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Revista de la Unión Matemática Argentina

versión On-line ISSN 1669-9637

Rev. Unión Mat. Argent. v.47 n.1 Bahía Blanca ene./jun. 2006

 

4-step Carnot spaces and the 2-stein condition

María J. Druetta

Dedicated to the memory of my goddaughter María Pía

Abstract: We consider the 2-stein condition on k-step Carnot spaces S. These spaces are a subclass in the class of solvable Lie groups of Iwasawa type of algebraic rank one and contain the homogeneous Einstein spaces within this class. They are obtained as a semidirect product of a graded nilpotent Lie group N and the abelian group R.
We show that the 2-stein condition is not satisfied on a proper 4-step Carnot spaces S.

2000 Mathematics Subject Classification. 53C30, 53C55

Key words and phrases: 2-stein spaces, k-step Carnot spaces, Einstein spaces, Lie groups of Iwasawa type, Damek-Ricci spaces

A Riemannian manifold M is said to be a 2 -stein space, if there exist functions μl, l = 1,2, defined on M such that

tr(Rl ) = μ |X |2l, l = 1,2, for all X ∈ TM. X l

Here, R X denote the Jacobi operator associated to X, defined by R Y = R (Y,X )X X for all Y ∈ T M, where R is the curvature tensor of M. Harmonic riemannian manifolds are necessarily 2 -stein.

A k -step Carnot space (k ≥ 2 ) is a simply connected solvable Lie group S, which is a semidirect product of a nilpotent Lie group N and the abelian group R. Assume that S and N have associated Lie algebras 𝔰 and 𝔫 , respectively. S has the left invariant metric induced by the one given on 𝔰 , where 𝔰 is a solvable metric Lie algebra 𝔰 with inner product ⟨,⟩ such that:

(i) 𝔰 = 𝔫 ⊕ RH with 𝔫 = [𝔰,𝔰] and H ⊥ 𝔫, |H | = 1.

(ii) 𝔫 has an orthogonal decomposition  ∑k -1 𝔫 = i=1 𝔫i into (k - 1) subspaces given by

𝔫 = {X ∈ 𝔫 : ad (X ) = iαX }, i = 1,...,k - 1, i H

for some positive constant α ∈ R.

Note that, since the adjoint representation adH is a derivation of 𝔫 the above decomposition defines a graded Lie algebra structure of 𝔫 , that is,

[𝔫i,𝔫j] ⊂ 𝔫i+j 1 ≤ i,j ≤ k - 1

with the convention 𝔫i = {0} for i > k - 1. In particular, 𝔫 is a (k - 1) -step nilpotent Lie algebra, that is, the (k - 1) -th derived algebra vanishes: 𝔫k-1 = [𝔫,...[𝔫, [𝔫,𝔫 ]]...] = {0 }.

Note, that up to a constant in the metric, we may suppose that α = 1. In this terminology, the 3 -Carnot spaces are the well known Carnot spaces and the case of a 2 -step Carnot space corresponds to the hyperbolic symmetric space  n RH . Moreover, due to a result of [6], any homogeneous Einstein space of Iwsawa type and rank one is a k - step Carnot space for some k ≥ 2.

We remark that the 2 -stein condition was studied in [1] and [3] on Lie groups S of Iwasawa type in the case of 2 -step nilpotent 𝔫 , and in particular, on Carnot spaces: in this class those which are 2 -stein are exactly the Damek-Ricci spaces (also the harmonic ones). Furthermore, in [7] is shown that a simply connected homogenous harmonic space is a Carnot space, which in turn is equivalent to S be a Damek-Ricci space.

1. k-step Carnot spaces

Assume that S is a k -step Carnot space. Let 𝔰 be the Lie algebra of S : that is 𝔰 is a solvable metric Lie algebra 𝔰 = 𝔫 ⊕ RH with H ⊥ 𝔫, |H | = 1, with 𝔫 = [𝔰,𝔰] a graded nilpotent Lie algebra as described above.

Let 𝔷 denote the center of 𝔫 and let 𝔳 be the orthogonal complement of 𝔷 with respect to the metric ⟨,⟩ restricted to 𝔫. Thus 𝔫 decomposes 𝔫 = 𝔷 ⊕ 𝔳 and note that adH : 𝔷 → 𝔷 hence adH : 𝔳 → 𝔳 since adH is symmetric.

For any Z ∈ 𝔷 the skew-symmetric linear operator jZ : 𝔳 → 𝔳 is defined by

⟨jZX, Y ⟩ = ⟨[X, Y ],Z⟩ for all X, Y ∈ 𝔳, all Z ∈ 𝔷.

The Levi Civita connection and the curvature tensor on S can be computed by,

2 ⟨∇X Y, Z⟩ = ⟨[X, Y ],Z⟩ - ⟨[Y,Z ],X ⟩ + ⟨[Z,X ],Y ⟩ R (X, Y) = [∇X ,∇Y ] - ∇ [X,Y]

for any X, Y, Z in 𝔰 .

Recall that since 𝔷 and 𝔳 are adH -invariant, they also have decompositions into eigenspaces as  ∑ 𝔷 = λ 𝔷λ,  ∑ 𝔳 = μ𝔳 μ with 𝔷λ = 𝔫λ ∩ 𝔷 , 𝔳 μ = 𝔫μ ∩ 𝔳 and the equality λ = μ may be possible. Moreover, 𝔫 is said to be 2 -step nilpotent if [𝔫,𝔫] = [𝔳,𝔳 ] ⊂ 𝔷. Here, {λ } and {μ } denote the eigenvalues of adH | 𝔷 and adH | 𝔳 , respectively, that we assume they are natural numbers.

1.1. Properties of the operator j , Z Z ∈ 𝔷 . We recall that for any Z ∈ 𝔷λ so that jZ ⁄= 0, jZ : 𝔳μ → 𝔳λ-μ → 𝔳μ and  ( )⊥ ( )⊥ j : ker j | → ker j | Z Z 𝔳μ Z 𝔳λ-μ isomorphically. Consequently, as follows from [5, Lemma1.1])

 ( | ) ( | ) tr j2Z|𝔳 = tr j2Z|𝔳 . μ λ-μ

1.2. The curvature formulas. By applying the connection formula, one obtains ∇H = 0 and if  * Z, Z ∈ 𝔷 and X, Y ∈ 𝔳 , then  * * 1 ∇Z Z = ∇Z *Z = ⟨[H, Z],Z ⟩H, ∇X Z = ∇Z X = - 2jZ X and ∇X Y = 12[X, Y ] + ⟨[H, X ],Y⟩H in case of 2 -step nilpotent 𝔫 . Consequently, by a direct computation we obtain the following formulas involving curvatures (see [2, Section 2]).

(i) RH = - ad2 H

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

RZ H = - λ2H and RX H = - μ2H.

(iii) If Z ∈ 𝔷λ, |Z | = 1, for any  * Z ∈ 𝔷 and X ∈ 𝔳, then

R Z * = λ (⟨Z,ad Z *⟩Z - ad Z *) and R X = - 1j2 (X ) - λad X. Z H H Z 4 Z H

(iv) Let μ1 denote the maximum eigenvalue of adH |𝔳 . If X ∈ 𝔳μ1, |X | = 1, then

 1 1 * RX Z = --[X, jZ X ] - μ1adH Z - -(adjZX ) X, Z ∈ 𝔷 4 4 RX Y = - 1-[X, ∇ 𝔫 Y] - 3j[X,Y]X - ∇ 𝔫 (∇𝔫 Y )𝔳 - μ1adH Y, Y ⊥ X in 𝔳. 2 X 4 X X

1.3. The Einstein condition. Assume that S is an Einstein space, that is Ric 2 (Y ) = c|Y| for all Y ∈ 𝔰 . Let  m {Yi}i=1 and {Zi}ki=1 be orthonormal bases of 𝔳 and 𝔷 , respectively, with k = dim 𝔷 and m = dim 𝔳 . We use the well known formulas that hold in a solvable metric Lie algebra 𝔰 = 𝔫 ⊕ RH ,

 m 1-∑ 2 Ric (Z ) = 4 |jZXi | - λ tradH for Z ∈ 𝔷λ, i=1 1∑k Ric(X ) = - -- |jZiX |2 - μtr adH for X ∈ 𝔳 μ, if 𝔫 is 2-step-nilpotent. 2 i=1

Note that the first Einstein condition becomes

 ( ) 1-2 2 tr - 4jZ = - tradH + λtr adH , for Z ∈ 𝔷λ, |Z | = 1.
(1)

In general, for a unit vector X ∈ 𝔳μ1, μ1 the maximun eigenvalue of adH |𝔳 , we have

 ∑k ∑m Ric (X ) = - 1 |j X |2 - μ tr ad + |(∇ 𝔫 Y )|2. 2 Zi 1 H X i𝔳 i=1 i=1
(2)

The last one is a direct computation by applying the above expressions of RX in (iv) to

 ∑k ∑m Ric(X ) = trR = ⟨R Z ,Z ⟩ + ⟨R Y ,Y ⟩, X X j j X i i j=1 i=1

having into account that the following equality holds,

 m k ∑ |[X, Y ]|2 = ∑ |j X |2. i Zi i=1 j=1

In fact, this follows by computing

∑m 2 ∑m ∑k 2 ∑m ∑k ⟨ ⟩2 |[X, Yi]| = ⟨[X, Yi],Zj⟩ = jZjX, Yi i=1 i=1 j=1 i=1 j=1 k ( m ) k | | ∑ ∑ ⟨ ⟩2 ∑ | |2 = jZjX, Yi = |jZjX | , j=1 i=1 j=1

since [X, Y ] ∈ 𝔷 i for any X ∈ 𝔳 . μ1

1.4. The 2-stein condition. For any X ∈ 𝔰 , the Jacobi operator RX associated to X is the symmetric endomorphism of 𝔰 defined by RX Y = R (Y, X )X. We said that S is a 2 -stein space, or equivalently S (or 𝔰 ) satisfies the 2 -stein condition, if there exist

tr(Rl ) = μl|X |2l, l = 1,2, for all X ∈ 𝔰. X

In particular, 2 -stein spaces are Einstein: Ric(X ) = trRX = μ1 |X |2 , μ1 a constant, for all X ∈ 𝔰 , and harmonic riemannian spaces are necessarily 2 -stein. The 2 -stein condition was studied on Carnot spaces in [1] and [3]: the 2 -stein Carnot spaces are exactly the Damek-Ricci spaces (also the harmonic ones) in this class.

The proposition below express the 2 -stein condition in an adequate form for our purposes. Let {μ } denote the set of eigenvalues of adH |𝔳 .

Proposition 1.1. Assume that 𝔰 satisfies the 2 -stein condition. If Z ∈ 𝔷λ is a unit vector, then

 | | 1 ∑ ( 1 | ) tr( ad4H )| ⊥ = tr (- RZ ∘ ad2H )| ⊥ + -- (λ - 2μ )2tr - --j2Z| . 𝔷∩𝔷λ⊕𝔳 𝔷∩𝔷λ⊕𝔳 2 μ 4 𝔳μ

In particular, if adH | = λ 𝔷 Id

 ( | ) ( | ) ∑ ( | ) tr ad4H| = tr - RZ ∘ ad2H| + 1- (λ - 2μ)2tr - 1-j2Z| . 𝔳 𝔳 2 μ 4 𝔳μ
(3)

Proof. It follows from the same argument as those used in Theorem 4.1 of [4] for k = 2 (See also [5, Proposition 1.2]). □

2. The 2-stein condition on 4-step Carnot spaces

We consider the case of a 4 -step Carnot space; that is, 𝔰 = 𝔫 ⊕ RH as above with 𝔫 = 𝔫1 ⊕ 𝔫2 ⊕ 𝔫3 . Hence, by the propeties of 𝔫i, [𝔫1,𝔫1] ⊂ 𝔫2, [𝔫1,𝔫2] ⊂ 𝔫3, [𝔫1,𝔫3] = 0, [𝔫2,𝔫2] = 0 = [𝔫2, 𝔫3], [𝔫3,𝔫3] = 0 and 𝔫3 ⊂ 𝔷. We observe that:

[𝔫,𝔫] ⊂ 𝔫 ⊕ 𝔫 2 3 and [[𝔫,𝔫],𝔫] ⊂ [𝔫 ⊕ 𝔫 ,𝔫 ⊕ 𝔫 ⊕ 𝔫 ] ⊂ 𝔫 . 2 3 1 2 3 3 Thus, 𝔫3 = 0 since [[[𝔫,𝔫 ],𝔫],𝔫] ⊂ [𝔫3,𝔫] = 0; that is 𝔫 is 3 -step nilpotent and 𝔫3 ⊂ 𝔷 . Since 𝔫 = 𝔷 ⊕ 𝔳 we have that 𝔳 ⊂ 𝔫1 ⊕ 𝔫2. Moreover, from the facts that 𝔷 and 𝔳 are adH -invariant, it follows that

𝔷 = 𝔫1 ∩ 𝔷 ⊕ 𝔫2 ∩ 𝔷 ⊕ 𝔫3 and 𝔳 = 𝔫1 ∩ 𝔳 ⊕ 𝔫2 ∩ 𝔳.

We set dim 𝔫i = ni, i = 1,2,3. Following with the notation introduced in the previous section we have,

Lemma 2.1. If S is a 4 -step Carnot space that satisfies the 2 -stein condition, then S is either a Damek-Ricci space or, up to scaling, the symmetric hyperbolic space  n S = RH (𝔫2 = 0 or 𝔫3 = 0) with n = dim S. If 𝔫3 ⁄= 0 the equalities 𝔷 = 𝔫3 , 𝔳 = 𝔫1 ⊕ 𝔫2 hold in the decomposition of 𝔫 = 𝔷 ⊕ 𝔳 .

Proof. (i ) 𝔫1 ∩ 𝔷 = 0, if the Einstein condition is satisfied.

If 0 ⁄= Z ∈ 𝔫1 ∩ 𝔷, ⟨jZX, Y ⟩ = ⟨[X, Y ],Z⟩ = 0 for all X, Y ∈ 𝔳 since [X, Y ] ∈ ∑ 𝔫i i≥2 and ⟨𝔫i,𝔫j⟩ = 0 if i ⁄= j. Hence, jZ = 0 and tr 2 (jZ) = 0, and the Einstein condition (1) implies that

 ( ) 1 2 2 tr - --jZ = - tradH + tradH = - (n1 + 4n2 + 9n3) + (n1 + 2n2 + 3n3 ) 4 = - (2n2 + 6n3).

Thus, n2 = n3 = 0, and consequently 𝔫 = 𝔫1. In this case 𝔫 is abelian and S corresponds to the symmetric hyperbolic space  n RH with n = n1 + 1 .

The previous fact implies that the eigenvalue μ = 1 is not achieved in 𝔷 , when 𝔫 is 3 -step properly. Thus,

𝔷 = 𝔫3 ⊕ 𝔫2 ∩ 𝔷 and 𝔳 = 𝔫1 ⊕ 𝔫2 ∩ 𝔳.

(ii ) 𝔫2 ∩ 𝔷 = 0 if the 2 -stein condition holds.

In fact, assume that there exists 0 ⁄= Z ∈ 𝔫2 ∩ 𝔷 . In this case, if |Z| = 1 condition (1) gives

 ( ) tr - 1-j2 = - trad2 + 2tradH = - (n1 + 4n2 + 9n3) + 2(n1 + 2n2 + 3n3) 4 Z H = n1 - 3n3.

Moreover, since Z ∈ 𝔫2 and ⟨jZ X, Y⟩ = ⟨[X, Y ],Z ⟩ = 0 for all X ∈ 𝔫2 ∩ 𝔳 ,Y ∈ 𝔳 , we have that jZ|𝔫2∩𝔳 = 0 and jZ : 𝔫1 → 𝔫1. Now, we apply the formula given in Proposition 1.1 for λ = 2, and μ = 1,2. We first note that,

 ( ) 1-∑ 2 1- 2|| 2 (λ - 2μ )tr - 4 jZ 𝔳μ = 0. μ=1,2

since jZ |𝔫2∩𝔳 = 0 and λ - 2μ = 0 for μ = 1. This implies that tr( 4 | ad H)|𝔷∩𝔷⊥ ⊕𝔳 = λ tr 2 | (- RZ ∘ adH )|𝔷∩𝔷⊥⊕𝔳 λ and consequently, equality holds in the Cauchy-Schwartz Inequality

( | )2 | | tr (- RZ ∘ ad2 )| ⊥ ≤ tr (- RZ )2| ⊥ tr (ad2 )| ⊥ H 𝔷∩ 𝔷λ⊕ 𝔳 𝔷∩𝔷λ⊕ 𝔳 H 𝔷∩𝔷λ⊕ 𝔳 4 || 4 || ( 4 || )2 = tr adH 𝔷∩𝔷⊥λ⊕𝔳tr adH 𝔷∩𝔷⊥λ⊕𝔳 = tr adH 𝔷∩𝔷⊥λ⊕𝔳 .

Therefore, - RZ = c ad2 H in 𝔷 ∩ 𝔷⊥ ⊕ 𝔳 λ with c = 1, by the Einstein condition. The equality  2 || RZ |𝔫3 = - adH 𝔫3 , implies that 𝔫3 = 0, since Z ∈ 𝔫2 ∩ 𝔷 and RZ |𝔫3 = - 6 Id. Hence, 𝔫 = 𝔫1 ⊕ 𝔫2 with 𝔫1 = 𝔳 , 𝔫2 = 𝔷 and S is a Carnot space. In this case S is a Damek-Ricci space (see [3, Theorem 3.1]).□

Assume that 𝔫 ⁄= 0. 3 If 𝔰 satisfies the 2 -stein condition then 𝔷 = 𝔫 , 3 and 𝔳 = 𝔫1 ⊕ 𝔫2.

The following proposition is basic for our purposes,

Proposition 2.2. If S is a 4 -step Carnot space that is 2 -stein, then n = 2n 1 2 and  1 2 - 4 jZ |𝔫2 = 3 Id for any unit vector Z ∈ 𝔷 . Equivalently,  2 |jZX | = 12 for all unit vectors Z ∈ 𝔷 and X ∈ 𝔫2 .

Proof. Next we will show that n1 = 2n2 ,

 ( ) ( ) tr - 1-j2|| = 3n and tr -1-j4|| = 9n 4 Z 𝔫2 2 16 Z 𝔫2 2

for any unit vector Z ∈ 𝔫3.

Let Z ∈ 𝔷 = 𝔫3 be a unit vector. The Einstein condition (1) applied to Z gives,

 ( ) tr - 1j2 = - (1n + 4n + 9n ) + 3(n + 2n + 3n ) (4 ) 4 Z 1 2 3 1 2 3 = 2(n1 + n2).

We compute separately the two terms in (3). First, using the definition of RZ |𝔳 = - 1j2Z - 3 adH |𝔳, 4 it follows that

 ( )| ( | ) ( | ) tr - RZ ∘ ad2H | = tr - RZ ∘ ad2H| + tr - RZ ∘ ad2H | 𝔳 ( ) 𝔫1 ( ( )𝔫2 ) 1- 2|| 1- 2 || = tr 4 jZ 𝔫1 + 3n1 + 4 tr 4 jZ 𝔫2 + 6n2 ( ) ( ) = tr 1-j2|| + 4tr 1-j2|| + 3n + 24n , 4 Z 𝔫1 4 Z 𝔫2 1 2

and applying (4),

 ∑ 2 ( ) ( ) ( ) 1- (λ - 2μ)2tr - 1-j2|| = 1tr - 1-j2|| + 1tr - 1-j2|| 2 4 Z 𝔳μ 2 4 Z 𝔫1 2 4 Z 𝔫2 μ=1 ( | ) 1- 1-2|| = 2tr - 4jZ| = n1 + n2 (m = n1 + n2). 𝔳

Thus, (3) gives

 | n1 + 24n2 = tr(ad4H|𝔳) ( 1 | ) ( 1 | ) = tr --jZ2| + 4tr --j2Z| + 3n1 + 24n2 + n1 + n2 4 𝔫1 4 𝔫2

and consequently,

 ( 1 | ) ( 1 | ) tr - --j2Z|𝔫1 + 4tr - --j2Z|𝔫2 = 3n1 + 9n2 = 3(n1 + 3n2). 4 4
(5)

Now, using that ( trj2 | ) = Z 𝔫1 trj2 | ) Z 𝔫2 (see 1.1), (4) and (5), that is

 ( | ) ( ) ( ) 1-2| 1- 2| 1- 2| 2tr - 4jZ|| = tr - 4 jZ|𝔫1 + tr - 4 jZ|𝔫2 = 2(n1 + n2) ( 𝔫1) ( ) ( ) 1 2| 1 2| 1 2| 5tr - --jZ|𝔫1 = tr - --jZ|𝔫1 + 4tr - --jZ|𝔫2 = 3(n1 + 3n2 ), 4 4 4

n = 2n 1 2 is obtained, since

3(n1 + 3n2) = 5(n1 + n2) ⇔ 2n1 = 4n2 ⇔ n1 = 2n2.

Finally, we will show the last assertion of the proposition. Using that RZ |𝔳 = - 14j2Z - 3 adH |𝔳 again, we compute

 ( )2 2 1- 2 tr(RZ |𝔳) = tr 4 jZ + 3 adH |𝔳

by developing  ( ) R2Z|𝔳 = - 14j2Z - 3 adH |𝔳 2. Thus,

 ( |) ( ) | ( ) tr R2Z | = tr 1-j4Z + 9tr ad2H| + 3tr jZ2∘ adH | 𝔳 16 𝔳 2 𝔳 ( 1 ) ( | | ) = tr --j4Z + 9 tr(ad2H |𝔫 ) + tr( ad2H|𝔫 ) 16 1 2 3-( 2|| 2|| ) + 2 tr(jZ 𝔫1 ∘ adH |𝔫1) + tr( jZ 𝔫2 ∘ adH |𝔫2)

and substituting the values of adH | = i 𝔫i Id, we have

 ( ) ( ) tr(R2 ||) = tr -1-j4 + 9(n + 4n ) + 3 tr(j2 || ) + 2tr(j2|| ) Z 𝔳 16 Z 1 2 2 Z 𝔫1 Z 𝔫2 ( ) ( | ) = tr -1-jZ4 + 9-tr j2Z | + 9(n1 + 4n2 ), 16 2 𝔫1

since tr( ) j2Z|𝔫1 = tr( ) j2Z|𝔫2 . Note, that the same argument used to show this equality, implies

 ( | ) ( | ) tr j4| = tr j4| . Z 𝔫1 Z 𝔫2
(6)

Therefore, the 2 -stein condition, tr 2 (R Z|𝔳) = tr( 4|| ) ad H 𝔳 , gives

 ( ) 1-- 4|| 9- ( 2 || ) 2tr 16 jZ 𝔫1 + 2 tr jZ 𝔫1 + 9(n1 + 4n2) = n1 + 16n2 or ( ) ( ) 2tr 1--j4|| + 9-tr j2 || = - (8n + 20n ). 16 Z 𝔫1 2 Z 𝔫1 1 2

By applying (4), the last equality is equivalent to

 ( 1 | ) ( 1 | ) 2tr ---j4Z| - 18tr - --j2Z| = - 4(2n1 + 5n2 ), or 16 𝔫1 4 𝔫1 ( 1 | ) = 2tr ---j4Z|𝔫1 - 18(n1 + n2) = - 4(2n1 + 5n2). 16

It follows from (6) that

 ( 1 | ) 2tr ---j4Z|𝔫 = 10n1 - 2n2 = 2(5n1 - n2), 16 1

and from the fact n1 = 2n2,

 ( ) ( ) -1- 4|| -1- 4|| tr 16 jZ 𝔫2 = tr 16 jZ 𝔫1 = 5n1 - n2 = 9n2.

The final assertion follows as claimed, since

 ( | ) ( ) 1- 2|| -1- 4|| tr - 4 jZ| = 3n2 and tr 16 jZ 𝔫2 = 9n2 𝔫2

implies that equality holds in the Cauchy-Schwartz Inequality

( ( ) )2 ( ) 1- 2|| 1-- 4|| tr - 4 jZ 𝔫2 ≤ n2tr 16 jZ 𝔫2 ,

which gives - 14 j2Z|𝔫 = 3 2 Id, or equivalently, |jZ X |2 = 12 for all unit vector X ∈ 𝔫2.

Let 𝔰 = 𝔫1 ⊕ 𝔫2 ⊕ 𝔫3 ⊕ RH be a 4 -step Lie algebra so that 𝔷 = 𝔫3, and 𝔳 = 𝔫1 ⊕ 𝔫2. In what follows let  k {Zi }i=1,  n {Yi} i=11 and  n {Xj }j2=1 be any orthonormal bases of 𝔷 , 𝔫1 and 𝔫2 , respectively. The following basic formulas are deduced from the hypothesis 𝔫3 ⁄= 0 and not assuming that the 2 -stein condition is satisfied. They are shown in [5, Proposition 2.3].

If X ∈ 𝔫2 and Y ∈ 𝔫1 are unit vectors, then

 n2 n3 ∑ 2 ∑ 2 |[Y, Xj]| = |jZkY | j=1 k=1
(7)
∑n2 |( ) | n∑1 || ∇ 𝔫 Y ||2 = 1- |[Y,Y ]|2 Xj 𝔳 4 i j=1 i=1
(8)

In order to show that the 2 -stein condition is not satisfied by 𝔰 , in case that 𝔰 is 4 -step properly, we need to compute Ric(Y ) and Ric(X ) for unit vectors Y ∈ 𝔫1 and X ∈ 𝔫 . 2

Lemma 2.3. If Y ∈ 𝔫1 is a unit vector then,

 n2 n1 1∑ 2 1∑ 2 tr RY = - n1 - 2n2 - 3n3 - 2 |[Y, Xj]| - 2 |[Y, Yi]| . j=1 i=1

Proof. Let Y ∈ 𝔫1 be a unit vector. We first note, tha it is a direct computation to see that

 1 ∇Y W = -[Y,W ] for W ∈ 𝔫1, W ⊥ Y 2 ∇ X = 1[Y,X ] + (∇ 𝔫X ) for X ∈ 𝔫 . Y 2 Y 𝔳 2

Therefore, for unit vectors Z ∈ 𝔷 , W ∈ 𝔫1 and X ∈ 𝔫2 using the definition of RY we obtain

 1 RY Z = - 3Z + -∇Y jZY, 2 1- RY W = - W - 2∇Y [W, Y ] - ∇ [W,Y]Y , 1 RY X = - 2X - ∇Y ∇X Y + -j[X,Y ]Y , RY H = - H. 2

Now, it is a strighforward computation using the definition of ∇ (.) to see that,

⟨RY Z,Z ⟩ = - 3 + 1⟨∇Y jZY,Z ⟩ = - 3 + 1|jZY |2, 2 4

 1- ⟨ ⟩ ⟨RY W, W ⟩ = - 1 - 2 ⟨∇Y [W, Y ],W ⟩ - ∇ [W,Y ]Y,W 1 1 = - 1 - -|[Y,W ]|2 + -⟨[Y,W ],[W, Y ]⟩ 4 2 = - 1 - 3|[Y,W ]|2 4

and

 ⟨ ⟩ ⟨RY X, X ⟩ = - 2 - ⟨∇Y ∇X Y,X ⟩ + 1- j[X,Y ]Y, X 2 1- = - 2 + ⟨∇X Y,∇Y X ⟩ + 2 ⟨[Y, X ],[X, Y ]⟩ 1 = - 2 + |∇X Y |2 + ⟨∇X Y,[Y,X ]⟩ ---|[X, Y ]|2 2 2 2 = - 2 + |∇X Y | - |[X, Y]| 3 2 2 = - 2 - -|[X,Y ]| + |(∇ 𝔫XY )𝔳| . 4

Next, by computing

 n n ∑3 1-∑ 3 2 tr RY |𝔷⊕RH = ⟨RY Zk, Zk⟩ - 1 = - 3n3 + 4 |jZkY | - 1 k=1 k=1 n∑1 3 n∑1 tr RY |𝔫1 = ⟨RY Yi,Yi⟩ = - (n1 - 1) - -- |[Y,Yi]|2 i=1 4 i=1 n∑2 n∑2 ∑n2 tr R | = ⟨R X ,X ⟩ = - 2n + ||(∇ Y ) ||2 - 3- |[X ,Y ]|2, Y 𝔫2 Y j j 2 Xj 𝔳 4 j j=1 j=1 j=1

it is immediate that

 n3 n1 1∑ 2 3-∑ 2 tr RY = - n1 - 2n2 - 3n3 + 4 |jZk Y| - 4 |[Y,Yi]| k=1 i=1 n∑2 |( ) |2 3∑n2 2 + | ∇Xj Y 𝔳| - -- |[Y, Xj ]| . j=1 4 j=1

By applying the equalities given by (7) and (8) we have,

 n2 n1 1-∑ 2 1-∑ 2 trRY = - n1 - 2n2 - 3n3 - 2 |[Y,Xj ]| - 2 |[Y,Yi]| , j=1 i=1

the expression stated in the lemma. □

Corollary 2.4. If 𝔰 = 𝔫1 ⊕ 𝔫2 ⊕ 𝔫3 ⊕ RH satisfies the Einstein condition, then for all unit vectors Y ∈ 𝔫1 and X ∈ 𝔫2 we have,

 ∑n1 1 ∑k (i) |(∇ 𝔫XYi)𝔳|2 - -- |jZiX |2 = n1 - 3n3 i=1 2 i=1 ∑n2 ∑n1 (ii) |[Y, X ]|2 + |[Y, Y ]|2 = 4 (n + 3n ). j i 2 3 j=1 i=1

Proof. (i) and (ii) are obtained by applying the Einstein condition tr(RX ) = - tr( ad2 H ) (2) to X ∈ 𝔫2, and tr(RY ) = - tr( ad2 H ) to Y ∈ 𝔫1 (Lemma 2.3), respectively. In fact, they are immediate since

 ∑k ∑m - 1- |jZ X |2 - 2(n1 + 2n2 + 3n3) + |(∇ 𝔫 Yi)|2 = - tr(ad2 ) 2 i=1 i i=1 X 𝔳 H - n1 - 4n2 - 9n3

and

 n n 1-∑ 2 2 1-∑ 1 2 2 - n1 - 2n2 - 3n3 - 2 |[Y, Xj]| - 2 |[Y, Yi]| = - tr(adH ) j=1 i=1 - n1 - 4n2 - 9n3,

implying the required formulas. □

Theorem 2.5. If S is a 4 -step Carnot space that satisfies the 2 -stein condition, then S is either, up to scaling, the hyperbolic space RHn (n = dim S) or a Damek-Ricci space.

Proof. Let S be a 4 -step Carnot space with associated Lie algebra 𝔰 . If 𝔫3 = 0 then either S is, up to scaling, the hyperbolic symmetric space RHn with n = n + 1 1 or n = n2 + 1 according to 𝔫2 = 0 or 𝔫1 = 0, or S is a Damek-Ricci space by [3, Theorem 3.1].

Assume that 𝔫3 ⁄= 0; we will show that S cannot be 2 -stein unless 𝔫1 = 0 = 𝔫2.

(i) Let Y ∈ 𝔫1 a unit vector. If U is the operator on 𝔰 defined by U (.) = R (.,Y)H + R (.,H )Y it follows that

 ( ) | 4 ( 2 ) 1- 2 || tr adH + RY ∘ adH - 2 U | ⊥= 0. 𝔰0
(9)

where 𝔰0 = span{Y, H } is the totally geodesic subalgebra of 𝔰 which corresponds to the symmetric hyperbolic space RH2 of constant curvature - 1. We show this expression (9) following the same argument developed in [5, Proposition 1.2]. In order to do that we need to revise some properties fulfilled by RY and U. First, note that

RY : 𝔷 → 𝔷 ⊕ 𝔫1 𝔫1 → 𝔷 ⊕ 𝔫1 and RY H = - H,

hence RY : 𝔫2 → 𝔫2, since it is symmetric.

Next, by using the definition of U, for any unit vectors Z ∈ 𝔷 , W ⊥ Y ∈ 𝔫1 and X ∈ 𝔫2 we compute:

U (Z ) = ∇Z ∇Y H - ∇Y ∇Z H - ∇ [Z,H ]Y = - ∇Z Y + 3 ∇Y Z + 3∇Z Y 5 = 5∇Z Y = - 2jZY,

U (W ) = ∇W ∇Y H - ∇Y ∇W H - ∇ [W,Y]H - ∇ [W,H ]Y = - ∇W Y + ∇Y W + 2[W, Y ] + ∇W Y 1- 3- = 2[Y,W ] - 2[Y,W ] = - 2 [Y, W ] and

U (X ) = ∇X ∇Y H - ∇Y ∇X H - ∇ [X,Y]H - ∇ [X,H ]Y = - ∇X Y + 2∇Y X + 3[X, Y ] + 2∇X Y = ∇X Y + 2∇Y X + 3 [X, Y ] = 3∇X Y + [X, Y ] = 3-[X, Y ] + 3(∇ 𝔫X Y )𝔳 + [X, Y ] = 5-[X, Y ] + 3(∇𝔫X Y )𝔳. 2 2

Thus, U : 𝔫2 → 𝔷 ⊕ 𝔫1 and 𝔷 ⊕ 𝔫1 → 𝔫2. This property, togheter with the fact RY : 𝔷 ⊕ 𝔫1 → 𝔷 ⊕ 𝔫1  and 𝔫2 → 𝔫2 (RY H = - H ) imply that (9) holds, since

 ( 2 2 2) || tr r RY - s adH ∘ U 𝔰⊥0 = 0.

(ii) Next, (9) it will be applied, computing separately each term and using (7) and (8). First, from the above computations we obtain

 ⟨U 2Z,Z ⟩ = |U (Z )|2 = 25-|jZY |2 for all Z ∈ 𝔷, |Z | = 1 4 ⟨ 2 ⟩ 2 9- 2 U W, W = |U (W )| = 4 |[Y,W ]| , |W | = 1, and ⟨ ⟩ 25 U 2X, X = |U (X )|2 = --|[X, Y]|2 + 9 |(∇ 𝔫X Y )𝔳|2, for X ∈ 𝔫2, |X | = 1. 4

Hence,

 ( 2|| ) n∑3 2 ∑n1 2 ∑n2 2 tr U 𝔰0⊥ = |U (Zk)|+ |U(Yi)|+ |U(Xj )| k=1n i=1 n j=1 n n 25∑ 3 2 9 ∑1 2 25 ∑2 2 ∑2 || 𝔫 ||2 = 4 |jZkY| + 4 |[Y,Yi]| + 4 |[Y,Xj]|+ 9 |(∇ XjY)𝔳| k=n11 i=1 n2 j=1 j=1 = 29∑ |[Y, Y]|2 + 225∑ |[Y,X ]|2, 4 i=1 i 4 j=1 j

which implies that

 1 ( | ) 9 n∑1 25 ∑n2 - --tr U 2| = - -- |[Y, Yi]|2 - --- |[Y,Xj ]|2 . 2 𝔰0⊥ 4 i=1 4 j=1

It is a straightforward computation to see that,

 2 | tr (RY ∘ ad H)|𝔰⊥0 = 9tr RY |𝔷 + tr (RY |𝔫1∩Y⊥) + 4tr(RY |𝔫2) ( n3 ) n1 ∑ 1- 2 3∑ 2 = 9 - 3n3 + 4 |jZkY | - (n1 - 1) - 4 |[Y, Yi]| ( k=1 i=)1 ∑n2 | |2 3 ∑n2 +4 - 2n2 + |∇Xj Y | - -- |[Y,Xj ]|2 j=1 4 j=1 ∑n2 ∑n1 = - 27n - (n - 1) - 8n - 3- |[Y, X ]|2 + 1- |[Y,Y ]|2, 3 1 2 4 j 4 i j=1 i=1

and (9) becomes

 ( ) | 0 = tr ad4 + (R ∘ ad2 ) - 1-U 2 || = (n - 1) + 16n + 81n H Y H 2 |𝔰⊥ 1 2 3 n 0 n 3-∑2 2 1∑ 1 2 - 27n3 - (n1 - 1) - 8n2 - 4 |[Y, Xj ]| + 4 |[Y, Yi]| j=1 i=1 9∑n1 25 ∑n2 - -- |[Y, Yi]|2 - --- |[Y,Xj ]|2 4 i=1 4 j=1 n2 n1 = 54n + 8n - 28 1∑ |[Y, X ]|2 - 8 1∑ |[Y, Y ]|2. 3 2 4 j 4 i j=1 i=1

Thus,

 n2 n1 ∑ 2 ∑ 2 7 |[Y,Xj ]| + 2 |[Y, Yi]| = 54n3 + 8n2. j=1 i=1
(10)

(iii) Now, from the equality given by Corollary 2.4 (ii), we obtain

 n2 n1 2 ∑ |[Y,X ]|2 + 2∑ |[Y, Y]|2 = 24n + 8n = 2 (12n + 4n ) j i 3 2 3 2 j=1 i=1
(11)

and the expression (10)- (11) gives

 n2 n2 ∑ 2 ∑ 2 5 |[Y, Xj]| = 30n3 ⇔ |[Y, Xj]| = 6n3 for a unit vector Y ∈ 𝔫1. j=1 j=1

Finally, it follows from (7) that

n3 ∑ 2 |jZkY | = 6n3 k=1

and we get a contradiction, since |j Y |2 = 12 Z for any unit vectors Z ∈ 𝔷 and Y ∈ 𝔫 . 1 In order to show this last assertion, note that  ⊥ jZ : 𝔫2 → ker(jZ |𝔫1) ⊂ 𝔫1 isomorphically (1.1) and by Proposition 2.2, a unit vector Y ∈ 𝔫1 is expressed as Y = √112jZX with X ∈ 𝔫2, |X | = 1 (Y = ∑m a √1-j X i=1 i 12 Z i where j2X = - |j X |2X , Z i Z i i |X | = 1, i in terms of the orthonormal basis {Xi} and {√112jZ Xi} : i = 1,...,m ). Hence  2 |jZY | = 12 and we have that ∑n3 |j Y |2 = 12n . k=1 Zk 3

3. References

[1] Benson P., Payne T., Ratcliff G.; Three-step harmonic solvmanifolds. Geometriae Dedicata 101, 2003, 103-127.        [ Links ]

[2] Dotti I, Druetta M.J.; Negatively curved homogeneous Osserman spaces. Differential Geometry and Applications 11, 1999, 163-178.        [ Links ]

[3] Druetta M.J.; On harmonic and 2-stein spaces of Iwasawa type. Differential Geometry and its Applications 18, 2003, 351-362.        [ Links ]

[4] Druetta M.J.; Carnot spaces and the k-stein condition. Advances in Geometry 6, 2006, 447-473.        [ Links ]

[5] Druetta M.J.; The 2-stein condition on generalized Carnot spaces, 2005 (pre-print).        [ Links ]

[6] Heber J.; Noncompact homogeneous Einstein spaces. Inventiones Math. 133, 1998, 279-352.        [ Links ]

[7] Heber J.; On harmonic and asymptotically harmonic homogeneous spaces. Geometric and Functional Analysis 16 (4), 2006, 869-890.        [ Links ]

María J. Druetta
CIEM-Conicet y FAMAF,
Universidad Nacional de Córdoba,
Córdoba 5000, Argentina
druetta@mate.uncor.edu

Recibido: 26 de enero de 2006
Aceptado: 1 de noviembre de 2006