On the Geometry of Homogeneous Spaces of Positive Curvature · On the Geometry of Homogeneous...

On the Geometry of Homogeneous

Spaces of Positive Curvature

Henrik Ekström

Master's thesis

2018:E8

Faculty of Science

Centre for Mathematical Sciences

Mathematics

CENTRUMSCIENTIARUMMATHEMATICARUM

Abstract

This Master’s thesis is a study of some of the geometric properties of Riemannianhomogeneous spaces. These are Riemannian manifolds M equipped with a transitivegroup of isometries G, meaning that the local geometry of the manifold is the sameat every point.

In Section 1.3 we see that such spaces are diffeomorphic to the quotient G/K,where G is a Lie group and K is the isotropy group at the identity element. Inthese cases the Lie algebra g of G has a reductive decomposition g = k⊕m, where kis the Lie algebra of K. This allows us to employ the classical curvature theory ofRiemannian submersions, developed by O’Neill in 1966. The relevant parts of thisare presented and derived in Section 1.1.

The geometric feature that we focus on are the sectional curvatures. Thesedescribe how curved the space is in some 2-dimensional subspace of the tangentspace at a given point. We are interested in those homogeneous spaces that havestrictly positive sectional curvatures.

An important property of such a space is the so called ’pinching constant’ i.e. thequotient of the maximal and minimal positive sectional curvatures. For the roundsphere this is 1 and the smaller it is, the more different the space is from being sucha sphere.

The Riemannian homogeneous spaces with positive curvature are completely clas-sified. In this thesis we will focus on the ones that are ’normal’. There exist elevendifferent such spaces and the sectional curvatures have, to our knowledge, been ex-plicitly calculated by others for four of them. We derive explicit formulae for thesectional curvature for six of the spaces in Chapter 3.

Throughout this work it has been my firm intention to give reference to the statedresults and credit to the work of others. All theorems, propositions, lemmas andexamples left unmarked are assumed to be too well known for a reference to be given.

Acknowledgements

I would like to thank my supervisor Sigmundur Gudmundsson for suggesting thisinteresting topic and for his many helpful comments whilst writing this thesis.

Henrik Ekstrom

Contents

1 Introduction 11.1 O’Neill’s Formulae for Riemannian Submersions . . . . . . . . . . . . 11.2 Basic Lie Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 Homogeneous Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.3.1 Normal Riemannian Homogeneous Spaces . . . . . . . . . . . 71.3.2 Riemannian Symmetric Spaces . . . . . . . . . . . . . . . . . . 9

2 The Normal Homogeneous Spaces of Positive Curvature 112.1 Pinching Constants and The Sphere Theorem . . . . . . . . . . . . . 11

3 Curvature Calculations 133.1 Useful Formulae . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133.2 Rank One Normal Symmetric Spaces of Positive Curvature . . . . . . 15

3.2.1 The Sphere Sn ∼= SO(n+ 1)/SO(n) . . . . . . . . . . . . . . . 153.2.2 The Complex Projective Space CP n ∼= U(n+ 1)/(U(1)×U(n)) 163.2.3 The Quaternionic Projective Space HP n ∼= Sp(n+1)/(Sp(n)×

Sp(1)) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183.3 Non-Symmetric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3.3.1 The Homogeneous Space SU(n+ 1)/SU(n) . . . . . . . . . . 223.3.2 The Homogeneous Space Sp(n+ 1)/Sp(n) . . . . . . . . . . . 253.3.3 The Homogeneous Space Sp(n+ 1)/(S1 × Sp(n)) . . . . . . . 28

3.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

Chapter 1

Introduction

The following thesis assumes the reader to be familiar with basic concepts fromRiemannian geometry. For an introduction to this interesting field of mathematicswe recommend the text [8].

1.1 O’Neill’s Formulae for Riemannian Submersions

In his very influential paper [13] from 1966, Barrett O’Neill not only defined Rie-mannian submersions but also derived their fundamental equations. This will beused extensively in this thesis, particularly in the setting of quotient groups. Wewill need the following definitions.

Definition 1.1. Let π : M → B be a differentiable map between differential man-ifolds M and B of dimensions m and b, with m ≥ b. Then π is a submersion if itsdifferential dπp is surjective for all points p ∈M .

Definition 1.2. Let π : (M, g)→ (B, h) be a smooth submersion between Rieman-nian manifolds. Then the vertical bundle V = Ker(dπ) is the kernel of the differentialdπ. The horizontal bundle is the orthogonal complement to V , i.e. H = (V)⊥.

At each point p ∈ M , we have TpM = Vp ⊕Hp. A vector field on a manifold Mis said to be vertical (or horizontal) if it is tangent (or orthogonal) to fibres π−1(b),for all b ∈ B. We also denote the projections of vector fields in C∞(TM) to thevertical and horizontal bundles by V and H, respectively.

Definition 1.3. Let (M, g) and (B, h) be Riemannian manifolds and p be a pointin M . A Riemannian submersion π : M → B is a mapping with a differential dπthat satisfies

i) dπ : TpM → Tπ(p)B is surjective for all p ∈M,

ii) dπ preserves lengths of horizontal vectors, i.e.g(X, Y ) = h(dπ(X), dπ(Y )) for all horizontal X and Y .

From now on we restrict ourselves to Riemannian submersions. O’Neill definestwo fundamental tensors describing submersions. The first, T , is basically the secondfundamental form of the fibres and the second, A, is the dual of T . For arbitraryvector fields E,F ∈ C∞(TM) on M , the tensors T and A are defined as

TEF = H∇VE(VF ) + V∇VE(HF ),

1

AEF = V∇HE(HF ) +H∇HE(VF ),

where ∇ is the Levi-Civita connection on (M, g).

Definition 1.4. A vector field X ∈ C∞(TM) on M is said to be basic if it ishorizontal and π-related to a vector field X on B i.e. dπ(X) = X.

The notion of basic vector fields is very useful, since there is a one-to-one corre-spondence between vector fields on B and basic vector fields on M .

Lemma 1.5. [13] For basic vector fields X, Y on (M, g) we have the following.

i) 〈X, Y 〉 = 〈X, Y 〉 ◦ π,

ii) H[X, Y ] is the basic vector field corresponding to [X, Y ],

iii) H∇XY is the basic vector field corresponding to ∇X Y .

Proof. [13] For all points p ∈ M , we have 〈X, Y 〉p = 〈dπ(X), dπ(Y )〉π(p) wheneverX is basic. It is apparent that dπ([X, Y ]) = [X, Y ] so the horizontal part of [X, Y ]must be the corresponding basic vector field. Similarly, using the first and secondstatements we see that for any horizontal vector field Z ∈ C∞(TM) we get

2〈∇XY,Z〉 = X〈Y, Z〉+Y〈Z,X〉−Z〈X, Y 〉−〈X, [Y, Z]〉+〈Y, [Z,X]〉+〈Z, [X, Y ]〉=(X〈Y , Z〉+ Y 〈Z, X〉 − Z〈X, Y 〉− 〈X, [Y , Z]〉+ 〈Y , [Z, X]〉+ 〈Z, [X, Y ]〉

)◦ π

= 2〈∇X Y , Z〉.

Lemma 1.6. [13] Let π : (M, g) → (B, h) be a Riemannian submersion. Then forbasic vector fields X, Y and a vertical vector field V on M , we have

i) AXY = −AYX,

ii) AXY = 12V [X, Y ],

iii) AXV = H∇XV = H∇VX.

Proof. [13] We know that A is vertical for two horizontal arguments, so for i) it isenough to show that 〈AXX, V 〉 = 〈∇XX, V 〉 = 0 for an arbitrary vertical V . Theproduct rule gives

〈∇XX, V 〉 = ∇X〈X, V 〉 − 〈X,∇XV 〉,

where the first term is zero and also the second, since the horizontal part of ∇XV iszero, as seen from dπ([X, V ]) = [dπ(X), dπ(V )] and dπ(V ) = 0.

With X, Y horizontal, the definition of A boils down to AXY = V∇XY , so ii)follows from

V [X, Y ] = V(∇XY −∇YX

)= AXY − AYX = 2AXY.

The third statement follows from the Koszul formula by considering

〈∇VX,E〉 − 〈∇XV,E〉 = 〈[X, V ], E〉,

which is zero if E is horizontal, since [X, V ] is vertical.

2

The following interesting result can be seen as some sort of dual to the famousGauss Equation, see [8].

Proposition 1.7. [13] For horizontal vector fields X, Y, Z,W , we have

g(R(X, Y )Z,W ) = h(R(X, Y )Z, W )− 2g(AXY,AWZ)

− g(AYZ,AXW )− g(AZX,AYW ).

Proof. Since we are dealing with tensors, we can assume that X, Y , and Z are basicwith vertical Lie brackets. We know that

R(X, Y )Z = ∇X∇YZ −∇Y∇XZ −∇[X, Y ]Z. (1.1)

Using the definition of A and noting that with horizontal arguments and H∇YZ =∇Y Z, we get

∇YZ = V∇YZ +H∇YZ = AYZ + ∇Y Z.

Furthermore,

∇X∇YZ = (V +H)∇X(AYZ + ∇Y Z

)= V∇X (AYZ) + AX∇Y Z + AX (AYZ) + ∇X∇Y Z.

The last term, together with the corresponding one from∇Y∇XZ, make up R(X, Y )Z.This since we let [X, Y ] be vertical, which implies

∇[X, Y ]Z = 0,

leaving us withR(X, Y )Z = ∇X∇Y Z − ∇Y ∇X Z.

Then, by using ii) and iii) of Lemma 1.6, we can write the remaining term from(1.1) as

∇[X, Y ]Z = H∇2AXYZ + V∇2AXYZ

= 2AZ (AXY ) + 2TAXYZ.

Now we combine all three equations, collect the horizontal terms and use i) ofLemma 1.6 to get

HR(X, Y )Z = R(X, Y )Z − 2AZ (AXY ) + AX (AYZ)− AY (AXZ)

g(R(X, Y )Z,H) = h(R(X, Y )Z, H)− 2g(AXY,AHZ)

− g(AYZ,AXH)− g(AXZ,AHY ).

Now we state the main theorem of this section: a comparison of the sectionalcurvatures of M and B.

Theorem 1.8. Let π : (M, g) → (B, h) be a Riemannian submersion and X, Y behorizontal vector fields on (M, g). With π(X) = X the sectional curvatures K of Mand K of B satisfy the following.

K(X, Y ) = K(X, Y )− 3|AXY |2

|X|2|Y |2 − g(X, Y )2. (1.2)

3

Proof. Using Proposition 1.7 we write

g(R(X, Y )Y,X) = h(R(X, Y )Y , X)− 2g(AXY,AXY )

+ g(AY Y,AXX)− g(AXY,AXY )

= h(R(X, Y )Y , X)− 3|AXY |2.

1.2 Basic Lie Theory

From now on this thesis will heavily depend on Lie theory, so we first go throughsome basics.

Definition 1.9. [11] Let g be the Lie algebra of G and a be a vector subspace of g.a is called a subalgebra of g if [a, a] ⊆ a.

Theorem 1.10. [11] If G is a Lie group with Lie algebra g, K is a Lie subgroup ofG with Lie algebra k, then k is a subalgebra of g.

Definition 1.11. For a Lie group G with Lie algebra g and a, b ∈ G, we define themap Ia : G→ G by

Ia : b 7→ aba−1

and view this as a map I : G→ AutG to the group of inner automorphisms of G

I : a 7→ Ia.

Its differential at the identity element is the map denoted by Ada : g→ g with

Ada = (dIa)e.

The adjoint representation of G is the above viewed as a homomorphism from G toAut g (the vector space of automorphisms on g), Ad : a 7→ Ada. If a matrix repre-sentation is used this can be expressed explicitly by using the matrix exponential.

AdaX =(dIa)eexp (tX)

=d

dta exp (tX)a−1

∣∣t=0

= aXa−1. (1.3)

At the Lie algebra level, the adjoint action of X on g is the Lie algebra homomor-phism from g to Aut g

adX : Y 7→ [X, Y ].

It can be shown that adX = d(Ad)e. With a matrix representation, the operator [, ]is the commutator

[X, Y ] = XY − Y X.

The adjoint action and representation are useful tools for transferring informationbetween the Lie group G and its Lie algebra g. Typically problems are simpler tosolve on the Lie algebra level.

4

1.3 Homogeneous Spaces

Intuitively, a Riemannian homogeneous space has the property that its geometry is’the same’ at each point i.e. geometric properties (e.g. curvature) of one point canbe extended to the whole space.

Irreducible homogeneous spaces can be of three different types: compact, Eu-clidean or non-compact. Their sectional curvatures are non-negative, zero, andnon-positive, respectively as described in the book [11]. We are only interested inthe compact case which arises from the use of compact groups.

In order to define Riemannian homogeneous spaces we will need the followingdefinitions.

Definition 1.12. A map φ : (M, g)→ (B, h) between Riemannian manifolds is saidto be isometric if for all vector fields X, Y ∈ C∞(TM) and points p ∈M we have

gp(Xp, Yp) = hφ(p)(dφp(Xp), dφp(Yp)

).

If φ also is a diffeomorphism, it is called an isometry.

Definition 1.13. A Riemannian metric g of a Lie group is called left-invariant ifthe left translations La : G→ G are isometries for all a ∈ G i.e. if

gb(X, Y ) = gLa(b)

((dLa)bX, (dLa)bY

)for all a, b ∈ G and X, Y ∈ TbG.

Definition 1.14. Let G be a Lie group and M be a differentiable manifold. A groupaction of G on M is a smooth map α : G×M →M such that

i) α(e, p) = p for all p ∈M,

ii) α(ab, p) = α(a, α(b, p)

)for all a, b ∈ G.

The action α : G×M →M is said to be transitive if for all p, q ∈M there exists anelement a ∈ G such that α(a, p) = q. The action is said to be effective if α(a, p) = pfor all p ∈M implies that a = e i.e. if the only element of G acting trivially on M isthe identity element e. A group action is said to be isometric if α(a, ·) is isometricfor all a ∈ G.

From now on we will assume the action to be left-translation i.e. α(g, p) = gp.After these preparations we are ready to define homogeneous spaces.

Definition 1.15. A homogeneous space is a differentiable manifold M together witha transitive action of a Lie group G.

We now turn our attention to coset spaces, which will play a major role in therest of this work. They allow for an alternative description of homogeneous spacesthat is most useful.

Definition 1.16. Let G be a group and K a subgroup of G. A coset space is theset of left cosets of K in G

G/K = {gK | g ∈ G}.

5

Theorem 1.17. [16] A coset space G/K of a Lie group G and a closed subgroup Khas a unique manifold structure such that the projection π : G→ G/K is a smoothsubmersion.

We assume that any coset space G/K mentioned henceforth is of the above typeand is endowed with this differentiable structure. We will refer to it as a cosetmanifold. On a coset manifold, the action of G by left-translation

α : G×G/K → G/K; (a, bK) 7→ abK,

is transitive. This is true, since given b1K, b2K ∈ G/K, the element a = b2b−11 gives

α(a, b1K) = b2K.

Definition 1.18. Let G be a Lie group acting transitively on a manifold M . Thenthe isotropy group Kp of a point p ∈M consists of the elements of G that fix p whenacting on M i.e.

Kp = {g ∈ G | gp = p} .

Theorem 1.19. [16],[11] Let G be a Lie group acting transitively on a smooth man-ifold M . Then the isotropy group Kp, fixing a point p in M , is a closed subgroup ofG and the map α : G/Kp →M defined by

α : gKp 7→ g · p

is a diffeomorphism. If M is connected, then the identity component of G actstransitively on M .

This means that rather than studying the pair (M, p) we may study (G,Kp),which is often simpler. We now define reductivity, a concept that connects homoge-neous spaces with the submersions discussed in Section 1.1.

Definition 1.20. Let G be a Lie group and K be a closed subgroup with Liealgebras g and k, respectively. A homogeneous space is said to be reductive if thereexists a complementary subspace m of k in g that is Ad(K)-invariant i.e. g = k⊕mwith Adk(m) ⊆ m for all k ∈ K.

The condition of m being Ad(K)-invariant implies that [k,m] ⊆ m and the con-verse statement is true if K is connected, see [2].

If π : G → G/K is the projection from G into a homogeneous space G/K andX ∈ g, one can see by

dπe(X) =d

dtπ(exp tX)

∣∣∣∣t=0

=d

dt

((exp tX)K

)∣∣∣∣t=0

that Ker(dπe) = k, so we get the following.

Proposition 1.21. The decomposition g = k⊕m corresponds to the decompositionof vectors in g into vertical and horizontal vectors (in k and m, respectively) discussedin section 1.1.

So far we have been discussing homogeneous spaces. We now define a metric toturn them into Riemannian homogeneous spaces. We need the following result.

Proposition 1.22. [2] There is a one-to-one correspondence between left-invariantmetrics on a Lie group G and the scalar products on its Lie algebra g.

6

Proof. Given a left-invariant metric g onG, the function g(X, Y ) : G→ R is constanton G for X, Y ∈ g. For such left-invariant vector fields, we have

ga(X, Y ) = g(Xa, Ya) = g(dLaXe, dLaYe) = g(Xe, Ye) = ge(X, Y )

and ge(, ) is a scalar product on g. Conversely, given ge(, ) on g, a left-invariantmetric on G can be defined by

ga(X, Y ) = ge(dLa−1X, dLa−1Y ).

Since the projection π : G→ G/K is surjective and k is the kernel of dπe, we getthe isomorphism g/k ∼= TeKG/K. In the reductive case, we can therefore identify mwith the tangent space of the quotient manifold m ∼= TeKG/K, see [2].

Definition 1.23. [2] Let (G/K, g) be a homogeneous space. A metric is said tobe G-invariant if the function τa : G/K → G/K defined by τa(bK) = abK (theanalogue of left-translation) is an isometry for all a ∈ G i.e.

g(X, Y ) = g(dτa(X), dτa(Y ))

for all X, Y ∈ TeKG/K.

The function τa(bK) = abK may be used to define an analogue of the adjointrepresentation for a homogeneous space.

Definition 1.24. Let G/K be a reductive homogeneous space. The isotropy repre-

sentation AdG/K : K → Aut(m) of G/K is given by

AdG/Kk (X) = (dτk)eK(X)

for all k ∈ K,X ∈ Tek(G/K).

Proposition 1.25. [2] Let G/K be a reductive homogeneous space. Then there is aone-to-one correspondence between G-invariant Riemannian metrics g on G/K and

AdG/K-invariant scalar products ge in m i.e. scalar products that satisfy

ge(X, Y ) = ge(AdG/Kk X,Ad

G/Kk Y ).

for all X, Y ∈ m and k ∈ K. We say that the scalar product is AdG(K)-invariantor simply Ad(K)-invariant.

1.3.1 Normal Riemannian Homogeneous Spaces

Normal Riemannian homogeneous spaces will be the object of study in the thirdsection of this work. They are spaces in which the scalar product in m is not onlyleft-invariant but also right-invariant.

Definition 1.26. A Riemannian metric g of a Lie group is called bi-invariant ifboth the left and right translations La, Ra : G→ G are isometries for all a ∈ G.

Proposition 1.27. [2] There is a one-to-one correspondence between bi-invariantmetrics on a Lie group G and Ad(G)-invariant scalar products on its Lie algebra g,i.e. scalar products that satisfy

g(AdaX,Ada Y ) = g(X, Y ) (1.4)

for all a ∈ G and X, Y ∈ g.

7

Proof. [2] We can express the function Ia, defined in Definition 1.11, as Ia = Ra−1 ◦La. For Xe ∈ g we get AdaXe = (dIa)eXe = dRa−1dLaXe = dRa−1Xa and then thestatement follows from the right-invariance of the metric and an argument analogousto the proof of Proposition 1.22.

Proposition 1.28. [2] The Ad(G)-invariance of the scalar product, Equation (1.4),is equivalent to the relation

g([X, Y ], Z) = g(X, [Y, Z]).

Proof. [2] Let γ : I → G be a curve in G with γ(0) = e, γ(0) = Y . AssumingAd(G)-invariance we get

g([X, Y ], Z) = g(− adY X,Z)

= g

(d

dt(−Adγ(t)X)

∣∣∣∣t=0

, Z

)=

d

dtg(−Adγ(t)X,Z

)∣∣∣∣t=0

=d

dtg(X,−Adγ(−t) Z

)∣∣∣∣t=0

= g (X,−[−Y, Z])

= g (X, [Y, Z]) .

Assuming the relation leads to Ad(G)-invariance by taking the same steps.

We now define the Killing form, a useful bilinear form which plays an importantrole in Lie theory.

Definition 1.29. The Killing form of a Lie algebra g is the function B : g× g→ Rdefined by

B(X, Y ) = trace(adX ◦ adY ).

Proposition 1.30. [2] Let g be a Lie algebra.

i) The Killing form is a symmetric bilinear form on g.

ii) If G is a Lie group with Lie algebra g, then the Killing form is Ad(G)-invarianti.e.

B(X, Y ) = B(AdaX,Ada Y ), for all X, Y ∈ g and a ∈ G.

Proof. The first statement follows from the facts that both the map ad : g→ Aut(g)given by ad : X 7→ adX and the trace are linear and that the trace is symmetric.For the second statement, we consider the composition of the adjoint action and theadjoint representation using the result for matrix representations of Lie algebras,see Equation (1.3), which states that AdaX = aXa−1.

adAdaX ◦Ada(Y ) = adaXa−1(aY a−1)

= [aXa−1, aY a−1]

= a[X, Y ]a−1

= Ada ◦ adX(Y ).

8

This allows us to write adAdaX = Ada ◦ adX ◦(Ada)−1 so we can confirm that

B(AdaX,Ada Y ) = trace(adAdaX ◦ adAda Y )

= trace(Ada ◦ adX ◦(Ada)−1 ◦ Ada ◦ adY ◦(Ada)

−1)

= trace(adX ◦ adY )

= B(X, Y ).

Definition 1.31. A Lie algebra is called semi-simple if its Killing form is non-degenerate.

The following results make it clear that the Killing form will be immensely usefulto us.

Theorem 1.32. [2] If G is a compact semi-simple Lie group, then its Killing formis negative definite.

If G is a compact semi-simple Lie group, then a scalar product on g defined asthe negative of the Killing form corresponds to a Riemannian metric on G.

Theorem 1.33. [2] A compact Lie group possesses a bi-invariant metric.

Proof. Let G be a Lie group with Lie algebra g. Then the negative of the Killingform is an Ad(G)-invariant bilinear form by Theorem 1.30. Proposition 1.27 thusensures the existence of a bi-invariant metric on G.

This result together with Theorem 1.27 ensures that there exists an Ad(G)-invariant scalar product on g. In particular, this scalar product is Ad(K)-invariantif K is a subgroup as above, which shows that any normal space is reductive. Themetric defined as the negative of the Killing form is called the standard homogeneousRiemannian metric.

1.3.2 Riemannian Symmetric Spaces

An interesting and well-behaved subset of the Riemannian homogeneous spaces arethe symmetric ones. All compact symmetric spaces are normal.

Definition 1.34. A Riemannian manifold (M, g) is said to be a symmetric space iffor each point p ∈M there exists an involution σp. This is an isometry σp : M →Mwith σp(p) = p and dσpX = −X, for all X ∈ TpM .

The involution σp reflects geodesics through p. This results in the following.

Proposition 1.35. [11] A symmetric space is geodesically complete.

Proof. Let γ be a geodesic in M defined on some closed interval I = [a, b] ⊂ R. Theinvolution at γ(b), σγ(b) reflects γ, extending it to the interval [a, b+a]. This processcan be repeated, so any geodesic can be extended to all of R.

Definition 1.36. The isometry group (I(M), ◦) of a Riemannian manifold is thegroup of all its isometries with the operation of composition.

Theorem 1.37. [11] Let (M, g) be a symmetric space and G be the connected com-ponent of its isometry group containing the identity element. Then G is a Lie groupwhich acts transitively on M .

9

Proof. Let p and q be points in M and γ be a geodesic such that γ(0) = p andγ(1) = q. Such a geodesic exists by the Hopf-Rinow Theorem (see [6]) since Mis geodesically complete. Then the involution at r = γ(1/2) is an isometry withσr(p) = q

From this it follows that every connected Riemannian symmetric space is homo-geneous. In the symmetric case, Theorem 1.19 becomes more specific and tells uswhat the Lie group G actually is.

Theorem 1.38. [11] Let (M, g) be a symmetric space and I(M) be its isometrygroup. If we denote by G the connected component of I(M) containing the identityelement e and let K be the isotropy group of G at e, then M is diffeomorphic toG/K.

10

Chapter 2

The Normal Homogeneous Spacesof Positive Curvature

The complete classification of homogeneous spaces with positive curvature has beensome time in the making, see [9]. In his paper [5] from 1961, Berger stated that heclassified all simply connected normal Riemannian homogeneous spaces with positivecurvature up to isomorphism. However, in 1999 an omission was noted by Wilkingand added in the article [17], namely the space (SU(3)× SO(3))/U∗(2).

For spaces that are not normal a classification was given in the article [15] byWallach in 1972 for even dimensions and for odd dimensions by Berard-Bergery in1976, see [3]. The latter shows that the infinite family of seven dimensional spacesW 7p,q, discussed by Aloff and Wallach in [1] from 1975, made the list complete. In

2015, the above results were compiled in the article [18] by Wilking and Ziller,together with a self-contained proof that these are the only such spaces.

In this thesis, we will consider case of normal homogeneous spaces of positivecurvature. The complete list of such spaces is presented in Table 2.1.

2.1 Pinching Constants and The Sphere Theorem

The standard n-dimensional sphere Sn has constant curvature +1. At each pointp ∈ Sn every two-plane of TpS

n has the same sectional curvature. We now define ameasure of how much a space deviates from being a sphere.

Definition 2.1. Let (M, g) be a compact Riemannian manifold with positive sec-tional curvature K. The pinching constant is defined as the quotient of the extremalvalues of its sectional curvatures:

δM =minK(σ)

maxK(σ),

where σ runs through all two-planes of TpM and p ∈M .

This means that the sectional curvature K satisfies

0 < δMKmax ≤ K ≤ Kmax.

The famous Sphere Theorem reads as follows.

11

Theorem 2.2. [6] Let M be a compact, simply connected, Riemannian manifoldwhose sectional curvature K satisfies

0 < hKmax < K ≤ Kmax.

If h = 1/4, then M is homeomorphic to a sphere.

The corresponding result for a weak inequality is slightly different. Then we havethe following.

Theorem 2.3. [4] Let M be a compact, simply connected, Riemannian manifoldwith its sectional curvature K satisfying

1

4Kmax ≤ K ≤ Kmax.

then either M is homeomorphic to a sphere or isometric to one of the compact rankone symmetric spaces CP n,HP n,OP 2.

This result together with the fact that e.g. CP n is a completely geodesic sub-manifold of HP n has the following consequence.

Corollary 2.4. The pinching constant for the compact rank one symmetric spacesCP n,HP n,OP 2 satisfy

δOP 2 ≤ δHPn ≤ δCPn ≤ 1

4.

We will see in the next chapter, by explicit calculations, that

δHPn = δCPn =1

4.

G K G/KNormal symmetric spaces of rank one:

SO(n+ 1) SO(n) Sn

U(n+ 1) U(1)×U(n) CPn

Sp(n+ 1) Sp(n)× Sp(1) HPn

F4 Spin(9) OP 2

Non-symmetric spaces:Sp(2) SU(2) B7

SU(5) Sp(2)× S1 B13

SU(3)× SO(3) U(2) W 71,1

SU(n+ 1) SU(n) S2n+1

Sp(n+ 1) Sp(n) S4n+3

Sp(n+ 1) Sp(n)× S1 CP 2n+1

Spin(9) Spin(7) S15

Table 2.1: All the pairs (G,K) makingG/K into a normal homogeneous space of positive curvature.The quotient G/K is diffeomorphic to the spaces in the third column.

12

Chapter 3

Curvature Calculations

In this section we calculate the sectional curvatures of quotient spaces using theformula by O’Neill (1.2). A closed expression for the sectional curvature is calculatedfor arbitrary orthogonal vectors in the complement vector space m. The extremesof this expression give the pinching constants for the spaces. Some authors, e.g.Puttmann in his article [14], strive to find the metric which produces the so calledoptimal pinching constant. When the space is isotropy irreducible (when its isotropyrepresentation is irreducible), the pinching constant with the standard homogeneousRiemannian metric is automatically the optimal one. This is true for all symmetricspaces. In the remaining cases, we will be content with calculating the pinchingconstant for the standard homogeneous Riemannian metric i.e. the negative of theKilling form.

In 1966, Eliasson calculated the sectional curvatures of the isotropy irreducibleseven-dimensional Berger space B7 = Sp(2)/SU(2) in his paper [7]. More recently,in 1999, the optimal pinching constant for the thirteen-dimensional Berger space,B13 = SU(5)/(S1×Sp(2)), and the seven-dimensional space introduced by Wallach,W 7

1,1 = (SU(3) × SO(3))/U(2), were calculated by Puttmann in [14]. All threespaces were found to have the optimal pinching constant of δ = 1/37.

3.1 Useful Formulae

Let G be a compact Lie group with Lie algebra g and K be a closed subgroup ofG. Let π : G → G/K be a Riemannian submersion. For independent X, Y ∈ g wedenote dπe(X) by X. By K(X, Y ) we mean the sectional curvature in G for theplane spanned by X and Y and by K(X, Y ) the sectional curvature in G/K.

We know from [8] that since we have a left-invariant metric with skew-symmetricad-operator, the Levi-Civita connection satisfies

∇XY =1

2[X, Y ].

This together with the Jacobi identity[X, [Y, Z]

]+[Y, [Z,X]

]+[Z, [X, Y ]

]= 0

results in

R(X, Y )Z = ∇X∇YZ −∇Y∇XZ −∇[X, Y ]Z

=1

4

[X, [Y, Z]

]− 1

4

[Y, [X,Z]

]− 1

2

[[X, Y ], Z

]13

=1

4

([X, [Y, Z]

]+[Y, [Z,X]

]+ 2[Z, [X, Y ]

])= −1

4

[[X, Y ], Z

],

which allows us to write the sectional curvature as

K(X, Y ) = −1

4·g([[X, Y ], Y

], X)

|X|2|Y |2 − g(X, Y )2.

We are dealing with compact groups, so by Theorem 1.33 we have a bi-invariantmetric. This means we can use the property

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

from Proposition 1.28 to write the sectional curvature as

K(X, Y ) =1

4· g([X, Y ], [X, Y ])

|X|2|Y |2 − g(X, Y )2.

If X and Y are orthonormal, we get

K(X, Y ) =1

4|[X, Y ]|2.

We also have, from Lemma 1.6, that for horizontal arguments

AXY =1

2V [X, Y ],

which put in O’Neill’s formula (1.2) gives us the simple expression

K(X, Y ) =1

4|[X, Y ]|2 + 3|AXY |2

=1

4|H[X, Y ]|2 + |V [X, Y ]|2. (3.1)

If all Lie brackets of the horizontal arguments are vertical only one term is non-zero:

K(X, Y ) = |V [X, Y ]|2. (3.2)

We are interested in finding the sectional curvatures for the metric given by thethe Killing form. Thus we need to consider arbitrary linear combinations of thebasis vectors. If {Xi}ni=1 is an orthonormal basis of an n-dimensional subspace m,we let X and Y be orthonormal vectors in m given by

X =n∑a=1

αaXa, Y =n∑a=1

xaXa.

This notation will be helpful when considering spaces with several types of basis.The coefficients of X will be denoted by Greek letters α, β, γ, . . . and those of Y byLatin x, y, z, . . . , making the calculations easier to follow. Their orthonormality isensured by imposing the restrictions

n∑a=1

α2a =

n∑a=1

x2a = 1,n∑a=1

αaxa = 0.

The results of the following calculations are summarised in Table 3.1.

14

3.2 Rank One Normal Symmetric Spaces of Positive Cur-vature

3.2.1 The Sphere Sn ∼= SO(n+ 1)/SO(n)

The first space we consider is the well known n-dimensional sphere. The specialorthogonal group SO(n + 1) acts transitively on Sn by isometries and the isotropygroup of this action is SO(n). We let SO(n) be the lower-right corner of SO(n+ 1)as follows. For A ∈ SO(n+ 1) we have

A =

(a vwt B

),

where B ∈ SO(n) and A · At = 1. In this case the standard bilinear homogeneousRiemannian metric is given by

g(X, Y ) =1

2· trace(XY t).

We let Xab denote skew-symmetric matrices:

Xab = Eab − Eba,

where Eab is an (n+ 1)× (n+ 1) matrix with (Eab)cd = δacδbd.As a basis for the Lie algebra so(n+1) of SO(n+1) we take {Xab}1≤a<b≤n+1. The

Lie algebra so(n) of the subgroup SO(n) is taken as the elements of so(n+ 1) withonly zeroes in the first row and first column. We denote the orthogonal complementto so(n) in so(n+ 1) by m. In the notation of Section 1.1, g = so(n+ 1), k = so(n)is the vertical part, and m is the horizontal part. Any X ∈ so(n+ 1) can be written

X = VX +HX

=

(0 00 Y

)+

(0 v−vt 0n

),

where Y ∈ so(n) and v ∈ Rn. Here 0n denotes the zero n × n matrix. The abovemeans that {X1a}n+1

a=2 is a basis for m. We let X, Y ∈ m be orthonormal vectorsgiven by

X =n+1∑a=2

αaX1a, Y =n+1∑b=2

xbX1b

with

n+1∑a=2

α2a =

n+1∑b=2

x2b = 1,n+1∑a=2

αaxa = 0.

Using the fact that EabEcd = δbcEad we can calculate the Lie bracket of any twobasis vectors as

[X1a, X1b] = δ1bXa1 + δa1X1b − δabX11 − δ11Xab

= −Xab.

15

We see that this is vertical, so we may use Equation (3.2) to calculate the sectionalcurvatures of Sn. We have that

V [X, Y ] =1

2

n+1∑a,b=2

αaxbV [X1a, X1b]

= −1

2

n+1∑a,b=2

αaxbXab.

In getting the length of this, the only surviving terms come from g(Xab, Xab) = 1and g(Xab, Xba) = −1. The addition of the latter terms takes care of the case wherethe indices equal, when Xaa = 0. This results in the following:

K(X, Y ) = |V [X, Y ]|2

=n+1∑

a,b,c,d=2

αaxbαcxd · g(Xab, Xcd)

=n+1∑a,b=2

(αaxbαaxb − αaxbαbxa)

=

(n∑a=2

α2a

)·

(n∑b=2

x2b

)−

(n∑a=2

αaxa

)·

(n∑b=2

αbxb

)= 1. (3.3)

The curvature is independent of the arguments, so the pinching constant for thesphere is

δSn = 1.

3.2.2 The Complex Projective Space CP n ∼= U(n+ 1)/(U(1)×U(n))

We now turn to the more complicated case of the unitary group U(n+ 1) with thesubgroup U(1) ×U(n). The subgroup is located within U(n + 1) in a way similarto the previous section. For A ∈ U(n+ 1)

A =

(b vwt B

),

where b ∈ U(1), B ∈ U(n) and A · A∗ = 1. As a metric we take

g(X, Y ) =1

2·Re trace(XY ∗).

Together with the skew-symmetric matrices we define the symmetric and diagonalmatrices

Xab = Eab − Eba, Yab = Eab + Eba, Da =√

2Eaa.

We note that Yaa =√

2Da. The calculations will occasionally produce terms of theform Yaa, but this will not be a problem so long as one remembers that g(Yaa, Yaa) =2.

16

We let u(n) be embedded in u(n + 1) in analogy with the previous section, andu(1) be the top-left element. With the notation from Section 1.1 we have g = u(n+1)and k = u(1)× u(n). Any X ∈ u(n+ 1) can be written as

X = VX +HX

=

(a 00 Y

)+

(0 v−v∗ 0n

),

where Y ∈ u(n), a ∈ iR and v ∈ Cn. We see that {X1a, iY1a}2≤a≤n+1 is an orthonor-mal basis for the complement m.

Let X, Y ∈ m be orthonormal vectors given by

X =n+1∑a=2

(αaX1a + βaiY1a), Y =n+1∑b=2

(xbX1b + ybiY1b),

with

n+1∑a=2

(α2a + β2

a) =n+1∑b=2

(x2b + y2b ) = 1,n+1∑a=2

(αaxa + βaya) = 0.

First, the Lie brackets are calculated using EabEcd = δbcEad.

[X1a, X1b] = −Xab,

[X1a, iY1b] = δab√

2iD1 − iYab,[iY1a, X1b] = −

(δab√

2iD1 − iYab),

[iY1a, iY1b] = −Xab.

We see that all Lie brackets are vertical so (3.2) can be used. We get

V [X, Y ] =n+1∑a,b=2

V [αaX1a + βaiYa, xbX1b + ybiY1b]

=n+1∑a,b=2

((αaxb + βayb)(−Xab) + (αayb − βaxb)(δab

√2iD1 − iYab)

).

Again, the length is calculated and only a few terms remain. We remember thatXaa = 0 is taken care of by the easily forgotten terms from g(Xab, Xba) = −1 asbefore. We now also get terms from g(Yab, Yba) = 1, the addition of which takes careof the extra diagonal terms needed to get g(Yaa, Yaa) = 2.

|V [X, Y ]|2

=n+1∑

a,b,c,d=2

((αaxb + βayb)(αcxd + βcyd)g(Xab, Xcd)

+ (αayb − βaxb)(αcyd − βcxd)(g(iYab, iYcd) + δabδcd2g(iD1, iD1)

))=

n+1∑a,b=2

((αaxb + βayb)(αaxb + βayb − αbxa − βbya)

17

+ (αayb − βaxb)(αayb − βaxb + αbya − βbxa))

+ 2n+1∑a,c=2

(αaya − βaxa)(αcyc − βcxc).

Here one does best by sorting terms by index and factorising. We find, and thereader may easily confirm in the opposite direction, that this can be written as

|V [X, Y ]|2 =n+1∑a,b=2

(αaαaxbxb + αaβaxbyb − αaxaαbxb − αayaβbxb

+ αaβaxbyb + βaβaybyb − βaxaαbyb − βayaβbyb+ αaαaybyb − αaβaxbyb + αayaαbyb − αaxaβbyb− αaβaxbyb + βaβaxbxb − βayaαbxb + βaxaβbxb

)+ 2

(n+1∑a=2

(αaya − βaxa)

)2

=n+1∑a,b=2

((α2

a + β2a)(x

2b + y2b )− (αaxa + βaya)(αbxb + βbyb)

+ (αaya − βaxa)(αbyb − βbxb))

+ 2

(n+1∑a=2

(αaya − βaxa)

)2

(3.4)

= 1− 0 + (1 + 2)

(n+1∑a=2

(αaya − βaxa)

)2

.

The sectional curvature is thus given by

K(X, Y ) = 1 + 3

(n+1∑a=2

(αaya − βaxa)

)2

. (3.5)

This takes values in the interval [1, 4]. The extremes are given by e.g.K(X12, X13) = 1 and K(X12, ¯iY 12) = 4, respectively. The pinching constant forCP n is thus

δCPn =1

4.

3.2.3 The Quaternionic Projective Space HP n ∼= Sp(n+1)/(Sp(n)×Sp(1))

The quaternions, a four-dimensional extension of the complex numbers, abide bythe following equation, famously carved into a bridge in Dublin by Hamilton:

i2 = j2 = k2 = ijk = −1.

The basis for u(n + 1) is easily extended to a basis of the Lie algebra sp(n + 1) ofthe symplectic group Sp(n+ 1) using the quaternions:

{Xab, iYab, jYab,, kYab}1≤a<b≤n+1 ∪ {iDa, jDa, kDa}1≤a≤n+1 .

18

Similarly, {X1a, iY1a, jY1a,, kY1a}2≤a≤n+1 is an orthonormal basis for the complement

m of sp(n)× sp(1), embedded in the same way as in the complex case. As a metricwe take

g(X, Y ) =1

2·Re trace(XY ∗),

where the quaternionic conjugation (a+ bi+ cj + dk)∗ = a− bi− cj − dk is used.We now turn to how a quaternion can be expressed as a scalar and a three-

dimensional vector. This is a compact notation that makes this section a naturalextension of previous ones. We denote a+ bi+ cj + dk by (a,~v), where ~v = (b, c, d).The multiplication can be written as

(a1, ~v1) ∗ (a2, ~v2) = (a1a2 − ~v1 · ~v2, a1 ~v2 + a2 ~v1 + ~v1× ~v2)

and we will frequently use that

Re(a,~v) = a, (a,~v)∗ = (a,−~v), and |(a,~v)|2 = a2 + ~v · ~v.

We let the Lie bracket of two matrices with quaternionic coefficients be given by

[q1X, q2Y ] = (q1∗q2)XY − (q2∗q1)Y X

and if we move imaginary coefficients out of the metric, we must take the conjugateand real part ’manually’ as follows.

g(q1X, q2Y ) = Re(q1∗q∗2) · g(X, Y ).

We let X, Y ∈ m be orthonormal vectors given by

X =n+1∑a=2

((αa,~0)X1a + (0, ~βa)Y1a

), Y =

n+1∑b=2

((xb,~0)X1b + (0, ~yb)Y1b

),

with

n+1∑a=2

(α2a + | ~βa|2) =

n+1∑b=2

(x2b + |~yb|2) = 1,n+1∑a=2

(αaxa + ~βa · ~ya) = 0.

We write out all the Lie brackets in our vector notation, remembering the non-commutativity of the cross product.

[(αa,~0)X1a, (xb,~0)X1b] = (αaxb,~0)(−Xab),

[(0, ~βa)Y1a, (0, ~yb)Y1b] = ( ~βa ·~yb,~0)(−Xab) + (0, ~βa×~yb)(δab√

2D1 + Yab),

[(αa,~0)X1a, (0, ~yb)Y1b] = (0, αa~yb)(δab√

2D1 − Yab),[(0, ~βa)Y1a, (xb,~0)X1b] = (0,−xb ~βa)(δab

√2D1 − Yab).

These are all vertical, so the vertical part of the Lie bracket is given by

V [X, Y ] =n+1∑a,b=2

V [(αa,~0)X1a + (0, ~βa)Y1a, (xb,~0)X1b + (0, ~yb)Y1b]

=n+1∑a,b=2

((αaxb + ~βa ·~yb,~0)(−Xab)− (0, αa~yb − xb ~βa − ~βa×~yb)Yab

19

+ (0, αa~yb − xb ~βa + ~βa×~yb)δab√

2D1

).

This has the following length.

|V [X, Y ]|2

= Re

n+1∑a,b,c,d=2

((αaxb + ~βa ·~yb,~0) ∗ (αcxd + ~βc · ~yd,~0)g(Xab, Xcd)

+ (0, αa~yb − xb ~βa − ~βa×~yb) ∗ (0,−αc ~yd + xd ~βc + ~βc× ~yd)g(Yab, Ycd))

+ 2 ·Re

n+1∑a,c=2

(0, αa ~ya − xa ~βa + ~βa× ~ya) ∗ (0,−αc~yc + xc ~βc − ~βc×~yc)g(D1, D1)

=n+1∑a,b=2

((αaxb + ~βa ·~yb)(αaxb + ~βa ·~yb − αbxa − ~βb · ~ya)

+ (αa~yb − xb ~βa − ~βa×~yb)·(αa~yb − xb ~βa − ~βa×~yb + αb ~ya − xa ~βb − ~βb× ~ya)

)+ 2

n+1∑a,c=2

(αa ~ya − xa ~βa + ~βa× ~ya)·(αc~yc + xc ~βc + ~βc×~yc)

Here we use the scalar ’quadruple product’ (a×b)·(c×d) = (a·c)(b·d) − (a·d)(b·c)to rewrite

~βa×~yb ·( ~βa×~yb + ~βb× ~ya) = | ~βa|2|~yb|2 − ( ~βa ·~yb)2 + ( ~βa · ~βb)(~ya ·~yb)− ( ~βa · ~ya)(~βb ·~yb).

Furthermore, we use the facts that the scalar ’triple product’ a·(b×c) is zero if anytwo of the three vectors equal and negated if two vectors are interchanged. Thisallows us to write

(αa~yb − xb ~βa)·(− ~βa×~yb − ~βb× ~ya) + (− ~βa×~yb)·(αa~yb − xb ~βa + αb ~ya − xa ~βb)= −αa~yb ·(~βb× ~ya) + xb ~βa ·(~βb× ~ya)− ( ~βa×~yb)· ~yaαb + ( ~βa×~yb)· ~βbxa= (αa ~ya − xa ~βa)·(~βb×~yb) + ( ~βa× ~ya)·(αb~yb − xb ~βb).

For later use we also show.

( ~βa× ~ya)·(~βb×~yb) = ( ~βa · ~βb)(~ya ·~yb)− ( ~βa ·~yb)(~ya · ~βb).

We are now ready to complete the calculation of |V [X, Y ]|2.

|V [X, Y ]|2

=n+1∑a,b=2

(α2ax

2b + αaxb ~βa ·~yb − αaxaαbxb − αaxb ~ya · ~βb

+ αaxb ~βa ·~yb + ( ~βa ·~yb)2 − xaαb ~βa ·~yb − (~ya · ~βb)( ~βa ·~yb)+ α2

a|~yb|2 − αaxb ~βa ·~yb + αaαb ~ya ·~yb − αaxa ~βb ·~yb− αaxb ~βa ·~yb + x2b | ~βa|2 − αbxb ~βa · ~ya + xaxb ~βa · ~βb+ | ~βa|2|~yb|2 − ( ~βa ·~yb)2 + ( ~βa · ~βb)(~ya ·~yb)− ( ~βa · ~ya)(~βb ·~yb)

20

+ (αa ~ya − xa ~βa)·(~βb×~yb) + ( ~βa× ~ya)·(αb~yb − xb ~βb))

+ 2

∣∣∣∣∣n+1∑a=2

(αa ~ya − xa ~βa + ~βa× ~ya)

∣∣∣∣∣2

=n+1∑a,b=2

((α2

a + | ~βa|2)(x2b + |~yb|2)− (αaxa + ~βa · ~ya)(αbxb + ~βb ·~yb)

+ (αa ~ya − xa ~βa)·(~βb×~yb) + ( ~βa× ~ya)·(αb~yb − xb ~βb)+ ( ~βa · ~βb)(~ya ·~yb)− (~ya · ~βb)( ~βa ·~yb)+ (αa ~ya − xa ~βa)(αb~yb − xb ~βb)

)+ 2

∣∣∣∣∣n+1∑a=2


∣∣∣∣∣2

=n+1∑a,b=2

((α2


+ (αa ~ya − xa ~βa + ~βa× ~ya)(αb~yb − xb ~βb + ~βb×~yb))

+ 2

∣∣∣∣∣n+1∑a=2


∣∣∣∣∣2

(3.6)

= 1 + 3

∣∣∣∣∣n+1∑a=2


∣∣∣∣∣2

.

We get

K(X, Y ) = 1 + 3

∣∣∣∣∣n+1∑a=2


∣∣∣∣∣2

,

This takes values in the interval [1, 4] and these extremes are given by e.g.

K((X12,~0), (X13,~0)

)= 1,

K((X12,~0),

(0, (Y12, 0, 0)

))= 4.

The pinching constant for HP n is

δHPn =1

4.

It is easy to see that restricting to one complex coefficient, e.g. i (by setting~βa = (βa, 0, 0) and similarly ~ya = (ya, 0, 0)), reduces the sectional curvature formulaabove to the result for the complex projective space, Equation (3.5). Setting allimaginary coefficients to zero reduces it further to the formula for the sectionalcurvature on Sn, Equation (3.3).

3.3 Non-Symmetric Spaces

Three of the spaces in Table 2.1 are bundles over projective spaces.

S1 ↪→ SU(n+ 1)/SU(n)→ CP n,

21

S3 ↪→ Sp(n+ 1)/Sp(n)→ HP n,

S2 ↪→ Sp(n+ 1)/(S1 × Sp(n))→ HP n.

This can be seen by

SU(n+ 1)/SU(n) ∼= S1 × SU(n+ 1)/S(U(n)×U(1))∼= S1 ×U(n+ 1)/U(n)×U(1)∼= S1 × CP n,

Sp(n+ 1)/Sp(n) ∼= Sp(1)× Sp(n+ 1)/(Sp(1)× Sp(n))∼= S3 ×HP n,

and

Sp(n+ 1)/(S1 × Sp(n)) ∼= S2 × Sp(n+ 1)/(S1 × S2 × Sp(n))∼= S2 × Sp(n+ 1)/(Sp(1)× Sp(n))∼= S2 ×HP n.

We calculate the sectional curvatures for these spaces using the same methods asbefore. The situation will be more complicated due to the new diagonal elements.

3.3.1 The Homogeneous Space SU(n+ 1)/SU(n)

The first non-symmetric homogeneous space we consider is SU(n + 1)/SU(n). Wewill work with the isometric U(n + 1)/U(n) for simplicity. Our metric is taken tobe

g(X, Y ) =1


and the matrices are the same as before.

Xab = Eab − Eba, Yab = Eab + Eba, Da =√

2Eaa.

We let m be the orthogonal complement of u(n) in u(n+ 1). This can be written

m =

{(a v−v∗ 0n

)∣∣∣∣ a ∈ iR, v ∈ Cn

}.

For X ∈ u(n+ 1) we have

X = VX +HX

=

(0 00 Y

)+

(a v−v∗ 0n

)where Y ∈ u(n). We take the set {iD1} ∪ {X1a, iY1a}2≤a≤n+1 as our orthonormalbasis for m and set

X =n+1∑a=2

(αaX1a + βaiY1a) + γ1iD1, Y =n+1∑b=2

(xbX1b + ybiY1b) + z1iD1,

22

with

n+1∑a=2

(α2a + β2

a) + γ21 =n+1∑b=2

(x2b + y2b ) + z21 = 1,n+1∑a=2

(αaxa + βaya) + γ1z1 = 0.

The Lie brackets will be the same as in section 3.2.2, with the addition of thefollowing.

1√2

[X1a, iD1] = iY1a,

1√2

[iY1a, iD1] = X1a,

[iD1, iD1] = 0.

All new brackets are horizontal, but one from before is now not completely vertical.

[X1a, iY1b] =(δab√

2iD1 − iYab),

H[X1a, iY1b] = δab√

2iD1,

V [X1a, iY1b] = −iYab.

The calculation of the vertical part of the Lie Bracket will be very similar to thecorresponding one for CP n so we only show the results, Equation (3.4), with theterm from δab

√2iD1 removed. We take the result from before the orthonormality

conditions on X and Y were used to simplify the expression; we will need to amendthose simplifications in accordance with our new conditions.

|V [X, Y ]|2 =n+1∑a,b=2

((α2

a + β2a)(x

2b + y2b )− (αaxa + βaya)(αbxb + βbyb)

)+

(n+1∑a=2

(αaya − βaxa)

)2

= (1− γ21)(1− z21)− (−γ1z1)2 +

(n+1∑a=2

(αaya − βaxa)

)2

= 1− (γ21 + z21) +

(n+1∑a=2

(αaya − βaxa)

)2

.

The horizontal part of the Lie bracket is given by

1√2H[X, Y ] =

1√2

n+1∑a=2

((αaz1 − γ1xa)[Xa1, D1] + (βaz1 − γ1ya)[Ya1, D1]

)+

1√2

n+1∑a,b=2

(αayb − βaxb)H[X1a, Y1b]

=n+1∑a=2

((αaz1 − γ1xa)iY1a + (βaz1 − γ1ya)X1a

)23

+1√2

n+1∑a,b=2

(αayb − βaxb)δab√

2iD1.

This has only one changing index, so no subtle trickery is needed in the calculationof its length.

1

2|H[X, Y ]|2 =

n+1∑a,b=2

((αaz1 − γ1xa)(αbz1 − γ1xb)g(iY1a, iY1b)

+ (βaz1 − γ1ya)(βbz1 − γ1yb)g(X1a, X1b))

+n+1∑a,c=2

(αaya − βaxa)(αcyc − βcxc)g(iD1, iD1)

=n+1∑a=2

((αaz1 − γ1xa)(αaz1 − γ1xa)

+ (βaz1 − γ1ya)(βaz1 − γ1ya))

+

(n+1∑a=2

(αaya − βaxa)

)2

=n+1∑a=2

((α2

a + β2a)z

21 + γ21(x2a + y2a)− 2(αaxa + βaya)γ1z1

)+

(n+1∑a=2

(αaya − βaxa)

)2

= (1− γ21)z21 + γ21(1− z21)− 2(−γ1z1)γ1z1

+

(n+1∑a=2

(αaya − βaxa)

)2

=(z21 + γ21

)+

(n+1∑a=2

(αaya − βaxa)

)2

.

We can now get the sectional curvature by using Equation (3.1).

K(X, Y ) = 1− (γ21 + z21) +

(n+1∑a=2

(αaya − βaxa)

)2

+2

4(z21 + γ21) +

2

4

(n+1∑a=2

(αaya − βaxa)

)2

= 1 +3

2

(n+1∑a=2

(αaya − βaxa)

)2

− 1

2

(γ21 + z21

). (3.7)

We see that

K(X, Y ) ∈[

1

2,5

2

]24

with the extremes given by e.g. K(D1, ·) = 1/2 and K(X12, ¯iY12) = 5/2, respectively.The pinching constant is thus

δSU(n+1)/SU(n) =1

5.

3.3.2 The Homogeneous Space Sp(n+ 1)/Sp(n)

The second non-symmetric homogeneous space we consider is Sp(n + 1)/Sp(n).The approach is the same as in the previous section, but we now expand on thecalculations for HP n. We take the metric

g(X, Y ) =1


with quaternionic conjugation and let X, Y ∈ m be given by

X =n+1∑a=2

((αa,~0)X1a + (0, ~βa)Y1a,

)+ (0, ~γ1)D1,

Y =n+1∑b=2

((xb,~0)X1b + (0, ~yb)Y1b

)+ (0, ~z1)D1,

with orthonormality ensured by

n+1∑a=2

(α2a + | ~βa|2) + |~γ1|2 =

n+1∑b=2

(x2b + |~yb|2) + |~z1|2 = 1,n+1∑a=2

(αaxa + ~βa · ~ya) + ~γ1 · ~z1 = 0.

The Lie brackets in this space not present in HP n are the following. We includeall combinations to avoid errors in signs with the cross product.

1√2

[(αa,~0)X1a, (0, ~z1)D1] = (0,−αa~z1)Y1a,

1√2

[(0, ~γ1)D1, (xb,~0)X1b] = (0, xb ~γ1)Y1a,

1√2

[(0, ~βa)Y1a, (0, ~z1)D1] = ( ~βa · ~z1,~0)X1a + (0, ~βa× ~z1)Y1a,

1√2

[(0, ~γ1)D1, (0, ~ya)Y1a] = −(~γ1 · ~ya,~0)X1a + (0, ~γ1× ~ya)Y1a,

1√2

[(0, ~γ1)D1, (0, ~z1)D1] = (0, ~γ1× ~z1)D1.

These are all horizontal. The ones from before that now have a horizontal part are

H[(0, ~βa)Y1a, (0, ~yb)Y1b] = δab(0, ~βa×~yb)√

2D1,

H[(αa,~0)X1a, (0, ~yb)Y1b] = δab(0, αa~yb)√

2D1,

H[(0, ~βa)Y1a, (xb,~0)X1b] = δab(0,−xb ~βa)√

2D1.

25

We see that |V [X, Y ]|2 for this space will be identical to the HP n case if we removethe term associated with D1 from Equation (3.6). We apply the new orthonormalityrelations and are left with

|V [X, Y ]|2

=n+1∑a,b=2

((α2



= (1− |γ1|2)(1− |z1|2)− (γ1 ·z1)2 +

∣∣∣∣∣n+1∑a=2


∣∣∣∣∣2

= 1− |γ1|2 − |z1|2 + |γ1|2|z1|2 − (γ1 ·z1)2 +

∣∣∣∣∣n+1∑a=2


∣∣∣∣∣2

.

The horizontal part of [X, Y ] consists of all horizontal terms from above.

1√2H[X, Y ] =

1√2H[ n+1∑a=2

((αa,~0)X1a + (0, ~βa)Y1a

)+ (0, ~γ1)D1,

n+1∑b=2

((xb,~0)X1b + (0, ~yb)Y1b

)+ (0, ~z1)D1

]=

n+1∑a=2

((0,−αa~z1 + xa ~γ1 + ~βa× ~z1 + ~γ1× ~ya)Y1a

+ ( ~βa · ~z1 − ~γ1 · ~ya,~0)X1a + (0, αa ~ya − xa ~βa + ~βa× ~ya)D1

)+ (0, ~γ1× ~z1)D1

We calculate its length, using the same technique as before.

1

2|H[X, Y ]|2

=n+1∑a=2

((−αa~z1 + xa ~γ1 + ~βa× ~z1 + ~γ1× ~ya)·(−αa~z1 + xa ~γ1 + ~βa× ~z1 + ~γ1× ~ya)

+ ( ~βa · ~z1 − ~γ1 · ~ya)·( ~βa · ~z1 − ~γ1 · ~ya))

+ (~γ1× ~z1)(~γ1× ~z1) +

∣∣∣∣∣n+1∑a=2


∣∣∣∣∣2

=n+1∑a=2

(α2a|~z1|2 − 2αaxa~z1 · ~γ1 + x2a|~γ1|2 − 2αa~z1 ·(~γ1× ~ya) + 2xa ~γ1 ·( ~βa× ~z1)

+| ~βa|2|~z1|2−( ~βa · ~z1)2+2( ~βa · ~γ1)(~z1 · ~ya)−2( ~βa · ~ya)(~z1 · ~γ1)+|~γ1|2|~ya|2−(~γ1 · ~ya)2

+ ( ~βa · ~z1)2 − 2( ~βa · ~z1)(~γ1 · ~ya) + (~γ1 · ~ya)2)

+ (~γ1× ~z1)2 +

∣∣∣∣∣n+1∑a=2


∣∣∣∣∣2

26

=n+1∑a=2

((α2

a + | ~βa|2)|~z1|2 + |~γ1|2(x2b + |~yb|2)− 2(αaxa + ~βa · ~ya)(~γ1 · ~z1)

− 2αa~z1 ·(~γ1× ~ya) + 2xa ~γ1 ·( ~βa× ~z1) + 2( ~βa · ~γ1)(~z1 · ~ya)− 2( ~βa · ~z1)(~γ1 · ~ya))

+ (~γ1× ~z1)2 +

∣∣∣∣∣n+1∑a=2


∣∣∣∣∣2

=n+1∑a=2

((α2

a + | ~βa|2)|~z1|2 + |~γ1|2(x2b + |~yb|2)− 2(αaxa + ~βa · ~ya)(~γ1 · ~z1)

+ 2αa ~ya ·(~γ1× ~z1)− 2xa ~βa ·(~γ1× ~z1) + 2( ~βa× ~ya)·(~γ1× ~z1))

+ (~γ1× ~z1)2 +

∣∣∣∣∣n+1∑a=2


∣∣∣∣∣2

= (1− |~γ1|2)|~z1|2 + |~γ1|2(1− |~z1|2)− 2(−~γ1 · ~z1)(~γ1 · ~z1)

+

∣∣∣∣∣n+1∑a=2

(αa ~ya − xa ~βa + ~βa× ~ya) + ~γ1× ~z1

∣∣∣∣∣2

= 2((~γ1 · ~z1)2−|~γ1|2|~z1|2

)+ |~γ1|2 + |~z1|2+

∣∣∣∣∣n+1∑a=2

(αa ~ya − xa ~βa + ~βa× ~ya) + ~γ1× ~z1

∣∣∣∣∣2

(3.8)

The sectional curvature of Sp(n+ 2)/Sp(n) therefore satisfies

K(X, Y )

= 1− |γ1|2 − |z1|2 + |γ1|2|z1|2 − (γ1 ·z1)2 +

∣∣∣∣∣n+1∑a=2


∣∣∣∣∣2

+ (~γ1 · ~z1)2 − |~γ1|2|~z1|2 +1

2

(|~γ1|2 + |~z1|2

)+

1

2

∣∣∣∣∣n+1∑a=2

(αa ~ya − xa ~βa + ~βa× ~ya) + ~γ1× ~z1

∣∣∣∣∣2

= 1 +

∣∣∣∣∣n+1∑a=2


∣∣∣∣∣2

− 1

2

(|γ1|2 + |z1|2

)+

1

2

∣∣∣∣∣n+1∑a=2

(αa ~ya − xa ~βa + ~βa× ~ya) + ~γ1× ~z1

∣∣∣∣∣2

.

The extremes of this are given by e.g.

K((0, (1, 0, 0))D1, (0, (0, 1, 0))D1

)=

1

2,

K((X12,~0), (0, (1, 0, 0))Y12

)=

5

2,

and the pinching constant is

δSp(n+1)/Sp(n) =1

5.

27

If one restricts to a single imaginary unit, we again see a reduction of the expres-sion above to the one for the sectional curvature for SU(n + 1)/SU(n), Equation(3.7).

3.3.3 The Homogeneous Space Sp(n+ 1)/(S1 × Sp(n))

The last non-symmetric homogeneous space studied in this paper is Sp(n+1)/(S1×Sp(n)). The construction is identical to Sp(n + 1)/Sp(n) with the exception thatone diagonal vector, chosen to be the i-component, is vertical. This forces us toview the i and j, k components apart on coefficients of D1. To facilitate this, weintroduce an alternative set of projections. Let H project a 3-vector to its j and kcomponents, and V project it to its i component.

Our two orthogonal vectors in m take the form

X =n+1∑a=2

((αa,~0)X1a + (0, ~βa)Y1a,

)+ (0, ~γ1)D1,

Y =n+1∑b=2

((xb,~0)X1b + (0, ~yb)Y1b

)+ (0, ~z1)D1,

with

~γ1 = (0, γj, γk), ~z1 = (0, zj, zk)

and

n+1∑a=2

(α2a + | ~βa|2) + |~γ1|2 =

n+1∑b=2

(x2b + |~yb|2) + |~z1|2 = 1,n+1∑a=2

(αaxa + ~βa · ~ya) + ~γ1 · ~z1 = 0.

We note that the cross product of the diagonal coefficients becomes vertical.

~γ1× ~z1 = (γjzk − γkzj, 0, 0) = V ~γ1× ~z1

The Lie brackets in this space are the same as in Sp(n + 1)/Sp(n) with theexception of the vertical terms

[(0, ~γ1)D1, (0, ~z1)D1] =√

2(0, ~γ1× ~z1)D1,

V [(0, ~βa)Y1a, (0, ~yb)Y1b] =√

2δab(0,V ( ~βa×~yb)

)D1,

V [(αa,~0)X1a, (0, ~yb)Y1b] =√

2δab(0, αaV ~yb)D1,

V [(0, ~βa)Y1a, (xb,~0)X1b] =√

2δab(0,−xbV ~βa)D1.

This leaves us with the following as the vertical part of the Lie bracket.

V [X, Y ]

=n+1∑a,b=2

((αaxb + ~βa ·~yb,~0)(−Xab)− (0, αa~yb − xb ~βa − ~βa×~yb)Yab

+√

2(0, αaV ~yb − xbV ~βa + V ( ~βa×~yb))δabD1

)+√

2(0, ~γ1× ~z1)D1.

28

The procedure in squaring this is identical to the HP n case and we get

|V [X, Y ]|2 =n+1∑a,b=2

((α2



+ 2

∣∣∣∣∣n+1∑a=2

(αaV ~ya − xaV ~βa + V ( ~βa× ~ya)

)+ ~γ1× ~z1

∣∣∣∣∣2

= (1− |~γ1|2)(1− |~z1|2)− (~γ1 · ~z1)2 +

∣∣∣∣∣n+1∑a=2


∣∣∣∣∣2

+ 2

∣∣∣∣∣Vn+1∑a=2

(αa ~ya − xa ~βa + ( ~βa× ~ya)

)+ ~γ1× ~z1

∣∣∣∣∣2

.

The horizontal terms will mimic those for the Sp(n + 1)/Sp(n) case, with onlythe j and k terms of the D1 coefficients. We get

1√2H[X, Y ] =

n+1∑a=2

((0,−αa~z1 + xa ~γ1 + ~βa× ~z1 + ~γ1× ~ya)Y1a + ( ~βa · ~z1 − ~γ1 · ~ya,~0)X1a

+ (0, αaH ~ya − xaH ~βa + H ( ~βa× ~ya))D1

)+ (0, ~γ1× ~z1)D1

Again, the change compared to the Sp(n + 1)/Sp(n) case is easily spotted, so wejust need to tweak the squared result in Equation (3.8).

1

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

((~γ1 · ~z1)2 − |~γ1|2|~z1|2

)+ |~γ1|2 + |~z1|2

+

∣∣∣∣∣Hn+1∑a=2


∣∣∣∣∣2

.

We can now get the sectional curvature using Equation (3.1).

K(X, Y ) = (1− |~γ1|2)(1− |~z1|2)− (~γ1 · ~z1)2 +

∣∣∣∣∣n+1∑a=2


∣∣∣∣∣2

+ 2

∣∣∣∣∣Vn+1∑a=2


)+ ~γ1× ~z1

∣∣∣∣∣2

+((~γ1 · ~z1)2 − |~γ1|2|~z1|2

)+

1

2

(|~γ1|2 + |~z1|2

)+

1

2

∣∣∣∣∣Hn+1∑a=2


∣∣∣∣∣2

= 1 +

∣∣∣∣∣n+1∑a=2


∣∣∣∣∣2

− 1

2

(|~γ1|2 + |~z1|2

)+

∣∣∣∣∣(2V +1

2H)( n+1∑

a=2


)+ ~γ1× ~z1

)∣∣∣∣∣2

.

29

The extremes of this are given by e.g.

K((1,~0)X12, (0, (0, 1, 0))D1

)=

1

2,

K((1,~0)X12, (0, (1, 0, 0))Y12

)= 4,

so the pinching constant is

δSp(n+1)/(S1×Sp(n)) =1

8.

3.4 Summary

We have derived explicit formulae for the sectional curvatures of six Riemannianhomogeneous spaces of which the first three are symmetric. The expression for thequaternionic projective space HP n simplifies to the expressions for CP n and Sn uponmaking the relevant restrictions, just as the expression for the sectional curvaturesof Sp(n + 1)/Sp(n) simplifies to the expression for SU(n + 1)/SU(n). Using thenotation of previous sections, the formulae are presented below.

M K(X,Y ) δ

Sn =SO(n+ 1)

SO(n)1 1

CPn =U(n+ 1)

U(1)×U(n)1 + 3

(n+1∑a=2

(αaya − βaxa)

)2

1/4

HPn =Sp(n+ 1)

Sp(1)× Sp(n)1 + 3

∣∣∣∣∣n+1∑a=2


∣∣∣∣∣2

1/4

SU(n+ 1)

SU(n)1 +

3

2

(n+1∑a=2

(αaya − βaxa)

)2

− 1

2

(γ21 + z21

)1/5

Sp(n+ 1)

Sp(n)

1 +

∣∣∣∣∣n+1∑a=2


∣∣∣∣∣2

− 1

2

(|γ1|2 + |z1|2

)

+1

2

∣∣∣∣∣n+1∑a=2

(αa ~ya − xa ~βa + ~βa× ~ya) + ~γ1× ~z1

∣∣∣∣∣2

1/5

Sp(n+ 1)

(S1 × Sp(n))

1 +

∣∣∣∣∣n+1∑a=2


∣∣∣∣∣2

− 1

2

(| ~γ1|2 + |~z1|2

)

+

∣∣∣∣∣(2V +1

2H)( n+1∑

a=2


)+ ~γ1× ~z1

)∣∣∣∣∣2

1/8

Table 3.1: Six Riemannian homogeneous spaces M , their sectional curvatures K(X,Y ) for or-thonormal horizontal vectors X,Y , and their pinching constants δ.

30

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31

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