Orbifold Lens Spaces that are Isospectral but not Isometric
aa r X i v : . [ m a t h . SP ] A ug ORBIFOLD LENS SPACES THAT ARE ISOSPECTRAL BUT NOTISOMETRIC
NAVEED SHAMS UL BARIA
BSTRACT . We answer Mark Kac’s famous question [K], “can one hear theshape of a drum?” in the negative for orbifolds that are spherical space forms.This is done by extending the techniques developed by A. Ikeda on Lens Spacesto the orbifold setting. Several results are proved to show that with certain re-strictions on the dimensionalities of orbifold Lens spaces we can obtain infin-itely many pairs of isospectral non-isometric Lens spaces. These results are thengeneralized to show that for any dimension greater than 8 we can have pairs ofisospectral non-isometric orbifold Lens spaces. C ONTENTS
1. Introduction 11.1. Acknowledgments 32. Orbifold Lens Spaces 32.1. Preliminaries 42.2. Orbifold Lens Spaces and their Generating Functions 83. Isospectral Non-isometric Lens Spaces 133.1. Odd-Dimensional Lens Spaces 133.2. Lens Spaces for General Integers 234. Examples 274.1. Examples for k = 2 k = 3 NTRODUCTION
Given a closed Riemannian manifold ( M, g ) , the eigenvalue spectrum of the as-sociated Laplace Beltrami operator will be referred to as the spectrum of ( M, g ) .The inverse spectral problem asks the extent to which the spectrum encodes thegeometry of ( M, g ) . While various geometric invariants such as dimension, vol-ume and total scalar curvature are spectrally determined, numerous examples ofisospectral Riemannian manifolds, i.e., manifolds with the same spectrum, show Keywords:
Spectral geometry Global Riemannian geometry Orbifolds Lens Spaces.2000
Mathematics Subject Classification:
Primary 58J53; Secondary 53C20. hat the spectrum does not fully encode the geometry. Not surprisingly, the earliestexamples of isospectral manifolds were manifolds of constant curvature includingflat tori ([M]), hyperbolic manifolds ([V]), and spherical space forms ([I1], [I2] and[Gi]). In particular, Lens spaces are quotients of round spheres by cyclic groupsof orthogonal transformations that act freely on the sphere. Lens spaces have pro-vided a rich source of isospectral manifolds with interesting properties. In additionto the work of Ikeda cited above, see the recent results of Gornet and McGowan[GoM].In this paper we generalize this theme to the category of Riemannian Orbifolds.A smooth orbifold is a topological space that is locally modelled on an orbitspace of R n under the action of a finite group of diffeomorphisms. Riemannian orbifolds are spaces that are locally modelled on quotients of Riemannian mani-folds by finite groups of isometries. Orbifolds have wide applicability, for example,in the study of 3-manifolds and in string theory.The tools of spectral geometry can be transferred to the setting of Riemannianorbifolds by using their well-behaved local structure (see [Chi], [S1] [S2]). As inthe manifold setting, the spectrum of the Laplace operator of a compact Riemann-ian orbifold is a sequence ≤ λ ≤ λ ≤ λ ≤ . . . ↑ ∞ where each eigenvalue isrepeated according to its finite multiplicity. We say that two orbifolds are isospec-tral if their Laplace spectra agree.The literature on inverse spectral problems on orbifolds is less developed thanthat for manifolds. Examples of isospectral orbifolds include pairs with boundary([BCDS] and [BW]); isospectral flat 2-orbifolds ([DR]); arbitrarily large finite fam-ilies of isospectral orbifolds ([SSW]); isospectral orbifolds with different maximalisotropy orders ([RSW]); and isospectral deformation of metrics on an orbifoldquotient of a nilmanifold ([PS]).In this article, we study the spectrum of orbifold Lens spaces, i.e., quotientsof round spheres by cyclic groups of orthogonal transformations that have fixedpoints on the sphere. Generalizing the work of Ikeda (see [I1], [I2] and [IY]) weconstruct the generating function for the spectrum and systematically constructisospectral orbifold Lens spaces. Section two introduces the orbifold Lens spacesand their generating functions. In section 3, we will develop the proofs of our maintheorems. We will first prove: Theorem 3.1.6. (i)
Let p ≥ (alt. p ≥ ) be an odd prime and let m ≥ (alt. m ≥ ) be any positive integer. Let q = p m . Then there exist at least two ( q − -dimensional orbifold lens spaces with fundamental groups of order p m which are isospectral but not isometric. (ii) Let p , p be odd primes such that q = p · p ≥ . Then there exist at leasttwo ( q − -dimensional orbifold lens spaces with fundamental groups oforder p · p which are isospectral but not isometric. (iii) Let q = 2 m where m ≥ is any positive integer. Then there exist at leasttwo ( q − -dimensional orbifold lens spaces with fundamental groups oforder m which are isospectral but not isometric. iv) Let q = 2 p , where p is an odd prime and p ≥ . Then there exist at leasttwo ( q − -dimensional orbifold lens spaces with fundamental groups oforder p which are isospectral but not isometric. To prove these results we proceed as follows:(1) Depending on the number of p i (alt. p , p ) divisors of q = p m (alt. q = p · p ), we reformulate the generating function in terms of rationalpolynomial functions.(2) Then we classify the number of generating functions that we will get byimposing different conditions on the domain values of these polynomialfunctions.(3) We prove sufficiency conditions on the number of generating functions thatwould guarantee isospectrality for non-isometric orbifold lens spaces.The techniques used to prove these results parallel similar techniques from themanifold lens space setting used in [I1].Generalizing this technique, we will get our second set of main results: Theorem 3.2.5.
Let W ∈ { , , , . . . } . (i) Let P ≥ (alt. P ≥ ) be any odd prime and let m ≥ (alt. m ≥ ) beany positive integer. Let q = P m . Then there exist at least two ( q + W − -dimensional orbifold lens spaces with fundamental groups of order P m which are isospectral but not isometric. (ii) Let P , P be two odd primes such that q = P · P ≥ . Then there existat least two ( q + W − -dimensional orbifold lens spaces with fundamentalgroups of order P · P which are isospectral but not isometric. (iii) Let q = 2 m where m ≥ is any positive integer. Then there exist at leasttwo ( q + W − -dimensional orbifold lens spaces with fundamental groupsof order m which are isospectral but not isometric. (iv) Let q = 2 P , where P ≥ is an odd prime. Then there exist at least two ( q + W − -dimensional orbifold lens spaces with fundamental groups oforder P which are isospectral but not isometric. A consequence of this theorem is that for every integer x ≥ , we can find a pairof isospectral non-isometric orbifold lens spaces of dimension x .In Section 4 we look at specific examples that show what the different generatingfunctions would look like and the types of orbifold lens spaces that correspond toeach generating function.1.1. Acknowledgments.
I would like to thank Carolyn Gordon very much forsome very helpful discussions and suggestions, and for reading an early version.2. O
RBIFOLD L ENS S PACES
In this section we will generalize the idea of manifold Lens spaces to orb-ifold Lens spaces. Manifold Lens spaces are spherical space forms where the n -dimensional sphere S n of constant curvature is acted upon by a cyclic group offixed point free isometries on S n . We will generalize this notion to orbifolds byallowing the cyclic group of isometries to have fixed points. For a more general efinition of Orbifolds see Satake [Sat] and Scott [Sc]. For details of spectral ge-ometry on Orbifolds, see Stanhope [S1] and E. Dryden, C. Gordon, S. Greenwaldand D. Webb in [DGGW]).To obtain our main results we will focus on a special subfamily of Lens spaces.Our technique will parallel Ikeda’s technique as developed in [I1].2.1. Preliminaries.
Let q be a positive integer that is not prime. Set q = ( q − if q is odd q if q is even . Throughout this article we assume that q ≥ and that q is not prime.For any positive integer n with ≤ n ≤ q − , we denote by e I ( q, n ) the set of n -tuples ( p , . . . , p n ) of integers. We define a subset e I ( q, n ) of e I ( q, n ) as follows: e I ( q, n ) = n ( p , . . . , p n ) ∈ e I ( q, n ) p i
6≡ ± p j ( mod q ) , ≤ i < j ≤ n , g.c.d. ( p , . . . , p n , q ) = 1 o . We introduce an equivalence relation in e I ( q, n ) as follows: ( p , . . . , p n ) is equiv-alent to ( s , . . . , s n ) if and only if there is a number l prime to q and there are num-bers e i ∈ {− , } such that ( p , . . . , p n ) is a permutation of ( e ls , . . . , e n ls n ) ( mod q ) .This equivalence relation also defines an equivalence relation on e I ( q, n ) .We set I ( q, n ) = e I ( q, n ) / ∼ and I ( q, n ) = e I ( q, n ) / ∼ .Let k = q − n . We define a map w of I ( q, n ) into I ( q, k ) as follows:For any element ( p , . . . , p n ) ∈ e I ( q, n ) , we choose an element ( q , . . . , q k ) ∈ e I ( q, k ) such that the set of integers n p , − p , . . . , p n , − p n , q , − q , . . . , q k , − q k o forms a complete set of incongruent residues ( mod q ) . Then we define w ([ p , . . . , p n ]) = [ q , . . . , q k ] It is easy to see that this map is a well defined bijection.The following proposition is similar to a result in [I1]:P
ROPOSITION
Let I ( q, n ) be as above. Then, | I ( q, n ) | ≥ q (cid:18) q n (cid:19) where (cid:0) q n (cid:1) = 1 if q n = 0 , and (cid:18) q n (cid:19) = q ! n !( q − n )! otherwise.Proof. Let I ( q, n ) be as above. Consider a subset e I ′ ( q, n ) of e I ( q, n ) as follows: e I ′ ( q, n ) = n ( p , . . . , p n ) ∈ e I ( q, n ) at least one of the p i is co-prime to q o t is easy to see that the equivalence relation on e I ( q, n ) induces an equivalencerelation on e I ′ ( q, n ) . Since we eliminate classes where none of the p i ’s is co-primeto q , we get | I ( q, n ) | ≥ | I ′ ( q, n ) | where I ′ ( q, n ) = e I ′ ( q, n ) / ∼ . Now consider a subset e I ′′ ( q, n ) of e I ′ ( q, n ) asfollows: e I ′′ ( q, n ) = n ( p , . . . , p n ) ∈ e I ′ ( q, n ) p < · · · < p n ≤ q o Then it is easy to see that any element of e I ′ ( q, n ) has an equivalent element in e I ′′ ( q, n ) . On the other hand, for any equivalence class in I ′ ( q, n ) , the number ofelements in e I ′′ ( q, n ) which belong to that class is at most n . Hence we have: | I ( q, n ) | ≥ | I ′ ( q, n ) | ≥ n e I ′′ ( q, n ) = 1 n (cid:18) q − n − (cid:19) = 1 q (cid:18) q n (cid:19) This proves the proposition. (cid:3) L EMMA
Let q = p m or q = p · p , where p, p , p are primes. Let D be theset of all non-zero integers mod q that are not co-prime to q . Then | D | is even if q is odd and | D | is odd if q is even.Proof. For q = p m .If q is odd, then p is an odd prime. qp = p m − which is an odd number. Thereforethe number of elements in D , ( p m − − is even.If q is even, then p = 2 . qp = 2 m − is even. So the number of elements in D , (2 m − − , is odd.For q = p · p . ( p = p ) If q is odd, then both p and p are odd primes. The number of elements in D is ( qp + qp −
2) = ( p + p − which is even since p + p is even.If q is even, then one of the p i ’s is and the other is an odd prime. Assume p = 2 . So, the number of elements in D is ( qp + qp −
2) = ( p + p −
2) =( p + 2 −
2) = p , which is odd.This proves the lemma. (cid:3) We will say that | D | = 2 r if | D | is even; and | D | = 2 r − if | D | is odd, where r is some positive integer. It is easy to see that if | D | is even, then exactly r membersof D are less than q . If | D | is odd, then r − members of D are strictly less than q and one member of D is equal to q (recall that for even q , we set q = q/ , andfor odd q , we set q = ( q − / ).With these results we now obtain a better lower bound for | I ( q, n ) | .P ROPOSITION
Let I ( q, n ) , I ′ ( q, n ) , e I ′ ( q, n ) and e I ′′ ( q, n ) be as in Proposi-tion 2.1.1. Let k = q − n . Then | I ( q, n ) | ≥ r X t = u n − t (cid:18) q − − rn − − t (cid:19)(cid:18) rt (cid:19) here u = r − k if r > k and u = 0 if r ≤ k , and r is as defined above.Proof. The number of ways in which we can assign values to the p i ’s in (1 = p , p , . . . , p n ) ∈ e I ′′ ( q, n ) such that t of the p i ’s are not co-prime to q is (cid:18) q − − rn − − t (cid:19)(cid:18) rt (cid:19) On the other hand for any equivalence class in I ′ ( q, n ) with t of the p i ’s notbeing co-prime to q , the number of elements which belong to that class is at most n − t . So the number of such possible classes is at least n − t (cid:18) q − − rn − − t (cid:19)(cid:18) rt (cid:19) Now if r > k , this would mean that n > q − r , or n − > q − − r . Thismeans that t cannot take any values less than r − k , since that would mean that weare choosing ( n − − t ) , a number larger than ( q − − r ) from q − − r andthat is not possible. So, the smallest value for t in this case can be r − k .On the other hand, if r ≤ k , then n ≤ q − r , or n − ≤ q − − r . Thismeans that it is possible for us to choose n -tuples in e I ′′ ( q, n ) with all values beingco-prime to q . Thus, the smallest value for t would be in this case.It is obvious that the maximum value t can take is r since (1 , p , . . . , p n ) cannothave more than r values that are not co-prime to q . Now, adding up all the degreesfor different values of t we get | I ( q, n ) | ≥ | I ′ ( q.n ) | ≥ r X t = u n − t (cid:18) q − − rn − − t (cid:19)(cid:18) rt (cid:19) where u = 0 if r ≤ k and u = r − k if r > k .This proves the proposition. (cid:3) D EFINITION q be a positive integer and γ a primitive q -th root of . We denote by Q ( γ ) the q -th cyclotomic field over the rational numberfield Q and denote by Φ q ( z ) the q -th cyclotomic polynomial Φ q ( z ) = q − X t =0 z t Let A be the set of residues mod q that are co-prime to q . We define a map ψ q,k of I ( q, k ) into Q ( γ )[ z ] as follows:For any equivalence class in I ( q, k ) , we take an element ( q , . . . , q k ) of e I ( q, k ) which belongs to that class. We define ψ q,k ([ q , . . . , q k ])( z ) = X l ∈ A k Y i =1 ( z − γ q i l )( z − γ − q i l ) This polynomial in Q ( γ )[ z ] is independent of the choice of elements whichbelong to the class [ q , . . . , q k ] . Therefore, the map is well-defined. ii) Given q = p m , we define B j = (cid:8) x ( modq ) ∈ Z + : p j | x, p j +1 ∤ x (cid:9) We define the maps α ( j ) q,k of I ( q, k ) into Q ( γ )[ z ] as follows: α ( j ) q,k ([ q , . . . , q k ])( z ) = X l ∈ B j k Y i =1 ( z − γ q i l )( z − γ − q i l ) (iii) Now assume q = p · p . We define the following sets of numbers that arenot co-prime to q . B = (cid:8) xp x = 1 , , . . . , ( p − (cid:9) and C = (cid:8) xp x = 1 , , . . . , ( p − (cid:9) We define maps α q,k and β q,k as follows: α q,k ([ q . . . . , q k ])( z ) = X l ∈ B k Y i =1 ( z − γ q i l )( z − γ − q i l ) and, β q,k ([ q . . . . , q k ])( z ) = X l ∈ C k Y i =1 ( z − γ q i l )( z − γ − q i l ) Since ( z − γ q i l )( z − γ − q i l ) = ( γ q i l z − γ − q i l z − , the following propositionis easy to see.P ROPOSITION
If we put ψ q,k ([ q , . . . , q k ])( z ) = k X i =0 ( − i a i z k − i α ( j ) q,k ([ q , . . . , q k ])( z ) = k X i =0 ( − i b i,j z k − i α q,k ([ q , . . . , q k ])( z ) = k X i =0 ( − i b i z k − i β q,k ([ q , . . . , q k ])( z ) = k X i =0 ( − i c i z k − i then we have (i) a i = a k − i , b i,j = b (2 k − i ) ,j , b i = b k − i and c i = c k − i (ii) a = | A | , b ,j = | B j | , b = | B | and c = | C | . .2. Orbifold Lens Spaces and their Generating Functions.
Let q be a positiveinteger and p , . . . , p n be n integers mod q such that g.c.d. ( p , . . . , p n , q ) = 1 . Wedenote by g the orthogonal matrix given by g = R ( p /q ) . . . R ( p n /q ) where R ( θ ) = (cid:18) cos 2 πθ sin 2 πθ − sin 2 πθ cos 2 πθ (cid:19) Then g generates the cyclic subgroup G = (cid:8) g l (cid:9) ql =1 of order q of the orthogonal group O (2 n ) .We define a Lens space L ( q : p , . . . , p n ) as follows: L ( q : p , . . . , p n ) = S n − /GL ( q : p , . . . , p n ) is a good smooth orbifold with S n − as its covering manifold.Let π be the covering projection of S n − onto S n − /Gπ : S n − → S n − /G Since the round metric of constant curvature one on S n − is G -invariant, it in-duces a Riemannian metric on S n − /G .We emphasize that, in contrast to the classical notion of Lens space, we do notrequire that the integers p i be coprime to q , and thus we allow the Lens spaces tohave singular points. In particular they are good orbifolds. Henceforth, the term”Lens space” will refer to this generalized definition.P ROPOSITION
Let L = L ( q : p , . . . , p n ) and L ′ = L ( q : s ′ , . . . , s ′ n ) beLens spaces. Then L is isometric to L ′ if and only if there is a number l coprimewith q and there are numbers e i ∈ {− , } such that ( p , . . . , p n ) is a permutationof ( e ls , . . . , e n ls n ) (mod q ) .Proof. Assume L is isometric to L ′ . Let φ be an isometry between L and L ′ . Then φ lifts to an isometry ˜ φ of S n − . In other words, φ is an orthogonal transformationthat conjugates G and G ′ where L = S n − /G and L ′ = S n − /G ′ .So ˜ φ takes g , a generator of G , to g ′ l , a generator of G ′ . This means thatthe eigenvalues of g and g ′ l are the same. This means that each p i is equiva-lent to some ls j or − ls j (mod q ) . That means ( p , . . . , p n ) is a permutation of ( e ls , . . . , e n ls n ) (mod q ) , for e i ∈ {− , } , ( i = 1 , . . . , n ) .Conversely, assume that there exists an integer l coprime with q and num-bers e i ∈ {− , } ( i = 1 , . . . , n ) such that ( p , . . . , p n ) is a permutation of ( e ls , . . . , e n ls n ) (mod q ) .Note that the isometry of S n − onto S n − defined by the map ( z , . . . , z i , . . . , z n ) ( z σ (0) , . . . , ¯ z σ ( i ) , . . . , z σ ( n ) ) where σ is a permutation, induces an isometry of L ( q : s , . . . , s n ) onto L ( q : s σ (1) , . . . , s σ ( i ) , . . . , s σ ( n ) ) . Since g ′ l is also a generator of G ′ , the Lensspace L ( q : ls , . . . , ls n ) is identical to L ( q : s , . . . , s n ) . ow the above isometry induces an isometry of L ( q : s , . . . , s n ) onto L ( q : e ls , . . . , e n ls n ) . This means that L ′ = L ( q : s , . . . , s n ) is isometric to L ( q : e ls , . . . , e n ls n ) . But ( e ls , . . . , e n ls n ) is simply a permutation of ( p , . . . , p n )(mod q ) . Therefore L ( q : p , . . . , p n ) is isometric to L ( q : e ls , . . . , e n ls n ) ,which, in turn, is isometric to L ( q : s , . . . , s n ) . This proves the converse. (cid:3) For any f ∈ C ∞ ( L ( q : p , . . . , p n )) , we define the Lapacian on the lense spaceas e ∆( π ∗ f ) = π ∗ (∆ f ) . We construct the generating function associated with theLaplacian on L ( q : p , . . . , p n ) analogous to the construction in the manifold case(see [I1], [I2] and [IY]).Let ˜∆ , ∆ and ∆ denote the Laplacian of S n − , L ( q : p , . . . , p n ) and R n ,respectively.D EFINITION λ , we define the eigenspaces e E λ and E λ as follows: e E λ = (cid:8) f ∈ C ∞ ( S n − ) e ∆ f = λf (cid:9) E λ = (cid:8) f ∈ C ∞ ( L ( q : p , . . . , p n )) ∆ f = λf (cid:9) The following lemma follows from the definitions of ∆ and smooth function.L EMMA
Let (cid:0) e E λ (cid:1) G be the space of all G -invariant functions of e E λ . Then dim( E λ ) = dim( e E ) G . Let ∆ be the Laplacian on R n with respect to the flat K¨ahler metric. Set r = P ni =1 x i , where ( x , x , . . . , x n ) is the standard coordinate system on R n .For k ≥ , let P k denote the space of complex valued homogenous polynomi-als of degree k on R n . Let H k be the subspace of P k consisting of harmonicpolynomials on R n , H k = (cid:8) f ∈ P k ∆ f = 0 (cid:9) Each orthogonal transformation of R n induces canonically a linear isomorphismof P k .P ROPOSITION
The space H k is O (2 n ) -invariant, and P k has the direct sumdecomposition: P k = H k ⊕ r P k − . The injection map i : S n − → R n inducesa linear map i ∗ : C ∞ ( R n ) → C ∞ ( S n − ) . We denote i ∗ ( H k ) by H k . P ROPOSITION H k is an eigenspace of e ∆ on S n − with eigenvalue k ( k +2 n − and P ∞ k =0 H k is dense in C ∞ ( S n − ) in the uniform convergence topology.Moreover, H k is isomorphic to H k . That is, i ∗ : H k ≃ −→ H k . For proofs of these propositions, see [BGM].Now by Corollary ?? and Proposition 2.2.5, we getC OROLLARY
Let L ( q : p , . . . , p n ) = S n − /G be a lens space and H kG be the space of all G -invariant functions in H k . Then dim E k ( k +2 n − = dim H kG oreover, for any integer k such that dim H kG = 0 , k ( k + 2 n − is an eigen-value of ∆ on L ( q : p , . . . , p n ) with multiplicity equal to dim H kG , and no othereigenvalues appear in the spectrum of ∆ . D EFINITION generating function F q ( z : p , . . . , p n ) associated to thespectrum of the Laplacian on L ( q : p , . . . , p n ) is the generating function associ-ated to the infinite sequence (cid:8) dim H kG (cid:9) ∞ k =0 , i.e., F q ( z : p , . . . , p n ) = ∞ X k =0 (cid:0) dim H kG (cid:1) z k By Corollary ?? we know that the generating function determines the spectrumof L ( q : p , . . . , p n ) . This fact gives us the following proposition:P ROPOSITION
Let L = L ( q : p , . . . , p n ) and L ′ = L ( q ′ : s , . . . , s n ) betwo lens spaces. Let F q ( z : p , . . . , p n ) and F q ′ ( z : s , . . . , s n ) be the gener-ating functions associated to the spectrum of L and L ′ , respectively. Then L isisospectral to L ′ if and only if F q ( z : p , . . . , p n ) = F q ′ ( z : s , . . . , s n ) . The following theorem ([I1] and [I2]) holds true for the orbifold lens spaces aswell.T
HEOREM
Let L ( q : p , . . . , p n ) be a lens space and F q ( z : p , . . . , p n ) thegenerating function associated to the spectrum of L ( q : p , . . . , p n ) . Then, on thedomain (cid:8) z ∈ C | z | < (cid:9) , F q ( z : p , . . . , p n ) = 1 q q X l =1 − z Q ni =1 ( z − γ p i l )( z − γ − p i l ) C OROLLARY
Let L ( q : p , . . . , p n ) be isospectral to L ( q ′ : s , . . . , s n ) .Then q = q ′ . Now let e L ( q, n ) be the family of all (2 n − -dimensional lens spaces withfundamental groups of order q , and let e L ( q, n ) be the subfamily of e L ( q, n ) definedby: e L ( q, n ) = (cid:8) L ( q : p , . . . , p n ) ∈ e L ( q, n ) p i
6≡ ± p j (mod q ) , ≤ i < j ≤ n (cid:9) The set of isometry classes of e L ( q, n ) is denoted by L ( q, n ) , and the set of isometryclasses of e L ( q, n ) is denoted by L ( q, n ) .By Proposition 2.2.1, the map L ( q : p , . . . , p n ) ( p , . . . , p n ) of e L ( q, n ) (cid:2) resp. e L ( q, n ) (cid:3) onto e I ( q, n ) (cid:2) resp. e I ( q, n ) (cid:3) induces a one-to-one map between L ( q, n ) and I ( q, n ) (cid:2) resp. L ( q, n ) and I ( q, n ) (cid:3) .The above fact, together with Proposition 2.1.3, gives us the following:P ROPOSITION
Retaining the notations as above, we get |L ( q, n ) | ≥ r X t = u n − t (cid:18) q − − rn − − t (cid:19)(cid:18) rt (cid:19) here u = r − k if r > k , and u = 0 if r ≤ k ; r is the number of residues mod q that are not co-prime to q and are less than or equal to q . Note that by Proposition 2.1.1, we also get that |L ( q, n ) | ≥ q (cid:18) q n (cid:19) Next, we will re-formulate the generating function F q ( z : p , . . . , p n ) in a formthat will help us find isospectral pairs that are non-isometric.P ROPOSITION
Let L ( q : p , . . . , p n ) be a lens space belonging to e L ( q, n ) , k = q − n , and let w be the map of I ( q, n ) onto I ( q, k ) defined in section 2.1.Then (i) If q = P m , where P is a prime, we have F q ( z : p , . . . , p n ) = 1 q ( (1 − z )(1 − z ) n + ψ q,k ( w ([ p , . . . , p n ]))( z )(1 − z )Φ q ( z ) + m − X j =1 α ( j ) q,k ( w ([ p , . . . , p n ]))( z )(1 − z )(Φ P m − j ( z )) P j (1 − z ) P j − ) (ii) If q = P · P , where P and P are primes, we have F q ( z : p , . . . , p n ) = 1 q ( (1 − z )(1 − z ) n + ψ q,k ( w ([ p , . . . , p n ]))( z )(1 − z )Φ q ( z )+ α q,k ( w ([ p , . . . , p n ]))( z )(1 − z )(Φ P ( z )) P (1 − z ) P − + β q,k ( w ([ p , . . . , p n ]))( z )(1 − z )(Φ P ( z )) P (1 − z ) P − ) where ψ q,k , α ( j ) q,k , α q,k and β q,k are as defined in definition 2.1.4 and Φ t ( z ) = P t − v =0 z v .Proof. We choose integers q , . . . , q k such that the set of integers { p , − p , . . . , p n , − p n , q , − q , . . . , q k , − q k } forms a complete set of residues mod q .(i) We write F q ( z : p , . . . , p n ) = 1 q " X l ∈ A (1 − z ) Q ni =1 ( z − γ p i l )( z − γ − p i l )+ m − X j =1 X l ∈ B j (1 − z ) Q ni =1 ( z − γ p i l )( z − γ − p i l ) ) ow, for any l ∈ A , we have Q ni =1 ( z − γ p i l )( z − γ − p i l ) = Q ki =1 ( z − γ q i l )( z − γ − q i l )Φ q ( z ) For l ∈ B j , we have Q ni =1 ( z − γ p i l )( z − γ − p i l ) = Q ki =1 ( z − γ q i l )( z − γ − q i l )(Φ P m − j ( z )) P j (1 − z ) P j − where Φ t ( z ) = P t − v =0 z v .Now, ( i ) follows from these facts.(ii) We write F q ( z : p , . . . , p n ) = 1 q " X l ∈ A (1 − z ) Q ni =1 ( z − γ p i l )( z − γ − p i l )+ X l ∈ B (1 − z ) Q ni =1 ( z − γ p i l )( z − γ − p i l ) + X l ∈ C (1 − z ) Q ni =1 ( z − γ p i l )( z − γ − p i l ) Again, for l ∈ A , Q ni =1 ( z − γ p i l )( z − γ − p i l ) = Q ki =1 ( z − γ q i l )( z − γ − q i l )Φ q ( z ) For l ∈ B , we have Q ni =1 ( z − γ p i l )( z − γ − p i l ) = Q ki =1 ( z − γ q i l )( z − γ − q i l )(Φ P ( z )) P (1 − z ) P − For l ∈ C , we have Q ni =1 ( z − γ p i l )( z − γ − p i l ) = Q ki =1 ( z − γ q i l )( z − γ − q i l )(Φ P ( z )) P (1 − z ) P − Now, ( ii ) follows from these facts. (cid:3) From Proposition 2.2.8 and Proposition 2.2.12, we get the following propositionP
ROPOSITION
Let L = L ( q : p , . . . , p n ) and L ′ = L ( q : s , . . . , s n ) belens spaces belonging to L ( q, n ) . Let k = q − n . (i) If q = P m , then L is isospectral to L ′ if ψ q,k ( w ([ p , . . . , p n ])) = ψ q,k ( w ([ s , . . . , s n ])) and α ( j ) q,k ( w ([ p , . . . , p n ])) = α ( j ) q,k ( w ([ s , . . . , s n ])) for j = 1 , . . . , m − . ii) If q = P · P , then L is isospectral to L ′ if ψ q,k ( w ([ p , . . . , p n ])) = ψ q,k ( w ([ s , . . . , s n ])) α q,k ( w ([ p , . . . , p n ])) = α q,k ( w ([ s , . . . , s n ])) and β q,k ( w ([ p , . . . , p n ])) = β q,k ( w ([ s , . . . , s n ]))
3. I
SOSPECTRAL N ON - ISOMETRIC L ENS S PACES
By applying Proposition 2.2.11 and Proposition 2.2.13 we will obtain our maintheorem in this Section for odd-dimensional lens spaces. Next, we will extend theresults to obtain even-dimensional pairs of lens spaces corresponding to every pairof odd-dimensional lens spaces.3.1.
Odd-Dimensional Lens Spaces.
From the results in Sections 2 and 3 we getthe following diagrams:For q = P m , L ( q, n ) ∼ −→ I ( q, n ) ∼ −→ w I ( q, k ) −−−→ τ ( m ) q,k Q m ( γ )[ z ] (3.1)where τ ( m ) q,k = ( ψ q,k , α (1) q,k , . . . , α ( m − q,k ) , and Q m ( γ )[ z ] denotes m copies of thefield of rational polynomials Q ( γ )[ z ] For q = P · P , L ( q, n ) ∼ −→ I ( q, n ) ∼ −→ w I ( q, k ) −−→ S (3) q,k Q ( γ )[ z ] (3.2)where S (3) q,k = ( ψ q,k , α q,k , β q,k ) .Now, from Proposition 2.2.13, if τ ( m ) q,k [resp. S (3) q,k ] is not one-to-one, then wewill have non-isometric lens spaces having the same generating function. Thiswould give us our desired results.We first calculate the values for the required coefficients of ψ q, , α ( j ) q, , α q, and β q, .Using Proposition 2.1.5 we can calculate the values of the various coefficientsof ψ q,k , α ( j ) q,k , α q,k and β q,k .First we will find coefficients for z and z for any given k , and from that we canfind the values for k = 2 .From the definitions of ψ q,k ([ q , . . . , q k ]) , in the notation of 2.1.5, it is easy tosee that a = k X i =1 X l ∈ A γ q i l + k X i =1 X l ∈ A γ − q i l = 2 k X i =1 X l ∈ A γ q i l imilarly, b ,j = 2 k X i =1 X l ∈ B j γ q i l b = 2 k X i =1 X l ∈ B γ q i l c = 2 k X i =1 X l ∈ C γ q i l Also, a = X l ∈ A h k + X ≤ i Let p be an odd prime and let q = p m where m is an integergreater than . Let q = q − . Let k = 2 and n = q − . Then the maps τ ( m ) q,k and S (3) q,k as defined in (3.1) and (3.2) (and hence the generating function) aredependent only on where the various q i ’s and their sums and differences reside. In a similar fashion we can find values of coefficients of higher powers of z when k > . These coefficients will contain terms that include higher sums anddifferences of the various q i ’s in the powers of γ .We will prove two propositions that will give us upper bounds on the number ofexpressions for τ ( j ) q,k and S (3) q,k , respectively, where k = 2 .P ROPOSITION Let p be an odd prime and let q = p m where m is an integergreater than . Let q = q − . Let k = 2 and n = q − . Then the number ofexpressions that τ ( j ) q, can have is at most m .Proof. We will find the number of τ ( j ) q, by considering the following cases: ase 1: q , q ∈ B j ( j = 1 , , . . . , ( m − where B j = (cid:8) x ( modq ) ∈ Z + : p j | x, p j +1 ∤ x (cid:9) If q , q ∈ B j , then either both of q ± q lie in B j or else one lies in B j and the other in some B k with j < k ≤ m − . Thus there are m − j possibilities. As j varies from to m − , we thus obtain ( m − 1) + ( m − 2) + · · · + 1 = m ( m − expressions.Case 2: q ∈ B j and q ∈ B t , B j = B t We may assume j < t . It follows that q ± q both lie in B j . Thus as j and t vary, we obtain (cid:0) m − (cid:1) = ( m − m − expressions.Case 3: q ∈ B j and q ∈ A , or vice versaHere we note that q ± q always belongs to A . Therefore, in this case wewill get ( m − possible expressions for τ ( j ) q, , one each for the case where q ∈ A and q ∈ B j ( j = 1 , , . . . , ( m − , or vice versa.Case 4: q , q ∈ A We will get possible expression if q ± q ∈ A . Then we will get possible expression each for the case when q + q ∈ A and q − q ∈ B j (or vice versa) for j = 1 , , . . . , ( m − . There are no other possibilitiesin this case. So the maximum number of possible expressions for τ ( j ) q, inthis case will be m − m .Case though Case are the only possible cases that occur for k = 2 .Adding up the numbers of all possible expressions for τ ( j ) q, from each casewe get the maximum number of possible expressions that τ ( j ) q, can have: m ( m − m − m − m − 1) + m = m − m + m − m + 2 + 2 m − m m m (cid:3) P ROPOSITION Let q = p · p , where p , p are distinct odd primes. Let q = q − . Let k = 2 and n = q − . Then the number of possible expressions for S (3) q, is at most .Proof. As in the previous proposition, we prove this result by considering all thepossible cases for q and q (where q ± q is not congruent to modq ) ).Case 1 q , q ∈ B ( or q , q ∈ C ) , where B = (cid:8) xp x = 1 , . . . , ( p − (cid:9) and C = (cid:8) xp x = 1 , . . . , ( p − (cid:9) .Then q ± q ∈ B ( or q ± q ∈ C , respectively ) . There are no otherpossibilities for this case.Case 2: q ∈ B and q ∈ C (or vice versa).We have just one possible expression in this case, when q ± q ∈ A . ase 3: q ∈ A, q ∈ B or q ∈ A, q ∈ C (or vice versa).We will get one expression each when q ± q ∈ A . Then we will getone possible expression for the case when q ∈ A, q ∈ B , and q + q ∈ A, q − q ∈ C , (or vice versa).We will get one more possible expression for the case when q ∈ A, q ∈ C , and q + q ∈ A, q − q ∈ B (or vice versa).So, in this case we get a possible expressions for S (3) q, .Case 4: q , q ∈ A .We will get one possible expression where q ± q ∈ A . We get anotherpossible expression where q + q ∈ A and q − q ∈ B (or vice versa).We get a third possible expression where q + q ∈ A and q − q ∈ C (orvice versa). We get a fourth possible expression where q + q ∈ B and q − q ∈ C (or vice versa).So, we get a total of possible expressions for S (3) q, in this case.Case through Case are the only possible cases than can occur for k = 2 .Adding up the number of all possible expressions for S (3) q, from each case we getthe maximum number of possible expressions for S (3) q, : (cid:3) It is important to note that in the above propositions the number of possibleexpressions is the maximum number of expressions that can happen. It is possiblethat for a given q = p m or q = p · p not all the expressions will occur. We willsee this in an example later.We now prove two similar propositions for even q of the form m and p , where m is a positive integer and p is a prime.P ROPOSITION Let q = 2 m where m ≥ . Let q = q , i.e., q = 2 m − . Let k = 2 and n = q − . Then the number of possible expressions that τ ( j ) q, can haveis at most ( m − .Proof. We proceed as in the previous propositions.Case 1: q , q ∈ B j ( j = 1 , , . . . , ( m − , where B j = (cid:8) x ( modq ) ∈ Z + :2 j | x, j +1 ∤ x (cid:9) We first note that the cases where q , q ∈ B m − or B m − will notoccur. Now when q , q ∈ B j , then one of the q + q or q − q willbelong to B j +1 and the other will belong to B t for t > j + 1 .Now, with q , q ∈ B j (where j < m − ), we get ( m − − j ) possibleexpressions for τ ( j ) q, .So, in this case, the total number of possible expressions for τ ( j ) q, are: ( m − 3) + ( m − 4) + · · · + 3 + 2 + 1 = ( m − m − ase 2: q ∈ B j and q ∈ B t , where B j = B t .We can assume that j < t . This would mean that q ± q ∈ B j always. So,as in Case of Proposition 3.1.2, we get that the total number of expres-sions for τ ( j ) q, will be ( m − m − .Case 3: q ∈ B j and q ∈ A (or vice versa).We notice that q ± q ∈ A always. So, just like in Case of Proposi-tion 3.1.2, we will get that the total number of possible expressions for τ ( j ) q, will be ( m − .Case 4: q , q ∈ A .In this case one of q + q or q − q will belong to B and the otherwill belong to one of the B j for j > . Therefore, for this case we willget ( m − possible expressions for τ ( j ) q, , one each for the case when q + q ∈ B (alt. q − q ∈ B ) and q − q ∈ B t (alt. q + q ∈ B t ) for t = 2 , , . . . , m − .Now, adding up all the possible expressions from the four cases above we getthe maximum number of possible expressions for τ ( j ) q, : ( m − m − m − m − m − 1) + ( m − m − m + 6 + m − m + 2 + 2 m − m − m − m + 1 = ( m − (cid:3) Our next proposition gives us the maximum number of expressions for S (3) q, when q = 2 p for some prime p .P ROPOSITION Let q = 2 p where p is an odd prime. Let q = q = p . Let k = 2 and n = q − . Then the number of possible expressions for S (3) q, is at most .Proof. As before we will analyze the different possible cases. Note that in thissituation we have B = (cid:8) , , , . . . , p − (cid:9) and C = { p } .Case 1: q , q ∈ B . We will have q ± q ∈ B always.Notice that in this case q , q cannot belong to C since C has only oneelement. So we get possible expression in this case for S (3) q, .Case 2: q ∈ B, q ∈ C . In this case q ± q ∈ A always.So, we get possible expression in this case for S (3) q, .Case 3: q ∈ A, q ∈ B or q ∈ A, q ∈ C .When q ∈ A and q ∈ C , then q ± q ∈ B always. So, we get possible expression for S (3) q, . When q ∈ A, q ∈ B , we will get possibleexpression for the situation when q ± q ∈ A . We will get another possibleexpression for S (3) q, where q + q ∈ A (alt. q − q ∈ A ) and q − q ∈ C (alt. q + q ∈ C ). o, there are a total of possible expressions for S (3) q, in this case.Case 4: q , q ∈ A . Then q ± q ∈ B always.So, we get possible expression for this case.Now, adding up all the possible expressions from the above four cases we getthe maximum number of possible expressions for S (3) q, to be . (cid:3) With these four propositions, we are now ready for our first main theorem.T HEOREM Let p ≥ (alt. p ≥ ) be an odd prime and let m ≥ (alt. m ≥ ) be any positive integer. Let q = p m . Then there exist at leasttwo ( q − -dimensional orbifold lens spaces with fundamental groups oforder p m which are isospectral but not isometric. (ii) Let p , p be odd primes such that q = p · p ≥ . Then there existsat least two ( q − -dimensional orbifold lens spaces with fundamentalgroups of order p · p which are isospectral but not isometric. (iii) Let q = 2 m where m ≥ be any positive integer. Then there exist at leasttwo ( q − -dimensional orbifold lens spaces with fundamental groups oforder m which are isospectral but not isometric. (iv) Let q = 2 p , where p ≥ is an odd prime. Then there exist at least two ( q − -dimensional orbifold lens spaces with fundamental groups of order p which are isospectral but not isometric.Proof. We first recall from Proposition 2.2.11 that |L ( q, n ) | ≥ r X t = r − n − t (cid:18) q − − rn − − t (cid:19)(cid:18) rt (cid:19) for k = 2 and r > . Thus for k = 2 and r > we have L ( q, n ) | ≥ n − ( r − (cid:18) q − r − n − − ( r − (cid:19)(cid:18) rr − (cid:19) + 1 n − ( r − (cid:18) q − r − n − − ( r − (cid:19)(cid:18) rr − (cid:19) + 1 n − r (cid:18) q − r − n − r − (cid:19)(cid:18) rr (cid:19) = 1 q − − r + 2 (cid:18) q − r − q − − − r + 2 (cid:19)(cid:18) rr − (cid:19) + 1 q − − r + 1 (cid:18) q − r − q − − − r + 1 (cid:19)(cid:18) rr − (cid:19) + 1 q − − r (cid:18) q − r − q − − r − (cid:19) since n = q − 2= 1 q − r (cid:18) q − r − q − r − (cid:19)(cid:18) rr − (cid:19) + 1 q − r − (cid:18) q − r − q − r − (cid:19)(cid:18) rr − (cid:19) + 1 q − r − (cid:18) q − r − q − r − (cid:19)(cid:18) rr (cid:19) = 1 q − r · · r ( r − q − r − · ( q − r − · r + 1( q − r − · ( q − r − q − r − · r ( r − q − r ) + r + ( q − r − (3.3)It is sufficient to show that the final expression in (3 . is greater than the number ofpossible expressions for the generating functions computed in Propositions 3.1.2–3.1.5 in order to establish the existence of isospectral pairs of non-isometric lensspaces.(i) For q = p m , we have a total of m possible expressions for τ ( j ) q, from Prop-osition 3.1.2. So, we will have isospectrality when (3 . is greater than orequal to m + 1 . That is r ( r − q − r ) + r + ( q − r − ≥ m + 1 ⇒ r ( r − 1) + 2 r ( q − r ) + ( q − r )( q − r − ≥ q − r )( m + 1) ⇒ r − r + ( q − r )[2 r + q − r − − m − ≥ ⇒ ( r − r ) + ( q − r )[ q + r − m − ≥ ⇒ r − r + q + q r − q m − q − q r − r + 2 m r + 3 r ≥ ⇒ q ( q − m − 3) + 2 r ( m + 1) ≥ ⇒ − q [(2 m + 3) − q ] ≥ − r ( m + 1) ⇒ q [(2 m + 3) − q ] ≤ r ( m + 1) . (3.4) o for any given m , we can choose p big enough so that m + 3 ≤ q .This would guarantee isospectrality. We can calculate r by r = ( p m − − ) in this case. Now if p ≥ , q ≥ m − > m + 3 for all m ≥ . Thisis easy to see since m > m + 7 for m ≥ as the left hand side growsexponentially greater than the right hand side. So, for all p ≥ and all m ≥ , (3 . will be true and we will get isospectral pairs of dimension ( q − 6) = 2 n − . Now for q = 3 m , we have m > m + 7 for m ≥ .So we will have isospectrality. We check cases m = 2 and m = 3 .When m = 2 , q = 9 , r = 1 , q = 4 . So L.H.S. of (3 . gives − 4] = 4(7) = 28 and R.H.S. of (3 . gives . So thesufficiency condition is not satisfied.When m = 3 , q = 27 , r = 4 , q = 13 . L.H.S. of (3 . gives − 13] = 13[8] = 104 and R.H.S. of (3 . gives .So the sufficiency condition is not satisfied.However, when we check individually all the possible expressions forthese cases we realize that they are less than m .For q = 3 , the only two expressions are for the cases when q ∈ A , q ∈ B , q ± q ∈ A , and q , q ∈ A , q + q ∈ A , q − q ∈ B . Noother possible expressions exist.However, there are only two classes in L ( q, n ) , i.e., |L ( q, n ) | = 2 . Thetwo classes are [1 , 2] = n ( p , p ) ∈ e L ( q, p , p ∈ A o [1 , 3] = n ( p , p ) ∈ e L ( q, p ∈ A, p ∈ B (alt. p ∈ B , p ∈ A ) o where n = 2 , A = { , , , , , } and B = { , } .Therefore, we do not obtain isospectral pairs.For q = 3 , there are expressions (instead of = 9 possible expres-sions). The case where q , q , q ± q ∈ B and the case where q , q ∈ B do not occur. This gives us less expressions than the estimated numberof . But the number of classes is greater than or equal to − − 4) + 4 + 13 − − 12 = 23 + 4 + 4 = 8 23 > from (3 . . This means we will have non-isometric isospectral lens spaces. This givesus our result that for p ≥ and m ≥ , we will get isospectral pairs thatare non-isometric.(ii) For q = p · p , r = p + p − .From (3 . and Proposition 3.1.3 we get the following sufficiency condi-tion: r ( r − q − r ) + r + ( q − r − ≥ ⇒ q (25 − q ) ≤ r (3.5) rom this we get that for q ≥ , we will always find non-isometric,isospectral lens spaces because (3 . will always be satisfied. We nowcheck for cases where q = 2 q + 1 < .For q < , and q = p · p with p , p being odd primes, there are onlythe following possibilities:(a) q = 3 · ; B = { , , , , , } , C = { , } .In this case we have instead of possible expressions. The casewhere q , q ∈ C = { , } is not possible, and the case where q , q ∈ A and q ± q ∈ A is also not possible since then q ≡− q ( modq ) . Therefore, we get less expressions. now for isospec-trality we use (3 . : − − 4) + 4 + (10 − − , which is not greater than . So the isospectrality condition is not met.(b) q = 3 · . In this case we have instead of expressions.Here B = { , , , } and C = { , } . In this case, the followingcases do not occur: q , q ∈ C ; q ∈ A, q ∈ C , q ± q ∈ A ; q , q , q ± q ∈ A ; q , q ∈ A , q + q ∈ A , q − q ∈ C . Sowe get less expressions than . To check for isospectrality we use (3 . and get − − + 3 + (7 − − = 5 , which is less than . So theisospectrality condition is not satisfied.For (a) and (b) it can be easily shown that |L ( q, n ) | is equal to and respectively. This means that there are no isospectral pairs in thesecases.(c) q = 3 · 11 = 33 . B = { , , , , , , , , , } and C = { , } . Here q = 16 and r = 6 . We check for isospectrality using (3 . : q (25 − q ) = 16(25 − 16) = 14424 r = 24(6) = 144 So (3 . is satisfied.(d) q = 5 · , B = { , , , , , } and C = { , , , } .Here q = 17 and r = 5 . Using (3 . we get q (25 − q ) = 17(25 − 17) = 13824 r = 24(5) = 120 So (3 . is not satisfied. However, we notice that in this case the actualnumber of expressions is instead of . So, we use (3 . to checkfor isospectrality. Plugging in r = 5 and q = 17 into (3 . we get > This implies that S (3) q, is not one-one and therefore, we will have non-isometric isospectral lens spaces in this case. e) Finally, we check q = 3 · 13 = 39 .Here q = 19 and r = 7 . Using (3 . we see q (25 − q ) = 19(25 − 19) = 11424 r = 24(7) = 168 So (3 . is satisfied and we will have isospectral pairs in this case.(a)-(e) are all the cases of q = p · p < , where p , p are oddprimes.Combining these results with the fact that for q ≥ , (3 . will al-ways be satisfied, we prove (iii).(iii) Let q = 2 m . We use Proposition 3.1.4 and (3 . to get a sufficiency condi-tion for isospectrality: r ( r − q − r ) + r + ( q − r − ≥ ( m − + 1 Here q = m = 2 m − and r = 2 m − . Therefore, q = 2 r in this case.Simplifying the above inequality, we get q [(2 m − m + 5) − q ] ≤ r ( m − m + 2) . But since q = 2 r , we get ( m − m + 3) ≤ q (3.6) If m ≥ , then m − m + 3 < m − . Further, the right hand side of (3 . grows exponentially bigger than the left hand side as m grows. For m = 3 , and , the actual number of expressions for τ ( j ) q, are , and respectively. Further, it can be easily shown that for m = 3 , and , |L ( q, n ) | is , and respectively. Therefore, for m = 3 , and we donot get isospectrality. This gives us (iii).(iv) Using Proposition 3.1.5 and (3 . we get the sufficiency condition forisospectrality for q = 2 p , where p is an odd prime ≥ . Note that inthis case q = q = p and r = q +24 . Now for isospectrality we should have r ( r − q − r ) + r + ( q − r − ≥ ⇒ q (15 − q ) ≤ r ⇒ p (15 − p ) ≤ p + 1) ⇒ ≤ p − p + 7 or ( p − p − ≥ Since p is positive, whenever p ≥ , we will have isospectrality. When q = 2 · , then |L ( q, n ) | = 6 = number of expressions for S (3) q, . So,we do not get isospectral pairs when p = 5 . This proves (iv). (cid:3) .2. Lens Spaces for General Integers. Let L = L ( q : p , . . . , p n ) = S n − /G and L ′ = L ( q : p , . . . , p n ) = S n − /G ′ be two isospectral non-isometric orbifold lens spaces as obtained in Section 3.1where G = h g i , G ′ = h g ′ i . g = R ( p /q ) . . . R ( p n /q ) and g ′ = R ( s /q ) . . . R ( s n /q ) We define ˜ g W + = R ( p /q ) . . . R ( p n /q ) I W and ˜ g ′ W + = R ( s /q ) . . . R ( s n /q ) I W where I W is the W × W identity matrix for some integer W . We can define ˜ G W + = h ˜ g W + i and ˜ G ′ W + = h ˜ g ′ W + i . Then ˜ G W + and ˜ G ′ W + are cyclic groups of order q . We define lens spaces ˜ L W + = S n + W − / ˜ G W + and ˜ L ′ W + = S n + W − / ˜ G ′ W + .Then, like Proposition 2.2.1, we get:P ROPOSITION Let ˜ L W + and ˜ L ′ W + be as defined above. Then ˜ L W + isisometric to ˜ L ′ W + iff there is a number l coprime with q and there are num-bers e i ∈ {− , } such that ( p , . . . , p n ) is a permutation of ( e ls , . . . , e n ls n )(mod q ) . The following lemma follows directly from this proposition.L EMMA Let L , L ′ , ˜ L W + and ˜ L ′ W + be as defined above. Then L is isometricto L ′ iff ˜ L W + is isometric to ˜ L ′ W + . We get the following theorem: HEOREM Let e L W +0 ( q, n, be the family of all (2 n + W − -dimensionalorbifold lens spaces with fundamental groups of order q that are obtained in themanner described above. Let ˜ L W + ∈ L W +0 ( q, n, (where L W +0 ( q, n, denotesthe set of isometry classes of e L W +0 ( q, n, ). Let F W + q ( z : p , . . . , p n , be thegenerating function associated to the spectrum of ˜ L W + . Then on the domain (cid:8) z ∈ C | z | < (cid:9) , F W + q ( z : p , . . . , p n , 0) = (1 + z )(1 − z ) W − · q q X l =1 Q ni =1 ( z − γ p i l )( z − γ − p i l ) Proof. Recall the definitions of ∆ , r , P k , H k , H k and H kG from Section ?? . Weextend the definitions for R n + W . That is, let ∆ be the Laplacian on R n + W withrespect to the Flat Riemannian metric; r = P n + Wi =1 x i , where ( x , . . . , x n + W ) is the standard coordinate system on R n + W ; for k ≥ , P k is the space ofcomplex valued homogenous polynomials of degree k in R n + W ; H k is the sub-space of P k consisting of harmonic polynomials on R n + W ; H k is the image of i ∗ : C ∞ ( R n + W ) −→ C ∞ ( S n + W − ) where i : S n + W − −→ R n + W is thenatural injection; and H k ˜ G is the space of all e G -invariant functions of H k .Then from Proposition 2.2.4 and Proposition 2.2.5, we get that H k is O (2 n + W ) -invariant; P k has the direct sum decomposition P k = H k ⊕ r P k − ; H k is aneigenspace of e ∆ on S n + W − with eigenvalue k ( k + 2 n + W − ; P ∞ k =0 H k isdense in C ∞ ( S n + W − ) in the uniform convergence topology and H k is isomor-phic to H k .This along with the results in Corollary 2.2.6, where dim E k ( k +2 n + W − =dim H k ˜ G W + , we get F q ( z : p , . . . , p n , 0) = ∞ X k =0 (dim H k ˜ G W + ) z k . Now ˜ G W + is contained in SO (2 n + W ) .Let χ k and ˜ χ k be the characters of the natural representations of SO (2 n + W ) on H k and P k respectively. Then dim H k e G W + = 1 (cid:12)(cid:12)(cid:12) e G W + (cid:12)(cid:12)(cid:12) X ˜ g W + ∈ e G χ k (˜ g W + ) = 1 q q X l =1 χ k (˜ g lW + ) (3.7)Proposition 2.2.4 gives χ k (˜ g lW + ) = ˜ χ k (˜ g lW + ) − ˜ χ k − (˜ g lW + ) (3.8)where ˜ χ t = 0 for t > .If W is even, then we can view the space P k as having a basis consisting of allmonomials of the form: z I · ¯ z J = ( z ) i · · · ( z n + v ) i n + v · (¯ z ) j · · · (¯ z n + v ) j n + v here W = 2 v and where I n + v + J n + v = i + · · · + i n + v + j + · · · + j n + v = k and i , . . . , i n + v , j , . . . , j n + v ≥ . Then, ˜ g lW + ( z I · ¯ z J ) = γ i p l + ··· + i n p n l − j p l −···− j n p n l ( z I · ¯ z J ) . If W is odd, (say W = 2 u + 1 ), then we get for basis of P k z I · ¯ z J · z tn +2 u +1 = ( z ) i · · · ( z n + u ) i n + u · (¯ z ) j · · · (¯ z n + u ) j n + u · ( z n +2 u +1 ) t where z n +2 u +1 = x n + W where ( x , y , . . . , x n + W − , y n + W − , x n + W ) is the stan-dard euclidean coordinate system on R n + W with z i = x i + iy i for i = 1 , , . . . , n + W − , and i , . . . , i n + u , j , . . . , j n + u , t ≥ and i + · · · + i n + u + j + · · · + j n + u + t = k = I n + u + J n + u + t . So, in that case ˜ g lW + ( z I · ¯ z J · z n +2 u +1 ) = γ i p l + ··· + i n p n l − j p l −···− j n p n l ( z I · ¯ z J · z n +2 u +1 ) So, for W , even case, we will get F W + q ( z : p , . . . , p n , 0) = 1 q ∞ X k =0 q X l =1 χ k (˜ g lW + ) z k = (1 − z ) q q X l =1 ∞ X k =0 ˜ χ k (˜ g lW + ) z k = (1 − z ) q q X l =1 ∞ X k =0 X I n + v + J n + v = k γ i p l + ··· + i n p n l − j p l −···− j n p n l z k = (1 − z ) q q X l =1 ∞ X k =0 X I n + v + J n + v = k ( γ p l z ) i · · · ( γ p n l z ) i n ( γ − p l z ) j · · · ( γ − p n l z ) j n · z i n +1 + ··· + i n + v + j n +1 + ··· + j n + v = (1 − z ) q q X l =1 n Y i =1 (1 + γ p i l z + γ p i l z + · · · )(1 + γ − p i l z + γ − p i l z + · · · )(1 + z + z + · · · ) W = (1 − z ) q q X l =1 Q ni =1 (1 − γ p i l z )(1 − γ − p i l z )(1 − z ) W on (cid:8) z ∈ C | z | < (cid:9) = (1 + z )(1 − z ) W − · q q X l =1 Q ni =1 ( z − γ p i l )( z − γ − p i l ) For W odd case, we get by similar calculations, F W + q ( z : p , . . . , p n ) = (1 − z ) q q X l =1 ∞ X k =0 X I n + u + J n + u + t = k γ i p l + ··· + i n p n l − j p l −···− j n p n l z k (1 − z ) q q X l =1 ∞ X k =0 X I n + u + J n + u + t = k ( γ p l z ) i · · · ( γ p n l z ) i n ( γ − p l z ) j · · · ( γ − p n l z ) j n · z i n +1 + ··· + i n + u + j n +1 + ··· + j n + u + t = (1 − z ) q q X l =1 n Y i =1 (1 + γ p i l z + γ p i l z + · · · )(1 + γ − p i l z + γ − p i l z + · · · )(1 + z + z + · · · ) W = (1 − z ) q q X l =1 Q ni =1 (1 − γ p i l z )(1 − γ − p i l z )(1 − z ) W on (cid:8) z ∈ C | z | < (cid:9) = (1 + z )(1 − z ) W − · q q X l =1 Q ni =1 ( z − γ p i l )( z − γ − p i l ) as before. (cid:3) C OROLLARY When L ( q : p , . . . , p n ) and L ( q : s , . . . , s n ) have the samegenerating function, then ˜ L W + and ˜ L ′ W + (as defined above) also have the samegenerating functionProof. This follows from the fact that F qW + ( z : p , . . . , p n , 0) = 1(1 − z ) W F q ( z : p , . . . , p n ) . (cid:3) The above results give us the following theorem.T HEOREM Let P ≥ (alt. P ≥ ) be any odd prime and let m ≥ (alt. m ≥ ) be any positive integer. Let q = P m . Then there exist atleast two ( q + W − -dimensional orbifold lens spaces with fundamentalgroups of order P m which are isospectral but not isometric. (ii) Let P , P be two odd primes such that q = P · P ≥ . Then there existat least two ( q + W − -dimensional orbifold lens spaces with fundamentalgroups of order P · P which are isospectral but not isometric. (iii) Let q = 2 m where m ≥ is any positive integer. Then there exist at leasttwo ( q + W − -dimensional orbifold lens spaces with fundamental groupsof order m which are isospectral but not isometric. (iv) Let q = 2 P , where P ≥ is an odd prime. Then there exist at least two ( q + W − -dimensional orbifold lens spaces with fundamental groups oforder P which are isospectral but not isometric. C OROLLARY Let x ≥ be any integer. Then there exist at leasttwo x -dimensional orbifold lens spaces with fundamental groups of order which are isospectral but not isometric. (ii) Let x ≥ be any integer. Then there exist at least two x -dimensional orb-ifold lens spaces with fundamental group of order which are isospectralbut not isometric. iii) Let x ≥ be any integer. Then there exist at least two x dimensional orb-ifold lens spaces with fundamental group of order which are isospectralbut not isometric. (iv) Let x ≥ be any integer. Then there exist at least two x dimensional orb-ifold lens spaces with fundamental group of order which are isospectralbut not isometric.Proof. (i) Let q = 25 and W ∈ { , , , , . . . } in ( i ) of the theorem.(ii) Let q = 33 and W ∈ { , , , , . . . } in ( ii ) of the theorem.(iii) Let q = 64 and W ∈ { , , , , . . . } in ( iii ) of the theorem.(iv) Let q = 14 and W ∈ { , , , , . . . } in ( iv ) of the theorem. (cid:3) When W is an odd numnber, we get even dimensional orbifold lens spaces thatare isospectral but not isometric. 4. E XAMPLES In this section we will look at some examples of isospectral non-isometric orb-ifold lens spaces by calculating their generating functions. We will also look at anexample that will suggest that our technique can be generalized for higher valuesof k = q − n . Recall that q = ( q − / for odd q and q = q/ for even q . Inthe previous sections we assumed k = 2 . The technique for getting examples forhigher values of k is similar, but as we shall see, the calculations for the differenttypes of generating functions becomes more difficult as k increases.In all the examples we will denote a lens space by L ( q : p , . . . , p n ) = S n − /G ,where G is the cyclic group generated by g = R ( p /q ) . . . R ( p n /q ) . Wewill write G = h g i .4.1. Examples for k = 2 . EXAMPLE q = 5 = 25 , q = q − = 12 , n = 10 , k = 2 , A = { , , , , , , , , , , , , , , , , , , , } , B = { , , , } .Let w ([ p , . . . , p ]) = [ q , q ] . Let γ = e πi/ and λ = e πi/ . a = | A | = 20 , b , = | B | = 4 , P l ∈ A γ l = 0 , P l ∈ B γ l = − , P l ∈ A λ l = − and P l ∈ B λ l =4 (from (4.1)).Case 1: q , q ∈ B and q ± q ∈ B . So, ψ , ([ q , q ])( z ) = 20 z + 20 z + 20 z + 20 z + 20 α (1)25 , ([ q , q ])( z ) = 4 z − z + 24 z − z + 4 This corresponds to the case where [ p , . . . , p ] = [1 , , , , , , , , , which corresponds to a manifold lens spaces.Case 2: Since there is only one B this case does not occur. ase 3: q ∈ B and q ∈ A ( alt. q ∈ A, q ∈ B ) . q ± q ∈ A always.So, ψ , ([ q , q ])( z ) = 20 z + 10 z + 40 z + 10 z + 20 α (1)25 , ([ q , q ])( z ) = 4 z − z + 4 z − z + 4 corresponding to [ p , . . . , p ] = [1 , , , , , , , , , and to [ s , . . . , s ] = [1 , , , , , , , , , and [ p , . . . , p ] = [ s , . . . , s ] . So, we get two isospectral non-isometric orbifolds: L = L (25 : 1 , , , , , , , , , and L = (25 : 1 , , , , , , , , , . We denote by P i the singularset of L i . Then P = (cid:8) (0 , , . . . , x , x , , , . . . , ∈ S x + x =1 (cid:9) and P = (cid:8) (0 , , . . . , x , x , , ∈ S x + x = 1 (cid:9) withisotropy groups h g i and h g i where g = R (5 p / . . . R (5 p / = R ( p / . . . R ( p / and g = R ( s / . . . R ( s / where g and g are generators of G and G , respectively with L = S /G and L = S /G . P and P are homeomorphic to S . Wedenote the two isotropy groups by H = h g i and H = h g i .Case 4: ( a ) q , q ∈ A and q ± q ∈ A . So, ψ , ([ q , q ])( z ) = 20 z + 40 z + 20 α (1)25 , ([ q , q ])( z ) = 4 z + 4 z + 4 z + 4 z + 4 corresponding to L = L (25 : 1 , , , , , , , , , 10) = S /G , where G = h g i L = L (25 : 1 , , , , , , , , , 11) = S /G , where G = h g i and L = L (25 : 1 , , , , , , , , , 12) = S /G , where G = h g i The isotropy groups for L , L and L are h g i , h g i and h g i , respec-tively. P , P and P are all homeomorphic to S . So, here we get isospectral orbifold lens spaces that are non-isometric. b ) q , q ∈ A and q + q ∈ B , q − q ∈ A ( alt. q + q ∈ A, q − q ∈ B ) . So, ψ , ([ q , q ])( z ) = 20 z + 30 z + 20 α (1)25 , ([ q , q ])( z ) = 4 z + 4 z + 14 z + 4 z + 4 corresponding to L = L (25 : 1 , , , , , , , , , 11) = S /G , where G = h g i and L = L (25 : 1 , , , , , , , , , 11) = S /G , where G = h g i Then, again, P and P are homeomorphic to S , and L and L haveisotropy groups h g i and h g i . EXAMPLE q = 3 = 27 ; q = 13 , k = 2 , n = 11 and A = { , , , , , , , , , , , , , , , , , } , B = { , , , , , } , B = { , } . Let w ([ p , . . . , p ]) = [ q , q ] . Let γ = e πi/ , λ = e πi/ and δ = e πi/ be primitive th , th and rd roots of unity, respectively. Here we getisospectral non-isometric pairs only in two cases:Case 1: q ∈ B and q ∈ A ( alt. q ∈ A and q ∈ B ) . q ± q ∈ A always. Sowe get, ψ , ([ q , q ])( z ) = 18 z + 36 z + 18 α (1)27 , ([ q , q ])( z ) = 6 z + 6 z + 12 z + 6 z + 6 α (2)27 , ([ q , q ])( z ) = 2 z − z − z + 2 corresponding to L = L (27 : 1 , , , , , , , , , , ,L = L (27 : 1 , , , , , , , , , , and L = L (27 : 1 , , , , , , , , , , If G = h g i , G = h g i , G = h g i are such that L = S /G , L = S /G and L = S /G , then P = (cid:8) (0 , , . . . , x , x , , x , x , , . . . , x , x , , ∈ S : isotropy group = h g i (cid:9)[ (cid:8) (0 , , . . . , x , x , , . . . , ∈ S : isotropy group = h g i (cid:9)P = (cid:8) (0 , , . . . , x , x , , . . . , x , x , . . . , , x , x , , ∈ S : isotropy group = h g i (cid:9)[ (cid:8) (0 , , . . . , x , x , , . . . , ∈ S : isotropy group = h g i (cid:9)P = (cid:8) (0 , , . . . , x , x , , . . . , x , x , . . . , x , x , , ∈ S : isotropy group = h g i (cid:9)[ (cid:8) (0 , , . . . , x , x , , . . . , ∈ S : isotropy group = h g i (cid:9) So all three orbifolds have the same isotropy type and all the singular setsare homeomorphic to S . ase 2: q + q ∈ B , q − q ∈ A ( alt. q + q ∈ A , q − q ∈ B ) . So we get, ψ , ([ q , q ])( z ) = 18 z + 36 z + 18 α (1)27 , ([ q , q ])( z ) = 6 z + 6 z + 6 α (2)27 , ([ q , q ])( z ) = 2 z + 4 z + 6 z + 4 z + 2 corresponding to L = L (27 : 1 , , , , , , , , , , 13) = S /G ; G = h g i L = L (27 : 1 , , , , , , , , , , 13) = S /G ; G = h g i L = L (27 : 1 , , , , , , , , , , 13) = S /G ; G = h g i Then, P = (cid:8) (0 , , x , x , , . . . , , x , x , , , x , x , , . . . , x , x , , ∈ S with isotropy group = h g i (cid:9)[ (cid:8) (0 , , . . . , , x , x , , . . . , ∈ S with isotropy group = h g i (cid:9)P = (cid:8) (0 , , x , x , , . . . , , x , x , , . . . , , x , x , , , x , x , , ∈ S with isotropy group = h g i (cid:9)[ (cid:8) (0 , , . . . , , x , x , , . . . , ∈ S with isotropy group = h g i (cid:9)P = (cid:8) (0 , , x , x , , , x , x , , . . . , , x , x , , . . . , , x , x , , ∈ S with isotropy group = h g i (cid:9)[ (cid:8) (0 , , . . . , , x , x , , . . . , ∈ S with isotropy group = h g i (cid:9) So, the singular sets are all homeomorphic to S and they all have the sameisotropy types. EXAMPLE q = 5 · , q = − = 17 , k = 2 , n = 15 . Here A = { , , , , , , , , , , , , , , , , , , , , , , , } , B = { , , , , , } and C = { , , , } . Here we get isospectral non-isometric pairs in three cases:Case 1: q ∈ A , q ∈ B , q ± q ∈ A . So we get, ψ , ([ q , q ])( z ) = 24 z + 6 z + 52 z + 6 z + 24 α , ([ q , q ])( z ) = 6 z + 4 z + 8 z + 4 z + 6 β , ([ q , q ])( z ) = 4 z − z + 4 z − z + 4 corresponding to L = L (35 : 1 , , , , , , , , , , , , , , 17) = S /G L = L (35 : 1 , , , , , , , , , , , , , , 17) = S /G and L are isospectral non-isometric orbifold lens spaces. G = h g i and G = h g i . P = (cid:8) (0 , . . . , , x , x , , . . . , , x , x , , . . . , ∈ S with isotropy group = h g i (cid:9)[ (cid:8) (0 , . . . , , x , x , , . . . , , x , x , , . . . , ∈ S with isotropy group = h g i (cid:9)P = (cid:8) (0 , . . . , , x , x , , . . . , , x , x , , . . . , ∈ S with isotropy group = h g i (cid:9)[ (cid:8) (0 , . . . , , x , x , , . . . , , x , x , , . . . , ∈ S with isotropy group = h g i (cid:9)P and P are both homeomorphic to S × S .Case 2: q , q ∈ A . ( a ) q ± q ∈ A . So we get, ψ , ([ q , q ])( z ) = 24 z − z + 52 z − z + 24 α , ([ q , q ])( z ) = 6 z + 4 z + 8 z + 4 z + 6 β , ([ q , q ])( z ) = 4 z + 4 z + 4 z + 4 z + 4 corresponding to L = L (35 : 1 , , , , , , , , , , , , , , 17) = S /G and L = L (35 : 1 , , , , , , , , , , , , , , 17) = S /G where G = h g i and G = h g i . Thus, P = (cid:8) (0 , . . . , , x , x , , . . . , , x , x , , . . . , , x , x , , . . . , ∈ S with isotropy group = h g i (cid:9)[ (cid:8) (0 , . . . , , x , x , , . . . , , x , x , , . . . , ∈ S with isotropy group = h g i (cid:9)P = (cid:8) (0 , . . . , , x , x , , . . . , , x , x , , . . . , , x , x , , . . . , ∈ S with isotropy group = h g i (cid:9)[ (cid:8) (0 , . . . , , x , x , , . . . , , x , x , , . . . , ∈ S with isotropy group = h g i (cid:9)P and P are homeomorphic to S × S . ( b ) q + q ∈ A , q − q ∈ B . So we get, ψ , ([ q , q ])( z ) = 24 z − z + 42 z − z + 24 α , ([ q , q ])( z ) = 6 z + 4 z + 8 z + 4 z + 6 β , ([ q , q ])( z ) = 4 z + 4 z + 14 z + 4 z + 4 corresponding to L = L (35 : 1 , , , , , , , , , , , , , , 17) = S /G and L = L (35 : 1 , , , , , , , , , , , , , , 17) = S /G here G = h g i and G = h g i . P = (cid:8) (0 , . . . , , x , x , , . . . , , x , x , , . . . , , x , x , , . . . , ∈ S with isotropy group = h g i (cid:9)[ (cid:8) (0 , . . . , , x , x , , . . . , , x , x , , . . . , ∈ S with isotropy group = h g i (cid:9)P = (cid:8) (0 , . . . , , x , x , , . . . , , x , x , , . . . , , x , x , , . . . , ∈ S with isotropy group = h g i (cid:9)[ (cid:8) (0 , . . . , , x , x , , . . . , , x , x , , . . . , ∈ S with isotropy group = h g i (cid:9)P and P are homeomorphic to S × S .Our final example for the case when k = 2 comes when q is even. EXAMPLE q = 2 · , q = = 7 , k = 2 and n = 5 . Here A = { , , , , , } , B = { , , , , , } and C = { } . Here we getisospctral non-isometric pairs in only one case:Case 1: q ∈ A , q ∈ B , q ± q ∈ A . a = 2 X l ∈ A γ l + 2 X l ∈ A λ l = 2(1) + 2( − 1) = 0 b = 2 X l ∈ B γ l + 2 X l ∈ B λ l = 2( − 1) + 2( − 1) = − c = 2 X l ∈ C γ l + 2 X l ∈ C λ l = 2( − 1) + 2(1) = 0 a = 2 | A | + 4 X l ∈ A γ l = 2(6) + 4(1) = 12 + 4 = 16 b = 2 | B | + 4 X l ∈ B γ l = 2(6) + 4( − 1) = 12 − c = 2 | C | + 4 X l ∈ C γ l = 2(1) + 4( − 1) = 2 − − So we get, ψ , ([ q , q ])( z ) = 6 z + 16 z + 6 α , ([ q , q ])( z ) = 6 z + 4 z + 8 z + 4 z + 6 β , ([ q , q ])( z ) = z − z + 1 corresponding to L = L (14 : 1 , , , , 7) = S /G and L = L (14 : 1 , , , , 7) = S /G here G = h g i and G = h g i . P = (cid:8) (0 , , x , x , x , x , , , , ∈ S with isotropy group = h g i (cid:9)[ (cid:8) (0 , . . . , , x , x ) ∈ S with isotropy group = h g i (cid:9)P = (cid:8) (0 , , x , x , , , x , x , , ∈ S with isotropy group = h g i (cid:9)[ (cid:8) (0 , . . . , , x , x ) ∈ S with isotropy group = h g i (cid:9)P and P are homeomorphic to S × S .4.2. Example for k = 3 . EXAMPLE q = 5 = 25 , q = − = 12 , k = 3 , n = 9 . Let w ([ p , . . . , p ]) = [ q , q , q ] . Here A = { , , , , , , , , , , , , , , , , , , , } and B = { , , , } .We will consider all the possible cases for the various possibilities of q i ’s, ( q i ± q j ) ’s and ( q ± q ± q ) ’s lying in A or B . Many of these possibilities will notoccur in our present example of q = 25 . However, these possibilities are statedbecause they may occur for higher values of q .Case 1: All the q i ’s ∈ B . This case does not happen for q = 25 since at most ofthe q i ’s can be in B at one time by the definition of I (25 , .Case 2: All the q i ’s ∈ A . Since k = 3 , we can’t have more than of the q i ± q j (1 ≤ i < j ≤ belonging to B . Also, at most only one of the q ± q ± q can be ≡ .Further, if one of the q ± q ± q is congruent to , then wecan’t have any other of the q ± q ± q belong to B . Also, at most ofthe q ± q ± q can be in B .Further, if one of the q ± q ± q is congruent to , then atmost of the q i ± q j ( for ≤ i < j ≤ can be in B . Similarly, if one ofthe q ± q ± q ∈ B , then at most of the q i ± q j can be in B . We notethat the above results hold true for all q = P , where P is any odd prime.We now look at the various sub-cases for Case 2. ( a ) All of the q ± q ± q ∈ A and all of the q i ± q j ∈ A (1 ≤ i < j ≤ .This case does not occur for q = 25 . ( b ) All of the q ± q ± q ∈ A and exactly one of the q i ± q j ∈ B (1 ≤ i < j ≤ . This case does not occur for q = 25 . ( c ) All of the q ± q ± q ∈ A and exactly two of the q i ± q j ∈ B (1 ≤ i < j ≤ . Again, this case does not occur for q = 25 . ( d ) exactly one of the q ± q ± q ∈ B and all of the q i ± q j ∈ A (1 ≤ i < j ≤ . This case does not occur for q = 25 . ( e ) exactly one of the q ± q ± q ≡ q ) and all of q i ± q j ∈ A (1 ≤ i < j ≤ . This case also does not occur for q = 25 . f ) All of the q ± q ± q ∈ A and exactly of the q i ± q j ∈ B (1 ≤ i < j ≤ . In this case, we get isospectral, non-isometric pairs sincewe get, ψ , ([ q , q , q ])( z ) = 20 z + 30 z + 30 z + 20 α (1)25 , ([ q , q , q ])( z ) = 4 z + 6 z + 30 z + 20 z + 30 z + 6 z + 4 corresponding to L = L (25 : 1 , , , , , , , , 11) = S /G and L = L (25 : 1 , , , , , , , , 12) = S /G where G = h g i and G = h g i . P = (cid:8) (0 , . . . , , x , x , , . . . , , x , x , , ∈ S with isotropy group = h g i (cid:9)P = (cid:8) (0 , . . . , , x , x , , . . . , , x , x , , ∈ S with isotropy group = h g i (cid:9) ( h ) exactly one of the q ± q ± q is congruent to and oneof the q i ± q j ( for ≤ i < j ≤ is in B . We again get isospectralnon-isometric pairs here since, ψ , ([ q , q , q ])( z ) = 20 z + 50 z + 10 z + 50 z + 20 α , ([ q , q , q ])( z ) = 4 z + 6 z + 10 z + 10 z + 10 z + 6 z + 4 corresponding to L = L (25 : 1 , , , , , , , , 10) = S /G , G = h g i L = L (25 : 1 , , , , , , , , 10) = S /G , G = h g i L = L (25 : 1 , , , , , , , , 10) = S /G , G = h g i L = L (25 : 1 , , , , , , , , 10) = S /G , G = h g i L = L (25 : 1 , , , , , , , , 10) = S /G , G = h g i L = L (25 : 1 , , , , , , , , 12) = S /G , G = h g i L = L (25 : 1 , , , , , , , , 12) = S /G , G = h g i L = L (25 : 1 , , , , , , , , 12) = S /G , G = h g i So, in this case we get a family of orbifold lens spaces that are isospec-tral but mutually non-isometric. P = (cid:8) (0 , . . . , , x , x , , . . . , , x , x ) ∈ S with isotropy group = h g i (cid:9)P = (cid:8) (0 , . . . , , x , x , , . . . , , x , x ) ∈ S with isotropy group = h g i (cid:9)P = (cid:8) (0 , . . . , , x , x , , . . . , , x , x ) ∈ S with isotropy group = h g i (cid:9)P = (cid:8) (0 , . . . , , x , x , , . . . , , x , x ) ∈ S with isotropy group = h g i (cid:9) = (cid:8) (0 , . . . , , x , x , , . . . , , x , x ) ∈ S with isotropy group = h g i (cid:9)P = (cid:8) (0 , . . . , , x , x , , . . . , , x , x , , ∈ S with isotropy group = h g i (cid:9)P = (cid:8) (0 , . . . , , x , x , , . . . , , x , x , , . . . , ∈ S with isotropy group = h g i (cid:9)P = (cid:8) (0 , . . . , , x , x , , . . . , , x , x , , ∈ S with isotropy group = h g i (cid:9) All of the P i ( for i = 3 , , . . . , are homeomorphic to S . ( h ) exactly one of the q ± q ± q is congruent to and oneof the q i ± q j ( for ≤ i < j ≤ is in B . Here we get, ψ , ([ q , q , q ])( z ) = 20 z + 50 z − z + 50 z + 20 α , ([ q , q , q ])( z ) = 4 z + 6 z + 10 z + 10 z + 10 z + 6 z + 4 corresponding to L = L (25 : 1 , , , , , , , , 10) = S /G , G = h g i L = L (25 : 1 , , , , , , , , 12) = S /G , G = h g i P = (cid:8) (0 , . . . , , x , x , , . . . , , x , x ) ∈ S with isotropy group = h g i (cid:9)P = (cid:8) (0 , . . . , , x , x , , . . . , , x , x , , . . . , ∈ S with isotropy group = h g i (cid:9)P and P are homeomorphic to S . a ) - h ) are all of the possible cases when all the q i ’s ∈ A .Case 3: Two of the q i ’s ∈ A and one of the q i ’s ∈ B . In this case we will have atmost one of the q ± q ± q congruent to .Also, we can have at most one of the q i ± q j (1 ≤ i < j ≤ in B .Further, it can be shown that at most two of the q ± q ± q can be in B . Also, if one of the q ± q ± q is congruent to , then atmost one of the remaining q ± q ± q can belong to B . In fact, it can beshown that exactly one of the remaining q ± q ± q must belong to B .All of these results can be shown to be true q = P , where P is any oddprime. Now we consider all the sub-cases for Case 3. ( a ) If all the q ± q ± q belong to A and exactly one of the q i ± q j (1 ≤ i < j ≤ belongs to B . This case does not occur for q = 25 . ( b ) exactly one of the q ± q ± q belongs to B and the remainingbelong to A . This case does not occur for q = 25 . ( c ) If all of the q ± q ± q belong to A and all of the q i ± q j (1 ≤ i 9) = S /G , G = h g i L = L (25 : 1 , , , , , , , , 11) = S /G , G = h g i L = L (25 : 1 , , , , , , , , 12) = S /G , G = h g i L = L (25 : 1 , , , , , , , , 12) = S /G , G = h g i L = L (25 : 1 , , , , , , , , 12) = S /G , G = h g i We get a family of orbifold lens spaces that are non-isometric andisospectral. P = (cid:8) (0 , . . . , , x , x , , . . . , ∈ S with isotropy group = h g i (cid:9)P = (cid:8) (0 , . . . , , x , x , , . . . , ∈ S with isotropy group = h g i (cid:9)P = (cid:8) (0 , . . . , , x , x , , . . . , ∈ S with isotropy group = h g i (cid:9)P = (cid:8) (0 , . . . , , x , x , , . . . , ∈ S with isotropy group = h g i (cid:9)P = (cid:8) (0 , . . . , , x , x , , . . . , ∈ S with isotropy group = h g i (cid:9) All the P i ’s ( i = 13 , . . . , are homeomorphic to S . ( d ) exactly two of the q ± q ± q belong to B and exactly one of the q i ± q j (1 ≤ i < j ≤ belongs to B .Here we get, ψ , ([ q , q , q ])( z ) = 20 z + 10 z + 50 z + 40 z + 50 z + 10 z + 20 α (1)25 , ([ q , q , q ])( z ) = 4 z − z + 10 z − z + 10 z − z + 4 corresponding to L = L (25 : 1 , , , , , , , , 12) = S /G , G = h g i L = L (25 : 1 , , , , , , , , 11) = S /G , G = h g i We have P = (cid:8) (0 , . . . , , x , x , , . . . , ∈ S with isotropy group = h g i (cid:9)P = (cid:8) (0 , . . . , , x , x , , . . . , ∈ S with isotropy group = h g i (cid:9) ( e ) One of the q ± q ± q is congruent to , and one one ofthe q ± q ± q is in B , and exactly one of the q i ± q j (1 ≤ i < j ≤ is in B .Here we get, ψ , ([ q , q , q ])( z ) = 20 z + 10 z + 50 z − z + 50 z + 10 z + 20 α (1)25 , ([ q , q , q ])( z ) = 4 z − z + 10 z − z + 10 z − z + 4 orresponding to L = L (25 : 1 , , , , , , , , 12) = S /G , G = h g i L = L (25 : 1 , , , , , , , , 12) = S /G , G = h g i L = L (25 : 1 , , , , , , , , 12) = S /G , G = h g i We have P = (cid:8) (0 , . . . , , x , x , , ∈ S with isotropy group = h g i (cid:9)P = (cid:8) (0 , . . . , , x , x , , . . . , ∈ S with isotropy group = h g i (cid:9)P = (cid:8) (0 , . . . , , x , x , , . . . , ∈ S with isotropy group = h g i (cid:9) There are no other sub-cases for Case 3.Case 4: One of the q i ’s ∈ A and two of the q i ’s ∈ B . In this case, q ± q ± q willalways belong to A , and exactly two of the q i ± q i (1 ≤ i < j ≤ willbelong to B . There are no other variations that will occur in this case.Here we get, ψ , ([ q , q , q ])( z ) = 20 z + 20 z + 40 z + 40 z + 40 z + 20 z + 20 α (1)25 , ([ q , q , q ])( z ) = 4 z − z + 20 z − z + 20 z − z + 4 corresponding to L (25 : 1 , , , , , , , , and we do not get isospec-tral pairs. Note that this lens space is a manifold.As this example illustrates, we can extend our technique for k = 2 to higher val-ues of k and we will get many examples of isospectral non-isometric orbifold lensspaces. At the same time, the example also illustrates the difficulty in accountingfor all the possible cases as the value of k is increased.R EFERENCES [BCDS] P. Buser, J. Conway, P. Doyle and K. Semmler, Some planar isospectral domains , Internat.Math. Res. Notices. (1994), 391ff., approx. 9 pp. (electronic).[BGM] M. Berger, P. Gaudachon and E. 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(1980), 2132.N AVEED S HAMS UL B ARI , 42/2 L ANE 21, DHA P HASE 7, K ARACHI , P AKISTAN - D IRECTOROF Q UANTITATIVE M ETHODS AT C ENTER FOR F INANCIAL T RAINING AND R ESEARCH (CFTR) IN K ARACHI , P AKISTAN E-mail address : [email protected]@yahoo.com