aa r X i v : . [ m a t h . R T ] F e b A PROBLEM ON ODD UNITARY GROUPS
SUBHA SANDEEP REPAKA
Abstract.
We study a problem concerning parabolic induction in certain p -adic unitarygroups. More precisely, for E/F a quadratic extension of p -adic fields the associated unitarygroup G = U( n, n + 1) contains a parabolic subgroup P with Levi component L isomorphicto GL n ( E ) × U ( E ). Let π be an irreducible supercuspidal representation of L of depth zero.We use Hecke algebra methods to determine when the parabolically induced representation ι GP π is reducible. Contents
1. Introduction 12. Preliminaries 43. Representation ρ of P
74. Calculation of N G ( P ) 75. Calculation of N G ( ρ ) 76. Structure of H ( G, ρ ) 87. Structure of H ( L, ρ ) 168. Calculation of simple H ( L, ρ )-modules 179. Final calculations to answer the question 1710. Answering the question 19References 191. Introduction
In this paper we solve a similar problem as the one which we did in [9]. In [9], we solved theproblem for U ( n, n ) over non- Archimedean local fields where as in this paper we are solvingthe same problem for U ( n, n + 1) over non- Archimedean local fields. Refer to the section 1in [9] for better understanding of what we are doing in this paper.Let G = U( n, n + 1) be the odd unitary group over non- Archimedean local field E and π is a smooth irreducible supercuspidal depth zero representation of the Siegel Levi component L ∼ = GL n ( E ) × U ( E ) of the Siegel parabolic subgroup P of G . The terms P, L, π, U( n, n + 1)are described in much detail later in the paper. We use Hecke algebra methods to determinewhen the parabolically induced representation ι GP π is reducible. Harish-Chandra tells us tolook not at an individual ι GP π but at the family ι GP ( πν ) as ν varies through the unramfiedcharacters of L ∼ = GL n ( E ) × U ( E ). The unramified characters of L and the functor ι GP arealso described in greater detail later in the paper. Date : February 16, 2021.2010
Mathematics Subject Classification.
Primary 22E50, Secondary 11F70.
Before going any further, let us describe how U( n, n + 1) over non-Archimedean local fieldslooks like. Let
E/F be a quadratic Galois extension of non-Archimedean local fields wherechar F = 2. Write − for the non-trivial element of Gal( E/F ). The group G = U( n, n + 1) isgiven by U( n, n + 1) = { g ∈ GL n +1 ( E ) | t gJ g = J } for J = Id n Id n where each block is of size n and for g = ( g ij ) we write g = ( g ij ).We write O E and O F for the ring of integers in E and F respectively. Similarly, p E and p F denote the maximal ideals in O E and O F and k E = O E / p E and k F = O F / p F denote theresidue class fields of O E and O F . Let | k F | = q = p r for some odd prime p and some integer r > E over F . One is the unramified extension andthe other one is the ramified extension. In the unramified case, we can choose uniformizers ̟ E , ̟ F in E, F such that ̟ E = ̟ F so that we have [ k E : k F ] = 2 , Gal( k E /k F ) ∼ = Gal( E/F ).As ̟ E = ̟ F , so ̟ E = ̟ E since ̟ F ∈ F . As k F = F q , so k E = F q in this case. In theramified case, we can choose uniformizers ̟ E , ̟ F in E, F such that ̟ E = ̟ F so that wehave [ k E : k F ] = 1 , Gal( k E /k F ) = 1. As ̟ E = ̟ F , we can further choose ̟ E such that ̟ E = − ̟ E . As k F = F q , so k E = F q in this case.We write P for the Siegel parabolic subgroup of G . Write L for the Siegel Levi componentof P and U for the unipotent radical of P . Thus P = L ⋉ U with L = ( a λ
00 0 t a − | a ∈ GL n ( E ) , λ ∈ E × , λλ = 1 ) and U = ( Id n u X − t u Id n | X ∈ M n ( E ) , u ∈ M n × ( E ) , X + t X + u t u = 0 ) . Note that L ∼ = GL n ( E ) × U ( E ) and U ( E ) ∼ = U ( O E ). Let P = L ⋉ U be the L -oppositeof P where U = ( Id n − t u X u Id n | X ∈ M n ( E ) , u ∈ M n × ( E ) , X + t X + u t u = 0 ) . Let K = GL n ( O E ) and K = Id n + ̟ E M n ( O E ). Note K = Id n + ̟ E M n ( O E ) is thekernel of the surjective group homomorphism( g ij ) −→ ( g ij + p E ) : GL n ( O E ) −→ GL n ( k E )Let π be a depth zero representation of L ∼ = GL n ( E ) × U ( E ). So π = λχ where λ is adepth zero representation of GL n ( E ) and χ is a depth zero character of U ( E ). We say π is adepth zero representation of the Siegel Levi component L of P if λ K = 0 and χ | U (1+ p E ) = 1.Let ( ρ, V ) be a smooth representation of the group H which is a subgroup of K . Thesmoothly induced representation from H to K is denoted by Ind KH ( ρ, V ) or Ind KH ( ρ ). Let PROBLEM ON ODD UNITARY GROUPS 3 us denote c - Ind KH ( ρ, V ) or c - Ind GP ( ρ ) for smoothly induced compact induced representationfrom H to K .The normalized induced representation from P to G is denoted by ι GP ( ρ, V ) or ι GP ( ρ ) where ι GP ( ρ ) = Ind GP ( ρ ⊗ δ / P ), δ P is a character of P defined as δ P ( p ) = k det ( Ad p ) | Lie U k F for p ∈ P and Lie U is the Lie-algebra of U . We work with normalized induced representations ratherthan induced representations in this paper as results look more appealing.Write L ◦ for the smallest subgroup of L containing the compact open subgroups of L . Wesay a character ν : L −→ C × is unramified if ν | L ◦ = 1. Observe that if ν is an unramifiedcharacter of L then ν = ν ′ β where ν ′ is an unramified character of GL n ( E ) and β is anunramified character of U ( E ). But as U ( E ) = U ( O E ), so β is trivial. Hence, ν can beviewed as an unramified character of GL n ( E ). Let the group of unramified characters of L be denoted by X nr ( L ).1.1. Question.
The question we answer in this paper is, given π an irreducible supercuspidalrepresentation of L of depth zero, we look at the family of representations ι GP ( πν ) for ν ∈ χ nr ( L ) and we want to determine the set of such ν for which this induced representation isreducible for both ramified and unramified extensions. By general theory, this is a finite set.Recall that π = λχ where λ is an irreducible supercuspidal depth zero representationof GL n ( E ) and χ is a supercuspidal depthzero character of U ( E ). Now λ | K contains anirreducible representation τ of K such that τ | K is trivial. So τ can be viewed as an irreduciblerepresentation of K /K ∼ = GL n ( k E ) inflated to K = GL n ( O E ). The representation τ iscuspidal by (a very special case of) A.1 Appendix [7]. Set ρ = τ χ which is a representationof K × U ( O E ). Further, we can view ρ = τ χ as a representation of GL n ( k E ) × U ( k E )inflated to K × U ( O E ).By the work of Green or as a very special case of the Deligne-Lusztig construction, irre-ducible cuspidal representations of GL n ( k E ) are parametrized by the regular characters ofdegree n extensions of k E . We write τ θ for the irreducible cuspidal representation τ thatcorresponds to a regular character θ .We now define the Siegel parahoric subgroup P of G which is given by: P = GL n ( O E ) M n × ( O E ) M n ( O E )M × n ( p E ) U ( O E ) M × n ( O E )M n ( p E ) M n × ( p E ) GL n ( O E ) T U( n, n + 1) . We have P = ( P ∩ U )( P ∩ L )( P ∩ U )(Iwahori factorization of P ). Let us denote ( P ∩ U )by P − , ( P ∩ U ) by P + , ( P ∩ L ) by P . Thus P = ( a λ
00 0 t a − | a ∈ GL n ( O E ) , λ ∈ O × E , λλ = 1 ) , P + = ( Id n u X − t u Id n | X ∈ M n ( O E ) , u ∈ M n × ( O E ) , X + t X + u t u = 0 ) , P − = ( Id n − t u X u Id n | X ∈ M n ( p E ) , u ∈ M n × ( p E ) , X + t X + u t u = 0 ) . SUBHA SANDEEP REPAKA
By Iwahori factorization of P we have P = ( P ∩ U )( P ∩ L )( P ∩ U ) = P − P P + . As ρ is a representation of K , it can also be viewed as a representation of P . This is because P ∼ = K .Let Z ( L ) denote the center of L . Hence Z ( L ) = ( aId n λ
00 0 a − Id n | a ∈ E × , λ ∈ E × , λλ = 1 ) . Let us set ζ = ̟ E Id n ̟ E − . Note that Z ( L ) P = ` n ∈ Z P ζ n , so we can extend ρ to a representation ‹ ρ of Z ( L ) P via ‹ ρ ( ζ k j ) = ρ ( j ) for j ∈ P , k ∈ Z . By standard Mackey theory arguments, we show in thepaper that π = c - Ind LZ ( L ) P ‹ ρ is a smooth irreducible supercuspidal depth zero representationof L . Also note that any arbitrary depth zero irreducible supercuspidal cuspidal representationof L is an unramified twist of π . To that end, we will answer the question which we posedearlier in this paper and prove the following result. Theorem 1.
Let G = U( n, n + 1) . Let P be the Siegel parabolic subgroup of G and L bethe Siegel Levi component of P . Let π = c - Ind LZ ( L ) P ‹ ρ be a smooth irreducible supercuspidaldepth zero representation of L ∼ = GL n ( E ) × U ( E ) where ‹ ρ ( ζ k j ) = ρ ( j ) for j ∈ P , k ∈ Z and ρ = τ θ for some regular character θ of l × with [ l : k E ] = n and | k F | = q . Consider thefamily ι GP ( πν ) for ν ∈ X nr ( L ) .(1) For E/F is unramified, ι GP ( πν ) is reducible ⇐⇒ n is odd, θ q n +1 = θ − q and ν ( ζ ) ∈{ q n , q − n , − } .(2) For E/F is ramified, ι GP ( πν ) is reducible ⇐⇒ n is even, θ q n/ = θ − and ν ( ζ ) ∈{ q n/ , q − n/ , − } . Acknowledgments : The author wishes to thank Alan Roche from University of Okla-homa, USA for suggesting the problem studied in this work and for many discussions andinsights. 2.
Preliminaries
Bernstein Decomposition.
Let G be the F -rational points of a reductive algebraicgroup defined over a non-Archimedean local field F . Let ( π, V ) be an irreducible smoothrepresentation of G . According to Theorem 3.3 in [6], there exists unique conjugacy class ofcuspidal pairs ( L, σ ) with the property that π is isomorphic to a composition factor of ι GP σ forsome parabolic subgroup P of G . We call this conjugacy class of cuspidal pairs, the cuspidalsupport of ( π, V ).Given two cuspidal supports ( L , σ ) and ( L , σ ) of ( π, V ), we say they are inertiallyequivalent if there exists g ∈ G and χ ∈ X nr ( L ) such that L = L g and σ g ≃ σ ⊗ χ . Wewrite [ L, σ ] G for the inertial equivalence class or inertial support of ( π, V ). Let B ( G ) denotethe set of inertial equivalence classes [ L, σ ] G . PROBLEM ON ODD UNITARY GROUPS 5
Let R ( G ) denote the category of smooth representations of G . Let R s ( G ) be the fullsub-category of smooth representations of G with the property that ( π, V ) ∈ ob ( R s ( G )) ⇐⇒ every irreducible sub-quotient of π has inertial support s = [ L, σ ] G .We can now state the Bernstein decomposition: R ( G ) = Y s ∈ B ( G ) R s ( G ) . Types.
Refer to section 2.2 in [9] for details.2.3.
Hecke Algebras.
Refer to section 2.3 in [9] for details.2.4.
Covers.
Refer to section 2.4 in [9] for details.
Proposition 1 (Bushnell and Kutzko, Theorem 8.3 [1]) . Let s L = [ L, π ] L ∈ B ( L ) and s = [ L, π ] G ∈ B ( G ) . Say ( K L , ρ L ) is an s L -type and ( K, ρ ) is a G -cover of ( K L , s L ) . Then ( K, ρ ) is an s -type. Note that in this paper K = P , K L = K ∩ L = P ∩ L = P and ρ L = ρ . Recallthe categories R s L ( L ) , R s ( G ) where s L = [ L, π ] L and s = [ L, π ] G . Note that πν lies in thecategory R s L ( L ) and ι GP ( πν ) lies in R s ( G ).Note that H ( G, ρ ) − M od is the category of H ( G, ρ )-modules and H ( L, ρ L ) − M od be thecategory of H ( L, ρ L )-modules.The functor ι GP was defined earlier. Note that the functor m L : R s L ( L ) −→ H ( L, ρ L ) − M od is given by m L ( πν ) = Hom K L ( ρ L , πν ). The representation πν ∈ R s L ( L ) being ir-reducible, it corresponds to a simple H ( L, ρ )-module under the functor m L . Let f ∈ m L ( πν ) , γ ∈ H ( L, ρ ) and w ∈ V . The action of H ( L, ρ ) on m L ( πν ) is given by ( γ.f )( w ) = R L π ( l ) ν ( l ) f ( γ ∨ ( l − ) w ) dl . Here γ ∨ is defined on L by γ ∨ ( l − ) = γ ( l ) ∨ for l ∈ L .Note that the functor m G : R s ( G ) −→ H ( G, ρ ) − M od is given by: m G ( ι GP ( πν )) = Hom K ( ρ, ι GP ( πν )) . Further the functor ( T P ) ∗ : H ( L, ρ L ) − M od −→ H ( G, ρ ) − M od is given by, for M an H ( L, ρ )-module, ( T P ) ∗ ( M ) = Hom H ( L,ρ ) ( H ( G, ρ ) , M )where H ( G, ρ ) is viewed as a H ( L, ρ )-module via T P . The action of H ( G, ρ ) on ( T P ) ∗ ( M ) isgiven by h ′ ψ ( h ) = ψ ( h h ′ )where ψ ∈ ( T P ) ∗ ( M ) , h , h ′ ∈ H ( G, ρ ).The importance of covers is seen from the following commutative diagram which we willuse in answering the question which we posed earlier in this paper. R s ( G ) m G −−−−→ H ( G, ρ ) − M od ι GP x ( T P ) ∗ x R s L ( L ) m L −−−−→ H ( L, ρ L ) − M od
SUBHA SANDEEP REPAKA
Depth zero supercuspidal representations.
Suppose τ is an irreducible cuspidalrepresentation of GL n ( k E ) inflated to a representation of GL n ( O E ) = K . Then let › K = ZK where Z = Z (GL n ( E )) = { λ n | λ ∈ E × } . As any element of E × can be written as u̟ nE forsome u ∈ O × E and m ∈ Z . So in fact, › K = < ̟ E n > K .Let ( λ, V ) be a representation of GL n ( E ) and 1 V be the identity linear transformation of V . As ̟ E n ∈ Z , so λ ( ̟ E n ) = ω λ ( ̟ E n )1 V where ω λ : Z −→ C × is the central characterof λ .Let e τ be a representation of ‹ K such that:(1) e τ ( ̟ E n ) = ω λ ( ̟ E n )1 V , (2) e τ | K = τ. Say ω λ ( ̟ E n ) = z where z ∈ C × . Now call e τ = e τ z . We have extended τ to e τ z whichis a representation of ‹ K , so that Z acts by ω λ . Hence λ | ‹ K ⊇ e τ z which implies thatHom ‹ K ( e τ z , λ | ‹ K ) = 0.By Frobenius reciprocity for induction from open subgroups,Hom ‹ K ( e τ z , λ | ‹ K ) ≃ Hom GL n ( E ) ( c - Ind GL n ( E ) ‹ K e τ z , λ ).Thus Hom GL n ( E ) ( c - Ind GL n ( E ) ‹ K e τ z , λ ) = 0. So there exists a non-zero GL n ( E )-map from c - Ind G ‹ K e τ z to λ . As τ is cuspidal representation, using Cartan decompostion and Mackey’s cri-teria we can show that c - Ind GL n ( E ) ‹ K e τ z is irreducible. So λ ≃ c - Ind GL n ( E ) ‹ K e τ z . As c - Ind GL n ( E ) ‹ K e τ z is irreducible supercuspidal representation of GL n ( E ) of depth zero, so λ is irreducible super-cuspidal representation of GL n ( E ) of depth zero.Conversely, let λ is a irreducible, supercuspidal, depth zero representation of GL n ( E ). So λ K = { } . Hence λ | K ⊇ K , where 1 K is trivial representation of K . This means λ | K ⊇ τ ,where τ is an irreducible representation of K such that τ | K ⊇ K . So τ is trivial on K .So λ | K contains an irreducible representation τ of K such that τ | K is trivial. So τ canbe viewed as an irreducible representation of K /K ∼ = GL n ( k E ) inflated to K = GL n ( O E ).The representation τ is cuspidal by (a very special case of) A.1 Appendix [7].So we have the following bijection of sets: ß Isomorphism classes of irreduciblecuspidal representations of GL n ( k E ) ™ × C × ←→ Isomorphism classesof irreduciblesupercuspidalrepresentations ofGL n ( E ) of depth zero . ( τ, z ) −−−−−−−−−−−−−−−−−−−−−−−−−−−−→ c − Ind GL n ( E ) ‹ K e τ z ( τ, ω λ ( ̟ E n )) ←−−−−−−−−−−−−−−−−−−−−−−−−−−−− λ Recall that π is an irreducible supercuspidal depth zero representation of L ∼ = GL n × U ( E ).So π = λχ where λ is an irreducible supercuspidal depth zero representation of GL n and χ is an irreducible supercuspidal depth zero character of U ( E ). From now on we denotethe representation τ χ by ρ . So ρ is an irreducible cuspidal representation of GL n ( k E ) × U ( k E ) inflated to K × U ( O E ) where K = GL n ( O E ). Recall that we can extend ρ to PROBLEM ON ODD UNITARY GROUPS 7 a representation ‹ ρ of Z ( L ) P = ` n ∈ Z P ζ n via ‹ ρ ( ζ k j ) = ρ ( j ) for j ∈ P , k ∈ Z . Alsoobserve that as λ = c - Ind GL n ( E ) ‹ K e τ , so π = λχ = c - Ind LZ ( L ) P ‹ ρ .3. Representation ρ of P Let V be the vector space associated with ρ . Now ρ is extended to a map ρ from P to GL ( V ) as follows. By Iwahori factorization, if j ∈ P then j can be written as j − j j + , where j − ∈ P − , j + ∈ P + , j ∈ P . Now the map ρ on P is defined as ρ ( j ) = ρ ( j ). Proposition 2. ρ is a homomorphism from P to GL ( V ) . So ρ becomes a representation of P .Proof. The proof goes in similar lines as Proposition 5 in [9]. (cid:3) Calculation of N G ( P )We set G = U( n, n + 1). To describe H ( G, ρ ) we need to determine N G ( ρ ) which is givenby N G ( ρ ) = { m ∈ N G ( P ) | ρ ≃ ρ m } . Further, to find out N G ( ρ ) we need to determine N G ( P ). To that end we shall calculate N GL n ( E ) ( K ). Let Z = Z (GL n ( E )). So Z = { λ n | λ ∈ E × } . Lemma 1. N GL n ( E ) ( K ) = K Z .Proof. This follows from the Cartan decomposition by a direct matrix calculation. (cid:3)
From now on let us denote K by K . Now let us calculate N G ( P ). Note that J = Id n Id n ∈ G . Indeed, J ∈ N G ( P ). The center Z ( P ) of P is given by Z ( P ) = ( uId n λ
00 0 u − Id n | u ∈ O × E , λ ∈ O × E , λλ = 1 ) . Recall the center Z ( L ) of L is given by Z ( L ) = ( aId n λ
00 0 a − Id n | a ∈ E × , λ ∈ E × , λλ = 1 ) . Proposition 3. N G ( P ) = h P Z ( L ) , J i = P Z ( L ) ⋊ h J i .Proof. We use Lemma 1 to prove this Proposition. The proof goes in the similar lines asProposition 6 in [9]. (cid:3) Calculation of N G ( ρ )5.1. Unramified case.
We have the following conclusion about N G ( ρ ) for the unramifiedcase:If n is even then N G ( ρ ) = Z ( L ) P and if n is odd then N G ( ρ ) = Z ( L ) P ⋊ h J i . Fordetails refer to section 5.1 in [9]. SUBHA SANDEEP REPAKA
Ramified case:
We have the following conclusion about N G ( ρ ) for ramified case:If n is odd then N G ( ρ ) = Z ( L ) P and if n is even then N G ( ρ ) = Z ( L ) P ⋊ h J i . Fordetails refer to section 5.2 in [9]. Lemma 2.
When n is odd in the unramified case or when n is even in the ramified case, wehave N G ( ρ ) = h P , w , w i , where w = J and w = ̟ E − Id n ̟ E Id n .Proof. The proof goes in the similar lines as Lemma 2 in [9]. (cid:3) Structure of H ( G, ρ )6.1.
Unramified case:
In this section, we determine the structure of H ( G, ρ ) for the un-ramified case when n is odd. Using cuspidality of ρ , it can be shown by Theorem 4.15in [7], that I G ( ρ ) = P N G ( ρ ) P . But from lemma 2, N G ( ρ ) = h P , w , w i . So I G ( ρ ) = P h P , w , w i P = P h w , w i P , as P is a subgroup of P . Let V be the vector space cor-responding to ρ . Let us recall that H ( G, ρ ) consists of maps f : G → End C ( V ∨ ) such thatsupport of f is compact and f ( pgp ′ ) = ρ ∨ ( p ) f ( g ) ρ ∨ ( p ′ ) for p, p ′ ∈ P , g ∈ G . In fact H ( G, ρ )consists of C -linear combinations of maps f : G −→ End C ( V ∨ ) such that f is supported on P x P where x ∈ I G ( ρ ) and f ( pxp ′ ) = ρ ∨ ( p ) f ( x ) ρ ∨ ( p ′ ) for p, p ′ ∈ P . We shall now show thereexists φ ∈ H ( G, ρ ) with support P w P and satisfies φ = q n + ( q n − φ . Let K (0) = U( n, n + 1) ∩ GL n +1 ( O E ) = { g ∈ GL n +1 ( O E ) | t gJ g = J } ,K (0) = { g ∈ Id n +1 + ̟ E M n +1 ( O E ) | t gJ g = J } , G = { g ∈ GL n +1 ( k E ) | t gJ g = J } . The map r from K (0) to G given by r : K (0) mod p E −−−−→ G is a surjective group homomorphismwith kernel K (0). So by the first isomorphism theorem of groups we have: K (0) K (0) ∼ = G .r ( P ) = P = GL n ( k E ) M n × ( k E ) M n ( k E )0 U ( k E ) M × n ( k E )0 0 GL n ( k E ) T G = Siegel parabolic subgroup of G .Now P = L ⋉ U , where L is the Siegel Levi component of P and U is the unipotent radicalof P . Here L = ( a λ
00 0 t a − | a ∈ GL n ( k E ) , λ ∈ k × E , λλ = 1 ) , U = ( Id n u X − t u Id n | X ∈ M n ( k E ) , u ∈ M n × ( k E ) , X + t X + u t u = 0 ) . Let V be the vector space corresponding to ρ . The Hecke algebra H ( K (0) , ρ ) is a sub-algebra of H ( G, ρ ).Let ρ be the representation of P which when inflated to P is given by ρ and V is also thevector space corresponding to ρ . The Hecke algebra H ( G , ρ ) looks as follows: PROBLEM ON ODD UNITARY GROUPS 9 H ( G , ρ ) = ß f : G → End C ( V ∨ ) (cid:12)(cid:12)(cid:12)(cid:12) f ( pgp ′ ) = ρ ∨ ( p ) f ( g ) ρ ∨ ( p ′ )where p, p ′ ∈ P , g ∈ G ™ . Now the homomorphism r : K (0) −→ G extends to a map from H ( K (0) , ρ ) to H ( G , ρ )which we again denote by r . Thus r : H ( K (0) , ρ ) −→ H ( G , ρ ) is given by r ( φ )( r ( x )) = φ ( x )for φ ∈ H ( K (0) , ρ ) and x ∈ K (0) . Proposition 4.
The map r : H ( K (0) , ρ ) −→ H ( G , ρ ) is an algebra isomorphism.Proof. Refer to Proposition 17 in [9] (cid:3)
Let w = r ( w ) = r ( Id n Id n ) = Id n Id n ∈ G . Clearly K (0) ⊇ P ∐ P w P = ⇒ r ( K (0)) ⊇ r ( P ∐ P w P ) = ⇒ G ⊇ r ( P ) ∐ r ( P w P ) = P ∐ P w P . So G ⊇ P ∐ P w P .Now Ind GP ρ = π ⊕ π , where π , π are distinct irreducible representations of G withdim π > dim π . Let λ = dim π dim π . By Proposition 3.2 in [4], there exists a unique φ in H ( G , ρ )with support P w P such that φ = λ + ( λ − φ . By Proposition 4, there is a unique element φ in H ( K (0) , ρ ) such that r ( φ ) = φ . Thus supp( φ )= P w P and φ = λ + ( λ − φ . Assupport of φ = P w P ⊆ K (0) ⊆ G , so φ can be extended to G and viewed as an elementof H ( G, ρ ). Thus φ satisfies the following relation in H ( G, ρ ): φ = λ + ( λ − φ . We shall now show that λ = q n . Recall that as ρ is an irreducible cuspidal representationof GL n ( k E ) × U ( k E ), so ρ = τ θ χ , where τ θ is an irreducible cuspidal representation ofGL n ( k E ) and χ is a cuspidal representation of U ( k E ). Note that here θ is a regular charcterof l × where [ l : k E ] = n and k E = F q so that l = F q n . From Proposition 8 in [9] we have, θ q n = θ − as θ Φ = θ q .As G = U( n, n + 1)( k E ), so the dual group G ∗ is given by G ∗ ∼ = U( n, n + 1)( k E ) (i.e G ∗ ∼ = G ).Note that θ corresponds to a semi-simple element s ∗ ∈ L ∗ in G ∗ . Then by Theorems 8.4.8and 8.4.9 in [2], we have λ = | c G ∗ ( s ∗ ) | p .Note that L ∗ ∼ = L . So s ∗ corresponds to s in L . Hence, we have λ = | c G ( s ) | p . We write s = α λ
00 0 t α − . Observe that λλ = 1 , λ ∈ k × E , α ∈ F × q n . More precisely, α is in the imageof F × q n under a fixed embedding F × q n ֒ → GL n ( F q ). This embedding arises when we let l act onthe basis of l over k E via multiplication. We can thus embed l in M n ( k E ) and l × in GL n ( k E )which we call the usual embedding. Note that θ is regular implies that F q n = F q ( α ). Ourgoal is to compute | c G ( s ) | p .By Proposition 3.19 in [3], we have Sylow p-subgroups of c G ( s ) are the sets of F q -pointsof the Unipotent radicals of the Borel subgroups of c G ( s ). By Proposition 2.2 in [3], we haveBorel subgroups of c G ( s ) are of the form B ∩ c G ( s ), where B is a Borel subgroup of G . AsSiegel parabolic subgroup P of G contains a Borel subgroup of G , so c P ( s ) = P ∩ c G ( s ) containsa Sylow p-subgroup of c G ( s ). Lemma 3. c P ( s ) = c L ( s ) ⋉ c U ( s ) . Proof.
Recall that P = L ⋉ U . Hence L ∩ U = ∅ and U E P . As L ∩ U = ∅ = ⇒ c L ( s ) ∩ c U ( s ) = ∅ .Note that c U ( s ) E ( c L ( s ) × c U ( s )). So it makes sense to talk of c L ( s ) ⋉ c U ( s ).Let x ∈ P ( s ) = ⇒ x ∈ P , sxs − = x . Note that as x ∈ P so x = lu for some l ∈ L , u ∈ U .Therefore, slus − = lu = ⇒ sls − sus − = lu. Let sls − = m and sus − = n . Now as s ∈ L , so sls − = m ∈ L . Note that sus − = n ∈ U as U E P . Therefore, we have mn = lu or m − l = nu − . But m − l ∈ L and nu − ∈ U ,so we have m − l, nu − ∈ L ∩ U . Recall that L ∩ U = e , so m = l, n = u . Therefore, sls − = l, sus − = u . So we have l ∈ c L ( s ) , u ∈ c U ( s ). Hence, x ∈ c L ( s ) ⋉ c U ( s ). So c P ( s ) ⊆ c L ( s ) ⋉ c U ( s ).Conversely, let x ∈ c L ( s ) ⋉ c U ( s ). So x = lu where l ∈ c L ( s ) and u ∈ c U ( s ). Hence sls − = l and sus − = u . Therefore, sxs − = slus − = sls − sus − = lu = x . So x ∈ c P ( s ). Hence c L ( s ) ⋉ c U ( s ) ⊆ c P ( s ). Therefore, c P ( s ) = c L ( s ) ⋉ c U ( s ). (cid:3) From lemma 3, we get | c P ( s ) | p = | c L ( s ) | p | c U ( s ) | p . Note that | c L ( s ) | p = 1. Therefore, | c P ( s ) | p = | c L ( s ) | p | c U ( s ) | p = | c U ( s ) | p . Lemma 4. | c U ( s ) | = | c U ( s ) | p = q n .Proof. Recall that the elements of U are of form m = Id u X − t u Id where x ∈ M n ( k E ) , u ∈ M n × ( k E ) , X + t X + u t u = 0. If m ∈ c U ( s ) then ms = sm . So wehave, α λ
00 0 t α − Id u X − t u Id = Id u X − t u Id α λ
00 0 t α − . From the above matrix relation, it follows that αu = λu, αX = X t α − , λ t u = t u t α − .Recall that X + t X + u t u = 0 , λλ = 1. Also recall that u ∈ M n × ( k E ) , α ∈ F × q n , k E ( α ) = l .As αu = λu , so if u = 0 then λ ∈ k E is an eigen value of α . So λ is a root of the minimalpolynomial of α over k E . But as the minimal polynomial is irreducible over k E [ x ], so this isa contradiction. So u = 0.So we have to find X such that X + t X = 0 , αX = X t α − . Let Ξ = M n ( k E ) and setΞ ǫ = { X ∈ Ξ | t X = ǫX } . Note that X ∈ Ξ can be written as X + t X + X − t X , so Ξ = Ξ ⊕ Ξ − .Let us set Ξ( α ) = { X ∈ Ξ | αX t α = X } and Ξ ǫ ( α ) = { X ∈ Ξ ǫ | αX t α = X } . Then wehave, Ξ( α ) = Ξ ( α ) ⊕ Ξ − ( α ). Let us choose γ ∈ k E such that γ = 0 and γ = − γ . Notethat, if X ∈ Ξ ( α ) then X = t X and αX t α = X . So t ( γX ) = − ( γX ) and α ( γX ) t α = γX .Therefore, γX ∈ Ξ − ( α ). We also have a bijection from c U ( s ) −→ Ξ ( α ) given by: PROBLEM ON ODD UNITARY GROUPS 11 Id X Id −→ X. Hence we have, | c U ( s ) | = | Ξ ( α ) | = | Ξ − ( α ) | . Let us now compute | Ξ( α ) | . So we want tofind the cardinality of X ∈ Ξ such that αX t α = X for a fixed α ∈ F × q n . Let φ : F q n ֒ → M n ( F q ) be the usual embedding take β to m β . Let f ( x ) be the minimal polynomial of α over k E = F q . So we have F q n ∼ = F q [ x ]
1. It can be further shown that φ and φ generate the Hecke algebra H ( G, ρ ). Let us denote the Hecke algebra H ( G, ρ ) by A . So A = H ( G, ρ ) = * φ i : G → End C ( ρ ∨ ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) φ i is supported on P w i P and φ i ( pw i p ′ ) = ρ ∨ ( p ) φ i ( w i ) ρ ∨ ( p ′ )where p, p ′ ∈ P , i = 0 , + where φ i satisfies the relation: φ i = q n + ( q n − φ i for i = 0 , Lemma 5. φ and φ are units in A .Proof. As φ i = q n + ( q n − φ i for i = 0 ,
1. So φ i ( φ i +(1 − q n )1 q n ) = 1 for i=0,1. Hence φ and φ are units in A . (cid:3) Lemma 6.
Let φ, ψ ∈ H ( G, ρ ) with support of φ, ψ being P x P , P y P respectively. Thensupp( φ ∗ ψ )=supp( φψ ) ⊆ (supp( φ ))(supp( ψ ))= P x P y P .Proof. The proof is same as that of Lemma 5 in [9]. (cid:3)
From B-N pair structure theory we can show that, P x P y P = P xy P ⇐⇒ l ( xy ) = l ( x ) + l ( y ). From lemma 6, we have supp( φ φ ) ⊆ P w P w P . But P w P w P = P w w P (since l ( w w ) = l ( w ) + l ( w )). Thus supp( φ φ ) ⊆ P w w P . Let ζ = w w , So ζ = ̟ E Id n ̟ − E Id n . As φ , φ are units in algebra A , so ψ = φ φ is a unit too in A and ψ − = φ − φ − . Nowas we have seen before that supp( φ φ ) ⊆ P w w P = ⇒ supp ( ψ ) ⊆ P ζ P = ⇒ supp ( ψ ) = ∅ or P ζ P . If supp( ψ )= ∅ = ⇒ ψ = 0 which is a contradiction as ψ is a unit in A . Sosupp( ψ ) = P ζ P . As ψ is a unit in A , we can show as before from B-N pair structure theorythat supp( ψ ) = P ζ P . Hence by induction on n ∈ N , we can further show from B-N pairstructure theory that supp( ψ n )= P ζ n P for n ∈ N .Now A contains a sub- algebra generated by ψ, ψ − over C and we denote this sub-algebraby B . So B = C [ ψ, ψ − ] where PROBLEM ON ODD UNITARY GROUPS 13 B = C [ ψ, ψ − ] = ß c k ψ k + · · · + c l ψ l (cid:12)(cid:12)(cid:12)(cid:12) c k , . . . , c l ∈ C ; k < l ; k, l ∈ Z ™ . Proposition 5.
The unique algebra homomorphism C [ x, x − ] −→ B given by x −→ ψ is anisomorphism. So B ≃ C [ x, x − ] .Proof. The proof is same as that of Proposition 18 in [9]. (cid:3)
Ramified case:
In this section we determine the structure of H ( G, ρ ) for the ramifiedcase when n is even. Recall I G ( ρ ) = P N G ( ρ ) P . But from lemma 2, N G ( ρ ) = h P , w , w i .So I G ( ρ ) = P h P , w , w i P = P h w , w i P , as P is a subgroup of P . Let V be the vectorspace corresponding to ρ . Let us recall that H ( G, ρ ) consists of maps f : G → End C ( V ∨ )such that support of f is compact and f ( pgp ′ ) = ρ ∨ ( p ) f ( g ) ρ ∨ ( p ′ ) for p, p ′ ∈ P , g ∈ G . Infact H ( G, ρ ) consists of C -linear combinations of maps f : G −→ End C ( V ∨ ) such that f issupported on P x P where x ∈ I G ( ρ ) and f ( pxp ′ ) = ρ ∨ ( p ) f ( x ) ρ ∨ ( p ′ ) for p, p ′ ∈ P . We shallnow show there exists φ ∈ H ( G, ρ ) with support P w P and satisfies φ = q n/ + ( q n/ − φ .Let K (0) = U( n, n + 1) ∩ GL n +1 ( O E ) = { g ∈ GL n +1 ( O E ) | t gJ g = J } ,K (0) = { g ∈ Id n +1 + ̟ E M n +1 ( O E ) | t gJ g = J } , G = { g ∈ GL n +1 ( k E ) | t gJ g = J } . The map r from K (0) to G given by r : K (0) mod p E −−−−→ G is a surjective group homomorphismwith kernel K (0). So by the first isomorphism theorem of groups we have: K (0) K (0) ∼ = G . r ( P ) = P = GL n ( k E ) M n × ( k E ) M n ( k E )0 U ( k E ) M × n ( k E )0 0 GL n ( k E ) T G = Siegel parabolic subgroup of G .Now P = L ⋉ U , where L is the Siegel Levi component of P and U is the unipotent radicalof P . Here L = ( a λ
00 0 t a − | a ∈ GL n ( k E ) , λ ∈ E × , λλ = 1 ) , U = ( Id n u X − t u Id n | X ∈ M n ( k E ) , u ∈ M n × ( k E ) , X + t X + u t u = 0 ) . Let V be the vector space corresponding to ρ . The Hecke algebra H ( K (0) , ρ ) is a sub-algebra of H ( G, ρ ).Let ρ be the representation of P which when inflated to P is given by ρ and V is also thevector space corresponding to ρ . The Hecke algebra H ( G , ρ ) looks as follows: H ( G , ρ ) = ß f : G → End C ( V ∨ ) (cid:12)(cid:12)(cid:12)(cid:12) f ( pgp ′ ) = ρ ∨ ( p ) f ( g ) ρ ∨ ( p ′ )where p, p ′ ∈ P , g ∈ G ™ . Now the homomorphism r : K (0) −→ G extends to a map from H ( K (0) , ρ ) to H ( G , ρ )which we again denote by r . Thus r : H ( K (0) , ρ ) −→ H ( G , ρ ) is given by r ( φ )( r ( x )) = φ ( x )for φ ∈ H ( K (0) , ρ ) and x ∈ K (0) . As in the unramified case, when n is odd, we can show that H ( K (0) , ρ ) is isomorphic to H ( G , ρ ) as algebras via r .Let w = r ( w ) = r ( Id n Id n ) = Id n Id n ∈ G . Clearly K (0) ⊇ P ∐ P w P = ⇒ r ( K (0)) ⊇ r ( P ∐ P w P ) = ⇒ G ⊇ r ( P ) ∐ r ( P w P ) = P ∐ P w P . So G ⊇ P ∐ P w P .Now G is a finite group. In fact, it is the special orthogonal group consisting of matricesof size (2 n + 1) × (2 n + 1) over finite field k E or F q . So G = SO n +1 ( F q ).According to the Theorem 6.3 in [4], there exists a unique φ in H ( G , ρ ) with support P w P such that φ = q n/ + ( q n/ − φ . Hence there is a unique element φ ∈ H ( K (0) , ρ ) such that r ( φ ) = φ . Thus supp( φ )= P w P and φ = q n/ + ( q n/ − φ . Now φ can be extended to G and viewed as an element in H ( G, ρ ) as P w P ⊆ K (0) ⊆ G . Thus φ satisfies the followingrelation in H ( G, ρ ): φ = q n/ + ( q n/ − φ . We shall now show there exists φ ∈ H ( G, ρ ) with support P w P satisfying the samerelation as φ .Recall that w = ̟ − E Id n ̟ E Id n , ̟ − E = − ̟ − E . So w = − ̟ − E Id n ̟ E Id n .Let η ∈ U( n, n + 1) be such that ηw η − = J ′ = − Id n Id n and η GL n ( O E ) M n × ( O E ) M n ( O E )M × n ( p E ) U ( O E ) M × n ( O E )M n ( p E ) M n × ( p E ) GL n ( O E ) η − = GL n ( O E ) M n × ( p E ) M n ( p E )M × n ( O E ) U ( O E ) M × n ( p E )M n ( O E ) M n × ( O E ) GL n ( O E ) . Recall that P looks as follows: P = GL n ( O E ) M n × ( O E ) M n ( O E )M × n ( p E ) U ( O E ) M × n ( O E )M n ( p E ) M n × ( p E ) GL n ( O E ) T G. Note that ηGη − = { g ∈ GL n +1 ( E ) | t gJ ′ g = J ′ } . Hence η P η − = GL n ( O E ) M n × ( p E ) M n ( p E )M × n ( O E ) U ( O E ) M × n ( p E )M n ( O E ) M n × ( O E ) GL n ( O E ) T ηGη − . Therefore η P η − is the opposite of the Siegel Parahoric subgroup of ηGη − . Let K ′ (0) = h P , w i . PROBLEM ON ODD UNITARY GROUPS 15
And let G ′ = { g ∈ GL n +1 ( k E ) | t gJ ′ g = J ′ } = { g ∈ GL n +1 ( k E ) | t gJ ′ g = J ′ } . Let r ′ : K ′ (0) −→ G ′ be the group homomorphism given by r ′ ( x ) = ( ηxη − ) modp E where x ∈ K ′ (0) . So we have r ′ ( K (0)) = ( ηK ′ (0) η − ) modp E = ( η h P , w i η − ) modp E . Let r ′ ( P ) = ( η P η − ) modp E = P ′ . We can see that r ′ ( w ) = ( ηw η − ) modp E = J ′ modp E = w ′ = − Id n Id n . So P ′ = r ′ ( P ) = ( η P η − ) modp E = GL n ( k E ) 0 0 M × n ( k E ) U ( k E ) 0M n ( k E ) M n × ( k E ) GL n ( k E ) T G ′ . Clearly P ′ isthe opposite of Siegel parabolic subgroup of G ′ . Hence r ′ ( K (0)) = h P ′ , w ′ i = G ′ , as P ′ is amaximal subgroup of G ′ and w ′ does not lie in P ′ . So r ′ is a surjective homomorphism ofgroups.Let V be the vector space corresponding to ρ . The Hecke algebra H ( K ′ (0) , ρ ) is a sub-algebra of H ( G, ρ ).Let ρ ′ be the representation of P ′ which when inflated to η P is given by η ρ and V is alsothe vector space corresponding to ρ ′ . Now the Hecke algebra H ( G ′ , ρ ′ ) looks as follows: H ( G ′ , ρ ′ ) = ß f : G ′ → End C ( V ∨ ) (cid:12)(cid:12)(cid:12)(cid:12) f ( pgp ′ ) = ρ ′∨ ( p ) f ( g ) ρ ′∨ ( p ′ )where p, p ′ ∈ P ′ , g ∈ G ′ ™ . Now the homomorphism r ′ : K ′ (0) −→ G ′ extends to a map from H ( K ′ (0) , ρ ) to H ( G ′ , ρ ′ )which we again denote by r ′ . Thus r ′ : H ( K ′ (0) , ρ ) −→ H ( G ′ , ρ ′ ) is given by r ′ ( φ )( r ′ ( x )) = φ ( x )for φ ∈ H ( K ′ (0) , ρ ) and x ∈ K ′ (0) . As in the unramified case when n is odd, we can show that H ( K ′ (0) , ρ ) is isomorphic to H ( G ′ , ρ ′ ) as algebras via r ′ .Clearly K ′ (0) ⊇ P ∐ P w P = ⇒ r ′ ( K ′ (0)) ⊇ r ′ ( P ∐ P w P ) = ⇒ G ′ ⊇ r ′ ( P ) ∐ r ′ ( P w P ) = P ′ ∐ P ′ w ′ P ′ . So G ′ ⊇ P ′ ∐ P ′ w ′ P ′ .Now G ′ is a finite group over the field K E or F q . Note that G ′ ∼ = Sp n ( k E ). Accordingto the Theorem 6.3 in [4], there exists a unique φ in H ( G ′ , ρ ′ ) with support P ′ w ′ P ′ suchthat φ = q n/ + ( q n/ − φ . Hence there is a unique element φ ∈ H ( K ′ (0) , ρ ) such that r ′ ( φ ) = φ . Thus supp( φ )= P w P and φ = q n/ + ( q n/ − φ . Now φ can be extendedto G and viewed as an element in H ( G, ρ ) as P w P ⊆ K ′ (0) ⊆ G . Thus φ satisfies thefollowing relation in H ( G, ρ ): φ = q n/ + ( q n/ − φ . Thus we have shown there exists φ i ∈ H ( G, ρ ) with supp( φ i )= P w i P satisfying φ i = q n/ + ( q n/ − φ i for i = 0 ,
1. It can be further shown that φ and φ generate the Heckealgebra H ( G, ρ ). Let us denote the Hecke algebra H ( G, ρ ) by A . So A = H ( G, ρ ) = * φ i : G → End C ( ρ ∨ ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) φ i is supported on P w i P and φ i ( pw i p ′ ) = ρ ∨ ( p ) φ i ( w i ) ρ ∨ ( p ′ )where p, p ′ ∈ P , i = 0 , + where φ i has support P w i P and φ i satisfies the relation: φ i = q n/ + ( q n/ − φ i for i = 0 , Lemma 7. φ and φ are units in A .Proof. As φ i = q n/ + ( q n/ − φ i for i = 0 ,
1. So φ i ( φ i +(1 − q n/ )1 q n/ ) = 1 for i=0,1. Hence φ and φ are units in A . (cid:3) As φ , φ are units in A which is an algebra, so ψ = φ φ is a unit too in A and ψ − = φ − φ − . As in the unramified case when n is odd, we can show that A contains sub-algebra B = C [ ψ, ψ − ] where B = C [ ψ, ψ − ] = ß c k ψ k + · · · + c l ψ l (cid:12)(cid:12)(cid:12)(cid:12) c k , . . . , c l ∈ C ; k < l ; k, l ∈ Z ™ . Further, as in the unramified case when n is odd, we can show that C [ ψ, ψ − ] ≃ C [ x, x − ]as C -algebras. 7. Structure of H ( L, ρ )In this section we describe the structure of H ( L, ρ ). Thus we need first to determine N L ( ρ ) = { m ∈ N L ( P ) | ρ m ≃ ρ } . We know from lemma 1 that N GL n ( E ) ( K ) = K Z , so we have N L ( P ) = Z ( L ) P . Since Z ( L ) clearly normalizes ρ and ρ is an irreducible cuspidal representation of P , so N L ( ρ ) = Z ( L ) P = ` n ∈ Z P ζ n .Define α ∈ H ( L, ρ ) by supp( α ) = P ζ and α ( ζ ) = 1 V ∨ . We can show that α n ( ζ n ) =( α ( ζ )) n for n ∈ Z and supp( α n ) = P ζ n P = P ζ n = ζ n P for n ∈ Z . Further we can showthat H ( L, ρ ) = C [ α, α − ]. For details refer to section 7 in [9]. Proposition 6.
The unique algebra homomorphism C [ x, x − ] −→ C [ α, α − ] given by x −→ α is an isomorphism. So C [ α, α − ] ≃ C [ x, x − ] . We have already shown before in sections 6.1 and 6.2 that B = C [ ψ, ψ − ] is a sub-algebraof A = H ( G, ρ ), where ψ is supported on P ζ P and B ∼ = C [ x, x − ]. As H ( L, ρ ) = C [ α, α − ] ∼ = C [ x, x − ], so B ∼ = H ( L, ρ ) as C -algebras. Hence H ( L, ρ ) can be viewed as a sub-algebra of H ( G, ρ ).Now we would like to find out how simple H ( L, ρ )-modules look like. Thus to understandthem we need to find out how simple C [ x, x − ]-modules look like. PROBLEM ON ODD UNITARY GROUPS 17 Calculation of simple H ( L, ρ ) -modules Recall that H ( L, ρ ) = C [ α, α − ]. Note that C [ α, α − ] ∼ = C [ x, x − ] as C -algebras. It can beshown by direct calculation that the simple C [ x, x − ]-modules are of the form C λ for λ ∈ C × ,where C λ is the vector space C with the C [ x, x − ]-module structure given by x.z = λz for z ∈ C λ .So the distinct simple H ( L, ρ )-modules(up to isomorphism) are the various C λ for λ ∈ C × .The module structure is determined by α.z = λz for z ∈ C λ .9. Final calculations to answer the question
Calculation of δ P ( ζ ) . Let us recall the modulus character δ P : P −→ R × > introducedin section 1. The character δ P is given by δ P ( p ) = k det ( Ad p ) | Lie U k F for p ∈ P , where Lie U is the Lie algebra of U . We have U = ( Id n u X − t u Id n | X ∈ M n ( E ) , u ∈ M n × ( E ) , X + t X + u t u = 0 ) , Lie U = ( u X − t u | X ∈ M n ( E ) , u ∈ M n × ( E ) , X + t X = 0 ) . Unramified case:
Recall ζ = ̟ E Id n ̟ − E Id n in the unramified case. So( Ad ζ ) Id n u X − t u Id n = ζ Id n u X − t u Id n ζ − = Id n ̟ E u ̟ E X − ̟ Et u Id n . Hence δ P ( ζ ) = k det ( Ad ζ ) | Lie U k F = k − ̟ n +2 n E k F = k − ̟ n +2 n F k F = q − n − n . Ramified case:
Recall ζ = ̟ E Id n − ̟ − E Id n in the ramified case. So( Ad ζ ) Id n u X − t u Id n = ζ Id n u X − t u Id n ζ − = Id n ̟ E u − ̟ E X ̟ Et u Id n . Hence δ P ( ζ ) = k det ( Ad ζ ) | Lie U k F = k ̟ n +2 n E k F = k ̟ n + n F k F = q − n − n . Understanding the map T P . Recall that there is an algebra embedding T P : H ( L, ρ ) −→H ( G, ρ ) such that the following diagram commutes: R [ L,π ] G ( G ) m G −−−−→ H ( G, ρ ) − M od ι GP x ( T P ) ∗ x R [ L,π ] L ( L ) m L −−−−→ H ( L, ρ ) − M od
Note that here T P ( α )( ζ ) = δ / P ( ζ )1 W ∨ with supp( T P ( α )) = P ζ P . For details refer tosection 9.2 in [9].9.3. Calculation of ( φ ∗ φ )( ζ ) . In this section we calculate ( φ ∗ φ )( ζ ). Let g i = q − n/ φ i for i = 0 , g i = q − n/ φ i for i = 0 , φ ∗ φ )( ζ ) would be useful in showing g ∗ g = T P ( α ) in both ramified and unramified cases. Lemma 8. supp ( φ ∗ φ ) = P ζ P = P w w P .Proof. The proof goes in the similar lines as Lemma 8 in [9]. (cid:3)
Proposition 7. ( φ ∗ φ )( ζ ) = φ ( w ) φ ( w ) . Proof.
The proof goes in the similar lines as Proposition 22 in [9]. (cid:3)
Relation between g , g and T P ( α ) . Unramified case:
Recall that H ( G, ρ ) = h φ , φ i where φ is supported on P w P and φ is supported on P w P respectively with φ i = q n + ( q n − φ i for i = 0 ,
1. In this sectionwe show that g ∗ g = T P ( α ), where g i = q − n/ φ i for i = 0 , Proposition 8. g g = T P ( α ) .Proof. The proof goes in the similar lines as Proposition 23 in [9]. (cid:3)
Ramified case:
We know that H ( G, ρ ) = h φ , φ i where φ is supported on P w P and φ is supported on P w P respectively with φ i = q n/ + ( q n/ − φ i for i = 0 ,
1. In thissection we show that g ∗ g = T P ( α ), where g i = q − n/ φ i for i = 0 , Proposition 9. g g = T P ( α ) .Proof. The proof goes in the similar lines as Proposition 24 in [9]. (cid:3)
Calculation of m L ( πν ) . Note that m L ( πν ) ∼ = C ν ( ζ ) . For details refer to section 9.5 in[9]. PROBLEM ON ODD UNITARY GROUPS 19
Answering the question
Recall the following commutative diagram which we described earlier.(CD) R [ L,π ] G ( G ) m G −−−−→ H ( G, ρ ) − M od ι GP x ( T P ) ∗ x R [ L,π ] L ( L ) m L −−−−→ H ( L, ρ ) − M od
Recall that πν lies in R [ L,π ] L ( L ). Note that from the above commutative diagram, it followsthat ι GP ( πν ) lies in R [ L,π ] G ( G ) and m G ( ι GP ( πν )) is an H ( G, ρ )-module. Recall m L ( πν ) ∼ = C ν ( ζ ) as H ( L, ρ )-modules. From the above commutative diagram, we have m G ( ι GP ( πν )) ∼ =( T P ) ∗ ( C ν ( ζ ) ) as H ( G, ρ )-modules. Thus to determine the unramified characters ν for which ι GP ( πν ) is irreducible, we have to understand when ( T P ) ∗ ( C ν ( ζ ) ) is a simple H ( G, ρ )-module.Using notation on page 438 in [5], we have γ = γ = q n/ for unramified case when n is odd and γ = γ = q n/ for ramified case when n is even. As in Proposition 1.6 of [5],let Γ = { γ γ , − γ γ − , − γ − γ , ( γ γ ) − } . So by Proposition 1.6 in [5], ( T P ) ∗ ( C ν ( ζ ) ) is asimple H ( G, ρ )-module ⇐⇒ ν ( ζ ) / ∈ Γ. Recall π = c - Ind LZ ( L ) P ‹ ρ where ‹ ρ ( ζ k j ) = ρ ( j ) for j ∈ P , k ∈ Z and ρ = τ θ for some regular character θ of l × with [ l : k E ] = n . Hence wecan conclude that ι GP ( πν ) is irreducible for the unramified case when n is odd ⇐⇒ ν ( ζ ) / ∈{ q n , q − n , − } , θ q n +1 = θ − q and ι GP ( πν ) is irreducible for the ramified case when n is even ⇐⇒ ν ( ζ ) / ∈ { q n/ , q − n/ , − } , θ q n/ = θ − .Recall that in the unramified case when n is even or in the ramified case when n is odd wehave N G ( ρ ) = Z ( L ) P . Thus I G ( ρ ) = P ( Z ( L ) P ) P = P Z ( L ) P .From Corollary 6.5 in [6] which states that if I G ( ρ ) ⊆ P L P then T P : H ( L, ρ ) −→ H ( G, ρ )is an isomorphism of C -algebras. As we have I G ( ρ ) = P Z ( L ) P in the unramified case when n is even or in the ramified case when n is odd, so H ( L, ρ ) ∼ = H ( G, ρ ) as C -algebras. So fromthe commutative diagram (CD), we can conclude that ι GP ( πν ) is irreducible for any unramifiedcharacter ν of L . That proves Theorem 1. References [1] Colin J. Bushnell and Philip C. Kutzko. Smooth representations of reductive p -adic groups: structuretheory via types. Proc. London Math. Soc. (3) , 77(3):582–634, 1998.[2] Roger W. Carter.
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Int. Math. Res. Not. , pages Art. ID 97957, 40, 2006.[5] Philip Kutzko and Lawrence Morris. Explicit Plancherel theorems for H ( q , q ) and SL ( F ). Pure Appl.Math. Q. , 5(1):435–467, 2009.[6] Philip C. Kutzko. Smooth representations of reductive p -adic groups: an introduction to the theory oftypes. In Geometry and representation theory of real and p -adic groups (C´ordoba, 1995) , volume 158 of Progr. Math. , pages 175–196. Birkh˝auser Boston, Boston, MA, 1998.[7] Lawrence Morris. Tamely ramified intertwining algebras.
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Journal ofAlgebra , 573:663–711, 2021.
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