Directory UMM :Journals:Journal_of_mathematics:OTHER:

M. J. Druetta

∗

P-SPACES OF IWASAWA TYPE AND ALGEBRAIC RANK ONE

Abstract. A Riemannian manifold is aP-space if the Jacobi operators along the geodesics are diagonalizable by a parallel orthonormal basis. We show that a solvable Lie group of Iwasawa type and algebraic rank one which is aP-space is a symmetric space of noncompact type and rank one. In particular, irreducible, non-flat homogeneous EinsteinP-spaces with nonpositive curvature and algebraic rank one, are rank one symmetric spaces of noncompact type.

Let M be a Riemannian manifold, R its curvature tensor and RXthe Jacobi operator

defined by RXY =R(Y,X)X,where X is a unit tangent vector. Following [3], we say

that M is aP-space if for every geodesicγin M the associated Jacobi operators Rγ′_(t)

are diagonalizable by a parallel orthonormal basis alongγ ,condition that is satisfied for symmetric spaces.

In this paper we study the homogeneousP-spaces of Iwasawa type and algebraic rank one and in particular, those with nonpositive curvature which are Einstein, since irreducible, non-flat homogeneous Einstein spaces with nonpositive curvature are rep-resented as Lie groups of Iwasawa type (see [6]). The geometry of Lie groups of Iwasawa type and algebraic rank one which at a first glance seems to be complicated, becomes very simple when they areP-spaces: they are Damek–Ricci spaces whose geometric structure is well known (Damek-Ricci spaces are defined in Section 1, fol-lowing [2], Chapter 4).

An outline of the paper is as follows. In Sections 1 and 2, we give the basic results concerning the Lie algebras of Iwasawa type and algebraic rank one, its geodesics in various directions and the expression of the Jacobi operator along them. Properties of its eigenvalues in the special case of parallel eigenvectors are obtained in Section 3. The geometric hypothesis involving condition(P)about the eigenvalues is strongly used in Section 4 to obtain algebraic properties of the solvable Lie algebra. This information allows us to show that an Iwasawa type Lie group of algebraic rank one satisfying condition (P)is a Damek-Ricci space. By applying a result from [2], we obtain the following:

THEOREM. If S is a solvable Lie group of Iwasawa type and algebraic rank one

which is anP-space, then S is a rank one symmetric space of noncompact type. The class of Riemannian manifolds obtained by considering Lie groups of Iwasawa type contains as a subclass the Damek-Ricci spaces, and more generally the irreducible,

∗_{Partially supported by Conicet, Conicor and Secyt (UNC). Part of this work was done during a stay of}

the author at ICTP, Trieste-Italy, by funds from IMPA (CNPq, Brazil).

non-flat homogeneous Einstein spaces with nonpositive curvature. As a consequence, we have

COROLLARY. An irreducible, non-flat homogeneous EinsteinP-space with

non-positive curvature and algebraic rank one is a symmetric space of noncompact type and rank one.

1. Lie algebras of Iwasawa type and algebraic rank one

A solvable Lie algebraswith inner producth,iis a metric Lie algebra of Iwasawa type if it satisfies the conditions

(i) s=n⊕awheren=[s,s] anda,the orthogonal complement ofn, is abelian. (ii) The operators adHare symmetric for all H ∈a.

(iii) For some H0∈a, adH0|nhas positive eigenvalues.

The simply connected Lie group S with Lie algebras and left invariant metric induced by the inner producth,i will be called of Iwasawa type. The Levi Civita connection and the curvature tensor associated to the metric can be computed by

2h∇XY,Zi = h[X,Y ],Zi − h[Y,Z ],Xi + h[Z,X ],Yi,

R(X,Y) = [∇X,∇Y]− ∇[X,Y ]

for any X,Y,Z ins.

For each unit vector X ins, RX,the Jacobi operator associated to X,is the

sym-metric endomorphism ofsdefined by RXY = R(Y,X)X.We will say that either the

metric Lie algebrassatisfies condition(P)or S is aP-space if for every geodesicγ in S the associated Jacobi operator Rγ′_(t) can be diagonalized by a parallel orthonormal

basis of Tγ (t)S. We note that condition(P)is equivalent to the fact RX◦R′_X =R′_X◦RX

for all X ∈ s(see Corollary 5 of [3] and note that S is a real analytic C∞-manifold). We recall that S is an Einstein space if

Ric(X)=trRX =c|X|2, c constant, for all X ∈s.

Ifs=n⊕ais a metric Lie algebra of Iwasawa type, letz denote the center ofn=[s,s] and letvbe the orthogonal complement ofzwith respect to the metrich,irestricted ton.Thusndecomposes asn= z⊕v, and for all H ∈ a adH : z → zand hence,

adH :v→vsince adH is symmetric. We recall thatnis said to be 2-step nilpotent if

[n,n]=[v,v]⊂z.

For any Z ∈zthe skew-symmetric linear operator jZ :v→vis defined by

hjZX,Yi = h[X,Y ],Zi for all X, Y ∈vand Z ∈z.

Equivalently, jZX =(adX)∗Z for all X ∈v, where(adX)∗denotes the adjoint of adX.

The operators jZ coincide with the usual one in the case of a 2-step nilpotentn(see

[5]) and their properties determine the geometry ofnands, as we will see.

We now assume thats=n⊕ais a metric Lie algebra of Iwasawa type and algebraic rank one; that is,a=RH where H, |H| =1,is chosen such that all eigenvalues of

adH|nare positive. S is called a Damek-Ricci space in the special case that adH|z=Id, adH|v=

1

2Id and j 2

Z = − |Z|

2_{Id for all Z ∈z(see [2], p. 78).}

We recall that since adH|nis a symmetric operator, nhas an orthogonal direct sum decomposition into eigenspacesnµ,for all eigenvaluesµof adH|n,which are in-variant by adHwith the property [nµ,nµ′]⊂nµ_+µ′ (by the Jacobi identity), whenever

µ+µ′is an eigenvalue of adH|n(see [7]). Moreover, sincezandvare adH-invariant, by

the same argument they also have decompositions into their eigenspaces asz=P λzλ,

andv=P µvµ.

1.1. Algebraic structure of the Lie algebras

The definition of the Lie algebra structure ons implies that, as a Lie algebra,sis the semidirect sum s = n+σ aof nand a = RH,by considering the R-algebra

homomorphismσ =ad: a →der n, H → (adH : n → n). Carrying this over to

the group level means that S = N ×τ A is a semidirect product of N and A = R

(considered in the canonical way), where

τ : A→AutN, τa: x →ax a−1, (dτa)e=Ad(a),

is given by a exp X a−1 = exp_n(Exp(tadH)X)for all X ∈ n, a = t, and Exp

de-notes the exponential map of matrices. Note that S is diffeomorphic tosunder the map(X,r) → exp_nX,rsince exp_n : n → N,the exponential map of N, is a diffeomorphism.

We assume thatnis 2-step nilpotent. In this case we have that for any Z ∈ zand X ∈ v, if Z∗ and Y are eigenvectors of adH restricted tozand v, with associated

eigenvaluesλandµ,respectively, then the product in S yields (exp_n(Z+X),r)·(exp_n(Z∗+Y),s) = (exp_n(Z+erλZ∗+1

2e

rµ_{[X,Y ]+X+e}rµ_Y),r+s).

In fact, note that by the definition of the product in S we have (exp_n(Z+X),r)·(exp_n(Z∗+Y),s) = exp_n(Z+X)τr(expn(Z

∗_+Y)),r+s

= exp_n(Z+X)exp_n Exp(r adH)Z∗+Exp(r adH)Y,r+s

= exp_n(Z+X)exp_n(erλZ∗+erµY),r+s ,

since exp_nX exp_nY =exp_n(X+Y+1₂[X,Y ])gives the multiplication law in N (see the Campbell-Hausdorff formula in [7]).

1.2. Global coordinates in S

We introduce global coordinates in S given byϕ =(x1, ...,xk,y1, ...,ym,r),defined

bases of eigenvectors of adH inzandv, with associated eigenvalues{λ1, ..., λk}and

{µ1, ..., µm}respectively, then

ϕ(x1, ...,xk,y1, ...,ym,r)= expn(x1Z1+...+xkZk+y1X1+...+ymXm),r

. Following the same argument as the given in [2], p. 82 for Damek-Ricci spaces, we see in the case of 2-step nilpotentnthat

∂ ∂xi

ϕ(x1,...,xk,y1,...,ym,r)

=e−rλi_Z

i(ϕ(x1, ...,xk,y1, ...,ym,r)), ∂

∂yi _ϕ(x

1,...,xk,y1,...,ym,r)

=e−rµi_X

i(ϕ(x1, ...,xk,y1, ...,ym,r))

+1 2

X

j,s

e−rλs_y

jhjZsXi,XjiZs(ϕ(x1, ...,xk,y1, ...,ym,r)),

∂ ∂r

_ϕ(x

1,...,xk,y1,...,ym,r)

=H(ϕ(x1, ...,xk,y1, ...,ym,r)),

where Zi,Xi and H on the right-hand side denote the left invariant vector fields on S

associated to the corresponding vectors ins.

1.3. Curvature formulas

By applying the connection formula given at the beginning of this section, one obtains ∇H =0 and if Z,Z∗∈ z, X,Y ∈vthen∇ZZ∗= ∇Z∗Z = h[H,Z ],Z∗iH, ∇_XZ =

∇ZX = −1₂jZX and

∇XY = 1₂[X,Y ]+ h[H,X ],YiH,in case of 2-step nilpotentn. Consequently, by a

direct computation, we obtain the following formulas (see [4], Section 2): (i) RH = −ad2_H.

(ii) If either Z∈zλ,|Z| =1,or X ∈vµ, |X| =1,

RZH= −λ2H and RXH = −µ2H.

(iii) If Z ∈zλ,|Z| =1,for any Z∗∈zand X ∈v, we have RZZ∗=λ hZ,adHZ∗iZ−adHZ∗ and RZX = −

1 4j

2

Z(X)−λadHX.

In the case thatnis 2-step nilpotent, we obtain (iv) If X∈vµ,|X| =1,for any Z ∈z,Y ∈v

RXZ =

1

4[X,jZX ]−µadHZ and RXY = − 3

2. Geodesics and associated Jacobi operators

Throughout this section and the following ones,s= n⊕a will denote a metric Lie algebra of Iwasawa type and algebraic rank one, wherea=RH,|H| =1,is chosen such that all eigenvalues of adH|nare positive, andn =z⊕vis expressed as in the previous section. LetγY denote the geodesic in S satisfyingγY(0)=e (the identity of

S)andγ_Y′(0)=Y.For any X∈s, the associated left invariant field along the geodesic γY will be denoted by X(t)=X(γY(t))=(dLγY(t))eX.

Next, we compute the geodesicγY with Y ∈n, an eigenvector of adH|n. LEMMA1. If Y ∈nis an eigenvector of adH|nwith eigenvalueα,then

γY(t)=

exp_n

_tanhαt α

Y,−1

αln(coshαt)

with associated tangent vector field γ_Y′(t)= 1

coshαtY(t)−tanhαt H(t).

Proof. Let S0 be the simply connected Lie group associated to s0, the Lie algebra

spanned by{Y,H}.Note that S0has global coordinatesϕ(x,r)=(expnx Y,r)and it is a totally geodesic subgroup of S with connection∇S0_{= ∇|}

S0 satisfying

∇YY =αH, ∇YH= −[H,Y ]= −αY, ∇H =0.

Since the coordinate fields associated toϕ are given by ∂ ∂x ϕ(x,r)

=e−rαY(ϕ(x,r)), ∂ ∂r ϕ(x,r)

=H(ϕ(x,r)), the Christoffel symbols are easily computed by the formulas

∇∂ ∂x ∂ ∂x _ϕ(x,r)

= αe−2rαH(ϕ(x,r)), ∇∂ ∂x ∂ ∂r _ϕ(x,r)

= −α ∂ ∂x _ϕ(x,r) , ∇∂ ∂r ∂ ∂r _ϕ(x,r)

= 0= ∇∂ ∂r ∂ ∂x _ϕ(x,r) .

Hence, we obtain the geodesicγY(t)=ϕ(x(t),r(t)),where x(t)and r(t)are solutions

of the differential equations

x′′−2αx′r′ = 0, r′′+αe−2rα(x′)2 = 0. Using thatγ_Y′(t)= x′(t)_∂x∂

γY(t)+r ′_(t) ∂

∂r

γY(t),

γ_Y′(t)=1,with x′(t)=e2αr(t), ∂

∂x _γ Y(t)

=e−αr(t)Y(t)and ∂ ∂r _γ Y(t)

we have the equivalent equations

x′(t) = e2αr(t), r′′(t)+α(1−(r′(t))2 = 0 whose solutions satisfy

x(t)= Z t

0

e2αr(u)du and r′(t)= −tanhαt. Therefore, we get

r(t)= −1

αln(coshαt), x(t)= 1

αtanhαt, and the expression ofγY andγ_Y′ follows as claimed.

PROPOSITION1. If Z ∈ zand X ∈ vare eigenvectors of adH with associated

eigenvaluesλandµ, respectively, then for any Y ∈v, we have (i) R_γ′

Z(t)Y(t)= 1

cosh2λt(dLγZ(t))e ·

RZY −sinh2λt ad2HY −sinhλt jZ ₁

2λId−adH

Y

(i i) R_γ′

X(t)Z(t)= 1

cosh2µt(dLγX(t))e ·

RXZ−λ2sinh2µt Z+(λ−

1

2µ)sinhµt jZX

(i i i) R_γ′

X(t)Y(t)= 1

cosh2µt(dLγX(t))e ·

RXY−sinh2µt ad2HY−sinhµt ₁

2µId−adH

[X,Y ]

in the case of 2-step nilpotentn, with Y⊥X inv.

Proof. (i) Let Z ∈zandγZ(t)be the associated geodesic. Since γ_Z′(t)= 1

coshλtZ(t)−tanhλt H(t), we have that

R(Y(t), γ_Z′(t))γ_Z′(t)= 1

cosh2λt(dLγZ(t))e

·RZY +sinh2λt RHY −sinhλt(R(Y,Z)H+R(Y,H)Z)

Using the Bianchi identity and the connection formulas we compute R(Y,Z)H+R(Y,H)Z = 2 R(Y,H)Z−R(Z,H)Y

= 2∇[H,Y ]Z − ∇[H,Z ]Y

= −jZ(adHY)+

1 2λjZY,

that substituted in the above expression, gives (i) as stated since RH = −ad2_H.

(ii)-(iii) Assume that X ∈ vis an eigenvector of adH with eigenvalueµ,and let

Y ⊥X inv.Using the expression ofγ_X′(t),in the same way as (i) we get R(Z(t), γ_X′(t))γ_X′(t)= 1

cosh2µt(dLγX(t))e

·RXZ+sinh2µt RHZ−sinhµt(R(Z,X)H+R(Z,H)X)

. Hence, the expression of R_γ′

X(t)Z(t)follows as claimed since

R(Z,X)H +R(Z,H)X = 2∇[H,Z ]X− ∇[H,X ]Z =2λ∇ZX−µ∇XZ

= −(λ−1 2µ)jZX. Finally, we have

R(Y(t), γ_X′(t))γ_X′(t)= 1

cosh2µt(dLγX(t))e

·RXY+sinh2µt RHY −sinhµt(R(Y,X)H+R(Y,H)X)

. In the same way as above, in the case of 2-step nilpotentn, we compute

R(Y,X)H+R(Y,H)X = 2∇[H,Y ]X− ∇[H,X ]Y

= −[H,[X,Y ]]+1

2µ[X,Y ] (hX,Yi =0), which completes the proof of the proposition.

3. Eigenvectors and eigenvalues along Jacobi operators

In this section we assume thatn=z⊕vis non-abelian and jZ is non-singular onvfor

all Z∈z. Note that ifλis an eigenvalue of adH|zand Z ∈zλthen jZ|vµ :vµ→vλ−µ

for any eigenvalueµof adH|v.In fact, for X ∈vµand Y ∈vµ′ hjZX,Yi = h[X,Y ],Zi 6=0⇒[X,Y ]6=0,

Hence, for any eigenvalueµof adH|v,λ−µis also an eigenvalue of adH|v,λ > µ and the symmetric operator j2_Z preservesvµ.Moreover, the map Z → [X,jZX ]

de-fines a symmetric operator onz(h[X,jZX ],Z∗i = hjZ∗X,j_ZXi = h[X,j_Z∗X ],Zi)

such that [X,jZX ] ∈ zλ for all X ∈ vµ since h[X,jZX ],Zi = |jZX|2 6= 0,

[X,jZX ]∈nλ(the Jacobi identity) andλ > µ.

Next, we describe the eigenvalues of the operators R_γ′

Z(t), RγX′(t)and the parallel vector fields along the geodesicsγZ(t)andγX(t)for some Z ∈zλand X ∈vµ.

LEMMA2. Let Z∈zλand X ∈vµbe unit vectors. We set Y = jZX |jZX|. (i) If X is an eigenvector of j_Z2,then R_γ′

Z(t)has an eigenvector U(t)=x(t)X(t)+ y(t)Y(t) with x(t)2+y(t)2=1 and associated eigenvalue aZ(t)satisfying

aZ(t)cosh2λt = hRZY,Yi −(λ−µ)2sinh2λt−(

1

2λ−µ)|jZX| x(t) y(t)sinhλt whenever y(t)6=0.

(ii) Assume thatnis 2-step nilpotent. If X satisfies [X,jZX ] = |jZX|2Z,then

R_γ′

X(t) has an eigenvector U(t) = x(t)Z(t)+y(t)Y(t), x(t)

2_+y(t)2 _{= 1,whose}

associated eigenvalue aX(t)is given by

aX(t)cosh2µt

= 1 4|jZX|

2_−λµ−λ2_sinh2_µt+(λ−1

2µ)|jZX| y(t)

x(t)sinhµt or aX(t)cosh2µt

= −3 4|jZX|

2_{−(λ−µ)(µ+(λ−µ)sinh}2_µt)+(λ−1

2µ)|jZX| x(t) y(t)sinhµt according to x(t)6=0 or y(t)6=0,respectively.

Proof. Note that by the properties of Z and X , the spaces span{X(t),Y(t)} and span{Z(t),Y(t)}are invariant under the symmetric operators R_γ′

Z(t) and RγX′(t), re-spectively. The assertion of the lemma follows from the expression of R_γ′

Z(t)and RγX′(t) given in Proposition 1 and using in each case the equalities

R_γ′

Z(t)(U(t))=aZ(t)(U(t))and RγX′(t)(U(t))=aX(t)U(t). Note that in the last case,

RXZ =

1 4|jZX|

2_{Z−λµZ and R}

XY = −

3 4|jZX|

2_{Y−µ(λ−µ)Y}

sincenis 2-step nilpotent.

PROPOSITION2. Let Z∈zλbe a unit vector.If X∈vµ,|X| =1,is an

(Y = jZX |jZX|), x

2

0+y02=1,is given by U(t)=x(t)X(t)+y(t)Y(t) where

x(t)=x0cos s(t)−y0sin s(t) , y(t)=x0sin s(t)+y0cos s(t), and

s(t)= 1 2|jZX|

Z t

0

du coshλu.

Proof. We first note that (dLγZ(t))e(span{X,jZX}) is invariant under ∇γ_Z′(t) since γ_Z′(t) = _cosh1_λtZ(t)−tanhλt H(t), ∇ZX = −1₂jZX,∇ZjZX = 1₂|jZX|2X and

∇H =0.

Hence the parallel vector field U alongγZ with U(0)=x0X+y0Y is given by U(t)=

x(t)X(t)+y(t)Y(t)satisfying the equation∇_γ′

Z(t)U(t)=0,which gives x′(t)X+x(t) 1

coshλt∇ZX+y

′_(t)Y+y(t) 1

coshλt∇ZY =0 for all real t since X and Y are left invariant. Thus,

x′(t)X−x(t) 1

2 coshλt jZX+y

′_{(t)Y −y(t)} 1

2 coshλt jZY =0, and x(t),y(t)are solutions of the differential equations

x′(t)+ |jZX|

2 coshλty(t)=0, y

′_(t)− |jZX|

2 coshλtx(t)=0 since jZY = − |jZX|X.By expressing these equations in the matrix form as

x′_(t) y′(t)

= 1

2 |jZX|

coshλt

0 −1

1 0

x(t) y(t)

, the solutions are given by

x(t) y(t)

=Exps(t)J

x0

y0

, where J =

0 −1

1 0

,

s(t)= 1₂|jZX|R₀t _coshdu_λu(Exp denotes the exponential map of matrices). The assertion

of the proposition follows since Exps J=

cos s −sin s sin s cos s

.

PROPOSITION3. Assume thatnis 2-step nilpotent. Let X ∈ vµ and Z ∈ zλ be

unit vectors satisfying [X,jZX ] = |jZX|2Z.Then the parallel vector field U along

the geodesicγX with U(0) = x0Z +y0Y (Y = _|j_jZX

ZX|), x

2

0+ y02 = 1,is given by

U(t)=x(t)Z(t)+y(t)Y(t) where

x(t)=x0cos s(t)−y0sin s(t) , y(t)=x0sin s(t)+y0cos s(t)and

s(t)= 1 2|jZX|

Z t

0

du coshµu.

Proof. Note that the parallel displacement U(t)of U(0)alongγX(t)is expressed as

U(t)=x(t)Z(t)+y(t)Y(t),since(dLγX(t))e(span{Z,jZX})is invariant under∇γ_X′(t).

In the same way as in Proposition 2, the equation∇γ′

X(t)U(t)=0 gives x′(t)Z −x(t) |jZX|

2 coshµtY+y

′_{(t)Y +y(t)} |jZX|

2 coshµtZ =0 for all real t. Hence, x(t)and y(t)are solutions of the differential equations

x′(t)+ |jZX|

2 coshµty(t)=0, y

′_(t)− |jZX|

2 coshµtx(t)=0, which are given by

x(t)=x0cos s(t)−y0sin s(t), y(t)=x0sin s(t)+y0cos s(t)

with s(t)=1₂|jZX|R₀t _coshdu_µu.

4. Condition(P)on Lie algebras of Iwasawa type and rank one

In this section we will assume that sis a metric Lie algebra of Iwasawa type with non-abeliannand algebraic rank one satisfying condition(P).We summarize in the proposition below some properties of the algebraic structure of the Lie algebran=z⊕v

(Similar ones were obtained in [4], Proposition 1.3). The following lemma will be useful in what follows.

LEMMA 3. If S0 is a totally geodesic subgroup of a Lie group S which is aP

-space, then S0is aP-space. Equivalently, a totally geodesic subalgebras0 of a Lie

algebrassatisfying condition(P)satisfies condition(P).

Proof. Letsands0denote the Lie algebras of S and S0,respectively, with associated

curvature tensors R and R0. Note that for a unit X ∈sthe symmetric operator R′_X =

∇_γ′

X(t)

R_γ′

X(t)

_t

=0 is defined alongγX(t)by R

′

γ_X′(t) = ∇γX′(t)

R_γ′

X(t)

on s(t) = (dLγX(t))es, and for any orthonormal parallel basis{ei(t)}satisfying Rγ_X′(t)ei(t) =

ai(t)ei(t),we have

R′_γ′

X(t)ei(t)= ∇γ

′

X(t)

R_γ′

X(t)ei(t)

−R_γ′

X(t)

∇_γ′

X(t)ei(t)

=a_i′(t)ei(t).

Let X ∈s0be a unit vector, and note thats0ands⊥₀ are invariant under the symmetric

operators RXand R′Xinssinces0is a totally geodesic subalgebra ofsand R0= R|s0.

Hence,s0ands⊥0 are also invariant by the skew-symmetric operator RX◦R′X−R′X◦RX.

Using that condition(P)is equivalent to RX◦RX′ −R′X◦RX =0 for all unit vectors

X ins, it follows that R0X◦ R′0X−R0X′ ◦R0X =0 (R0Xand R0X′ are the restrictions

PROPOSITION4. Lets=n⊕RH,|H| =1,be a metric Lie algebra of Iwasawa type that satisfies condition(P).Then

(i) adH|z=λId.

(ii) Ifnis non-abelian,λ > µ1, the maximum eigenvalue of adH|v. In particular

nλ=zand [nµ₁,v]⊆z.

(iii) For any Z ∈zthe linear map jZ :v→vis an isomorphism with the properties

jZ|vµ :vµ→ vλ−µand j 2

Z

vµ :vµ →vµ(isomorphically). In particular, ifµis an

eigenvalue of adH|v,thenλ−µis also an eigenvalue of adH|v.

Proof. (i) Lets0 =z⊕RH be the subalgebra ofswith associated simply connected

Lie group S0. It follows from the above lemma that S0 is a P-space since s0 is a

totally geodesic subalgebra ofs.Moreover, if Z ⊥Z∗are eigenvectors of adH|zwith associated eigenvaluesλandλ∗,respectively, the Lie algebra spanned by{Z,Z∗,H} is a three-dimensional totally geodesic subalgebra ofs0,whose associated Lie group is

aP-space, by the previous lemma. Then we can assume thats=s0and it is spanned

by{Z,Z∗,H}.Using Lemma 7 and the Remark on page 73 of [3], it follows that the condition RX◦R′_X−R′_X◦RX =0 implies that

(∇ZRic) (Z,H)=(∇Z∗Ric) (Z∗,H),

where the Ricci tensor associated to S is defined by Ric(X,Y)=tr(V → R(V,X)Y). We compute

Ric(Z,Z∗) = Ric(Z,H)=Ric(Z∗,H)=0, Ric(Z,Z)= −λ(λ+λ∗), Ric(Z∗,Z∗) = −λ∗(λ+λ∗)and Ric(H,H)= −(λ2+λ∗2).

As a consequence, it is easy to see that

(∇ZRic) (Z,H) = −Ric(∇ZZ,H)−Ric(Z,∇ZH)

= λ (Ric(Z,Z)−Ric(H,H))=λλ∗(λ∗−λ) and similarly,

(∇Z∗Ric_{) (}Z∗,H)=λλ∗(λ−λ∗).

Henceλ∗=λand there is a unique eigenvalue of adH|z.

(ii) Assume thatnis non-abelian. Then we have that [nµ1,nµ]6=0 for some

eigen-valueµof adH|v,which implies thatµ1+µ =λsinceµ1is the maximum. Hence, λ > µ1 ≥µfor all eigenvaluesµof adH inv,and it also follows that [nµ1,v] ⊆z

from the definition ofµ1.

(iii) We recall that jZ|nµ : nµ → nλ−µ. Hence jZ2 preservesnµ for allµ(see

the beginning of Section 3). Moreover, ker jZ is invariant by adH since the condition

jZX =0 implies that

hjZ([H,X ]),Yi = h[[H,X ],Y ],Zi = −h[[X,Y ],H ],Zi − h[[Y,H ],X ],Zi

= h[X,Y ],[H,Z ]i + hjZ([H,Y ]),Xi

We will show that jZX 6= 0 for all unit Z in zand X ∈ v. If jZX = 0,then by

the previous remark we can assume that X ∈ nµ and|X| = 1. Thens0, the Lie

algebra spanned by{Z,X,H},is a totally geodesic subalgebra ofssince∇ZZ =λH,

∇XX =µH ,∇ZX = ∇XZ =0 and∇ZH = −λZ,∇XH = −µX,∇H =0.Hence, s0satisfies condition(P)and applying the same argument used to show thatλ=λ∗

in (i) above, we obtain thatλ = µ, contradicting (ii). Consequently, λ−µ is an eigenvalue of adH|vwheneverµis an eigenvalue since jZ|nµ :nµ→nλ−µ. Assertion

(iii) follows since dimvµ≤dimvλ−µ≤dimvµ( jZ is an isomorphism).

Next we will show that under the hypothesis of condition(P),the number of eigen-values of adH|ncan be reduced to at most two, namelyλand

1

2λ, attained inzandv,

respectively. For any subspaceuof the Lie algebrasandγX (X ∈s)a fixed geodesic

in S,we will denote byu(t)=(dLγX(t))eu.

REMARK 1. If s0⊕uis a direct sum decomposition into subspaces of the Lie

algebrassuch thats0(t)andu(t)are invariant under∇γ_X′(t)and Rγ_X′(t)for all t,then

e(t) = e1(t)+e2(t),expressed according to the decomposition s0(t)⊕u(t),is a

parallel eigenvector of R_γ′

X(t) alongγX if and only if e1(t)and e2(t)are also parallel eigenvectors of R_γ′

X(t).

In the case of an orthogonal direct sum decomposition, ifs0(t)is invariant under

∇_γ′

X(t) and RγX′(t),thenu(t)=s0(t)

⊥_{is also invariant under∇}

γ′

X(t) and RγX′(t) since ∇_γ′

X(t)is skew-symmetric and RγX′(t)is symmetric.

PROPOSITION5. Ifs=n⊕RH, |H| =1,is a metric Lie algebra of Iwasawa type

with non-abelian nilpotentnsatisfying condition(P),then the eigenvalues of adH|n areλand1₂λ,when restricted tozandv, respectively. Thennis 2-step nilpotent. Proof. Note that adH|z = λId, by Proposition 4. Next we will show that adH|v =

1

2λId. For this purpose, we fix an eigenvalueµof adH|vand assume thatµ 6=

1

2λ.

If Z ∈zis a unit vector, it follows from the definition of∇_γ′

Z(t)and the expression of R_γ′

Z(t)given by Proposition 1 (i) that for any eigenvalueµ ∗_{of ad}

H|v,vµ∗(t)⊕vλ−µ∗(t)

(orv1

2λ(t)in caseµ

∗₌ 1

2λ) is invariant under Rγ_Z′(t)and∇γ_Z′(t)sincevµ∗⊕vλ−µ∗is

RZ-invariant and∇γ′

Z(t)X(t)= −

1

2 coshλt jZX(t).

Assume that the condition(P)is satisfied and let{ei(t): i =1, ...,dims}be an

or-thonormal parallel basis that diagonalizes R_γ′

Z(t).Therefore, ei(0)has a non-zero com-ponent e∈vµ⊕vλ−µfor some i =1, ...,dims(otherwise{ei(0): i =1, ...,dims}

would be a basis ofz⊕P

µ∗_6=µ,λ−µvµ∗⊕RH ).It follows from the previous remark

that e(t),the parallel displacement of e alongγZ(t),is a parallel eigenvector of Rγ_Z′(t)

with e(t)∈vµ⊕vλ−µ(t).

Now, we choose a basis{Xi : i = 1, ...,m = dimvµ}ofvµ that diagonalizes

the symmetric operator j_Z2 : vµ → vµ. Hence {Xi, jZXi : i = 1, ...,m}is an

orthogonal basis ofvµ⊕vλ−µby Proposition 4, andvλ⊕vλ−µ= ⊕m_i=1span{Xi,jZXi},

eigenvectors of RZ).Applying Remark 1 again, we can choose a unit X ∈ vµ such

that j_Z2X = − |jZX|2X (X = Xi for some i =1, ...,m so that e = Pei,ei 6= 0)

and a parallel eigenvector U(t)of Rγ′

Z(t) along the geodesicγZ(t)satisfying U(t) = x(t)X(t)+y(t)Y(t)(Y = jZX

|jZX|),with x(t)

2_+y(t)2_=1.

We set U(0)=x0X+y0Y and let aZ(t)be the associated eigenvalue of Rγ_Z′(t).

If x06=0 and y06=0,we have that RZX =aZ(0)X, RZY =aZ(0)Y.Therefore,

hRZX,Xi =aZ(0)= hRZY,Yiand consequentlyµ= 1₂λsince

1 4|jZX|

2_−λµ=1

4|jZX|

2_{−λ(λ−µ).}

Assume that x0 6= 0 and y0 =0,hence x0 = 1, y0 = 0 and aZ(0)= hRZX,Xi.It

follows from Lemma 2 (applied at t=0)and Proposition 2 that aZ(0)= hRZY,Yi −(

1

2λ−µ)|jZX|tlim→0

x(t) y(t)sinhλt

where x(t)=cos s(t),y(t)=sin s(t)with s′(t)= 1₂|jZX|_cosh1_λt. We compute

limt→0

_x(t) y(t)sinhλt

= limt→0

_(sinhλt)′ (tan s(t))′

=λlimt→0

cosh2λt cosh2s(t)

1

2|jZX|

= 2λ |jZX|

(s(t)→0), which substituted in the above expression gives

aZ(0)=

1 4|jZX|

2_{−λ(2λ−3µ).}

From the equality aZ(0)= hRZX,Xi,we obtainµ=1₂λand get a contradiction. The

same argument is used in the case x0=0,y0=1.

Now we observe that the conditions adH|z=λId and adH|v=

1

2λId imply that the

eigenspaces associated to adH|narenλ=zandn1

2λ=v. Thus, [v,v]=[n12λ,n12λ]⊆

nλ=z, showing thatnis 2-step nilpotent.

EXAMPLE1. Consider the four-dimensional metric Lie algebrasof Iwasawa type

and algebraic rank one with nilpotent non-abeliann=z⊕v. Hence,sis spanned by an orthogonal basis{Z,X,jZX,H}with unit vectors Z ∈z, X∈vand Lie bracket

[Z,X ] = 0=[Z,jZX ], [X,jZX ]= |jZX|2Z

[H,Z ] = λZ, [H,X ]= 1

2λX, [H,jZX ]= 1 2λjZX.

Note that j2_ZX = − |jZX|2X. We show that the Lie group S associated tosis aP

-space if and only if|jZX| = λ. Thus, up to scaling, S is the 2-complex hyperbolic

For this purpose we see that

RZ+X ◦R′Z+X(H)−R′Z+X◦RZ+X(H)=0⇔ |jZX| =λ.

Note that from the connection formulas, ∇ZZ =λH, ∇XX = 1₂λH, ∇jZXjZX =

1

2λ|jZX|

2_{H, ∇}

ZX = −1₂jZX, ∇H =0,

we get

RZX =

1 2

₁ 2|jZX|

2_−λ2

X, RZ(jZX) =

1 2

₁ 2|jZX|

2_−λ2

jZX,

RXZ =

1 2

₁ 2|jZX|

2_−λ2

Z, RXjZX = −

1 4

3|jZX|2+λ2

jZX,

RjZXZ = 1 2|jZX|

2

₁ 2|jZX|

2_−λ2

Z, RjZXX = −

1 4|jZX|

2_3|j

ZX|2+λ2

X.

Hence, taking into account that RZ+X(·)=RZ(·)+RX(·)+R(·,Z)X+R(·,X)Z,it

is a direct computation to see that RZ+X(H) =

1 4

3λjZX−5λ2H

, (1)

RZ+X(jZX) = − ₁

2|jZX|

2₊3

4λ

2

jZX+

3 4λ|jZX|

2_{H ,}

RZ+X(Z−X) = ₁

2|jZX|

2_−λ2_(Z−X),

Recall that, by definition, R′_Z+X(·)=

[∇Z+X,RZ+X](·)−R(·,∇Z+X(Z+X))(Z+X)−R(·,Z+X)∇Z+X(Z+X),

and by a straightforward computation, using the connection formulas and the definition of R,we obtain the following expressions of R′_Z+X

R′_Z+X(H) = 1 2λ

|jZX|2−λ2

(Z−X), (2)

R′_Z+X(jZX) = −

1 2|jZX|

2_(|j

ZX|2−λ2)(Z−X),

since

[∇Z+X,RZ+X](H)=λ ₁

2|jZX|

2_+λ2_Z+1

4λ

|jZX|2+

7 2λ

R(H,∇Z+X(Z+X))(Z +X)= −

1 4λ|jZX|

2_(Z−X),

R(H,Z +X)∇Z+X(Z+X)=

1 2λ

(3λ2+1 2|jZX|

2_)Z+(3

4λ

2_{+ |j}

ZX|2)X

, and

[∇Z+X,RZ+X](jZX)= −

1 4|jZX|

2

(|jZX|2+

9 2λ

2_{)Z +(|j}

ZX|2+3λ2)X

,

R(jZX,∇Z+X(Z+X))(Z+X) = −

3 8λ

2_|j

ZX|2(Z−X),

R(jZX,Z+X)∇Z+X(Z+X) =

1 4|jZX|

2_(|j

ZX|2−5λ2)Z

−1 4|jZX|

2_(|j

ZX|2(3|jZX|2+

5 2λ

2_)X.

Finally, we get

[RZ+X,R′Z+X](H)=

1 8λ

5|jZX|2+λ2

(|jZX|2−λ2)(Z−X), since

RZ+X ◦R′Z+X(H) =

1 2λ

|jZX|2−λ2 1

2|jZX|

2_−λ2

(Z−X)and R′_Z+X◦ RZ+X(H) =

3 4λR

′

Z+X(jZX)−

5 4λ

2_R′

Z+X(H)

= −1 8λ

3|jZX|2+5λ2

(|jZX|2−λ2)(Z−X),

which are computed using (1) and (2) above. The assertion follows as claimed.

THEOREM1. If S is a Lie group of Iwasawa type and algebraic rank one which is

aP-space, then S is a rank one symmetric space of noncompact type.

Proof. Lets=n⊕RH, |H| =1,be the metric Lie algebra of Iwasawa type associated to S.Ifnis abelian, thens=z⊕RH withλas unique eigenvalue of adH|z; thus S is, up to scaling, the real hyperbolic space.

Assume that nis non-abelian, then by Proposition 5 nis 2-step nilpotent. We show that |jZX|2 = λ2 for unit vectors Z ∈ z and X ∈ v. For this

pur-pose let X 6= 0, |X| = 1, be a vector in v. Let {Z1, ...,Zk} be an

orthonor-mal basis ofzdiagonalizing the symmetric operator Z → [X,jZX ] on z. Hence,

hjZiX,jZlXi =δil jZiX

2

and{jZ1X, ...,jZkX}is an orthogonal basis of jzX . More-over, since [X,jZiX ] =

jZiX

2

Zi (i = 1, ...,k),it follows that Zi and jZiX are eigenvectors of RX|zand RX|jzX,respectively (see Section 1, 1.3), and consequently,

z⊕jzX = ⊕ki=1span{Zi,jZiX}where span{Zi,jZiX}(t)is invariant under Rγ_X′(t)and

∇_γ′

Assume that condition(P) is satisfied and let{ej(t) : j = 1, ...,dims}be an

orthonormal parallel basis diagonalizing R_γ′

X(t).Applying the same argument as used in the previous proposition, for each i = 1, ...,k, we can choose 1 ≤ ji ≤ dims

such that Ui(t),the non-zero component of eji(t)in span{Zi,jZiX}(t)is a parallel eigenvector of R_γ′

X(t)alongγX(t) by Remark 1. We set Z = Zi, Y = jZX |jZX| , U = Ui(0)=x0Z+y0Y with x₀2+y₀2=1,and note that the unit Z ∈zsatisfies [X,jZX ]=

|jZX|2Z.By applying Proposition 3, the parallel eigenvector U alongγXwith U(0)=

U is given by U(t)=x(t)Z(t)+y(t)Y(t), with

x(t)=x0cos s(t)−y0sin s(t), y(t)=x0sin s(t)+y0cos s(t)

and s(t) = 1₂|jZX| 1

cosh1₂λt, whose associated eigenvalue aX(t) satisfies aX(0) =

hRXU,Ui =x2₀hRXZ,Zi +y₀2hRXY,Yi,

aX(t)cosh2

1 2λt =

1 4|jZX|

2₋1

2λ

2_−λ2_sinh21

2λt+ 3 4λ|jZX|

y(t) x(t)sinh

1 2λt and

aX(t)cosh2

1 2λt = −

3 4|jZX|

2₋1

4λ

2_cosh21

2λt + 3 4λ|jZX|

x(t) y(t)sinh

1 2λt for all real t,by Lemma 2.

If x0 6= 0 and y0 6= 0,then aX(0) = hRXZ,Zi = hRXY,Yi (Z and Y are

eigenvectors of RX)and we get|jZX|2=1₄λ2since

1 4|jZX|

2₋1

2λ

2_{= −}3

4|jZX|

2₋1

4λ

2_.

If x0 =1 and y0 = 0,it follows from Proposition 3 that x(t) = cos s(t), y(t) =

sin s(t)and in the same way that in the previous proposition,

limt→0

_x(t) y(t)sinh

1 2λt

=1

2λlimt→0 



cosh1₂λt

s′_(t) cosh2_s(t)



= λ |jZX|

.

Substituting this limit in the last expression of aX(t)above, we get

aX(0) = −

3 4|jZX|

2₋1

4λ

2₊3

4λ|jZX|limt→0 _x(t)

y(t)sinh 1 2λt

= −3 4|jZX|

2₊1

2λ

2_,

and from the equalityhRXZ,Zi =aX(0)it follows that|jZX|2=λ2.

The same condition is obtained in the case x0=0 and y0 =1,since in this case

x(t)= −sin s(t), y(t)=cos s(t)and aX(0)= hRXY,Yiwith aX(0)computed as

aX(0)=

1 4|jZX|

2₋1

2λ

2₋3

4λ|jZX|limt→0

sinh1₂λt tan s(t)

! =1

4|jZX|

2₋5

4λ

Finally, using the same argument on each span{Zi,jZiX}(i = 1, ...,k),it follows that for any unit vector X ∈ v, an orthonormal basis{Z1, ...,Zk}can be chosen so

that [X,jZiX ]= jZiX

2

Zi and{jZiX : i =1, ...,k}is an orthogonal basis of jzX satisfying eitherjZiX

2

=1₄λ2orjZiX

2

=λ2for i =1, ...,k.Hence, (∗) 1

4λ

2_|Z|2_{≤ |j}

ZX|2≤λ2|Z|2 for all Z ∈zand X ∈v, |X| =1,

since jZX =Pki=1aijZiX is an orthogonal sum for any Z= Pk

i=1aiZi inz.

Next we show that the conditionjZiX

2

=1₄λ2in the above basis is not possible. In fact, for a such Zi, it follows from(∗)above that1₄λ2= −hjZ2iX,Xiis the minimum eigenvalue of the symmetric operator−j_Z2

i with X as associated eigenvector. Thus −j_Z2

iX =

1

4λ2X and s0, the Lie algebra spanned by{Zi,X,jZiX,H},is a totally geodesic subalgebra ofs. Applying Lemma 3, the associated Lie group tos0is aP

-space, which is not possible by Example 1. The conditionjZiX

2

= λ2 for all i = 1, ...,k,implies that|jZX|2 = λ2|Z|2

for all Z ∈ zand X ∈ v,|X| = 1,or equivalently j_Z2 = −λ2|Z|2Id for all Z ∈ z. Since adH|z=λId and adH|v=

1

2λId it follows that S is, up to scaling of the metric,

a Damek-Ricci space. Theorem 2 of [2], Section 4.3 implies that S is a rank one symmetric space of noncompact type.

COROLLARY 1. If M is an irreducible, non-flat homogeneous EinsteinP-space

with nonpositive curvature and algebraic rank one, then M is a symmetric space of noncompact type and rank one.

Proof. Since M is irreducible and non-flat, by applying Corollary 1 of [8] it follows that M is a simply connected homogeneous space with nonpositive curvature. Hence, M can be represented as a solvable Lie group S of algebraic rank one with left invariant metric of nonpositive curvature, whose associated metric Lie algebrasdecomposes as an orthogonal direct sums=n⊕awheren=[s,s] andais one-dimensional (see [1]). By applying Proposition 4.9 and Theorem 4.10 of [6], S is isometric to a Lie group of Iwasawa type and algebraic rank one. It follows from Theorem 1 that M is a symmetric space of noncompact type and rank one.

References

[1] AZENCOTTR. ANDWILSONE., Homogeneous manifolds with negative curva-ture I, Trans. Amer.Math. Soc. 215 (1976), 323–362.

[2] BERNDT J., TRICERRI F. AND VANHECKE L.,Generalized Heisenberg groups and Damek-Ricci harmonic spaces, Lecture Notes in Math. 1598, Springer, Berlin 1995.

[3] BERNDT J., VANHECKE L., Two natural generalizations of locally symmetric spaces, Differential Geometry and its Applications 2 (1992), 57–80.

[4] DOTTI I. AND DRUETTA M.J., Negatively curved homogeneous Osserman spaces, Differential Geometry and Applications 11 (1999), 163–178.

[5] EBERLEIN P., Geometry of 2-step nilpotent groups with a left invariant metric, Ann. Sci. Ecole. Norm. Sup. 27 (1994), 611–660.

[6] HEBERJ., Noncompact homogeneous Einstein spaces, Invent. Math. 133 (1998), 279–352.

[7] HELGASON S., Differential geometry, Lie groups, and symmetric spaces, Aca-demic Press, New York 1978.

[8] WOLFJ., Homogeneity and bounded isometries in manifolds of negative curva-ture, Illinois J. Math. 8 (1964), 14–18.

AMS Subject Classification: 53C30, 53C55.

Maria J. DRUETTA

Facultad de Matem´atica, Astronom´ıa y F´ısica Universidad Nacional de C´ordoba

Ciudad Universitaria

5000 C´ordoba, ARGENTINA e-mail:druetta@mate.uncor.edu

eigenvectors of RZ).Applying Remark 1 again, we can choose a unit X ∈ vµ such that j_Z2X = − |jZX|2X (X = Xi for some i =1, ...,m so that e = Pei,ei 6= 0)

and a parallel eigenvector U(t)of Rγ′

Z(t) along the geodesicγZ(t)satisfying U(t) = x(t)X(t)+y(t)Y(t)(Y = jZX

|jZX|),with x(t)

2_+y(t)2_=1.

We set U(0)=x0X+y0Y and let aZ(t)be the associated eigenvalue of Rγ_Z′(t). If x06=0 and y06=0,we have that RZX =aZ(0)X, RZY =aZ(0)Y.Therefore, hRZX,Xi =aZ(0)= hRZY,Yiand consequentlyµ= 1₂λsince

1 4|jZX|

2_−λµ=1 4|jZX|

2_{−λ(λ−µ).}

Assume that x0 6= 0 and y0 =0,hence x0 = 1, y0 = 0 and aZ(0)= hRZX,Xi.It

follows from Lemma 2 (applied at t=0)and Proposition 2 that aZ(0)= hRZY,Yi −(

1

2λ−µ)|jZX|tlim→0 x(t)

y(t)sinhλt

where x(t)=cos s(t),y(t)=sin s(t)with s′(t)= 1₂|jZX|_coshλ1 _t. We compute

limt→0

_x(t)

y(t)sinhλt

= limt→0

_(sinhλt)′ (tan s(t))′

=λlimt→0

cosh2λt cosh2s(t)

1 2|jZX|

= 2λ

|jZX|

(s(t)→0),

which substituted in the above expression gives aZ(0)=

1 4|jZX|

2_{−λ(2λ−3µ).}

From the equality aZ(0)= hRZX,Xi,we obtainµ=1₂λand get a contradiction. The

same argument is used in the case x0=0,y0=1.

Now we observe that the conditions adH|z=λId and adH|v=

1

2λId imply that the eigenspaces associated to adH|narenλ=zandn1

2λ=v. Thus, [v,v]=[n12λ,n12λ]⊆

nλ=z, showing thatnis 2-step nilpotent.

EXAMPLE1. Consider the four-dimensional metric Lie algebrasof Iwasawa type and algebraic rank one with nilpotent non-abeliann=z⊕v. Hence,sis spanned by an orthogonal basis{Z,X,jZX,H}with unit vectors Z ∈z, X∈vand Lie bracket

[Z,X ] = 0=[Z,jZX ], [X,jZX ]= |jZX|2Z

[H,Z ] = λZ, [H,X ]= 1

2λX, [H,jZX ]= 1 2λjZX.

Note that j2_ZX = − |jZX|2X. We show that the Lie group S associated tosis aP

-space if and only if|jZX| = λ. Thus, up to scaling, S is the 2-complex hyperbolic

For this purpose we see that

RZ+X ◦R′Z+X(H)−R′Z+X◦RZ+X(H)=0⇔ |jZX| =λ.

Note that from the connection formulas, ∇ZZ =λH, ∇XX = 1₂λH, ∇jZXjZX =

1 2λ|jZX|

2_{H, ∇}

ZX = −1₂jZX, ∇H =0,

we get

RZX =

1 2

₁

2|jZX| 2_−λ2

X,

RZ(jZX) =

1 2

₁

2|jZX| 2_−λ2

jZX,

RXZ =

1 2

₁

2|jZX| 2_−λ2

Z,

RXjZX = −

1 4

3|jZX|2+λ2

jZX,

RjZXZ = 1 2|jZX|

2

₁

2|jZX| 2_−λ2

Z,

RjZXX = − 1 4|jZX|

2_3|j

ZX|2+λ2

X.

Hence, taking into account that RZ+X(·)=RZ(·)+RX(·)+R(·,Z)X+R(·,X)Z,it

is a direct computation to see that RZ+X(H) =

1 4

3λjZX−5λ2H

,

(1)

RZ+X(jZX) = −

₁

2|jZX| 2₊3

4λ 2

jZX+

3 4λ|jZX|

2_{H ,}

RZ+X(Z−X) =

₁

2|jZX|

2_−λ2_(Z−X),

Recall that, by definition, R′_Z+X(·)=

[∇Z+X,RZ+X](·)−R(·,∇Z+X(Z+X))(Z+X)−R(·,Z+X)∇Z+X(Z+X),

and by a straightforward computation, using the connection formulas and the definition of R,we obtain the following expressions of R′_Z+X

R′_Z+X(H) = 1 2λ

|jZX|2−λ2

(Z−X),

(2)

R′_Z+X(jZX) = −

1 2|jZX|

2_(|j

ZX|2−λ2)(Z−X),

since

[∇Z+X,RZ+X](H)=λ

₁

2|jZX|

2_+λ2_Z+1 4λ

|jZX|2+

7 2λ

R(H,∇Z+X(Z+X))(Z +X)= −

1 4λ|jZX|

2_(Z−X),

R(H,Z +X)∇Z+X(Z+X)=

1 2λ

(3λ2+1 2|jZX|

2_)Z+(3 4λ

2_{+ |j}

ZX|2)X

,

and

[∇Z+X,RZ+X](jZX)= −

1 4|jZX|

2

(|jZX|2+

9 2λ

2_{)Z +(|j}

ZX|2+3λ2)X

,

R(jZX,∇Z+X(Z+X))(Z+X) = −

3 8λ

2_|j

ZX|2(Z−X),

R(jZX,Z+X)∇Z+X(Z+X) =

1 4|jZX|

2_(|j

ZX|2−5λ2)Z −1

4|jZX| 2_(|j

ZX|2(3|jZX|2+

5 2λ

2_)X.

Finally, we get

[RZ+X,R′Z+X](H)=

1 8λ

5|jZX|2+λ2

(|jZX|2−λ2)(Z−X), since

RZ+X ◦R′Z+X(H) =

1 2λ

|jZX|2−λ2

1

2|jZX| 2_−λ2

(Z−X)and

R′_Z+X◦ RZ+X(H) =

3 4λR

′

Z+X(jZX)−

5 4λ

2_R′ Z+X(H)

= −1

8λ

3|jZX|2+5λ2

(|jZX|2−λ2)(Z−X),

which are computed using (1) and (2) above. The assertion follows as claimed. THEOREM1. If S is a Lie group of Iwasawa type and algebraic rank one which is aP-space, then S is a rank one symmetric space of noncompact type.

Proof. Lets=n⊕RH, |H| =1,be the metric Lie algebra of Iwasawa type associated to S.Ifnis abelian, thens=z⊕RH withλas unique eigenvalue of adH|z; thus S is,

up to scaling, the real hyperbolic space.

Assume that nis non-abelian, then by Proposition 5 nis 2-step nilpotent. We show that |jZX|2 = λ2 for unit vectors Z ∈ z and X ∈ v. For this

pur-pose let X 6= 0, |X| = 1, be a vector in v. Let {Z1, ...,Zk} be an

orthonor-mal basis ofzdiagonalizing the symmetric operator Z → [X,jZX ] on z. Hence, hjZiX,jZlXi =δil

jZiX

2

and{jZ1X, ...,jZkX}is an orthogonal basis of jzX .

More-over, since [X,jZiX ] =

jZiX

2

Zi (i = 1, ...,k),it follows that Zi and jZiX are eigenvectors of RX|zand RX|jzX,respectively (see Section 1, 1.3), and consequently, z⊕jzX = ⊕ki=1span{Zi,jZiX}where span{Zi,jZiX}(t)is invariant under Rγ_X′(t)and ∇_γ′

Assume that condition(P) is satisfied and let{ej(t) : j = 1, ...,dims}be an

orthonormal parallel basis diagonalizing R_γ′

X(t).Applying the same argument as used in the previous proposition, for each i = 1, ...,k, we can choose 1 ≤ ji ≤ dims

such that Ui(t),the non-zero component of eji(t)in span{Zi,jZiX}(t)is a parallel eigenvector of R_γ′

X(t)alongγX(t) by Remark 1. We set Z = Zi, Y =

jZX

|jZX| , U = Ui(0)=x0Z+y0Y with x₀2+y₀2=1,and note that the unit Z ∈zsatisfies [X,jZX ]= |jZX|2Z.By applying Proposition 3, the parallel eigenvector U alongγXwith U(0)=

U is given by U(t)=x(t)Z(t)+y(t)Y(t), with

x(t)=x0cos s(t)−y0sin s(t), y(t)=x0sin s(t)+y0cos s(t)

and s(t) = 1₂|jZX| 1

cosh1₂λt, whose associated eigenvalue aX(t) satisfies aX(0) = hRXU,Ui =x2₀hRXZ,Zi +y₀2hRXY,Yi,

aX(t)cosh2

1 2λt =

1 4|jZX|

2₋1 2λ

2_−λ2_sinh21 2λt+

3 4λ|jZX|

y(t)

x(t)sinh

1 2λt and

aX(t)cosh2

1 2λt = −

3 4|jZX|

2₋1 4λ

2_cosh21 2λt +

3 4λ|jZX|

x(t)

y(t)sinh

1 2λt for all real t,by Lemma 2.

If x0 6= 0 and y0 6= 0,then aX(0) = hRXZ,Zi = hRXY,Yi (Z and Y are

eigenvectors of RX)and we get|jZX|2=1₄λ2since

1 4|jZX|

2₋1 2λ

2_{= −}3 4|jZX|

2₋1 4λ

2_.

If x0 =1 and y0 = 0,it follows from Proposition 3 that x(t) = cos s(t), y(t) = sin s(t)and in the same way that in the previous proposition,

limt→0

_x(t)

y(t)sinh

1 2λt

=1

2λlimt→0





cosh1₂λt

s′_(t)

cosh2_s(t)



=

λ |jZX|

.

Substituting this limit in the last expression of aX(t)above, we get

aX(0) = −

3 4|jZX|

2₋1 4λ

2₊3

4λ|jZX|limt→0

_x(t)

y(t)sinh

1 2λt

= −3

4|jZX| 2₊1

2λ 2_,

and from the equalityhRXZ,Zi =aX(0)it follows that|jZX|2=λ2.

The same condition is obtained in the case x0=0 and y0 =1,since in this case x(t)= −sin s(t), y(t)=cos s(t)and aX(0)= hRXY,Yiwith aX(0)computed as

aX(0)=

1 4|jZX|

2₋1 2λ

2₋3

4λ|jZX|limt→0

sinh1₂λt tan s(t)

!

=1 4|jZX|

2₋5 4λ

Finally, using the same argument on each span{Zi,jZiX}(i = 1, ...,k),it follows that for any unit vector X ∈ v, an orthonormal basis{Z1, ...,Zk}can be chosen so

that [X,jZiX ]=

jZiX

2

Zi and{jZiX : i =1, ...,k}is an orthogonal basis of jzX

satisfying eitherjZiX

2

=1₄λ2orjZiX

2

=λ2for i =1, ...,k.Hence,

(∗) 1

4λ

2_|Z|2_{≤ |j}

ZX|2≤λ2|Z|2 for all Z ∈zand X ∈v, |X| =1,

since jZX =Pki=1aijZiX is an orthogonal sum for any Z=

Pk

i=1aiZi inz.

Next we show that the conditionjZiX

2

=1₄λ2in the above basis is not possible. In fact, for a such Zi, it follows from(∗)above that1₄λ2= −hjZ2iX,Xiis the minimum eigenvalue of the symmetric operator−j_Z2

i with X as associated eigenvector. Thus −j_Z2

iX = 1

4λ2X and s0, the Lie algebra spanned by{Zi,X,jZiX,H},is a totally geodesic subalgebra ofs. Applying Lemma 3, the associated Lie group tos0is aP -space, which is not possible by Example 1.

The conditionjZiX

2

= λ2 for all i = 1, ...,k,implies that|jZX|2 = λ2|Z|2

for all Z ∈ zand X ∈ v,|X| = 1,or equivalently j_Z2 = −λ2|Z|2Id for all Z ∈ z. Since adH|z=λId and adH|v=

1

2λId it follows that S is, up to scaling of the metric, a Damek-Ricci space. Theorem 2 of [2], Section 4.3 implies that S is a rank one symmetric space of noncompact type.

COROLLARY 1. If M is an irreducible, non-flat homogeneous EinsteinP-space with nonpositive curvature and algebraic rank one, then M is a symmetric space of noncompact type and rank one.

Proof. Since M is irreducible and non-flat, by applying Corollary 1 of [8] it follows that M is a simply connected homogeneous space with nonpositive curvature. Hence, M can be represented as a solvable Lie group S of algebraic rank one with left invariant metric of nonpositive curvature, whose associated metric Lie algebrasdecomposes as an orthogonal direct sums=n⊕awheren=[s,s] andais one-dimensional (see [1]). By applying Proposition 4.9 and Theorem 4.10 of [6], S is isometric to a Lie group of Iwasawa type and algebraic rank one. It follows from Theorem 1 that M is a symmetric space of noncompact type and rank one.

References

[1] AZENCOTTR. ANDWILSONE., Homogeneous manifolds with negative curva-ture I, Trans. Amer.Math. Soc. 215 (1976), 323–362.

[2] BERNDT J., TRICERRI F. AND VANHECKE L.,Generalized Heisenberg groups and Damek-Ricci harmonic spaces, Lecture Notes in Math. 1598, Springer, Berlin 1995.

[3] BERNDT J., VANHECKE L., Two natural generalizations of locally symmetric spaces, Differential Geometry and its Applications 2 (1992), 57–80.

[4] DOTTI I. AND DRUETTA M.J., Negatively curved homogeneous Osserman spaces, Differential Geometry and Applications 11 (1999), 163–178.

[5] EBERLEIN P., Geometry of 2-step nilpotent groups with a left invariant metric, Ann. Sci. Ecole. Norm. Sup. 27 (1994), 611–660.

[6] HEBERJ., Noncompact homogeneous Einstein spaces, Invent. Math. 133 (1998), 279–352.

[7] HELGASON S., Differential geometry, Lie groups, and symmetric spaces, Aca-demic Press, New York 1978.

[8] WOLFJ., Homogeneity and bounded isometries in manifolds of negative curva-ture, Illinois J. Math. 8 (1964), 14–18.

AMS Subject Classification: 53C30, 53C55.

Maria J. DRUETTA

Facultad de Matem´atica, Astronom´ıa y F´ısica Universidad Nacional de C´ordoba

Ciudad Universitaria

5000 C´ordoba, ARGENTINA e-mail:druetta@mate.uncor.edu