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Revista de la Unión Matemática Argentina
versión Online ISSN 16699637
Rev. Unión Mat. Argent. v.47 n.1 Bahía Blanca ene./jun. 2006
4step Carnot spaces and the 2stein condition
María J. Druetta
Dedicated to the memory of my goddaughter María Pía
Abstract: We consider the 2stein condition on kstep 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 2stein condition is not satisfied on a proper 4step Carnot spaces S.
2000 Mathematics Subject Classification. 53C30, 53C55
Key words and phrases: 2stein spaces, kstep Carnot spaces, Einstein spaces, Lie groups of Iwasawa type, DamekRicci spaces
A Riemannian manifold is said to be a stein space, if there exist functions defined on such that

Here, denote the Jacobi operator associated to defined by for all where is the curvature tensor of Harmonic riemannian manifolds are necessarily stein.
A step Carnot space () is a simply connected solvable Lie group which is a semidirect product of a nilpotent Lie group and the abelian group Assume that and have associated Lie algebras and , respectively. has the left invariant metric induced by the one given on , where is a solvable metric Lie algebra with inner product such that:
(i) with and
(ii) has an orthogonal decomposition into subspaces given by

for some positive constant
Note that, since the adjoint representation ad is a derivation of the above decomposition defines a graded Lie algebra structure of , that is,

with the convention for In particular, is a (step nilpotent Lie algebra, that is, the (th derived algebra vanishes:
Note, that up to a constant in the metric, we may suppose that In this terminology, the Carnot spaces are the well known Carnot spaces and the case of a step Carnot space corresponds to the hyperbolic symmetric space Moreover, due to a result of [6], any homogeneous Einstein space of Iwsawa type and rank one is a  step Carnot space for some
We remark that the stein condition was studied in [1] and [3] on Lie groups of Iwasawa type in the case of step nilpotent , and in particular, on Carnot spaces: in this class those which are stein are exactly the DamekRicci 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 be a DamekRicci space.
Assume that is a step Carnot space. Let be the Lie algebra of : that is is a solvable metric Lie algebra with with a graded nilpotent Lie algebra as described above.
Let denote the center of and let be the orthogonal complement of with respect to the metric restricted to Thus decomposes and note that ad hence ad since ad is symmetric.
For any the skewsymmetric linear operator is defined by

The Levi Civita connection and the curvature tensor on can be computed by,
for any in .
Recall that since and are adinvariant, they also have decompositions into eigenspaces as with , and the equality may be possible Moreover, is said to be step nilpotent if Here, and denote the eigenvalues of and , respectively, that we assume they are natural numbers.
1.1. Properties of the operator . We recall that for any so that and isomorphically. Consequently, as follows from [5, Lemma1.1])

1.2. The curvature formulas. By applying the connection formula, one obtains and if and , then and in case of step nilpotent . Consequently, by a direct computation we obtain the following formulas involving curvatures (see [2, Section 2]).
(i) ad
(ii) If either or

(iii) If for any and then

(iv) Let denote the maximum eigenvalue of . If then
1.3. The Einstein condition. Assume that is an Einstein space, that is Ric for all . Let and be orthonormal bases of and , respectively, with and . We use the well known formulas that hold in a solvable metric Lie algebra ,
Note that the first Einstein condition becomes
 (1) 
In general, for a unit vector the maximun eigenvalue of , we have
 (2) 
The last one is a direct computation by applying the above expressions of in (iv) to

having into account that the following equality holds,

In fact, this follows by computing
since for any
1.4. The 2stein condition. For any , the Jacobi operator associated to is the symmetric endomorphism of defined by We said that is a stein space, or equivalently (or ) satisfies the stein condition, if there exist

In particular, stein spaces are Einstein: Rictr a constant, for all , and harmonic riemannian spaces are necessarily stein. The stein condition was studied on Carnot spaces in [1] and [3]: the stein Carnot spaces are exactly the DamekRicci spaces (also the harmonic ones) in this class.
The proposition below express the stein condition in an adequate form for our purposes. Let denote the set of eigenvalues of .
Proposition 1.1. Assume that satisfies the stein condition. If is a unit vector, then

In particular, if Id
 (3) 
Proof. It follows from the same argument as those used in Theorem 4.1 of [4] for (See also [5, Proposition 1.2]). □
2. The 2stein condition on 4step Carnot spaces
We consider the case of a step Carnot space; that is, as above with . Hence, by the propeties of and We observe that:
and Thus, since that is is step nilpotent and . Since we have that Moreover, from the facts that and are adinvariant, it follows that

We set Following with the notation introduced in the previous section we have,
Lemma 2.1. If is a step Carnot space that satisfies the stein condition, then is either a DamekRicci space or, up to scaling, the symmetric hyperbolic space or with If the equalities , hold in the decomposition of .
Proof. () if the Einstein condition is satisfied.
If for all since and if Hence, and tr and the Einstein condition (1) implies that
Thus, and consequently In this case is abelian and corresponds to the symmetric hyperbolic space with .
The previous fact implies that the eigenvalue is not achieved in , when is step properly. Thus,

() if the stein condition holds.
In fact, assume that there exists . In this case if condition (1) gives
Moreover, since and for all ,, we have that and Now, we apply the formula given in Proposition 1.1 for and We first note that,

since and for This implies that tr(tr and consequently, equality holds in the CauchySchwartz Inequality
Therefore, ad in with by the Einstein condition. The equality , implies that since and Id Hence, with , and is a Carnot space. In this case is a DamekRicci space (see [3, Theorem 3.1]).□
Assume that If satisfies the stein condition then and
The following proposition is basic for our purposes,
Proposition 2.2. If is a step Carnot space that is stein, then and Id for any unit vector . Equivalently, for all unit vectors and .
Proof. Next we will show that ,

for any unit vector
Let be a unit vector. The Einstein condition (1) applied to gives,
and applying (4),
Thus, (3) gives
and consequently,
 (5) 
Now, using that trtr (see 1.1), (4) and (5), that is
is obtained, since

Finally, we will show the last assertion of the proposition. Using that again we compute

by developing Thus,
and substituting the values of Id, we have
since trtr Note, that the same argument used to show this equality, implies
 (6) 
Therefore, the stein condition, trtr gives
By applying (4), the last equality is equivalent to
It follows from (6) that

and from the fact

The final assertion follows as claimed, since

implies that equality holds in the CauchySchwartz Inequality

which gives Id or equivalently, for all unit vector □
Let be a step Lie algebra so that and In what follows let and be any orthonormal bases of , and , respectively. The following basic formulas are deduced from the hypothesis and not assuming that the stein condition is satisfied. They are shown in [5, Proposition 2.3].
If and are unit vectors, then
 (7) 
 (8) 
In order to show that the stein condition is not satisfied by , in case that is step properly, we need to compute Ric and Ric for unit vectors and
Lemma 2.3. If is a unit vector then,

Proof. Let be a unit vector. We first note, tha it is a direct computation to see that
Therefore, for unit vectors , and using the definition of we obtain
Now, it is a strighforward computation using the definition of to see that,

and
Next, by computing
it is immediate that
By applying the equalities given by (7) and (8) we have,

the expression stated in the lemma. □
Corollary 2.4. If satisfies the Einstein condition, then for all unit vectors and we have,
Proof. (i) and (ii) are obtained by applying the Einstein condition trtrad (2) to and trtrad to (Lemma 2.3), respectively. In fact, they are immediate since
and
implying the required formulas. □
Theorem 2.5. If is a step Carnot space that satisfies the stein condition, then is either, up to scaling, the hyperbolic space or a DamekRicci space.
Proof. Let be a step Carnot space with associated Lie algebra . If then either is, up to scaling, the hyperbolic symmetric space with or according to or or is a DamekRicci space by [3, Theorem 3.1].
Assume that we will show that cannot be stein unless
(i) Let a unit vector. If is the operator on defined by it follows that
 (9) 
where span is the totally geodesic subalgebra of which corresponds to the symmetric hyperbolic space of constant curvature 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 and First, note that

hence since it is symmetric.
Next, by using the definition of for any unit vectors , and we compute:
Thus, and This property, togheter with the fact and imply that (9) holds, since

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

It is a straightforward computation to see that,
and (9) becomes
Thus,
 (10) 
(iii) Now, from the equality given by Corollary 2.4 (ii), we obtain
 (11) 
and the expression (10)(11) gives

Finally, it follows from (7) that

and we get a contradiction, since for any unit vectors and In order to show this last assertion, note that isomorphically (1.1) and by Proposition 2.2, a unit vector is expressed as with ( where in terms of the orthonormal basis and Hence and we have that □
[1] Benson P., Payne T., Ratcliff G.; Threestep harmonic solvmanifolds. Geometriae Dedicata 101, 2003, 103127. [ Links ]
[2] Dotti I, Druetta M.J.; Negatively curved homogeneous Osserman spaces. Differential Geometry and Applications 11, 1999, 163178. [ Links ]
[3] Druetta M.J.; On harmonic and 2stein spaces of Iwasawa type. Differential Geometry and its Applications 18, 2003, 351362. [ Links ]
[4] Druetta M.J.; Carnot spaces and the kstein condition. Advances in Geometry 6, 2006, 447473. [ Links ]
[5] Druetta M.J.; The 2stein condition on generalized Carnot spaces, 2005 (preprint). [ Links ]
[6] Heber J.; Noncompact homogeneous Einstein spaces. Inventiones Math. 133, 1998, 279352. [ Links ]
[7] Heber J.; On harmonic and asymptotically harmonic homogeneous spaces. Geometric and Functional Analysis 16 (4), 2006, 869890. [ Links ]
María J. Druetta
CIEMConicet 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