Spherical Functions on 2-adic Ramified Hermetian Spaces
aa r X i v : . [ m a t h . R T ] O c t Spherical Functions on 2-adic RamifiedHermitian Spaces
Dror Ozeri
[email protected] of Mathematics, Technion - Israel Institute of Technology, Haifa 32000
Abstract
Y. Hironaka introduced the spherical functions on the p-adic space of Hermi-tian matrices. For the space of 2 × Contents
Calculation of L ( x , , c , z ) : 22 L ( x , , c , z ) on the diagonal elements . . . . . . . . . . . 247.2 Calculating L ( x , , c , z ) on the non diagonal elements: . . . . . . . . 277.2.1 RP case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277.2.2 RU case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 L ( x , c ∗ , c , z ) L ( c ∗ , c , x , z ) on the diagonal representatives. . . . . . . . 508.2.1 The representatives with l − l < s . . . . . . . . . . . . . . 518.2.2 The representative with l − l = s : . . . . . . . . . . . . . . 548.3 The diagonal representatives with l − l > s . . . . . . . . . . . . . 55 • E - Local p-adic field. • E ∗ = E − { } . • O E The ring of integers of E . • O ∗ E - The units of O E . • p - A uniformizer of E . • E - The residue field of the field E . • | | E - The p-adic absolute value on E normalized such that | p | E = | E | − . • v E - The valuation map corresponding to | · | E . • F - A quadratic extension of E . • N = N F / E , Tr = Tr F / E - The Norm and Trace maps.1 v - A uniformizer of F . • q = F . • | | = | | F - The p-adic absolute value of F normalized by | v | = q − . • v F - The valuation map corresponding to | · | F . • x x - The conjugation map of the field extension. • s = v F [( v ) · v − − ] - The number is defined only if F / E is ramified. • l = [ s ] - The largest integer k such that k ≤ s . • D ∈ + p s O E , r ∈ O ∗ E - Fixed elements such that D = + rp s / ∈ N ( F ) . • c ∗ - The non trivial character of E ∗ / N ( F ∗ ) . • A - The characteristic function of a set A . • G = GL ( F ) . • K = GL ( O F ) . • For a group A and a normal subgroup B we denote by [ x ] A / B to be the coset xB of an element x ∈ A . Let E be a p-adic local field and F a quadratic extension of E . Let O F be thering of integers of F , G = GL ( F ) , and K = GL ( O F ) . Set A ∗ = A t . Let X = { A ∈ GL ( F ) | A ∗ = A } . We note that the group G acts on X by g · x = gxg ∗ .We denote by C ¥ ( K \ X ) the space of complex-valued K -invariant functions on X and S ( K \ X ) ⊆ C ¥ ( K \ X ) the subspace of all compactly supported K invariant functionson X . Let H ( G , K ) be the Hecke algebra of G with respect to K : the space of allcompactly supported K -bi-invariant complex valued functions on G .Let dg be the Haar measure on G normalized by ´ G K dg = H ( G , K ) acts on C ¥ ( K \ X ) and on S ( K \ X ) by the convolution product: f · f ( x ) = ˆ G f ( g ) f ( g − · x ) dg . Under this action we call f ∈ C ¥ ( K \ X ) a spherical function on X if f is a common H ( G , K ) -eigenfunction. 2ecall that in a quadratic extension we have the following maps: N : F ∗ → E ∗ , Tr : F → E , where: N ( x ) = xx , Tr ( x ) = x + x . Let ( s , s ) ∈ C , c , c characters of E ∗ / N ( F ∗ ) , x ∈ X . Hironaka introduced inH[1-4] the following function : L ( x , c , c , s , s ) = ˆ K ′ (cid:213) i = c i ( d i ( k · x )) | d i ( k · x ) | s i dk , where dk is the Haar measure on K normalized such that ´ K dk = d ( y ) = y , , d ( y ) = det ( y ) and K ′ = { k ∈ K | (cid:213) | i = d i ( k · x ) | 6 = } .It is known that this integral converges absolutely for Re ( s ) , Re ( s ) > q s , q s .We transform the variables s = ( s , s s ) ∈ C to new variables z = ( z , z ) by thefollowing equations: s = z − z − s = − z + . It is known that this function is indeed a spherical function [H1] and for any f ∈ H ( G , K ) we have [ f · L ( ∗ , c , c , z )]( x ) = ˜ f ( z ) × L ( x , c , c , z ) , where ˜ f ( z ) is defined to be the Satake transform:˜ f ( z ) = ˆ G f ( g ) F z ( g ) dg and F z ( g ) = | a | z − | a | z + where a , a are determined by the Iwasawa de-composition: g = k (cid:18) a a (cid:19) (cid:18) ∗ (cid:19) , k ∈ K . By abuse of notation we denote L ( x , c , c , z ) = L ( x , c , c , s ( z )) .If the field extension is ramified, it is known that s = v F [ v − ( v ) − ] is an invariantof the field extension F / E (see [FV] Ch3 p. 70 ) .3ironaka computed L ( x , c , c , z ) for the following cases [H5, H2]:The case where the field extension F / E is unramified, the case where the fieldsextension F / E is ramified and | | = F / E is ramified , | | E < s = , G = GL n in Case 1.In this paper we complete Hironaka’s calculation of Case 3 for the case of generaluniformizer p of E and general s . We compute the spherical functions defined aboveand provide the general functional equations. Our computation is by brute force andapplies properties of the Norm and Trace maps in a quadratic extension of local fields.Chapter 3 of this work will be dedicated to the summary of the facts we need fromthe local field theory. In Chapter 4 we compute useful (and interesting) p-adic integralsthat we use in Chapters 7 and 8 . In Chapter 5 we explicate convenient representativesof K \ X , following Jacobowitz [J] . In Chapter 6 we state the main theorems of thiswork regarding the computation of the spherical functions and the functional equation.In Chapters 7 and 8 we prove the main theorem of Chapter 6.It is our hope that the functional equations can be used in the future to calculatethe spherical functions for general GL n ( F ) case as was shown in [H4, O] with theCasselman-Shalika basis method [CS].Spherical functions on the Hermitian spaces X are related to the concept of localdensities [H4, H5, HS] and calculation of the spherical functions for GL n can be usedto calculate the local densities , which are important in several aspects.Integrals of the form ´ O F | a + Tr ( bx ) + cN ( x ) | zF dx appear naturally in many places.It is hoped that their computation will have further applicationsI would like to thank my advisors Dr. Omer Offen and Ass. Prof. Moshe Baruchfrom the Technion for introducing me to this subject and for their endless help andsupport for the last 2 years. Let E be a p-adic local field, and F / E a quadratic extension. We normalize theabsolute value on F by | v | = [ F ] − .We denote by v F , v E the corresponding valuation4aps.We recall without proofs facts and properties from local class field theory, proofscan be found in ([FV], Chapter 3).We distinguish between the following cases of extension: Case 1 : The unramified case F / E is an unramified extension. We denote | F | = q , | ¯ E | = q . For convenience,we take p = v . Case 2 : The tamely ramified case F / E is totally ramified and Char ( ¯ E ) = . We denote q = | ¯ E | = | F | . For conve-nience, we take p = N ( v ) . Case 3 : The wildly ramified case F / E is totally ramified and Char ( ¯ E ) = . We denote q = | ¯ E | = | F | . For conve-nience, we take p = N ( v ) . Theorem 3.1.
Let F / E be a wildly ramified extension, then :1. The number s = v F [ v − v ) − ] does not depend on the choice of the uniformizer v . < s ≤ v E ( ) .
3. If s is even then s = v E ( ) . Proof of Theorem 3.1 could be found in [FV] ,p. 75.Theorem 3.1 motivates us to distinguish between two types of extensions:
Ramified
Prime (RP)
The extension F / E is wildly ramified extension and the invariant s is even. It canbe shown [FV] that if F / E is RP then F = E ( √ p ′ ) , where p ′ is some uniformizer of E . 5 amified Unit (RU)
The extension F / E is wildly ramified extension and the invariant s is odd. Again,it can be shown that if F / E is RU then F = E ( √ + dp k + ) for some d ∈ O ∗ E . Onecan take a uniformizer on F to be v ′ = + √ + dp k + p k ( note that | v ′ v ′ | E = q − ). Aquick calculation shows that s = v E ( ) − ( k + ) > . Example 3.1.
Consider the extension F / E where F = Q ( √ ) , E = Q .Since | | F = |√ · √ | F = − = ⇒ |√ | F = √ − , but 2 is known to be the uni-formizer of Q , hence this extension is ramified. ( | v | F > | p | F ) . We take v = √ s = v F ( s ( √ ) √ − ) = v F ( −√ √ − ) = v F ( − − ) = v F ( − ) = Example 3.2.
Consider the extension F / E where F = Q ( √− ) , E = Q .By one of the definitions of the absolute value | · | F : | a + √− b | F = q | a + b | E , a , b ∈ E . Therefore: | + √− | F = p | | E = √ − . So F / E is ramified extension.One can take v = + √− . We calculate: s = v F ( − √− + √− − ) = v F ( − √− + √− ) = v F ( − + √− ) = v F ( ) − v F ( + √− ) = − = . We conclude that F / E is a ramified unit extension. Note that F = E ( p + ( − ) · . For i > l i , F (resp. l i , E ) the natural map l i , F : ( + p i O F ) → F l i , F ( x ) = v − i ( x − ) mod v O F and l i , E ( x ) = p − i ( x − ) mod p O F . Denote the residue map by l , K : O K → K , ( K = E , F ) .6e summarize the properties of the norm by the following diagrams: Case 1-Unramified :
The following diagrams commute: F ∗ N F / E (cid:15) (cid:15) v F / / Z × (cid:15) (cid:15) E ∗ v E / / Z O FN F / E (cid:15) (cid:15) l , F / / ¯ F N F / E (cid:15) (cid:15) O E l , E / / ¯ E (3.1)1 + v i O FN F / E (cid:15) (cid:15) l i , F / / ¯ F Tr F / E (cid:15) (cid:15) + p i O E l i , E / / ¯ E , i ≥ Case 2- Tamely ramified E ∗ = F ∗ N F / E (cid:15) (cid:15) v F / / Z id (cid:15) (cid:15) E ∗ v E / / Z O FN F / E (cid:15) (cid:15) l , F / / ¯ E = ¯ F x x (cid:15) (cid:15) O E l , E / / ¯ E + v i O FN F / E (cid:15) (cid:15) l i , F / / ¯ E = F × (cid:15) (cid:15) + p i O E l i , E / / ¯ E And N ( + v i − O F ) = N ( + v i O F ) .7 ase 3- wildly ramified Let h ∈ O ∗ F to be such that vv = + hv s .Denote k = l , F ( h ) .The following diagrams commute: F ∗ N F / E (cid:15) (cid:15) v F / / Z id (cid:15) (cid:15) E ∗ v E / / Z O FN F / E (cid:15) (cid:15) l , F / / ¯ E = ¯ F x x (cid:15) (cid:15) O E l , E / / ¯ E (3.2)For i < s : 1 + v i O FN F / E (cid:15) (cid:15) l i , F / / ¯ E = F x x (cid:15) (cid:15) + p i O E l i , E / / ¯ E (3.3)1 + v s O FN F / E (cid:15) (cid:15) l s , F / / ¯ E = F x x − k · x (cid:15) (cid:15) + p s O E l s , E / / ¯ E (3.4)(Note that the homomorphism x x − k · x is an additive homomorphism withkernel of size 2) 8or j > + v s + j O FN F / E (cid:15) (cid:15) l s + j , F / / ¯ E = F × k (cid:15) (cid:15) + p s + j O E l i , E / / ¯ E (3.5)and N ( + v s + i O F ) = N ( + v s + i + O F ) if i > ∤ i . Proofs for commutativity could be found in [FV] p. 68-73.We will use throughout this paper the following corollaries ( Case 3):
Corollary 3.1. N ( + v s + O F ) = + p s + O E .Proof. From (3.5) we conclude that since the map x x × k is a surjective homomor-phism between the additive groups F → E and form the commutativity of diagram 3.5that for i > s , every ball of radius q − i in 1 + p s + O E contains an element that is a norm,thus N ( + v s + O F ) is dense in 1 + p s + O E . From compactness of 1 + v s + O F wehave N ( + v s + O F ) = + p s + O E . Corollary 3.2. [ + p s O E : N ( + v s O F )] = . Proof.
Follows from commutativity of (3.4) and the last corollary
Corollary 3.3.
Let x , y ∈ p i O ∗ E , If | x − y | E < q − s − i then x ∈ N ( F ) ⇐⇒ y ∈ N ( F ) .Proof. Since p is a norm we can assume w.l.o.g that i =
0. We have from the lastcorollary and the commutativity of (3.1) that N ( O ∗ E ) is a union of cosets of 1 + p s + O E ,thus if | x − y | E < q − s then they are in the same coset and we have x ∈ N ( O ∗ F ) ⇐⇒ y ∈ N ( O ∗ F ) . Example.
Consider F / E from example 3.1. Note that F = E = F (The field with twoelements) . Note that s =
2, and k = + O FN F / E (cid:15) (cid:15) l , F / / F x x − x (cid:15) (cid:15) + Z l , E / / F Since the image of the map: y : F → F , y ( x ) x − x is { } we conclude from thecommutativity of the diagram that: N ( + O F ) = + Z . .2 The Trace In the wildly ramified case (Case 3) the following holds: Tr F / E ( v i O F ) = p j ( i ) O E , j ( i ) = s + + [( i − − s ) / ] .Proof could be found in [FV], p. 71.Explicitly, RP: s = l , l = n E ( ) Tr ( v i O F ) = p l + i O E Tr ( v i − O F ) = p l + i O E RU: s = l + Tr ( v i O F ) = p l + + i O E Tr ( v i − O F ) = p l + i O E Lemma 4.1.
Let H and G be compact abelian topological groups with Haar measures m H , m G and f : H → G a surjective (continuous) homomorphism, then the push-forwardmeasure m H ∗ = m H ◦ f − is an invariant Haar measure on G and m H ∗ = m H ( H ) m G ( G ) · m G .Proof. Because the characters span a dense subset in L ( G , m G ) it is sufficient to showthat ´ G c d m H ∗ = c . But ´ G c d m H ∗ = ´ H c ( f ( x )) d m H ( x ) = , since c ( f ( x )) is an non trivial character of H . We obtain the multiplicative factorbetween the measures from integrating over the trivial character. Remark . The conditions of the previous lemma can be further generalized.Take m E , m F to be the Haar measures on E , F normalized by m E ( O E ) = m F ( O F ) = . Lemma 4.2.
For n ≥ m F [ N − ( + p n O ∗ E )] = m F ( N − [( + p n O E ) − ( + p n + O E )]) = ase q − q n + Case ( q − ) q n + Case ( q − ) q − ( n + ) i f n < s ( q − ) q − ( s + ) i f n = s ( q − ) q − ( n + ) i f n > sProof. In Case 1, since N ( O ∗ F ) = O ∗ E we conclude that the push-forward of the mul-tiplicative Haar measure of O ∗ F is an multiplicative Haar measure of O ∗ E . In O ∗ F theadditive and multiplicative measures coincide, so we get that m F ∗ = m F ( O ∗ F ) m E ( O ∗ E ) m E = − q − − q − m E = ( + q − ) m E . Since m E ( + p n O ∗ E ) = q − n ( − q − ) , we get that: m F ∗ ( + p n O ∗ E ) = ( + q − ) · q − n ( − q − ) = q − q n + In Case 2, we have N ( + v O ) = + p O E . Since this case is ramified, we havethat m F ( O ∗ F ) = m E ( O ∗ E ) . So m F [ N | − + v O F ( + p n O ∗ E )] = q − n ( − q − ) ( N | + v O F is therestriction of the norm to 1 + v O F ) . But since the norm induces the square mapbetween the groups e N : O ∗ F + v O F → O ∗ E + p O E . The size of the kernel of the this map is 2. We have that : N − ( + p n O ∗ E ) = a · N | − + v O F ( + p n O ∗ E ) ˙ ∪ N | − + v O F ( + p n O ∗ E ) , where a is a non trivial representative of the kernel of e N . So: m F ∗ ( + p n O ∗ E ) = m F [ N | − + v O F ( + p n O ∗ E )] = ( q − ) q n + . In Case 3 we have N ( + v s + O F ) = + p s + O E . So if we restrict the norm to1 + v s + O F we have m ∗ = m E .From the diagrams of Section 3.1 that the norm map induces an homomorphism ofthe finite groups: e N : 1 + v O F + v s + O F → + p O E + p s + O E . The size of the kernel of this map is 2. 11e have that for n > s : N − ( + p n O ∗ E ) = a [ N | − + v s + O F ( + p n O ∗ E )] ˙ ∪ N | − + v s + O F ( + p n O ∗ E ) , ( N | + v s + O F is the restriction of the norm to 1 + v s + O F ), where a is a representativeelement of the kernel of e N . But since N | + v s + O F is surjective on 1 + p s + O E we canuse Lemma 4.1 to obtain: m F [ N − ( + p n O ∗ E )] = m E ( + p n O ∗ E ) = ( − q − ) q − n . For n = s , we can conclude from (3.5) that N − ( + p s O ∗ E ) = q − ∪ i = a i ( + v s + O F ) , where the a i are representatives of the cosets that are the preimages of e N . So m F ( N − ( + p s O ∗ E )) = ( q − ) q − ( s + ) .For n < s the norm induces an isomorphism between the finite groups O ∗ F + v n + O F and O ∗ E + p n + O E , therefore m F ( N − ( + p n O ∗ E )) = m E ( + p n O ∗ E ) = ( q − ) q − ( n + ) , al-together we obtain the result above. Corollary 4.1. (case 1) ˆ x ∈ O F | + N ( x ) | s dx = q − q − q + q − q (cid:18) q − s − − q − s − (cid:19) (4.1) Proof.
Note that − − = z · z for some z ∈ O ∗ F . We substitute x = z y : ˆ x ∈ O F | + N ( x ) | s dx = ˆ y ∈ O F | − N ( y ) | s dy = ˆ x ∈ O F | ( − ) + N ( y ) | s dy . The integrand is equal to 1 for values of y such that N ( y ) = − mod p O E . From (3.1) q + N F / E ) cosets of O ∗ F are map to the coset of −
1. It is easyto see that the integrand is also equal to 1 in the set v O F . Therefore: m F ( − + O ∗ E ) = ( q − ) − q − q | {z } m F ([ O ∗ F − N − ( − + p O E )] + q |{z} m F ( v O F ) = q − q − q .
12t follows immediately from Lemma 4.2 that the integral is equal to the followinggeometric sum: ˆ x ∈ O F | + N ( x ) | s dx = q − q − q + (cid:229) ≤ n q − q q − ns − n = q − q − q + q − q (cid:18) q − s − − q − s − (cid:19) For the next lemmas, let c ∗ be a non trivial character of E ∗ that is trivial on N ( F ) .We will define c ∗ ( ) = Lemma 4.3. (Case 3 ) Let < m < s + , q ∈ O ∗ E ,then ´ x ∈ O F c ∗ ( + qp m N ( x )) dx = Proof.
First, we show that the integrand is constant on the (additive) cosets of group O F v s + − m O F :If x ∈ v s + − m O F then c ∗ ( + qp m N ( x )) = x , y ∈ O ∗ F . Assume that x = yu ,where u ∈ + v s + − m O F (that is, they are inthe same coset of O F v s + − m O F ). We know (see Section 3.1) that N ( u ) ∈ + p s + − m O E .So N ( u ) = + p s + − m n , n ∈ O F . Then :1 + qp m N ( x ) = + qp m N ( y ) N ( u ) = + qp m N ( y ) + d , d = p s + qn . Recall from Section 3.1 , that for g , g ∈ O ∗ E if | g − g | < q − s ,then g ∈ N ( F ) ifand only if g ∈ N ( F ) , since | d | < q − s we conclude that: c ∗ ( + qp m N ( x )) = c ∗ ( + qp m N ( y )) . We conclude that the integrand is constant on the cosets of O F v s + − m O F . Since s + − m < s +
1, we conclude from the (3.1),(3.3) that the norm map definesa bijection between the groups : e N : H = O F v s + − m O F → O E p s + − m O E = H ′ . For x ∈ O F , we denote by [ x ] H the coset of x in H (and similarly for the rest of thegroup in this proof).Since the integrand is well defined over H , integrating over O F is equivalent tosumming over H : ˆ x ∈ O F c ∗ ( + qp m N ( x )) dx = m F ( v s + − m O F ) S [ h ] H ∈ H c ∗ ( + qp m N ( h ))
13e use the bijection e N to calculate the sum over H ′ : S [ h ] H ∈ H c ∗ ([ + qp m N ( h )] H ) = S [ h ′ ] h ′ ∈ H ′ c ∗ ( + qp m h ′ ) The group O E p s + − m O E is isomorphic to the multiplicative group: U = + p m O F + p s + O F by themap ˜ y : ([ x ] H ′ ) [ + p m q x ] U , so we get another bijection:˜ y : O F p s + − m O F → + p m O F + p s O F , The previous bijection , ˜ y , tells us that summing over H ′ is equivalent to summingover the multiplicative group U by u = ˜ y ( h ) , so S [ h ′ ] H ′ ∈ H ′ c ∗ ( + qp m h ′ ) = S [ u ] U ∈ U c ∗ ( u ) . This sum surely vanish, since it is a summation of a non trivial character of U . Lemma 4.4. (Case 3) Let < m < s , q ∈ O ∗ E then ´ x ∈ O ∗ F c ∗ ( + qp m N ( x )) dx = Proof.
From Lemma 4.3: ˆ x ∈ O F c ∗ ( + qp m N ( x )) dx = ˆ x ∈ v O F c ∗ ( + qp m N ( x )) dx = . By substituting x = v y , we get: ˆ x ∈ v O F c ∗ ( + qp m N ( x )) dx = q − ˆ y ∈ O F c ∗ ( + qp m + N ( y )) dy By Lemma 4.3 we conclude that the last integral vanishes.
Lemma 4.5. (Case 3) Let F / E a ramified extension of case 3 and h ∈ O ∗ F then ifRe ( s ) > − , and s = − z + z − we have ˆ x ∈ O F | h + N ( x ) | s dx = q z − q z − + c ∗ ( − h ) · ( q s ( z − z )+ z − q ( z − z ) s + z − ) q z − q z (4.2)14 roof. Suppose − h / ∈ N ( F ) . Let − n ∈ + p s O E be an element that is not a norm,note that − n = ( − m ) · e and e ∈ N ( F ∗ ) , that is e = a a for some a ∈ O ∗ F . By makingcoordinate transformation a y = x , we can assume w.l.o.g that − h ∈ + p s O E .In the coset v O F , it is easy to see that the integrand is equal to 1 .The norm map induces an isomorphism f N : O ∗ F + v O F → O ∗ E + p O E (The square map x → x of the finite group F ∗ = E ∗ ) and so q − O ∗ F + v O F do not map by f N to the coset 1 + p O E (the coset of − h ) and so N ( x ) − ( − h ) ∈ O ∗ E and the integrand onthose cosets is equal to 1 .So far we have: ˆ x ∈ O F | h + N ( x ) | s dx = q − + ˆ x ∈ O ∗ F | h + N ( x ) | s dx = q − |{z} m F ( v O F ) + ( q − ) q − | {z } m F [ O ∗ F − ( + v O F )] + ˆ x ∈ + v O F | h + N ( x ) | s dx . Then for any 1 ≤ i < s we have that the norm map induce a isomorphism f N : 1 + v i O F + v i + O F → + p i O E + p i + OE (The square map of the additive groups F = E ). So, for each i , q − + v i O F + v i + O F do not map to the coset 1 + p i + O E (the coset of − h ) and the value of the integrand is q − is . We have: ˆ x ∈ O F | h + N ( x ) | s dx = q − + q − q + s − (cid:229) i = q − is ( q − ) q − i − + ˆ x ∈ + v s O F | ( − h ) − N ( x ) | s dx . Since − h / ∈ N ( + v s O F ) and [( − h ) + p s O E ] ∩ N ( + v s O F ) = f , we have thatin 1 + v s O F the integrand is equal to q − s · s .Altogether: ˆ x ∈ O F | h + N ( x ) | s dx = ( − q − ) s − (cid:229) j = q ( − s − ) j + q s · ( − s − ) = q z − q z − + q ( z − z ) s + z − − q s ( z − z )+ z q z − q z . − h ∈ N ( F ) . Then − h = N ( n ) , n ∈ O ∗ F by the substitution n y = x andLemma (4.2): ˆ x ∈ O F | h + N ( x ) | − z + z − dx = ˆ y ∈ O F | − + N ( y ) | − z + z − dy =( − q − ) s − (cid:229) j = q ( z − z ) j + ¥ (cid:229) j = s + ( − q − ) q ( z − z ) j + q s ( z − z ) = q z − q z − + q ( z − z ) s + z − q s ( z − z )+ z − q z − q z . Representative of the K-Orbits (Ramified dyadic)5 Classification of K-Orbits
We summarize results obtained by Jacobowitz [J] that classified Hermitian latticesover local fields ( O F -module equipped with a Hermitian product h· , ·i L . An Hermitianlattice with basis { x l } can be represented by an Hermitian matrix: L lm = h x l , x m i L and equivalence classes of lattices correspond to orbits of Hermitian matrices underthe action of K given by k · x = kxk ∗ , k ∈ K , x ∈ X .The following definitions are also presented on [J]. We translate them in term ofmatrices: Definition 5.1.
An hermitian matrix is called v i modular if for every primitive vector x ∈ M n , (that is, a vector x = ( x i ) , ∃ i , ≤ i ≤ n s . t x i ∈ O ∗ F ) there is a vector w ∈ M n , such that w ∗ Ax = v i . Definition 5.2. The norm ideal: nL The ideal of O F generated by elements h v , v i L , v ∈ M , n ( O F ) Definition 5.3. The scalar ideal sL The Ideal generated by h v , w i L , v , w ∈ M , n ( O F ) Definition 5.4. dL = det ( L ) ( mod N ( F ) ) ∈ E ∗ / N ( F ) .16 efinition 5.5. v i -modular hyperbolic matrix: H ( i ) ≈ (cid:18) v i v i (cid:19) . The norm ideal, the scalar ideal, dL and v i − modularity are all invariants of thelattice under the action of K .Jacobowitz investigated the ramified non-dyadic case relevant for our problem.It was shown : Every Hermitian matrix is equivalent to the direct sum of v i modular 2 × × In the case of RP : n · H ( i ) = v s + i O F In the case of RU: n · H ( i ) = v s − + i O F If L is v i modular then: n · H ( i ) ⊆ n · L We conclude the following representatives of K \ X in the case : n · H ( i ) = n · L : RP . (cid:18) (cid:19) , . (cid:18) p s − p s r (cid:19) v -modular matrices:3 . (cid:18) vv (cid:19) RU . (cid:18) (cid:19) v -modular matrices:2 . p s + vv p s + r ! , . (cid:18) vv (cid:19) Now, suppose n · H ( i ) ⊂ n · L = v m O F :The other v i − modular planes: (see[J] p. 459)17 P < m < s . (cid:18) p m
11 0 (cid:19) , . (cid:18) p m p s − m r (cid:19) v -modular matrices:0 < m < s + . (cid:18) p m vv (cid:19) , . (cid:18) p m vv p s − m + r (cid:19) RU < m < s − . (cid:18) p m
11 0 (cid:19) , . (cid:18) p m p s − m r (cid:19) v -modular matrices:0 < m < s + . (cid:18) p m vv (cid:19) , . (cid:18) p m vv p − m + s + r (cid:19) We now deal with the case when the lattice is a sum of two p i -modular matrices:We have the following representatives:(Both RP and RU) (cid:18) p l e p l e (cid:19) , l ≥ l , e i ∈ { , D } , e = i f l − l ≤ s (see [J] p. 463)The representative we presented (beside the diagonals) are the 1 and v modularrepresentatives , up to multiplying by the scalar p a ( a ∈ Z ) , this are all the representa-tives of K \ X .We denote by [ K \ X ] to be the set of representatives that were have presented.18 The spherical functions for Case 3: L ( x , c , c , s , s ) = ´ K ′ c ( d ( k · x )) · | d ( k · x ) | s c ( d ( k · x )) · | d ( k · x ) | s dk Substitute: s = − z + z − s = − z + We denote : z = ( z , z ) . For any a , a , z z ∈ C h ( a , a ) , ( b , b ) i = h ( a , a ) , z i = a z + a z By abuse of notation we will some times denote: L ( x , c , c , z , z ) = L ( x , c , c , s ( z ) , s ( z )) From now on we will denote by s ∈ { s , s } = S ⊂ Aut ( C ( q ± z , q ± z )) where : s = Id s ( q z ) = q z , s ( q z ) = q z . Theorem 6.1.
Let x ∈ [ K \ X ] andL ( x , c , c , s , s ) = ˆ K ′ (cid:213) i = d i ( c i ( k · x )) | d i ( k · x ) | s i dkas was defined in Section 2 then L ( x , c , c , z , z ) is equal to the following : RP: L ( x , c ∗ , c , z , z ) = x = (cid:18) p l e p l e (cid:19) with l − l > s If x = (cid:18) p l e p l e (cid:19) with l − l > s then: (Hironaka ) L ( x , c ∗ , c , z ) = q l − l c ∗ ( e ) c ( e e ) + q − q sz ( q z − q z − ) × (cid:229) s ∈ S s ( q h ( l − s , l ) , z i q z − q z ) . L ( (cid:18) p l e p l e (cid:19) , , c , z ) = q l − l c ( e e ) q − sz ( q − + )( q z − q z ) × [ c ∗ ( − e e ) q < ( l + s , l ) , z > ( q z − q z − ) + q h ( l , l + s ) , z i ( q z − q z − )] L ( (cid:18) (cid:19) , , c , z ) = c ( − )( − q − ) q l − s · z q − + [ q s ( z + z ) ( q z + q z ) q z − q z ] L ( (cid:18) vv (cid:19) , , c , z ) == c ( − )( − q − ) q l + q − sz q s ( z + z ) [ q z + z q z − q z ] L ( (cid:18) p s − p s r (cid:19) , , c , z ) = c ( − D ) q − s · z + l q s · ( z + z ) < m < s L ( (cid:18) p m
11 0 (cid:19) , , c , z ) == c ( − ) q − + q m − sz ( − q − ) q z − q z (cid:229) s s ( q h ( m , s − m + ) , z i ) < m < s + L ( (cid:18) p m vv (cid:19) , , c , z ) == c ( − ) q − q − + q m − sz q z − q z [ (cid:229) s s ( q h ( m , + s − m ) , z i − q h ( s + − m , m + ) , z i− )] < m < s L ( (cid:18) p m − p s − m r (cid:19) , , c , z ) = c ( − D ) q m − sz q − + [ (cid:229) s s ( q h ( m , − m + s ) , z i q z − q z )] < m < s + L ( (cid:18) p m vv − p s − m r (cid:19) , , c , z ) == c ( − D ) q m − − s z q − + [ (cid:229) s s ( q h ( m , + s − m ) , z i q z − q z )] RU L ( (cid:18) (cid:19) , , c , z ) = q c ( − )( − q − ) q − sz [ q ( s + )( z + z ) q z − q z ] L ( (cid:18) vv (cid:19) , , c , z ) = c ( − )( − q − ) q l + q − s · z + q − q ( s + )( z + z ) [ q z + q z q z − q z ] L ( x , p s + vv − p s + r ! , c , z ) = c ( − D ) q − s · z + s [ q ( s + )( z + z ) ] For the following representatives: (cid:18) p m
11 0 (cid:19) , (cid:18) p m vv (cid:19) , (cid:18) p m − p s − m r (cid:19) (cid:18) p m vv − p s − m r (cid:19) the result is the same as RP Case..And any a ∈ E ∗ , L ( ax , c , s , s ) = | a | s + s c ( a ) · c ( a ) L ( x , c , c , s , s ) (6.1)So one can calculate the spherical function for any K orbit in X .The proof of the theorem will be given in chapters 7 and 8. A corollary that follows from Theorem (6.1) is the following functional equations:21efine ˜ L ( x , c , c , z ) = q sz L ( x , c , c , z ) Then :˜ L ( x , , c , z , z ) = − c ∗ ( − ) ˜ L ( x , , c ∗ c , z , z ) ˜ L ( x , c ∗ , c , z , z ) = q s ( z − z ) q z − q z − q z − q z − ˜ L ( x , c ∗ , c , z , z ) And actually if we define: c = w − w c = w − We have the following functional equations :If w w − = L ( x , w , w , z , z ) = − c ∗ ( − ) ˜ L ( x , c ∗ w , c ∗ w , z , z ) If w w − = L ( x , w , w , z , z ) = q s ( z − z ) q z − q z − q z − q z − ˜ L ( x , c ∗ w , c ∗ w , z , z ) Remark . The previous transformation of characters comes form the fact that if wedefine q ( a ) = w ( a ) | a | z and q ( a ) = w ( a ) | a | z then for p = (cid:18) a xd (cid:19) we have d ( p · x ) = c ( d ( p · x )) · | d ( p · x ) | s · c ( d ( p · x )) · | d ( p · x ) | s = q ( a ) · q ( d ) · d P ( p ) · d ( x ) , where d P ( p ) is the modular character of the Borel subgroup, this property isrelated to the principal series representation of GL ( F ) . L ( x , , c , z ) : We will compute the integral over cosets of the Iwahori subgroup, BB = (cid:18) a bc d (cid:19) ∈ K , b ∈ v O F . B has the factorization: B = N − A N + ( v O F ) = (cid:18) t (cid:19) (cid:18) a a (cid:19) (cid:18) y (cid:19) = (cid:18) a a ya t a ty + a (cid:19) ,Where a , a ∈ O ∗ F , t ∈ O F , y ∈ v O F .22he Haar measure on B , m B is taken to be dt × da × da × dy , where: dt − the Haar measure on O F normalized to be 1, da , da - the Haar measure of O ∗ F normalized to be 1 and dy - the Haar measure on v O F normalized to be q − .Overall we have m B ( B ) = q − .Here is a list of coset representatives for B \ K : b i = (cid:18) r i (cid:19) if 0 ≤ i ≤ q and r i are representative of ¯ F , q = | ¯ F | and b q + = (cid:18) (cid:19) . So: Bb i = (cid:18) a a y + a r i a t a t ( y + r i ) + a (cid:19) if 1 ≤ i ≤ q and Bb q + = (cid:18) a y a a ty + a a t (cid:19) .Instead of integrating on K , we integrated over the different cosets of B using theHaar measure of B .We want to normalize the Haar measure to be m G ( K ) = qq + .Let A = (cid:18) a c ¯ c b (cid:19) be a fixed Hermitian matrix and C = k · A · k t then using thefactorization on k one can parametrize C as follows: d ( C ) = ( N ( a ) · [ a + Tr ( ¯ c ( y + r i )) + bN ( y + r i )] k ∈ Bb i i < q + N ( a ) · [ aN ( y ) + Tr ( ¯ cy ) + b ] k ∈ Bb q + (7.1)In the first step of the proof we show that we can actually integrate over K instead of K ′ , from now on we will assume that the characters of F ∗ are defined on 0, say: c i ( ) = . Lemma 7.1.
Let A ∈ X be a fixed Hermitian matrix. Then m K measure of the set { k ∈ K | d ( k · A ) = } is .Proof. By (7.1), it is sufficient to show that the inverse image of a point of the traceand norm maps is of measure zero. Note that the trace and norm maps: Tr : O F → Tr ( O F ) , N : O ∗ F → N ( O ∗ F ) are surjective homomorphisms, and that the Haar measureof Tr ( O F ) , N ( O ∗ F ) (being an open sets) is the induced measure of O E , O ∗ E (respectively).In particular the measure of a singleton is 0. We use Lemma 4.1 to deduce that themeasure of the inverse images is also 0. 23 emma 7.2. L ( x , c , c , s , s ) = ˆ K c ( d ( k · x )) · | d ( k · x ) | s c ( d ( k · x )) · | d ( k · x ) | s dk = q c ( det ( x )) | det ( x ) | s q + [ ˆ O F | a + Tr ( ¯ ct ) + bN ( t ) | s c ( a + Tr ( ¯ ct ) + bN ( t )) dt + ˆ v O F | aN ( y ) + Tr ( cy ) + b | s c ( aN ( y ) + Tr ( cy ) + b ) dy ] . Proof.
For any f ∈ L ( X ) : ˆ K f ( k · x ) dk = qq + q + (cid:229) i = ˆ k = bb i ∈ Bb i f ( k · x ) db Using (7.1) and the Haar measure of B , We have: L ( x , c , c , s , s ) = q c ( det ( x )) | det ( x ) | s q + [ q (cid:229) i = ´ y ∈ v O F c ( a + Tr ( ¯ c ( y + r i )) + bN ( y + r i )) · | a + Tr ( ¯ c ( y + r i )) + bN ( y + r i ) | s dy + ˆ y ∈ v O F c ( aN ( y ) + Tr ( ¯ cy ) + b ) · | aN ( y ) + Tr ( ¯ cy ) + b | s dy ] . The last equation follows from the facts that | N ( a ) | = c ( N ( a )) = | det ( k · x ) | = | det ( x ) | .Notice that performing the first sum is equivalent to integrating over O F . Alto-gether we obtain our formula . L ( x , , c , z ) on the diagonal elements We use Lemma 7.2 to calculate the various cases.On the diagonal representatives: x ∼ (cid:18) p l e p l e (cid:19) L ( x , , c , s , s ) = ( l + l ) − s c ( e e ) q − + [ ˆ O F | p l e + p l e N ( t ) | s dt + ˆ v O F | e p l + e p l N ( y ) | s dy ] = q ( l + l ) − s c ( e e ) · q − l s q − + [ ˆ O F | p l − l e e + N ( t ) | s dt + ˆ v O F | + e e p l − l N ( y ) | s dy ] . The second integrand is the constant 1 since | N ( y ) | < y ∈ v O F .Note that for any t ∈ O F − v l − l O F , we have | p l − l e e + N ( t ) | s = | t | s and sowe can calculate this integral easily on this space: L ( x , , c , s , s ) = q ( l + l ) − s · q − l s c ( e e ) q − + [ l − l − (cid:229) j = q ( − s − ) j ( − q − )+ q ( l − l )( − s ) ˆ v l − l O F | e e + N ( t ) | s dt + q − ] . After the substitution v l − l y = t , we get: L ( x , , c , s , s ) = q ( l + l ) − s · q − l s c ( e e ) q − + [( − q − ) − q ( − s − )( l − l ) − q ( − s − ) ++ q ( l − l )( − s − ) ˆ O F | e e + N ( y ) | s dy + q ] . L ( x , , c , z ) = q l − l · q l z + l z c ( e e ) q − + [ q z − q z − − q ( z − z )( l − l )+ z + q ( z − z )( l − l )+ z − q z − q z + q ( l − l )( z − z ) ˆ O F | e e + N ( t ) | s dt + q − ] . (7.2)There are two cases: 25 ase 1: − e e / ∈ N ( F ) : We have by Lemma 4.5 : ˆ O F | e e + N ( t ) | dt = q z − q z − + q ( z − z ) s + z − − q s ( z − z )+ z q z − q z . We set this into Eq 7.2: L ( x , , c , z ) = q l − l c ( e e ) q − sz q − + [ q z + ( l + s ) z + l z q z − q z ++ q z ( s + l )+ z l + z − − q z ( s + l )+ z l + z q z − q z + − q z − + ( l + s ) z + l z q z − q z = q l − l c ( e e ) q − sz q − + [ q h ( s + l , l ) , z i ( q z − − q z ) + q h ( l , l + s ) , z i ( q z − q z − ) . Case 2: − e e ∈ N ( F ) : We have ˆ O F | e e + N ( t ) | s dt = q z − q z − − q s ( z − z )+ z − + q ( z − z ) s + z q z − q z . Setting this into Eq 7.2: L ( x , , c , z ) = q l − l c ( e e ) q − + [ q z + ( l + s ) z + l z q z − q z ++ q z ( s + l )+( l − s ) z + z − q z ( s + l )+( l − s ) z + z − q z − q z + − q z − + ( l + s ) z + l z q z − q z ] = q l − l c ( e e ) q − sz ( q − + )( q z − q z ) [ q h ( l + s , l ) , z i ( q z − q z − ) + q h ( l , l + s ) , z i ( q z − q z − )] . .2 Calculating L ( x , , c , z ) on the non diagonal elements: Let x = (cid:18) a v c v c b (cid:19) . Using Lemma 7.2 : L ( x , , c , s , s ) = q c ( det ( x )) | det ( x ) | s q + [ ˆ O F | a + Tr ( v c t ) + bN ( t ) | s dt + ˆ v O F | aN ( y ) + Tr ( v c y ) + b | s dy ] . In the following section we will make use of Lemma 4.1 and 3.2 (coordinate sub-stitution).
Recall : l = s , Tr ( v i O F ) = p l + i O E , Tr ( v i − O F ) = p l + i O E . x ∼ (cid:18) (cid:19) L ( x , , c , s , s ) = c ( − ) q − + [ ˆ x ∈ O F | Tr ( t ) | s dt + ˆ y ∈ v O F | Tr ( y ) | s dy ] . We substitute u = Tr ( t ) , v = Tr ( y ) L ( x , , c , s , s ) = q l c ( − ) q − + [ ˆ u ∈ p l O E | u | s du + ˆ v ∈ p l + O E | v | s dv ] . The integrals above are simple and reduces to geometric sums: L ( x , , c , s , s ) = · c ( − ) q − + [ ¥ (cid:229) i = q − i ( − q − ) q ( l + i )( − s ) + ¥ (cid:229) i = q − i ( − q − ) q ( l + i )( − s ) ] = ( − )( − q − ) q − ls q − + [ ¥ (cid:229) i = q ( − s − ) i + ¥ (cid:229) i = q ( − s − ) i ] = c ( − )( − q − ) q − ls q − + [ − q − s − + q − s − − q − s − ] . After simplifying this expression and performing the transformation s z , we ob-tain: L ( x , , c , z ) = c ( − )( − q − ) q l − lz q − + [ q s ( z + z ) ( q z + q z ) q z − q z ] . x ∼ (cid:18) vv (cid:19) L ( x , , c , s , s ) = c ( − ) q − s q − + [ ˆ x ∈ O F | Tr ( v t ) | s dt + ˆ y ∈ v O F | Tr ( v y ) | s dy ] . c ( − ) q − s + q − + [ ˆ u ∈ v O F | Tr ( u ) | s du + ˆ u ∈ v O F | Tr ( u ) | s du ] . Similarly to the previous calculations, we substitute v = Tr ( t ) , y = Tr ( u ) , anduse Lemma 4.1. Note that the image of both transformations coincides Im ( v O F ) = Im ( v O F ) = p l + O E , but the multiplicative factors (The “Jacobians”) are different: L ( x , , c , s , s ) = c ( − ) q − s + q − + [ q l ˆ u ∈ p l + O E | v | s dv + q l − ˆ u ∈ p l + O E | y | s dy ] = c ( − ) q − s + q l q − + [ ¥ (cid:229) j = q − l − i ( − q − ) q ( l + i )( − s ) + q − ¥ (cid:229) j = q − l − i ( − q − ) q ( l + i )( − s ) ] . s z : L ( x , , c , z ) = c ( − )( − q − ) q z + q − + q l ( z − z + ) [ ¥ (cid:229) j = q i ( z − z ) + q − ¥ (cid:229) j = q i ( z − z ) ] = c ( − )( − q − ) q z + q − + q l ( z − z + ) [ q z q z − q z + q z − q z − q z ] = c ( − )( − q − ) q q − + q − sz q l ( z + z + ) [ q z + z q z − q z + q z + z − q z − q z ] . After simplifying this expression, we get: L ( x , , c , z ) = c ( − )( − q − ) q l + q − sz q s ( z + z ) [ q z + z q z − q z ] . x ∼ (cid:18) p s − p s r (cid:19) , + p s r = D / ∈ N ( F ∗ ) L ( x , , c , s , s ) = c ( − D ) q − + [ ˆ O F | p s + Tr ( t ) − p s r N ( x ) | s dt + ˆ v O | p s N ( y )+ Tr ( y ) − p s r | s dy ] . Because | Tr ( y ) | , | p s N ( y ) | < v O F . L ( x , , c , s , s ) = c ( − D ) q − + [ ˆ O ∗ F | p s + Tr ( t ) − p s r N ( t ) | s dt + q − s · s − + q − s · s − ] . (7.3)Since | N ( t ) | = t ∈ O ∗ F , and N ( t − ) Tr ( t ) = Tr ( t − ) , one can verify : | p s + Tr ( t ) − p s r N ( t ) | s = | p s N ( t ) + Tr ( t ) − p s r | s Setting this into (7.3): 29 ( x , , c , s , s ) = c ( − D ) q − + [ ˆ O ∗ F | p s N ( t ) + Tr ( t ) − p s r | s dt + q − s · s − + q − s · s − ] . We substitute u = t L ( x , , c , s , s ) = c ( − D ) q − + [ ˆ O ∗ F | p s N ( u ) + Tr ( u ) − p s r | s du + q − s · s − ] . (7.4)We use the following identity: p s N ( u ) + Tr ( u ) = ( v s u + ¯ v − s )( v s u + v − s ) − p − s = N ( v s u + ¯ v − s ) − p − s Setting this identity into (7.4): L ( x , , c , s , s ) = q c ( − D ) q + [ ˆ O ∗ F | N ( v s u + ¯ v − s ) − p − s − p s r | s du + q − s · s − ] = c ( − D ) q − + [ ˆ O ∗ F | p s N ( v s u + ¯ p − s ) − p − s − p s r | s du + q − s · s − ] = c ( − D ) q − + [ ˆ O ∗ F q − s · s | p − s [ N ( v s u + ) − ( + p s r )] | s du + q − s · s − ] . Since 1 + p s r is not a norm, we have [ N ( v s u + ) − ( + p s r )] = p s h , h ∈ O ∗ E (see Section 3.1 ), we have: L ( x , , c , s , s ) = c ( − D ) q − + [ ˆ O ∗ F q − s · s du + q − s · s − ] = c ( − D ) q − + [ q − s · s ( − q − ) + q − s · s − ] = c ( − D ) q − + [ q − s · s + q − s · s − ] . Substituting s z : 30 ( x , , c , z ) = c ( − D ) q − s · z + s q s · ( z + z ) . x = (cid:18) p m
11 0 (cid:19) < m < s L ( x , , c , s , s ) = c ( − ) q − + [ ˆ O F | p m + Tr ( t ) | s dt + ˆ v O | p m N ( y ) + Tr ( y ) | s dy ] . We have that | Tr ( t ) | < | p m | (see Section (3.2)), and so the first integrand is constant: L ( x , , c , s , s ) = c ( − ) q − + [ q − s m + ˆ v O | p m N ( y ) + Tr ( y ) | s dy ] . (7.5)We use the following identity: p m N ( y ) + Tr ( y ) = ( v − m + v m y )( ¯ v − m + v m ¯ y ) − p − m = N ( v − m + v m y ) − p − m . We set it into (7.5): L ( x , , c , s , s ) = c ( − ) q − + [ q − s m + q ms ˆ v O F | N ( + p m y ) − | s dy ] . We substitute u = p m y : L ( x , , c , s , s ) = c ( − ) q − + [ q − s m + q ms + m ˆ v m + O F | N ( + u ) − | s du ] . We substitute t = + u : L ( x , , c , s , s ) = c ( − ) q − + [ q − s m + q ms + m ˆ + v m + O F | N ( t ) − | s dt ] .
31e use Lemma 4.2. The last integral splits to 3 different sums: L ( x , , c , s , s ) = c ( − ) q − + [ q − s m + q ms + m [ ¥ (cid:229) i = s + ( − q − ) q − is − i +( q − ) q − s − q − s · s + s − (cid:229) i = m + q − is − i ( − q − )] . After calculating the sums and substituting s z , we have: L ( x , , c , z ) = c ( − ) q − + [ q m ( z − z + ) + { ( − q − )[ q ( z − z )( s + ) − q z − z + ( q ( m + )( z − z ) − q s ( z − z ) ) − q z − z ]++( q − ) q m ( z − z + ) q s ( z − z ) − + z − q s ( z − z ) − + z q z − q z } ] . Simplifying this expression once more, we get: L ( x , , c , z ) = c ( − ) q − + q m − sz ( − q − ) q z − q z (cid:229) s s ( q h ( m , s − m + ) , z i ) . x = (cid:18) p m vv (cid:19) < m < s + L ( x , , c , s , s ) = c ( − ) q − s q − + [ ˆ O F | p m + Tr ( v t ) | s dt + ˆ v O F | p m N ( y ) + Tr ( v y ) | s dy ] . Once again, note that | p m + Tr ( v t ) | s = q − s m for any t ∈ O F . (see Section 3.2).We substitute u = v y : L ( x , , c , s , s ) = c ( − ) q − s q − + [ q − s m + q ˆ v O F | p m − N ( u ) + Tr ( u ) | s du ] . We use the identity: p m − N ( u ) + Tr ( u ) = N ( v − m + ¯ v m − u ) − p − m to get:32 ( x , , c , s , s ) = c ( − ) q − s q − + [ q − ms + q +( m − ) s ˆ v O F | N ( + p m − u ) − | s du ] Substitute y = p m − u : L ( x , , c , s , s ) = c ( − ) q − s q − + [ q − ms + q m − +( m − ) s ˆ v + ( m − ) O F | N ( + y ) − | s dy ] . Substitute t = + y : L ( x , , c , s , s ) = c ( − ) q − s q − + [ q − ms + q m − +( m − ) s ˆ + v m O F | N ( t ) − | s dt ] . We calculate the last integral by using Lemma 4.2 . L ( x , , c , s , s ) = c ( − ) q − s q − + { q − ms + q m +( m − )+( m − ) s × [ ¥ (cid:229) i = s + ( − q − ) q − is − i + ( q − ) q − s − q − s · s + s − (cid:229) i = m q − is − i ( − q − )] } = Substituting s z and simplifying the expression, we get: L ( x , , c , z ) = c ( − ) q − s q − + { q − ms + q m +( m − )( z − z ) × [( − q − )( q ( z − z )( s + )+ z q z − q z + q ( z − z ) m + z − q ( z − z ) s + z q z − q z )+( q − ) q s ( z − z ) − ] } = c ( − ) q z − + m q − + [ q z − z + q ( − m )( z − z ) [ ( q − ) q ( z − z )( s + )+ z − q z − q z + q − ) q ( z − z ) m + z − − ( q − ) q ( z − z ) s + z − q z − q z ++ ( q − ) q s ( z − z ) − + z − ( q − ) q s ( z − z ) − + z q z − q z ] . We extract and simplify the different elements in this expression: L ( x , , c , z ) = c ( − ) q − sz − + m q − + × [ q m z +( − m + s ) z q z − q z + q ( s + − m ) z + m z q z − q z + − q ( m + ) z +( − m + s ) z − − q ( s + − m ) z +( + m ) z − q z − q z ] Note that q sz L ( x , , c , z ) is an antisymmetric function. We can express L ( x , , c , z ) as follows: L ( x , , c , z ) = c ( − ) q − q − + q m − sz q z − q z [ (cid:229) s s ( q h ( m , + s − m ) , z i − q h ( m + , s + − m ) , z i− )] x ∼ (cid:18) p m − p s − m r (cid:19) , 0 < m < s L ( x , , c , s , s ) = q c ( − D ) q + [ ˆ O F | p m + Tr ( t ) − p s − m r N ( t ) | s dt + ˆ v O F | p m N ( y )+ Tr ( y ) − p s − m r | s dy ] . Because | Tr ( O F ) | ≤ | p s | (see Section 3.2) and | p s − m | < | p | m , we have thatthe first integrand is constant: L ( x , , c , s , s ) = q c ( − D ) q + [ q − m · s + ˆ v O F | p m N ( y ) + Tr ( y ) − p s − m r | s dy ] . (7.6)We use the identity p m N ( y ) + Tr ( y ) = N ( v − m + v m y ) − p − m and set it into Eq 7.6:34 ( x , , c , s , s ) = c ( − D ) q − + [ q − m · s + ˆ v O F | N ( v − m + v m y ) − p − m − p s − m r | s dy ] = c ( − D ) q − + [ q − m · s + q ms ˆ v O F | N ( + p m y ) − ( + p s r ) | s dy ] . Substituting t = + p m y : L ( x , , c , s , s ) = c ( − D ) q − + [ q − m · s + q ms + m ˆ + v m + O F | N ( t ) − ( + p s r ) | s dt ] . (7.7)We calculate the last integral by decomposing 1 + v m + O F = s − S j = m + ( + v j O ∗ F ) ∪ ( + v s O F ) . From Section 3.1, we deduce that for any t ∈ + p i O ∗ F and 2 m + ≤ i < s the valueof | N ( t ) − ( + p s r ) | s is q − is .Because 1 + p s r / ∈ N ( F ) , in the subgroup 1 + v s O F the integrand is simply q − s · s . Altogether, we have: L ( x , , c , s , s ) = c ( − D ) q − + [ q − m · s + q ms + m ( s − (cid:229) i = m + ( − q − ) q − i q − i s + q − s q − s · s )] = c ( − D ) q − + [ q − m · s + q ms + m ( s − (cid:229) i = m + q ( − s − ) i + q s ( − s − ) )] We simplify the sum and substitute s z to get: L ( x , , c , s , s ) = c ( − D ) q − + [ q m ( z − z + )+ z − q m · ( z − z + )+ z q z − q z + q ms + m ( q ( z − z )( m + ) − q ( z − z ) s − q z − z + q s ( z − z ) )] = ( − D ) q m q − sz q − + [ q m z +( − m + s ) z q z − q z + q ( s + − m ) z + m z q z − q z ] And after further manipulations: L ( x , , c , z ) = c ( − D ) q m − sz q − + [ (cid:229) s s ( q h ( m , − m + s ) , z i q z − q z )] . x ∼ (cid:18) p m pp − p s + − m r (cid:19) , 0 < m < s + . L ( x , , c , z ) = c ( − D ) q − s q − + [ ´ O F | p m + Tr ( ¯ v t ) − p s + − m r N ( t ) | s dt + ´ v O F | p m N ( y ) + Tr ( ¯ v y ) − p s + − m r | s dy ] Similarly to the previous calculations the first integrand is also constant, so L ( x , , c , s , s ) = c ( − D ) q − s q − + [ q − ms + ˆ v O F | p m − N ( ¯ v y )+ Tr ( ¯ v y ) − p s + − m r | s dy ] . We substitute u = v y : L ( x , , c , s , s ) = c ( − D ) q − s q − + [ q − ms + q ˆ v O F | p m − N ( u ) + Tr ( u ) − p s + − m r | s du ] . (7.8)We use the identity: p m − N ( u ) + Tr ( u ) = N ( v − m + v m − u ) − p − m and set it toEq 7.8: L ( x , , c , s , s ) = c ( − D ) q − s q − + [ q − ms + q ˆ v O F | N ( v − m + v m − u ) − p − m − p s + − m r | s du ] = c ( − D ) q − s q − + [ q − ms + q +( m − ) s ˆ v O F | N ( + v m − u ) − ( + p s r ) | s du ] .
36e substitute t = + v m − u : L ( x , , c , s , s ) = c ( − D ) q − s q − + [ q − ms + q ( m − ) s + m − ˆ + v m O F | N ( t ) − ( + p s r ) | s dt ] . The calculation of the integral is similar to the calculation of the integral in Eq 7.7: L ( x , , c , s , s ) = c ( − D ) q − s q − + [ q − ms + q ( m − )( z − z )+ m ( s − (cid:229) i = m q ( − s − ) i + q − s q − s · s )] . We substitute s z : L ( x , , c , z ) = c ( − D ) q z − + m − ( s + ) z q − + [ q m z +( + s − m ) z + z q z − q z − q ( s − m + ) z + m z q z − q z ] . = c ( − D ) q m − − s z q − + [ (cid:229) s s ( q h ( m , + s − m ) , z i q z − q z )] . Recall that l = s − , Tr ( v i O F ) = p l + + i O E and Tr ( v i − O F ) = p l + i O E . x ∼ p s + vv − p s + r ! L ( x , , c , s , s ) = c ( − D ) q − s q − + [ ´ O F | p s + + Tr ( ¯ v t ) − p s + r N ( t ) | s dt + ´ v O F | p s + N ( y )+ Tr ( ¯ v y ) − p s + r | s dy ] Similarly to previous calculation, the second integrand is constant because | p s + r | > | p s + N ( y ) + Tr ( ¯ v y ) | (see Section 3.1) . 37 ( x , , c , s , s ) = c ( − D ) q − s q − + [ ˆ O F | p s + + Tr ( ¯ v t ) − p s + r N ( t ) | s dt + q ( s + )( − s ) − ] . We have that: ˆ O F | p s + + Tr ( ¯ v t ) − p s + r N ( t ) | s dt = ˆ O ∗ F | p s + + Tr ( ¯ v t ) − p s + r N ( t ) | s dt + . ˆ v O F | p s + + Tr ( ¯ v t ) − p s + r N ( t ) | s dt = ˆ O ∗ F | p s + + Tr ( ¯ v t ) − p s + r N ( t ) | s dt + q ( s + )( − s ) − . So : L ( x , , c , s , s ) = c ( − D ) q − s q − + [ ˆ O ∗ F | p s + + Tr ¯ ( v y ) − p s + r N ( y ) | s dy + q ( s + )( − s ) − ] = c ( − D ) q − s q − + [ ˆ O ∗ F | p s + N ( y ) + Tr ¯ ( v y ) − p s + r | s dy + q ( s + )( − s ) − ] We substitute u = y : L ( x , , c , s , s ) = c ( − D ) q − s q − + [ ˆ O ∗ F | p s + N ( u )+ Tr ( v u ) − p s + r | s du + q ( s + )( − s ) − ] . (7.9)We use the identity: p s + N ( u ) + Tr ( pu ) = N (( v s + u + v − s ) − p − s − and set it toEq 7.9: L ( x , , c , z ) = q c ( − D ) q − s q + [ ˆ O ∗ F | N (( v s + u + v − s ) − p − s + − p s + r ) | s du + q ( s + )( − s ) − ] = c ( − D ) q − s q + [ ˆ O ∗ F | p − s N (( v s u + ) − p − s − p s + r ) | s du + q ( s + )( − s ) − ]= c ( − D ) q − s q − + [ ˆ O ∗ F | p − s [ N (( v s u + ) − ( + p s r )] | s du + q ( s + )( − s ) − ] But, as in the RP case, we have: N ( v s u + ) − ( + p s r ) = p s h , h ∈ O ∗ E L ( x , , c , z ) = c ( − D ) q − s q − + [ ˆ O ∗ F | p + s h | s du + q ( s + )( − s ) − ] = c ( − D ) q − s q − + [ q ( s + )( − s ) ( − q − ) + q ( s + )( − s ) − ] Substitute s z : L ( x , , c , z ) = c ( − D ) q − s · z + s [ q ( s + )( z + z ) ] . x ∼ (cid:18) (cid:19) L ( x , , c , s , s ) = c ( − ) q − + [ ´ O F | Tr ( t )) | s dt + ´ v O F | Tr ( y ) | s dy ] Similarly to the RP case , we use 3.2 to the substitution u = Tr ( t ) , v = Tr ( y ) . Wehave: L ( x , , c , s , s ) = c ( − ) q − + [ q l + ˆ p l + O E | u | s du + q l ˆ p l + O E | v | s dv ] = Calculation on the integrals above results in geometric sums:39 ( x , , c , s , s ) = c ( − ) q − + [ q ¥ (cid:229) i = q − ( l + i ) s · q − i ( − q − )+ ¥ (cid:229) i = q − ( l + i ) s · q − i ( − q − )] = c ( − ) q − + [ ¥ (cid:229) i = q − ( l + i ) s · q − i ( − q − )( q + )] = q ( − q − ) c ( − ) q − ls [ ¥ (cid:229) i = q ( − s − ) i ] . We substitute s z : L ( x , , c , z ) = q c ( − )( − q − ) q − sz q l ( z + z ) [ q z + z q z − q z ] = q c ( − )( − q − ) q − sz [ q ( s + )( z + z ) q z − q z ] x ∼ (cid:18) vv (cid:19) L ( x , , c , s , s ) = c ( − ) q − s + q − [ ˆ O F | Tr ( v t )) | s dt + ˆ v O F | Tr ( v y ) | s dy = c ( − ) q − s + + q − [ ˆ v O F | Tr ( t )) | s dt + ˆ v O F | Tr ( y ) | s dy ] We substitute u = Tr ( t ) , v = Tr ( y ) : (See Section 3.2 and Lemma 4.1) L ( x , , c , s , s ) = q l c ( − ) q − s + + q − [ ˆ p l + O E | u | s du + ˆ p l + O E | v | s dv ] = c ( − ) q − s + + q − [ ¥ (cid:229) i = q − ( l + i ) s · q − i ( − q − ) + ¥ (cid:229) i = q − ( l + i ) s · q − i ( − q − )] = We compute the geometric sums and simplify:40 ( x , , c , s , s ) = c ( − ) q − s + q − ls ( − q − ) + q − [ q ( − s − ) − q − s − + q ( − s − ) − q − s − ] = Substitute s z : L ( x , , c , z ) = c ( − ) q q l ( z − z + ) ( − q − ) + q − [ q z + z q z − q z + q z q z − q z ] = c ( − )( − q − ) q l + q − s · z + q − q l ( z + z ) q z + z [ q z + q z q z − q z ] L ( x , c ∗ , c , z ) In the following two subsections we prove that on most of the representatives thespherical function vanish.For this section, A i , B i , C i , .. will denote constants that are resulted from differentcoordinate transformation. Recall that for x ∼ (cid:18) a v c v c b (cid:19) , we have from 7.2 : L ( x , , c , s , s ) = q c ( det ( x )) | det ( x ) | s q + [ ´ O F | a + Tr ( v c t )+ bN ( t ) | s dt + ´ v O F | aN ( y )+ Tr ( v c y )+ b | s dy ] . We will denote : I = ´ O F | a + Tr ( v c t ) + bN ( t ) | s dtI = ´ v O F | aN ( y ) + Tr ( v c y ) + b | s dy . .1.1 Common Representatives for RP and RU1. x ∼ (cid:18) v a v a (cid:19) L ( x , c ∗ , c , s ) = q c ( ) q + [ ˆ O F c ∗ ( Tr ( v a t )) | Tr ( v a t ) | s dt + ˆ v O F c ∗ ( Tr ( v a t )) | Tr ( v a t ) | s dt ] After substitution u = Tr ( v a t ) (see Lemma 3.2, Lemma 4.1) we have: ˆ v k O F c ∗ ( Tr ( v a t )) | Tr ( v a t ) | s dt = A ˆ p h O E c ∗ ( u ) | u | s du = A ¥ (cid:229) j = h ˆ p j O ∗ E c ∗ ( u ) | u | s du Where h is defined by Tr ( ¯ v a + k O F ) = p h O E . But: ´ p j O ∗ E c ∗ ( u ) | u | s du = q − js − j ´ O ∗ E c ∗ ( y ) dy ∗ =
0, since c ∗ is a non trivial charac-ter on O ∗ E , the integral on each term vanishes, so L ( x , c ∗ , c , z ) = x = (cid:18) p m − p s − m r (cid:19) , 0 < m < s • We show that I = I = ˆ O F c ∗ ( p m + Tr ( t ) − p s − m r N ( t )) | p m + Tr ( t ) − p s − m r N ( t ) | s dt Note that | Tr ( t ) − p s − m r N ( t ) | < | p m | (see Section 3.3) and so the absolute valueis constant: I = A ˆ O F c ∗ ( p m + Tr ( t ) − p s − m r N ( t )) dt We use the identity: p m + Tr ( t ) − p s − m r N ( t ) = p m + p m − s r − p s − m r N ( t − p m − s r ) tohave: 42 = ˆ O F c ∗ ( p m + p m − s r − p s − m r N ( t − p m − s r )) dt Denote by A = c ∗ ( p m − s r ) , then we have: I = A ˆ O F c ∗ ( + p s r − p ( s − m ) r N ( t − p m − s r )) dt = A ˆ O F c ∗ ( + p s r − N ( p s − m r t − )) dt . We Substitute u = p s − m r t −
1, we have: I = A ˆ + v ( s − m ) O F c ∗ ( + p s r − N ( u )) du . We substitute n = N ( u ) , note that N ( + v ( s − m ) O F ) = + p s − m + s O E (see 3.1).We make use of Lemma 4.1 :we have I = A ˆ + p s − m + s O E c ∗ [( + p s r ) − t ] dt . We substitute s = ( + p s r ) − t . Note that p s r − p s − m + s O E = p s r ( + p s − m O E ) ,so for the substitution p s r y = s we get: I = A ˆ p s r ( + p s − m O E ) c ∗ ( s ) ds = A ˆ + p s − m O E c ∗ ( y ) dy Because 1 + p s − m O F contains non norm elements, the character c ∗ is a non trivialcharacter of the group 1 + p s − m O E and the integral vanish. • Now we show that: 43 = ˆ v O F c ∗ ( p m N ( y ) + Tr ( y ) − p s − m r ) | p m N ( y ) + Tr ( y ) − p s − m r | s dy = I = ˆ v O c ∗ ( p m N ( y ) + Tr ( y ) − p s − m r ) | p m N ( y ) + Tr ( y ) − p s − m r | s dy = B ˆ + v m + O F c ∗ ( N ( t ) − ( + p s r )) | N ( t ) − ( + p s r ) | s dt ] The Norm induces an homomorphism : e N : 1 + v m + O F + v s + O F → + p m + O F + p s + O F On each coset the | N ( t ) − ( + p s r ) | s is a constant function. (By our corollaries inSection 3.1)Note that q − s < | N ( t ) − ( + p s r ) | E ≤ q − m − since N ( t ) ∈ + p m + O E and 1 + p s r / ∈ N ( F ∗ ) .We integrate on the different coset of the form a i ( + v s + O F ) ⊂ + v m + O F I = ˆ + v m + O F c ∗ ( N ( t ) − ( + p s r )) | N ( t ) − ( + p s r ) | s dy = (cid:229) i B i ˆ a i ( + v s + O F ) c ∗ ( N ( t ) − ( + p s r )) dt Now we show that each integral in the different terms vanish: I , i = ˆ a i ( + v s + O F ) c ∗ ( N ( t ) − ( + p s r )) | dt . We Substitute u = N ( t ) (note that N ( + v s + O F ) = + p s + O E ), by making useof Lemma 4.1 we get: I , i = C ˆ N ( a i )( + p s + O E ) c ∗ ( u − ( + p s r )) du
44e substitute h = u − ( + p s r ) . Since we know that q − s < | h | E ≤ q − m − , wededuce that we can present the integral in the following form: I , i = C ˆ p k z i + p s + O E c ∗ ( h ) d h , where z i ∈ O ∗ F and m + ≤ k < s . We substitute again y = p − k z − i · h to get: I , i = C ˆ + p s + − k O E c ∗ ( y ) dy = . Since 1 + p s + − k O E contains a non-norm elements, c ∗ is a non-trivial character andthe integral vanish. x = (cid:18) p m vv − p s − m r (cid:19) A similar proof to the previous representative show that L ( x , c ∗ , s , s ) = . x = (cid:18) p m
11 0 (cid:19) • We prove that I = ´ O F c ∗ ( p m + Tr ( t )) | p m + Tr ( t ) | s dt = u = Tr ( t ) . We know that (see Section 3.2) Tr ( O F ) = p h O F , s ≤ h ≤ s + ˆ O F c ∗ ( p m + Tr ( t )) | p m + Tr ( t ) | s dt = A ˆ p h O E c ∗ ( p m + u ) | p m + u | s du . Since m < h the absolute value is constant. We have: I = A ˆ p s O E c ∗ ( p m + t ) dt = A ˆ + p h − m O E c ∗ ( t ) dt = h − m < s + + p h − m O E contains a non normelements and hence c ∗ is a non-trivial character and the integral vanishes.45 We show that I = . We have already shown in Section 7.2.1 I = ˆ v O F c ∗ ( p m N ( y )+ Tr ( y )) | p m N ( y )+ Tr ( y ) | s dy ] = B ˆ + v m + O F c ∗ ( N ( t ) − ) | N ( t ) − | s dt . For every 2 m + ≤ k we show that the integral vanish on the set N − ( + p k O ∗ E ) ∩ ( + v m + O F ) :For k < s + N − ( + p k O ∗ E ) ∩ ( + v m + O F ) can be represented as aunion of cosets: N − ( + p k O ∗ E ) ∩ ( + v m + O F ) = ∪ i a i [ + v s + O F ] . On each coset the absolute value is constant ans so: ˆ N − ( + p k O ∗ E ) ∩ ( + v m + O F ) c ∗ ( N ( t ) − ) | N ( t ) − | s dt = (cid:229) i C i ˆ a i [ + v s + O F ] c ∗ ( N ( t ) − ) dt We substitute N ( t ) − = u (use Lemma 4.1 and Section 3.1) , then on each term inthe sum: ˆ a i [ + v s + O F ] c ∗ ( N ( t ) − ) dt = A i ˆ N ( a i )[ + p s + O E ] − c ∗ ( u ) du But we know that | u | E = q − k and so we can represent the domain of integration as: N ( a i )( + p s + O E ) − = p k h i + p s + O E , h i ∈ O ∗ E We have: ˆ N ( a i )[ + p s + O E ] c ∗ ( u − ) du = ˆ p k h i + p s + O E c ∗ ( u ) du We substitute s = p − k h − i u and get: ˆ p k h i + p s + O E c ∗ ( u ) du = ˆ + p s + − k O E c ∗ ( s ) ds = + p s + − k O E contains a non-norm elements c ∗ is a non-trivial character andthe integral vanish.For s + ≤ k we can represent N − ( + p k O ∗ E ) ∩ + v m + O F = ∪ i a i ( + v k + O F ) Now we integrate on coset of the form a i ( + v k + O F ) , similarly the absolute valueis constant: ˆ N − ( + p k O ∗ E ) ∩ ( + v m + O F ) c ∗ ( N ( t ) − ) | N ( t ) − | s dt = (cid:229) i D i ˆ a i [ + v k + O F ] c ∗ ( N ( t ) − ) dt . We substitute N ( t ) − = u . On each term : ˆ a i [ + v k + O F ] c ∗ ( N ( t ) − ) dt = ˆ N ( a i )( + p k + O E ) c ∗ ( u ) du But | u | E = q − k so one can represent the domain of integration as: N ( a i )( + p k + O E ) = p k h i + p k + O E , h i ∈ O ∗ E We have by making the substitution y = p − k h − i : ˆ p k h i + p k + O E c ∗ ( u ) du = D ˆ + p O E c ∗ ( y ) dy = . We showed that ´ N − ( + p k O ∗ E ) ∩ ( + v m + O F ) c ∗ ( N ( t ) − ) | N ( t ) − | s dt = m + ≤ k , and so : ˆ + v m + O F c ∗ ( N ( t ) − ) | N ( t ) − | s dt = ⇒ I = . x = (cid:18) p m vv (cid:19) -A similar proof to the previous will show that L ( x , c ∗ , s , s ) = . .1.2 RP representative: x = (cid:18) p s − p s r (cid:19) (RP) L ( x , , c , z ) = c ( − D ) q − + [ ˆ O F c ∗ ( p s + Tr ( t ) − p s r N ( t )) | p s + Tr ( t ) − p s r N ( t ) | s dt + ˆ v O c ∗ ( p s N ( y ) + Tr ( y ) − p s r ) | p s N ( y ) + Tr ( y ) − p s r | s dy ] . • We show that I = I = ˆ O F c ∗ ( p s + Tr ( t ) − p s r N ( t )) | p s + Tr ( t ) − p s r N ( t ) | s dt (8.1)Setting the following identity to (8.1): − rp s N ( t )+ Tr ( u ) = ( v s a t + a − v − s )( v s a t + a − v − s )+ p − s r = − p − s r N ( p s r t + )+ p − s r I = ˆ O F c ∗ ( p s − p − s r N ( p s r t + ) + p − s r ) | p s − p − s r N ( p s r t + ) + p − s r | s dt Simplifying this expression, we have: I = A ˆ O F c ∗ ( + p s r − N ( p s r t + )) | + p s r − N ( p s r u + ) | s dt . We Substitute u = + p s r t : I = A ˆ + p s O F c ∗ ( + p s r − N ( u )) | + p s r − N ( u ) | s dt . Since N ( u ) ∈ + p s O E and 1 + p s r / ∈ N ( F ∗ ) we have that | N ( u ) − ( + p s r ) | = q − · s · s : I = B ˆ + p s O F c ∗ ( + p s r − N ( u )) du . n = N ( u ) : I = C ˆ + p s O E c ∗ (( + p s r ) − n ) d n . Substitute y = + p s r − n : I = D ˆ p s r + p s O E c ∗ ( y ) dy . Substitute t = p − s r − y : I = E ˆ + p s O E c ∗ ( y ) dy = . • We Show that I =
0, since the absolute value in the integral is constant, we havethat: I = ˆ v O c ∗ ( p s N ( y ) + Tr ( y ) − p s r ) | p s N ( y ) + Tr ( y ) − p s r | s dy = A ˆ v O F c ∗ ( p s N ( y ) + Tr ( y ) − p s r ) dy We use the identity: p s N ( y ) + Tr ( y ) = ( v − s + v s y )( v − s + v s y ) − p − s = N ( v − s + v s y ) − p − s . Setting it into the integral we get: ˆ v O F c ∗ ( p s N ( y ) + Tr ( y ) − p s r ) dy = ˆ v O F c ∗ ( N ( v − s + v s y ) − p − s − p s r ) dy = ˆ v O F c ∗ ( N ( + v s y ) − ( + p s r )) dy We substitute t = + v s y to get: I = A ˆ + v s + O F c ∗ ( N ( t ) − ( + p s r )) dx We substitute u = N ( x ) and use Lemma 4.1 :49 = A ˆ + p s + O E c ∗ ( u − ( + p s r ) du (8.2)We substitute n = u − ( + p s r ) I = A ˆ − p s r + p s + O E c ∗ ( n ) d n . We substitute − p s r · z = n to get: I = A ˆ + p O E c ∗ ( z ) d z = . Since 1 + p O E contains a non-norm, c ∗ is a non-trivial character and the integralvanish. x ∼ p s + vv p s + r ! Showing that L ( x , c ∗ , s , s ) = L ( c ∗ , c , x , z ) on the diagonal representatives. x ∼ (cid:18) p l e p l e (cid:19) By Lemma 7.2: L ( x , c ∗ , c , s , s ) = q l − l q l z + l z c ( e e ) c ∗ ( e ) q − + [ I + I ] . Where: I = ´ x ∈ O F c ∗ ( p l − l e e + N ( x )) | p l − l e e + N ( x ) | s dxI = ´ y ∈ v O F c ∗ ( + e e p l − l N ( y )) | + e e p l − l N ( y ) | s dy . .2.1 The representatives with l − l < s Note that I = C ˆ y ∈ v O F c ∗ ( + e e p l − l N ( y )) dy = D ˆ y ∈ O F c ∗ ( + e e p l − l + N ( y )) dy By Lemma 4.3 this integral vanishes.Showing I = I = ˆ t ∈ O F c ∗ ( p l e + p l e N ( t )) | p l e + p l e N ( t ) | s dt = q − l s ˆ x ∈ O F c ∗ ( p l − l e e + N ( t )) | p l − l e e + N ( t ) | s dt This is most complicated and tricky, we prove it with careful steps:
Step 1 :We Show that If 0 ≤ m < sI = A ˆ x ∈ O F c ∗ ( p m e e + N ( t )) | p m e e + N ( t ) | s dx = ∪ i v i O ∗ F : I = A (cid:229) i ˆ v i O ∗ F c ∗ ( p m e e + N ( t )) | p m e e + N ( t ) | s dt For i ≤ m the absolute value is constant, so on each term. We substitute t = u − toget: ˆ O ∗ F c ∗ ( p m − i e e + N ( t )) dt = B ˆ O ∗ F c ∗ ( + p m − i e e N ( u )) du By Lemma 4.4 this integral vanishes. 51o we have: I = A (cid:229) i ≥ m ˆ v i O ∗ F c ∗ ( p m e e + N ( t )) | p m e e + N ( t ) | s dt = ˆ v m O F c ∗ ( p m e e + N ( t )) | p m e e + N ( t ) | s dt On the space v m + O F the absolute value is constant again, Substitute u = v − m − t : ˆ v m + O F c ∗ ( p m e e + N ( t )) dt = C ˆ O F c ∗ ( e e + p N ( u )) du = I = A ˆ v m O ∗ F c ∗ ( p m e e + N ( t )) | p m e e + N ( t ) | s dt . Substitute u = v − m t to get: I = C ˆ x ∈ O ∗ F c ∗ ( e e + N ( t )) | e e + N ( t ) | s dt (8.3)Observe that the integrand is not a locally constant function. Step 2:
We compute the integral in (8.3) on cosets of the group: O ∗ F + v O F ˆ O ∗ F c ∗ ( e e + N ( t )) | e e + N ( t ) | s dt = S i ˆ a i ( + v O F ) c ∗ ( e e + N ( t )) | e e + N ( t ) | s dt . We substitute on each term u = a − i t .Note that the integrand is invariant to the substitution, it follows that:52 = D ˆ + v O F c ∗ ( e e + N ( t )) | e e + N ( t ) | s dt Let z ∈ + v O F . we prove: J = ˆ + v O F c ∗ ( z + N ( t )) | z + N ( t ) | s dx = . We decompose the integration on spaces such that the absolute value would beconstant.We have that (up to a measure zero subset): N ( + v O F ) ⊆ + p O E = ∪ j ≥ ( + p j O ∗ E ) . Observe that: ∪ j ≥ ( + p j O ∗ E ) = ∪ j ≥ ( − z + p j O ∗ E ) . Now we calculate J on the spaces N − ( − z + p j O ∗ E ) : J = ¥ (cid:229) j = ˆ N − ( + p j O ∗ E ) c ∗ ( z + N ( t )) | z + N ( t | s dt If 0 < j ≤ s , the inverse image N − ( − z + p j O ∗ E ) is a disjoint union of coset ∪ k a k , j ( + v s + O F ) .For s < j , N − ( − z + p j O ∗ E ) can be represented as union of b k , j ( + v j + O F ) We compute the integral on each coset: J = s (cid:229) j = ˆ N − ( + v O F ) c ∗ ( z + N ( t )) | z + N ( t ) | s dt + (cid:229) j > s ˆ N − ( − z + p j O ∗ E ) c ∗ ( z + N ( t )) | z + N ( t ) | s dt = s (cid:229) j = (cid:229) k ˆ a k , j ( + v s + O F ) c ∗ ( z + N ( t )) | z + N ( t ) | s dt + (cid:229) j > s (cid:229) k ˆ b k , j ( + v j + O F ) c ∗ ( z + N ( t )) | z + N ( t ) | s dt . t ∈ a k , j ( + v s + O F ) we have that | z + N ( t ) | E = q − j , after thesubstitution: n = z + N ( t ) , each one of the domains of integration a k , j ( + p s + O F ) can be represented : p j h i , j + p s + O E , h i , j ∈ O ∗ F And so we have that: ˆ a k , j ( + v s + O F ) c ∗ ( z + N ( t )) | z + N ( t ) | s dt = D i , j ˆ p j h i , j + p s + O E c ∗ ( n ) | n | s d n . Substitute u = p − j h − i , j n to get: ˆ p j h i , j + p s + O E c ∗ ( n ) | n | s d n = E i , j ˆ x ∈ + p s + − j O E c ∗ ( t ) dt =
0. Because c ∗ is a non trivial character on 1 + p s + − j O E .A similar trick will show that the integral ´ b k , j ( + v j + O F ) c ∗ ( z + N ( t )) | z + N ( t ) | s dt = . We deduce that: J = = ⇒ I = I = I = = ⇒ L ( x , c ∗ , c , z ) = l − l = s : Suppose l − l = s In this case the I = ˆ y ∈ v O F c ∗ ( + e e p l − l N ( y )) | + e e p l − l N ( y ) | s dy = I + I = I − ˆ O ∗ F c ∗ ( p l − l e e + N ( t )) dt + ˆ O ∗ F c ∗ ( p l − l e e + N ( t )) dt + I t t − , we have:: ˆ O ∗ F c ∗ ( p l − l e e + N ( t )) dt = ˆ O ∗ F c ∗ ( p l − l e e N ( t ) + ) dt . By Lemma 4.3, the sum of the integrals vanish: ˆ O ∗ F c ∗ ( p l − l e e N ( t ) + ) dt + I = ˆ O F c ∗ ( + p l − l e e N ( t )) dt = I + I = I − ˆ O ∗ F c ∗ ( p l − l e e + N ( t )) dt . But: I − ˆ O ∗ F c ∗ ( p l − l e e + N ( t )) dt = C ˆ v O F | p l − l e e + N ( t ) | c ∗ ( p l − l e e + N ( t )) dt = D ˆ O F | p l − l − e e + N ( t ) | c ∗ ( p l − l − e e + N ( t )) dt The last integral vanish by Subsection 8.2.1 ( l − l < s ). l − l > s This is the only case where the spherical function do not vanish and c = c ∗ : L ( x , c ∗ , c , s , s ) = q l − l q l z + l z c ( e e ) c ∗ ( e ) q − + [ I + I ] I = ˆ O F c ∗ ( p l − l e e + N ( t )) | p l − l e e + N ( t ) | s dtI = ˆ y ∈ v O F c ∗ ( + e e p l − l N ( y )) | + e e p l − l N ( y ) | s dy + e e p l − l N ( y ) is a norm if y ∈ v O F it could be shown easily that I = q − . We integrate I on the spaces ∪ i ≥ v i O ∗ F : • If l − l − s > i and t ∈ v i O ∗ F then p l − l e e + N ( t ) is a norm and ˆ v i O ∗ F c ∗ ( p l − l e e + N ( t )) | p l − l e e + N ( t ) | s dx = q − is − i ( − q − ) . So I = l − l − s − (cid:229) j = ( − q − ) q − js − j + (cid:229) i ≥ l − l − s ˆ v i O ∗ F c ∗ ( p l − l e e + N ( t )) | p l − l e e + N ( t ) | s dt • For i = l − l − s , we have that if t ∈ v i O ∗ F then | p l − l e e + N ( t ) | s = q − i · s : ˆ x ∈ v i O ∗ F c ∗ ( p l − l e e + N ( t )) | p l − l e e + N ( t ) | s dt = q − i · s ˆ x ∈ v i O ∗ F c ∗ ( p l − l e e + N ( t )) dt Substituting u = v i t − , we get: q − i · s ˆ x ∈ v i O ∗ F c ∗ ( p l − l e e + N ( t )) dt = q − i · s − i ˆ O ∗ F c ∗ ( + p s e e N ( u )) du . (8.4)By Lemma 4.3: ˆ O F c ∗ ( + p s e e N ( t )) dt = ⇒ ˆ O ∗ F c ∗ ( + p s e e N ( t )) dt = − ˆ v O c ∗ ( + p s e e N ( t )) dt = − q − (8.5)56ubstituting (8.5)to Eq 8.4 , we get that for i = l − l − s : ˆ v i O ∗ F c ∗ ( p l − l e e + N ( t )) | p l − l e e + N ( t ) | s dt = − q − i · s − i − . So : I = l − l − s − (cid:229) j = ( − q − ) q − js − j − q − ( l − l − s ) · s − ( l − l − s ) − + ˆ v l − l − s + O F c ∗ ( p l − l e e + N ( t )) | p l − l e e + N ( t ) | s dt Note that: ˆ v l − l − s + O F c ∗ ( p l − l e e + N ( t )) | p l − l e e + N ( t ) | s dt = C ˆ O F c ∗ ( e e + p − s + N ( t )) | e e + p − s + N ( t )) | s dt = C ˆ O F c ∗ ( p s − e e + N ( t )) | e e p s − + N ( t )) | s dt We apply the case of l − l < s to get that: C ˆ O F c ∗ ( p s − e e + N ( t )) | e e p s − + N ( t )) | s dt = I = ˆ O F c ∗ ( p l − l e e + N ( t )) | p l − l e e + N ( t ) | s dt = l − l − s − (cid:229) j = ( − q − ) q − js − j ] − q − ( l − l − s ) · s − ( l − l − s ) − =[( − q − ) − q ( l − l − s ) · ( z − z ) − q z − z − q i ( z − z + ) − ] . L ( x , c ∗ , c , s , s ) = q l − l q l z + l z c ( e e ) c ∗ ( e ) q − + I After simplifying: L ( x , c ∗ , c , s , s ) = q l − l q l z + l z c ( e e ) c ∗ ( e ) q − + × [ q − + ( − q − ) q z − q ( l − l − s ) · ( z − z )+ z q z − q z − q ( l − l − s )( z − z ) − ] After further simplification: L ( x , c ∗ , c , s , s ) = q l − l q l z + l z c ( e e ) c ∗ ( e ) q − + × [ q z − q z − q z + − q z + q ( l − l − s ) · ( z − z )+ z q z − q z + − q ( l − l − s )( z − z ) − + z q z − q z ]= q l − l c ( e e ) c ∗ ( e ) q sz ( q z − q z − ) q − + × ( q l z + l z − sz − q l z + l z − sz ) q z − q z We represent the function in a more symmetric form: L ( x , c ∗ , c , s , s ) = q l − l c ∗ ( e ) c ( e e ) + q − q sz ( q z − q z − ) × (cid:229) s ∈ S s ( q h ( l − s , l ) , z i q z − q z ) . References [CS] W. Casselman: The unramified principal series of p-adic groups I . Thespherical function, Compositio. Math. (1988), 203-223.[H2] Y. Hironaka: Spherical functions of Hermitian and symmetric forms II,Japan. J. Math. (1989), 15-51.[H3] Y. Hironaka: Spherical functions of Hermitian and symmetric forms III,Tohoku Math.J. (1988), 651-671.[H4] Y. Hironaka: Spherical functions and local densities of hermitian forms,J. Math. Soc. Japan. (1999).[H5] Y. Hironaka: Spherical functions of hermitian and symmetric forms over2-adic fields„ Comment. Math. Univ. St. Pauli, (1990), 157-193.[HS] Y. Hironaka and F. Sato, Local Densities of Representations of QuadraticForms over p-Adic Integers (The Non-Dyadic Case). J. Number Theory (2000), no. 1, 106-136.[J] R. Jacobowitz, "Hermitian forms over local fields," American Journal ofMathematics, vol. (1962), p. 441-465.[M] I.G MacDonald, “Symmetric Functions and Hall Polynomials”, Clare-don, Oxford (1979) ,298-299.[O] O. Offen : Relative spherical functions on p-adic symmetric spaces (threecases). Pacific J. Math.215