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 Rγ0(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).
55
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 Rγ0(t) can be diagonalized by a parallel orthonormal basis of Tγ (t)S. We note that condition(P)is equivalent to the fact RX◦R0X =R0X◦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=12Id and j2Z = − |Z|2Id 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,nµ0]⊂nµ+µ0 (by the Jacobi identity), whenever µ+µ0is 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 = expn(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) → expnX,r
since expn : 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
(expn(Z+X),r)·(expn(Z∗+Y),s)
= (expn(Z+erλZ∗+1
2erµ[X,Y ]+X+erµY),r+s).
In fact, note that by the definition of the product in S we have (expn(Z+X),r)·(expn(Z∗+Y),s)
= expn(Z+X)τr(expn(Z∗+Y)),r+s
= expn(Z+X)expn Exp(r adH)Z∗+Exp(r adH)Y ,r+s
= expn(Z+X)expn(erλZ∗+erµY),r+s ,
since expnX expnY =expn(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 as follows. If{Z1, ...,Zk}and{X1, ...,Xm}(k =dimz, m =dimv) are orthonormal
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−rλiZi(ϕ(x1, ...,xk,y1, ...,ym,r)),
∂
∂yi
ϕ(x
1,...,xk,y1,...,ym,r)
=e−rµiXi(ϕ(x1, ...,xk,y1, ...,ym,r)) +1
2 X
j,s
e−rλsyjhjZsXi,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 = −ad2H.
(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
4j2Z(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
4j[X,Y ]X−µadHY.
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)andγY0(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)=
expn
tanhαt α
Y,−1
αln(coshαt)
with associated tangent vector field γY0(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= ∇|S
0 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
x00−2αx0r0 = 0, r00+αe−2rα(x0)2 = 0.
Using thatγY0(t)= x0(t)∂x∂
γY(t)+r0(t) ∂r∂ γY(t),
γY0(t)
=1,with x0(t)=e2αr(t), ∂
∂x γ
Y(t)
=e−αr(t)Y(t)and ∂
∂r γ
Y(t)
=H(t),
we have the equivalent equations
x0(t) = e2αr(t), r00(t)+α(1−(r0(t))2 = 0 whose solutions satisfy
x(t)= Z t
0
e2αr(u)du and r0(t)= −tanhαt.
Therefore, we get
r(t)= −1
αln(coshαt), x(t)= 1
αtanhαt, and the expression ofγY andγY0 follows as claimed.
PROPOSITION1. If Z ∈ zand X ∈ vare eigenvectors of adH with associated eigenvaluesλandµ, respectively, then for any Y ∈v, we have
(i) Rγ0
Z(t)Y(t)= 1
cosh2λt(dLγZ(t))e
·
RZY −sinh2λt ad2HY −sinhλt jZ
1
2λId−adH
Y
(i i) Rγ0
X(t)Z(t)= 1
cosh2µt(dLγX(t))e
·
RXZ−λ2sinh2µt Z+(λ−1
2µ)sinhµt jZX
(i i i) Rγ0
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 γZ0(t)= 1
coshλtZ(t)−tanhλt H(t), we have that
R(Y(t), γZ0(t))γZ0(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 = −ad2H. (ii)-(iii) Assume that X ∈ vis an eigenvector of adH with eigenvalueµ,and let Y ⊥X inv.Using the expression ofγX0(t),in the same way as (i) we get
R(Z(t), γX0(t))γX0(t)= 1
cosh2µt(dLγX(t))e
·
RXZ+sinh2µt RHZ−sinhµt(R(Z,X)H+R(Z,H)X)
. Hence, the expression of Rγ0
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), γX0(t))γX0(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 ∈vµ0
hjZX,Yi = h[X,Y ],Zi 6=0⇒[X,Y ]6=0,
and thusµ+µ0=λsince [X,Y ]∈nµ+µ0and has non-zero component inzλ.
Hence, for any eigenvalueµof adH|v,λ−µis also an eigenvalue of adH|v,λ > µ and the symmetric operator j2Z preservesvµ.Moreover, the map Z → [X,jZX ] de- fines a symmetric operator onz(h[X,jZX ],Z∗i = hjZ∗X,jZXi = h[X,jZ∗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 Rγ0 Z(t), Rγ0
X(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 =|jjZX
ZX|. (i) If X is an eigenvector of jZ2,then Rγ0
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 Rγ0
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−λ2sinh2µt+(λ−1
2µ)|jZX| y(t)
x(t)sinhµt or aX(t)cosh2µt
= −3
4|jZX|2−(λ−µ)(µ+(λ−µ)sinh2µt)+(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 Rγ0
Z(t) and Rγ0
X(t), re- spectively. The assertion of the lemma follows from the expression of Rγ0
Z(t)and Rγ0
X(t)
given in Proposition 1 and using in each case the equalities Rγ0
Z(t)(U(t))=aZ(t)(U(t))and Rγ0
X(t)(U(t))=aX(t)U(t).
Note that in the last case, RXZ = 1
4|jZX|2Z−λµZ and RXY = −3
4|jZX|2Y−µ(λ−µ)Y sincenis 2-step nilpotent.
PROPOSITION2. Let Z∈zλbe a unit vector.If X∈vµ,|X| =1,is an eigenvec- tor of jZ2,then the parallel vector field U along the geodesicγZwith U(0)=x0X+y0Y
(Y = |jjZX
ZX|), x02+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 ∇γ0
Z(t) since γZ0(t) = cosh1λtZ(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∇γ0
Z(t)U(t)=0,which gives x0(t)X+x(t) 1
coshλt∇ZX+y0(t)Y+y(t) 1
coshλt∇ZY =0 for all real t since X and Y are left invariant. Thus,
x0(t)X−x(t) 1
2 coshλt jZX+y0(t)Y −y(t) 1
2 coshλt jZY =0, and x(t),y(t)are solutions of the differential equations
x0(t)+ |jZX|
2 coshλty(t)=0, y0(t)− |jZX|
2 coshλtx(t)=0 since jZY = − |jZX|X.By expressing these equations in the matrix form as
x0(t) y0(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|Rt
0 du
coshλu(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 = |jjZX
ZX|), x02+ 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∇γ0
X(t). In the same way as in Proposition 2, the equation∇γ0
X(t)U(t)=0 gives x0(t)Z −x(t) |jZX|
2 coshµtY+y0(t)Y +y(t) |jZX|
2 coshµtZ =0 for all real t. Hence, x(t)and y(t)are solutions of the differential equations
x0(t)+ |jZX|
2 coshµty(t)=0, y0(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|Rt
0 du coshµu.
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 R0X =
∇γ0
X(t)
Rγ0
X(t)
t=0 is defined alongγX(t)by R0
γX0(t) = ∇γ0
X(t)
Rγ0
X(t)
on s(t) = (dLγX(t))es, and for any orthonormal parallel basis{ei(t)}satisfying Rγ0
X(t)ei(t) = ai(t)ei(t),we have
R0γ0
X(t)ei(t)= ∇γ0
X(t)
Rγ0
X(t)ei(t)
−Rγ0
X(t)
∇γ0
X(t)ei(t)
=ai0(t)ei(t).
Let X ∈s0be a unit vector, and note thats0ands⊥0 are invariant under the symmetric operators RXand R0Xinssinces0is a totally geodesic subalgebra ofsand R0= R|s0. Hence,s0ands⊥0 are also invariant by the skew-symmetric operator RX◦R0X−R0X◦RX. Using that condition(P)is equivalent to RX◦R0X −R0X◦RX =0 for all unit vectors X ins, it follows that R0X◦ R00X−R0X0 ◦R0X =0 (R0Xand R0X0 are the restrictions of RX and R0Xtos0,respectively). Thus S0is aP-space.
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|v
µ :vµ→ vλ−µand jZ2
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◦R0X−R0X◦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
= λhjZX,Yi − h[H,Y ],jZXi =0 for all Y ∈v.
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 12λ, 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∇γ0
X(t)and Rγ0
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 Rγ0
X(t) alongγX if and only if e1(t)and e2(t)are also parallel eigenvectors of Rγ0
X(t).
In the case of an orthogonal direct sum decomposition, ifs0(t)is invariant under
∇γ0
X(t) and Rγ0
X(t),thenu(t)=s0(t)⊥is also invariant under∇γ0
X(t) and Rγ0
X(t) since
∇γ0
X(t)is skew-symmetric and Rγ0
X(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= 12λ.
If Z ∈zis a unit vector, it follows from the definition of∇γ0
Z(t)and the expression of Rγ0
Z(t)given by Proposition 1 (i) that for any eigenvalueµ∗of adH|v,vµ∗(t)⊕vλ−µ∗(t) (orv1
2λ(t)in caseµ∗= 12λ) is invariant under Rγ0
Z(t)and∇γ0
Z(t)sincevµ∗⊕vλ−µ∗is RZ-invariant and∇γ0
Z(t)X(t)= −2 coshλt1 jZX(t).
Assume that the condition(P)is satisfied and let{ei(t): i =1, ...,dims}be an or- thonormal parallel basis that diagonalizes Rγ0
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 Rγ0
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 jZ2 : vµ → vµ. Hence {Xi, jZXi : i = 1, ...,m}is an orthogonal basis ofvµ⊕vλ−µby Proposition 4, andvλ⊕vλ−µ= ⊕mi=1span{Xi,jZXi}, where each span{Xi,jZXi}(t)is invariant under∇γ0
Z(t)and Rγ0
Z(t)(Xi and jZXi are
eigenvectors of RZ).Applying Remark 1 again, we can choose a unit X ∈ vµ such that jZ2X = − |jZX|2X (X = Xi for some i =1, ...,m so that e = P
ei,ei 6= 0) and a parallel eigenvector U(t)of Rγ0
Z(t) along the geodesicγZ(t)satisfying U(t) = x(t)X(t)+y(t)Y(t)(Y = |jjZX
ZX|),with x(t)2+y(t)2=1.
We set U(0)=x0X+y0Y and let aZ(t)be the associated eigenvalue of Rγ0
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|lim
t→0
x(t) y(t)sinhλt
where x(t)=cos s(t),y(t)=sin s(t)with s0(t)= 12|jZX|coshλt1 . We compute limt→0
x(t) y(t)sinhλt
= limt→0
(sinhλt)0 (tan s(t))0
=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=12λId imply that the eigenspaces associated to adH|narenλ=zandn1
2λ=v. Thus, [v,v]=[n1 2λ,n1
2λ]⊆ 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 j2ZX = − |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 space CH2.
For this purpose we see that
RZ+X ◦R0Z+X(H)−R0Z+X◦RZ+X(H)=0⇔ |jZX| =λ.
Note that from the connection formulas,
∇ZZ =λH, ∇XX = 12λH, ∇jZXjZX = 12λ|jZX|2H, ∇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|jZX|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 4λ2
jZX+3
4λ|jZX|2H , RZ+X(Z−X) =
1
2|jZX|2−λ2
(Z−X), Recall that, by definition, R0Z+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 R0Z+X
R0Z+X(H) = 1 2λ
|jZX|2−λ2
(Z−X), (2)
R0Z+X(jZX) = −1
2|jZX|2(|jZX|2−λ2)(Z −X), since
[∇Z+X,RZ+X](H)=λ 1
2|jZX|2+λ2
Z+1 4λ
|jZX|2+7 2λ2
X,
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
4λ2+ |jZX|2)X
, and
[∇Z+X,RZ+X](jZX)= −1 4|jZX|2
(|jZX|2+9
2λ2)Z +(|jZX|2+3λ2)X
,
R(jZX,∇Z+X(Z+X))(Z+X) = −3
8λ2|jZX|2(Z−X), R(jZX,Z+X)∇Z+X(Z+X) = 1
4|jZX|2(|jZX|2−5λ2)Z
−1
4|jZX|2(|jZX|2(3|jZX|2+5 2λ2)X.
Finally, we get
[RZ+X,R0Z+X](H)= 1 8λ
5|jZX|2+λ2
(|jZX|2−λ2)(Z−X), since
RZ+X◦ R0Z+X(H) = 1 2λ
|jZX|2−λ21
2|jZX|2−λ2
(Z−X)and R0Z+X◦ RZ+X(H) = 3
4λR0Z+X(jZX)−5
4λ2R0Z+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
2and{jZ1X, ...,jZkX}is an orthogonal basis of jzX . More- over, since [X,jZiX ] =
jZiX
2Zi (i = 1, ...,k),it follows that Zi and jZiX are eigenvectors of RX|zand RX|j
zX,respectively (see Section 1, 1.3), and consequently, z⊕jzX = ⊕ki=1span{Zi,jZiX}where span{Zi,jZiX}(t)is invariant under Rγ0
X(t)and
∇γ0
X(t)for all i=1, ...,k.
