A Group of Automorphisms of the Rooted Dyadic Tree and Associated Gelfand Pairs

A Group of Automorphisms of the Rooted Dyadic Tree and Associated Gelfand Pairs

DANIELED'ANGELI(*) - ALFREDODONNO(*)

ABSTRACT- Inthis paper we study the groupGof automorphisms of the rooted dyadic tree generated by translation by 1 and multiplication by an odd integerq, showing thatGadmits a self-similar presentation and it is isomorphic to the Baumslag-Solitar groupBSq. Moreover, we show that the actionofGoneach level of the tree gives rise to a Gelfand pair.

1. Introduction.

LetTbe the infinite binary rooted tree. Denote byL_nthen-th level of T, which consists of 2ⁿ vertices. The root ofT canbe identified with the groupZ, each vertex, say at levelLn, canbe regarded as a coset of 2ⁿZin Z. Finally, the boundary@Tcorresponds to the ring of dyadic integersZ₂ (for more details see [F]).

LetGbe the group of automorphisms ofTgenerated by the translation by 1 and by the multiplication by an odd integer q for each vertex inT.

Denote byaandbsuch automorphisms, respectively. The actionofGonT isself-similar: we will directly prove that these automorphisms admit the following form

aˆ(1;a)e; bˆ(b;ba^h);

withqˆ2h‡1, according to the fact that every self-similar group can be embedded ina wreath product. By using the self-similarity, we deduce that G is isomorphic to the Baumslag-Solitar group BSqˆ hs;t:t ¹stˆs^qi, introduced in [BS]. We mention that self-similar groups are closely con-

(*) Indirizzo degli A.: Section de MatheÂmatiques, Universite de GeneÁve, 2-4, Rue du LieÁvre, Case Postale 64, 1211 GeneÁve 4, Suisse.

E-mail: daniele.dangeli@unige.ch alfredo.donno@unige.ch

nected with automata. For example, Bartholdi and SÏunik in [BSÏ] obtained the same result in the more general context of a family of solvable groups generated by finite automata.

LetG_n be the finite homomorphic image ofGacting faithfully onL_n. Also denote byK_nG_n the parabolic subgroup which stabilizes a fixed vertex in Ln. This way, Ln will be identified with the corresponding homogeneous spaceGn=Kn. We thenprove that (Gn;Kn) is a Gelfand pair for everyn1. In particular, we show, by direct computations involving characters, that the decomposition of the corresponding permutation representation into irreducible G representations is multiplicity-free.

Actually the result can also be obtained from the general theory of re- presentations of semidirect products developed in [CST]. Finally, a for- mula for the relative spherical functions is given.

The idea for this work has beeninspired by Professor A. FigaÁ-Tala- manca whom we thank for his interest and continuous encouragement.

2. Preliminaries.

LetTbe the binary infinite rooted homogeneous tree, that is the tree in which each vertex has two children. For everyn1, denote byLnthen-th level ofT, formed by 2ⁿ vertices. Identify each vertex ofLn with a word

x₀. . .x_{n 1} of lengthninthe alphabetXˆ f0;1g. LetX be the set of all

words of finite length in the alphabetX. The root will be represented by the empty word ;. We will denote by Tx the subtree of T rooted at the vertexxand isomorphic toT.

The setLn can be endowed with an ultrametric distanced, defined in the following way: ifxˆx₀. . .x_{n 1}andyˆy₀. . .y_{n 1}, then

d(x;y)ˆn maxfi:xkˆyk; 8kig:

We observe thatdˆd⁰=2, whered⁰denotes the usual geodesic distance onT.

Inthis way (Ln;d) becomes anultrametric space, inparticular a metric space, onwhich the automorphisms groupAut(T) acts isometrically. Note that the diameter of (L_n;d) is exactlyn.

Each automorphism g2Aut(T) canbe represented by its labelling.

The labelling of g2Aut(T) canbe realized as follows: givena vertex

xˆx0. . .xn 12T, we associate withxa permutationgx2S2(S2denotes

the symmetric group on2 elements) that gives the actionofgonthe two

childrenofx. Formally, the actionofgonthe vertex labelled with the word xˆx0. . .xn 1 is

x^gˆx^g₀^;x^g₁^x0. . .x^g_n^x0...₁^{xn 2}:

It will be used alsog(x) to indicate the elementx^g. LetGAut(T). We define the stabilizer of x2T as Stab_G(x)ˆ fg2G:g(x)ˆxg and the stabilizer of the n-th level as Stab_G(n)ˆT

x2LnStab_G(x). Observe that StabG(n) is a normal subgroup ofGof finite index for alln1.

Anautomorphism g2Stab_G(1) canbe identified with the elements g_i;iˆ0;1 that describe the actionofgonthe respective subtreesT_i. So we get the following embedding

W:StabG(1) !Aut(T)Aut(T) that associates togthe pair (g₀;g₁).

The following definitions can be found in [BGSÏ].

Gisspherically transitive if the actionofGonL_n is transitive for everyn.

Inthis case the subgroupsStab_G(x);x2L_n are conjugate each other.

GisfractalifStab_G(x)j_TxG, for eachx2T. This is equivalent to require that the mapW:Stab_G(1) !GGis a subdirect embedding, that is, surjective oneach factor.

Clearly, a fractal group is spherically transitive if and only if it is transitive on the first level ofT.

Gisself-similarif for everyg2G,x2X, there existsgx2G,x⁰2X such thatg(xw)ˆx⁰gx(w) for allw2X.

A self-similar group G canbe embedded inthe wreath product GoS2ˆG^2jS2, so that anelementg2Gis represented asgˆ(g0;g1)s, whereg_i2Gdescribe the actiononthe subtreeT_i, an ds2S₂corresponds to the actiononL₁. With this notation, the product between two elements gˆ(g0;g1)sandhˆ(h0;h1)tis

ghˆ(g₀;g₁)(h₀;h₁)^sstˆ(g₀h_s ¹₍₀₎;g₁h_s¹₍₁₎)st:

Observe that inthe product gh the automorphism g acts before the automorphismh, to have coherence with the wreath recursion. Consider again the binary tree but from another point of view. Observe that the

infinite sequence x0x1x2. . . canbe regarded as the dyadic number x02⁰‡x12‡x22²‡. . ..

Let us identify the root of the tree with the ring Z, the vertices of the first level with 2Z and 2Z‡1 respectively. Again, the sons of 2Z are 4Zand 4Z‡2, whereas the sons of 2Z‡1 are 4Z‡1 an d 4Z‡3.

Actually, the 2ⁿvertices of then-th level are, for alln1, the cosets of 2ⁿZ inZ.

Inthis way the boundary@T of the tree is identified with the dyadic closure ofZ, which will be denoted byZ₂.

Fig. 1. ± The dyadic tree.

3. Some automorphisms of the dyadic tree.

3.1 ±The automorphism a.

Under this identification of the boundary ofTwith the ringZ2of dyadic integers, we introduce the automorphismaofTgivenby the translationby 1. Actually, this is the automorphism that generates the Adding Machine (see, for instance, [BGN]). If we consider the action ofaonthe levelL_n, thenthe order of a is 2ⁿ. Therefore a admits the following self-similar

form:

aˆ(1;a)e;

…1†

whereeis the nontrivial permutation ofS₂.

The labelling ofais giveninthe following figure.

Fig. 2. ± Labelling ofa.

It is clear that the actionofafixes the root and so it preserves each level Ln, onwhich it acts transitively for everyn1.

Let us verify, by induction onn, that the actionof the automorphisma givenin(1) is the translationby 1. Onthe first level the actionofais ob- viously transitive, since it exchanges the vertices corresponding to the cosets 2Zand 2Z‡1, that means to perform the sum of 1 modulo 2Z.

Assume the claim true forn 1 and let us prove it forn. First consider anelement of the form 2ⁿZ‡k, withkeven. From the labelling of a it follows that the first letter 0 is changed in 1 (that corresponds to add 1), thenaacts as the identity. We have:

2ⁿZ‡k7 !^e 2ⁿZ‡k‡17 !¹ 2ⁿZ‡k7 !^:2 2^{n 1}Z‡k 2 7 !^id 2^{n 1}Z‡k

27 !² 2ⁿZ‡k7 !^‡1 2ⁿZ‡k‡1:

Infact, after applyinge, the element 2ⁿZ‡kis inthe subtreeT1. The translation by 1 and the following division by 2 are made to identify the subtree T₁ with the whole tree T. Now, a acts trivially. Finally, multi-

plication by 2 and translation by 1 give the image of the element 2ⁿZ‡kin T1, which is 2ⁿZ‡k‡1, as desired.

A similar argument can be developed whenaacts onanelement of the form 2ⁿZ‡k, withkodd. First we have the actionofe, which is the sum by

1, thenagainthe actionofa. We have:

2ⁿZ‡k7 !^e 2ⁿZ‡k 17 !^:2 2^{n 1}Z‡k 1 2 7 !^a 2^{n 1}Z‡k 1

2 ‡1ˆ2^{n 1}Z‡k‡1

2 7 !² 2ⁿZ‡k‡1:

Inthis case, after applyinge, the element 2ⁿZ‡kis inthe subtreeT₀. The divisionby 2 is made to identifyT₀with the whole treeT. Nowaacts againasa. By induction, this action is the translation by 1. Finally, mul- tiplicationby 2 gives the image of the element 2ⁿZ‡k in T0, which is 2ⁿZ‡k‡1.

3.2 ±The automorphism b.

Let us introduce the automorphismbofTgivenby multiplicationby an odd numberqˆ2h‡1 different from 1. Sinceqis odd,bpreserves each level and in particular it fixes the vertex 2ⁿZfor everyn1. Moreoverq has dyadic norm 1, so it is invertible in the ring of dyadic integers, but its inverse does not belong to Z (except inthe case qˆ 1). So, it is not possible to identify the inverse of this automorphism with the multi- plication by an integer. This is clear by considering that the order of the multiplicationbyqonL_n is not the same for alln1.

OnL₁ multiplicationbyqis trivial.

OnL₂ we haveb²ˆ1 for eachqodd.

OnL3 we still haveb²ˆ1 for eachqodd. This is true because 8j(q² 1) for each qodd.

To compute the order of the multiplicationbyqoneach levelLn, with n3, we have to study the equation q^2t 1(2ⁿ). Consider the following decomposition:

q^2t 1ˆ(q^{2t 1}‡1)(q^{2t 2}‡1)(q^{2t 3}‡1) (q²‡1)(q‡1)(q 1):

Since q²1(4) for every q odd, we get that each of the factors above, except the last two, are divisible by 2 but are not divisible by higher powers of 2. Moreover, the fact that 8j(q² 1) for eachqodd, implies that for all

n3 we have 2ⁿjq²ⁿ² 1, and so q^{2n 2}1(2ⁿ) for all n3. Actually, there exists someqwhose order, at the levelLn, is strictly less then2^{n 2}. Infact, from the previous argument it follows that if

q²1(2^k) but q²61(2^k‡1);

then, at the levelLn, the order of multiplicationbyqis exactly 2^{n k‡1}, for all nk. Also, we have k3. To describe the actionof b onLn,n5k, we distinguish two cases.

If q1(4), then2^{k 1}jq 1 an d so bˆ1 oneach level L_n, n5k. If q 1(4), then2^{k 1}jq‡1 an d sob²ˆ1.

As anexample, let us consider the caseqˆ3. Nowq² 1 is divided by 8 but not by 16 and this implies that the order of the multiplication by 3 is 2^{n 2}onLnfor eachn3. InL1 one hasbˆ1, inL2 b²ˆ1.

We can now investigate the order of the orbits of vertices ofL_n under the actionofb.

Let us consider the sphere of center 2ⁿZand radiusrwith respect to the ultrametric distance already defined. It contains 2^{r 1}vertices forr1.

The first vertex from the left belonging to this sphere is the element 2ⁿZ‡2^{n r}. Assume that 2^lis the period of this element under the action of b, so that we get

2^{n r}q^2l 2^{n r}(2ⁿ):

This impliesq^2l 1(2^r). Therefore, if the order ofbonL_nis 2^{n k‡1}for every nk, it must be

lˆr k‡1;

with rk. So, the orbit of the element 2ⁿZ‡2^{n r} is exactly of length 2^{r k‡1}.

It is easy to verify that the remaining vertices ofL_n belonging to the sphere of radiusrhave the form

2ⁿZ‡2^{n r}‡t2^{n r‡1}; tˆ1;. . .;2^{r 1} 1:

We want to prove that they have the same order 2^{r k‡1}. Consider the equation

(2^{n r}‡t2^{n r‡1})q^2l 2^{n r}‡t2^{n r‡1}(2ⁿ):

Dividing by 2^{n r}, we get

q^2l‡2tq^2l 1‡2t(2^r);

from which

(q^2l 1)(2t‡1)0(2^r):

Since 2t‡1 is odd, we have againthe equationq^2l1(2^r) an d so lˆr k‡1, ifqhas order 2^{n k‡1}modulo 2ⁿ.

Hence, the sphere of radiusrdecomposes, under the multiplication by q, in 2^{k 2} orbits of length 2^{r k‡1}, for all rk. For r5k, from what ob- served, the sphere of radiusrdecomposes into orbits of length 1 or 2.

PROPOSITION3.1. The automorphismbadmits the self-similar repre- sentation

bˆ(b;ba^h);

where qˆ2h‡1.

PROOF. Let us prove it by induction onn.

Fornˆ1, multiplicationbyqcoincides with the identity, as in the self- similar form ofb.

Assume the result true forn 1 and let us show it forn. First consider anelement of the form 2ⁿZ‡k, withkeven. We have:

2ⁿZ‡k7 !^:2 2^{n 1}Z‡k

27 !^b 2^{n 1}Z‡qk

2 7 !² 2ⁿZ‡qk:

Infact the element 2ⁿZ‡k belongs to T0. Divisionby 2 is made to identify the subtreeT0 with the whole treeT. Now,bacts againasband this action is, by induction, the multiplication byq. Finally, multiplication by 2 gives the imagine of 2ⁿZ‡kinT₀, that is 2ⁿZ‡qk, as desired.

A similar argument can be developed whenbacts onanelement of the form 2ⁿZ‡k, withkodd. Inthis case, we have:

2ⁿZ‡k7 !¹ 2ⁿZ‡k 17 !^:2 2^{n 1}Z‡k 1

2 7 !^b 2^{n 1}Z‡q(k 1) 2 7 !^ah 2^{n 1}Z‡q(k 1)

2 ‡hˆ2^{n 1}Z‡q(k 1)

2 ‡q 1

2 ˆ2^{n 1}Z‡qk 1 2 7 !² 2ⁿZ‡qk 17 !^‡1 2ⁿZ‡qk:

Inthis case, the element 2ⁿZ‡kis inthe subtreeT1. The translation by 1 and the following division by 2 are made to identifyT1with the whole

treeT. Now,bacts asba^h. By induction, the action ofbis multiplicationby qand the following action ofa^his the translation byh. The multiplication by 2 and the sum of 1 give the final image of the element 2ⁿZ‡kinT₁, that

is 2ⁿZ‡qk. p

3.3 ±The group Gq.

It is possible to verify that the automorphism group GˆG_q of the dyadic tree generated by the translation by 1 and by the multiplication by anodd integerqˆ2h‡1 is a homomorphic image of the solvable Baum- slag-Solitar group

BSqˆ hs;t:t ¹stˆs^qi:

Infact, this relationholds for every levelL_n, as we are going to prove.

We have

aˆ(1;a)e; bˆ(b;ba^h); b ¹ˆ(b ¹;a ^hb ¹) and so

b ¹abˆ(a^h;a ^hb ¹ab)e:

Fornˆ1 the relationis satisfied for anyqodd, because both the elements b ¹abanda^qexchange the vertices inL1.

Suppose, now, that the result is true forn 1 and let us show it forn.

We have

b ¹abˆ(a^h;a ^hb ¹ab)eˆ(a^h;a ^ha^q)eˆ(a^h;a ^h‡2h‡1)eˆ

ˆ(a^h;a^h‡1)eˆa^2h‡1ˆa^q;

where the second equality follows by induction, and so the relation is proved.

Actually, the two groups coincide. In fact, for q6ˆ 1, the automor- phismsaandbthat we defined have infinite order. On the other hand, the following lemma holds.

LEMMA 3.2. Let BSqˆ hs;t:t ¹stˆs^qithe Baumslag-Solitar group.

Then the order of at least one of s or t in any proper homomorphic image of BS_qmust be finite.

PROOF. Consider the factor group BSq=N, where N is any proper normal subgroup ofBSq. Ifs^kort^hbelong toNfor somek;hnot trivial, then the claim is true. Using the following generalized relations

t ^kst^kˆs^qk; t ¹s^ktˆs^kq; t ^ks^mt^kˆs^mqk

and suitable conjugations it is possible to write any word with occurrences of bothsandtas a word of the forms^mt^k. Suppose that such anelement belongs to N, this implies that t ^ks^mt^2kˆs^mqkt^k2N. So the element s^mt^k(s^mqkt^k) ¹ˆs^m(1‡qk)is inNand this completes the proof. p

So we get the following

THEOREM3.3. The automorphisms group GˆG_q of the dyadic tree generated by the automorphism a which is the sum of 1 and the auto- morphism b which is the multiplication by q is isomorphic to the solvable Baumslag-Solitar group:

BS_qˆ ha;b:b ¹abˆa^qi:

Note that the groupGis a self-similar group whose actiononthe treeT is spherically transitive. Since

a²ˆ(a;a); bˆ(b;ba^h); ba ^2hˆ(ba ^h;b);

Gis a fractal group.

REMARK 3.4. The caseqˆ 1 yields the infinite dihedral group. In fact, the group we get is

Gˆ ha;b:b ¹abˆa ¹i;

and this is exactly the presentation of the infinite dihedral group (see, for example, [BGN]). It is obvious that inthis case we have b²ˆ1 and the generators forGare

aˆ(1;a)e; bˆ(b;ba ¹):

REMARK3.5. In[BSÏ] L. Bartholdi and Z. SÏunik obtained these groups of automorphisms as groups generated by the automatonSm;ninthe more general case of the multiplication byminthe ringZnofn-adic integers, with (m;n)ˆ1.

4. Gelfand pairs 4.1 ±Some definitions

We present now some basic elements of the theory of finite Gelfand pairs that will be applied inwhat follows (see, for example, [CST1], [CST2]

and [D] for many applications).

Let G be a finite group and let KG be a subgroup, denote XˆG=Kˆ fgK:g2Gg the associated homogeneous space. In this way Gacts transitively onXandKis the stabilizer of the elementx0K2X.

The space L(G)ˆ ff :G !Cg is analgebra with respect to the con- volution

(f₁f₂)(g)ˆX

h2G

f₁(gh)f₂(h ¹);

for all f1;f22L(G) an d g2G. The subspace of L(G) consisting of the functionsK-invariant to the right (i.e.f(gk)ˆf(g) for allg2G,k2K) is a subalgebra that canbe identified withL(X)ˆ ff :X !Cg.

This subalgebra canbe endowed with a Hilbert space structure by the scalar product hf₁;f₂i ˆ P

x2Xf₁(x)f₂(x), for all f₁;f₂2L(X). Analo- gously, the subspace ofL(X) consisting of the functionsK-invariant is a subalgebra that we identify with L(KnG=K)ˆ ff :G !C:f(kgk⁰)ˆ

ˆf(g); for allk;k⁰2K;g2Gg. The group G acts on L(X) in the fol- lowing way: gf(x)ˆf^g(x)ˆf(g ¹x).

We will call (G;K) aGelfand pairif the algebraL(KnG=K) is commu- tative. The following are equivalent:

(1) (G;K) is a Gelfand pair;

(2) the decompositionof the spaceL(X) into irreducible submodules under the action of Gis multiplicity-free, i.e. each irreducible submodule occurs with multiplicity 1;

(3) givenanirreducible representationVofG, the dimension of the subspace ofK invariant vectorsV^Kˆ fv2V:kvˆv 8k2Kgis less or equal to 1, and it is 1 if and only ifV L(X).

The reader is referred to [CST1] for details.

A particular example of Gelfand pair is given by thesymmetric Gelfand pairs. A finite group G and a subgroup KG constitute a symmetric Gelfand pair if for everyg2Gthe conditiong ¹2KgKis satisfied. Infact one can directly verify that this condition implies that the algebra

L(KnG=K) is commutative. Moreover, it is easy to verify that the condition g ¹2KgKfor everyg2Gis equivalent to require that, for allx;y2X, the pairs (x;y) an d (y;x) are in the same orbit under the diagonal action ofGon XX.

For example, if we consider the finite dihedral groupD_nˆC_n^jC₂, it is easy to verify that (Dn;C2) is a symmetric Gelfand pair. The condition g ¹2C2gC2, for everyg2Dn, is trivially satisfied for this group.

EXAMPLE4.1. Suppose that the finite groupGacts ona finite metric space (X;d) isometrically and with a 2-points homogeneous action, i.e. in such a way that for all x;y;x⁰;y⁰2X such that d(x;y)ˆd(x⁰;y⁰) there existsg2Gsuch thatgxˆx⁰andgyˆy⁰. LetK Gbe the stabilizer of anelementx02X. It follows from the previous argument that (G;K) is a symmetric Gelfand pair.

We canobserve that inthis case theK-orbits (which canbe identified with the double cosets ofK inG) are the spheres

L_jˆ fx2X:d(x₀;x)ˆjg:

Hence, a functionf 2L(X) isK-invariant if and only if it is constant on the spheresL_j.

If (G;K) is a Gelfand pair andL(X)ˆLⁿ

iˆ0V_iis a decompositionofL(X) into irreducible submodules, then for each iˆ0;1;. . .;n there exists a unique (up to normalization) bi-K-invariant functionf_iwhoseG-translates generate theV_i's. In particular, we will require that these functions take value exactly 1 onthe element x₀2X stabilized by K. The functions f_i,

iˆ0;1;. . .;nare calledspherical functionsand they form a basis for the

algebraL(KnG=K). So, the number ofK-orbits under the action ofGonX is equal to the number of spherical functions. A different basis for the al- gebraL(KnG=K) is given by the characteristic functions of theK-orbits.

A spherical functionfcanbe also defined as a bi-K-invariant function onGsatisfying the following properties:

(1) ff ˆ((ff)(1_G))ffor everyf 2L(KnG=K);

(2) f(1G)ˆ1.

As anexample, the functionf₀1 is a spherical function: this corre- sponds to the fact that the trivial representation always occurs in the de- compositionof the spaceL(X) into irreducible submodules.

4.2 ±Gelfand pairs associated with G.

From now on, we will apply the theory of Gelfand pairs to the groups we introduced in Section 3.

Letqˆ2h‡1. Then, the multiplication byqhas order 2^{n k‡1}for each levelLn,nk. Withnfixed, the groupGnˆG=StabG(n) which acts onthe n-th level ofT, is givenby

GnˆC2^njC₂^{n k‡1};

where hai ˆC₂ⁿ and hbi ˆC₂n k‡1, since a²ⁿ;b^{2n k‡1} 2Stab_G(n), and the actionis defined asb ¹abˆa^q. Observe thatC₂ⁿ acts transitively onL_n. MoreoverC₂^{n k‡1} ˆStabGn(2ⁿZ) for allnk. This parabolic subgroup will be denoted byKn.

We want to show that (G_n;K_n) are Gelfand pairs. In order to prove it we use the characterizationof Gelfand pairs about the decompositionof the space L(L_n) into irreducible submodules, by showing that this decom- positionismultiplicity-free.

We distinguish two cases.

CASEq1(4).

For everyn5kwe haveG_n ˆC₂ⁿ,K_nˆ f1g and so in the decomposi- tionof the spaceL(L_n) all the irreducible representations (of dimension 1) ofC2ⁿoccur with multiplicity 1. The pairs (C2ⁿ;1) are not symmetric (except inthe casenˆ1) Gelfand pairs.

Let nownk. We have

G_nˆC₂^njC₂n k‡1:

The spaceL(Ln) decomposes under the action ofC2ⁿas L(Ln)ˆ²M^{n 1}

jˆ0

V_j; …2†

where the representationV_jis associated with the characterx_j defined by x_j(a)ˆv^j, withvˆe^2pi2n. We want to study how the automorphismbacts on theVj's. By using the relations holding in the group we get

a(bV_j)ˆba^qV_jˆv^qjbV_j

and sobVj ˆVqj. We want to understand which of these subspaces are in- variant under this action. The equation qjj(2ⁿ) gives j(q 1)0(2ⁿ), that impliesj0(2^{n k‡1}). TheV_j's such thatjsatisfies this property are exactly 2^{k 1}and they correspond to the vertices of the spheres of radiusr,

withr5k. They are infact irreducible submodules of dimension1 inthe representation ofGnonL(Ln). More generally theVj's such thatj0(2^{n r}) butj60(2^{n r‡1}) correspond to the vertices of the sphere of radiusrinLn

and in fact they are 2^{r 1} innumber. They are the V_j's with jˆ2^{n r}‡t2^{n r‡1},tˆ0;. . .;2^{r 1} 1. Since we know that the order ofbon the sphere of radius ris 2^{r k‡1}, we deduce that theseVj's will form 2^{k 2} irreducible submodules forGnof dimension 2^{r k‡1}. As anexample, the orbit

V₂^{n r}7 !^b V_q2^{n r}7 !^b V_q22^{n r}7 !^b . . .7 !^b V_q2r k‡1 12^{n r}

generates the irreducible representation forGn

V2^{n r}Vq2^{n r}V_q²₂^{n r}. . .V_q2r k‡1 12^{n r}: The matrices corresponding to this representation are

a7 !

v^{2n r} 0 0

0 v^{q2n r} ...

... ...

0 0 v^{q2r k‡}0^{1 12n r} 0

BB B@

1 CC CA;

b7 !

0 0 1

1 0 0

0 ... ... ...

0 0 1 0

0 BB

@

1 CC A:

Let us callcthe character of this representation. We have c(a)ˆx₂n r(a)‡x_q2n r(a)‡ ‡x_q2r k‡1 12^{n r}(a)ˆ0;

since the 2ⁿ-th roots of 1 which occur inthe sum are pairwise opposite.

Moreoverc(b)ˆ0. Denote byc_d,dˆ1;. . .;2^{k 2}the characters of the 2^{k 2} irreducible representations associated with the sphere of radiusr. Inorder to prove that they are pairwise non isomorphic, we present a suitable element ofG_nonwhich the charactersc_dtake different values, for eachd.

Let us consider the elementa^{2r k‡1}2G_n. We observe that (v^{2n r})^{2r k‡1}ˆv^{2n k‡1}ˆe^{2pi2n k‡12n} :

Moreover we have 2^{n k‡1}q2^{n k‡1}(2ⁿ), since 1q(2^{k 1}). The same computationholds for the following powers ofq. Hence, all the powers ofv which occur inthe diagonal of the matrix associated witha^{r k‡1}correspond to the same angle2p2^{n k‡1}

2ⁿ ˆ 2p 2^{k 1}.

Let us consider now any element 2^{n r}‡t2^{n r‡1} inthe sphere of radius r. We have

(v^{2n r‡t2n r‡1})^{2r k‡1} ˆv^{2n k‡1‡t2n k‡2} and

q(2^{n r}‡t2^{n r‡1})2^{r k‡1}2^{n k‡1}‡t2^{n k‡2}(2ⁿ);

as one can easily prove by using the property q1(2^{k 1}). We want to study under which condition the elements 2^{n r}‡t2^{n r‡1} and 2^{n r}‡s2^{n r‡1} belong to the same orbit under the multiplication by q.

We get the equation

2^{n k‡1}‡t2^{n k‡2}2^{n k‡1}‡s2^{n k‡2}(2ⁿ);

which gives ts(2^{k 2}).

So, the elements

V₂^{n r};V₂n r‡2^{n r‡1};V₂n r‡22^{n r‡1};. . .;V₂n r‡(2^k2 1)2^{n r‡1}

belong to the 2^{k 2}different orbits associated to the sphere of radiusr. The angles corresponding to the respective powers ofvare

2p

2^{k 1}‡2u 2p

2^{k 1}; uˆ0;1;. . .;2^{k 2} 1:

Foruˆ2^{k 2} 1 we get the angle 2p

2^{k 1} and so the angles associated with these orbits are pairwise different.

We have shownthat the charactersc_dtake different values on the same elementa^{2r k‡1}and so the associated representations are not isomorphic, as required.

CASEq 1(4).

Fornˆ1 the groupG1is the cyclic groupC2and we get the symmetric Gelfand pair (C2;1).

For 2nk 1 we haveb²ˆ1 an d so G_n ˆC₂^njC₂;

which is the dihedral group of order 2^n‡1. The pairs (Gn;Kn) are symmetric Gelfand pairs (as we observed above).

Let nk. Let the V_j's be the spaces givenin(2), with the relation bV_j ˆV_qj. In this case the subspaces which are invariant under the action ofbareV₀andV₂n1, as one can get from the equationqjj(2ⁿ) using that

q1(2) butq61(4). The spheres of radius 15r5kdecompose into 2^{r 2} irreducible submodules of dimension 2 and each of them is the direct sum V_jV _j, withj0(2^{n r}) butj60(2^{n r‡1}) and they are non isomorphic.

Letrk. We distinguish two subcases.

Let qˆ4h 1 with hˆ2x‡1. Inthis case necessarily kˆ3, since (q 1)(q‡1)ˆ8(2x‡1)(4x‡1), and so the sphere of radius r decom- poses exactly into two orbits under the action ofb. Analogously to the case q1(4), we want to write a certain power ofaonwhich the characters of these two representations take different values to deduce that they are non isomorphic.

Let us consider the elementa^{2r 3}. We observe that (v^{2n r})^{2r 3}ˆv²ⁿ³ ˆe^2pi22nn3ˆe^ip4: Moreover

(v^{q2n r})^{2r 3}ˆv^q2n3ˆe^2piq22nn3 ˆe^iqp4: Nowqp

4 ˆ4hp 4

p

4p p

4(2p), sincehis odd. The same holds for the fol- lowing powers ofq.

Viceversa, the elementv ^{2n r} raised to the power 2^{r 3} gives the angle p

4. As before, one can show that the elementv ^{q2n r} raised to the same power 2^{r 3}gives the anglep‡p

4. The same argument can be developed for the following powers ofq. It is clear that the characters associated with these two representations take opposite nonzero values on the element a^{2r 3}2G_n and this completes the proof in this first case.

Now letqˆ4h 1, withhˆ2x. Observe that q² 1ˆ(4h 2)(4h)ˆ8h(2h 1)

and so 2^{k 3}jh. Inthis case we consider the elementa^{2r k}. We have (v^{2n r})^{2r k} ˆv^{2n k} ˆe^{2pi22n kn} ˆe^2pi2k:

Moreover

(v^{q2n r})^{2r k}ˆv^{q2n k} ˆe^{2piq22nn k} ˆe^2pqi2k: Now observe that

2pq

2^k ˆ2p(4h 1)

2^k ˆ2p42^{k 3}m 2^k

2p

2^k ˆmp 2p

2^k p 2p 2^k(2p):

The last equality is true since the expression ofhimplies thatmis odd. A similar argument can be developed for the following powers ofq. So the angles2p

2^k andp 2p

2^k alternate in the orbit containingV₂^{n r}.

Consider now any element 2^{n r}‡t2^{n r‡1}, withtˆ0;. . .;2^{r 1} 1. Its 2^{r k}-th power gives anangle equal to2p

2^k if the equation 2^{n k}2^{n k}‡t2^{n k‡1}(2ⁿ);

which impliest0(2^{k 1}), is satisfied.

Onthe other hand the 2^{r k}-th power gives anangle equal top 2p 2^k if the equation

2^{n k}‡2^{n k}‡t2^{n k‡1}2^{n 1}(2ⁿ);

which gives 1‡t2^{k 2}(2^{k 1}), is satisfied.

The values of t modulo 2^{r 1} that satisfy these two equations are 22^{r 1}

2^{k 1} ˆ2^{r k‡1}, as many as the order of an orbit. So the elements whose 2^{r k}th power gives 2p

2^k or p 2p

2^k are exactly those which satisfy the two equations.

Consider now the two equations

2^{n k}‡t2^{n k‡1}2^{n k}‡s2^{n k‡1}(2ⁿ) and

2^{n k}‡t2^{n k‡1}‡2^{n k}‡s2^{n k‡1}(2ⁿ)2^{n 1}(2ⁿ):

Their solutions are, respectively,

ts(2^{k 1}) an d t2^{k 2} 1 s(2^{k 1}):

So, if we fix a value oftmodulo 2^{r 1}not satisfying the first two equations, it is possible to find, by using these two new equations, 2^{r k‡1} values which will form a new orbit, whose corresponding angle will be different from 2p

2^k andp 2p 2^k.

This way we get 2^{k 2}orbits inthe decompositionof the sphere of radius r, all of them corresponding to different angles and so the corresponding 2^{k 2} submodules have characters which take different values ona^{2r k} and so they are non isomorphic.

Consider now the general case of any oddqto show that the Gelfand pairs considered in this paper are non symmetric for every nk. We

recall that the pair (Gn;Kn) is symmetric if the conditiong ¹2KngKn is satisfied for everyg2Gn.

Let us choosegˆaand considerKnˆ hb:b^{2n k‡1}ˆ1i. We canrestrict our attentiononthe elements of the formb ⁱabⁱ. Infact the elements of the form bⁱab^j with i6ˆ j give elements with occurrences of both a and b which are clearly different froma ¹. Sinceb ⁱabⁱˆa^qi, we have to study the equation

qⁱ 1(2ⁿ);

…3†

withiˆ0;. . .;2^{n k‡1} 1. It is clear that inthe caseiˆ0 the equationis

satisfied only in the trivial casenˆ1, whenG_nˆC₂. Also inthe case al- ready developedqˆ 1 this equationis satisfied byiˆ1.

Consider now the equation (3). If 2ⁿ divides qⁱ‡1, then2^n‡1 divides (qⁱ‡1)(qⁱ 1)ˆq²ⁱ 1. So one has q²ⁱ1(2^n‡1). The smallest nonzero exponent satisfying this equation is, as we know, 2^{n k‡2}. But it must be 2i2^{n k‡2} 2. This completes the proof. Collecting all the results ob- tained in this section we have the following

THEOREM 4.2. Let q be an odd integer such that q²1(2^k) but q²61(2^k‡1), then, for every nk, GnˆG=Stab_G(n)ˆC2^njC₂^{n k‡1} and KnˆStab_Gn(2ⁿZ)ˆC₂^{n k‡1}. The associated pairs (Gn;Kn) are non sym- metric Gelfand pairs.

For n5k and q1(4) we get G_n ˆC₂ⁿ and K_nˆ f1g, which are non symmetric (except in the case nˆ1) Gelfand pairs.

For15n5k and q 1(4) we get GnˆD2ⁿ and KnˆC2; which are symmetric Gelfand pairs. For nˆ1we have the symmetric Gelfand pair (C₂;1).

4.3 ±Final remarks.

In[CST] the authors show ina different way that the pairs (G_n;K_n) introduced above are Gelfand pairs by using the general theory of the Gelfand pairs associated to the semidirect products.

Let GˆN^jHbe a finite group and let KN be a subgroup of N invariant under the action ofHand such that (N;K) is a Gelfand pair. Set XˆN=K and let L(X)ˆLⁿ

iˆ0

V_i be the multiplicity-free decompositionof L(X) into irreducibleN submodules and letf_i2Vibe the corresponding spherical functions. The mapnK7 !nKHis a bijectionbetweenN=Kand

G=KH. The actionof G on XˆG=KH canbe defined as nh(n⁰KH)ˆ

ˆnhn⁰h ¹KHand the induced action onL(X) is, as usual,f^g(x)ˆf(g ¹x).

It is easy to check that ifV_iis any irreducibleN-invariant submodule in L(X), thenhV_iis stillN-invariant and irreducible, for everyh2H, inother wordsHpermutes theV_i's.

So the decompositionof L(X) under the action of G into irreducible submodules is

L(X)ˆM^r

jˆ0

Wj; …4†

where Wj ˆ L

i:Vi2Gj

Vi and Gj denotes, for jˆ0;. . .;r, the H-orbits on

fV₀;V₁;. . .;Vng. The W_j's are pairwise non isomorphic because the re-

strictions toNof the representationsW_j andW_j⁰ decompose into non iso- morphic submodules forj6ˆj⁰.

Hence the decomposition (4) is multiplicity-free and this implies that (G;KH) is a Gelfand pair. The corresponding spherical functions can be easily computed from thef_i's and they are given by

Fjˆ 1 jG_jj

X

i:Vi2Gj

f_iˆ 1 jHj

X

h2H

hf_i:

Inthe case of the groups introduced above we haveNˆC2ⁿ,Kˆ f1gand HˆC₂n k‡1. TheV_i's are the submodules inwhich decomposes the space L(L_n) under the action ofC₂ⁿ whose spherical functions are exactly the characters of C2ⁿ. The Wj's are the orbits of the spaces Vi's under the automorphismbwithbV_iˆV_qi.

The previous argument guarantees that (C₂^njC₂^{n k‡1};C₂^{n k‡1}) is a Gel- fand pair and yields the associated spherical functions.

InSection4.2 we preferred to give anexplicit calculationof the irre- ducible subspaces ofL(Ln) under the action ofGnusing only congruences modulo 2ⁿand identifying the spacesVj's with the vertices of then-th level of the dyadic tree.

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Manoscritto pervenuto in redazione il 7 luglio 2007.