ = R ∪∆+ ∪∆− disjoint union.

So (4.2) is satisfied.To check (4.3) we let α ∈ RK and β ∈ ∆+ and assume that α+ β ∈ R. Then α(Z) = 0 since

Z ∈ C ⊂ t = {X ∈ h| α(X) = 0 for all α ∈ RK}.

Hencei(α+ β)(Z) = iα(Z) + iβ(Z) = iβ(Z) > 0,

so α+ β ∈ ∆+. Next we let α ∈ ∆+ as well and assume that α+ β ∈ R. Then clearly

i(α+ β)(Z) = iα(Z) + iβ(Z) > 0,

so α + β ∈ ∆+. Since we did this for arbitrary α ∈ RK ∪ ∆+ and β ∈ ∆+, it follows that (4.3)holds as well.

32

We have just constructed a function

Φ : the set of Weyl chambers→ the subsets of ∆ satisfying (4.2) and (4.3).

We now define

Ψ : the subsets of ∆ satisfying (4.2) and (4.3)→ the set of Weyl chambers

by Ψ(∆+) = {X ∈ t′ | iα(X) > 0 for all α ∈ ∆+}. We need to show that Ψ(∆+) is nonempty forany given ∆+ which satisfies (4.2) and (4.3). In order to do this we let R+

K be a set of positiveroots in RK . We then let R+ = R+

K ∪∆+. It is easy to see using (4.2), (4.3) and the fact that R+K

is a set of positive roots in RK , that R+ is a set of positive roots in R. By Proposition 1.7, wemay choose a set

Π = {α1, α2, . . . , αn}

of simple roots in R+. Since R+ = R+K ∪∆+, a simple root αi belongs to either R+

K or ∆+. Nowlet V = spanR(R). Then Π is a basis of V . It is clear that V ⊆ (hC)∗. The vector space W := V ∗

is of dimension n, so W is isomorphic to Rn, with the isomorphism F given by

F (X) = (α1(X), α2(X), . . . , αn(X)).

We may assume that the simple roots are ordered so that αi ∈ ∆+ for 1 ≤ i ≤ m for some1 < m < n and αi ∈ R+

K for m < i ≤ n. Let Y = (y1, y2, . . . , yn) ∈ Rn be such that yi > 0 for all1 ≤ i ≤ m and yi = 0 for m < i ≤ n. Since F is an isomorphism, there exists X ∈ W such thatF (X) = Y , or equivalently

(α1(X), α2(X), . . . , αm(X), αm+1(X), . . . , αn(X)) = (y1, y2, . . . , ym, 0, . . . , 0).

Every element in RK can be written as a linear combination of αm+1, . . . αn, so it follows thatα(X) = 0 for all α ∈ RK . Recall that every root takes purely imaginary values on h. If wecombine this with the fact that αi(X) > 0 for all 1 ≤ i ≤ m we must have that X = iH for someH ∈ h. Let α ∈ ∆+. From Proposition 1.7 we know that α can be written as

α =n∑i=1

aiαi

where ai ∈ Z+ and ai > 0 for at least one 1 ≤ i ≤ m. Using this, we see that

iα(H) = α(X) =n∑i=1

aiαi(X) > 0.

We have now shown that the element H ∈ h satisfies α(H) = 0 for all α ∈ RK and that iα(H) > 0for all α ∈ ∆+. This implies that H ∈ t and that H ∈ Ψ(∆+), which shows that Ψ(∆+) isnonempty.

We see that(1− t)X + tY ∈ Ψ(∆+)

for all X,Y ∈ Ψ(∆+) and all t ∈ [0, 1]. From this we see that Ψ(∆+) is a connected subset of t′,so Ψ(∆+) ⊆ D, where D is a Weyl chamber. Let α ∈ ∆+ and let X ∈ D. Since iα(Z) > 0 forall Z ∈ Ψ(∆+), it follows that iα(X) > 0 since D is connected. This holds for all α ∈ ∆+ and allX ∈ D which implies that D ⊆ Ψ(∆+). We therefore have that Ψ(∆+) = D.

The following calculations show that Ψ is an inverse to Φ.

Φ(Ψ(∆+)) = {α ∈ ∆ | iα(Z) > 0 for Z ∈ Ψ(∆+)}= {α ∈ ∆ | iα(Z) > 0 for Z ∈ t′ such that iβ(Z) > 0 for all β ∈ ∆+}

We see directly that ∆+ ⊆ Φ(Ψ(∆+)) and since these two sets must have the same number ofelements, it follows that Φ(Ψ(∆+)) = ∆+.

Ψ(Φ(C)) = {X ∈ t′ | iα(X) > 0 for all α ∈ Φ(C)}

33

= {X ∈ t′ | iα(X) > 0 for all α ∈ ∆ such that iα(Z) > 0 for all Z ∈ C}.

It is clear that C ⊆ Ψ(Φ(C)) and since Ψ(Φ(C)) is a Weyl chamber, we must have that C =Ψ(Φ(C)). So Φ is a bijection between the set of Weyl chambers and the set of subsets of ∆satisfying (4.2) and (4.3).

Combining Theorems 4.12 and 4.14 gives us the following result.

Theorem 4.15. [4] There is bijection between the set of G-invariant complex structures on M andthe set of Weyl chambers in t′.

4.3 The Construction of the Kahler Metrics

In this section we will see that for each G-invariant complex structure on a generalized flagmanifold M , there exists a family of G-invariant Kahler metrics on M .

Let M = G/K be a generalized flag manifold and let J be a G-invariant complex structure onM . As before we have

gC = hC ⊕∑α∈R

gα,

kC = hC ⊕∑α∈RK

gα

and we let ∆ = R\RK . We then define

mC =∑α∈∆

gα.

It follows directly that gC = kC ⊕mC.Before moving on with the construction of the metrics, we state and prove some preliminary

results.

Lemma 4.16. Let (kC)⊥ = {X ∈ gC | B(X,Y ) = 0 for all Y ∈ kC}. Then mC = (kC)⊥.

Proof. Recall from Chapter 1 that B(gα, gβ) = 0 for α, β ∈ R ∪ {0} such that α + β 6= 0. Letα ∈ ∆ and β ∈ RK ∪ {0}. Then α + β 6= 0, so we have that B(gα, gβ) = 0. Since every elementin kC is a linear combination of elements in gα, α ∈ RK ∪ {0} and every element in mC is a linearcombination of elements in gα, α ∈ ∆, it follows that

mC ⊆ (kC)⊥.

Conversely, suppose that Z = Z1 + Z2 ∈ (kC)⊥, where Z1 ∈ kC and Z2 ∈ mC. Then for allX ∈ kC we have

0 = B(Z,X) = B(Z1, X) + B(Z2, X) = B(Z1, X)

where we used Z2 ∈ mC ⊆ (kC)⊥. But this implies that for any element Y = Y1 + Y2 ∈ gC, whereY1 ∈ kC and Y2 ∈ mC, we have

B(Z1, Y ) = B(Z1, Y1) + B(Z1, Y2) = 0.

This contradicts the nondegeneracy of B on gC unless Z1 = 0. This means that Z ∈ mC and sinceZ ∈ (kC)⊥ was chosen arbitrarily, it follows that

(kC)⊥ ⊆ mC.

Proposition 4.17. [4] For all α ∈ R we may choose basis elements Eα ∈ gα, such that θ(Eα) =−E−α and [Eα, E−α] = Hα, for all α ∈ R.

34

Proof. [4] For each α ∈ R we choose basis elements Eα ∈ gα, such that

[Eα, E−α] = Hα, B(Eα, E−α) = 1,

for all α ∈ R. We know from Proposition 1.6 that such a choice of basis elements is possible. FromLemma 4.9 we know that θ(Eα) ∈ g−α and since the gα are of dimension 1, we must have that

θ(Eα) = cαE−α

for some cα ∈ C.Let us begin by considering θ(Hα). For any H = H1 + iH2 ∈ hC, we have

B(Hα, θ(H)) = α(θ(H)) = α(H1)− iα(H2),

which impliesB(Hα, θ(H)) = −iα(H2)− α(H1) = −α(H) = B(−Hα, H). (4.5)

As in Section 2 of Chapter 1, we see that

ad(θ(X))(Y ) = (θ ◦ ad(X) ◦ θ)(Y ),

and if we combine this with the fact that θ is conjugate-linear, we get that

B(θ(X), θ(Y )) = Tr(θ ◦ ad(X) ◦ ad(Y ) ◦ θ)

= Tr(ad(Y ) ◦ θ ◦ θ ◦ ad(X))

= B(X,Y ).

This implies thatB(Hα, θ(H)) = B(θ(Hα), θ(θ(H))) = B(θ(Hα), H).

Using this expression for B(Hα, θ(H)) in (4.5), we obtain

B(θ(Hα), H) = B(−Hα, H).

Since H was chosen arbitrarily, this holds for all H ∈ hC. Using that B is non-degenerate on hC

we get thatθ(Hα) = −Hα, for all α ∈ R. (4.6)

We may use (4.6) to obtain the following:

−Hα = θ(Hα)= θ([Eα, E−α])= [θ(Eα), θ(E−α)]= [cαE−α, c−αEα]= −cαc−α[Eα, E−α]= −cαc−αHα.

This implies that cαc−α = 1. Moreover, we have the following calculation

Eα = θ(θ(Eα))= θ(cαE−α)= cαθ(E−α)= cαc−αEα,

which implies that cαc−α = 1.So the coefficients cα satisfy {

cαc−α = 1cαc−α = 1

35

and it is easy to see that this implies that cα ∈ R, for all α ∈ R.Now define for each α ∈ R,

E′α =1|cα|

12Eα.

Then[E′α, E

′−α] =

1|cαc−α|

12

[Eα, E−α] = Hα

and

θ(E′α) =1|cα|

12θ(Eα)

=cα

|cα|12E−α

=cα

|cαcαc−α|12E−α

=cα|cα|

E′−α

= sign(cα)E′−α.

So we have now found elements E′α ∈ gα, such that

θ(E′α) = ±E′−α, [E′α, E′−α] = Hα

We now drop the prime ′ and simply call these elements Eα.Assume that there exists α ∈ R such that θ(Eα) = E−α. We want to show that this assumption

leads to a contradiction. In order to do this we determine

g ∩ (gα ⊕ g−α) = {Z ∈ gα ⊕ g−α | θ(Z) = Z }.

Let Z = X + Y , where X = aEα, Y = bE−α, a, b ∈ C and Z 6= 0. Applying θ to Z gives us

θ(Z) = θ(X) + θ(Y )

= aθ(Eα) + bθ(E−α)

= aE−α + bEα.

So θ(Z) = Z if and only ifaE−α + bEα = aEα + bE−α

which is equivalent to b = a. This implies that θ(Z) = Z if and only if

Z = aEα + aE−α

for some a ∈ C. Now recall that B(Eα, E−α) = 1 and B(Eα, Eα) = B(E−α, E−α) = 0. We thensee that

B(Z,Z) = B(aEα + aE−α, aEα + aE−α) = 2|a|2 > 0.

This means that B is positive definite on g∩(gα⊕g−α). However, since g is compact and semisimple,B is negative definite on g. Hence we have obtained a contradiction and we conclude that therecannot exist any α ∈ R such that θ(Eα) = E−α. So the only possibility is that

θ(Eα) = −E−α for all α ∈ R

which is exactly what we wanted to show.

We let m = {X ∈ mC | θ(X) = X}. Then it is clear that g = k⊕m. From Lemma 4.16 we seethat m = (k)⊥.

Lemma 4.18. The Killing form B is nondegenerate on m.

36

Proof. Assume that there exist X ∈ m such that B(X,Y ) = 0 for all Y ∈ m. Since B(X,Y ) = 0for all Y ∈ k and g = k⊕m, we get that for Y = Y1 + Y2 ∈ g where Y1 ∈ k and Y2 ∈ m,

B(X,Y ) = B(X,Y1) + B(X,Y2) = 0.

The killing form B is nondegenerate on g since g is semisimple. We must therefore have thatX = 0, which implies that B is nondegenerate on m.

Since Ad(k) is an automorphism for all k ∈ K we have that

B(Ad(k)(X),Ad(k)(Y )) = B(X,Y )

for all X,Y ∈ g and all k ∈ K. So for X ∈ m and Y ∈ k,

B(Ad(k)(X), Y ) = B(X,Ad(k−1)(Y )) = 0,

which implies that Ad(k)(X) ∈ m for all k ∈ K and all X ∈ m. So the decomposition

g = k⊕m

is reductive.Suppose now thatM = G/K is a Kahler manifold withG-invariant Kahler metric g. This means

that g is Hermitian and ω(X,Y ) = g(JX, Y ) is closed. Let I0 be the Ad(K)-invariant complexstructure on m which corresponds to J and 〈, 〉 the Ad(K)-invariant scalar product on m whichcorresponds to g. Recall from Proposition 3.13 that g is Hermitian if and only if 〈I0(X), I0(Y )〉 =〈X,Y 〉 for all X,Y ∈ m. Also recall the 2-form

ω0(X,Y ) = 〈(I0(X), Y 〉

on m and the result that ω is closed if and only if

ω0([X,Y ]m, Z) + ω0([Y, Z]m, X) + ω0([Z,X]m, Y ) = 0, (4.7)

for all X,Y, Z ∈ m.Since B is nondegenerate on m, there exists a linear map φ : m→ m, such that

ω0(X,Y ) = B(φ(X), Y ),

for all X,Y ∈ m. By using that ω0(X,Y ) = −ω0(Y,X), we see that

B(φ(X), Y ) = ω0(X,Y ) = −ω0(Y,X) = −B(φ(Y ), X) = −B(X,φ(Y )).

So φ is skewsymmetric with respect to B. We extend φ to g by letting φ(X) = 0 for X ∈ k. Wewill continue to call this extension φ.

This is a good place to write out what we intend to do. We will begin by showing that φ is aderivation on g. From Proposition 1.14 this implies that φ = ad(Zφ), for some Zφ ∈ g. We canthen obtain an expression for 〈, 〉 acting on the basis elements of m, which depends on the roots in∆, the set ∆+ and the element Zφ. Using that 〈, 〉 is an inner product we will see that Zφ mustbelong to the Weyl chamber which corresponds to J .

Proposition 4.19. [4] The linear map φ : g→ g is a derivation of g.

Proof. [4] Our plan is to show that

B(φ([X,Y ]), Z) = B([φ(X), Y ], Z) + B([X,φ(Y )], Z),

for all X,Y, Z ∈ g and then use the fact that B is nondegenerate on g.Let X,Y, Z ∈ m. We rewrite (4.7) by using that ω0(X,Y ) = B(φ(X), Y ):

B(φ([X,Y ]m), Z) + B(φ([Y, Z]m), X) + B(φ([Z,X]m), Y ) = 0

37

which is equivalent with

B(φ([X,Y ]m), Z) = −B(φ([Y,Z]m), X)− B(φ([Z,X]m), Y ),

and

−B(φ([Y,Z]m), X)− B(φ([Z,X]m), Y ) = B([Y,Z]m, φ(X)) + B([Z,X]m, φ(Y ))= B([Y,Z], φ(X)) + B([Z,X], φ(Y ))= B(Y, [Z, φ(X)]) + B(Z, [X,φ(Y )])= B([Z, φ(X)], Y ) + B([X,φ(Y )], Z)= B(Z, [φ(X), Y ]) + B([X,φ(Y )], Z).

Combining the above equalities we see that

B(φ([X,Y ]), Z) = B(φ([X,Y ]m), Z) = B([φ(X), Y ], Z) + B([X,φ(Y )], Z) (4.8)

for all X,Y, Z ∈ m. We continue to let X,Y ∈ m but now let Z ∈ k. Then

B([φ(X), Y ], Z) + B([X,φ(Y )], Z) = B(φ(X), [Y, Z])− B([φ(Y ), X], Z)= B(φ(X), [Y, Z]) + B(φ(Y ), [Z,X])= ω0(X, [Y,Z]) + ω0(Y, [Z,X])= 0= B(φ([X,Y ]m), Z)= B(φ([X,Y ]), Z),

where we used Corollary 3.16. So

B(φ([X,Y ]), Z) = B([φ(X), Y ], Z) + B([X,φ(Y )], Z), (4.9)

for all X,Y ∈ m and Z ∈ k. Since g = k⊕m it follows from (4.8) and (4.9) that

B(φ([X,Y ]), Z) = B([φ(X), Y ], Z) + B([X,φ(Y )], Z) (4.10)

for all X,Y ∈ m and Z ∈ g.The identity ω0([Z,X], Y ) = −ω0(X, [Z, Y ]) for X,Y ∈ m and Z ∈ k, translates into

B(φ([Z,X]), Y ) = −B(φ(X), [Z, Y ]).

Using this we obtain

B(φ([Z,X]), Y ) = −B(φ(X), [Z, Y ])= −B(Z, [Y, φ(X)])= −B([Y, φ(X)], Z)= B([Z, φ(X)], Y )

for X,Y ∈ m and Z ∈ k. So

B(φ([Z,X]), Y ) = B([Z, φ(X)], Y ), (4.11)

for all X,Y ∈ m and all Z ∈ k. This equality actually holds for all Y ∈ g, since

B(φ([Z,X]), Y ) = B(φ([Z,X]), Ym)= B([Z, φ(X)], Ym)= B([Z, φ(X)], Y ).

Now for the general case we let X,Y, Z ∈ g and write X = X1 + X2, Y = Y1 + Y2, whereX1, Y1 ∈ k and X2, Y2 ∈ m. We then get, using (4.10) and (4.11), that

B(φ([X,Y ]), Z) = B(φ([X1 +X2, Y1 + Y2], Z))

38

= B(φ([X1, Y1]), Z) + B(φ([X1, Y2]), Z) + B(φ([X2, Y1]), Z)+ B(φ([X2, Y2]), Z)

= B([X1, φ(Y2)], Z) + B([φ(X2), Y1], Z) + B([φ(X2), Y2], Z)+ B([X2, φ(Y2)], Z)

= B([X1 +X2, φ(Y2)], Z) + B([φ(X2), Y1 + Y2], Z) == B([φ(X), Y ], Z) + B([X,φ(Y )], Z).

Using that the Killing form is nondegenerate on g we get

φ([X,Y ]) = [φ(X), Y ] + [X,φ(Y )]

for all X,Y ∈ g.

From Proposition 1.14 we know that there exists a unique Zφ ∈ g, such that φ = ad(Zφ). Hencewe may write

ω0(X,Y ) = B(φ(X), Y ) = B(ad(Zφ)(X), Y ) = B(Zφ, [X,Y ]),

for all X,Y ∈ m.

Proposition 4.20. [4] The element Zφ is in the subspace t.

Proof. Recall that t = Z(kC) ∩ h and that

hC = g0 = {X ∈ gC | [X,H] = 0 for all H ∈ hC}.

That Zφ ∈ Z(kC) follows directly from the fact that φ(X) = 0 for all X ∈ k and φ(X) =ad(Zφ)(X). In particular we have that Zφ commutes with every element in hC, since hC ⊂ kC .This implies that Zφ ∈ hC. We know that Zφ ∈ g and we just saw that Zφ ∈ hC so we mustnecessarily have that Zφ ∈ hC ∩ g = h. It follows directly from this that Zφ ∈ Z(kC) ∩ h = t.

We are now ready to prove the following theorem:

Theorem 4.21. [4] Given a generalized flag manifold M , with G-invariant complex structure Jand G-invariant Kahler metric g, the corresponding Ad(K)-invariant scalar product 〈, 〉 on m mustsatisfy

〈Eα, Eβ〉 =

{1εαα(Zφ) if β = −α

0 if β 6= −α,

where εα = ±i for α ∈ ∆±. Moreover, the element Zφ belongs to the Weyl chamber that correspondsto J .

Proof. [4] Let ∆+ be the subset of ∆ corresponding to J , which satisfies (4.2) and (4.3). We let

m+ =∑α∈∆+

gα, m− =∑α∈∆−

gα.

Since m± satisfies a± = kC ⊕m± we know from Lemma 4.4 that

m± = {X ∈ mC | I0(X) = ±iX }.

Let Eα, for α ∈ ∆, be the generators for the gα’s, which satisfies

[Eα, E−α] = Hα, θ(Eα) = −E−α.

We have that I0(Eα) = εαEα, where

εα =

{i if α ∈ ∆+

−i if α ∈ ∆−

39

Let α, β ∈ ∆. By definition,

ω0(Eα, Eβ) = 〈I0(Eα), Eβ〉 = εα〈Eα, Eβ〉. (4.12)

However we showed earlier that ω0(X,Y ) = B(Zφ, [X,Y ]) for X,Y ∈ mC, so

ω0(Eα, Eβ) = B(Zφ, [Eα, Eβ ]).

Recall from Chapter 1 that

B(gα, gβ) = 0 if α+ β 6= 0 for α, β ∈ R ∪ {0}.

We also know from Chapter 1 that [Eα, Eβ ] ∈ gα+β . Since Zφ ∈ t ⊂ hC = g0, we have that

B(Zφ, [Eα, Eβ ]) = 0 if α+ β 6= 0,

which implies that ω0(Eα, Eβ) = 0 if β 6= −α. Let us now consider the case when β = −α. In thiscase

ω0(Eα, E−α) = B(Zφ, [Eα, E−α])= B(Zφ, Hα)= α(Zφ).

We have now shown the following:

ω0(Eα, Eβ) =

{α(Zφ) if β = −α0 if β 6= −α

(4.13)

From (4.12) we get that

〈Eα, Eβ〉 =1εαω0(Eα, Eβ).

We combine this with (4.13) to obtain the following:

〈Eα, Eβ〉 =

{1εαα(Zφ) if β = −α

0 if β 6= −α.(4.14)

By assumption 〈, 〉 is a scalar product on m, so if we let

〈X,Y 〉C = 〈X, θ(Y )〉

for X,Y ∈ mC, then 〈, 〉C will be an inner product on mC. Let α ∈ ∆+. Then Eα 6= 0, so

〈Eα, Eα〉C > 0.

If we recall that θ(Eα) = −E−α, we see that

−〈Eα, E−α〉 = 〈Eα, θ(Eα)〉 = 〈Eα, Eα〉C > 0.

Using (4.14) we get

− 1εαα(Zφ) > 0, (4.15)

since we assumed α ∈ ∆+ we know that εα = i. So (4.15) is equivalent to

iα(Zφ) > 0

and since α ∈ ∆+ was chosen arbitrarily, this holds for all α ∈ ∆+. We know from Proposition4.20 that Zφ ∈ t and we just showed that iα(Zφ) > 0 for all α ∈ ∆+. This implies that Zφ belongsto some Weyl chamber. Because of the inequalities iα(Zφ) > 0 for all α ∈ ∆+, it is clear that thisWeyl chamber must be precisely the one that corresponds to ∆+.

40

Suppose now that we are only given a generalized flag manifold M with a G-invariant complexstructure J . We would then like to show that there exists a G-invariant Kahler metric on M . Inorder to do this we will construct an Ad(K)-invariant inner product 〈, 〉 on m which satisfies

〈I0(X), I0(Y )〉 = 〈X,Y 〉

for all X,Y ∈ m and such that the corresponding 2-form ω0, is closed i.e. satisfies (4.7). By ourprevious results, this implies the existence of a G-invariant Kahler metric on M .

We continue to let ∆+ be the subset of ∆ = R\RK which corresponds to J and we also letmC, m and m± be defined as before. Let ZJ be any element in the Weyl chamber correspondingto ∆+. By using that m = {X ∈ mC | θ(X) = X}, we get that every X ∈ m can be written as

X =∑α∈∆+

aαEα − aαE−α

where aα ∈ C. We begin by defining a bilinear form 〈, 〉 on the basis elements Eα, for α, β ∈ ∆.Bases on our results from Theorem 4.21 we define 〈, 〉 in the following way:

〈Eα, Eβ〉 =

{1ε αα(ZJ) if β = −α

0 if β 6= −α,

where εα = ±i for α ∈ ∆±.

Proposition 4.22. [4] The bilinear form 〈, 〉 is an inner product on m.

Proof. We begin by noting that

〈Eα, E−α〉 =1εαα(ZJ) ∈ R

for all α ∈ ∆, since iα(ZJ) > 0 for all α ∈ ∆+ and iα(ZJ) < 0 for all α ∈ ∆−. This is because wechose ZJ to be in the Weyl chamber corresponding to ∆+. Moreover,

〈E−α, Eα〉 =1ε−α

(−α(ZJ)) =1εαα(ZJ) = 〈Eα, E−α〉,

where we used that −ε−α = εα. Next we let

X =∑α∈∆+

aαEα − aαE−α, Y =∑β∈∆+

bβEβ − bβE−β

be two elements in m. Then

〈X,Y 〉 = 〈∑α∈∆+

aαEα − aαE−α,∑β∈∆+

bβEβ − bβE−β〉

=∑α∈∆+

(−aαbα〈Eα, E−α〉 − aαbα〈E−α, Eα〉)

=∑α∈∆+

−(aαbα + aαbα)〈Eα, E−α〉

=∑α∈∆+

−(aαbα + aαbα)1εαα(ZJ)

=∑α∈∆+

(aαbα + aαbα)iα(ZJ).

First we note thataαbα + aαbα = aαbα + aαbα

41

which implies that aαbα + aαbα ∈ R for all α ∈ ∆+ and we already know that iα(ZJ) ∈ R. Itfollows that 〈, 〉 is real valued on m×m. If we let Y = X we see that

〈X,X〉 =∑α∈∆+

(aαaα + aαaα)iα(ZJ)

=∑α∈∆+

2|aα|2iα(ZJ).

It is clear that |aα|2 ≥ 0 and iα(ZJ) > 0 for all α ∈ ∆+. Hence

〈X,X〉 ≥ 0

with equality if and only if aα = 0 for all α ∈ ∆+, which is equivalent to X = 0. So 〈, 〉 is bilinear,positive definite and it is obvious that it is symmetric, since 〈Eα, Eβ〉 = 〈Eβ , Eα〉, for all α, β ∈ ∆.So 〈, 〉 is an inner product on m.

We directly move on with the next proposition.

Proposition 4.23. [4] The inner product 〈, 〉 satisfies

〈I0(X), I0(Y ))〉 = 〈X,Y 〉

for all X,Y ∈ m.

Proof. We know that I0(Eα) = ±iEα, depending on whether α ∈ ∆±. So for β 6= −α

〈I0(Eα), I0(Eβ)〉 = 〈±iEα,±iEβ〉 = 0

and for β = −α we have

〈I0(Eα), I0(E−α)〉 = 〈±iEα,∓iE−α〉 = 〈Eα, E−α〉.

This means that 〈I0(X), I0(Y )〉 = 〈X,Y 〉 whenever X = Eα, Y = Eβ , for α, β ∈ ∆. Let

X =∑α∈∆

aαEα, Y =∑β∈∆

bβEα

be two elements in m. Then

〈I0(X), I0(Y )〉 =∑α,β∈∆

aαbβ〈I0(Eα), I0(Eβ)〉

=∑α,β∈∆

aαbβ〈Eα, Eβ〉

= 〈X,Y 〉.

We proceed by showing the Ad(K)-invariance.

Proposition 4.24. [4] The inner product 〈, 〉 is Ad(K)-invariant.

Proof. We begin by noting that for α ∈ ∆ we have

〈Eα, E−α〉 =1εαα(ZJ)

= −εαB(Hα, ZJ)= −εαB([Eα, E−α], ZJ)= −B([εαEα, E−α], ZJ)= −B([I0(Eα), E−α], ZJ).

42

Recall that B(gα, gβ) = 0 for α, β ∈ R ∪ {0} such that α+ β 6= 0. This means that for β 6= −α wehave that

−B([I0(Eα), Eβ ], ZJ) = −εαB([Eα, Eβ ], ZJ) = 0 = 〈Eα, Eβ〉,

where we also used that [Eα, Eβ ] ∈ gα+β and that ZJ ∈ t ⊂ hC = g0. These calculations implythat

〈Eα, Eβ〉 = −B([I0(Eα), Eβ ], ZJ).

For all α, β ∈ ∆. Now letX =

∑α∈∆

aαEα, Y =∑β∈∆

bβEβ

be two elements in m. Then

−B([I0(X), Y ], ZJ) = −B([I0(∑α∈∆

aαEα),∑β∈∆

bβEβ ], ZJ)

=∑α,β∈∆

aαbβ(−B([I0(Eα), Eβ ], ZJ))

=∑α,β∈∆

aαbβ〈Eα, Eβ〉

= 〈X,Y 〉.

Hence we have that〈X,Y 〉 = −B([I0(X), Y ], ZJ)

for all X,Y ∈ m.By construction ZJ belongs to the Weyl chamber corresponding to ∆+, so in particular

ZJ ∈ t = Z(kC) ∩ h.

It is easy to see that Z(k) is the Lie algebra of C(K) and from this we get that exp(tZJ) ∈ C(K)for all t ∈ R. Using this we see that for k ∈ K

Ad(k)(ZJ) =d

dt(Ik(exp(tZJ)))|t=0 =

d

dt(exp(tZJ))|t=0 = ZJ .

We also know that I0 is Ad(K)-invariant, that is I0(Ad(k)(X)) = Ad(k)(I0(X)), for all k ∈ K andall X ∈ m. Recall that Ad(k) is a Lie algebra automorphism for all k ∈ K. If we combine thiswith Lemma 1.12, the Ad(K)-invariance of I0 and the fact that Ad(k)(ZJ) = ZJ for all k ∈ K, weobtain the following identity:

B([I0(Ad(k)(X)),Ad(k)(Y )], ZJ) = B([Ad(k)(I0(X)),Ad(k)(Y )], ZJ)= B(Ad(k)([I0(X), Y ]), ZJ)

= B([I0(X), Y ],Ad(k−1)(ZJ))= B([I0(X), Y ], ZJ),

for all k ∈ K and all X,Y ∈ m. But this implies that

〈Ad(k)(X),Ad(k)(Y )〉 = −B([I0(Ad(k)(X)),Ad(k)(Y )], ZJ)= −B([I0(X), Y ], ZJ)= 〈X,Y 〉

for all k ∈ K and all X,Y ∈ m. This proves that the inner product 〈, 〉 is Ad(K)-invariant.

Finally we will show that ω0 satisfies (4.7). In order to do this we need an additional result.Recall the number Nα,β defined for α, β ∈ R such that α+ β 6= 0 by

[Eα, Eβ ] = Nα,βEα+β

43

Lemma 4.25. [9] If α, β, γ ∈ R satisfies α+ β + γ = 0 then

Nα,β = Nβ,γ = Nγ,α.

Proof. [9] Using the Jacobi identity and the assumption that α+ β + γ = 0, we see that

0 =[Eα, [Eβ , Eγ ]] + [Eγ , [Eα, Eβ ]] + [Eβ , [Eγ , Eα]]= [Eα, Nβ,γEβ+γ ] + [Eγ , Nα,βEα+β ] + [Eβ , Nγ,αEγ+α]= Nβ,γHα +Nα,βHγ +Nγ,αHβ

SoNβ,γHα +Nα,βHγ +Nγ,αHβ = 0 (4.16)

By definition Hγ is the unique element in hC such that B(Hγ , H) = γ(H) for all H ∈ hC. But

γ(H) = −(α+ β)(H) = −B(Hα, H)− B(Hβ , H) = B(−(Hα +Hβ), H).

By uniqueness we must therefore have that −Hγ = Hα +Hβ . If we use this in (4.16) we get

(Nβ,γ −Nα,β)Hα + (Nγ,α −Nα,β)Hβ = 0.

Suppose that the vectors Hα and Hβ are not linearly independent. Then there exists some complexnumber c such that Hβ = cHα. Hence

β(H) = B(Hβ , H) = B(cHα, H) = cα(H)

for all H ∈ hC and therefore β = cα. But cα /∈ R unless c = −1, 0, 1. If c = 1 then α = β so wehave 2α + γ = 0 or equivalently γ = −2α. But −2α /∈ R so c 6= 1. If c = 0 we get β = 0 and 0 isnot a root so c 6= 0. If c = −1 we get γ = 0 which is also a contradiction. So there cannot existsuch a number c. This means that

(Nβ,γ −Nα,β)Hα + (Nγ,α −Nα,β)Hβ = 0

if and only ifNβ,γ = Nα,β = Nγ,α.

Proposition 4.26. [4] The bilinear form ω0 is closed.

Proof. Let α, β, γ ∈ ∆ and consider

ω0([Eα, Eβ ]m, Eγ) + ω0([Eβ , Eγ ]m, Eα) + ω0([Eγ , Eα]m, Eβ). (4.17)

Suppose first that α+ β /∈ R. Then [Eα, Eβ ] = 0 or [Eα, Eβ ] = Hα. In either case [Eα, Eβ ]m = 0,which implies that ω0([Eα, Eβ ]m, Eγ) = 0.

We now suppose that α+ β ∈ R. In the case when α+ β ∈ RK we clearly have that

[Eα, Eβ ]m = Nα,β(Eα+β)m = 0.

If α+ β ∈ ∆ we instead get that

ω0([Eα, Eβ ]m, Eγ) = ω0(Nα,β(Eα+β)m, Eγ)= ω0(Nα,βEα+β , Eγ)= 〈Nα,βI0(Eα+β), Eγ〉= 〈Nα,βεα+βEα+β , Eγ〉

=

{Nα,β(α+ β)(ZJ) if α+ β + γ = 00 otherwise .

44

We cannot have that α+ β + γ = 0 when α+ β /∈ ∆, because this would imply that γ /∈ ∆. If wecombine this with the fact that ω0([Eα, Eβ ]m, Eγ) = 0 when α+ β /∈ ∆, we get that

ω0([Eα, Eβ ]m, Eγ) =

{Nα,β(α+ β)(ZJ) if α+ β + γ = 00 otherwise

for all α, β, γ ∈ ∆.Clearly the other two terms in (4.17) can also be expressed in this way.

ω0([Eβ , Eγ ]m, Eα) =

{Nβ,γ(β + γ)(ZJ) if α+ β + γ = 00 otherwise

ω0([Eγ , Eα]m, Eβ) =

{Nγ,α(γ + α)(ZJ) if α+ β + γ = 00 otherwise

.

It then follows that (4.17) vanishes when α + β + γ 6= 0. We now assume α + β + γ = 0. In thiscase we know from Lemma 4.25 that Nα,β = Nβ,γ = Nγ,α. Using this we see that

ω0([Eα, Eβ ]m, Eγ) + ω0([Eβ , Eγ ]m, Eα) + ω0([Eγ , Eα]m, Eβ)= Nα,β(α+ β)(ZJ) +Nβ+γ(β + γ)(ZJ) +Nγ,α(γ + α)(ZJ)= Nα,β(α+ β + β + γ + γ + α)(ZJ)= Nα,β((α+ β + γ)(ZJ) + (α+ β + γ)(ZJ))= 0,

since α+ β + γ = 0.We now consider the general case, so let

X =∑α∈∆

aαEα, Y =∑β∈∆

bβEβ , Z =∑γ∈∆

cγEγ

be elements in m. Then[X,Y ]m =

∑α,β∈∆

aαbβ [Eα, Eβ ]m

so

ω0([X,Y ]m, Z) = ω0(∑α,β∈∆

aαbβ [Eα, Eβ ]m,∑γ∈∆

cγEγ)

=∑

α,β,γ∈∆

aαbβcγω0([Eα, Eβ ]m, Eγ).

In the same way we get

ω0([Y,Z]m, X) =∑

α,β,γ∈∆

aαbβcγω0([Eβ , Eγ ]m, Eα)

ω0([Z,X]m, Y ) =∑

α,β,γ∈∆

aαbβcγω0([Eγ , Eα]m, Eβ).

This means that

ω0([X,Y ]m, Z) + ω0([Y,Z]m, X) + ω0([Z,X]m, Y )

=∑

α,β,γ∈∆

aαbβcγω0([Eα, Eβ ]m, Eγ) +∑

α,β,γ∈∆

aαbβcγω0([Eβ , Eγ ]m, Eα)

+∑

α,β,γ∈∆

aαbβcγω0([Eγ , Eα]m, Eβ)

=∑

α,β,γ∈∆

aαbαcα(ω0([Eα, Eβ ]m, Eγ) + ω0([Eβ , Eγ ]m, Eα) + ω0([Eγ , Eα]m, Eβ))

45

= 0.

Since X,Y, Z were chosen arbitrarily, this concludes the proof.

Combining these propositions with our earlier results, we may prove the following theorem.

Theorem 4.27. [1], [4] Given a generalized flag manifold M with a G-invariant complex structureJ there exists a G-invariant Kahler metric on M .

Proof. By Propositions 4.22, 4.23, 4.24 and 4.26 the scalar product 〈, 〉 is Ad(K)-invariant, satisfies〈I0(X), I0(Y )〉 = 〈X,Y 〉 for all X,Y ∈ m and the corresponding bilinear form ω0 is closed. ByPropositions 2.12, 3.13 and 3.17 this implies the existence of a G-invariant Kahler metric on M .

Remark 4.28. Note that this metric is not uniquely defined since ZJ was chosen arbitrarily fromthe Weyl chamber corresponding to J . So we could have chosen another element WJ in this Weylchamber and we would have then ended up with a different metric.

46

Appendix A

Differentiable Manifolds

Let M be a differentiable manifold. Following [8] we will always assume that M is Hausdorff andsecond-countable.

Lemma A.1. [14] Let M and N be two differentiable manifolds. If ϕ : M → N is a smooth,bijective and everywhere non-singular map, then ϕ is a diffeomorphism.

Proof. [14] Since ϕ is assumed to be smooth, bijective and everywhere non-singular we see thatϕ is a diffeomorphism if and only if dϕx is surjective for all x ∈ M . Assume for a contradictionthat this is not the case, so there exists some x ∈M such that dϕx is not surjective. Since ϕ wasassumed to be everywhere non-singular this implies

m = dim(M) < dim(N) = n.

Let (V, y) be a chart on N such that y(V ) = Rn. Let {(Ui, xi) | i = 1, 2 . . .} be a countable atlason M . Since M is assumed to be second-countable it is possible to find such an atlas.

Consider the function y ◦ ϕ. This function has range Rn since ϕ was assumed to be surjective.For each i xi(Ui) is an open subset of Rm. Let λn be the Lebesgue measure on Rn. Then, sincem < n and xi(Ui) ⊂ Rm ⊂ Rn, it follows that λn(xi(Ui)) = 0. Since ϕ was assumed to be smooth,the map

y ◦ ϕ ◦ x−1i : xi(Ui) ⊂ Rn → Rn

is smooth. Since smooth functions from Rn to Rn takes sets of measure 0 to sets of measure 0, wehave

λn((y ◦ ϕ ◦ x−1i )(xi(Ui))) = 0.

Since M = ∪∞n=1Ui we have that

Rn = (y ◦ ϕ)(M)= ∪∞n=1(y ◦ ϕ)(Ui)

= ∪∞n=1(y ◦ ϕ ◦ x−1i )(xi(Ui)).

But this implies that

λn(Rn) = λn(∪∞n=1(y ◦ ϕ ◦ x−1i )(xi(Ui))) ≤

∞∑n=1

λn((y ◦ ϕ ◦ x−1i )(xi(Ui))) = 0

which is clearly false.

47

48

Bibliography

[1] Alekseevsky, D.V. and Perelomov, A.M. Invariant Kahler-Einstein metrics on compact ho-mogeneous spaces, Funktsional. Anal. i Prilozhen., 20(3); Engl. transl. in Funct. Anal. Appl.,20(3), 171–182, (1986).

[2] A. Arvanitoyeorgos, An Introduction to Lie Groups and the Geometry of Homogeneous Spaces,Student Mathematical Library 22, AMS (2003).

[3] A. Arvanitoyeorgos, New Invariant Einstein Metrics On Generalized Flag Manifolds, Trans.Amer. Math. Soc. 337 (1993), pp. 981–995.

[4] M. Bordemann, M. Forger, H. Romer, Homogeneous Kahler manifolds: paving the way towardssupersymmetric sigma-models, Commun. Math. Phys. 102 (1986), 605–647.

[5] A. Borel and F. Hirzebruch, Characteristic classes and homogeneous spaces I, Amer. J. Math.80 (1958), 458-538.

[6] D. Bump Lie groups, Graduate Texts in Mathematics, vol. 225. Springer, (2004)

[7] A. Frohlicher, Zur DifferentialGeometrie der Komplexen Structuren, Math. Ann. 129 (1955),50-95.

[8] S. Gudmundsson, An Introduction to Riemannian Geometry, Lund University (2012).http://www.matematik.lu.se/matematiklu/personal/sigma/Riemann.pdf

[9] S. Helgason, Differential Geometry, Lie Groups and Symmetric Spaces, Academic Press(1978).

[10] A W. Knapp, Lie Groups Beyond an Introduction, Birkhauser (2002).

[11] S. Kobayashi, K. Nomizu, Foundations of Differential Geometry vol. II, John Wiley and Sons(1969).

[12] M. Nishiyama, Classification of invariant complex structures on irreducible compact simplyconnected coset spaces, Osaka J. Math. 21 (1984), 39-58.

[13] H. C. Wang, Closed manifolds with homogeneous complex structure, Amer. J. Math, vol. 76(1954), pp. 1-32.

[14] F. Warner, Foundations of Differential Manifolds and Lie Groups, Springer-Verlag, New York(1983).

[15] K. Yang, Almost complex homogeneous spaces and their submanifolds, World Scientific (1987).

[16] K. Yano, M. Kon, Structures on Manifolds, Series in Pure Mathematics 3, World Scientific(1984).

49

50

Master’s Theses in Mathematical Sciences 2013:E17ISSN 1404-6342

LUNFMA-3077-2013

MathematicsCentre for Mathematical Sciences

Lund UniversityBox 118, SE-221 00 Lund, Sweden

http://www.maths.lth.se/