jonas nordstrom msc

8/13/2019 Jonas Nordstrom MSc

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Abstract

In this thesis we investigate the existence of complex-valued harmonic morphisms

on Lie groups and homogeneous Hadamard manifolds. The Lie groups that we areinterested in have a particular decomposition of their Lie algebra. This decomposi-tion allows us to define harmonic morphisms to Rn, n2.

Any homogeneous Hadamard manifold is isometric to a solvable Lie group Switha left-invariant metric. We give sufficient conditions for the Lie algebra ofSto havea decomposition that allows us to define complex-valued harmonic morphisms.


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Contents

Introduction 1

1 Lie Groups and Lie Algebras 31.1 Lie Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Lie Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3 Semi-direct Products . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.4 Lie group/algebra decompositions . . . . . . . . . . . . . . . . . . . . 71.5 Curvature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2 Homogeneous Spaces 112.1 Homogeneous Hadamard Manifolds . . . . . . . . . . . . . . . . . . . 122.2 Symmetric Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142.3 Damek-Ricci Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

3 Root Decompositions 17

3.1 Almost Normal Operators . . . . . . . . . . . . . . . . . . . . . . . . 173.2 Homogeneous Hadamard manifolds . . . . . . . . . . . . . . . . . . . 22

4 Harmonic Morphisms 254.1 Conformal maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254.2 Foliations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264.3 Harmonic Morphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . 284.4 Harmonic Morphisms on Lie Groups . . . . . . . . . . . . . . . . . . 294.5 Harmonic morphisms by Conformal Foliations . . . . . . . . . . . . . 324.6 A New Construction . . . . . . . . . . . . . . . . . . . . . . . . . . . 364.7 Application to Homogeneous Hadamard Manifolds . . . . . . . . . . . 39

4.8 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

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Introduction

S. Gudmundsson and M. Svensson have recently studied the existence of complex-valued harmonic morphisms from Riemannian Lie groups G and symmetric spaces.They have shown that ifg is the Lie algebra ofGand the dimensiondof the quotientalgebrag/[g, g] is at least 3 thenGcan be equipped with a left-invariant Riemannianmetric admitting complex-valued harmonic morphisms.

Any symmetric space of non-compact type is a homogeneous Hadamard manifoldi.e. a Riemannian homogeneous space of non-positive curvature. Each homogeneousHadamard manifold can be presented as a solvable Lie group Sequipped with a left-invariant Riemannian metric given by a Euclidean scalar product on the Lie algebras ofS.

The main aim of this Masters thesis is to present the structure theory of homoge-neous Hadamard manifolds and to construct complex-valued harmonic morphismson solvable Riemannian Lie groups. We are particularly interested in the cases whend = 1, 2. In our Theorem 4.32 we generalize a result of Gudmundsson and Svens-son. This gives manynew examples of complex-valued harmonic morphisms fromsolvable Lie groups ofanydimension greater than 3.

The thesis is divided into 4 chapters. The first chapter deals with Lie groups andthere associated Lie algebras. We also present some results on decompositions ofLie groups specifically, the Levi-, Cartan- and Iwasawa-decompositions. Finally wederive a formula for the curvature on a Riemannian Lie group.

Chapter 2 concerns homogeneous Hadamard manifolds. We define the homoge-neous Hadamard manifolds and show that any homogeneous Hadamard manifoldis isometric to a solvable Lie group with a left-invariant metric. We show that inthis case ad is an almost normal operator. Finally we give examples in the form ofsymmetric spaces and Damek-Ricci spaces.

Chapter 3 deals with the linear algebra of almost normal operators. We applythis to the Lie algebras of homogeneous Hadamard manifolds and find a root-spacedecomposition thereof.

In the last chapter we define harmonic morphisms and construct examples on Liegroups. We give a sufficient condition for the existence of harmonic morphisms onhomogeneous Hadamard manifolds.

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Chapter 1

Lie Groups and Lie Algebras

We introduce Lie groups and their relation to Lie algebras. We also study semi-simple, solvable and nilpotent Lie groups/algebras. We also study the Levi-, Cartan-

and Iwasawa-decomposition of Lie groups and algebras.

1.1 Lie Groups

Definition 1.1. A smooth manifold G equipped with a operation :G G Gissaid to be a Lie group if

(i) (G, ) is a group, and(ii) (g, h) g h1, g, h G is smooth.In this thesis we assume, unless otherwise stated, that all Lie groups are con-

nected. For a Lie group G and a point g G we define the left-translationLg :G G by Lg :hg h.Definition 1.2. A vector field Xon a Lie group G is said to be left-invariant if

Xg = d(Lg)e(Xe),

for all gG. The set of all left-invariant vector fields is denoted byL(T G).For any two vector fields X, Y L(T G) we define the commutator

[X, Y]g(f) =Xg(Y(f))

Yg(X(f)),

where f :G R is any smooth function. [X, Y] is also a left-invariant vector field.Definition 1.3. A subgroup Hof a Lie group G is said to be a

(i) Lie subgroup if it is an immersed submanifold ofG(ii) closed subgroup if it is closed as a subset ofG.

Proposition 1.4. [16, Theorem 6.9]LetG be a Lie group. IfHis a closed subgroupofG, thenHis an embedded submanifold ofG

Definition 1.5. For a Lie group G and two subgroups H1, H2 we define the com-

mutator subgroup (H1, H2) as the subgroup ofG generated by elements of theformghg1h1, gH1, hH2.

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Definition 1.6. A Lie group S is said to be solvable if there is an n Z+ suchthat the sequence Si defined by

S1 = (S, S) and Si+1 = (Si, Si)

satisfies Sn ={e}.Definition 1.7. A Lie group N is said to be nilpotent if there is an n Z+ suchthat the sequence Ni defined by

N1 = N and Ni+1 = (N, Ni)

satisfies Nn= {e}.A map : G Hbetween Lie groupsG and His said to be a homomorphism

if(g1 g2) =(g1) (g2). If, in addition is bijective is said to be an isomor-phism. Anautomorphism on G is an isomorphism from G toG, the group of allautomorphisms on G is denoted by Aut(G). An inner automorphism is a mapig :G G defined by

ig :h g h g1,where gG. The group of all inner automorphisms is denoted by Inn(G).

The setZ(G) ={gG|gh = hg for all hG}.

is called the center ofG.

1.2 Lie AlgebrasDefinition 1.8. A Lie algebrais a vector spaceVequipped with a skew-symmetricbilinear operation [, ] :V VV, called the Lie bracket, satisfying

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

for all X, Y , ZV.A linear map: g hbetween Lie algebras gand his said to be a homomor-

phismif([X, Y]) = [(X), (Y)] for all X, Y g. If in addition is a bijection itis said to be an isomorphism. A homomorphism : g

gis said to be an endo-

morphism. The set of all endomorphism on g is denoted by End(g). A bijectiveendomorphism is said to be an automorphism. The set of all automorphisms ong is denoted by Aut(g)

Given a Lie group G the set of left-invariant vector fieldsL(T G) on G is a Liealgebra gwith the Lie bracket defined by the commutator. gis called the Lie algebraofG.

Proposition 1.9. [16, Theorem 6.6] LetG be a Lie group with Lie algebrag. Foreach subalgebrah ofgthere exists a unique Lie subgroup HofGwith Lie algebrah.

Proposition 1.10. [16, Theorem 8.6] LetG andHbe Lie groups with Lie algebras

g and h and : g h be a homomorphism. IfG is simply connected then thereexists a unique homomorphism :G Hsuch thatde = .

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Proposition 1.11. [16, Corollary 2.8.7] For any Lie group G there is a unique Liealgebrag, and to each Lie algebrag there is at least one Lie group G with Lie algebrag. IfG1 andG2 are simply connected with Lie algebrasg1 andg2 isomorphic thenG1 andG2 are isomorphic.

Definition 1.12. Let g be a Lie algebra, we define the derived sequence gi andthe lower central sequence gi by

g1 = [g, g], gi+1 = [gi, gi], and

g1 =g, gi+1 = [g, gi].

gis said to be solvableifgn = 0 for some n andnilpotentifgn= 0 for some n. If[g, g] = 0 then gis said to be abelian.

Definition 1.13. Given a Lie algebra gwe call the set

Z(g) =

{X

g

|[X, g] = 0

}the centerofg.Lemma 1.14. For any Lie algebrag the quotient algebrag/[g, g] is abelian.

Proof. The Lie bracket on g/[g, g] is defined by

[A+ [g, g], B+ [g, g]] = [A, B] + [g, g],

for A, Bg. But [A, B][g, g] so[A, B] + [g, g] = [g, g].

Even ifg/[g, g] always is abelian there does not need to exist any abelian subal-

gebraa such that g= a [g, g].Proposition 1.15. [16, Proposition 11.9]Letgbe a Lie algebra. Thengis solvableif and only if [g, g] is nilpotent.

Proposition 1.16. [10, Lemma 6.1] Ifnis a nilpotent Lie algebra withdim(n)2then

dim([n, n])dim(n) 2Proof. n is nilpotent therefore [n, n] n. Suppose that dim([n, n]) = dim(n) 1that is

n= RX [n, n],for some X n. Suppose Z= [X1, X2][n, n] for X1, X2n. Then

X1 =a1X+ Y1 and X2=a2X+ Y2,

where Y1, Y2[n, n]. ThusZ=a1[X, Y2] + a2[Y1, X] + [Y1, Y2] [n, [n, n]].

We have shown that [n, n][n, [n, n]], and obviously [n, [n, n]][n, n], so[n, [n, n]] = [n, n].

Since n is nilpotent this is only possible if [n, n] = 0. Thus either dim([n, n])dim(n) 2 or n is abelian in which case dim([n, n])dim(n) 2 is also true sincedim(n)2.

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Theorem 1.17. [16, Theorem 10.9][16, Theorem 11.13] LetG be a Lie group withLie algebrag then

(i) G is nilpotent if and only ifg is nilpotent(ii) G is solvable if and only ifg is solvable.

We define

Ad :G Aut(g) Adg :g gby

Adg(X) = d(ig)e(X),

where ig is the inner automorphismig(h) =ghg1, and

ad = dAde : gEnd(g) adX :g g.In fact adX(Y) = [X, Y] where [

,

] is the Lie bracket ofg, [17, Proposition 3.47].

Proposition 1.18. [16, Corollary 7.12] LetG be a Lie group with Lie algebrag. IfX g andgG then

(i) Adexp(X)=eadX , and

(ii) g exp(X)g1 = exp(Adg(X)).

A Lie algebra g is said to be unimodular if trace(adX) = 0 for all X g. Anynilpotent Lie algebra is unimodular.

The subgroup of Aut(g) with Lie algebra adg is denoted by Int(g).

Proposition 1.19. [16, Corollary 7.14(b)] LetG be a Lie group. IfZ(G) ={e},then

Ad :G Int(g)is a Lie group isomorphism.

Since g is a vector space this means that a Lie group with trivial center is iso-morphic to a group of matrices.

Proposition 1.20. [16, Lemma 7.15(b)(c)] LetG be a Lie group with Lie algebrag and leta be a subalgebra andm a subspace ofg. Thenada(m)m if and only ifAdA(m)m, whereA is the subgroup ofG with Lie algebraa.

Proof. () Let Ha and X m thenadH(X) =dAde(H)(X)

=d

dt[Adexp(tH)(X)]t=0m

since AdA(m)m.() We have

Adexp(tH)(X) =eadH(X) m

since ada(m

)m

. Thus AdU(m

)m

in a neighbourhood U ofe in A. Since A isconnected and Ad :A Aut(g) is a homomorphism AdA(m)m.

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1.3 Semi-direct Products

Let H1 and H2 be Lie groups then the product H1 H2 with(h1, h2)

(k1, k2)

1 = (h1

k11 , h2

k12 )

is a Lie group where H1 and H2 are closed normal subgroup. H1 H2 is called thedirect-product ofH1 andH2. The Lie algebra ofH1 H2 equals h1 h2 where bothh1 and h2 are ideals. The semi-direct product generalizes this.

Definition 1.21. Let g be a Lie algebra. An element D End(g) is said to be aderivation if

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

for all X, Yg. The set of all derivations on g is denoted by Der(g), this is also aLie algebra.

In particular adX is a derivation for all X

g.

Proposition 1.22. [15, Proposition 1.22] Leta andn be Lie algebras and supposethere is a homomorphism : a Der(n). Then there exists a unique Lie bracketon the vector spaceg = a n, retaining the Lie brackets ona andn and [A, X] =(A)(X) forAa andX n. We callg= a n thesemi-direct productofaandn.

It is obvious that a is a subalgebra and nan ideal.

Theorem 1.23. [15, Theorem 1.102] LetA andNbe simply connected Lie groupswith Lie algebrasa andn and let : aDer(n) be a homomorphism. Then thereexist a unique action : A

N

N such thatA N is a simply connected Lie

group with Lie algebrag= a n.When n is an ideal and (A) = adA, we omit the subscript in the semi-direct

product. Similarly, when N is a normal subgroup and (a)(n) =a n, we will omitthe subscript .

1.4 Lie group/algebra decompositions

Given a Lie algebra gthen we call the sum of all solvable ideals ofgthe radical ofg. This is the unique maximal solvable ideal ofg.

Proposition 1.24. [16, p.254] Letg be a Lie algebra, then there exist a decompo-sitiong= h r, whereh is a semi-simple subalgebra andris the radical ofg. Thisis called the Levi-decompositionofg.

Definition 1.25. A Lie algebrag is said to besimpleif it is non-abelian and has nonon-trivial ideals, gis said to besemi-simpleif it is isomorphic to a direct-productof simple Lie algebras. A Lie group G is said to be semi-simple is its Lie algebrais semi-simple.

Let g be a Lie algebra and let B be its Killing form. A Cartan involution isan endomorphism: g g such that 2 =Ig and

B(X, Y) =

B(X, (Y))

is positive definite.

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Proposition 1.26. [15, Corollary 6.18] Any real semi-simple Lie algebra has aCartan involution.

Ifgis a semi-simple Lie algebra then a Cartan involution gives rise to a Cartan-decomposition

g= k pwhere has eigenvalue 1 on k and1 on p. They satisfy

[k, k] k, [k, p] p, [p, p] k.Theorem 1.27. [15, Theorem 6.31] Let G be a semi-simple Lie group with Liealgebragand let be a Cartan-involution andk andp the eigenspaces of1 and1.LetK be the subgroup ofG with Lie algebrak. Then

(i) K is closed(ii) K contains the centerZ(G) ofG(iii) K is compact if and only ifZ(G) is finite(iv) whenZ(G) is finite, Kis a maximal compact subgroup ofG.

Proposition 1.28. [15, Proposition 6.43] Letgbe a semi-simple Lie algebra. Thenthere exists a decomposition into subalgebras

g= k a n,wherea is a abelian, n is nilpotent andanis solvable with[a n, a n] =n. Thisis said to be anIwasawa-decompositionofg.

The subalgebras are obtained by a root decomposition but that would lead tofar.

Theorem 1.29. [15, Proposition 6.46] LetG be a semi-simple Lie group and let

g= k a nbe an Iwasawa-decomposition of its Lie algebra g. Let A and N be the connectedsubgroups ofG with Lie algebrasa andn, then

K A NG, (k,a,n) =kanis a diffeomorphism andA andNare simply connected.

1.5 Curvature

The type of metrics normally defined on a Lie group are the left-invariant metrics.Together with the left-invariant vector fields they determine, locally, the geometryof the Lie group.

Definition 1.30. A metric g on a Lie group G is said to be left-invariant if theleft-translations Lg : G G are isometries for all g G. A Lie group with aleft-invariant metric is said to be a Riemannian Lie group.

If we assume that g is a left-invariant metric, then

g(XY , Z) =12

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

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for X, Y , Z g. Let{Zi} be an orthonormal frame of left-invariant vector fieldsthen

XY =i

g(XY , Zi)Zi

=1

2

i

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

=1

2

i

(g([X, Y], Zi) g(adYX, Zi) g(adXY, Zi))Zi

=1

2([X, Y] adYX adXY).

This a formula for the Levi-Civita connection for left-invariant vector fields.Let g be the Lie algebra of G and, = ge(, ). For orthonormal vectors

X, Y

g we have the following formula for the sectional curvature

K(X, Y) = R(X, Y)Y, X=XYY YXY [X,Y]Y , X

= YY , XX + XY , YX [X,Y]Y , X

= YY , XX + XY , XY XY , [X, Y]

[X,Y]Y , X=|XY|2 YY , XX YX, [X, Y]

[X, Y], [X, Y] [X,Y]Y , X

=|XY|

2

YY , XX |[X, Y]|2

X, Y[X, Y] [X,Y]Y=|XY|2 YY , XX |[X, Y]|2 X, [Y, [Y, X]]=|XY|2 YY , XX |[X, Y]|2

ad2YX, X

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Chapter 2

Homogeneous Spaces

Homogeneous Hadamard manifolds, that we are interested in in this thesis, arehomogeneous spaces, as it turns out they are in fact Lie groups.

Let Mbe a manifold and (G, ) be a Lie group. A group action is a C-mapa: G MMsuch that for all g, h G and xM

a(g h, x) =a(g, a(h, x)) and a(e, x) =xHere e denotes the identity element in G. The action is said to be transitiveif

M= {a(g, x0)|gG}for anyx0G and it is said to be simple if given xM

a(g, x) =a(h, x)

implies that g = h.

Theorem 2.1. [17, Theorem 3.58] LetG be a Lie group andHa closed subgroup,then there is a uniqueC-structure onG/Hsuch that thehomogeneous projec-tion

: G G/H (g) =gHis a smooth submersion. In that case the manifold M = G/H is said to be ahomogeneous space.

The group of isometries of a Riemannian manifold Mis denoted by I(M). I(M)is a Lie group, not necessarily connected. The component ofI(M) containing the

identity element is denoted by I0(M).Definition 2.2. A Riemannian manifold with a transitive group of isometries issaid to be a Riemannian homogeneous space.

Theorem 2.3. [17, Theorem 3.62] LetG be a Lie group anda: G MM be agroup action on a manifoldM. LetpMand setK={gG|a(g, p) =p}, thenthe map :G/KM defined by

(gK) =a(g, p)

is a diffeomorphism.

We see that a Riemannian homogeneous space is a homogeneous space. Thesubgroup K is called the isotropy subgroup ofG at p.

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2.1 Homogeneous Hadamard Manifolds

Definition 2.4. A connected, simply connected complete Riemannian manifoldMwith non-positive sectional curvature is said to be a Hadamard manifold. IfM is

also a Riemannian homogeneous space it is said to be a homogeneous Hadamardmanifold.

Theorem 2.5. [3, Hadamard-Cartan p. 15]For a complete Riemannian manifoldMwith non-positive curvature the (geodesic) exponential map expp : TpM M,forpM, is a covering map.

Since Hadamard manifolds are simply connected they are there own universalcovering spaces, thus any Hadamard manifold is diffeomorphic to Rm. In, [18,Corollary 1(d)], Wolf proved that any transitive group of isometries on a homoge-neous Hadamard manifold with no Euclidean de Rham factor has trivial center.

Theorem 2.6. [12, p. 24][1, Proposition 2.5] LetM be a homogeneous Hadamardmanifold andGIo(M)be a connected closed subgroup that operates transitively onM. Then there is a solvable closed subgroupSofG, that operates simply transitivelyonM.

Corollary 2.7. [12, p. 24][1, Corollary 2.6] Any homogeneous Hadamard manifold(M, g) is isometric to a solvable Riemannian Lie group (S, h).

Proof. Let Sbe a simply transitive group of isometries ofM and let pM, themap P : S M defined by P(s) = s(p) is a diffeomorphism with P(e) = p. Ifs S then s: M M is an isometry so

gs(p)(dsp(Xp), dsp(Yp)) =gp(Xp, Yp).Let the scalar product, on s be given by

U, V =gp(dPe(U), dPe(V)),for U, V s. Also s P =P Ls and therefore

dsp dPe= dPs (dLs)e.Let the scalar product on s induce a left-invariant metric h on S then

hs(Us, Vs) =hs((dLs)e(Ue), (dLs)e(Ve))

=he(Ue, Ve)=gp(dPe(Ue), dPe(Ve))

=gs(p)(dsp(dPe(Ue)), dsp(dPe(Ve)))

=gs(p)(dPs((dLs)e(Ue)), dPs((dLs)e(Ve)))

=gs(p)(dPs(Us), dPs(Vs))

i.e. Pis an isometry.

Let g be a Lie algebra with a scalar product. As shown in Section 1.5 the Levi-Civita connection on g is given by

UV =12

([U, V] adU(V) adV(U)),

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for U, V g, and the sectional curvature satisfiesK(U, V) =|UV|2 UU , VV |adU(V)|2

ad2V(U), U

.

In what follows we will use some operator notation, for definitions see Section

3.1.

Lemma 2.8. Letg be a Lie algebra with a scalar product. Then

Re(adA)X=XA and Im(adA)X= AXforX[g, g] andA[g, g].Proof. Define the operators

HA: [g, g] [g, g] andGA : [g, g] [g, g]by

HAX= AX and GAX= XA.Let Y[g, g], then from the compatibility of the metric we have

HAX, Y =AX, Y=X, AY=X, HAY

i.e. HA is skew-adjoint. Since A[g, g] we getGAX, Y =XA, Y

=A, XY=A, YX=YA, X=X, GAY

i.e. GA is self-adjoint. Thus since

adAX= AX XA= HAX GAXthe lemma follows.

Proposition 2.9. [1, Theorem 5.2][19, Proposition 1.3] Lets= an be a solvable

Lie algebra with non-positive curvature, where n = [s,s] and a = n. Then n isnilpotent anda is an abelian subalgebra ofs.

Proof. Thatn is nilpotent follows from Proposition 1.15. For anyAa and U swe have

AA, U= adAA, U =A, adAU= 0since adA(U)na. Now ifXker(adA), then we have that

K(A, X) =|AX|2 0,which implies that

AX=

XA= 0. Since [X, U]

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Therefore

adAU, X = AU UA, X= AU , X UA, X= U, AX U, XA= U, AX+ XA=0.

Thus adA(U)X for all A a, all X ker(adA) and all U s. This implies thatker(adA)adA(s). Also since dim(ker(ad)) + dim(adA(s)) = dim(s) we in fact haves = ker(adA) + adA(s). Now since adA(s) n we must have a = n adA(s) =ker(adA). Thus [A, a] = 0 for all Aa i.e. [a, a] = 0.

IfA, Ba and U n then

AB, U= 12

B, adAU 12

A, adBU= 0,

so obviously K(A, B) = 0. Further,

K(A, X) = XX, AA + AX, XA [A,X]X, A

= X, AXA +

X, [A,X]A

= X, Im(adA)(Re(adA)X) + Re(adA)(adAX)= X, Im(adA)(Re(adA))X+ Re(adA)(Im(adA)X+ Re(adA)X)=

X, (Re(adA)

2 + [Re(adA), Im(adA)])X .Since the curvature is non-positive Re(adA)

2 + [Re(adA), Im(adA)] must be positivesemi-definite, i.e, adA : n n is almost normal for any A a. The fact thata isabelian implies that

(adA adB)(X) (adB adA)(X) =[A, [B, X]] [B, [A, X]]=[[A, B], X]

=0,

i.e., all the operators adA: n n, Aa commute.

2.2 Symmetric Spaces

In this section we introduce the non-compact irreducible globally symmetric spaces.They are examples of homogeneous Hadamard manifolds.

Definition 2.10. A Riemannian manifold (M, g) is said to be a globally symmetricspace if there for each point p M exist an isometry p : M M such thatp(p) =p and

d(p)p = idTpM.

Theorem 2.11. [11, p. 205] Every global symmetric space(M, g) is a Riemannianhomogeneous space.

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Proof. Let :I Mbe a geodesic in (M, g). Let p= (t0), t0I, then p is a geodesic with

(p

)(t0) =(t0) and

d

dt

[p

]t0 =

d

dt

[]t0.

Thus the two geodesics coincide when they are both defined. So every geodesic maybe extended to infinite length, i.e., Mis complete.

Letp, qM. Then since (M, g) is complete there exist a geodesic: [t0, t1] Msuch that (t0) =p and (t1) =q. Let r= (

t0+t12

) then r(p) =q, so the action istransitive. Thus the group of isometries I(M) is transitive.

In fact, for any global symmetric space Mthere exists a semi-simple subgroup GofI(M) that acts transitively on M, [11, p. 198].

Definition 2.12. A simply connected global symmetric space is said to be irre-

ducible if it is not a product of global symmetric spaces.

Theorem 2.13. [11, Theorem V.3.1(ii)]All irreducible non-compact global symmet-ric spaces are homogeneous Hadamard manifolds.

Proposition 2.14. [15, Proposition 6.43][15, Lemma 6.45(a)] Let M = G/K bean irreducible non-compact global symmetric space where G is semi-simple. LetS=NAbe a solvable Lie group that is obtained from the Iwasawa-decompositionofG, ands be its Lie algebra with the corresponding scalar product. Then

(i) s= a n, wherea is abelian andn= [s,s] is nilpotent,(ii) all the operatorsadA, Aa, are diagonalizable overR and commute sincea

is abelian.

2.3 Damek-Ricci Spaces

Damek-Ricci spaces are a generalization of the rank one symmetric spaces. Theyare also examples of homogeneous Hadamard manifolds.

Definition 2.15. Let v and z be real vector spaces and : vv z a skew-symmetric bilinear map. We define a scalar product, on n= v z such that vand z are orthogonal and define

J :z End(v), JZ :v vby

JZ(U), V =(U, V), ZDefine a Lie bracket on nby

[U+ Z, V + W] =(U, V).

The Lie algebra is said to be a generalized Heisenberg algebra if

J2Z=

Z, Z

Iv.

The dimension ofv must be at least 2 since is skew-symmetric.

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Example 2.16. Let v= span{X, Y} and z= RZand define by(X, Y) =Z.

If the scalar product is such that X, Y , Z is an orthonormal basis for n, then n is ageneralized Heisenberg algebra.

Definition 2.17. Let a be a one-dimensional real vector space, n be a generalizedHeisenberg algebra and set s= a n. Choose an element A a and define a scalarproduct on s by

U+ Z+ rA,V + W+ sA= U+ Z, V + Wn+ rsand a Lie bracket by

[U+ Z+ rA,V + W+ sA] = [U, V]n+1

2(rV sU) + (rW sZ).

The connected, simply connected Lie group with Lie algebra s and left-invariantmetric induced by the scalar product on s is said to be a Damek-Ricci space.

Proposition 2.18. [6, Proposition 2, p. 85] Any Damek-Ricci space is a homoge-neous Hadamard manifold.

Proposition 2.19. The Lie algebras of a Damek-Ricci space satisfies(i) s= a n, wherea is abelian andn= [s,s] nilpotent(ii) all the operatorsadA, Aa, are symmetric.

Proof. (i) By definition.

(ii) For V + W n and rAa, r R we haveadrA(V + W) =r

1

2V + rW

which implies

adrA(V + W), U+ Z =r 12

V, U + r W, Zalso

adrA(V + W), U+ Z =V + W, adrA(U+ Z)

=r

1

2 V, U + r W, Z=adrA(V + W), U+ Z

so adrA|n is symmetric.

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Chapter 3

Root Decompositions

As we saw in Chapter 2, if s is a solvable Lie algebra with non-positive sectionalcurvature then adA

|nis almost normal for anyA

a. Thus in this chapter we present

the analysis of almost normal operators and apply it to homogeneous Hadamardmanifolds.

3.1 Almost Normal Operators

To give the almost normal operators a position in the operator hierarchy we beginby introducing self-adjoint, skew-adjoint and normal operators.

Definition 3.1. Let V be a vector space equipped with a scalar product and letA: VVbe a linear map. Ais said to be self-adjoint ifA=A.

By the Spectral Theorem, ifA is self-adjoint then V has a orthonormal basis ofeigenvectors to A.

IfA : V V is a diagonalizable operator on a linear space Vwe can define ascalar product on V such that the eigenvectors are orthonormal. With respect tothis scalar product Ais self-adjoint. Thus on a real vector spaceV a operatorA isdiagonalizable if and only if there is a scalar product onVsuch thatAis self-adjoint.

Definition 3.2. Let V be a vector space equipped with a scalar product and letA: VVbe a linear map. Ais said to be normal ifAA= AA.Definition 3.3. An operator A is said to be semi-simple if each A-invariant sub-space has a complementary A-invariant subspace.

An operator A is semi-simple if and only if there is a scalar product on V suchthat A is normal, [13] p. 265. If V is a complex vector space then an operatorA is semi-simple if and only if A is diagonalizable over C, [13] p. 317. So on acomplex vector space V an operator A : V V is diagonalizable if and only ifthere is a scalar product on V such that A is normal, in this case the eigenvectorsare orthogonal.

Finally an operatorA is skew-adjointifA=A. Obviously, any skew-adjointoperator is normal.

Given an operatorA we define the self-adjoint and the skew-adjoint part ofA by

Re(A) =12(A+ A) Im(A) =12 (A A)

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Definition 3.4. Let V be a vector space with a scalar product. An operator A :V V is said to be almost normal if

|Av|2 |Im(A)v|2

for all vV.Proposition 3.5. [1, p.331] Let V be a vector space with a scalar product andA: VV an operator. A: V V is almost normal if and only if

N(A) = Re(A)2 + [Re(A), Im(A)]

ispositive semi-definite, i.e.v, N(A)v 0 for allvV.Proof. We notice that

AA + Im(A)2 = Re(A)2 + [Re(A), Im(A)].

Thus

v, N(A)v = v, (AA + Im(A)2)v 0if and only if

Av,Av Im(A)v, Im(A)v .

Since Re(A)2 always is positive semi-definite, all normal operators are almostnormal since then [Re(A), Im(A)] = 0. IfA is almost normal then A+ivI, v

R,

is also almost normal but A does not have to be almost normal.

Lemma 3.6. [1, Lemma 4.3] Let A be almost normal and L be an A-invariantsubspace of V. Then A1 = A|L is almost normal. If A1 is skew-adjoint then theorthogonal complementL inV is also A-invariant.

Proof. Let P be the orthogonal projection, ker(P)P(V), ofV onto L and R=I Pthe orthogonal projection onto L. We decompose Aas

A= I AI= (P+ R)A(P+ R) =P AP+ P AR+ RAP+ RAR,

or equivalently in matrix form

A=

A1 ZW A2

,

whereA1 = P AP,Z=P AR,W =RAP= 0 sinceLisAinvariant, andA2 = RAR.Thus

N(A) =

N(A1) 14ZZ A1Z+ 12(Im(A1)Z+ ZIm(A2))

ZA1 12(ZIm(A1) + Im(A2)Z) N(A2) + 34ZZ

.

Since this matrix is positive semi-definite N(A1) 14ZZ must be positive semi-definite. Therefore A1 is almost normal. IfA1 is skew-adjoint then N(A1) = 0 soZZ is negative semi-definite but it must be positive semi-definite so Z= 0.

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For an operator Alet

V0= {vV|(A iI)kv = 0 for some k Z+ and R, }and

V1 = {vV|(A I)kv= 0 for some k Z+ and C with Re()= 0}.We say that Ais almost semi-simple ifA|V0 is semi-simple.Lemma 3.7. [1, p. 332] An operatorH :VV is almost semi-simple if and onlyif there is a scalar product onV such thatH is almost normal.

Proof. () Suppose that H is almost semi-simple, then H|V0 is semi-simple.Choose any scalar product on V0 such that H|V0 is skew-adjoint and hence nor-mal.

Let{

v1, . . . , vn}

be a basis for V1 such that the matrixA={

ai,j}ni,j=1 for H

|V1 is

upper triangular, such a basis can always be found in the complex case [13, Corollaryp.203]. Let wi=ivi, for i= 1, . . . , nwith 1> > n be a new basis for V1. Thematrix Afor H|V1 in the basis wi is given by

A= E1AE=

a1,121

a1,2 . . . n1

1a1,n1 n1 a1,n

0 a2,2 . . . n1

2a2,n1 n2 a2,n

... ...

. . . ...

...0 0 . . . an1,n1 nn1 an,n0 0 . . . 0 an,n,

where E is the diagonal matrix with entries i. Choose the scalar product on V1 insuch a way that wi is an orthogonal basis then Re(H|V1)= 0 and we can choose Ein such a way that [Re(H|V1), Im(H|V1)] is arbitrarily small. Finally let the scalarproduct be such that V0 and V1 are orthogonal then N is almost normal.

() Pick an element v V0 with k > 1. Let vj = (AiI)kjv. Supposethat v1= 0. Now since Av1 = iv1, we haveAz,w = i z, w = z,Aw forz, wspan{v1}. By Lemma 3.6 span{v1} is A-invariant. Thus

w= (A iI)(v2 v2, v1 v1v1, v1 ) span{v1},

and therefore 0 =w, v1= (A iI)v2, v1 =v1, v1which is a contradiction, so v1 = 0. Induction over j shows that Av = iv. ThusAV0 is skew-adjoint which implies that it is semi-simple.

For the next theorem we need that any operatorA on a finite dimensional vectorspace Vcan be decomposed as A= S+ N where S is semi-simple, Nnilpotent i.e.Nk = 0 for some k1, and they are both given by polynomials in A, [13, Theorem6.13]. A nilpotent operator has at least one eigenvector and it must have eigenvalue0.

Let A= S+ Nbe as above and assume that Av=v, v

= 0, then

Nv= (anAn + + a0I)v= (ann + + a0)v

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sov is an eigenvector for Nand thus N v= 0. Therefore Sv = Av = v and is aneigenvalue for S. Suppose instead that is an eigenvalue for Swith eigenspace Vthen

SAv = ASv= Av

so that AV V, i.e. V is A-invariant. Therefore (A I)|V is nilpotent andthere is at least one vector w such that

(A I)w= 0,thus is an eigenvalue for A. We have shown that is an eigenvalue for A if andonly if it is an eigenvalue for S.

If V is complex and all eigenvalues for Sare distinct then A is diagonalizableand A = S.

Lemma 3.8. [11, Proposition III.1.1] Let V be an n-dimensional complex vector

space and letH be a real vector space of commuting operators onV. For eachH Hand eachC letV= {vV| k Z+ such that(H I)kv= 0}.

Then there exist1, . . . , m C, the weights ofH, such thatV =V1 . . . Vm.

Furthermore, eachV isH-invariant.

Proof. For each eigenvalue (H

I)

|Vis nilpotent this implies that

V =V1+ . . . + Vm.

Let v be such that(H I)kv= 0

then we see that {1, . . . , m}. Suppose that = 1. Write v =m

i=1 vi whereviVi, clearly

m

i=1

(H 1I)kvi= 0.

Each Vi is (H 1I)-invariant so(A 1I)kvi= 0,

fori= 1, . . . , m. Ni=AiIis nilpotent onVi so that is has eigenvalue 0. Further(H 1I)vi=((Ni+ iI) 1I)vi

=(Ni+ (i 1)I)vi=0,

this implies that 1 i is a eigenvalue for Ni ifvi= 0 i.e. 1 =i. Thus vi = 0 if1< i so the sum is direct.20


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SupposevVi and GH then(H iI)k(Gv) =G(H iI)kv

=0,

since the operators in H commute, so GvVi .Theorem 3.9. [16, Theorem 11.14]LetV be ann-dimensional complex vector spaceand letH a real vector space of commuting operator onV. For any: H C let

V={vV| k Z+ such that(H (H)I)kv = 0 H H}then there existsroots 1, . . . mH such that

V =V1 . . . Vm.

Furthermore, each root spaceV isH-invariant.Proof. The proof is done by induction over dim(V). If dim(V) = 1 then any vectoris an eigenvector to any operator. Suppose that the theorem is true for dim(V) 1,then letHH with at least two distinct weights 1, . . . , n, this may be done sinceif all H H only has one weight H then V =V, where : HHand we aredone. Otherwise the theorem is true on each Vi since they must have a smallerdimension than Vand since each Vi is H-invariant, so H is a commuting family ofoperator on each Vi.

We also need to show that each i is linear. Let G, HH and vVi then(H+ G

i(H)I

i(G)I)

2kv =((H

i(H)I) + (G

i(G)I))2kv

=2kl

(H i(H)I)l(G i(G)I)2klv

=0,

since (H i(H)) and (G i(G)) commute. Thus we have i(H+ G) =i(H) +i(G).

Since the operators commute, each V is H-invariant.

For a family H of commuting almost normal operators, let U0 be the collectionroot spaces where (H)

iR for all H

H.

For all other = + idivide them into equivalence classes where i j ifand only if Ker(i) = Ker(j). Let Ui be the sum of all equivalent root spaces.The spaces Ui are H-invariant and pairwise orthogonal since if i and j are notequivalent then there is an H H such that i(H) = 0 and j(H)= 0. ThereforeH|Vi is skew-adjoint which implies that Vi is H-invariant, so VjVi .

IfV is a real vector space we consider the root space decomposition

VC =VC1 . . . VCmand put

V, = (V

C + V

C)

V,

these are called the generalized root spaces.

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3.2 Homogeneous Hadamard manifolds

Let nC be root spaces, n, be the generalized root spaces as in Section 3.1 and n0the root spaces with vanishing real part.

Proposition 3.10. [1, Proposition 5.5] Let s be a Lie algebra with non-positivecurvature andn= [s,s], then

(i) n0Z(n), and(ii) n0 is orthogonal to [n, n].

Proof. Let nk be the lower central sequence ofn i.e. n1 =n and nk+1 = [nk, n] fork1, since n is nilpotent we have nr = 0 for some r. We prove by induction on jthat

(a) nrj[n0, n] for 0jr 1, and(b) n0nrj for 0jr 2.The statements are obvious for j = 0. Suppose they are true for j , we prove that

they are true for j+ 1.(a) Let X n0, Y nr(j+1) (nrj) and Z n, then [X, Y] [n, n0] since

X n0 and [X, Y]nrj since Y nr(j+1). Therefore [X, Y]nrj [n, n0] = 0by the induction hypothesis.

Now since Ynrj and [Y, Z]nrj we haveadYY, Z = Y, [Y, Z] = 0,

which implies that adYYa.Let Aa, then

adXX, A= X, [X, A]= X, adAX .By definition adA|n0 is skew-adjoint so

X, adAX =adAX, Xi.e.X, adAX= 0 which implies adXX n.

By the formula for the Levi-Civita connection in Section 1.5 we haveXX =adXX andYY =adYY. Thus we find that

K(X, Y) = |XY|2.The negative curvature implies that

12

(adXY + adYX) =XY = 0,

or equivalently adXY =adYXSince

adYX, Z= X, [Y, Z] = 0,we have adYX a and because of the assumption, (b), that Xnrj.

ThereforeY, [X, Z]= adXY, Z =adYX, Z= 0.

This implies that nr(j+1)

(nrj

) is orthogonal to [n0, n]. Since the inductionassumption states that nrj is orthogonal to [n0, n], nr(j+1) is orthogonal to [n0, n].

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(b) Let X nr(j+1) n0 and Z, W n. Then [X, Z]nrj [n, n0] = 0 by theinduction assumption (a), therefore

adXW, Z =W, [X, Z] = 0.This means that adX(n) a however X n0 implies that adXX n so thatadXX= 0.

ThusK(X, Z) =|XZ|2

and the non-positive curvature gives us that

1

2(adXZ adZX) = XZ= 0.

We find that

[Z, W], X

=

W, adZ(X)

,

which implies that X is orthogonal to [n, n]. But X nr(j+1) [n, n] so X = 0,i.e. nr(j+1) n0 = 0.

Sincenr(j+1) isa invariant, by induction on nk using the Jacobi identity, it mustbe the sum of all n, where = 0. Hence n0 has to be orthogonal to nr(j+1).

The statement follows by setting j=r 1 in (a) and j =r 2 in (b).If we leta0be the subalgebra ofasuch that all adA,A a0are skew-adjoint. Then

s= s0 s+ where s0=a0+ n0 and s+ = s0. From [2, Theorem 4.6] a homogeneousHadamard manifold has no Euclidean factor in its de Rham decomposition ifs0 = 0.

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Chapter 4

Harmonic Morphisms

In this chapter we introduce harmonic morphisms, they are maps between Rieman-nian manifolds that preserve harmonic functions. The type of harmonic morphisms

we are interested in are the so called complex valued harmonic morphisms i.e. theimage is a subset ofC. To construct such complex valued harmonic morphisms it isenough to be able to construct a harmonic morphism to Rn, n2.

4.1 Conformal maps

Definition 4.1. Let f : (M, g) R be a smooth function, we define the gradientgrad(f) :M T M by

gx(grad(f)x, Xx) = dfx(Xx)

for all Xx

TxM.

For a smooth map : (M, g) N we callVx = ker(dx) the vertical spaceandHx=Vx the horizontal space at xM.Definition 4.2. Let : (M, g) (N, h) be a smooth map. Then is said to behorizontally weakly conformal if for each xM either

(i) xiscritical i.e. dx = 0, or(ii) x is regular i.e. dx= 0 and dx|Hx :Hx T(x)N is surjective and there

exist a number (x) such that

h(dx(Xx), dx(Yx)) = (x)g(Xx, Yx)

for all Xx, Yx Hx.The function : M R is obviously non-negative, hence there exist a non-

negative function : M R with 2 = . is called the dilation of.Proposition 4.3. [5, Example 2.4.6] Let : (M, g) Rn be a smooth map thenis horizontally weakly conformal if and only if

g(gradi, gradj) = ij,

where = (1, . . . , n).

If : (M, g) C is a smooth map then is horizontally weakly conformal if andonly if (, ) =gC(grad, grad) = 0,

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where gC is the complex bilinear extension of the metric g and

grad() = grad(Re()) + igrad(Im()).

Definition 4.4. Let : (M, g)

(N, h) be a horizontally weakly conformal map.Then is said to be horizontally homothetic if

PH(grad()) = 0,

at all regular points. PH is the projection onto the horizontal distribution.

This means that for a curve :I M with (t) H(t), is constant, sinced

ds[( )(s)]s=t=d(t)((t))

=g(grad()(t),(t))

=0.

If is constant equal to 1 then is said to be a Riemannian submersion.

Definition 4.5. For a function f : (M, g) R we define the Laplacian (f) :M R by

(f)x=mi=1

(X2i(f)x dfx(XiXi))

for an orthonormal frame{Xi}ofT M aroundx. If (f) = 0 then fis said to be aharmonic function.

If : (M, g) Rn then we define () : M Rn by() = ((1), . . . , (n)).

4.2 Foliations

Definition 4.6. A partitionF ofM into connected subsets L, called the leavesofF, is said to be a foliation of dimension q and codimension n= m qif allL are immersed submanifolds of dimension qofM.

If : M N is a smooth submersion then the fibres of are the leaves of asmooth foliationF. Not all foliations are produced in this way, however for eachpoint there exists a neighbourhood U, such thatFUis the foliation of a submersiondefined onU.

Definition 4.7. LetMbe a manifold, a smooth subbundleVof the tangent bundleT M with dim(Vx) = q for all x M is said be a distribution of dimensionq. If [V, W] C(V) for all V, W C(V) then the distribution is said to beintegrable.

Proposition 4.8. [17, Theorem 1.60] LetV be a distribution on a manifold M,thenV is integrable if and only if there is a foliationF of M where the tangentbundle of each leafL is given by restrictingV to L.

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Definition 4.9. LetVbe a distribution on (M, g) and let E, F C(T M). Wedefine the second fundamental form BV by

BV(E, F) =1

2

(PH(

PV(E)PV(F)) + PH(

PV(F)PV(E))),

where PH and PVare the orthogonal projections onto the horizontal and verticaldistributions, respectively.

The mean curvature vector field V is defined by

V=1

qtrace(BV),

where q= dim(V).Given V Vdefine the Lie derivativeLVg:H H C(M) by

(LVg)(X, Y) =V(g(X, Y)) g([V, X], Y) g([V, Y], X).Definition 4.10. A distributionVon a Riemannian manifold (M, g) is said to be

(i) minimal, if for all xM, V(x) = 0.(ii) totally geodesic if for all xM, BVx = 0.(iii)umbilic, if for all xM, BVx (V, V) is independent ofVVx, with|V|= 1.(iii) conformal if for all xM,

(LVg)(X, Y) =(V)g(X, Y),for all X, Y Hx and all V Vx for some function .

(iv) Riemannian if for all x

M, (

LVg)(X, Y) = 0

Proposition 4.11. [5, Proposition 2.5.8] LetV be a distribution on (M, g) andH=V. ThenV is conformal if and only ifH is umbilic.Proof. Let X, Y Hx and V Vx. We notice that

(LVg)(X, Y) =V(g(X, Y)) g([V, X], Y) g([V, Y], X)=g(VX, Y) + g(X, VY)

g(VX, Y) + g(XV , Y) g(VY , X) + g(YV , X)

=g(XV , Y) + g(YV , X)= g(XY + YX, V)= 2g(BH(X, Y), V).

() Suppose thatV is conformal then2g(BH(X, Y), V) =(V)g(X, Y).

IfVi is an orthonormal basis forV thenBH(X, X) =

i

g(BH(X, X), Vi)Vi

= 1

2i

(Vi)g(X, X)Vi

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= 12

i

(Vi)Vi

which is independent ofX,|X| = 1, soHis umbilic.() IfH is umbilic then BH(X, X) is independent of X,|X| = 1. Say thatBH(X, X) =V0 then

(LVg)(X, X) =2g(V0, V) =(V),for all X H. IfX, Y Hwith|X| =|Y| = 1 and XY then|X+ Y| = 2 sothat

2V0=BH(X+ Y, X+ Y)

=BH(X, X) + BH(Y, Y) + 2BH(X, Y)

=V0+ V0+ 2BH(X, Y)

i.e. BH(X, Y) = 0 this implies that (LVg)(X, Y) = 0 soVis conformal.

4.3 Harmonic Morphisms

For a smooth map : (M, g) (N, h) we denote thetension fieldofby(), see[5, Definition 3.2.4] for a definition of the tension field. is said to be a harmonicmapif() = 0, [5, Theorem 3.3.3].

Definition 4.12. Let (M, g) and (N, h) be Riemannian manifolds. A smooth map: (M, g)(N, h) is said to be a harmonic morphism if for every open subsetU ofN, with 1(U) non-empty, and any harmonic function f : (U, h)

R the

composition f : (1(U), g) R is a harmonic function.Theorem 4.13. [9, p. 123][14, Theorem 5.1] A smooth map : (M, g) (N, h) isa harmonic morphism if and only if is a horizontally weakly conformal harmonicmap.

Proposition 4.14. [5, Example 4.2.8] A smooth map : (M, g) Rn is a har-monic morphism if and only if each component i, i = 1, . . . , n is a harmonicfunction and

g(grad(i), grad(j)) = ij,

where is a function onM.

Theorem 4.15. [4, Theorem 5.2] Let : (M, g) (N, h)be a smooth non-constanthorizontally weakly conformal map. Then is a harmonic morphism if and only if atevery regular point, the mean curvature vector fieldVof the fibres and the gradientof the dilation of are related by

(n 2)PH(grad ln ) + (m n)V= 0.HerePH denotes the orthogonal projection ofT M ontoH.Corollary 4.16. [4, Theorem 5.2]Let : (M, g) (N, h)be a smooth non-constanthorizontally weakly conformal map if

(i) n= 2, then is a harmonic morphism if and only if the fibres are minimal.

(ii) n= 2 and is horizontally homothetic, then is a harmonic morphism ifand only if the fibres are minimal.28


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In particular any Riemannian submersion with minimal fibres is a harmonic mor-phism. For manifolds of non-positive curvature we have the following

Proposition 4.17. [8, Proposition 3.1] Let(M, g) be a Riemannian manifold withnon-positive sectional curvature, and : (M, g)

(N, h) a surjective Riemannian

submersion. If the fibres of are totally geodesic then(i) the horizontal distribution is integrable,(ii) Nhas non-positive sectional curvature,(iii) IfMhas negative sectional curvature then so doesN.

Definition 4.18. Let (M, g) be a Riemannian manifold. Then a foliation of codi-mensionn on Mis said toproduce harmonic morphisms if for each point pMthere exists around p a locally defined harmonic morphisms : U (N, h) to someRiemannian manifold Nof dimension n.

Theorem 4.19. [5, Corollary 4.7.7] Let (M, g) be a Riemannian manifold and

Fbe a foliation onMof codimensionn. If(i) n = 2, thenF produces harmonic morphisms if and only ifF is conformal

with minimal leaves,(ii) n= 2, thenF produces harmonic morphism if and only ifF is conformal

and the1-form:C(T M) R defined by(E) = (n 2)g(H, E) (m n)g(V, E),

is closed. HereH andVdenote the mean curvature vector fields of the horizontaland the vertical distributions, respectively.

The 1-form isclosed if the 2-form d defined byd(E, F) =E((F)) F((E)) ([E, F])

satisfies d= 0.

4.4 Harmonic Morphisms on Lie Groups

A subspace V ofCn is said to be isotropic if (u, v) = 0 for all u, v V, where(, ) : CnCn C is the complex bilinear extension of the canonical scalar producton Rn. IfVis isotropic thenvV implies v /V. Since (v, v) =|v| = 0, this impliesthat

dimC(V) n

2

,

where [] is the integer part. Hence a maximal isotropic subspace V satisfies

dimC(V) =n

2

.

Proposition 4.20. [10, Proposition 2.5]Let : (M, g) Rn,n2, be a harmonicmorphism and letVbe an isotropic subspace ofCn. Thenv : (M, g)C definedby

v(x) = ((x), v)

is a harmonic morphism for anyvV.

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Proof. Denote the dilation of by . Then

(v) =(n

i=1ivi)

=ni=1

(i)vi

=0

so v is a harmonic map and

(v, v) =n

i,j=1

gC(grad(ivi), grad(jvj))

=

ni,j=1 v

ivjg(grad(i), grad(j))

=n

i,j=1

vivj2ij

=2(v, v)

=0,

so v is horizontally weakly conformal.

Thus if we have a harmonic morphism to Rn we also obtain a complex-valued

harmonic morphism. The next theorem gives sufficient conditions for the construc-tion of complex valued harmonic morphisms even if : GRn is not a harmonicmorphism.

Lemma 4.21. [10, Proposition 5.1] LetG be a simply connected Lie group with Liealgebrag. Leta= g/[g, g] and supposen= dim(a)= 0. Then there exist a naturalsurjective homomorphism : G Rn and a left-invariant Riemannian-metric onG such that is a Riemannian submersion.

Proof. Botha and Rn are abelian and thus there exists a Lie algebra isomorphism : a Rn, that induces a Lie group isomorphism : A Rn, where A is theconnected simply connected Lie group with Lie algebra a. By equipping a with asuitable scalar product becomes an isometry. The scalar product on a induces aleft-invariant metric on Asuch that is an isometry.

Equip g with any scalar product such that the surjective Lie algebra homomor-phism :g a,

(X) =X+ [g, g],

is a Riemannian submersion. This induces a metric on G such that the uniquehomomorphism, Proposition 1.10, : G A is a Riemannian submersion. Thesurjective homomorphism = :GRn is a Riemannian submersion.

Theorem 4.22. [10, Proposition 5.2] LetG be a simply connected Lie group and : (G, g) Rn be a group homomorphism which is also a Riemannian submersion.30


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For the canonical basis{e1, . . . , en} ofRn let{X1, . . . , X n} be the orthonormal basisforHe withde(Xi) =ei. Define C by

= (trace(adX1), . . . , trace(adXn))

and putN() = {w Cn|(, w) = 0}.

If there is a maximal isotropic subspaceW ofCn such that

V =W N() ={0}thenv :G C, vV, defined by

v(x) = ((x), v)

is a globally defined complex-valued harmonic morphism.

Proof. The map : GRn

is a group homomorphism so

(LpexpG(tX)) = (p) + (expG(tX)),

for each X g. Further,(expG(tX)) = expR

n

(de(tX)) =tde(X),

since expRn

= idRn. Thus

Xk()(p) = dk

dtk[(Lp(exp

G(tX)))]t=0 = 0,

for all k2, in particular X2

()(p) = 0 and

dp(X) =X()(p) = de(X).

Let{Xn+1, . . . , X N} be an orthonormal basis forVe = [n, n]. The Laplacian for is given by

()(p) =Ni=1

X2i()(p) dp(XiXi)

= N

i=1

dp(XiXi)

= dp(Ni=1

XiXi)

= de(Ni=1

XiXi)

= de(N

i,j=1

XiXi, Xj Xj)

= N

i,j=1XiXi, Xj d(Xj)

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= n

j=1

(trace(adXj))ej

= ,which implies that

(v)p = (, v)=0

and

(v, v) =n

i,j=1

gC(grad(ivi), grad(jvj))

=

ni,j=1

vivjg(grad(i), grad(j))

=n

i,j=1

vivjij

=(v, v)

=0.

Corollary 4.23. Suppose thatG satisfies the conditions of Theorem 4.22. If= 0,then is a harmonic morphism.

We have that m = dimC(W) = [n2 ] and l = dimC(N()) n 1 and obviouslynm + l dimC(WN()) which implies that dimC(W N())m 1. Ifn4then dimC(W N()) 1. If n = 3 and Rn then redefine the basis so thate1 = /|| and{e1, e2, e3} is an orthonormal basis. Then W= spanC{e2+ie3} is amaximal isotropic subspace ofC with N() W={0}.Corollary 4.24. [10, Proposition 6.2] LetNbe a connected, simply connected andnilpotent Lie group withdim(N) 2. Then there exists a left-invariant metric onN, such that there exist globally defined complex-valued harmonic morphisms.

Proof. The Lie algebra, n, ofN is nilpotent so adZ : n n is nilpotent for eachZn. Therefore trace(adZ) = 0 which implies that in Theorem 4.22 is a harmonicmorphism and because of Proposition 1.16 n2.

The same conclusion as in Corollary 4.24 is true for any unimodular Lie algebrag with dim(g/[g, g]) 2 since trace(adX) = 0 for all X g for a unimodular Liealgebra.

4.5 Harmonic morphisms by Conformal Foliations

Another method for producing harmonic morphism is by conformal foliations. In

this section be give sufficient conditions for their existence. It is unknown what thetarget space of these harmonic morphisms is.

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Lemma 4.25. LetVbe a real vector space with a scalar product and letL: V Vbe a linear operator then the following are equivalent:

(i) there exists an orthonormal basis{Xi} ofV such that

LXi, Xi LXj, Xj = 0and

LXi, Xj + LXj , Xi = 0for alli, j= 1, . . . , n, i=j ,

(ii)LZ,Z is independent ofZV with|Z| = 1,(iii) there exist a(L) R such thatL= (L)I+ Im(L).

Remark 4.26. In fact this means that L is in the Lie algebra of the group ofconformal actions onV.

Proof. (i)

(ii). Let Z

V with|

Z|

= 1 be given, then

Z=ni=1

ciXi,

withn

i=1 |ci|2 = 1. We find that

LZ,Z =n

i,j=1

cicj LXi, Xj

=n

i=1 |

ci

|2

LXi, Xi

+

n

i


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=1

2

ni,j=1

cj(LXj, Xi + Xj, LXi)Xi

=n

i=1

LXi, X

iciX

i+

i=j(LX

j, X

i+

Xj

, LXi

)Xi

= LX1, X1 Wi.e.LX1, X1= (L).

(iii) (ii). Im(L) is skew-adjoint soIm(L)Z, Z =Z, Im(L)Z

i.e.Im(L)Z, Z = 0.Theorem 4.27. Let G be a Riemannian Lie group with Lie algebra g. Suppose

there is an orthogonal decompositiong

=a

k

msuch that(i) [a k, a k] =a k

(ii) [a,m] k m(iii) [k m, k m]k.Further for allAa assume that(iv)adA(X), X is constant forXm with|X| = 1 and(v) trace(adX) = 0 for allXm.

Then the distributionV=a k is integrable and produces harmonic morphisms.Proof. (1) From (i) and (iii) we have

[

V,

V] = [a

k, a

k]

a

k

Vso the integrability is satisfied.(2) Conformality: LetX, Ym be orthonormal,{Zi}vi=1a orthonormal basis for

k and{Ai}mni=v a orthonormal basis fora then

BH(X, Y) =1

2PV(XY + YX)

=1

4

vi=1

XY + YX, Zi Zi+14

mni=v

XY + YX, Aj Aj

=1

4

v

i=1

(

[X, Y] + [Y, X], Zi

+ 2

[Zi, X], Y

+ 2

[Zi, Y], X

)Zi

+1

4

mni=v

([X, Y] + [Y, X], Aj + 2 [Aj , X], Y + 2 [Aj, Y], X)Aj

=1

2

vi=1

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

+1

2

mni=v

([Aj, X], Y + [Aj , Y], X)Aj

=1

2

mni=v

([Aj, X], Y + [Aj , Y], X)Aj

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=0

because of (iv) and

BH(X, X) =1

2V(X

X+X

X)

=1

2

i

XX, Zi Zi+12

j

XX, Aj Aj

=i

([Zi, X], X)Zi+j

([Aj , X], X)Aj

=j

([Aj , X], X)Aj

=j

([Aj , Y], Y)Aj

=BH(Y, Y)

because of (iii) and (iv).(3) Closed form: Let Aa and let{Xj}nj=1 be a orthonormal frame for m then

(A) =n 2

n

nj=1

PV(XjXj), A

+m nm n

vi=1

PH(ZiZi), A +mni=v

PH(AiAi), A

=n

2

nj

[A, Xj ], Xj=(n 2)(A)

for : a R thus (A) :G R is constant.Let Z k then

(Z) =n 2

n

nj=1

PV(XjXj), Z

+m nm n

v

i=1

PH

(

ZiZi), Z

+

mn

i=v

PH(AjAj), Z=

n 2n

j

[Z, Xi], Xi

=0.

Finally let X m then

(X) =n 2

n

nj=1

PV(XjXj), X

+m

n

m n v

i=1PH(ZiZi), X +

mni=v

PH(AiAi), X35


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=v

i=1

[X, Zi], Zi +mni=v

[X, Ai], Ai

=trace(adX)

=0.

Thus we find thatd(E, F) = 0.

(4) Minimality:

V, X

=

vi=1

PH(ZiZi), X +mni=v

PH(AiAi), X

=v

i=1

[X, Zi], Zi

+

mn

i=v

[X, Ai], Ai

=trace(adX)

=0

because of (v).

4.6 A New Construction

Lemma 4.28. LetG be a Riemannian Lie group with metricg and letNbe a closednormal subgroup. Then there is a unique left-invariant metric h on the Lie group

G/N such that is a Riemannian submersion.

Proof. As stated in Theorem 2.1 : G G/N is a submersion. Since N is anormal subgroup

(g1N) (g2N) = (g1 g2)Nis a well-defined group operation. Thus d(Lg1N) = d(Lg2N) for g1N=g2N.

The vertical space of ate isVe= n. LetU He= n i.e. the horizontal spaceof at eG. Then

dg(d(Lg)e(U)) =d

dt[(g exp(tU))]t=0

=ddt

[(g exp(tU))N]t=0

=d

dt[(gN) (exp(tU)N)]t=0

=d(LgN)eN(U).

We define the metric heN on Te(G/N) by

heN(X, Y) =ge(de(X), de(Y)),

for X, Y

TeN(G/N), and extend it to TgN(G/N) by

hgN(d(LgN)e(X), d(LgN)e(Y)) =heN(X, Y).

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Let U, V He thengg(d(Lg)e(U), d(Lg)e(V)) =ge(U, V)

=heN(de(U), de(V))

=hgN(d(LgN)e(de(U)), d(LgN)e(de(V)))=hgN(dg(d(Lg)e(U)), dg(d(Lg)e(V))).

It is obvious that this metric is unique.

Definition 4.29. A Lie group G with Lie algebra gis said to be exponential if

exp : g Gis a diffeomorphism. In this case the Lie algebra is also said to be exponential.

Obviously an exponential groups has to be simply connected however there existsimply connected Lie groups that are not exponential groups, [15, p. 63]. Forsolvable groups there are however sufficient and necessary conditions.

Theorem 4.30. [15, Theorem 1.104] Let N be a simply connected nilpotent Liegroup with Lie algebran thenexp :nN is a diffeomorphism.Theorem 4.31. [7, Theorem 3] LetSbe a simply connected solvable Lie group withLie algebras then the following are equivalent

(i) exp : s S is a diffeomorphism(ii) adX :s s has no purely imaginary eigenvalues for anyX s.(iii) All roots: sC ofadS :s s, Ss, are of the form= + i where

andare real valued and proportional.

The next theorem generalizes [10, Theorem 12.1].

Theorem 4.32. LetG = NAbe a simply connected Riemannian Lie group whereAandNare simply connected Lie groups with Lie algebrasa andkm, respectivelyand suppose that there is a closed connected subgroup K of N with Lie algebra k.Assume further that there is an orthogonal decompositiong= a k m such that

(i) [a, k] k(ii) [a,m] m(iii) [k m, k m]k(iv) adH|m=(H)Im+ Im(adH|m) for allH a(v) trace(adX) = 0 for allX

m.

Then there exist a harmonic morphism :GRm wherem= dim(m).Remark 4.33. IfNis an exponential group theKmust be a closed simply connectedsubgroup ofN.

Proof. Kis a normal subgroup of bothNandG. The Lie groupN/Kis connectedand since [km, km]k it has an abelian Lie algebra, hence it is abelian. Let Nhave the induced metric fromGand suppose that N/Khas the unique left-invariantmetric such that the homogeneous projection

: N N/K (n) =nK

is a Riemannian submersion. Since Nis simply connected andK connectedN/K isa simply connected abelian Lie group, hence it is isometric to Rm. Define the map

37


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:GNby :n an and put = :GN/K. We will show that is a harmonic morphism.

We have that 1(n K) = n K A = n A K. The tangent space ofG at eis gand the vertical and horizontal spaces areVe=a +k, because of condition (ii),andHe=m. These spaces at a point n aare given by the left translate i.e.

Vna = d(Lna)e(a+k) andHna= d(Lna)e(m).Let X m then

dna(d(Lna)e(X)) =d

dt

(Lna(exp(tX)))

t=0

=d

dt

(n a exp(tX))

t=0

=d

dt(n exp(tAda(X)) a)t=0=d

dt

(n exp(tAda(X)))K

t=0

Since A is connected condition (ii) implies that AdA(m) m i.e. Ad : AAut(m) which implies that = dAde : aEnd(m). Thus for each H a, (H) isgiven by (H) = adH, i.e.,

(H)(X) = [H, X] =(H)X+ Im((H))X.

So for all aA the map Ada: m mis conformal with dilation (a). IfX, Y mthen

gN/KnK (dna(d(Lna)e(X)), dna(d(Lna)e(Y)))

=gN/KnK (d(LnK)eK(Ada(X)), d(LnK)eK(Ada(Y)))

=(a)2gN/KnK (d(LnK)eK(X), d(LnK)eK(Y))

=(a)2gN/KeK (X, Y)

=(a)2gNe (X, Y)

=(a)2gGe(X, Y)

=(a)2gGna(d(Lna)e(X), d(Lna)e(Y))

Any horizontal curve passing through n a is given by (t) =n exp(tAda(X)) a.Thus we find that the dilation of along is given by ((t)) =(a). Since thisdoes not depend on t, is constant along horizontal curves.

Thus is homothetic and we only need to show that has minimal fibres. Since1(n K) =n K A and since Ln is an isometry on G it is enough to show thatthe fibre 1(e K) is minimal. Since 1(e K) = K A is a subgroup ofG theLevi-Civita connection is the same in every point depending only on the Lie algebra.Thus we only have to show that is minimal at eG. Let{Hj}be a orthonormalbasis for a and{Zi} an orthonormal basis for k. Let V be the mean curvaturevector field along 1(e K), then for Xm, we have

Ve , X

=j

HjHj, X + i

ZiZi, X

38


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=1

2

j

([Hj , Hj], X + [X, Hj ], Hj + [X, Hj ], Hj)

+1

2i

(

[Zi, Zi], X

+

[X, Zi], Zi

+

[X, Zi], Zi

)

=j

[X, Hj], Hj + tracek(adX)

=trace(adX)

=0,

because of (vi). Thus the fibres are minimal and is a harmonic morphism.

4.7 Application to Homogeneous Hadamard Manifolds

Lemma 4.34. Letn, be a generalized root space ofnand suppose thatadA|n, isnormal for allH a. Then

adH|n, =(H)In,+ Im(adH)|n,for allHa.Proof. By definition n, = (n

C nC)n. Since adA|n, is normal adCH|n, =

adH|nCnC is also normal. Thus adCH|n, it is diagonalizable over C with orthogonaleigenvectors{X1,, . . . , X n,} for nC and{X1,, . . . , X n,} for nC . IfXnC then

adH(X) =

nk=1

adH(X), Xk, Xk,+n

k=1adH(X), Xk, Xk,

=n

k=1

X, adH(Xk,) Xk,+n

k=1

X, adH(Xk,) Xk,

=n

k=1

X, (H)Xk, Xk,

=n

k=1

(H) X, Xk, Xk,

=(H)X.

IfYnC then in the same way we find thatadH(Y) =(H)Y.

Therefore

Re(adH)(X) =1

2((H) + (H))X=(H)X

for all X nC . The same is true for any YnC . AsRe(adH)

|nCnC =(A)I

this must also be the case for Re(adH)|n, .39


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Corollary 4.35. LetSbe a simply connected solvable Riemannian Lie group, corre-sponding to a homogeneous Hadamard manifold, with Lie algebras = a+n. Supposethere exist a generalized root spacen, ofn, withdim(n,)2 such that

(i) adA|n, is normal for allA a,(ii) [n, n]n,, wheren, is the orthogonal complement inn, and(iii) adA(n

,)n, for allAa.

Then there exists a complex valued harmonic morphism onS.

Proof. We check the conditions of Theorem 4.32.(i) with k= n, this is satisfies.(ii) by construction this is satisfied for n,.(iii) is obvious since [n, n]n,.(iv) follows from Lemma 4.34.(v) follows from the fact that n is nilpotent.

4.8 Examples

We will construct examples of Lie algebras that satisfy the conditions of Theorem4.32 but are not covered by [10, Theorem 12.1].

Let nbe a nilpotent Lie algebra anda a one dimensional vector space, let Ha.We define adH :n n in such a way that s= a + n is a Lie algebra with n= [s,s].Let{X1, . . . , X n} be a basis for nand set

adH(Xi) =cijXj,

and use the Jacobi identity to find the requirements oncij for sto be a Lie algebra.Since n is nilpotent condition (v) in Theorem 4.32 is always satisfied.

Example 4.36. Our first example is a solvable Lie algebra s where n = [s,s] isabelian and dim(a) = 1. Let span{X1, . . . , X n, Z , W } be an orthogonal basis for nwith all brackets zero.

Let H a and define adH by

adH(X1) =n

j=1

c1jXj

adH(X2) =n

j=1c2jXj

...

adH(Xn) =n

j=1

cnjXj

adH(Z) =sZ+ tW

adH(W) = tZ+ sW.Then s is a Lie algebra for any real cij , s and t. Set k = span{X1, . . . , X n} andm= span{Z, W}then the assumptions of Theorem 4.32 are satisfied.

If n 2, cij = 0 for j < i and for some i < j we have cii = cjj and cij= 0or ift= 0, then adH is not diagonalizable and we get new examples of harmonicmorphisms not covered by [10, Theorem 12.1].

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In the following examples we try to find new examples where nis not abelian westart with the case dim(n) = 4. The only nilpotent non-abelian Lie algebras withdim(n) = 4, up to isomorphism, are

L4,1 = span{X1, . . . , X 4} with [X4, X3] =X2, [X4, X2] =X1and

L3,1 R = span{X1, . . . , X 4} with [X4, X3] =X2.Example 4.37. We start with n = L4,1. There are

53

43

= 6 combinations of

[H, [Xi, Xj]] that we need to check.(i) H, X1, X2

0 = [H, 0] = [H, [X1, X2]] =[X1, [H, X2]] [X2, [H, X1]]=[X1, c2jXj] [X2, c1jXj]=c14X1,

so c14 = 0.(ii) H, X1, X3

0 = [H, 0] = [H, [X1, X3]] =[X1, [H, X3]] [X3, [H, X1]]=[X1, c3jXj] [X3, c1jXj]=c14X2

(iii) H, X1, X4

0 = [H, 0] = [H, [X1, X4]] =[X1, [H, X4]] [X4, [H, X1]]=[X1, c4jXj] [X4, c1jXj]= c12X1 c13X2,

so c12 =c13 = 0.(iv) H, X2, X3

0 = [H, 0] = [H, [X2, X3]] =[X2, [H, X3]] [X3, [H, X2]]=[X2, c3jXj] [X3, c2jXj]= c34X1+ c24X2.

so c24 =c34 = 0

(v) H, X2, X4

c1jXj = [H, X1] = [H, [X2, X4]] =[X2, [H, X4]] [X4, [H, X2]]=[X2, c4jXj] [X4, c2jXj]= c44X1 c22X1 c23X2,

soc11 =c44 c22,c12 =c23, c13= 0 and c14= 0.(vi) H, X3, X4

c2jXj = [H, X2] = [H, [X3, X4]] =[X3, [H, X4]] [X4, [H, X3]]=[X3, c4jXj]

[X4, c3jXj]

= c44X2 c32X1 c33X2,

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soc21 =c32,c22 = c44 c33, c23= 0 and c24= 0.We must have

c33+ 2c44 0 0 0

c21 c33+ c44 0 0

c31 c21 c33 0c41 c42 c43 c44

i.e.

adH(X1) =(c33+ 2c44)X1

adH(X2) =c21X1+ (c33+ c44)X2

adH(X3) =t31X1+ c21X2+ c33X3

adH(X4) =c41X1+ c42X2+ c43X3+ c44X4

in the basis{Xi}4i=1. For these Lie algebras we have [n, n] = span{X1, X2}. Wecheck what is required for the conditions of Theorem 4.32 to be satisfied.

Let m = span{X3, X4}, k = span{X1, X2}. Then conditions (i) and (iii) areobviously satisfied. While

adH(X3) =c31X1+ c21X2+ c33X3

adH(X4) =c41X1+ c42X2+ c43X3+ c44X4

so that we must have c31 = c21 = c41 = c42 = c43 = 0 and c33 = c44 in order tosatisfy (ii) and (iv).

Thus adHmust be given by

adH(X1) =3c44X1

adH(X2) =2c44X2adH(X3) =c44X3

adH(X4) =c44X4.

So in this case there are no new examples of harmonic morphisms of type we arelooking for.

Example 4.38. Let n= L3,1 R.(i) H, X1, X2

0 = [H, 0] = [H, [X1, X2]] =[X1, [H, X2]] [X2, [H, X1]]=[X1, c2jXj] [X2, c1jXj]=0.

(ii) H, X1, X3


so c14 = 0.(iii) H, X1, X4

0 = [H, 0] = [H, [X1, X4]] =[X1, [H, X4]] [X4, [H, X1]]42


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=[X1, c4jXj] [X4, c1jXj]= c13X2,

so c13 = 0.

(iv) H, X2, X3


so c24 = 0(v) H, X2, X4

0 = [H, 0] = [H, [X2, X4]] =[X2, [H, X4]] [X4, [H, X2]]=[X2, c4jXj]

[X4, c2jXj]

= c23X2,soc23 = 0.

(vi) H, X3, X4

c2jXj = [H, X2] = [H, [X3, X4]] =[X3, [H, X4]] [X4, [H, X3]]=[X3, c4jXj] [X4, c3jXj]= c44X2 c33X2

Soc21 = 0,c22 = c44 c33, c23= 0 and c24 = 0We must

c11 c12 0 00 c33+ c44 0 0

c31 c32 c33 c34c41 c42 c43 c44

.i.e.

adH(X1) =c11X1+ c12X2

adH(X2) =(c33+ c44)X2

adH(X3) =c31X1+ c32X2+ c33X3+ c34X4

adH(X4) =c41X1+ c42X2+ c43X3+ c44X4

in the basis{Xi}4i=1. We set m= span{X3, X4} and k= span{X1, X2}.Let m = span{X3, X4}, k = span{X1, X2}. Then conditions (i) and (iii) are

obviously satisfied. While

adH(X3) =c31X1+ c32X2+ c33X3+ c34X4

adH(X4) =c41X1+ c42X2+ c43X3+ c44X4

so we needc31=c32= c41=c42= 0, c34=c43 andc33=c44 in order to satisfy (ii)and (iv).

Thus adHmust be given by

adH(X1) =c11X1+ c12X2

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adH(X2) =2c44X2

adH(X3) =c44X3+ c34X4

adH(X4) = c34X3+ c44X4.

For c11 = 2c44 and c12= 0 or c34= 0 we will in fact get new examples of the typewe are looking for.

Example 4.39. Let n= span{X1, . . . , X 5} with [X5, X4] =X1 and [X3, X2] =X1.With a= RHwe must have that.

(i) H, X1, X2


so c13 = 0.(ii) H, X1, X3

0 = [H, 0] = [H, [X1, X3]] =[X1, [H, X3]] [X3, [H, X1]]=[X1, c3jXj] [X3, c1jXj]= c12X1,

so c12 = 0.(iii) H, X1, X4

0 = [H, 0] = [H, [X1, X4]] =[X1, [H, X4]]

[X4, [H, X1]]

=[X1, c4jXj] [X4, c1jXj]=c15X2,

so c15 = 0.(iv) H, X1, X5

0 = [H, 0] = [H, [X1, X5]] =[X1, [H, X5]] [X5, [H, X1]]=[X1, c5jXj] [X5, c1jXj]= c14X2,

so c14 = 0.(v) H, X2, X3

c1jXj = [H, X1] = [H, [X2, X3]] =[X2, [H, X3]] [X3, [H, X2]]=[X2, c3jXj] [X3, c2jXj]= c33X1 c22X1,

soc11 =c22 c33 and c12=c13= c14 = c15 = 0.(vi) H, X2, X4

0 = [H, 0] = [H, [X2, X4]] =[X2, [H, X4]] [X4, [H, X2]]

=[X2, c4jXj] [X4, c2jXj]= c43X1+ c25X144


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=(c25 c43)X1,so c43 =c25.

(vii) H, X2, X5

0 = [H, 0] = [H, [X2, X5]] =[X2, [H, X5]] [X5, [H, X2]]=[X2, c5jXj] [X5, c2jXj]= c53X1 c24X1= (c24+ c53)X1,

so c53 = c24.(viii)H, X3, X4

0 = [H, 0] = [H, [X3, X4]] =[X3, [H, X4]] [X4, [H, X3]]=[X3, c4jXj]

[X4, c3jXj]

=c42X1+ c35X1

=(c42+ c35)X1,

so c42 = c35.(ix) H, X3, X5

0 = [H, 0] = [H, [X3, X5]] =[X3, [H, X5]] [X5, [H, X3]]=[X3, c5jXj] [X5, c3jXj]=c52X1 c34X1=(c52 c34)X1,

so c52 =c34.(x) H, X4, X5

c1jXj = [H, X1] = [H, [X4, X5]] =[X4, [H, X5]] [X5, [H, X4]]=[X4, c5jXj] [X5, c4jXj]= c55X1 c44X1,

soc11 =c44 c55 and c12=c13= c14 = c15 = 0.So adA must be of the form

c22

+ c33

0 0 0 0c21 c22 c23 c24 c25c31 c32 c33 c34 c35c41 c35 c25 c44 c45c51 c34 c24 c54 c55

,

where c22+ c33=c44+ c55.If we set k = span{X1, X2, X3} and m = span{X4, X5}. Then conditions (i) is

obviously satisfied. While we need c24=c25= c34=c35= 0 to satisfy (ii). Since

adH(X1) =(c22+ c33)X1

adH(X2) =c21X1+ c22X2+ c23X3adH(X3) =c31X1+ c32X2+ c33X3

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adH(X4) =c41X1+ c44X4+ c45X5

adH(X5) =c51X1+ c54X4+ c55X5

We need that c41 = c51 = 0, c45 =c54 and c44 = c55 in order to satisfy (iii) and(iv).

Thus adH is given by

adH(X1) =2c55X1

adH(X2) =c21X1+ c22X2+ c23X3

adH(X3) =c31X1+ c32X2+ c33X3

adH(X4) =c55X4+ c45X5

adH(X5) = c45X4+ c55X5.In order toget new examples not covered by [10, Theorem 12.1], we need that

either c22

=c33

(=c55

) or c22

= 0 and c33

=c55

, and that at least one ofc21

, c23

, c31

, c32is non-zero or c45= 0.

46


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Bibliography

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