versión On-line ISSN 1669-9637
Rev. Unión Mat. Argent. v.48 n.1 Bahía Blanca ene./jun. 2007
Abstract. In this paper, we study a qualitative uncertainty principle for completely solvable Lie groups.
2000 Mathematics Subject Classification. 22E25, 22E27.
Keywords and phrases. Completely solvable Lie groups, uncertainty principle, orbit method.
Let be a connected, simply connected, and completely solvable Lie group, with Lie algebra . Let be the dual of . The equivalence classes of irreducible unitary representations of is parameterized by the coadjoint orbits via the Kirillov bijective map
We recall that if and , then there exists an analytic subgroup of and a unitary character of , such that the induced representation is equivalent to Moreover the push forward of a Plancherel measure in is a measure equivalent to a Lebesguian measure on convenient set of representatives in for
Let in and set its Fourier transform, let and . By Bénédicks theorem [1, Theorem 2], if and then a.e. Here, denote Lebesgue measure on . That is, for the qualitative uncertainty principle holds.
In this note we prove that a completely solvable Lie group has the qualitative uncertainty principle. In  we showed the theorem for nilpotent Lie groups, by induction on the dimension of . To prove the theorem we apply induction, for this, we need an explicit description of the dual space of as well as an explicit description of Plancherel measure on . For our approach we use a result of B.N. Currey , which is a generalization of a result of L. Pukanszky. Let be a locally compact group. Denote a fixed Haar measure on by and the corresponding Plancherel measure on by
Let and ,
Let be a connected, simply connected, and completely solvable Lie group, with the Lie algebra . Let be its dual. Since is completely solvable, there exists a chain of ideals of
such that the dimension of is , for all . We fix an ordered basis of such that is spanned by the vectors , Let be the dual basis of We fix a Lebesgue measure on and a right invariant Haar measure on such that where
Let be the modular function such that for all , . Let be a co-adjoint orbit in and . The bilinear form defines a skew-symmetric and nondegenerate bilinear form on . Since the map induces an isomorphism between and the tangent space of at , the bilinear form defines a nondegenerate 2-form on this tangent space. If is the dimension of we note that
is a canonical measure on . Lemma 3.2.2 in  says that there exists a nonzero rational function on such that
and there exists a unique measure on such that
for all Borel function on . B.N. Currey [3,] gave an explicit description of the measure with the help of the coadjoint orbits. We recall the theorem proved by B.N. Currey which is a essential tool to prove our main theorem.
Theorem 2.1. Let be a connected, simply connected and completely solvable Lie group. There exists a Zariski open subset in , a subset of , a subset of , for each a real valued rational function , non vanishing on , and real analytic functions in the variables such that the following hold.
- If denotes the number of elements of , for each , the set
is a non empty open subset in .
- Define by , where if and otherwise. Let for , define by if and otherwise. Then for each , the mapping is a diffeomorphism of with the coadjoint orbit of .
- Define as the subspace spanned by the vectors and the subspace spanned by the vectors Then the set
- For each , let such that . Then the mapping , defined by , is a diffeomorphism.
If the subset is empty, then and the coordinates for are obtained by identifying with , which is the parametrization of in the nilpotent case. If is not empty and the number of elements in . From , for each is a non empty Zariski open subset and (disjoint union). Set . from  we have:
is a rational nonsingular function on .
Let From , there is a Zariski open subset of and a rational function such that is the graph of . From , the projection of into parallel to defines a diffeomorphism of with .
Summarizing: let be connected, simply connected and completely solvable Lie group. Let be a Jordan-Holder basis of . Then, there is a finite family of disjoint open subsets of and there is a subspace of such that for each , the orbits in are parameterized by a Zariski open subset of The union of this open sets determines an open dense subset of whose complement has Plancherel measure zero.
Consider the group
We use the notation
Matrix multiplication is:
and the inverse is
The Lie algebra of is the set of matrices
We choose as ordered base
We have . Thus the group is not nilpotent.
Let the dual basis of . Let The orbits of in are: the upper half plane , the lower half plan and the points . Here, and , so that and is spanned by the vector . The Zariski open sets and are the half planes of and . Since there are two orbits, the set
has exactly two points. We have and . The Zariski open set or of , reduces to a point.
We must consider two cases(see ):
- All the orbits in general position are saturated with respect to That is, for each Then, we may and will choose a basis of
where the last vector of the basis does not depend on . We apply the previous setting to Let the index set for , then is a subset of let denote the number elements of . For each , the set is nonempty open subset of . Let and then is the subspace spanned by . We apply the inductive hypothesis to , hence, there is a Zariski open subset and a rational function such that is the graph of . Let denote the projection of on . From [5,lemma 3.2], the measure on in terms of the measure on and is
- If some orbit in general position is not saturated with respect to , we can still obtain a basis of such that the last vector of the basis does not depend on , and for certain with . In this case since , we have . Moreover . The Plancherel measure can be written as .
Proof 5.1. We proceed by induction on the dimension of . The result is true if the dimension of is one, since (see [1,theorem2]). Assume that the result is true for all completely solvable Lie groups of dimension . Suppose that are finite. From [4, lemma 1.6], is finite. To conclude, it remains to show that is finite. We can assume that is contained in (It suffices to take as the finite union of ).We consider tho cases
- We suppose that for all That is, all the orbits in general position are saturated with respect to . For , let be the restriction of to , then is irreducible. From [6, proposition 2.5] we have:
where is the Lebesgue measure on and is the Lebesgue measure on . From the formula (1) and the definition of we conclude that the map is an isomorphism which respect to the measures and , then
By induction hypothesis almost everywhere on for almost everywhere which implies that almost everywhere on by using the theorem of Fubini.
- Some orbit is not satured with respect to That is, for some . For , we choose an extension defined by . From this we have
Then for almost everywhere , is finite. By inductive hypothesis almost everywhere on for almost everywhere in , which implies that almost everywhere on by Fubini's theorem.
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University of Mohammed premier,
Recibido: 13 de febrero de 2006
Aceptado: 20 de julio de 2007