THE GEOMETRY OF HARMONIC MORPHISMS

8/13/2019 THE GEOMETRY OF HARMONIC MORPHISMS

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THE GEOMETRY OF HARMONIC MORPHISMS

bySigmundur Gudmundsson

Submitted in accordance with the regulations for the degree ofPh.D. in Pure Mathematics.

University of Leeds,Department of Pure Mathematics,

April 1992.

The candidate confirms that the work submitted is his own and that

appropriate credit has been given where reference has been made to the work ofothers.


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.

This thesis is dedicated to my parents

Gudmundur and Olafina


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ABSTRACT.

The objects under consideration are harmonic morphisms : (M, g)

(N, h)

between Riemannian manifolds. They are maps which pull back germs of real

valued harmonic functions onNto germs of harmonic functions on M. They have

been characterized as being those harmonic maps which are horizontally conformal,

so they form a special class of harmonic maps.

We generalize ONeills fundamental curvature equations for Riemannian sub-

mersions to the case of horizontal conformality. These equations relate the cur-

vature tensors on M and N and give necessary conditions for the existence of

horizontally conformal maps : (M, g) (N, h). From these conditions we derivesome new non-existence results for harmonic morphisms.

We observe a connection linking harmonic morphisms between space forms and

isoparametric systems. From this we obtain a classification of a natural class of

harmonic morphisms between open subsets of space forms.

We study the connection between horizontally conformal submersions : (M, g) (N, h) with minimal fibres and minimal submanifolds. We find conditions on such

maps, which imply the equivalence of the minimality of submanifolds L ofN andtheir inverse images 1(L) in M. We then use this to give examples of minimal

foliations of real hypersurfaces in (C)m for any m 2.We describe how the classical theory of multivalued holomorphic functions can

be generalized to harmonic morphisms : (Mm, g) (N2, h) onto surfaces. Wegive examples of such maps for the cases ofM3 = R3 andM3 =S3.

We use a result of L.Berard Bergery to give many new examples of homogeneous

harmonic morphisms : (G/K,g)

(G/H, g) between reductive homogeneous

spaces. We then show that some well known examples from 3-manifolds to surfaces

can be obtained in this way.


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CONTENTS

CHAPTER 0. INTRODUCTION . . . . . . . . . . . . . . . . . . p. 1

0.1. Background . . . . . . . . . . . . . . . . . . . . . . . . . . p. 10.2. Main Results . . . . . . . . . . . . . . . . . . . . . . . . . p. 2

0.3. Acknowledgements . . . . . . . . . . . . . . . . . . . . . . . p. 5

CHAPTER 1. HARMONIC MORPHISMS . . . . . . . . . . . . . . p. 6

1.1. Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . p. 6

1.2. Basic Properties . . . . . . . . . . . . . . . . . . . . . . . . p. 7

1.3. Examples . . . . . . . . . . . . . . . . . . . . . . . . . . p. 11

CHAPTER 2. GEOMETRIC CONSTRAINTS . . . . . . . . . . . p. 14

2.1. Horizontal Conformality . . . . . . . . . . . . . . . . . . . . p. 14

2.2. Necessary Curvature Conditions . . . . . . . . . . . . . . . . p. 17

CHAPTER 3. CONSTANT CURVATURE . . . . . . . . . . . . . p. 24

3.1. Known Results . . . . . . . . . . . . . . . . . . . . . . . . p. 24

3.2. Examples of Higher Codimension . . . . . . . . . . . . . . . . p. 25

3.3. A Higher Dimensional Classification . . . . . . . . . . . . . . p. 28

CHAPTER 4. MINIMAL SUBMANIFOLDS . . . . . . . . . . . . p. 35

4.1. Horizontally Conformal Maps and Minimal Submanifols . . . . . . p. 354.2. Examples . . . . . . . . . . . . . . . . . . . . . . . . . . p. 37

CHAPTER 5. MULTIVALUED HARMONIC MORPHISMS . . . . . p. 40

5.1. Locally Defined Harmonic Morphisms . . . . . . . . . . . . . . p. 40

5.2. The Covering Construction . . . . . . . . . . . . . . . . . . p. 41

5.3. Multivalued Harmonic Morphisms from R3 . . . . . . . . . . . p. 45

5.4. Multivalued Harmonic Morphisms fromS3 . . . . . . . . . . . p. 54

CHAPTER 6. HOMOGENEOUS HARMONIC MORPHISMS . . . . . p. 59

6.1. Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . p. 59

6.2. Existence for Reductive Homogeneous Spaces . . . . . . . . . . p. 61

6.3. Applications . . . . . . . . . . . . . . . . . . . . . . . . . p. 62

APPENDIX A . . . . . . . . . . . . . . . . . . . . . . . . . . p. 65

LIST OF REFERENCES . . . . . . . . . . . . . . . . . . . . . p. 67


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CHAPTER 0. INTRODUCTION.

0.1. Background.

In this section we very briefly mention some known results related to this work.

For a more detailed account see chapter 1.

In the late 70s Fuglede and Ishihara published two papers [Fug-1] and [Ish],

where they discuss their results on harmonic morphisms or mappings preserving

harmonic functions. They characterize non-constant harmonic morphisms :

(M, g)

(N, h) between Riemannian manifolds as those harmonic maps, which

are horizontally conformal. This means that we are dealing with a special class of

harmonic maps.

As in many areas of Differential Geometry the theory becomes especially rich if

we restrict our attention to the case, when one of the manifolds involved is a surface.

It is observed in [Fug-1], as a direct consequence of the Cauchy-Riemann equations,

that every holomorphic function f : U Cm C is a harmonic morphism. Aregular fibre of such a function is of course a minimal submanifold ofCm.

This is generalized by Baird and Eells in [Bai-Eel], where they show that a

regular fibre of a harmonic morphism : (Mm, g) (N2, h) from any Riemannianmanifold to a surface is minimal in M. Furthermore, they prove that ifn 3, thena harmonic morphism : (Mm, g) (Nn, h) has minimal fibres if and only if it ishorizontally homothetic.

It is noted in [Bai-Eel], that the dilation of a horizontally homothetic harmonic

morphism: (Mm, g) (Nn, h) of codimension 1 is an isoparametric function onM. Conversely, Baird proves in [Bai-1] that totally umbilic isoparametric foliations

of hypersurfaces in space forms give rise to harmonic morphisms.

In a series of papers [Bai-2] and [Bai-Woo-1-2-3-4], Baird and Wood study har-

monic morphisms from 3-dimensional manifolds to surfaces. They show that any

non-constant harmonic morphism : (M3, g) (N2, h) determines a conformalfoliation of geodesics on M3.

For the cases whenM3 = R3, S3 orH3 they prove that any submersive harmonic

morphism from an open subset ofM3 to a Riemann surface N2 can be represented

by two meromorphic functions on the surface N2. This is used to classify such


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maps which are either globally defined on M3 or have an isolated singularity.

So far, less attention has been given to harmonic morphisms between manifolds

of higher dimensions. The only such result related to this work is given by Kasueand Washio in [Kas-Was]. They prove that if n 3 and : Rm (Nn, h) isa harmonic morphism with totally geodesic fibres, then Nn = Rn and is an

orthogonal projection followed by a homothety.

0.2. Main Results.

In this section we state the most important results presented in this thesis. Some

of them have appeared in [Gud-1], [Gud-2], [Gud-Woo] and [Bai-Gud].By exploiting the fact that any non-constant harmonic morphism is horizontally

conformal we obtain:

Proposition 2.1.6. Letm > n 3 and: (Mm, g) (Nn, h) be a non-constantharmonic morphism with totally geodesic fibres. If the horizontal distributionH isintegrable, then the resulting foliationFH is spherical.

We generalize ONeills well known fundamental curvature equations for Rie-

mannian submersions to the case of horizontal conformality. As a direct conse-

quence we have the following:

Corollary 2.2.6. Letm > n 2 and(Mm, g), (Nn, h) be two Riemannian man-ifolds withKM(H)0 andKN 0. If : M N is a horizontally homotheticmap, then

(1) KN 0 andKM(H) 0,

(2) the dilation: M R+ is constant, and(3) the horizontal distributionH is integrable.

For space forms this leads to the following non-existence result:

Theorem 3.3.1. Letm > n2 and(M, N)=(Sm,Rn), (Sm, Hn) or(Rm, Hn).If U is an open subset of M, then there exists no horizontally homothetic map

: U N.

We show that if (M, g) has constant curvature, then Proposition 2.1.6 implies:

Corollary 3.3.3. Let m > n 2, (Mm, g), (Nn, h) be simply connected spaceforms andUan open and connected subset ofM. Let: U Nbe a horizontally


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projection as a limit case. These examples are both surjective and have connected

fibres, no two of which are parallel as oriented line segments. Then in Theorem

5.3.5 we characterize these examples as the only harmonic morphisms from anopen subset ofR3 to a closed Riemann surface satisfying the above conditions (up

to isometries ofR3 and conformal transformations of the surface).

We use a result of L.Berard Bergery to give many new examples of homogeneous

harmonic morphisms : (G/K,g) (G/H, g) between reductive homogeneousspaces. We then show that some well known examples from 3-manifolds to surfaces

can be obtained in this way.

0.3. Acknowledgements.

First and foremost I thank my supervisor John C. Wood for his encouragement,

extremely useful ideas and constructive criticism. It has been a great privilege to

work with him. I also thank his family for their warm hospitality while I stayed at

their home in Bures-Sur-Yvette during their stay in France.

I thank Paul Baird and Chris Wood for many useful discussions and their con-

tinuing interest in this work.This research was supported by grants from The Foreign and Commonwealth

Office and The Committee of Vice-Chancellors and Principals of the Universities

of the United Kingdom and a loan from The Icelandic Governments Student Loan

Fund (LIN).

Last but not least I would like to thank my best friend Gudrun for all her patience

and understanding, whilst I was so often lost in mathematical orbit.


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CHAPTER 1. HARMONIC MORPHISMS.

The main purpose of this chapter is to set up the necessary framework for this

thesis. We introduce the notions of harmonic maps and harmonic morphisms be-

tween Riemannian manifolds. We then make the reader familiar with the basic

facts and examples needed later on.

1.1. Definitions.

Throughout we assume, unless otherwise stated, that all our objects such as man-

ifolds, metrics and maps are smooth, that is, in the C-category. By (Mm, g, )and (Nn, h, N) we denote two Riemannian manifolds with their Levi-Civita con-nections. The positive integers m and n denote the dimensions of Mm and Nn

respectively.

For a map: (Mm, g) (Nn, h) its energy densityis a function e:M R+0,given by

e:= 1

2|d|2 = 1

2

mi=1

|d(Xi)|2,

where{

X1

,...,Xm}

is any local orthonormal frame for the tangent bundle T M of

M. For any compact subset ofM, the energyof on is given by

E(, ) :=

e dM,

where dM is the volume element on M. A map : (M, g) (N, h) is calledharmonicif it is a critical point of the energy functional E( , ) for every compactsubset of M. This means that for every compact subset and 1-dimensional

family of maps t : (M, g)

(N, h), such that = 0 and (x) = t(x) for all

x M and all t, we haved

dtE(t, )|t=0 = 0.

IfN= R, then a harmonic map : (M, g) R is called aharmonic functiononM.For a detailed account on harmonic maps we refer the reader to [Eel-Lem-1-2-3].

We now define the objects of central interest for this work, the harmonic mor-

phisms.

Definition 1.1.1. A map : (M, g)(N, h) is called a harmonic morphism if forany harmonic function f :U R, defined on an open subset U ofN with1(U)non-empty,f : 1(U) R is a harmonic function.


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Thus, harmonic morphisms are maps which pull back germs of harmonic func-

tions on N to germs of harmonic functions on M. An alternative description of

non-constant harmonic morphisms : (M, g) (N, h) is that they map Brownianmotions on Mto Brownian motions on N. For this see [Lev], [Ber-Cam-Dav] and

[Dar].

1.2. Basic Properties.

First of all we mention the following composition law for harmonic morphisms:

Lemma 1.2.1. (The First Composition Law) If 1 : (M, g) (N, h) and 2 :(N , h) (N, h) are harmonic morphisms, so is the composition2 1 : (M, g) (N, h).

Proof. This follows directly from Definition 1.1.1.

The next result was proved independently by Fuglede and Ishihara in [Fug-1]

and [Ish].

Proposition 1.2.2. Let : (Mm, g)

(Nn, h) be a harmonic morphism. If

m < n, then is a constant map.

One can think of a Brownian motion on a Riemannian manifold (M, g) as a

mathematical description of a particle travelling at random along M. If one does,

then the last proposition becomes intuitively clear, as follows: Ifwas a Brownian

motion on M, then a non-constant harmonic morphism : (Mm, g) (Nn, h)would map to a Brownian motion() onN. The corresponding particle would

therefore stay inside the image (M) of. Ifm < n, then(M) has measure zeroin N, so () cannot describe a random motion on N.

Constant maps are trivially harmonic morphisms but not very interesting, so

from now on we will assume that all our maps are non-constant. For our harmonic

morphisms: (Mm, g) (Nn, h) this means that we will always have m n. Bythe codimensionof we mean the non-negative integer codim() :=m n.

The following result, proved independently in [Fug-1] and [Ish], serves as one of

the most useful devices when dealing with the geometry of harmonic morphisms.

Theorem 1.2.3. A non-constant map : (Mm, g) (Nn, h) with m n is aharmonic morphism if and only if it is a harmonic map, and horizontally conformal.


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Since the notion of horizontal conformality is still not a standard feature in the

literature, we now give its definition.

Definition 1.2.4. Form n, a non-constant map : (Mm, g) (Nn, h) andx MputVx := Ker dx TxM andHx :=Vx TxM. IfC :={x M| dx = 0}and Mm := M C, then : (M, g)(N, h) is said to be horizontally (weakly)conformalif there exists a function : M R+ such that

2(x)g(X, Y) =h(d(X), d(Y)),

for all X, Y Hx, andx M. The functionis then extended to the whole ofM

by putting |C 0. The extended function: M R+0 is called the dilationof.

Note that it follows from the last definition that if : (Mm, g) (Nn, h) ishorizontally conformal, then dx : TxM T(x)N is of rank n on Mm, and 0 onC.

IfM2 andN2 are connected Riemann surfaces, then : M2

N2 is a harmonic

morphism if and only if isholomorphic. Harmonic morphisms can therefore bethought of as generalizing holomorphic maps between Riemann surfaces to higherdimensions.

It follows directly from Definition 1.2.4 that the function2 :M R+0 is smooth.By grad(2) we shall denote the gradient of2, which is a smooth section ofT M.

On ( Mm, g),V :={Vx| x M} andH :={Hx| x M} are smooth distributionsor subbundles ofTM, the tangent bundle ofM. They are called the verticaland

horizontal distributionsdefined by . ByV andH we also denote the projectionsontoVx andHx at each point x M. On M we have the unique orthogonaldecomposition of the gradient of2 into its vertical and horizontal parts, given by

grad(2) = gradV(2) + gradH(

2).

Definition 1.2.5. A non-constant map : (M, g) (N, h) is said to behorizontally

homotheticif it is horizontally conformal and gradH(2) 0 on M.For the horizontally homothetic case we have the following lemma due to Fuglede,

see [Fug-2].


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Lemma 1.2.6. If: (M, g) (N, h) is a horizontally homothetic map, then isa submersion, that is M=M.

The horizontal homothety is therefore equivalent to2 :M R+0 being constantalong horizontal curves in (M, g). Since the horizontal homothety will play a major

role in our work, we now mention some of its geometric consequences. The following

Theorem 1.2.7 is due to Baird and Eells and was first published in [Bai-Eel]. It

turns out to be one of our main tools throughout this thesis.

Theorem 1.2.7. Let m > n 2 and : (Mm, g) (Nn, h) be a horizontallyconformal submersion. If

(a) n= 2, then is a harmonic map if and only if has minimal fibres,

(b) n 3, then two of the following conditions imply the other,(1) is a harmonic map,

(2) has minimal fibres,

(3) is horizontally homothetic.

For further understanding of the horizontal homothety we now draw attention

to our observation of a duality between horizontally conformal submersions and

weakly conformal immersions.

For m > n 2, a non-constant map i : (Nn, h) (Mm, g) is called a weaklyconformal immersionif there exists a function : N R+0, such that

2(x)h(X, Y) =g(di(X), di(Y)),

for all X, Y TxN and x N. The function is called the conformal factor

of i. It follows immediately from the definition, that 2

: N R+

0 is a smoothfunction. For weakly conformal immersions just as for the horizontally conformal

submersions, the harmonicity is in fact a very geometric condition:

Theorem 1.2.8. Let m > n 2 and i : (Nn, h) (Mm, g) be a non-constantweakly conformal immersion with conformal factor: N R+0. If(a) n= 2, theni is a harmonic map if and only if i(N) is minimal in(M, g),

(b) n 3, then two of the following conditions imply the other,(1) i is a harmonic map,

(2) i(N) is minimal in(M, g),

(3) i is homothetic, that is, is constant.


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Proof. This is just a combination of Proposition page 119 of [Eel-Sam] and Example

3.3 of [Bai-Eel].

1.3. Examples.

For m 3 we shall always denote by Sm, Rm and Hm the m-dimensionalsimply connected space forms of constant sectional curvature +1, 0 and1, i.e.the sphere, the euclidean and hyperbolic spaces. The following examples are all

maps between open subsets of these spaces. It is fairly easy to see that they are

horizontally homothetic and have totally geodesic fibres, so by Theorem 1.2.7 they

are all harmonic morphisms. For6 and7 the dilations are constant, in contrast

to1 to5. The horizontal distributionHis integrable for 1 to 6 but for 7 it isnon-integrable.

Example 1. Using the standard model for Sm Rm+1 we have Sm S0 =

({(cos(s), sin(s) e) RRm| s (0, ) ande Sm1}, < , >Rm+1). Let1 :Sm S0 Sm1 be the projection along the longitudes onto the equatorial hypersphere,

given by 1 : (cos(s), sin(s)e) e. For e Sm1 the fibre of 1 over e isparametrized w.r.t. arclength bye(s) = (cos(s), sin(s) e), wheres (0, ). Alongthe fibres we have 21(s) = 1/ sin

2(s). The level hypersurfaces of 1 are parallel

small spheresSm1| sin(s)| with constant sectional curvatures KSm1| sin(s)|= 1/ sin2(s).

Example 2. Let 2 : Rm R0 Sm1 be the radial projection, given by 2 :

x x/|x|. Fore Sm1 the fibre of2 over e is parametrized w.r.t. arclengthby

e(s) = s

e, where s

R+. Along the fibres we have 2

2(s) = 1/s2. The

level hypersurfaces are parallel spheres Sm1s with constant sectional curvatures

KSm1s = 1/s2.

Example 3. Using the standard Poincare model for Hm we have Hm H0 =

(Bm1 (0){0}, 4(1|x|2)2 Rm), whereBm1 (0) := {x Rm| |x|


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Example4. Using the standard upper half space model forHm we have Hm =

(Rm1 R+, 1x2m Rm). Let 4 : Hm Rm1 be the projection onto Rm1

followed by a homothety, given by 4 : (p, x) p, where R {0}. Forp Rm1 the fibre of4overp is parametrized w.r.t. arclength byp(s) = (p, es),where s R. Along the fibres we have24(s) = 2e2s. The level hypersurfaces of4 are parallelaffine subspaces R

m1es :={(p, es) Rm1 R+| p Rm1} with

constant sectional curvatures KRm1es

= 0.

Example5. ForHm = (Rm2 RR+, 1x2m Rm) andH

m1 = (Rm2 {0}R+, 1x2m

Rm1) define 5 : H

m

Hm1 by 5 : (p, x, y)

(p, 0,x

2 + y2).

The fibre over (p, 0, r) is the semicircle in Rm2RR+ with centre (p, 0, 0), radiusrand parallel to the coordinate plane {(0, a , b)|a, b R}. The fibre is parametrizedw.r.t. arclength by(p,r)(s) = (p, r tanh(s), r/ cosh(s)). Geometrically this map isa projection along the geodesics ofHm orthogonal to Hm1. Along the fibre 25(s) =

1/ cosh2(s). The level hypersurfaces of5are Hm1s := {(p, r tanh(s), r/ cosh(s))

Hm| p Rm2, r R+} = Rm2 spanR+{(0, tanh(s), 1/ cosh(s))}. They arepar-allelhyperbolic hyperplanes with constant sectional curvatures KHm1s =

1/ cosh2(s).

Example 6. Let 6 : Rm Rm1 be the orthogonal projection followed by a

homothety, given by 6 : (x1,...,xm) (x1,...,xm1), where R {0}. ForpRm1 the fibre of6 over p is parametrized w.r.t. arclength byp(s) = (p, s),where s R. The dilation is constant26(s) = 2. The horizontal distribution isobviously integrable, and its integral submanifolds are the parallelaffine hyperplanes

Rm1s := {(p, s) Rm| p Rm1}.

Example7. Let F = C, H or Ca, that is, the complex numbers, the quaternions

or the Cayley numbers. Put (m, n) := (2 dim F 1, dim F) = (3, 2), (7, 4) or(15, 8). Define : FF = Rm+1 FR = Rn+1 by : (x, y) (2xy, |x|2 |y|2).The restrictions 7 of to S

m Rm+1 are the well known Hopf maps. They areharmonic morphisms 7 : S

m Sn with constant dilation 7 = 2. The fibres aretotally geodesic and therefore isometric to Smn Sm. It is well known that thehorizontal distributions are nowhere integrable.


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CHAPTER 2. GEOMETRIC CONSTRAINTS.

In his very important paper [ONe], ONeill studies the geometry of Riemannian

submersions : (M, g) (N, h). He derives equations which relate the sectionalcurvatures of the two manifolds involved. These give necessary conditions for the

existence of such maps between M andN.

Riemannian submersions are special cases of horizontally conformal maps, namely

those with constant dilation 1. It is therefore very natural for us to generalizeONeills work to the case of horizontal conformality.

2.1. Horizontal Conformality.

Let :M Nbe a submersion. A vector field EonMis said to beprojectableif there exists a vector field E on N, such that d(Ex) = E(x) for all xM. Inthis case E and E are called -related. A horizontal vector field Y on (M, g) is

called basic, if it is projectable. It is a well known fact, that ifZ is a vector field

onN, then there exists a unique basic vector field ZonM, such thatZand Zare

-related. The vector field Z is called the horizontal liftofZ.

The fundamental tensors of a submersion were introduced in [ONe]. They play

a similar role to that of the second fundamental form of an immersion. ONeills

definition is as follows:

Definition 2.1.1. Let EandFbe two vector fields on M and: (M, g) N be asubmersion. Then the fundamental tensorsof are given by:

TEF := HV

EVF+ VV

EHF , andAEF := HHEVF+ VHEHF .

It is easily seen that for x M, X Hx and V Vx the linear operatorsTV, AX: TxM TxMare skew-symmetric, that is

g(TVE, F) =g(E, TVF) and g(AXE, F) =g(E, AXF)

for allE, F

TxM. We also see that the restriction ofTto the vertical distribution

T|VVis exactly the second fundamental form of the fibres of . SinceTV is skew-symmetric we get: has totally geodesic fibres if and only ifT 0.

For the special case when is horizontally conformal we have the following.


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Proposition 2.1.2. Let : (Mm, g)(Nn, h) be a horizontally conformal sub-mersion with dilation andX,Ybe horizontal vectors, then

AXY =12{V[X, Y] 2g(X, Y)gradV( 1

2)}.

Proof. ExtendX,Y to basic vector fields, and let {Vn+1, . . ,Vm} be a local orthonor-mal frame for the vertical distribution. Then by using the well known formula for

the Levi-Civita connection, we get

AXY =

VXY =

m

i=n+1g(

XY , Vi)Vi

=1

2

mi=n+1

{X g(Y, Vi) =0

+Y g(Vi, X) =0

Vig(X, Y)

g(X, [Y, Vi]) =0

+ g(Y, [Vi, X]) =0

+g(Vi, [X, Y])}Vi

=1

2{V[X, Y] gradV(g(X, Y))},

since Vi,[X, Vi],[Y, Vi] are all vertical, because X andY are basic. Now

gradV(g(X, Y)) = gradV(1

2h(X,Y))

=h(X, Y)gradV(1

2)

=2g(X, Y)gradV(1

2),

and hence the result.

We see that the skew-symmetric part of A|HH measures the obstruction tointegrability of the horizontal distributionH, and is therefore of special interest. IfH is integrable, then we have a horizontal foliation on (M, g), which we denote byFH.

We will now show that the horizontal conformality of has some very important

geometric consequences forFH. The next proposition was observed by Wood in[Woo].

Proposition 2.1.3. Let : (M, g)(N, h) be a horizontally conformal submer-sion with integrable horizontal distributionH. Then the horizontal foliationFH istotally umbilic in(M, g).


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Proof. LetX, Y be two local horizontal vector fields. Since the horizontal distribu-

tion is integrable we haveV[X, Y] = 0, so from Proposition 2.1.2 we obtain

AXY = 2

2g(X, Y)gradV(1

2).

The second fundamental formL of a leafL FH is given by

L(X, Y) = VXY =AXY = 2

2g(X, Y)gradV(

1

2),

so L is an totally umbilic submanifold of (M, g).

Definition 2.1.4. Let

Fbe a totally umbilic foliation on a Riemannian manifold

(M, g). A leafL F is called spherical if the mean curvature vector HL ofL isparallel in the normal bundle (L) ofL in (M, g). The foliation Fis calledsphericalif each leafL Fis spherical.

Lemma 2.1.5. Let(M, g)be a Riemannian manifold andf :M R be a functiononM. IfX, Yare vector fields onM, then

g(

Xgrad(f), Y) =g(

Ygrad(f), X).

Proof.

g(Xgrad(f), Y) g(Ygrad(f), X)=X(Y(f)) g(grad(f), XY) Y(X(f)) + g(grad(f), YX)=X(Y(f)) Y(X(f)) [X, Y](f) = 0.

For harmonic morphisms we now get the following result:

Proposition 2.1.6. Letm > n 3 and: (Mm, g) (Nn, h) be a non-constantharmonic morphism with totally geodesic fibres. If the horizontal distributionH isintegrable, then the foliationFH is spherical.

Proof. It follows from Theorem 1.2.7 that is horizontally homothetic, and then

by Lemma 1.2.6 that is a submersion. IfL is a leaf inFHthen the corresponding

mean curvature vector field HL is given by

HL= 2

2gradV(

1

2) =

2

2grad(

1

2).


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Let X H andV Vbe two local vector fields. Then

g(X

HL

, V) =

2

2g(

Xgrad( 1

2), V)

= 2

2g(Vgrad( 12 ), X)

= 0,

since is horizontally homothetic with totally geodesic fibres. This means thatHL

is parallel in the normal bundle.

Remark 2.1.7. It follows from Theorem 1.2.7 that we could obtain the same result

for n= 2 by assuming that is horizontally homothetic.

2.2. Necessary Curvature Conditions.

Before generalizing ONeills fundamental curvature equations to the horizontally

conformal case, we now state two standard results used in our proofs. Lemma 2.2.1

describes how the curvature tensor of (M, g) changes when the metric is changed

by a conformal factor. This formula can for example be found on page 90 in [Gro-

Kli-Mey].

Lemma 2.2.1. Let m 2 and (Mm, g, , R), (Mm, g, , R) be two Riemann-ian manifolds with their Levi-Civita connections and the corresponding curvature

tensors. Ifg=2g, then

g(R(X, Y)Z, H) =

1

2 g(R(X, Y)Z, H)

+2

2

g(X, Z)g(Ygrad( 12 ), H) g(Y, Z)g(Xgrad( 12 ), H)

+ g(Y, H)g(Xgrad( 12 ), Z) g(X, H)g(Ygrad( 12 ), Z)

+4

4

(g(X, H)g(Y, Z) g(Y, H)g(X, Z)) |grad(1

2)|2

+ g(X(1

2)Y Y(1

2)X, H(

1

2)Z Z(1

2)H)

.

The second fact needed is given in Lemma 2.2.2 and is simply one of ONeills

original equations. For this see Theorem 2.{4} of [ONe].


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Lemma 2.2.2. Letm > n2 and (Mm, g, , R), (Nn, h, N, RN) be two Rie-mannian manifolds with their Levi-Civita connections and the corresponding cur-

vature tensors. If: (Mm

, g) (Nn

, h) is a Riemannian submersion, then

g(R(X, Y)Z, H) =h(RN(X, Y)Z, H) +1

4

g(V[X, Z], V[Y, H])

g(V[Y, Z], V[X, H]) + 2g(V[X, Y], V[Z, H])

.

We now state our promised fundamental curvature equations for horizontally

conformal submersions.

Theorem 2.2.3. Let m > n 2 and (Mm, g, , R), (Nn, h, N, RN) be twoRiemannian manifolds with their Levi-Civita connections and the corresponding

curvature tensors. Let: (M, g) (N, h)be a horizontally conformal submersion,with dilation: M R+ and letRV be the curvature tensor of the fibres of. IfX,Y ,Z,H are horizontal andU, V,W, F vertical vectors, then

g(R(U, V)W, F) =g(RV(U, V)W, F)

+ g(TUW , TVF) g(TVW , TUF), (1)

g(R(U, V)W, X) =g((UT)VW , X) g((VT)UW , X) (2)

g(R(U, X)Y, V) =g((UA)XY , V) + g(AXU, AYV) g((XT)UY , V) g(TVY , TUX) (3)

+ 2

g(AXY , U)g(V, gradV(

1

2 ))

g(R(X, Y)Z, U) =g((XA)YZ, U) g((YA)XZ, U) g(TUZ, V[X, Y]), (4)

g(R(X, Y)Z, H) = 1

2h(RN(X, Y)Z), H) +

1

4

g(V[X, Z], V[Y, H])

g(V[Y, Z], V[X, H]) + 2g(V[X, Y], V[Z, H])

+2

2

g(X, Z)g(Ygrad( 12 ), H) g(Y, Z)g(Xgrad( 12 ), H)


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+ g(Y, H)g(Xgrad( 12 ), Z) g(X, H)g(Ygrad( 12 ), Z)

(5)

+

4

4 (g(X, H)g(Y, Z) g(Y, H)g(X, Z))|grad(1

2 )|2

+ g(X(1

2)Y Y( 1

2)X, H(

1

2)Z Z( 1

2)H)

.

Proof.

(1) This is simply the Gauss equation for the fibres.

(2) The proof is exactly the same as in [ONe].

(3) Extend XandYto basic vector fields, then

VR(U, X)Y = VUVXY + VUHXY VXVUY VXHUY V[U, X]Y .

From this follows:

g(R(U, X)Y, V)

=g(

U(AXY), V) + g(TU(

HXY), V)

g(X(TUY), V) g(AX(HUY), V) g(TV[U, X]Y , V)=g((UA)XY , V) + g(AUX

Y , V)

g((XT)UY , V) g(TUXY , V)=g((UA)XY , V) g(AY(HXU), V)

2g(Y, XU)g(gradV(1

2), V)

g((

XT)

U

Y , V)

g(T

V

Y , T

U

X),

which is the same as stated.

(4) Extend X, Y andZto basic vector fields, then

VR(X, Y)Z= VXVYZ+ VXHYZ VYVXZ

VYHXZ VH[X, Y]Z VV[X, Y]Z, so

g(R(X, Y)Z, U)

=g(X(AYZ), U) + g(AX(HYZ), U) g(Y(AXZ), U)


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g(AY(HXZ), U) g(AH[X, Y]Z, U) g(TV[X, Y]Z, U)=g((XA)YZ, U) + g(A

XY

Z, U) + g(AY(XZ), U)+ g(AX(YZ), U) g((YA)XZ, U) g(AYXZ, U) g(AX(YZ), U) g(AY(XZ), U) g(A[X, Y]Z, U) g(TUZ, V[X, Y])

=g((XA)YZ, U) g((YA)XZ, U) g(TUZ, V[X, Y]).

(5) The map : (M, g) (N, h) can be factorised, = idM, where idM :(M, g)

(M, 2g) and : (M, 2g)

(N, h) are defined by idM(p) = p and

(p) = (p). Now Lemma 2.2.1 applies to idM and Lemma 2.2.2 to . If we put

the formula of Lemma 2.2.1 into that of Lemma 2.2.2, the statement of (5) follows,

by using the fact that 2g(X, Y) =h(d(X), d(Y)) for all X,Y horizontal.

We now show that Theorem 2.2.3 has some interesting consequences for the

sectional curvatures involved. To make the computations easier to follow, we do

this in two steps.

Corollary 2.2.4. Let m > n 2 and : (Mm, g) (Nn, h) be a horizontallyconformal submersion, with dilation: M R+ andX, Y horizontal. Then

g(R(X, Y)Y, X) = 1

2h(RN(X, Y)Y , X) 3

4|V[X, Y]|2

+2

2

g(X, Y)g(Ygrad( 12 ), X) g(Y, Y)g(Xgrad( 12 ), X)

+ g(Y, X)g(Xgrad( 12 ), Y) g(X, X)g(Ygrad( 12 ), Y)

+4

4

|X Y|2|grad(1

2)|2 + |X( 1

2)Y Y( 1

2)X|2

.

Proof. Put Z = Y and H = X into the equation in Theorem 2.2.3.(5), and the

result follows immediately.

This last formula seems to be too complicated to be of any use in the general

case. But when assuming that : (M, g)(N, h) is horizontally homothetic, we

get the following necessary curvature condition.

Theorem 2.2.5. Let m > n 2 and : (Mm, g) (Nn, h) be a horizontallyhomothetic map, with dilation : M R+. If X and Y are horizontal vectors,


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such that|X| = |Y| = 1 andg(X, Y) = 0, then

KM(X

Y) =2

KN(X

Y)

3

4 |V[X, Y]

|2

4

4|gradV(

1

2)|2.

Proof. Since : (M, g) (N, h) is horizontally homothetic, it follows from Lemma1.2.6, that it is a submersion. The equation gradH(

12 )0 implies that X( 12 ) =

Y( 12 ) = 0 and that

g(Zgrad( 12 ), Z) = g(gradV(1

2), ZZ)

for Z {X, Y}. Then Corollary 2.2.4 and g(X, Y) = 0 give

KM(X Y) =g(R(X, Y)Y, X)=

1

2h(RN(X, Y)Y , X) 3

4|V[X, Y]|2

+2

2g(gradV(

1

2), VXX+ VYY) +

4

4|gradV( 1

2)|2

=2 KN(XY) 34|V[X, Y]|2

4

4|gradV(1

2)|2,

since by Proposition 2.1.2,VZZ=AZZ= 2

2g(Z, Z)gradV(

12 ) for Z {X, Y}.

Corollary 2.2.6. Letm > n 2 and(Mm, g), (Nn, h) be two Riemannian man-ifolds withKM(H)0 andKN 0. If : M N is a horizontally homotheticmap, then

(1) KN 0 andKM(H) 0,(2) the dilation: M R+ is constant, and(3) the horizontal distributionH is integrable.

Proof. It follows from Lemma 1.2.6 that if such a map exists, then it is a submer-

sion. LetXandYbe horizontal vectors, such that |X| = |Y| = 1 andg(X, Y) = 0.For the above cases we then obtain from Theorem 2.2.5

0

KM(X

Y)

2

KN(X

Y) =

3

4 |V[X, Y]

|2

4

4|gradV(

1

2

)

|2

0,

from which the result follows.

Another necessary curvature condition is given by the following:


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Proposition 2.2.7. Letm > n2 and : (Mm, g)(Nn, h) be a horizontallyconformal submersion with constant dilation and totally geodesic fibres. Further let

X H andV V, such that|X| = 1 and|V| = 1, then

KM(X V) = |AXV|2.

Proof. Since has totally geodesic fibres we have T 0. The dilation of is constant, so it follows from Proposition 2.1.2 that the fundamental tensor A is

skew-symmetric. Hence AXX= 0 and

V((VA)XX) = V(VAXX AVXX AXVX) = 0.

The result now immediately follows from Theorem 2.2.3.(3).


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CHAPTER 3. CONSTANT CURVATURE.

In this chapter we assume that (M, g) and (N, h) are simply connected space

forms. We study non-constant harmonic morphisms : U N from open con-nected subsetsU ofM. The horizontal conformality of is obviously independent

of homothetic changes of the metrics concerned. The same is true for the har-

monicity of. Since we are assuming that our manifolds have constant sectional

curvatures, we can without loss of generality restrict our attention to the cases

when (M, g) and (N, h) are the standard spheres, euclidean or hyperbolic spaces

withKM, KN

{1, 0, 1

}.

3.1. Known Results.

First of all we remind the reader that for any m 3 we have the followingexamples of harmonic morphisms of codimension 1, given in section 1.3:

1 : Sm S0 Sm1,

2 : Rm R0 Sm1,

3 :Hm H0 Sm1,

4 :Hm Rm1,

5 :Hm Hm1,

6 : Rm Rm1,

and the harmonic morphism of codimension m n:

7 :Sm Sn,

with (m, n) = (3, 2), (7, 4) or (15, 8).

For harmonic morphisms from simply connected space forms there exist two

very interesting classification results. The first one is due to Baird and Wood, see

[Bai-Woo-1-2].

Theorem 3.1.1. Let (M3, g) = S3,R3 orH3 and (N2, h) be a surface. Further

letUbe an open and connected subset ofM3.

(i) If : M3 N2 is a non-constant harmonic morphism, then up to isometriesofM3, is one of4, 5, 6 or7, followed by a weakly conformal map.


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(ii) If : U N2 is a harmonic morphism with an isolated singularity, then upto isometries ofM3, is a restriction of one of1, 2 or3, followed by a weakly

conformal map.

In [Kas-Was], Kasue and Washio were able to generalize to higher dimensions in

the case ofM= Rm, when assuming that has totally geodesic fibres:

Theorem 3.1.2. Letm > n 3 and : Rm (Nn, h) be a non-constant har-monic morphism with totally geodesic fibres. ThenN= Rn and is an orthogonal

projection, followed by a homothety.

The reader should note that both Theorem 3.1.1.(i) and Theorem 3.1.2 are of

global nature, that is, (M, g) is complete. In contrast, most of our results will be

local, that is, concerning harmonic morphisms :U Ndefined on an arbitraryopen and connected subset U ofM.

3.2. Examples of Higher Codimension.

The examples1 to6 are all of codimension 1. The First Composition Law for

harmonic morphisms given in Lemma 1.2.1 allows us to compose these maps and

thereby construct examples of arbitrary codimension. In this way we define the

following maps 1, 2,..., 6, which we shall call the standard harmonic morphisms.

1 :Sm Sm(n+1) 1 1 Sn+1 S0 1 Sn,

2 : Rm Rm(n+1) 6 6 Rn+1 R0 2 Sn,

3 :Hm Hm(n+1) 5 5 Hn+1 H0 3 Sn,

4 :Hm 5 5 Hn+1 4 Rn,

5 :Hm 5 5 Hn,

6 : Rm 6 6 Rn.

We now study some special properties of these maps. To be able to do so we

need the next lemma. For two basic vector fields XandY on (M, g) we denote by

XYthe horizontal lift ofNXY.


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Lemma 3.2.1. If : (Mm, g)(Nn, h) is a horizontally conformal submersionandX,Yare basic vector fields onM, then

HXY = XY +2

2{X( 1

2)Y + Y(1

2)X g(X, Y)gradH(1

2)}.

Proof. We choose a local orthonormal frame{Zi|i = 1, . . ,n} on (N, h) and lift Zihorizontally to Zi. Then{Zi|i = 1, . . ,n} is a local orthonormal frame for thehorizontal distribution of (M, g), so we get

HXY =n

i=1g(XY , Z i)Zi =2

n

i=1g(XY , Zi)Zi

=2

2

ni=1

X(g(Y, Zi)) + Y(g(Zi, X)) Zi(g(X, Y))

g(X, [Y, Zi]) + g(Y, [Zi, X]) + g(Zi, [X, Y])

Zi

=2

2

ni=1

X(

1

2)h(Y , Zi) +

1

2X(h(Y , Zi))

+ Y(1

2)h(Zi, X) +

1

2Y(h(Zi, X))

Zi(12

)h(X, Y) 12

Zi(h(X, Y))

+ 1

2{h(X, [Y , Zi]) + h(Y , [Zi, X]) + h(Zi, [X, Y])}

Zi.

We then apply the differential d and obtain

d(HXY)

=

n

i=1 h(

N

XY ,

Zi)

Zi+

2

2{X(1

2 )

n

i=1 h(

Y ,

Zi)

Zi

+ Y(1

2)

ni=1

h(X, Zi)Zi h(X, Y)ni=1

Zi(1

2)Zi}

= NXY +2

2{X( 1

2)Y + Y(

1

2)X g(X, Y)d(gradH( 1

2))},

from which the result immediately follows.

As already mentioned in section 1.3, 1 to 6 are all horizontally homothetic

with totally geodesic fibres. They also have integrable horizontal distributions.

The following two lemmas show that this is also true for their compositions 1 to

6.


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Lemma 3.2.2. (The Second Composition Law) If 1 : (M, g) (N, h) and2 :(N , h) (N, h) are two horizontally homothetic maps, with totally geodesic fibres,

so is the composition=2 1 : (M, g) (N, h).Proof. Fori= 1, 2 let i denote the dilation ofi and

1 2 the pull-back of2 via

1, given by (1 2)(x) := 2 1(x) for all xM. If is the dilation of, then

2 =21(1 2)

2. IfXis a horizontal vector, then

X(2) =X(21) 0

(1 2)2 + 21X((

1 2)

2)

=21d1(X)(22) = 0,

so is horizontally homothetic.

The vertical distributionV of splits naturally into two orthogonal parts,V=V1 W, whereV1 := Ker d1 andW :=V1 V. It is easily seen, thatV2 =Ker d2 = d1(W). Ifx M and V = V1+W Vx such that V1 (V1)x andW Wx, then we extend Vonto a small neighbourhood ofx, such that W is abasic vector field w.r.t 1. Let X Hx be any horizontal vector w.r.t. , then

g(VV , X) =g(V1V1, X) 0

+g(V1W+ WV1, X) + g(WW , X)

= 2 g(TV1W , X) 0

+1

2h(d1(WW), d1(X))

= 1

2h(Nd1(W)d1(W), d1(X)) = 0

by Lemma 3.2.1 and the fact that 2 has totally geodesic fibres.

Since the second fundamental formFof a fibre F of is symmetric, it followsfrom

F(U, V) = 14{F(U+ V, U+ V) F(U V, U V)},

that has totally geodesic fibres.

Lemma 3.2.3. (Third Composition Law) Let 1 : (M, g) (N , h) and 2 :(N , h) (N, h) be two submersions. If1 and2 have integrable horizontal distri-

bution, so has the composition= 2 1 : (M, g) (N, h).Proof. For i = 1, 2 letVi andHi denote the vertical and horizontal distributionsof i, respectively. Let X, Y be two local vector fields on N, and X, Y their


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horizontal lifts via 2. Further let X, Y H H1 be the horizontal lifts of X,Y via 1. The integrability ofH1 impliesV1[X,Y] = 0, soV[X, Y] H1. But

nowd1(V[X, Y]) = V2[d1(X), d1(Y)] = V2[X, Y] = 0, sinceH2 is integrable, soV[X, Y] = 0.

3.3. A Higher Dimensional Classification.

First of all we have a non-existence result. It explains why the following cases

are missing in the list of examples, given in the last section 3.2.

Theorem 3.3.1. Letm > n2 and(M, N)=(Sm

,Rn

), (Sm

, Hn

) or(Rm

, Hn

).If U is an open subset of M, then there exists no horizontally homothetic map

: U N.

Proof. It follows from Lemma 1.2.6 that if such a map exists, then it is a submer-

sion. LetXandYbe horizontal vectors, such that |X| = |Y| = 1 andg(X, Y) = 0.For the above cases we then obtain from Theorem 2.2.5

0< KM(X Y) 2 KN(XY) = 3

4 |V[X, Y]|2 4

4|gradV(1

2 )|2 0.

This contradicts the existence of.

It was noted in (5.6) of [Bai-Eel] that a non-constant dilation : M R+of a horizontally homothetic harmonic morphism : (Mm, g) (Nm1, h) is anisoparametric function, or equivalently,FH is an isoparametric foliation of codi-mension 1. For basic facts on isoparametric systems, see Appendix A.

Proposition 3.3.2. Let (M, g) be a Riemannian manifold of constant sectional

curvature, andFH be a totally umbilic foliation on M. Then the following areequivalent:

(1)FH is spherical, and(2)FH is isoparametric.

Proof. Let

V be the full distribution orthogonal to

FH, L be a leaf in

FH and

HL be the mean curvature vector field ofL. ThenL(X, Y) = g(X, Y)HL, andconsequently the shape operatorSV satisfiesSV = g(V, HL)idTMfor any normalfieldV V.


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Suppose that (1) holds. To show that L is isoparametric we must prove that

its normal bundle (L) =V|L is flat and that the principal curvatures ofL in the

direction of any parallel normal field are constant.Let V and Vbe two local normal fields along L. Then the two shape operators

SVand SVare both multiples of the identity so they commute, i.e.

[SV, SV] =SV SV SV SV = 0.

It then follows from Proposition 2.1.2 of [Pal-Ter] that (L), the normal bundle of

L in (M, g), is flat.

Let Xbe a local horizontal vector field and W be a parallel normal field along

L, then

Xg(W, HL) =g(XW , HL) + g(W,XHL) = 0,

since both W and HL are parallel in the normal bundle. This means that L has

constant principal curvatures in the direction of any parallel normal field, soFH isisoparametric.

Conversely, suppose that (2) holds. Let Vbe a local parallel normal field along

L. Then

g(XHL, V) =X g(HL, V) g(HL, XV) = 0,

since V is parallel andg(V, HL) is constant along L because L is isoparametric.This means that HL is parallel in the normal bundle, soFH is spherical.

Corollary 3.3.3. Let m > n 2, (Mm, g), (Nn, h) be simply connected spaceforms andUbe an open and connected subset ofM. Let : U

Nbe a horizontally

homothetic harmonic morphism with totally geodesic fibres and integrable horizontal

distribution. ThenFH is a totally umbilic isoparametric foliation onU.

Proof. This follows directly from Proposition 2.1.6, Remark 2.1.7 and Lemma 3.3.2.

Corollary 3.3.4. Letm > n 3 and(M, N) = (Sm, Sn), (Rm, Sn) or(Hm, Sn).Then there exists no harmonic morphism : M N with totally geodesic fibres

and integrable horizontal distribution.

Proof. Suppose that such a map : M Nexists. Then by Theorem 1.2.7 andLemma 1.2.6is a submersion so by Corollary 3.3.3 it would define a totally umbilic


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isoparametric foliationFH on the whole ofM without focal varieties. Let L be aleaf inFH, then the restriction |L : L Sn is a homothety, so L has constant

positive curvature. Such a foliation must have focal varieties, which contradicts thehypothesis.

Theorem 3.3.5. Letm > n 2,(M, N)=(Sm, Sn),(Rm, Sn),(Hm, Sn),(Hm,Rn),(Hm, Hn) or (Rm,Rn), andi : Ui N be the corresponding standard harmonicmorphism. Further letUbe an open and connected subset ofM and: U N bea horizontally homothetic harmonic morphism. If has totally geodesic fibres and

integrable horizontal distribution, then up to isometries ofM andN, = i

|Ui

.

Proof. It follows from Corollary 3.3.3 that determines a totally umbilic isopara-

metric foliationFU on Uwithout singularities. By Appendix A this foliation is upto isometries ofM uniquely determined. This means that there exists a foliation

preserving isometric embedding : (U, FU) (Ui, FUi), whereFUi is the foliationgiven by i : Ui N. Sinceis foliation preserving, there exists a map : N N,such that the following diagram commutes.

U Ui

iN

NThe maps and iare horizontally conformal and is an isometry. This implies,

that is conformal. If2 is the corresponding conformal factor, then2 :U

R+, the pullback of2 via , satisfies

2 2 =2i .

Now2i and2 are horizontally constant, so2 is. But being a pull-back

2

is also vertically constant. This means that 2 is constant, so is a homothety. If

N=Sn orHn this means that 2 = 1, but when N= Rn, can be any element

ofR+.

For m > n

2 Theorems 3.3.1 and 3.3.5 give a classification for horizontally

homothetic harmonic morphisms: U (Mm, g) (Nn, h) with totally geodesicfibres and integrable horizontal distribution between open and connected subsets

of simply connected space forms. For m > n 3 we have the following:


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Corollary 3.3.6. Let m > n 3 and (M, N)=(Sm, Sn), (Rm, Sn), (Hm, Sn),(Hm,Rn), (Hm, Hn) or(Rm,Rn), andi : Ui Nbe the corresponding standard

harmonic morphism. Further let U be an open and connected subset of M and: U Nbe a harmonic morphism with totally geodesic fibres. If has integrablehorizontal distribution, then up to isometries ofM andN, = i|Ui .

Proof. It follows from Theorem 1.2.7 that the map : U N is horizontallyhomothetic. The result then follows from Theorem 3.3.5.

To see that the condition of integrability is in general necessary, consider the

following harmonic morphisms.

8 : R4 R1 6 R3 R0 2 S2,

9 : R4 R1 2 S3 S0 7 S2.

They are both horizontally homothetic with totally geodesic fibres. The correspond-

ing vertical foliations are fundamentally different, so the maps must be different. It

is the non-integrability of the horizontal distribution of 9

that makes this possible.

In the special case of (M, N) = (Rm,Rn) the condition of integrability of the

horizontal distribution is automatically satisfied, so we get:

Corollary 3.3.7. Letm > n3, U be an open and connected subset ofRm and : U Rn be a harmonic morphism, with totally geodesic fibres. Then is anorthogonal projection, followed by a homothety.

Proof. Since is a harmonic morphism with totally geodesic fibres and n

3,

it follows from Theorem 1.2.7, that it is horizontally homothetic. It then follows

directly from Theorem 2.2.5, that its horizontal distribution H is integrable andthe dilation is constant. The result then follows from Theorem 3.3.5.

The reader should compare Corollary 3.3.7 with the result of Kasue and Washio,

given in Theorem 3.1.2.

If has codimension 1, then its fibres are automatically totally geodesic, which

implies:

Corollary 3.3.8. Letm 3, (M, N) = (Sm, Sm1), (Rm, Sm1), (Hm, Sm1),(Hm,Rm1), (Rm,Rm1), or(Hm, Hm1) andUbe an open and connected subset


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of M. If : U N is a horizontally homothetic harmonic morphism, then has constant dilation, or up to isometries of M andN, is one of the standard

harmonic morphisms.

Proof. If the dilation is not constant, there exists a point x U, such thatgrad(2) = 0 on an open neighbourhoodW Uofx. It follows from gradH(2) = 0that H is integrable onWand its integral manifolds are the level hypersurfaces of.Since is of codimension 1 it follows from Theorem 1.2.7, that its fibres are totally

geodesic. We can therefore apply Theorem 3.3.5 onW. Hence up to isometries|Wis the restriction of one of the standard maps i. Then by the unique continuation

principle of Sampson (see [Sam]) for harmonic maps this is in fact the case on the

whole ofU.

When has constant dilation and totally geodesic fibres, we get the following:

Theorem 3.3.9. Let m > n 2, (Mm, g), (Nn, h) be simply connected spaceforms, and U an open and connected subset of M. If : U N is a harmonicmorphism with constant dilation and totally geodesic fibres, then(M, N) = (Sm, Sn)

or(Rm,Rn).

Proof. The cases (Sm,Rn), (Sm, Hn) and (Rm, Hn) are excluded by Theorem 3.3.1.

Let X, Y H and V Vbe local vector fields, such that|X| = |Y| = |V| = 1 andg(X, Y) = 0. By Theorem 2.2.5 and Proposition 2.2.7 we get:

(1) KM(X V) = |AXV|2, and(2) KM(X Y) =2 KN(XY) 34 |V[X, Y]|2.

Equation (1) excludes the cases (Hm, Sn), (Hm,Rn) and (Hm, Hn). IfM = Rm,

then follows from (1) that AXV = 0. The linear operatorAX is skew-symmetric,

so AXY = 1

2V[X, Y] = 0, which is equvalent to the horizontal distribution beingintegrable. This makes the case (Rm, Sn) impossible by (2).

In Corollary 3.3.7 we have already dealt with (M, N) = (Rm,Rn). If (M, N) =

(Sm, Sn) we have the following theorem:

Theorem 3.3.10. Let m > n 2 and : Sm Sn be a harmonic morphismwith constant dilation. If has totally geodesic fibres, then up to isometries ofSm

andSn it is one of the Hopf maps.


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Proof. The map has constant dilation so, up to a homothety ofSn, it is a Rie-

mannian submersion with totally geodesic fibres. The result then follows from

Escobales classification of such maps between spheres, see [Esc].

For the local case we conjecture the following:

Conjecture 3.3.11. Let m > n 2, U be an open and connected subset of Smand : U Sn be a harmonic morphism with constant dilation. If has totallygeodesic fibres, then up to isometries ofSm andSn, is a restriction of one of the

Hopf maps.


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CHAPTER 4. MINIMAL SUBMANIFOLDS.

Throughout this chapter we shall assume that m > n 2 and that : (Mm, g) (Nn, h) is a horizontally conformal submersion. We are mainly interested in the

following question: Under what extra conditions on are the inverse images of

minimal submanifolds ofNminimal in M?

4.1. Horizontally Conformal Maps and Minimal Submanifolds.

As before we denote byH andV the horizontal and vertical distributions of : (M, g) (N, h). Let L be a submanifold of N. Since is a submersionK :=1(L) is a submanifold ofM. For x Kwe define

(H1)x :=TxK Hx and (H2)x :=TxK.

This means that along Kwe get the following orthogonal decompositions

T K= V H1,H = H1 H2 and T M= V H =T K H2.

By H1 and H2 we shall also denote the orthogonal projections onto the correspond-ing subbundles ofT M along K.

ByKwe denote the second fundamental formK :T K T K H2 ofK inM, and similarily by L the second fundamental form L :T L T L (L) ofLinN. The corresponding mean curvature vector fields are denoted byHKand HL.

Definition 4.1.1. Let : (M, g)(N, h) be a horizontally conformal submersion,L be a submanifold ofN and K = 1(L) M. The map is called normallyhomotheticalong K if gradH2(

12 ) = 0 onK.

Note that if is horizontally homothetic, i.e. gradH( 12 ) = 0, then is auto-

matically normally homothetic along K for any submanifold L ofN.

Lemma 4.1.2. Let : (Mm, g) (Nn, h)be a horizontally conformal submersionwith minimal fibres and letLbe a submanifold ofN. Then the following conditions

are equivalent:

(1) K=1(L) is minimal inM, and

(2) HL = 12 d(gradH2(

12 )).


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Proof. Choose a local orthonormal frame{ Xi| i = 1, . . ,l} for T L, then the nor-malized horizontal lifts{Xi| i = 1, . . ,l} form a local orthonormal frame forH1.

Further choose a local orthonormal frame {Vr| r= n + 1, . . ,m} for V. For the meancurvature vector HK ofK in Mwe now get

(m(n l))HK=l

i=1

K(Xi, Xi) +m

r=n+1

K(Vr, Vr) =0, since fibres minimal

=2l

i=1

H2(XiXi),

and from Lemma 3.2.1

=2l

i=1

H2(XiXi+2

2{2Xi( 1

2)Xi g(Xi, Xi)gradH(1

2)})

=2l

i=1

{H2(XiXi) 2

2g(Xi, Xi)gradH2(

1

2)}.

If we now apply the differential d then

(m

(n

l))d(HK) =2

l

i=1

L(Xi, Xi)

2

2

l

i=1

h(Xi, Xi)d(gradH2

(1

2))

=2lHL l2

2 d(gradH2(

1

2)),

from which the the statement immediately follows.

Corollary 4.1.3. Let: (M, g) (N, h) be a horizontally conformal submersionwith minimal fibres andL be a submanifold ofN. If is normally homothetic along

1(L), then the two following conditions are equivalent:

(1) 1(L) is minimal inM, and

(2) L is minimal inN.

Proof. This follows directly from Proposition 4.1.2.

For harmonic morphisms we get the following version of Corollary 4.1.3.

Theorem 4.1.4. Let m > n 2, : (Mm, g) (Nn, h) be a horizontallyhomothetic harmonic morphism andL be a submanifold ofN. Then the following

conditions are equivalent.

(1) L is minimal inN, and

(2) 1(L) is minimal inM.


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Proof. The map is horizontally homothetic, so by Lemma 1.2.6 it is a horizontally

conformal submersion. It then follows from Theorem 1.2.7 that has minimal

fibres, so Corollary 4.1.3 applies.

Note that Theorem 4.1.4 can be applied to any of the standard harmonic mor-

phisms 1,..., 6 defined in section 3.2 and to the Hopf maps 7. They could

therefore be used to construct minimal submanifolds in space forms.

4.2. Examples.

In this section we find horizontally conformal submersions : (C)m C

which are in general not horizontally homothetic. They have minimal, non-totally

geodesic fibres. We then use these maps to construct minimal foliations F :={F1({et| t R})| S1} of real hypersurfaces of (C)m.

LetFbe the minimal foliation ofC given by

F := { l = {et| t R} | S1 C}.

For a holomorphic function f : U C from an open subset U ofCm we definef :U C, by

f := 2C

f ,

where gradC(f) = (f/z1,...,f/zn).

Lemma 4.2.1. Let f : U C be a holomorphic submersion on an open subsetU ofCm. If f(U) R {0}, then the foliationFf ={f1(l)| S1} of real

hypersurfaces inU is minimal.

Proof. The function f = u +iv : U C is holomorphic, so by [Fug-1] it ishorizontally conformal with dilation , satisfying 12 = |gradC(f)|2. Then oneeasily checks that

grad(1

2) = 2 gradC(

1

2) = 2 gradC(|gradC(f)|2).

The horizontal space

Hx at x

U is given by

Hx = spanR

{grad(u), grad(v)

} =

spanC{gradC(f)}, so

gradH(1

2) =

C

|gradC(f)|2 gradC(f).


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Then applying the differential df gives

df(gradH

(1

2)) =

C

|gradC(f)|2 gradC(f) gradC(f)t

=C

= 2< gradC(f), grad(1

2)>=f f.

It then follows from f(U) R {0}, that

df(gradH(1

2)/f) =f,

so the vector fieldX:= gradH( 12 )/fis basic. It then follows that the vector field

X :=df(gradH(1

2)/f)

satisfies X(p) = p, so the elements ofF are the integral curves of X. HencegradH(

12 ) is tangential to f

1(l) for all S1, so f is normally homotheticalong f1(l) for all S1. It then follows from Corollary 4.1.2 that Ff is aminimal foliation ofU (C)m.

For our examples we need the following lemma.

Lemma 4.2.2. Let f : U Cm C, and g : V Cn C be holomorphicfunctions, such thatf(U) R+ andg(V) R+. Ifh: U V C is given byh(z, w) :=f(z)g(w), thenh(U V) R+.

Proof. This follows directly from the fact that

h= |g|2f+ 2|gradC(f)|2|gradC(g)|2 + |f|2g.

Examples 4.2.3. Define F : (C)m C byF(z) := a0mi=1 z

kii , where a0 C

and ki > 1. Further let FF :={F1(l)| S1}. Then the foliation FF of realhypersurfaces in (C)m is minimal.

Proof. Iff : C C is given by f :z a0zk then f(z) = 2k3(k1)|a0|2|z|2k4 R+ for all z C. The result then follows from Lemmas 4.2.1 and 4.2.2.


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CHAPTER 5. MULTIVALUED HARMONIC MORPHISMS.

5.1. Locally Defined Harmonic Morphisms.

In [Bai-Woo-1-2], Baird and Wood classify locally defined harmonic morphisms

from 3-dimensional simply connected space forms (M3, gM) to Riemann surfaces

N2, in terms of meromorphic functions defined on the surface. This description

implies that globally there exist only very few such examples, see Theorem 3.1.1.

For later use we describe this local classification in the case that M3 = R3:

Letg, h: N2 C{}be two meromorphic functions satisfying the followingconditions:

(1) Ifg(z0) is finite, then h(z0) is finite, and (5.1.1)

(2) Ifg(z0) = , then limzz0h(z)/g2(z) is finite. (5.1.2)For such meromorphic functions g , h: N2 C{}, z N2 andx R3 we havethe following equation:

G(z, x) := (1 g2(z))x1+ i(1 + g2(z))x2+ 2g(z)x3 2h(z) = 0. (5.1.3)

To make sense of this equation at a pole z0 N2 ofg we must divide through byg2(z) and treat it as a limit.

By a local solution of this equation we mean a map : U N2 defined onan open subset U ofR3, satisfying G((x), x) = 0. The above mentioned clas-

sification says: Every local solution to equation (5.1.3) is a harmonic morphism,

and every locally defined harmonic morphism is a local solution to (5.1.3) for some

meromorphic pair (g, h) as above.

For the cases of (M3, gM) =S3 orH3 there exists a similar description for locally

defined harmonic morphisms. For M3 =S3 see section 5.4.


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5.2. The Covering Construction.

In complex analysis the analytic continuation or gluing together of locally

defined holomorphic functions leads to the construction of Riemann surfaces. In thissection we generalize this to harmonic morphisms from an arbitrary Riemannian

manifold (Mm, gM) to a surface (N2, gN). Proposition 5.2.2 and Theorem 5.2.5

resulted from joint work with J.C.Wood, see also [Gud-Woo].

Let the product manifold Nn Mm be equiped with the product metric g =gN gM.

Definition 5.2.1. LetG : (Nn

Mm, g)

(Pp, gP) be a map. Forz

Nandx

M

we denote by Gz : M P and Gx : N P the maps given by Gz : yG(z, y)and Gx : w G(w, x), respectively. G is said to be a harmonic morphism ineach variable separately ifGz : (M, gM) (P, gP) and Gx : (N, gN) (P, gP) areharmonic morphisms for every z N andx M.

Iff :U Pp Ris a function andG : Nn Mm Pp is a smooth map, thenit is easily verified that

NM(f G)(z, x) = N(f Gx)(z) + M(f Gz)(x)

for all (z, x) NM. Thus ifGis a harmonic morphism in each variable separately,then it is a harmonic morphism as a map from the product manifold.

Proposition 5.2.2. Let(Mm, gM), (N2, gN) and (P

2, gP) be Riemannian man-

ifolds with dim N2 = dim P2 = 2, and let G : N2 Mm P2 be a harmonicmorphism in each variable separately. Ifw is a fixed point onP2 anddG

= 0 on

G1(w), then any smooth local solution: U N2, z = (x) to the equation

G(z, x) =w, (5.2.1)

defined on an open subsetU ofMm, is a harmonic morphism.

Proof. Let z = (x) be a solution to (5.2.1) on U Mm through x0 and putz0 =(x0). To prove that is a harmonic morphism we show that it is horizontally

conformal and harmonic.

Since Gx is conformal and non-constant in a neighbourhood of (z0, x0) and our

problem is local, we can without loss of generality assume that N2 and P2 are


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connected and oriented and that Gx is holomorphic. Then we may choose local

complex coordinatesz and w onN2 andP2 in neighbourhoods ofz0 andw respec-

tively, and normal coordinates x= (x1,...,xn) centred at the point x0 Mm

. In aneighbourhood ofx0 a local solution: x z satisfies

G(z(x), x) =w.

Differentiating with respect to xi gives:

G

z

z

xi+

G

xi= 0. (5.2.2)

Note thatdG = 0 onG1(w), so equation (5.2.2) implies thatdGx =G/z= 0,for ifG/z= 0 then by (5.2.2) G/xi= 0 for alli, sodG= 0, contradicting the

hypothesis. Hencez

xi= ( G

z)1 G

xi.

Since Gzis a harmonic morphism it is horizontally conformal, that ism

i=1(G/xi)2 =

0 at x0, thus at that point,mi=1

(z

xi )2 = 0. (5.2.3)

Differentiating (5.2.2) with respect to xi gives:

2G

z 2(

z

xi)2 +

G

z

2z

x2i+

2G

x2i= 0.

Summing using (5.2.3), we have at x0,

G

z

m

i=1

2z

x2i+

m

i=1

2G

x2i= 0.

SinceG/z= 0 and the last term vanishes we conclude thatmi=1

2z

x2i= 0. (5.2.4)

At the pointx0equations (5.2.3) and (5.2.4) are the conditions for horizontal confor-

mality and harmonicity respectively. This implies that is a harmonic morphism.

In general there is not a unique solution to equation (5.2.1); we need the following

concept:


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Definition 5.2.3. Let C(Nn) be the set of all closed subsets ofNn. By a multivaluedharmonic morphism from (Mm, gM) to (N

n, gN) we shall mean a mapping :

Mm

C(Nn

), such that any smooth map : U Nn

defined on an open subsetU of Mm satisfying (x) (x) for all x U is a harmonic morphism. Such amap will be called a branchof .

Note that Proposition 5.2.2 says that the set-valued mapping : Mm C(N2)with (x) = {z N2| G(z, x) =w} is a multivalued harmonic morphism.

Definition 5.2.4. We call : Mm C(N2) given by (x) := {z N2| G(z, x) =w}the multivalued harmonic morphismdefined by equation (5.2.1).

For a map G: N2 Mm P2 and a point w P2, define

Mm = Mmw := {(z, x) N M| G(z, x) =w},

and let = 1|M : M N and = 2|M : M Mbe the restrictions of thenatural projections 1 : N M N and 2 : N M M to M. We call theclosed subset E :=

{(z, x)

Mm

| dGx(z, x) = 0

} of Mm the envelopeofG and

its image E = (E) in Mm its geometric envelope. Further let F :={(z, x)Mm| dGz(z, x) = 0}.

The next theorem explains how Mm is the Riemannian covering manifold of

the multivalued harmonic morphism defined by equation (5.2.1).

Theorem 5.2.5. LetG : (N2 Mm, g) (P2, gP) be a harmonic morphism ineach variable separately. Suppose that for some w P2, dG= 0 along Mm :=

G1

(w). Then, with notations as above,(1) Mm is anm-dimensional minimal submanifold of(N2 Mm, g),(2) : Mm Mm is a local diffeomorphism except onE,(3) : ( Mm, g) (N2, gN) is a harmonic morphism with critical setF,(4) any local solution : U Mm N2 of equation (5.2.1), necessarily a

harmonic morphism by Proposition 5.2.2, satisfies= on1(U).

Proof. First note that the tangent space ofM at (z, x) is given by

T(z,x) M= {(Z, X) TzN2 TxMm| dG(Z, X) = 0}= {(Z, X) TzN2 TxMm| dGz(X) + dGx(Z) = 0}.


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(1) SinceG is a harmonic morphism, it is horizontally conformal, so the fact that

dG = 0 on Mm implies that dG is surjective. It follows from the implicit function

theorem that Mm

is an m-dimensional submanifold of N2

Mm

. That Mm

isminimal is a consequence of Theorem 1.2.7.

(2) SinceMm and Mm have the same dimension, we only have to show that d

is surjective outside E. Let (z, x) Mm \ Eand letX TxMm be non-zero. SinceGx is a harmonic morphism, dGx= 0 means that dGx is non-singular. Let Z :=(dGx)1 dGz(X) TzN2, then dGx(Z) + dGz(X) = 0, so (Z, X) T(z,x) Mmandd(Z, X) =X. Thus d is surjective.

(3) We look first at points (z, x) F. There the tangent space ofMm is {(0, X) TzN

2 TxMm}, so clearlyd= 0 on F.On the other hand let (z, x) Mm \ F, then we have 1(z) ={(z, x)

Mm|G(z, x) =w}, so that the vertical space V(z,x) at (z, x) Mm with respect to is given by V(z,x) ={(0, X)T(z,x) Mm| X VGzx }. Hence H(z,x) ={(Z, X)T(z,x) M

m| X HGzx }. Given Z TzN2, there exists exactly one X HGzx suchthat dGz(X) + dGx(Z) = 0, namely X= (dGz|HGzx )1 dGx(Z). From this it isclear that d|H(z,x) :H(z,x) TzN2 is given by

(Z, X) = (Z, (dGz|HGzx )1 dGx(Z)) Z.

SincedGz|HGzx anddGx are conformal, this shows that is horizontally conformal.As regards the harmonicity of , note that is the composition of the inclusion

mapi and the projection 2,

: M

m i

Mm

N2 2

N2

.

Now the composition law for the the tension field (see [Eel-Sam]) is:

() = traced2(di,di) + d2((i)).

The first term is zero since2 is totally geodesic; also Mm is minimal inN2 Mm

so that(i) = 0. Hence is harmonic, and since horizontally conformal, a harmonic

morphism.

(4) To say that is a local solution on U Mm, means thatG((x), x) =w forall x U. Thus ((x), x) Mm and ((x), x) =(x), i.e. = on 1(U).


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By Theorem 5.2.5 the local solutions : U N2 defined by equation (5.2.1) havebeen glued together to get a globally defined harmonic morphism : ( Mm, g)

(N2

, gN).

5.3. Multivalued Harmonic Morphisms from R3.

As already mentioned in section 5.1, any locally defined harmonic morphism from

an open subset U ofR3 to a Riemann surface N2, is given by two meromorphic

functionsg, h: N2 C {}. For such a pair (g, h) satisfying conditions (5.1.1)and (5.1.2) we define a meromorphic map : N2 (C {})3 by

:= 12h

(1 g2, i(1 + g2), 2g).

ThenG : (N2 R3, g) C{} withG(z, x) :=< (z), x >C is a harmonic mor-phism in each variable separately. Theorem 5.2.5 therefore allows us to construct

the Riemannian covering manifold

R3 := {(z, x) N2 R3| G(z, x) = 1},

with suitable interpretation at the poles of . Furthermore it gives a globally

defined harmonic morphism : R3 N2 by: (z, x) z. Note, that G(z, x) = 1is equivalent to equation (5.1.3).

Proposition 5.3.1. The harmonic morphism : (R3, g) (N2, gN)defined aboveis a bundle map for a trivialR-bundle overN2.

Proof. For each z

N2 we have

3i=1

2i(z) = 0 or equivalently

|Re (z)|2 = |Im(z)|2 and = 0.

The fibre 1(z) of is given by < (z), x >C= 1 or equivalently

= 1 and = 0.

Reand Imare linearily independent, so Lz :=(1(z)) is an affine line in R3.

Ifc, :N2 R3 are given by

c:= Re

|Re |2 and := Re Im|Re Im| ,


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then Lz is parametrized w.r.t. arclength by t c(z) +t (z). The claim thenfollows from the fact that l : N2 R R3 given by l : (z, t)(z, c(z) +t (z))

is a global bundle homeomorphism.

For each z N2, c(z) is the point of the oriented affine line Lz nearest to theorigin. The point(z) S2 R3 is the direction ofLz. Obviously c(z) (z)for every z N2. This means that the map s : N2 R3 R3 given by s : z((z), c(z)) has values inT S2, the tangent bundle ofS2. IdentifyS2 with C {}via the stereographic projection from the south pole (0, 0, 1), whose inverse1is given by

1 :z ( 2z1 + |z|2 ,

1 |z|21 + |z|2 ).

It was shown in [Bai-Woo-1] that the meromorphic pair (g, h) represents (, c) via

, i.e.

() =g and d(c) =h.

Using this it can be shown that

=1(g) = ( 2g1 + |g|2 ,1 |g|

2

1 + |g|2 ) CR = R3

and

c= 2 ( h h g2

(1 + |g|2)2 , h g+ h g(1 + |g|2)2 ) CR = R

3.

Example 5.3.2. (Orthogonal Projection) Let M3 = R3, N2 = C and g, h : CC be given by g : z 0 and h : z z/2. Then equation (5.1.3) becomes

x1+ ix2 z = 0. This has a global solution the single-valued harmonic morphismz = x1 +ix2 which is an orthogonal projection. This means that the covering

manifoldR3 = {(z, (x1, x2, x3)) CR3| x1 + ix2 =z} is a 3-dimensional subspaceofR5. The envelope E R3 is the empty set. The harmonic morphism : R3 Ccan be thought of as an orthogonal projection R3 R2, where the metric on thehorizontal spaces has been multiplied by the constant factor 2.

Example 5.3.3. (Radial Projection) Let M3 = R3, N2 = C {} = S2 andg, h : C {} C {} be given by g : z z and h : z 0. Then equation(5.1.3) becomes

(x1 ix2)z2 + 2x3z (x1+ ix2) = 0. (5.3.1)


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Solving this equation gives two local solutions 1 z : R3 {0} S2 givenbyx x/|x| which are radial projection and its negative, well-known harmonic

morphisms. We can think of equation (5.3.1) as defining a multivalued harmonicmorphismz(x), 2-valued away from 0, with these local solutions as branches.

It is easily seen that the covering manifold is

R3 = {(v, x) S2 R3| x= tv for some t R},

that is, the tautological bundle over S2; this has an explicit trivialization S2 R R3 given by (v, t) (v,tv). The harmonic morphism : R3 S2 is theprojection map of this bundle. The projection : R3 R3 is a double cover with(v, x) = (v, x) except on the envelope E = S2 {0}. This means that thegeometric envelope E is the single point{0}. Thus on passing fromR3 to R3 theorigin 0 R3 is blown up to an S2.

Example 5.3.4. (The Outer Disk Family) LetM3 = R3, N2 = C {}= S2 andg, h: C {} C {} be given by g: z z andh: z irz for some r R+.

Equation (5.1.3) is now:

(1 z2)x1+ i(1 + z2)x2 2zx3 2irz = 0. (5.3.2)

We can think of this as defining a multivalued harmonic morphism z(x) which is

2-valued except on the zero set of the discriminant of this quadratic equation in

z where the two values coincide. This set gives the geometric envelope E={xR3| x21+ x22 = r2 and x3 = 0}. It is the circle of radiusr in the (x1, x2)-plane,whose centre is the origin 0 R3. In this case E=1(E), and Eis also a circle.OutsideEequation (5.3.2) has the two solutions:

zr =(x3+ ir)

x21+ x

22+ x

23 r2 + 2irx3

x1 ix2 . (5.3.3)

If we choose to be the principal square root on C R0 (

ei :=

ei/2,

where (, )), then we obtain two harmonic morphisms z+r and zr definedon R3

Dr

, where Dr

:={

xR3

|x2

1+ x2

2r2 and x

3= 0

} is the disk in the

(x1, x2)-plane with Eas its boundary. For every (x1, x2, x3) R3 Dr we have

z+r (x1, x2, x3) =zr (x1, x2, x3)


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so z+r and zr are, up to isometries ofR

3 Dr and S2, the same map. We call{z+r : R3 Dr S2| r R+} the outer disk family.

The map (, c) :S2

T S2

(CR)2

is given by

(, c) :1(z) 2z

1 + |z|2 ,1 |z|21 + |z|2

, 2irz

1 + |z|2 , 0

.

In spherical polar coordinates (, t) [0, 2] [/2, /2] (ei cos t, sin t) S2,this reads

(, c) : (, t) ((ei cos t, sin t), (irei cos t, 0)).

The covering manifold R3 is therefore parametrized by

(,t,s) ((ei cos t, sin t), (irei cos t, 0) + s(ei cos t, sin t))

where [0, 2], t [/2, /2] and s R. The fibre over the point z+r at(ei cos t, sin t) S2 is given by

s

L+(,t,s) := (irei cos t, 0) + s(ei cos t, sin t), s

R+, (5.3.4)

and the fibres ofzr are given by the same formula withs R. Note that z +r andzr map a point (ire

i cos t, 0) + s(ei cos t, sin t) R3 Dr to (ei cos t, sin t) S2so they are both surjective.

Note that as r 0, L(,t,s) s(ei cos t, sin t), which shows that as r 0the foliation of half-lines given by the fibres of z+r approaches the corresponding

foliation for the radial projection.

Let (x, 0) be a point in CR, such that|x| r. Then setting t= cos1(|x|/r),we see that the fibres fu and fl of z+r in the upper and lower half spaces, with

boundariesfu =fl = (x, 0) are orthogonal to the radius from (0, 0) to (x, 0) and

make an angle t with the (x1, x2)-plane. As|x| increases from 0 to r, t decreasesfrom/2 to 0. The fibres through a point (x, 0) with |x| > rlie in the (x1, x2)-planeand are tangent to the envelope E. Note that the direction of the fibres changes

discontinuously as we cross the diskDr.

Keepingx fixed and lettingr , we have thatt /2. In this sense the foli-ation ofz+r approaches the corresponding one for the orthogonal projection. Thus

the outer disk family interpolates between the radial and orthogonal projections.


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It follows from Proposition 5.3.1 that the covering manifold R3 is homeomorphic

to S2 R. This can also be seen directly as follows: The map L+ : S2 R+

R

3

Dr given by (5.3.4) is a homeomorphism, and so is L

:S2

R

R3

Drgiven by the same formula. Extending these maps continuously to S2 {0} andglueing them together along this manifold gives a homeomorphismL: S2R R3.We can thus think ofR3 as being obtained by glueing the two copies ofR3 Dracross Dr in an analogous way to that used in Riemann surface theory, the disk

Dr playing the role of a cut joining two branch points. IndeedDr = E and Dr

is a Seifert surface in the sence of [Rol] where this procedure of glueing across a

Seifert surface to obtain a branched covering is discussed. Note finally that the

globally defined harmonic morphism : R3 S2 is, via the homeomorphism L,just natural projection S2 R S2.

Note that the solution z+r : R3 Dr S2 of (5.3.2) satisfies the following

properties:

(1) it is surjective,

(2) it has connected fibres, (5.3.5)

(3) no two fibres are parallel as oriented line segments.

This is also true for the radial projections of Example 5.3.3. The following result

shows that, up to equivalence, any harmonic morphism satisfying these conditions

is one of these two examples.

Theorem 5.3.5. Let : U N2 be a harmonic morphism from an open subset ofR3 to a closed Riemann surface, satisfying conditions (5.3.5). ThenN2 is confor-

mally equivalent to S2 and, up to isometries ofR3 and conformal transformations

ofS2, is a restriction of radial projection or a solution to (5.3.2).

Proof. Condition (2) implies that is submersive, otherwise by [Bai-Woo-2], it

would be locally of the form where is submersive and of the form zzk. Such a composition clearly does not have connected fibres. From the locally

defined harmonic morphism : U N2 we obtain two meromorphic functionsg, h : N2

C

{}. From the interpretation of g as giving, via stereographic

projection, the direction of the fibres, condition (3) tell us that 1 g : N2 S2is injective and holomorphic, so bijective. Thus, up to composition with a conformal

transformation,N2 =S2 and gis the identity map g(z) =z . Since h then represents


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the holomorphic vector fieldc on S2,h(z) must be given by a quadratic polynomial

in z. This means that h can be written in the form

h(z) =< v(z),p >C,

where v(z) := (1 z2, i(1 + z2), 2z) for some p ofC3 and C denote the innerproduct on C3 given by< z, w >:=z1w1 + z2w2 + z3w3. Then equation (5.1.3) can

be written in the neat form

< v(z), x p >C= 0. (5.3.6)

Choose r R+0, [0, 2] and [/2, /2] such that Imp= r(cos cos , sin cosand let A, B SO(3) be given by

A=

cos sin 0 sin cos 0

0 0 1

andB =

sin 0 cos 0 1 0

cos 0 sin

.

Then equation (5.3.6) is equivalent to

< B A v(z), B A(x Re(p)) + iB A(Im (p))>C= 0. (5.3.7)

If we make the following isometric changes of coordinates on R3 andS2:

y:=B A(x Re (p)) and w:= aeiz b

beiz+ a,

where a := cos((2 )/4) and b:= sin((2 )/4) then it is not difficult to seethat equation (5.3.7) is equivalent to

< v(w), y+ (0, 0, ir)>C= 0,

which is equation (5.3.2) forr >0, or (5.3.1) for r = 0. In the latter case it is clear

that must be a restriction of radial projection or its negative.

Whether a solution to equation (5.3.2) on a given domain U satisfies conditions

(1),(2) and (3) of (5.3.5) depends on the domain. For example if we choose a

different square root in Example 5.3.4, for instance the one defined onC

R+

0, thenwe obtain different harmonic morphisms

z+r , zr : R

3 Cr S2


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whereCr := {x R3| r2 x21 + x22 and x3 = 0}. We call {z+r : R3 Cr S2| rR+} the inner disk family. The map z+1 : R3 C1 S2 is the map described in

Example 2.9 of [Ber-Cam-Dav], and in [Bai-1], [Bai-Woo-1] as the disk example.Note that, in contrast to the outer disk family, the image z+r (R

3 Cr) S2 ofz+r is the upper hemisphere, so z

+r is not surjective. The direction of the fibres

of z+r changes continuously when crossing the disk Dr but discontinuously when

crossing Cr, this being another possible Seifert surface for the branched covering

: R3 R3. More generally, we can find solutions to (5.3.5) on R3 Sr for anySeifert surface Sr, i.e. any surface with boundary the envelope E, the resulting

harmonic morphisms having very different properties according as Sr is bounded

or not.

On the way we have shown that, ifN2 = S2, g(z) = z and h(z) is a quadratic

polynomial inz , then (5.1.3) has solutions satisfying conditions (5.3.5). This is not

true, for example ifN2 = S2, g(z) = z and h(z) = zk, where k 3. In this case(5.1.3) can have no solution which is a surjective harmonic morphism to S2, since

hcannot represent a vector field atz = ; indeed only quadratic polynomials inzdefine vector fields globally onS2.

Example 5.3.6. Let M3 = R3, N2 = C and g, h : C C be given by g : z zand h: zzk/2 for some k 3. Then G : C R3 C is given by G : (z, x)zk+(x1ix2)z2 +2x3z(x1 +ix2). The projectionis therefore ak-fold covering ofR3 except on the envelope E. As before one could use the map (, c) : C T S2 toget a parametrization ofR3 which we know by Proposition 5.3.1 is homeomorphic

to C

R = R3.

We are mainly interested in the geometric envelope E R3, for which we givea complete parametrization for any k 3. As before the envelope E is given bysolving two simultaneous equations:

G(z, x) zk (x1 ix2)z2 2x3z+ (x1+ ix2) = 0 (5.3.8)

andG

z

(z, x)

kzk1

2(x1

ix2)z

2x3 = 0. (5.3.9)

By eliminatingzk andzk1 between the two equations we get

(k 2)w z2 + 2x(k 1) z kw = 0. (5.3.10)


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It immediately follows, that ifw = 0 then both x = 0 andz = 0, soEhas only one

point in common with the x-axis and that is (0, 0) CR = R3. From now on we

assume that w= 0. Solving (5.3.10) gives

z =x(k 1) x2(k 1)2 + k(k 2)|w|2

(k 2) w . (5.3.11)

It then follows from (5.3.9) that zk1 =(2x+ 2w z)/k R since w z R.Hence there exist n {1, 2,...,k 1} and r R {0}, such that w = rein( k1 ).This means that Ek1n=1 Pn where Pn is the 2-plane in C R given by Pn :=spanR

{(0, 1), (ein(

k1 ), 0)

}. For n

{1, 2,..., 2(k

1)

}define n : C

C

R

CCR by

n : (z,w,x) (ein( k1 ) z, (1)n ein( k1 ) w, (1)n x),

then{n| n {1, 2,..., 2(k 1)}} is an isometry subgroup of C C R andan easy calculation shows that a point p E if and only if n(p) E for alln {1, 2,..., 2(k 1)}. This means that if one knows the part of the geometricenvelope in the plane Pk1 = spanR{(0, 1), (1, 0)} C R, then one obtains therest by applying the maps n to E Pk1. In Pk1 both x and w are real, so thecorrespondingz given by (5.3.11) is also real. For such points lett := z andy := w.

Equations (5.3.8) and (5.3.9) now become

2t t2 12 2t

xy

=

tkktk1

,

or equivalently

xy

(t) =

1

2(1 + t2)

(k 2)tk+1 ktk1

2(1 k)tk

,

which parametrizes E Pk1 completely. ThusEconsists of 2(k 1) such curvesmeeting at the origin, and so is not a manifold, not even topologically.

5.4. Multivalued Harmonic Morphisms fromS3.

In [Bai-Woo-1] harmonic morphisms defined locally on S3 to a Riemann sur-

face are classified in terms of two meromorphic functions on the surface. We now

describe this classification, which can also be found in [Bai-3].


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Let f, g : N2 C be meromorphic functions and : N2 C4 {0} be themeromorphic map, given by

:= (1 + f g,i(1 f g), f g, i(f+ g)).

Then every local solution : U S3 N2 to the equation

< (z), x >C= 0, (5.4.1)

where x S3 R4 and z N2, is a harmonic morphism. To make sense of thislast equation at the poles off and g, we must divide it by f(z) and/or g(z) and

treat it as a limit.

Conversely, every locally defined harmonic morphism : U N2 from an openand connected subset U ofS3, is a solution of equation (5.4.1) for some pair (f, g)

of meromorphic functions defined locally on the Riemann surface, as above.

As in the R3-case equation (5.4.1) defines a Riemannian covering manifold

S3 :=

{(z, x)

N2

S3

| < (z), x >C= 0

},

with suitable interpretation at the poles of. Furthermore it gives a globally defined

harmonic morphism : S3 N2 with : (z, x) z and a map : S3 S3 givenby: (z, x) x.

Proposition 5.4.1. For any meromorphic functions f, g on a Riemann surface

N2, the harmonic morphism : S3 N2 given by : (z, x) z is a principalS1-bundle overN2.

Proof. The fibres of are closed geodesics ofS3 all of length 2, indeed for a fixed

z N2, 1(z) ={z} Cz where Cz is the great circle ofS3 given by equation(5.4.1). Such fibres have a natural orientation, so that, together with Re (z) and

Im(z), they give an oriented basis for{z} S3. We define the action ofei S1on each fibre ofS3 as rotation through +. Thus : S3 N2 is a principalS1-bundle.

Regarding the degree of the bundle : S3 N2, let us form the associatedR2-bundle, : R4 N2, given as for S3 by equation (5.4.1) but with x R4; thisis naturally oriented and : S3 N2 is its unit circle bundle.


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Lemma 5.4.2. For two meromorphic functions f, g : N2 C let : S3 N2and: R3 N2 be the correspondingS1-bundle and its associatedR2-bundle over

N2

. Further define a map

:N2

C4

\ {0}, by

:=

1 fg

, i(1 +f

g), f+

1

g, i(f 1

g)

.

Then the two sections Re and Im form an oriented basis for : R4 N2,away from poles of.

Proof. First note that from equation (5.4.1) it follows that Re and Imform an

oriented basis for the horizont

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THE GEOMETRY OF HARMONIC MORPHISMS