A reducibility problem for even unitary groups: The depth zero case
aa r X i v : . [ m a t h . R T ] A ug A REDUCIBILITY PROBLEM FOR EVEN UNITARY GROUPS: THEDEPTH ZERO CASE
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 ) contains a parabolic subgroup P with Levi component L isomorphic toGL n ( E ). Let π be an irreducible supercuspidal representation of L of depth zero. We useHecke algebra methods to determine when the parabolically induced representation ι GP π isreducible. Contents
1. Introduction 12. Preliminaries 63. Representation ρ of P N G ( P ) 115. Calculation of N G ( ρ ) 146. Structure of H ( G, ρ ) 197. Structure of H ( L, ρ ) 278. Calculation of simple H ( L, ρ )-modules 289. Final calculations to answer the question 3010. Answering the question 42References 431. Introduction
A central feature of reductive p -adic groups is that there are irreducible representationsthat never occur as subrepresentations of parabolically induced representations. These arethe supercuspidal representations. Suppose Π is an irreducible representation of a reductive p -adic group G . By insights of Harish-Chandra and others it is known that there is a parabolicsubgroup of G and an irreducible supercuspidal representation π of a Levi component of theparabolic subgroup such that Π occurs in the representation obtained from π via parabolicinduction.Thus, a core problem in p -adic representation theory is to understand when and howparabolically induced representations decompose, especially when the inducing representationis supercuspidal. The problem of when the parabolically induced representation decomposeis what we study in a very special situation when G = U( n, n ) is even unitary group over Date : August 4, 2020.2010
Mathematics Subject Classification.
Primary 22E50, Secondary 11F70. non- Archimedean local field E and π is a smooth irreducible supercuspidal depth zero rep-resentation of the Siegel Levi component L ∼ = GL n ( E ) of the Siegel parabolic subgroup P of G . The terms P, L, π, U( n, n ) are described in much detail later in the paper. We use Heckealgebra methods to determine when the parabolically induced representation ι GP π is reducible.Harish-Chandra tells us to look not at an individual ι GP π but at the family ι GP ( πν ) as ν variesthrough the unramfied characters of L ∼ = GL n ( E ). Note that the functor ι GP and unramifiedcharacters of L are also described in greater detail later in the paper.The approach taken in solving the above problem in this paper is similar to the approachtaken by Kutzko and Morris in [6] for the groups Sp n , SO n , SO n +1 over non-Archimedeanlocal fields, but the computations for the even unitary group U( n, n ) over non-Archimedeanlocal fields done in this paper are more involved than in [6].Before going any further, let me describe how U( n, n ) 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 ) isgiven by U( n, n ) = { g ∈ GL n ( E ) | t gJ g = J } for J = ñ ô 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 themaximal ideals in O E and O F and k E = O E / p E and k F = O F / p F denote the residue classfields 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 = ( ñ a t a − ô | a ∈ GL n ( E ) ) and U = ( ñ X ô | X ∈ M n ( E ) , X + t X = 0 ) . Let P = L ⋉ U be the L -opposite of P where U = ( ñ X ô | X ∈ M n ( E ) , X + t X = 0 ) . REDUCIBILITY PROBLEM FOR EVEN UNITARY GROUPS: THE DEPTH ZERO CASE 3
Let K = GL n ( O E ) and K = 1 + ̟ E M n ( O E ). Note K = 1 + ̟ E M n ( O E ) is the kernelof the surjective group homomorphism( g ij ) −→ ( g ij + p E ) : GL n ( O E ) −→ GL n ( k E )We say π is a depth zero representation of the Siegel Levi component L of P if π K = 0. Let( ρ, V ) be a smooth representation of the group H which is a subgroup of K . The smoothlyinduced representation from H to K is denoted by Ind KH ( ρ, V ) or Ind KH ( ρ ). Let us denote c - Ind KH ( ρ, V ) or c - Ind GP ( ρ ) for smoothly induced compact induced representation from 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. Let the group of unramified charactersof 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 ). 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.As π is an irreducible supercuspidal depth zero representation of L which is isomorphic toGL n ( E ), so π | K contains an irreducible representation ρ of K such that ρ | K is trivial.So ρ can be viewed as an irreducible representation of K /K ∼ = GL n ( k E ) inflated to K =GL n ( O E ). The representation ρ is cuspidal by (a very special case of) A.1 Appendix [10].We now define the Siegel parahoric subgroup P of G which is given by: P = ñ GL n ( O E ) M n ( O E )M n ( p E ) GL n ( O E ) ô T U( n, n ) . 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 = ( ñ a t a − ô | a ∈ GL n ( O E ) ) , P + = ( ñ X ô | X ∈ M n ( O E ) , X + t X = 0 ) , P − = ( ñ X ô | X ∈ M n ( O E ) , X + t X = 0 ) .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 . In section 3, we show that ρ can be extended to a representation ρ of P . If j ∈ P has Iwahori factorization j = j − j j + where j − ∈ P − , j ∈ P , j + ∈ P + then ρ is defined by ρ ( j ) = ρ ( j ) for j ∈ P .In section 4, we compute the normalizer of P in G which is given by N G ( P ) and it isindependent of whether we are working in ramified or unramified case. The explicit calculation SUBHA SANDEEP REPAKA of N G ( P ) was not done in [6]. By the work of Green or as a very special case of the Deligne-Lusztig construction, irreducible cuspidal representations of GL n ( k E ) are parametrized bythe regular characters of degree n extensions of k E . We write τ θ for the irreducible cuspidalrepresentation ρ that corresponds to a regular character θ .Let Z ( L ) denote the center of L . Hence Z ( L ) = ( ñ a a − ô | a ∈ E × ) . let N G ( ρ ) = { m ∈ N G ( P ) | ρ ≃ ρ m } . We shall show in section 5 that in the unramified case 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 . In the ramified case we show, if n is oddthen N G ( ρ ) = Z ( L ) P and if n is even then N G ( ρ ) = Z ( L ) P ⋊ h J i . Also the explicitcalculation of N G ( ρ ) for the ramified and unramified cases was not done in [6]. We use N G ( ρ ) to calculate the structure of the Hecke Algebra H ( G, ρ ) for unramified( n is odd) andramified( n is even) cases in section 6.Let w = J and w = ñ ̟ E − ̟ E ô . Now w w = ζ = ñ ̟ E ̟ E − ô . The Hecke algebra H ( G, ρ ) in the unramified case and n is odd is given by: 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 , H ( G, ρ ) in the ramified case and n is even is given by: 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 , H ( G, ρ )for G being Sp n , SO n , SO n +1 .Let N L ( ρ ) = { m ∈ N L ( P ) | ρ m ≃ ρ } . We use N L ( ρ ) to dermine the structure of the Hecke Algebra H ( L, ρ ) which is independentof whether we are in the ramified or unramfied case. Let V be vector space correspondingto ρ . We shall show that there exists an element α in H ( L, ρ ) such that supp( α ) = P ζ and α ( ζ ) = 1 V ∨ . We show in section 7 that the Hecke algebra H ( L, ρ ) = C [ α, α − ] which isisomorphic to the C -algebra C [ x, x − ] which was again done by root systems in [6]. Then we REDUCIBILITY PROBLEM FOR EVEN UNITARY GROUPS: THE DEPTH ZERO CASE 5 determine the structure of simple C [ α, α − ]-modules in section 8. The computation of simple C [ α, α − ]-modules was not done explicitly in [6].The categories R s L ( L ) , R s ( G ) where s L = [ L, π ] L and s = [ L, π ] G are described later in thispaper. The category H ( G, ρ ) − M od is the category of H ( G, ρ )-modules and H ( L, ρ ) − M od be the category of H ( L, ρ )-modules. The functor ι GP was defined earlier.For s L = [ L, π ] L , the representation πν ∈ R s L ( L ). The functor m L : R s L ( L ) −→ H ( L, ρ ) − M od given by m L ( πν ) = Hom P ( ρ , πν ) is an equivalance of categories. The represen-tation πν ∈ R s L ( L ) being irreducible, it corresponds to a simple H ( L, ρ )-module underthe functor m L . The simple H ( L, ρ )-module m L ( πν ) is calculated in section 9. 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 . For s = [ L, π ] G , the representation ι GP ( πν ) ∈ R s ( G ). The functor m G : R s ( G ) −→ H ( G, ρ ) − M od is given by m G ( ι GP ( πν )) = Hom P ( ρ, ι GP ( πν )) is an equivalance of categories.There is an algbera embedding t P : H ( L, ρ ) −→ H ( G, ρ ) described in section 9. The map t P induces the functor ( t P ) ∗ : H ( L, ρ L ) − M od −→ H ( G, ρ ) − M od which 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 ).From Corollary 8.4 in [3], the functors m L , m G , Ind GP , ( t P ) ∗ fit into the following commu-tative diagram: R [ L,π ] G ( G ) m G −−−−→ H ( G, ρ ) − M od
Ind GP x ( t P ) ∗ x R [ L,π ] L ( L ) m L −−−−→ H ( L, ρ ) − M od
We further show in section 9 that there is an algebra embedding T P : H ( L, ρ ) −→ H ( G, ρ )given by T P ( φ ) = t P ( φδ − / P ) for φ ∈ H ( L, ρ ) so 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
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 ). SUBHA SANDEEP REPAKA
Define g i = q − n φ i for i = 0 , n is odd and g i = q − n/ φ i for i = 0 , n is even. We shall show in section 9 that g ∗ g = T P ( α )for both the cases. From this result and Propn. 1.6 in [7] Theorem 1 follows.As Z ( L ) P = ` n ∈ Z P ζ n , so we can extend ρ to a representation f ρ of Z ( L ) P via f 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 f 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 ) . Let P be the Siegel parabolic subgroup of G and L be the SiegelLevi component of P . Let π = c - Ind LZ ( L ) P f ρ be a smooth irreducible supercuspidal depth zerorepresentation of L ∼ = GL n ( E ) where f ρ ( ζ k j ) = ρ ( j ) for j ∈ P , k ∈ Z and ρ = τ θ for someregular character θ of l × with [ l : k E ] = n and | k F | = q . Consider the family 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/ , − } . Let E ′ /F ′ be a quadratic Galois extension of non-Archimedean local fields where char F ′ = 0. Write − for the non-trivial element of Gal( E ′ /F ′ ). The group G ′ = U( n, n ) is givenby G ′ = U( n, n ) = { g ∈ GL n ( E ′ ) | t gJ g = J } for J = ñ ô where each block is of size n and for g = ( g ij ) we write g = ( g ij ). Let P ′ be the Siegel parabolic subgroup of G ′ and let L ′ be the Siegel Levi component of P ′ . Let π ′ be a smooth irreducible supercuspidal representation of L ′ ∼ = GL n ( E ′ ). In [5], Goldbergcomputed the reducibility points of ι G ′ P ′ ( π ′ ) for the ramified extension case by computing thepoles of certain L -functions attached to the representations of GL n ( E ′ ). Note however thatthe base field F ′ was assumed to be of characteristic 0 in [5], whereas I assumed characteristicof F = 2. In [5] no restriction on the depth of the representation π ′ was there, while inthis paper I have assumed depth of the representation π to be zero. I have computed myresults for both ramified and unramified cases where as in [5], computations were done onlyfor the ramified case. The final results obtained in [5] were in terms of matrix coefficents of π ′ whereas my results are in terms of the unramified characters of L . Acknowledgments : This work is a part of author’s thesis. He wishes to thank his advisorAlan Roche from University of Oklahoma, USA for suggesting the problem studied in thiswork and guidance. Further, he wishes to thank C.G.Venketasubramanian and Arnab Mitraat IISER Tirupati, India for their interest and suggestions to improve the paper.2.
Preliminaries
Bernstein Decomposition.
Let ( π, V ) be an irreducible smooth representation of G .According to Theorem 3.3 in [8], there exists unique conjugacy class of cuspidal pairs ( L, σ ) REDUCIBILITY PROBLEM FOR EVEN UNITARY GROUPS: THE DEPTH ZERO CASE 7 with the property that π is isomorphic to a composition factor of ι GP σ for some parabolicsubgroup P of G . We call this conjugacy class of cuspidal pairs, the cuspidal support 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 .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.
Let K be a compact open subgroup of G . Let ( ρ, W ) be an irreducible smoothrepresentation of K and ( π, V ) be a smooth representation of G . Let V ρ be the ρ -isotopicsubspace of V . V ρ = X W ′ W ′ where the sum is over all W ′ such that ( π | K , W ′ ) ≃ ( ρ, W ).Let H ( G ) be the space of all locally constant compactly supported functions f : G → C .This is a C - algebra under convolution ∗ . So for elements f, g ∈ H ( G ) we have( f ∗ g )( x ) = Z G f ( y ) g ( y − x ) dµ ( y ) . Here we have fixed a Haar measure µ on G . Let ( π, V ) be a representation of G . Then H ( G ) acts on V via hv = Z G h ( x ) π ( x ) vdµ ( x )for h ∈ H ( G ) , v ∈ V . Let e ρ be the element in H ( G ) with support K such that e ρ ( x ) = dim ρµ ( K ) tr W ( ρ ( x − )) , x ∈ K. We have e ρ ∗ e ρ = e ρ and e ρ V = V ρ for any smooth representation ( π, V ) of G . Let R ρ ( G )be the full sub-category of R ( G ) consisting of all representations ( π, V ) where V is generatedby ρ -isotopic vectors. So ( π, V ) ∈ R ρ ( G ) ⇐⇒ V = H ( G ) ∗ e ρ V . We now state the definitionof a type. Definition 1.
Let s ∈ B ( G ). We say that ( K, ρ ) is an s -type in G if R ρ ( G ) = R s ( G ).2.3. Hecke algebras.
Let K be a compact open subgroup of G . Let ( ρ, W ) be an irreduciblesmooth representation of K . Here we introduce the Hecke algebra H ( G, ρ ). H ( G, ρ ) = f : G → End C ( ρ ∨ ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) supp( f ) is compact and f ( k gk ) = ρ ∨ ( k ) f ( g ) ρ ∨ ( k )where k , k ∈ K, g ∈ G . SUBHA SANDEEP REPAKA
Then H ( G, ρ ) is a C -algebra with multiplication given by convolution ∗ w.r.t some fixedHaar measure µ on G . So for elements f, g ∈ H ( G ) we have( f ∗ g )( x ) = Z G f ( y ) g ( y − x ) dµ ( y ) . The importance of types is seen from the following result. Let π be a smooth representationin R s ( G ). Let H ( G, ρ ) − M od denote the category of H ( G, ρ )-modules. If (
K, ρ ) is an s -typethen m G : R s ( G ) −→ H ( G, ρ ) − M od given by m G ( π ) = Hom K ( ρ, π ) is an equivalence ofcategories.2.4. Covers.
Let K be a compact open subgroup of G . Let ( ρ, W ) be an irreducible rep-resentation of K . Then we say ( K, ρ ) is decomposed with respect to (
L, P ) if the followinghold:(1) K = ( K ∩ U )( K ∩ L )( K ∩ U ).(2) ( K ∩ U ) , ( K ∩ U ) ker ρ .Suppose ( K, ρ ) is decomposed with respect to (
L, P ). We set K L = K ∩ L and ρ L = ρ | K L .We say an element g ∈ G intertwines ρ if Hom K g ∩ K ( ρ g , ρ ) = 0. Let I G ( ρ ) = { x ∈ G | x intertwines ρ } . We have the Hecke algebras H ( G, ρ ) and H ( L, ρ L ). We write H ( G, ρ ) L = { f ∈ H ( G, ρ ) | supp( f ) ⊆ KLK } . We recall some results and constructions from pages 606-612 in [3]. These allow us totransfer questions about parabolic induction into questions concerning the module theory ofappropriate Hecke algebras.
Proposition 1.
Let ( K, ρ ) decompose with respect to ( L, P ) .Then(1) ρ L is irreducible.(2) I L ( ρ L ) = I G ( ρ ) ∩ L .(3) There is an embedding T : H ( L, ρ L ) −→ H ( G, ρ ) such that if f ∈ H ( L, ρ L ) has support K L zK L for some z ∈ L , then T ( f ) has support KzK .(4) The map T induces an isomorphism of vector spaces: H ( L, ρ L ) ≃ −→ H ( G, ρ ) L . Definition 2.
An element z ∈ L is called ( K, P )-positive element if:(1) z ( K ∩ U ) z − ⊆ K ∩ U . (2) z − ( K ∩ U ) z ⊆ K ∩ U. Definition 3.
An element z ∈ L is called strongly ( K, P )-positive element if:(1) z is ( K, P ) positive.(2) z lies in center of L .(3) For any compact open subgroups K and K ′ of U there exists m > m ∈ Z suchthat z m Kz − m ⊆ K ′ .(4) For any compact open subgroups K and K ′ of U there exists m > m ∈ Z suchthat z − m Kz ⊆ K ′ . Proposition 2.
Strongly ( K, P ) -positive elements exist and given a strongly positive element z ∈ L , there exists a unique function φ z ∈ H ( L, ρ L ) with support K L zK L such that φ z ( z ) isidentity function in End C ( ρ ∨ L ) . REDUCIBILITY PROBLEM FOR EVEN UNITARY GROUPS: THE DEPTH ZERO CASE 9 H + ( L, ρ L ) = f : L → End C ( ρ ∨ L ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) supp( f ) is compact and consistsof strongly ( K, P )-positive elementsand f ( k lk ) = ρ ∨ L ( k ) f ( l ) ρ ∨ L ( k )where k , k ∈ K L , l ∈ L . The isomorphism of vector spaces T : H ( L, ρ L ) −→ H ( G, ρ ) L restricts to an embedding ofalgebras: T + : H + ( L, ρ L ) −→ H ( G, ρ ) L ֒ → H ( G, ρ ). Proposition 3.
The embedding T + extends to an embedding of algebras t : H ( L, ρ L ) −→ H ( G, ρ ) ⇐⇒ T + ( φ z ) is invertible for some strongly ( K, P ) -positive element z , where φ z ∈ H ( L, ρ L ) has support K L zK L with φ z ( z ) = 1 . Definition 4.
Let L be a proper Levi subgroup of G . Let K L be a compact open subgroup of L and ρ L be an irreducible smooth representation of K L . Let K be a compact open subgroupof G and ρ be an irreducible, smooth representation of K . Then we say ( K, ρ ) is a G -coverof ( K L , ρ L ) if(1) The pair ( K, ρ ) is decomposed with respect to (
L, P ) for every parabolic subgroup P of G with Levi component L .(2) K ∩ L = K L and ρ | L ≃ ρ L .(3) The embedding T + : H + ( L, ρ L ) −→ H ( G, ρ ) extends to an embedding of algebras t : H ( L, ρ L ) −→ H ( G, ρ ). Proposition 4.
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. Recall the categories R s L ( L ) , R s ( G ) where s L = [ L, π ] L and s = [ L, π ] G . Also recall H ( G, ρ ) − M od is the category of H ( G, ρ )-modules. Let H ( L, ρ L ) − M od be the categoryof H ( L, ρ L )-modules. The functors ι GP , m G were defined earlier. Let π ∈ R s L ( L ). Then thefunctor m L : R s L ( L ) −→ H ( L, ρ L ) − M od is given by m L ( π ) = Hom K L ( ρ L , π ). The functor( T P ) ∗ : H ( L, ρ L ) − M od −→ H ( G, ρ ) − M od is defined later in this paper.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
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 ( π, V ) be a representation of GL n ( E ) and 1 V be the identity linear transformation of V . As ̟ E n ∈ Z , so π ( ̟ E n ) = ω π ( ̟ E n )1 V where ω π : Z −→ C × is the central characterof π .Let e τ be a representation of f K such that:(1) e τ ( ̟ E n ) = ω π ( ̟ E n )1 V , (2) e τ | K = τ. Say ω π ( ̟ 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 f K , so that Z acts by ω π . Hence π | e K ⊇ e τ z which implies thatHom e K ( e τ z , π | e K ) = 0.By Frobenius reciprocity for induction from open subgroups,Hom e K ( e τ z , π | e K ) ≃ Hom GL n ( E ) ( c - Ind GL n ( E ) e K e τ z , π ).Thus Hom GL n ( E ) ( c - Ind GL n ( E ) e K e τ z , π ) = 0. So there exists a non-zero GL n ( E )-map from c - Ind G e K e τ z to π . As τ is cuspidal representation, using Cartan decompostion and Mackey’s cri-teria we can show that c - Ind GL n ( E ) e K e τ z is irreducible. So π ≃ c - Ind GL n ( E ) e K e τ z . As c - Ind GL n ( E ) e K e τ z is irreducible supercuspidal representation of GL n ( E ) of depth zero, so π is irreducible super-cuspidal representation of GL n ( E ) of depth zero.Conversely, let π is a irreducible, supercuspidal, depth zero representation of GL n ( E ).So π K = { } . Hence π | K ⊇ K , where 1 K is trivial representation of K . This means π | K ⊇ τ , where τ is an irreducible representation of K such that τ | K ⊇ K . So τ istrivial on K . So π | K contains an irreducible representation τ of K such that τ | K istrivial. So τ can be 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[10].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 ) e K e τ z ( τ, ω π ( ̟ E n )) ←−−−−−−−−−−−−−−−−−−−−−−−−−−−− π From now on we denote the representation τ by ρ . So ρ is an irreducible cuspidalrepresentation of GL n ( k E ) inflated to K = GL n ( O E ).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 5. ρ is a homomorphism from P to GL ( V ) . So ρ becomes a representation of P .Proof. Let P , = ( ñ a t a − ô | a ∈ K = 1 + ̟ E M n ( O E ) ) . Clearly, P , ∼ = K . Now let us define P = P − P , P + . We can observe clearly that P isa subgroup of U( n, n ) ∩ GL n ( O E ). We have the following group homomorphism: REDUCIBILITY PROBLEM FOR EVEN UNITARY GROUPS: THE DEPTH ZERO CASE 11 φ : P mod p E −−−−−→ P ( k E ).Here P ( k E ) is the Siegel parabolic subgroup of { g ∈ GL n ( k E ) | t gJ g = J } . Now P ( k E ) = L ( k E ) ⋉ U ( k E ), where L ( k E ) , U ( k E ) are the Levi component and unipotent radical of theSiegel parabolic subgroup respectively. L ( k E ) = ( ñ a t a − ô | a ∈ GL n ( k E ) ) ,U ( k E ) = ( ñ X ô | X ∈ M n ( k E ) , X + t X = 0 ) . Observe that φ is a surjective homomorphism. Now let us find the inverse image of U ( k E ).Let j ∈ P and j = j − j j + be the Iwahori factorization of j , where j + ∈ P + , j − ∈ P − , j ∈ P .So φ ( j ) ∈ U ( k E ) ⇐⇒ j ∈ P , . Therefore P is the inverse image of U ( k E ) under φ . So wehave P (cid:30) P ∼ = P ( k E ) (cid:30) U ( k E ) ∼ = L ( k E ) ∼ = GL n ( k E ). As ρ ( j ) = ρ ( j ), so ρ is a representationof P which is lifted from representation ρ of P that is trivial on P . (cid:3) Calculation of N G ( P )We set G = U( n, n ). To describe H ( G, ρ ) we need to determine N G ( ρ ) which is given by 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. By the Cartan decomposition, any g ∈ GL n ( E ) can be written as g = k ̟ l E . . . ̟ l E . . . ... ... . . .
00 0 . . . ̟ l n E k where k , k ∈ K and for certain l , l . . . l n ∈ Z with l l . . . l n .So we only need to determine the matrices ̟ l E . . . ̟ l E . . . ... ... . . .
00 0 . . . ̟ l n E that normalize K . Let A be one such matrix which normalizes K . So ABA − ∈ K for all B ∈ K . Let the matrix A be of form ̟ l E . . . ̟ l E . . . ... ... . . .
00 0 . . . ̟ l n E for certain l , l . . . l n ∈ Z with l l . . . l n . Now matrix A − looks like ̟ − l E . . . ̟ − l E . . . ... ... . . .
00 0 . . . ̟ − l n E . Let the matrix B be of form ( b ij ) i,j n . So matrix B looks like: b b . . . b n b b . . . b n ... ... . . . ...b n b n . . . b nn where b ij ∈ O E for 1 i, j n . Now ABA − ∈ K for all B ∈ K . And that implies ̟ l i − l j E b ij ∈ O E for 1 i, j n . Choose a matrix B in K such that b ii = 1 for 1 i n , b ij = 0 for 1 i, j n, i = 1 , j = 2 , i = j and b = 1. So we have ̟ l − l E ∈ O E . As onlypositive integral powers of ̟ E lie in O E . Hence l > l . Similarly we can show that l > l .So l = l . We can show in a similar fashion that l = l , l = l , . . . , l n − = l n . Let us call l = l = l = · · · = l n = r for some r ∈ Z . Hence any matrix ̟ l E . . . ̟ l E . . . ... ... . . .
00 0 . . . ̟ l n E in N GL n ( E ) ( K ) is of the form ̟ rE . . . ̟ rE . . . ... ... . . .
00 0 . . . ̟ rE for some r ∈ Z . So N GL n ( E ) ( K ) consists of all the matrices in g ∈ GL n ( E ) such that g = M ′ ̟ rE . . . ̟ rE . . . ... ... . . .
00 0 . . . ̟ rE M ′′ where M ′ , M ′′ ∈ K , r ∈ Z . But we can see that ̟ rE . . . ̟ rE . . . ... ... . . .
00 0 . . . ̟ rE ∈ Z (GL n ( E )).Let M = M ′ M ′′ u − for some u ∈ O × E and let u̟ rE = a for some a ∈ E × . So now any matrixin N GL n ( E ) ( K ) is of form g ∈ GL n ( E ) such that REDUCIBILITY PROBLEM FOR EVEN UNITARY GROUPS: THE DEPTH ZERO CASE 13 g = M a . . . a . . . ... ... . . .
00 0 . . . a where a ∈ E × , M ∈ K . So we have N GL n ( E ) ( K ) = ZK = K Z . (cid:3) From now on let us denote K by K . Now let us calculate N G ( P ). Note that J = ñ ô ∈ G . Indeed, J ∈ N G ( P ). The center Z ( P ) of P is given by Z ( P ) = ( ñ u u − ô | u ∈ O × E ) . Recall the center Z ( L ) of L is given by Z ( L ) = ( ñ a a − ô | a ∈ E × ) . Proposition 6. N G ( P ) = h P Z ( L ) , J i = P Z ( L ) ⋊ h J i .Proof. It easy to see that N G ( P ) N G ( Z ( P )). Now suppose g = ñ A BC D ô ∈ N G ( Z ( P )),where A, B, C, D ∈ M n ( E ). Let us choose u ∈ O × E such that u = u − . Now such a u exists in O × E . Because if u = u − for all u ∈ O × E then u = u − for all u ∈ O × E . But O × E ∩ F × = O × F .Therefore u = u − for all u ∈ O × F or u = 1 for all u ∈ O × F which is a contradiction.As ñ A BC D ô ∈ N G ( Z ( P )) , ñ A BC D ô ñ u u − ô ñ A BC D ô − = ñ v v − ô for some v ∈ O × E . The left and right hand sides must have the same eigenvalues. So u = v or v − . Let u = v . Then we have ñ A BC D ô ñ u u − ô ñ A BC D ô − = ñ v v − ô = ⇒ ñ Au Bu − Cu Du − ô = ñ Av BvCv − Du − ô . As u = v , so Au = Av, Du − = Dv − . Now as u = v − (i.e v = u − ), from the abovematrix relation we can see that Bu − = Bv , Cu = Cv − for arbitrary matrices B and C .So this would imply that B = C = 0. In a similar way, we can show that if u = v − then A = D = 0. Hence any element of N G ( Z ( P )) is of the form ñ A D ô or ñ BC ô with A, B, C, D ∈ GL n ( E ). As N G ( P ) N G ( Z ( P )), so any element which normalizes P is alsoof the form ñ A D ô or ñ BC ô with A, B, C, D ∈ GL n ( E ). If ñ A D ô normalizes P then ñ A D ô ñ a t a − ô ñ A − D − ô ∈ P for all a ∈ K .= ⇒ ñ AaA − D t a − D − ô ∈ P for all a ∈ K .Hence AaA − , D t a − D − ∈ K for all a ∈ K . So this implies that A, D ∈ N GL n ( E ) ( K ) = ZK = KZ from lemma 1 and also t ( AaA − ) − = D t a − D − for all a ∈ K . If t ( AaA − ) − = D t a − D − for all a ∈ K then t A − t a − t A = D t a − D − for all a ∈ K = ⇒ A = t D − (i.e D = t A − ). And as A ∈ ZK , so A = zk for some z ∈ Z, k ∈ K . Hence ñ A D ô = ñ zk t ( zk ) − ô .Similarly, we can show that if ñ BC ô normalizes P then ñ BC ô = ñ z ′ k ′ t ( z ′ k ′ ) − ô for some z ′ ∈ Z, k ′ ∈ K .If ñ A D ô ∈ N G ( P ), we have shown that it looks like ñ zk t ( zk ) − ô and if ñ BC ô ∈ N G ( P ), we have shown that it looks like ñ z ′ k ′ t ( z ′ k ′ ) − ô where z, z ′ ∈ Z , k, k ′ ∈ K . Weknow that J ∈ N G ( P ) and as ñ BC ô J = ñ B C ô , so N G ( P ) = Æ J ; ñ zk t ( zk ) − ô | z ∈ Z, k ∈ K ∏ = P Z ( L ) ⋊ h J i . (cid:3) Calculation of N G ( ρ )5.1. Unramified case:
We now calculate N G ( ρ ) in the unramified case. This will help indetermining the structure of H ( G, ρ ).Recall that if ρ is an irreducible cuspidal representation of GL n ( k E ) then there exists aregular character θ of l × (where l is a degree n extension of k E ) such that ρ = τ θ . We have k E = F q . So l = F q n .Let Γ = Gal( l/k E ). The group Γ is generated by the Frobenius map Φ given by Φ( λ ) = λ q for λ ∈ l . Here Φ n ( λ ) = λ q n = λ (since l × is a cyclic group of order q n −
1) = ⇒ Φ n = 1.Let us look at the action of Γ on Hom( l × , C × ). For γ ∈ Γ and θ ∈ Hom( l × , C × ), γ acts on θ by γ.θ ( λ ) = θ ( γ ( λ )). Here γ.θ is also represented by θ γ .We say a character θ is regular character if stab Γ ( θ ) = { γ ∈ Γ | θ γ = θ } = 1. So if θ isregular character of l × then θ γ = θ = ⇒ γ = 1. And also for two regular characters θ and θ ′ we have τ θ ≃ τ θ ′ ⇐⇒ there exists γ ∈ Γ such that θ γ = θ ′ . REDUCIBILITY PROBLEM FOR EVEN UNITARY GROUPS: THE DEPTH ZERO CASE 15
As we are in the unramified case, so Gal( k E /k F ) ∼ = Gal( E/F ). Let ι : GL n ( k E ) −→ GL n ( k E ) be a group homomorphism given by: ι ( g ) = t g − . Let us denote τ θ ◦ ι by τ θι . So τ θι ( g ) = τ θ ( ι ( g )) = τ θ ( t g − ) for g ∈ GL n ( k E ). We also denote τ θ ( g ) for τ θ ( g ) for g ∈ GL n ( k E ).It can be observed clearly as θ is a character of l × , so θ ( λ m ) = θ m ( λ ) for m ∈ Z , λ ∈ l × .Let τ ∨ θ be the dual representation of τ θ . Let V be the vector space corresponding to τ θ which is finite dimensional. Choose a basis { v , v , . . . v n } of the vector space V . The dualbasis { v ∗ , v ∗ , . . . v ∗ n } for the dual space V ∗ of V can be constructed such that v ∗ i ( v j ) = δ ij for 1 i, j n . Suppose with respect to the above basis { v , v , . . . v n } , τ θ ( g − ) representsmatrix A and with respect to the dual basis { v ∗ , v ∗ , . . . v ∗ n } , τ ∨ θ ( g ) represents matrix B ,then A = t B .From Propn. 3.5 in [9] we have τ θ ≃ τ θ q and from Propn. 3.4 in [9] we have τ ∨ θ ≃ τ θ − . Proposition 7.
Let θ be a regular character of l × . Then τ ιθ ≃ τ θ ⇐⇒ θ γ = θ − q for some γ ∈ Gal( l/k E ) .Proof. = ⇒ As τ ιθ ≃ τ θ , so χ τ ιθ ( g ) = χ τ θ ( g ) for g ∈ GL n ( k E ). But χ τ ιθ ( g ) = χ τ θ ( t g − ), since χ τ ιθ ( g ) = χ τ θ ( ι ( g )) for g ∈ GL n ( k E ). As we know from the above discussion that τ ∨ θ ( g ) =( τ θ ( g − )) t , so trace ( τ ∨ θ ( g )) = trace ( τ θ ( g − )) t . Now trace ( τ θ ( g − )) = trace ( τ θ ( g − )) t as thetrace of the matrix and it’s transpose are same. So we have trace ( τ θ ( g − )) = trace ( τ ∨ θ ( g )). Letus choose h ∈ GL n ( k E ) such that h − t g − h = g − . So, χ τ ∨ θ ( g ) = χ τ θ ( g − ) = χ τ θ ( h − t g − h ) = χ τ θ ( t g − ). Let us denote τ ηθ ( g ) for τ θ ( η ( g )) where η : g −→ t g − is a group automorphism ofGL n ( k E ). Hence χ τ ηθ ( g ) = χ τ θ ( t g − ). But we have already shown before that χ τ θ ( t g − ) = χ τ ∨ θ ( g ). so χ τ ∨ θ ( g ) = χ τ ηθ ( g ). This implies τ ∨ θ ≃ τ ηθ . Hence τ ιθ = τ ηθ ≃ τ ∨ θ ≃ τ ∨ θ q ≃ τ θ − q (since τ ∨ θ ≃ τ ηθ , τ θ ≃ τ θ q , τ ∨ θ ≃ τ θ − ). Now from the hypothesis of Propn. we know that τ ιθ ≃ τ θ , sothis implies τ θ ≃ τ θ − q (since τ ιθ ≃ τ θ − q ). But as θ is a regular character θ γ = θ − q for some γ ∈ Γ = Gal( l/k E ) where [ l : k E ] = n . ⇐ = Now we can reverse the arguments and show that if θ γ = θ − q for some γ ∈ Γ =Gal( l/k E ) then τ ιθ ≃ τ θ − q . (cid:3) Proposition 8. If θ is a regular character of l × such that θ γ = θ − q for some γ ∈ Γ then n is odd. Conversely, if n = 2 m + 1 is odd and θ is a regular character of l × then θ Φ m +1 = θ − q .Proof. = ⇒ Suppose θ is a regular character of l × such that θ γ = θ − q for some γ ∈ Γ. Weknow that Γ = < Φ > where Φ : l −→ l is the Frobenius map given by Φ( λ ) = λ q for λ ∈ l . Now Φ n ( λ ) = λ q n = λ for λ ∈ l = ⇒ Φ n = 1. Now we have ( θ − q ) γ = ( θ γ ) − q .Hence ( θ ) γ = ( θ γ ) γ = ( θ − q ) γ = ( θ γ ) − q = ( θ − q ) − q = θ q . Now θ q = θ Φ because for λ ∈ l × , θ q ( λ ) = θ ( λ q ) = θ (Φ( λ )) = θ Φ ( λ ). As θ is a regular character and ( θ ) γ = θ Φ , so γ = Φ. Let Φ be a generator of Γ and γ = Φ. So γ is also a generator of Γ.Hence order of γ = order of Φ = ⇒ ng.c.d (2 ,n ) = n = ⇒ g.c.d (2 , n ) = 1. So n is odd. ⇐ = Suppose n is odd. Let n = 2 m + 1 where m ∈ N . NowHom( l × , C × ) ∼ = l × .So Hom( l × , C × ) is a cyclic group of order ( q n − d of ( q n − l × , C × ) of order d . As ( q n + 1) is a divisor of ( q n − θ in Hom( l × , C × ) of order ( q n + 1). Hence θ q n +1 = 1 = ⇒ θ q n = θ − = ⇒ θ q n +1 = θ − q = ⇒ θ q m +2 = θ − q = ⇒ θ ( q ) m +1 = θ − q = ⇒ θ Φ m +1 = θ − q = ⇒ θ γ = θ − q ,where γ = Φ m +1 ∈ Γ. Now we claim that θ is a regular character in Hom( l × , C × ). Suppose θ γ = θ for some γ ∈ Γ. Let γ = Φ k for some k ∈ N . So we have θ Φ k = θ . But θ Φ = θ q , hence θ q k = θ . Thatimplies θ q k − = 1. As θ has order ( q n + 1), so ( q n + 1) | ( q k − l = 2 k , so we have( q n + 1) | ( q l −
1) . If l < n then it is a contradiction to the fact that ( q n + 1) | ( q l − l > n . Now by applying Euclidean Algorithm for the integers l, n we have l = nd + r for some 0 r < n and d > r, d ∈ Z . Now d = 0, because if d = 0 then l = r andthat means l < n which is a contradiction. So d ∈ N . As we have ( q n + 1) | ( q l −
1) = ⇒ ( q n + 1) | (( q l −
1) + ( q n + 1)) = ⇒ ( q n + 1) | ( q l + q n ) = ⇒ ( q n + 1) | q n ( q r .q n ( d − + 1).Now as q n and ( q n + 1) are relatively prime, so ( q n + 1) | ( q r .q n ( d − + 1) = ⇒ ( q n + 1) | (( q r .q n ( d − + 1) − ( q n + 1)) = ⇒ ( q n + 1) | q n ( q r .q n ( d − − q n and ( q n + 1) are relativelyprime, so ( q n + 1) | ( q r .q n ( d − − q n + 1) | ( q r + 1)if d is odd or ( q n + 1) | ( q r −
1) if d is even. But degree of ( q n + 1) is greater than degree of( q r + 1) as r < n . So r has to be equal to 0 and l = 2 k = nd + r = nd . And that implies2 | nd . But n is odd so 2 | d . Now this means that d is even and hence ( q n + 1) | ( q r − q n + 1) | ( q r −
1) is not possible because r = 0. So we have 2 k = nd = ⇒ n | k .But as n is odd this implies n | k . And this further implies k = np for some p ∈ N . So γ = Φ k = Φ np = 1 = ⇒ θ is regular character. (cid:3) Combining Propn. 7 and Propn. 8, we have the following Propn.
Proposition 9.
Let θ is a regular character of l × . Then τ ιθ ≃ τ θ ⇐⇒ n is odd. We know that ρ is an irreducible cuspidal representation of K . But K ∼ = P . So ρ canbe viewed as a representation of P . Now let us compute N G ( ρ ), where N G ( ρ ) = { m ∈ N G ( P ) | ρ ≃ ρ m } . Let m ∈ N G ( P ). Hence m is either J or m is of the form ñ zk t ( zk ) − ô for some z ∈ Z, k ∈ K . Proposition 10. If m = ñ zk t ( zk ) − ô for some z ∈ Z, k ∈ K then ρ m ≃ ρ .Proof. As ρ is an irreducible cuspidal representation of K , so K normalizes ρ . Clearly, Z normalizes ρ . Thus ZK normalizes ρ . As ρ can also be viewed as a representation of P ,so ρ m ≃ ρ where m = ñ zk t ( zk ) − ô for some z ∈ Z, k ∈ K . (cid:3) Proposition 11. If m = J = ñ ô then ρ m ≃ ρ only when n is odd.Proof. We know that ι : a −→ t a − is a group homomorphism of GL n ( k E ). Now ι : a −→ t a − can be inflated to a group homomorphism of GL n ( O E ). Further, ι can be viewed as a grouphomomorphism from P to P given by: ι Çñ a t a − ôå = ñ t a − a ô REDUCIBILITY PROBLEM FOR EVEN UNITARY GROUPS: THE DEPTH ZERO CASE 17 where a ∈ GL n ( O E ). Let g = ñ a t a − ô . If m = J then ρ m ( g ) = ρ ( J gJ − ) = ρ Çñ t a − a ôå = ρ ( ι ( g )) = ρ ι ( g ). So ρ m ( g ) = ρ ι ( g ) for g ∈ P = ⇒ ρ m = ρ ι . But from the hypothesis of Propn., we knowthat ρ m ≃ ρ . So we have ρ ≃ ρ ι . Now from Propn. 9, ρ ≃ ρ ι = ρ m ⇐⇒ n is odd. (cid:3) Thus we have the following conclusion about N G ( ρ ) for the unramified case: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 .5.2. Ramified case:
Now that we have calculated N G ( P ), let us calculate N G ( ρ ) for theramified case which would help us in determining the structure of H ( G, ρ ) in the ramifiedcase.As in section 5.1, ρ = τ θ for some regular character θ of l × (where l is a degree n extensionof k E ). We have k E = F q . So l = F q n .Let Γ = Gal( l/k E ). The group Γ is generated by Frobenius map Φ given by Φ( λ ) = λ q for λ ∈ l . Here Φ n ( λ ) = λ q n = λ (since l × is a cyclic group of order q n −
1) = ⇒ Φ n = 1.For γ ∈ Γ and θ ∈ Hom( l × , C × ), γ acts on θ by γ.θ ( λ ) = θ ( γ ( λ )). Here γ.θ is alsorepresented by θ γ .As we are in the ramified case, so Gal( k E /k F ) = 1. So g = g for g ∈ k E . Let ι : GL n ( k E ) −→ GL n ( k E ) be a group homomorphism given by: ι ( g ) = t g − = t g − . Let us denote τ θ ◦ ι by τ θι .So τ θι ( g ) = τ θ ( ι ( g )) = τ θ ( t g − ) = τ θ ( t g − ) for g ∈ GL n ( k E ). We also denote τ θ ( g ) for τ θ ( g )for g ∈ GL n ( k E ). But τ θ ( g ) = τ θ ( g ) = τ θ ( g ). It can be observed clearly as θ is a character of l × , so θ ( λ m ) = θ m ( λ ) for m ∈ Z , λ ∈ l × . Proposition 12.
Let θ be a regular character of l × . Then τ ιθ ≃ τ θ ⇐⇒ θ γ = θ − for some γ ∈ Gal( l/k E ) .Proof. = ⇒ As τ ιθ ≃ τ θ , so χ τ ιθ ( g ) = χ τ θ ( g ) for g ∈ GL n ( k E ). But χ τ ιθ ( g ) = χ τ θ ( t g − ), since χ τ ιθ ( g ) = χ τ θ ( ι ( g )) for g ∈ GL n ( k E ). As we know from the above discussion that τ ∨ θ ( g ) =( τ θ ( g − )) t , so trace ( τ ∨ θ ( g )) = trace ( τ θ ( g − )) t . Now trace ( τ θ ( g − )) = trace ( τ θ ( g − )) t as thetrace of the matrix and it’s transpose are same. So we have trace ( τ θ ( g − )) = trace ( τ ∨ θ ( g )). Letus choose h ∈ GL n ( k E ) such that h − t g − h = g − . So, χ τ ∨ θ ( g ) = χ τ θ ( g − ) = χ τ θ ( h − t g − h ) = χ τ θ ( t g − ).Let us denote τ ηθ ( g ) for τ θ ( η ( g )) where η : g −→ t g − is a group automorphism ofGL n ( k E ). Hence χ τ ηθ ( g ) = χ τ θ ( t g − ). But we have already shown before that χ τ θ ( t g − ) = χ τ ∨ θ ( g ). so χ τ ∨ θ ( g ) = χ τ ηθ ( g ). This implies τ ∨ θ ≃ τ ηθ . Hence τ ιθ = τ ηθ ≃ τ ∨ θ ≃ τ θ − (since τ ∨ θ ≃ τ ηθ , τ ∨ θ ≃ τ θ − ). Now from the hypothesis of Propn. we know that τ ιθ ≃ τ θ , so this implies τ θ ≃ τ θ − (since τ ιθ ≃ τ θ − ).But as θ is a regular character θ γ = θ − for some γ ∈ Γ = Gal( l/k E )where [ l : k E ] = n . ⇐ = Now we can reverse the arguments and show that if θ γ = θ − for some γ ∈ Γ =Gal( l/k E ) then τ ιθ ≃ τ θ − . (cid:3) Proposition 13. If θ is a regular character of l × such that θ γ = θ − for some γ ∈ Γ then n is even. Conversely, if n = 2 m is even and θ is a regular character of l × then θ Φ m = θ − .Proof. = ⇒ Suppose θ is a regular character of l × such that θ γ = θ − for some γ ∈ Γ. Weknow that Γ = < Φ > where Φ : l −→ l is the Frobenius map given by Φ( λ ) = λ q for λ ∈ l .Now Φ n ( λ ) = λ q n = λ for λ ∈ l = ⇒ Φ n = 1. So for λ ∈ l × we have θ γ ( λ ) = θ γ ( γ ( λ )) = θ − ( γ ( λ )) = θ (( γ ( λ )) − ) = θ ( γ ( λ − )) = θ γ ( λ ) − = θ − ( λ − ) = θ (( λ ) − ) − ) = θ ( λ ). So thisimplies θ γ = θ . As θ is a regular character, so we have γ = 1. Now for λ ∈ l × we have θ Φ ( λ ) = θ (Φ( λ )) = θ ( λ q ) = θ q ( λ ). That implies θ Φ = θ q . As γ = 1 = ⇒ γ = 1 or γ has order2. If γ = 1 as θ γ = θ − = ⇒ θ = θ − = ⇒ θ ( λ ) = θ − ( λ ) for λ ∈ l × = ⇒ θ ( λ ) = θ ( λ − ) = ⇒ θ ( λ ) = 1 = ⇒ ( θ ( λ )) = 1 = ⇒ θ ( λ ) = {± } for λ ∈ l × .Let q be an odd prime power. So for λ ∈ l × we have θ Φ ( λ ) = θ q ( λ ) = ( θ ( λ )) q = θ ( λ )(since θ ( λ ) = {± } ). So this implies θ Φ = θ and that further implies Φ = 1 as θ is a regularcharacter = ⇒ n = 1 which contradicts our assumption that cardinality of Γ is greater than 1.Now suppose q is a prime power of 2. As the characteristic of k E = 2 that implies +1 = − k E . So θ ( λ ) = ± λ ∈ l × . So we have for λ ∈ l × , θ Φ ( λ ) = θ q ( λ ) = ( θ ( λ )) q = 1(since θ ( λ ) = 1). And this implies θ Φ = θ and that further implies Φ = 1 as θ is a regularcharacter = ⇒ n = 1 which contradicts our assumption that cardinality of Γ is greater than 1.Hence γ = 1 or γ has order 2, since γ = 1. Now Γ has order n and γ ∈ Γ has order 2. So2 | n = ⇒ n is even. ⇐ = Suppose n is even. Let n = 2 m where m ∈ N . NowHom( l × , C × ) ∼ = l × .So Hom( l × , C × ) is a cyclic group of order ( q n −
1) = ( q m − d of ( q m − l × , C × ) of order d . As ( q m + 1) is a divisorof ( q m − θ in Hom( l × , C × ) of order ( q m + 1). So θ q m +1 =1 = ⇒ θ q m = θ − = ⇒ θ Φ m = θ − (since θ Φ = θ q ). Hence we have θ γ = θ − , where γ = Φ m .Now we claim that the character θ is regular. Suppose θ γ = θ for some γ ∈ Γ. Let γ = Φ k for some k ∈ Z . Then we have θ Φ k = θ = ⇒ θ q k = θ = ⇒ θ q k − = 1. As θ has order( q m + 1) that means ( q m + 1) | ( q k − k = md + r where r, d ∈ Z , r < m and d >
0. If d = 0 then k = r < m which contradicts the fact that( q m + 1) | ( q k − d >
1. Now as ( q m + 1) | ( q k −
1) = ⇒ ( q m + 1) | (( q k −
1) + ( q m + 1)) = ⇒ ( q m + 1) | ( q k + q m ) = ⇒ ( q m + 1) | ( q md + r + q m ) = ⇒ ( q m + 1) | q m ( q m ( d − r + 1). But as q m and ( q m + 1) are relatively prime, so this implies ( q m + 1) | ( q m ( d − r + 1). But ( q m + 1) | ( q m ( d − r +1) = ⇒ ( q m +1) | (( q m ( d − r +1) − ( q m +1)) = ⇒ ( q m +1) | q m ( q m ( d − r − q m and ( q m + 1) are relatively prime, so this implies ( q m + 1) | ( q m ( d − r − q m + 1) | ( q r + 1) if d is odd and ( q m + 1) | ( q r − d is even.As r < m , the above conditions are possible only when r = 0. If r = 0, then k = md . So if d is odd then ( q m + 1) | ⇒ ( q m + 1) is either 1 or 2. If ( q m + 1) = 1 then q = 0 which isa contradiction. So let ( q m + 1) = 2 then we have q = 1 which is again a contradiction as q is a prime power. So d has to be even. Let d be even and is greater than 2. So d can takevalues 4 , , , . . . . But as k = md , so k can take values 4 m, m, m, . . . . That is k can takevalues 2 n, n, n, . . . which is a contradiction as k < n . So d = 2. Hence k = 2 m = n . SoΦ k = Φ n = 1 = ⇒ γ = 1. So θ is a regular character. (cid:3) Combining Propn. 12 and Propn. 13 we have the following Propn.
Proposition 14.
Let θ be a regular character of l × . Then τ ιθ ≃ τ θ ⇐⇒ n is even. We know that ρ is an irreducible cuspidal representation of K . But K ∼ = P . So ρ canbe viewed as a representation of P . Now let us compute N G ( ρ ), where N G ( ρ ) = { m ∈ REDUCIBILITY PROBLEM FOR EVEN UNITARY GROUPS: THE DEPTH ZERO CASE 19 N G ( P ) | ρ ≃ ρ m } . Let m ∈ N G ( P ). Hence m is either J or m is of the form ñ zk t ( zk ) − ô for some z ∈ Z, k ∈ K . Proposition 15. If m = ñ zk t ( zk ) − ô for some z ∈ Z, k ∈ K then ρ m ≃ ρ .Proof. The proof is similar to the proof of Propn. 10. (cid:3)
Proposition 16. If m = J = ñ ô then ρ m ≃ ρ only when n is even.Proof. The proof is similar to the proof of Propn. 11. (cid:3)
So 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 . 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 = ñ ̟ E − ̟ E ô .Proof. Let ζ = w w . So ζ = ñ ̟ E ̟ E − ô . We can clearly see that w = 1. So w = w − and w = w − ζ = w ζ . From the hypothesis of lemma, we have N G ( ρ ) = Z ( L ) P ⋊ h J i .As any element in E × can be written as u̟ nE for some n ∈ Z , u ∈ O × E , so Z ( L ) = Z ( P ) h ζ i .So Z ( L ) P = h P , ζ i . Hence N G ( ρ ) = h P , ζ i ⋊ J . But J = w , w = w ζ . So N G ( ρ ) = h P , w , w i . (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 4.15 in[10], 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 ) ∩ GL n ( O E ) = { g ∈ GL n ( O E ) | t gJ g = J } ,K (0) = { g ∈ ̟ E M n ( O E ) | t gJ g = J } , G = { g ∈ GL n ( 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 = ñ GL n ( k E ) M n ( k E )0 GL n ( k E ) ô 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 G . Here L = ( ñ a t a − ô | a ∈ GL n ( k E ) ) , U = ( ñ X ô | X ∈ M n ( k E ) , X + t X = 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) . Proposition 17.
The map r : H ( K (0) , ρ ) −→ H ( G , ρ ) is an algebra isomorphism.Proof. To prove that the map r is an isomorphism of algebras, we have to show that r is ahomomorphism of algebras and is a bijective map.In order to show that the map r is a homomorphism, we need to show that it is C -linearand it preserves convolution. It is obvious that the map r is C -linear. Let us now show thatthe map preserves convolution.If x ∈ K (0) and φ , φ ∈ H ( K (0) , ρ ) then( φ ∗ φ )( x ) = Z K (0) φ ( y ) φ ( y − x ) dy. Now Z K (0) φ ( y ) φ ( y − x ) dy = X y ∈ P /K (0) φ ( xy − ) φ ( y ) . Hence r ( φ ∗ φ )( r ( x )) = ( φ ∗ φ )( x )= X y ∈ P /K (0) φ ( xy − ) φ ( y )= X y ∈ P /K (0) ( r ( φ )( r ( xy − )))( r ( φ )( r ( y )))= X r ( y ) ∈ P / G ( r ( φ )( r ( x )( r ( y )) − ))( r ( φ )( r ( y ))) REDUCIBILITY PROBLEM FOR EVEN UNITARY GROUPS: THE DEPTH ZERO CASE 21 = ( r ( φ ) ∗ r ( φ ))( r ( x )) . So we have r ( φ ∗ φ )( r ( x )) = ( r ( φ ) ∗ r ( φ ))( r ( x )). But r is a surjective group homomor-phism from K (0) to G . Hence r ( φ ∗ φ )( y ) = ( r ( φ ) ∗ r ( φ ))( y ) for y ∈ G which would implythat r ( φ ∗ φ ) = ( r ( φ ) ∗ r ( φ )). Hence r is a homomorphism of algebras.In order to show that r is bijective map, we first show here that the map r is a one-one map.Let φ , φ ∈ H ( K (0) , ρ ) , y ∈ G . Suppose r ( φ )( y ) = r ( φ )( y ). As r is surjective map from K (0) to G , so there exists x ∈ K (0) such that r ( x ) = y . So r ( φ )( r ( x )) = r ( φ )( r ( x )) = ⇒ φ ( x ) = φ ( x ). As r is a surjective map from K (0) to G , so when y spans over G , x spansover K (0). So φ ( x ) = φ ( x ) for x ∈ K (0) = ⇒ φ = φ . So r is a one-one map.Now we show that r is a surjective map from H ( K (0) , ρ ) to H ( G , ρ ). Let ψ ∈ H ( G , ρ ), then ψ : G −→ End C V ∨ is a map such that ψ ( pgp ′ ) = ρ ∨ ( p ) ψ ( g ) ρ ∨ ( p ′ ) for p, p ′ ∈ P , g ∈ G . As r is asurjective map from K (0) to G , so ψ ◦ r makes sense. Now let us call ψ ◦ r as φ . So φ is a mapfrom K (0) to End C V ∨ . Let p, p ′ ∈ P , k ∈ K (0), so φ ( pkp ′ ) = ( ψ ◦ r )( pkp ′ ) = ψ ( r ( pkp ′ )) = ψ ( r ( p ) r ( k ) r ( p ′ )) = ρ ∨ ( r ( p )) ψ ( r ( k )) ρ ∨ ( r ( p ′ )) = ρ ∨ ( p )( ψ ◦ r )( k ) ρ ∨ ( p ′ ) = ρ ∨ ( p ) φ ( k ) ρ ∨ ( p ′ ). So φ ∈ H ( K (0) , ρ ). Let y ∈ G . So there exits x ∈ K (0) such that r ( x ) = y . Now consider ψ ( y ) = ψ ( r ( x )) = ( ψ ◦ r )( x ) = φ ( x ) = r ( φ )( r ( x )) = r ( φ )( y ). So ψ ( y ) = r ( φ )( y ) for y ∈ G = ⇒ ψ = r ( φ ). Hence r is a surjective map.As r is both one-one and surjective map, hence it is a bijective map. (cid:3) Let w = r ( w ) = r ( ñ ô ) = ñ ô ∈ 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 Propn. 3.2 in [6], there exists a unique φ in H ( G , ρ ) withsupport P w P such that φ = λ + ( λ − φ . By Propn. 17, 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 . From Lemma3.12 in [9], λ = q n . Hence φ = q n + ( q n − φ . As support of φ = P w P ⊆ K (0) ⊆ G , so φ can be extended to G and viewed as an element of H ( G, ρ ). 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 . Let η = ñ ̟ E ô . Now we can check that ηw η − = w . Recall that P looksas follows: P = ñ GL n ( O E ) M n ( O E )M n ( p E ) GL n ( O E ) ô T G. Lemma 3. η P η − = P . Proof. P = ñ GL n ( O E ) M n ( O E )M n ( p E ) GL n ( O E ) ô \ G = ⇒ η P η − = η ñ GL n ( O E ) M n ( O E )M n ( p E ) GL n ( O E ) ô η − T ηGη − . It is easy to show that η ñ GL n ( O E ) M n ( O E )M n ( p E ) GL n ( O E ) ô η − = ñ GL n ( O E ) M n ( O E )M n ( p E ) GL n ( O E ) ô . Now we claim that ηGη − = G . To prove this let us consider G ′ = { g ∈ GL n ( E ) | t gJ g = λ ( g ) J for some λ ( g ) ∈ F × } . Now η ∈ G ′ clearly, as t ηJ η = ̟ E J = ̟ F J . And λ : G ′ −→ F × is a homomorphismof groups with kernel G . So G E G ′ . As η ∈ G ′ and G E G ′ , so ηGη − = G . Hence η P η − = P . (cid:3) As P ⊆ K (0) and w ∈ K (0), so K (0) ⊇ P ∐ P w P = ⇒ ηK (0) η − ⊇ η P η − ∐ η P w P η − .But from lemma 3, we know that η P η − = P and η P w P η − = ( η P η − )( ηw η − )( η P η − ) = P w P (since ηw η − = w ). So ηK (0) η − ⊇ P ∐ P w P .Let r ′ be homomorphism of groups given by the map r ′ : ηK (0) η − −→ G such that r ′ ( x ) =( η − xη ) modp E for x ∈ ηK (0) η − . Observe that r ′ is a surjective homomorphism of groupsbecause r ′ ( ηK (0) η − ) = ( η − ηK (0) η − η ) modp E = K (0) modp E = G . The kernel of grouphomomorphism is ηK (0) η − . Now by the first isomorphism theorem of groups we have ηK (0) η − ηK (0) η − ∼ = K (0) K (0) ∼ = G . Also r ′ ( η P η − ) = ( η − η P η − η ) modp E = P modp E = P . Let ρ berepresentation of P which when inflated to P is given by ρ . The Hecke algebra of ηK (0) η − which we denote by H ( ηK (0) η − , ρ ) is a sub-algebra of H ( G, ρ ).The map r ′ : ηK (0) η − −→ G extends to a map from H ( ηK (0) η − , ρ ) to H ( G , ρ ) which wegain denote by r ′ . Thus r ′ : H ( ηK (0) η − , ρ ) −→ H ( G , ρ ) is given by r ′ ( φ ( r ′ ( x )) = φ ( x )for φ ∈ H ( ηK (0) η − , ρ ) and x ∈ ηK (0) η − . The proof that r ′ is an isomorphism goes in the similar lines as Propn. 17. We can observethat r ′ ( w ) = w ∈ G , where w is defined as before in this section. As we know from ourprevious discussion in this section, that 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 ( ηK (0) η − , ρ ) suchthat 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 ⊆ ηK (0) η − ⊆ 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 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 4. φ and φ are units in A . REDUCIBILITY PROBLEM FOR EVEN UNITARY GROUPS: THE DEPTH ZERO CASE 23
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 5.
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. As supp( φ )= P x P and supp( ψ )= P y P , so if z ∈ supp( φ ∗ ψ ) then ( φ ∗ ψ )( z ) = R G φ ( zr − ) ψ ( r ) dr = 0. So there exists r ∈ G such that φ ( zr − ) ψ ( r ) = 0. Because if φ ( zr − ) ψ ( r ) = 0 for r ∈ G then R G φ ( zr − ) ψ ( r ) = 0 = ⇒ ( φ ∗ ψ )( z ) = 0 which is a con-tradiction. So φ ( zr − ) ψ ( r ) = 0 for some r ∈ G . As φ ( zr − ) = 0 = ⇒ zr − ∈ supp( φ ) = P x P and ψ ( r ) = 0 = ⇒ r ∈ supp( ψ ) = P y P . Hence ( zr − )( r ) = z ∈ (supp( φ ))(supp( ψ )) =( P x P )( P y P ) = P x P y P . Hence supp( φ ∗ ψ )= supp( φψ ) ⊆ (supp( φ ))(supp( ψ ))= P x P y P . (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 5, 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 = ñ ̟ E ̟ − E ô . 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 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 18.
The unique algebra homomorphism C [ x, x − ] −→ B given by x −→ ψ is anisomorphism. So B ≃ C [ x, x − ] .Proof. It is obvious that the map is an algebra homomorphism and is surjective as { ψ n | n ∈ Z } spans B . Now we show that the kernel of map is 0. Suppose c k ψ k + · · · + c l ψ l = 0 with c k , . . . , c l ∈ C ; l > k > l, k ∈ Z . Let x ∈ supp ψ s = P ζ s P where 0 k s l . As doublecosets of a group are disjoint or equal, so ψ s ( x ) = 0 and ψ i ( x ) = 0 for 0 k i l, i = s .Hence c k ψ k ( x ) + · · · + c l ψ l ( x ) = 0 would imply that c s = 0. In a similar way we canshow that c k = c k +1 = . . . = c l = 0. So { ψ k , ψ k +1 , . . . , ψ l } is a linearly independent setwhen 0 k < l ; k, l ∈ Z . Now suppose if k < c k ψ k + · · · + c l ψ l = 0 with c k , . . . , c l ∈ C ; k, l ∈ Z . Let us assume without loss of generality that k < l . Multiplyingthroughout the above expression by ψ − k , we have c k + · · · + c l ψ l − k = 0. Now repeating theprevious argument we have c k = c k +1 = . . . = c l = 0. So again { ψ k , ψ k +1 , . . . , ψ l } is a linearlyindependent set when k < k < l ; k, l ∈ Z . Hence B ≃ C [ x, x − ]. (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 vector space 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 ) ∩ GL n ( O E ) = { g ∈ GL n ( O E ) | t gJ g = J } ,K (0) = { g ∈ ̟ E M n ( O E ) | t gJ g = J } , G = { g ∈ GL n ( 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 = ñ GL n ( k E ) M n ( k E )0 GL n ( k E ) ô 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 G . Here L = ( ñ a t a − ô | a ∈ GL n ( k E ) ) , U = ( ñ X ô | X ∈ M n ( k E ) , X + t X = 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 ( ñ ô ) = ñ ô ∈ 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 × n over finite field k E or F q . So G = SO n ( F q ). REDUCIBILITY PROBLEM FOR EVEN UNITARY GROUPS: THE DEPTH ZERO CASE 25
According to the Theorem 6.3 in [6], 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 .We know that w = ñ ̟ − E ̟ E ô , ̟ − E = − ̟ E . So w = ñ − ̟ − E ̟ E ô . Let η = ñ ̟ E ô . So, ηw η − = J ′ = ñ −
11 0 ô . Recall that P looks as follows: P = ñ GL n ( O E ) M n ( O E )M n ( p E ) GL n ( O E ) ô T G. Now η ñ GL n ( O E ) M n ( O E )M n ( p E ) GL n ( O E ) ô η − = ñ GL n ( O E ) M n ( p E )M n ( O E ) GL n ( O E ) ô ,ηGη − = G ′ = { g ∈ GL n ( E ) | t gJ ′ g = J ′ } . Hence η P η − = ñ GL n ( O E ) M n ( p E )M n ( O E ) GL n ( O E ) ô T G ′ . Therefore η P η − is the opposite of the Siegel Parahoric subgroup of G ′ . Let K ′ (0) = h P , w i . And let G ′ = { g ∈ GL n ( k E ) | t gJ ′ g = J ′ } = { g ∈ GL n ( 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 ′ = ñ −
11 0 ô . So P ′ = r ′ ( P ) =( η P η − ) modp E = ñ GL n ( k E ) 0M n ( k E ) GL n ( k E ) ô T G ′ . Clearly P ′ is the opposite of Siegel parabolicsubgroup of G ′ . Hence r ′ ( K (0)) = h P ′ , w ′ i = G ′ , as P ′ is a maximal subgroup of G ′ and w ′ does not lie in P ′ . So r ′ is a surjective homomorphism of groups.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. In fact, it is the symplectic group consisting of matrices of size2 n × n over finite field k E or F q . So G ′ = Sp n ( F q ).According to the Theorem 6.3 in [6], 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 beextended to G and viewed as an element in H ( G, ρ ) as P w P ⊆ K ′ (0) ⊆ G . Thus φ satisfiesthe following 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 , g ∈ G, 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 6. 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. REDUCIBILITY PROBLEM FOR EVEN UNITARY GROUPS: THE DEPTH ZERO CASE 27 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 .Now that we have calculated N L ( ρ ), we determine the structure of H ( L, ρ ). Using thecuspidality of ρ , it can be shown by A.1 Appendix [10] that I L ( ρ ) = P N L ( ρ ) P . As N L ( ρ ) = Z ( L ) P , so I L ( ρ ) = P Z ( L ) P P = Z ( L ) P . Let V be the vector space of ρ .The Hecke algebra H ( L, ρ ) consists of C -linear combinations of maps f : L −→ End C ( V ∨ )such that each map f is supported on P x P where x ∈ I L ( ρ ) = Z ( L ) P and f ( pxp ′ ) = ρ ∨ ( p ) f ( x ) ρ ∨ ( p ′ ) for p, p ′ ∈ P . It is clear that Z ( L ) P = a n ∈ Z P ζ n . So the Hecke algebra H ( L, ρ ) consists of C -linear combinations of maps f : L −→ End C ( V ∨ )such that each map f is supported on P x P where x ∈ P ζ n with n ∈ Z and f ( pxp ′ ) = ρ ∨ ( p ) f ( x ) ρ ∨ ( p ′ ) for p, p ′ ∈ P .Let φ , φ ∈ H ( L, ρ ) with supp( φ ) = P z and supp( φ ) = P z respectively with z , z ∈ Z ( L ). As ρ is an irreducible cuspidal representation of P . So if f ∈ H ( L, ρ ) withsupp( f ) = P z where z ∈ Z ( L ) then from Schur’s lemma f ( z ) = c V ∨ for some c ∈ C × .Hence φ ( z ) = c V ∨ and φ ( z ) = c V ∨ where c , c ∈ C × .We have supp( φ φ ) ⊆ (supp( φ ))(supp( φ )) = P z P z = P z z . The proof goes inthe similar lines as lemma 5.We assume without loss of generality that vol P = vol P − = vol P + = 1. Thus we havevol P = 1. Lemma 7.
Let φ , φ ∈ H ( L, ρ ) with supp ( φ ) = P z and supp ( φ ) = P z where z , z ∈ Z ( L ) . Also let φ ( z ) = c V ∨ and φ ( z ) = c V ∨ where c , c ∈ C × . Then ( φ ∗ φ )( z z ) = φ ( z ) φ ( z ) = c c V ∨ .Proof. ( φ ∗ φ )( z z ) = Z L φ ( z z y − ) φ ( y ) dy = Z P φ ( z z z − p − ) φ ( z p ) dy = Z P φ ( z p − ) φ ( pz ) dy = Z P φ ( z ) ρ ∨ ( p − ) ρ ∨ ( p ) φ ( z ) dy = Z P φ ( z ) φ ( z ) dy = Z P c c V ∨ dy = c c Vol( P )1 V ∨ = c c V ∨ = φ ( z ) φ ( z ) . (cid:3) As supp( φ ∗ φ ) = supp( φ φ ) ⊆ P z z , so supp( φ ∗ φ ) = ∅ or P z z . If supp( φ ∗ φ ) = ∅ then it means that ( φ ∗ φ ) = 0. This contradicts ( φ ∗ φ )( z z ) = c c = 0. Sosupp( φ ∗ φ ) = P z z .This implies that φ is invertible and φ − be it’s inverse. Thus supp( φ − ) = P z − and φ − ( z − ) = c − V ∨ .Define α ∈ H ( L, ρ ) by supp( α ) = P ζ and α ( ζ ) = 1 V ∨ . Proposition 19. (1) α n ( ζ n ) = ( α ( ζ )) n for n ∈ Z . (2) supp ( α n ) = P ζ n P = P ζ n = ζ n P for n ∈ Z . Proof.
Using lemma 7 over and over we get, 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 (cid:3) We know that H ( L, ρ ) consists of C -linear combinations of maps f : L −→ End C ( V ∨ )such that each map f is supported on P x P where x ∈ P ζ n with n ∈ Z and f ( pxp ′ ) = ρ ∨ ( p ) f ( x ) ρ ∨ ( p ′ ) for p, p ′ ∈ P . So from Propn. 19, H ( L, ρ ) is generated as a C -algebra by α and α − . Hence H ( L, ρ ) = C [ α, α − ]. Proposition 20.
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.8. Calculation of simple H ( L, ρ ) -modules The following Propn. is taken from Propn. 3.11 in [1].
Proposition 21. If A is a commutative ring with identity and S is a multiplicative closedsubset of A. If A is a principal ideal domain then S − A is also a principal ideal domain. Andalso if I is an ideal in S − A then there exists an ideal J in A such that I = J S − A . Lemma 8. C [ x, x − ] is a principal ideal domain.Proof. Let A = C [ x ] and S = { x n | n ∈ N ∪ { }} in Propn. 21. Clearly, S is a multiplicativeclosed subset of A and A is a principal ideal domain. Now we have S − A = C [ x, x − ]. Hence C [ x, x − ] is a principal ideal domain. (cid:3) Lemma 9.
Any maximal ideal in C [ x, x − ] is of the form ( x − λ ) C [ x, x − ] where λ ∈ C × .Proof. Suppose I be a proper ideal in C [ x, x − ]. From Propn. 21, we know that I is of the form J C [ x, x − ] where J is an ideal in C [ x ]. As C [ x ] is a principal ideal domain so J = p ( x ) C [ x ]for some p ( x ) ∈ C [ x ] and deg p ( x ) >
0. Let λ ∈ C be a root of p ( x ). So ( x − λ ) | p ( x ).This would imply p ( x ) C [ x ] ⊆ ( x − λ ) C [ x ]. Hence I = p ( x ) C [ x, x − ] ⊆ ( x − λ ) C [ x, x − ]. REDUCIBILITY PROBLEM FOR EVEN UNITARY GROUPS: THE DEPTH ZERO CASE 29
But I is a maximal ideal in C [ x, x − ]. So I = ( x − λ ) C [ x, x − ]. So any maximal ideal in C [ x, x − ] is of the form ( x − λ ) C [ x, x − ] where λ ∈ C . But if λ = 0 then ( x − λ ) = x and( x − λ ) C [ x, x − ] = x C [ x, x − ] = C [ x, x − ] which is not a maximal ideal. So λ ∈ C × . (cid:3) The following Propn. is taken from exercise problem 9 on page 356 in [4].
Proposition 22.
Let R be a commutative ring with identity. An R -module M is simple ⇐⇒ M ∼ = R/I for some maximal ideal I in R . From Propn. 22, every simple C [ x, x − ]-module is isomorphic to C [ x, x − ]-module C [ x,x − ]( x − λ ) C [ x,x − ] for some λ ∈ C × .The following Propn. is taken from Propn. 3.11 in [1]. Proposition 23. A is a commutative ring with identity and S is a multiplicative closed subsetof A . Let J be an ideal in A . Then we have S − AJS − A ∼ = AJ as S − A -modules. Let A = C [ x ] and S = { x n | n ∈ N ∪ { }} in the Propn. 23. So we have S − A = C [ x, x − ].Then Propn. 23 says that C [ x,x − ]( x − λ ) C [ x,x − ] ∼ = C [ x ]( x − λ ) C [ x ] as C [ x, x − ]-modules, where λ ∈ C × . Proposition 24. C [ x ]( x − λ ) C [ x ] ∼ = C λ as C [ x ] -modules, where λ ∈ C × and C λ is the ring C with C [ x ] -module structure given by x.z = λz for z ∈ C λ .Proof. The C [ x ]-module structure of C [ x ]( x − λ ) C [ x ] is given by p ( x ) .q ( x ) = p ( λ ) q ( x ) where p ( x ) , q ( x ) ∈ C [ x ]. The map φ : C [ x ]( x − λ ) C [ x ] −→ C λ is defined as φ ( p ( x )) = p ( λ ) for p ( x ) ∈ C [ x ]. We shall now check that φ is a C [ x ]-modulehomomorphism. Let p ( x ) , q ( x ) ∈ C [ x ]. Now let us consider φ ( p ( x ) + q ( x )) = φ (( p + q )( x ))= ( p + q )( λ )= p ( λ ) + q ( λ )= φ ( p ( x )) + φ ( q ( x )) . Now let us look at φ ( p ( x ) .q ( x )) = φ ( p ( λ ) q ( x ))= φ ( p ( λ ) q ( x ))= p ( λ ) q ( λ )= p ( λ ) φ ( q ( x ))= p ( x ) .φ ( q ( x )) . So φ is a homomorphism of C [ x ]-modules. Let z ∈ C , then there exists a polynomial p ( x ) ∈ C [ x ] such that p ( λ ) = z . Hence φ ( p ( x )) = p ( λ ) = z . So φ is surjective map.Suppose if φ ( p ( x )) = φ ( q ( x )) where p ( x ) , q ( x ) ∈ C [ x ] then p ( λ ) = q ( λ ). This implies that( p − q )( λ ) = 0 = ⇒ ( x − λ ) | ( p − q )( x ) = ⇒ ( x − λ ) | ( p ( x ) − q ( x )) = ⇒ p ( x ) = q ( x ). So φ is one-one map. Hence φ is an isomorphism of C [ x ]-modules. Hence the module structure of ring C [ x ] over C [ x ]( x − λ ) is preserved for C λ . Therefore the C [ x ]-module structure of C λ is givenby x.z = λz where z ∈ C λ . (cid:3) So from Propn. 24, we have C [ x ]( x − λ ) ∼ = C λ as C [ x ]-modules for λ ∈ C × . This means that C [ x ]( x − λ ) ∼ = C λ as C [ x, x − ]-modules for λ ∈ C × . Recall that C [ x,x − ]( x − λ ) C [ x,x − ] ∼ = C [ x ]( x − λ ) C [ x ] as C [ x, x − ]-modules for λ ∈ C × . Therefore C [ x,x − ]( x − λ ) C [ x,x − ] ∼ = C λ as C [ x, x − ]-modules for λ ∈ C × with the C [ x, x − ]-module structure on C λ given by x.z = λz where z ∈ C λ .As H ( L, ρ ) = C [ α, α − ], so the simple H ( L, ρ )-modules are same as the simple C [ α, α − ]-modules. We have shown before that C [ α, α − ] ∼ = C [ x, x − ] as algebras. So the distinct simple H ( L, ρ )-modules(up to isomorphism) are the various C λ for λ ∈ C × . The module structureis 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 = ( ñ X ô | X ∈ M n ( E ) , X + t X = 0 ) , Lie U = ( ñ X ô | X ∈ M n ( E ) , X + t X = 0 ) . Unramified case:
Recall ζ = ñ ̟ E ̟ − E ô in the unramified case. So( Ad ζ ) ñ X ô = ζ ñ X ô ζ − = ñ ̟ E X ô . Hence δ P ( ζ ) = k det ( Ad ζ ) | Lie U k F = k ̟ dim F (Lie U )) E k F = k ̟ n E k F = k ̟ n F k F = q − n . Ramified case:
Recall ζ = ñ ̟ E − ̟ − E ô in the ramified case. So( Ad ζ ) ñ X ô = ζ ñ X ô ζ − = ñ − ̟ E X ô . Hence δ P ( ζ ) = k det ( Ad ζ ) | Lie U k F = k − ̟ dim F (Lie U )) E k F REDUCIBILITY PROBLEM FOR EVEN UNITARY GROUPS: THE DEPTH ZERO CASE 31 = k ̟ n E k F = k ̟ n F k F = q − n . Understanding the map T P . Let us denote the set of strongly ( P , P )-positive elementsby I + . Thus I + = { x ∈ L | x P + x − ⊆ P + , x − P − x ⊆ P − } . where P + = P ∩ U, P − = P ∩ U . We have H + ( L, ρ ) = { f ∈ H ( L, ρ ) | supp f ⊆ P I + P } . Note ζ ∈ I + , so H + ( L, ρ ) = C [ α ]. The following discussion is taken from pages 612-619in [3]. Let W be space of ρ . Let f ∈ H + ( L, ρ ) with support of f being P x P for x ∈ I + .The map F ∈ H ( G, ρ ) is supported on P x P and f ( x ) = F ( x ). The algebra embedding T + : H + ( L, ρ ) −→ H ( G, ρ )is given by T + ( f ) = F .Recall support of α ∈ H + ( L, ρ ) is P ζ . Let T + ( α ) = ψ , where ψ ∈ H ( G, ρ ) has support P ζ P and α ( ζ ) = ψ ( ζ ) = 1 W ∨ . As T + ( α ) = ψ is invertible, so from Propn. 3 we can concludethat T + extends to an embedding of algebras t : H ( L, ρ ) −→ H ( G, ρ ) . Let φ ∈ H ( L, ρ ) and m ∈ N is chosen such that α m φ ∈ H + ( L, ρ ). The map t is thengiven by t ( φ ) = ψ − m T + ( α m φ ). For φ ∈ H ( L, ρ ), the map t P : H ( L, ρ ) −→ H ( G, ρ )is given by t P ( φ ) = t ( φδ P ), where φδ P ∈ H ( L, ρ ) and is the map φδ P : L −→ End C ( ρ ∨ )given by ( φδ P )( l ) = φ ( l ) δ P ( l ) for l ∈ L . As α ∈ H ( L, ρ ) we have t P ( α )( ζ ) = t ( αδ P )( ζ )= T + ( αδ P )( ζ )= δ P ( ζ ) T + ( α )( ζ )= δ P ( ζ ) ψ ( ζ )= δ P ( ζ )1 W ∨ . Let H ( L, ρ )-Mod denote the category of H ( L, ρ )-modules and H ( G, ρ )-Mod denote thecategory of H ( G, ρ )-modules. The map t P induces a functor ( t P ) ∗ given by( t P ) ∗ : H ( L, ρ ) − Mod −→ H ( G, ρ ) − Mod . 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 ).Let τ ∈ R [ L,π ] L ( L ) then functor m L : R [ L,π ] L ( L ) −→ H ( L, ρ ) − M od is given by m L ( τ ) =Hom P ( ρ , τ ). The functor m L is an equivalence of categories. Let f ∈ m L ( τ ) , γ ∈ H ( L, ρ )and w ∈ W . The action of H ( L, ρ ) on m L ( τ ) is given by ( γ.f )( w ) = R L τ ( l ) f ( γ ∨ ( l − ) w ) dl .Here γ ∨ is defined on L by γ ∨ ( l − ) = γ ( l ) ∨ for l ∈ L . Let τ ′ ∈ R [ L,π ] G ( G ) then the functor m G : R [ L,π ] G ( G ) −→ H ( G, ρ ) − M od is given by m G ( τ ′ ) = Hom P ( ρ, τ ′ ). The functor m G isan equivalence of categories. From Corollary 8.4 in [3], the functors m L , m G , Ind GP , ( t P ) ∗ fitinto the following commutative diagram: R [ L,π ] G ( G ) m G −−−−→ H ( G, ρ ) − M od
Ind GP x ( t P ) ∗ x R [ L,π ] L ( L ) m L −−−−→ H ( L, ρ ) − M od If τ ∈ R [ L,π ] L ( L ) then from the above commutative diagram, we see that ( t P ) ∗ ( m L ( τ )) ∼ = m G ( Ind GP τ ) as H ( G, ρ )-modules. Replacing τ by ( τ ⊗ δ / P ) in the above expression, ( t P ) ∗ ( m L ( τ ⊗ δ / P )) ∼ = m G ( Ind GP ( τ ⊗ δ / P )) as H ( G, ρ )-modules. As
Ind GP ( τ ⊗ δ / P ) = ι GP ( τ ), we have( t P ) ∗ ( m L ( τ ⊗ δ / P )) ∼ = m G ( ι GP ( τ )) as H ( G, ρ )-modules.Our aim is to find an algebra embedding T P : H ( L, ρ ) −→ H ( G, ρ ) such that the followingdiagram 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
Let τ ∈ R [ L,π ] L ( L ) then m L ( τ ) ∈ H ( L, ρ )- Mod. The functor ( T P ) ∗ is defined as below:( T P ) ∗ ( m L ( τ )) = ß ψ : H ( G, ρ ) → m L ( τ ) (cid:12)(cid:12)(cid:12)(cid:12) hψ ( h ) = ψ ( T P ( h ) h ) where h ∈ H ( L, ρ ) , h ∈ H ( G, ρ ) ™ . From the above commutative diagram, we see that ( T P ) ∗ ( m L ( τ )) ∼ = m G ( ι GP ( τ )) as H ( G, ρ )-modules. Recall that ( t P ) ∗ ( m L ( τ ⊗ δ / P )) ∼ = m G ( ι GP ( τ )) as H ( G, ρ )-modules. Hence wehave to find an algebra embedding T P : H ( L, ρ ) −→ H ( G, ρ ) such that ( T P ) ∗ ( m L ( τ )) ∼ =( t P ) ∗ ( m L ( τ ⊗ δ / P )) as H ( G, ρ )-modules.
Proposition 25.
The map T P is given by T P ( φ ) = t P ( φδ − / P ) for φ ∈ H ( L, ρ ) so that wehave ( T P ) ∗ ( m L ( τ )) = ( t P ) ∗ ( m L ( τ ⊗ δ / P )) as H ( G, ρ ) - modules.Proof. Let W be space of ρ . The vector spaces for m L ( τ δ / P ) and m L ( τ ) are the same. Let f ∈ m L ( τ ) = Hom P ( ρ , τ ) , γ ∈ H ( L, ρ ) and w ∈ W . Recall the action of H ( L, ρ ) on m L ( τ )is given by ( γ.f )( w ) = Z L τ ( l ) f ( γ ∨ ( l − ) w ) dl. REDUCIBILITY PROBLEM FOR EVEN UNITARY GROUPS: THE DEPTH ZERO CASE 33
Let f ′ ∈ m L ( τ δ / P ) = Hom P ( ρ , τ δ / P ) , γ ∈ H ( L, ρ ) and w ∈ W . Recall the action of H ( L, ρ ) on m L ( τ δ / P ) is given by( γ.f ′ )( w ) = Z L ( τ δ / P )( l ) f ′ ( γ ∨ ( l − ) w ) dl = Z L τ ( l ) δ / P ( l ) f ′ ( γ ∨ ( l − ) w ) dl. Now f ′ is a linear transformation from space of ρ to space of τ δ / P . As δ / P ( l ) ∈ C × , so δ / P ( l ) f ′ ( γ ∨ ( l − ) w ) = f ′ ( δ / P ( l ) γ ∨ ( l − ) w ). Hence we have( γ.f ′ )( w ) = Z L τ ( l ) f ′ ( δ / P ( l ) γ ∨ ( l − ) w ) dl = Z L τ ( l ) f ′ ( δ / P ( l ) γ ( l ) ∨ w ) dl. Further as δ / P ( l ) ∈ C × , so δ / P ( l )( γ ( l )) ∨ = ( δ / P γ )( l ) ∨ . Therefore( γ.f ′ )( w ) = Z L τ ( l ) f ′ (( δ / P γ )( l ) ∨ w ) dl = ( δ / P γ ) .f ′ ( w ) . Hence we can conclude that the action of γ ∈ H ( L, ρ ) on f ′ ∈ m L ( τ δ / P ) is same as theaction of δ / P γ ∈ H ( L, ρ ) on f ′ ∈ m L ( τ ). So we have ( T P ) ∗ ( m L ( τ )) = ( t P ) ∗ ( m L ( τ ⊗ δ / P ))as H ( G, ρ )- modules. (cid:3)
From Propn. 25, T P ( α ) = t P ( αδ − / P ). So we have T P ( α ) = t P ( αδ − / P )= t ( αδ − / P δ P )= t ( αδ / P )= T + ( αδ / P ) . Hence T P ( α )( ζ ) = T + ( αδ / P )( ζ )= δ / P ( ζ ) T + ( α )( ζ )= δ / P ( ζ ) α ( ζ )= δ / P ( ζ )1 W ∨ . Thus T P ( α )( ζ ) = δ / P ( ζ )1 W ∨ with supp( T P ( α )) = supp( t P ( α )) = P ζ P .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.From now on, we assume without loss of generality that vol P = vol P − = vol P + = 1. Thuswe have vol P = 1. Lemma 10. supp ( φ ∗ φ ) = P ζ P = P w w P .Proof. We first claim that supp( φ ∗ φ ) ⊆ P w P w P . Suppose z ∈ supp( φ ∗ φ ) then( φ ∗ φ )( z ) = R G φ ( zr − ) φ ( r ) dr = 0. This would imply that there exists an r ∈ G suchthat φ ( zr − ) φ ( r ) = 0. As φ ( zr − ) φ ( r ) = 0, this means that φ ( zr − ) = 0 , φ ( r ) = 0. But φ ( zr − ) = 0 would imply that zr − ∈ P w P and φ ( r ) = 0 would imply that r ∈ P w P . So z = ( zr − )( r ) ∈ ( P w P )( P w P ) = (supp φ )(supp φ ) = P w P w P . Hence supp( φ ∗ φ ) ⊆ P w P w P . Let us recall P , P + , P − . P = ( ñ a t a − ô | a ∈ GL n ( O E ) ) , P + = ( ñ X ô | X ∈ M n ( O E ) , X + t X = 0 ) , P − = ( ñ X ô | X ∈ ̟ E M n ( O E ) , X + t X = 0 ) . It is easy observe that w P − w − ⊆ P + , w P w − = P , w − P + w ⊆ P − .Now we have P w P w P = P w P − P P + w P = P w P − w − w P w − w w w − P + w P ⊆ PP + P w w P − P = P w w P = P ζ P . So P w P w P ⊆ P w w P = P ζ P . On the contrary, as 1 ∈ P , so P ζ P = P w w P ⊆ P w P w P . Hence we have P w P w P = P w w P = P ζ P . Therefore supp( φ ∗ φ ) ⊆ P w P w P = P w w P = P ζ P . This implies supp( φ ∗ φ ) = ∅ or P ζ P . But if supp( φ ∗ φ ) = ∅ then ( φ ∗ φ ) = 0 which is a contradiction. Thus supp( φ ∗ φ ) = P ζ P . (cid:3) For r ∈ Z let K − ,r = ( ñ X ô | X ∈ M n ( p rE ) , X + t X = 0 ) , K + ,r = ( ñ X ô | X ∈ M n ( p rE ) , X + t X = 0 ) . Proposition 26. ( φ ∗ φ )( ζ ) = φ ( w ) φ ( w ) . Proof.
From Lemma 10, supp( φ ∗ φ ) = P ζ P = P w w P . So now let us consider( φ ∗ φ )( ζ ) = ( φ ∗ φ )( w w )= Z G φ ( y ) φ ( y − ζ ) dy = Z P w P φ ( y ) φ ( y − ζ ) dy. We know that P w P = ∐ z ∈ P w P / P z P . Let y = zp ∈ z P . So we have φ ( y ) φ ( y − ζ ) = φ ( zp ) φ ( p − z − ζ )= φ ( z ) ρ ∨ ( p ) ρ ∨ ( p − ) φ ( z − ζ )= φ ( z ) φ ( z − ζ ) . REDUCIBILITY PROBLEM FOR EVEN UNITARY GROUPS: THE DEPTH ZERO CASE 35
Hence ( φ ∗ φ )( ζ ) = X z ∈ P w P / P φ ( z ) φ ( z − ζ )Vol P = X z ∈ P w P / P φ ( z ) φ ( z − ζ )Let α : P /w P w − ∩ P −→ P w P / P be the map given by α ( x ( w P w − ∩ P )) = xw P where x ∈ P . We can observe that the map α is bijective. So P /w P w − ∩ P is in bijectionwith P w P / P .Hence ( φ ∗ φ )( ζ ) = X x ∈ P /w P w − ∩ P φ ( xw ) φ ( w − x − ζ ) . From Iwahori factorization of P we have P = P − P P + = K − , P K + , . Therefore w P w − = w P = w K w − , P w K + , = K + , P K − , . So P ∩ w P w − = P ∩ w P = K + , P K − , . Let β : P /w P w − ∩ P −→ K + , /K + , be the map given by β ( x ( P ∩ w P )) = x + K + , where x ∈ P and x = x + px − , x + ∈ P + , p ∈ P , x − ∈ P − . We can observe that the map β isbijective. So P /w P w − ∩ P is in bijection with K + , /K + , .Therefore ( φ ∗ φ )( ζ ) = X x + ∈ K + , /K + , φ ( x + w ) φ ( w − x − ζ )= X x + ∈ K + , /K + , ρ ∨ ( x + ) φ ( w ) φ ( w − x − ζ ) . As ρ ∨ is trivial on P + and x + ∈ P + so we have( φ ∗ φ )( ζ ) = P x + ∈ K + , /K + , φ ( w ) φ ( w − x − ζ ) . The terms in above summation which do not vanish are the ones for which w − x − ζ ∈ P w P = ⇒ x − ∈ w P w P ζ − = ⇒ x + ∈ ζ P w − P w − = ⇒ w − x + w ∈ w P w − P . It isclear w P w − P = ( w P )( P ). As w P = w K w − , P w K + , = K − , P K + , − , so w P w − P =( w P )( P ) = K − , P K + , − P K − , . Hence we have w − x + w ∈ K − , P K + , − P K − , = ⇒ w − x + w = k − p k + k ′− where k − ∈ K − , , k + ∈ K + , − , k ′ − ∈ K − , , p ∈ P . Hence wehave p k + = k − − w − x + w k ′ − − . Now as w − x + w ∈ K − , , k − − ∈ K − , , k ′ − − ∈ K − , , so k − − w − x + w k ′ − − ∈ K − , and p k + ∈ P K + , − . But we know that K − , ∩ P K + , − = 1 = ⇒ p k + = 1 = ⇒ w − x + w = k − k ′− ∈ K − , = ⇒ x + ∈ w K − , w − = K + , . As x + ∈ K + , , soonly the trivial coset contributes to the above summation. Hence( φ ∗ φ )( ζ ) = φ ( w ) φ ( w − ζ ) = φ ( w ) φ ( w ) . (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 27. g g = T P ( α ) . Proof.
Let us choose ψ i ∈ H ( G, ρ ) for i = 0 , ψ i ) = P w i P for i = 0 ,
1. So φ i is a scalar multiple of ψ i for i = 0 ,
1. Hence φ i = λ i ψ i where λ i ∈ C × for i = 0 ,
1. Let ψ i ( w i ) = A ∈ Hom P ∩ wi P ( w i ρ ∨ , ρ ∨ ) for i = 0 , W be the space of ρ . So A = 1 W ∨ .From Propn. 26, we have ( ψ ∗ ψ )( ζ ) = ψ ( w ) ψ ( w ) = A = 1 W ∨ . Now let ψ i satisfies thequadratic relation given by ψ i = aψ i + b where a, b ∈ R for i = 0 ,
1. As ψ i = aψ i + b = ⇒ ( − ψ i ) = ( − a )( − ψ i ) + b , so a can be arranged such that a >
0. We can see that 1 ∈ H ( G, ρ )is defined as below: 1( x ) = ( , if x / ∈ P ; ρ ∨ ( x ) if x ∈ P .Let us consider ψ i (1) = R G ψ i ( y ) ψ i ( y − ) dy for i = 0 ,
1. Now let y = pw i p ′ where p, p ′ ∈ P for i = 0 ,
1. So we have ψ i (1) = Z P w i P ψ i ( pw i p ′ ) ψ i ( p ′ − w − i p − ) d ( pw i p ′ )= Z P w i P ρ ∨ ( p ) ψ i ( w i ) ρ ∨ ( p ′ ) ρ ∨ ( p ′ − ) ψ i ( w − i ) ρ ∨ ( p − ) d ( pw i p ′ )= Z P w i P ρ ∨ ( p ) ψ i ( w i ) ψ i ( w − i ) ρ ∨ ( p − ) d ( pw i p ′ )= Z P w i P ρ ∨ ( p ) ψ i ( w i ) ψ i ( w i ) ρ ∨ ( p − ) d ( pw i p ′ )= Z P w i P ρ ∨ ( p ) A ρ ∨ ( p − ) d ( pw i p ′ )= Z P w i P A ρ ∨ ( p ) ρ ∨ ( p − ) d ( pw i p ′ )= A vol( P w i P )= 1 W ∨ vol( P w i P ) . So ψ i (1) = 1 W ∨ vol( P w i P ) for i = 0 ,
1. We already know that ψ i = aψ i + b where a, b ∈ R and for i = 0 ,
1. Now evaluating the expression ψ i = aψ i + b at 1, we have ψ i (1) = aψ i (1) + b ψ i (1) = 0 as support of ψ i is P w i P for i = 0 ,
1. Wehave seen before that ψ i (1) = 1 W ∨ vol( P w i P ) for i = 0 , ∈ P , ρ ∨ (1) = 1 W ∨ .So ψ i (1) = aψ i (1) + b ⇒ W ∨ vol( P w i P ) = 1 W ∨ b for i = 0 ,
1. Comparing coefficients of1 W ∨ on both sides of the equation 1 W ∨ vol( P w i P ) = 1 W ∨ b for i = 0 , b = vol( P w i P ).As φ i = λ i ψ i for i = 0 ,
1, hence φ i = λ i ψ i = λ i ( aψ i + b ) = ( λ i a )( λ i ψ i )+ λ i b = ( λ i a ) φ i + λ i b for i = 0 ,
1. But φ i = ( q n − φ i + q n for i = 0 ,
1. So φ i = ( λ i a ) φ i + λ i b = ( q n − φ i + q n for i = 0 ,
1. As φ i and 1 are linearly independent, hence λ i a = ( q n −
1) for i = 0 ,
1. Therefore λ i = q n − a for i = 0 ,
1. As a > , a ∈ R , so λ i > , λ i ∈ R for i = 0 ,
1. Similarly, as φ i and 1are linearly independent, hence λ i b = q n = ⇒ λ i = q n b for i = 0 , P w i P = ∐ x ∈ P / P ∩ wi P xw i P = ⇒ vol( P w i P ) = [ P w i P : P ]vol P = [ P w i P : P ] =[ P : P ∩ w i P ] for i = 0 ,
1. Hence b = vol( P w i P ) = [ P : P ∩ w i P ] for i = 0 ,
1. Now as λ = λ = q n b = ⇒ λ = λ = q n/ b / = q n/ [ P : P ∩ w P ] / . Therefore REDUCIBILITY PROBLEM FOR EVEN UNITARY GROUPS: THE DEPTH ZERO CASE 37 φ φ = ( λ ψ )( λ ψ )= λ ψ ψ = q n ψ ψ [ P : P ∩ w P ] . We have seen before that, P = K − , P K + , and P ∩ w P = K − , P K + , . So[ P : P ∩ w P ] = | K + , K + , | = |{ X ∈ M n ( k E ) | X + t X = 0 }| = ( q n )( q ) ( n )( n − = ( q n )( q n − n )= q n . Hence ( φ φ )( ζ ) = q n ( ψ ψ )( ζ )[ P : P ∩ w P ]= q n ( ψ ψ )( ζ ) q n = q n − n W ∨ . Recall g i = q − n/ φ i for i = 0 ,
1. We know that φ i = ( q n − φ i + q n for i = 0 ,
1. So for i = 0 , g i = q − n φ i = q − n (( q n − φ i + q n )= (1 − q − n ) φ i + 1= (1 − q − n ) q n/ g i + 1= ( q n/ − q − n/ ) g i + 1 . So g g = ( q − n/ φ )( q − n/ φ ) = q − n φ φ = ⇒ ( g g )( ζ ) = q − n ( φ φ )( ζ ) = q − n q n − n W ∨ = q − n W ∨ . From the earlier discussion in this section we have T P ( α )( ζ ) = δ / P ( ζ )1 W ∨ . Fromsection 9.1, we have δ P ( ζ ) = q − n . Hence δ / P ( ζ ) = q − n . Therefore ( g g )( ζ ) = T P ( α )( ζ ).So ( g g )( ζ ) = T P ( α )( ζ ). We have supp( T P ( α )) = P ζ P . As supp( g i ) = P w i P , Lemma 10gives supp( g g ) = P ζ P . Therefore g g = T P ( α ). (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 28. g g = T P ( α ) . Proof.
Let us choose ψ i ∈ H ( G, ρ ) for i = 0 , ψ i ) = P w i P for i = 0 ,
1. So φ i is a scalar multiple of ψ i for i = 0 ,
1. Hence φ i = λ i ψ i where λ i ∈ C × for i = 0 ,
1. Let ψ i ( w i ) = A i ∈ Hom P ∩ wi P ( w i ρ ∨ , ρ ∨ ) for i = 0 , W be the space of ρ . So A i = 1 W ∨ for i = 0 ,
1. From section 5.1 on page 24 in [6], we can say that A = A . From Propn. 26, wehave ( ψ ∗ ψ )( ζ ) = ψ ( w ) ψ ( w ) = A A = A = 1 W ∨ . Now let ψ i satisfies the quadraticrelation given by ψ i = a i ψ i + b i where a i , b i ∈ R for i = 0 ,
1. As ψ i = a i ψ i + b i = ⇒ ( − ψ i ) =( − a i )( − ψ i ) + b i , so a i can be arranged such that a i > i = 0 ,
1. We can see that1 ∈ H ( G, ρ ) is defined as below: 1( x ) = ( , if x / ∈ P ; ρ ∨ ( x ) if x ∈ P .Let us consider ψ (1) = R G ψ ( y ) ψ ( y − ) dy . Now let y = pw p ′ where p, p ′ ∈ P . So wehave ψ (1) = Z P w P ψ ( pw p ′ ) ψ ( p ′ − w − p − ) d ( pw p ′ )= Z P w P ρ ∨ ( p ) ψ ( w ) ρ ∨ ( p ′ ) ρ ∨ ( p ′ − ) ψ ( w − ) ρ ∨ ( p − ) d ( pw p ′ )= Z P w P ρ ∨ ( p ) ψ ( w ) ψ ( w − ) ρ ∨ ( p − ) d ( pw p ′ )= Z P w P ρ ∨ ( p ) ψ ( w ) ψ ( w ) ρ ∨ ( p − ) d ( pw p ′ )= Z P w P ρ ∨ ( p ) A ρ ∨ ( p − ) d ( pw p ′ )= Z P w P A ρ ∨ ( p ) ρ ∨ ( p − ) d ( pw p ′ )= A vol( P w P )= 1 W ∨ vol( P w P ) . So ψ (1) = 1 W ∨ vol( P w P ). We already know that ψ = a ψ + b where a , b ∈ R . Nowevaluating the expression ψ = a ψ + b at 1, we have ψ (1) = a ψ (1) + b ψ (1) = 0 as support of ψ is P w P . We have seen before that ψ (1) = 1 W ∨ vol( P w P )and as 1 ∈ P , ρ ∨ (1) = 1 W ∨ . So ψ (1) = a ψ i (1)+ b ⇒ W ∨ vol( P w P ) = 1 W ∨ b .Comparing coefficients of 1 W ∨ on both sides of the equation 1 W ∨ b = 1 W ∨ vol( P w P ) we get b = vol( P w P ).As φ = λ ψ , hence φ = λ ψ = λ ( a ψ + b ) = ( λ a )( λ ψ ) + λ b = ( λ a ) φ + λ b .But φ = ( q n/ − φ + q n/ . So φ = ( λ a ) φ + λ b = ( q n/ − φ + q n/ . As φ and 1 arelinearly independent, hence λ a = ( q n/ − λ = q n/ − a . As a > , a ∈ R , so λ > , λ ∈ R . Similarly, as φ and 1 are linearly independent, hence λ b = q n/ = ⇒ λ = q n/ b . REDUCIBILITY PROBLEM FOR EVEN UNITARY GROUPS: THE DEPTH ZERO CASE 39
Now P w P = ∐ x ∈ P / P ∩ w P xw P = ⇒ vol( P w P ) = [ P w P : P ]vol P = [ P w P : P ] = [ P : P ∩ w P ]. Hence b = vol( P w P ) = [ P : P ∩ w P ]. Now as λ = q n/ b = ⇒ λ = q n/ b / = q n/ [ P : P ∩ w P ] / .We have seen before that, P = K − , P K + , and P ∩ w P = K − , P K + , . So[ P : P ∩ w P ] = | K + , K + , | = |{ X ∈ M n ( k E ) | X + t X = 0 }| = q ( n )( n − = q n − n . So λ = q n/ [ P : P ∩ w P ] / = q n/ q n − n . Let us consider ψ (1) = R G ψ ( y ) ψ ( y − ) dy . Now let y = pw p ′ where p, p ′ ∈ P . So wehave ψ (1) = Z P w P ψ ( pw p ′ ) ψ ( p ′ − w − p − ) d ( pw p ′ )= Z P w P ρ ∨ ( p ) ψ ( w ) ρ ∨ ( p ′ ) ρ ∨ ( p ′ − ) ψ ( w − ) ρ ∨ ( p − ) d ( pw p ′ )= Z P w P ρ ∨ ( p ) ψ ( w ) ψ ( w − ) ρ ∨ ( p − ) d ( pw p ′ )= Z P w P ρ ∨ ( p ) ψ ( w ) ψ ( − w ) ρ ∨ ( p − ) d ( pw p ′ )= Z P w P ρ ∨ ( p ) ψ ( w ) ρ ∨ ( − ψ ( w ) ρ ∨ ( p − ) d ( pw p ′ )= ρ ∨ ( − Z P w P A ρ ∨ ( p ) ρ ∨ ( p − ) d ( pw p ′ )= ρ ∨ ( − A vol( P w P )= ρ ∨ ( − W ∨ vol( P w P ) . So ψ (1) = 1 W ∨ vol( P w P ). We already know that ψ = a ψ + b where a , b ∈ R .Now evaluating the expression ψ = a ψ + b at 1, we have ψ (1) = a ψ (1) + b ψ (1) = 0 as support of ψ is P w P . We have seen before that ψ (1) =1 W ∨ vol( P w P ) and as 1 ∈ P , ρ ∨ (1) = 1 W ∨ . So ψ (1) = a ψ i (1) + b ⇒ ρ ∨ ( − W ∨ vol( P w P ) = 1 W ∨ b . Comparing coefficients of 1 W ∨ on both sides of the equation1 W ∨ b = 1 W ∨ ρ ∨ ( − P w P ) we get b = ρ ∨ ( − P w P ).As φ = λ ψ , hence φ = λ ψ = λ ( a ψ + b ) = ( λ a )( λ ψ ) + λ b = ( λ a ) φ + λ b .But φ = ( q n/ − φ + q n/ . So φ = ( λ a ) φ + λ b = ( q n/ − φ + q n/ . As φ and 1 are linearly independent, hence λ a = ( q n/ − λ = q n/ − a . As a > , a ∈ R , so λ > , λ ∈ R . Similarly, as φ and 1 are linearly independent, hence λ b = q n/ = ⇒ λ = q n/ b .Now P w P = ∐ x ∈ P / P ∩ w P xw P = ⇒ vol( P w P ) = [ P w P : P ]vol P = [ P w P : P ] = [ P : P ∩ w P ]. Hence b = vol( P w P ) = [ P : P ∩ w P ]. Now as λ = q n/ b = ⇒ λ = q n/ b / = q n/ [ P : P ∩ w P ] / .We have seen before that P = K − , P K + , , w P = K − , P K + , − . So P ∩ w P = K − , P K + , . Hence [ P : P ∩ w P ] = | K − , K − , | = |{ X ∈ M n ( k E ) | X = t X }| = q ( n )( n +1)2 = q n n . So λ = q n/ [ P : P ∩ w P ] / = q n/ q n n ( ρ ( − / . Hence ( φ φ )( ζ ) = ( λ ψ )( λ ψ )( ζ )= ( λ λ )( ψ ψ )( ζ )= q n/ q n − n q n/ q n n ( ρ ( − / ∨ W = q n − n ∨ W ( ρ ( − / = q n − n ∨ W ( ρ ( − / . As − ∈ Z ( P ) and ρ ∨ is a representation of P , so ρ ∨ ( −
1) = ω ρ ∨ ( −
1) where ω ρ ∨ is thecentral character of P . Now 1 = ω ρ ∨ (1) = ( ω ρ ∨ ( − , so ρ ∨ ( −
1) = ω ρ ∨ ( −
1) = ±
1. Wehave seen before that λ = q n/ − a and a ∈ R , a >
0, so λ >
0. But we know that λ = q n/ [ P : P ∩ w P ] / = q n/ q n n ( ρ ( − / , hence ρ ∨ ( −
1) = 1.Recall g i = q − n/ φ i for i = 0 ,
1. We know that φ i = ( q n/ − φ i + q n/ for i = 0 ,
1. So for i = 0 , g i = q − n/ φ i = q − n/ (( q n/ − φ i + q n/ ) REDUCIBILITY PROBLEM FOR EVEN UNITARY GROUPS: THE DEPTH ZERO CASE 41 = (1 − q − n/ ) φ i + 1= (1 − q − n/ ) q n/ g i + 1= ( q n/ − q − n/ ) g i + 1 . So g g = q − n/ φ φ = ⇒ ( g g )( ζ ) = q − n/ ( φ φ )( ζ ) = q − n/ q n − n ∨ W ( ρ ( − / = q − n ∨ W . Fromthe earlier discussion in this section we have T P ( α )( ζ ) = δ / P ( ζ )1 W ∨ . From section 9.1, wehave δ P ( ζ ) = q − n . Hence δ / P ( ζ ) = q − n / . Therefore ( g g )( ζ ) = T P ( α )( ζ ). So ( g g )( ζ ) = T P ( α )( ζ ). We have supp( T P ( α )) = P ζ P . As supp( g i ) = P w i P , Lemma 10 gives supp( g g ) = P ζ P . Therefore g g = T P ( α ). (cid:3) Calculation of m L ( πν ) . Note πν lies in R [ L,π ] L ( L ). Recall m L is an equivalence ofcategories. As πν is an irreducible representation of L , it follows that m L ( πν ) is a simple H ( L, ρ )-module. In this section, we identify the simple H ( L, ρ )-module corresponding to m L ( πν ). Calculating m L ( πν ) will be useful in answering the question in next section.From section 2.5, we know that π = c - Ind L f P f ρ , where › P = h ζ i P , f ρ ( ζ k j ) = ρ ( j ) for j ∈ P , k ∈ Z . Let us recall that ν is unramified character of L from section 1. Let V bespace of πν and W be space of ρ . Recall m L ( πν ) = Hom P ( ρ , πν ). Let f ∈ Hom P ( ρ , πν ).As P is a compact open subgroup of L and ν is an unramified character of L , so ν ( j ) = 1 for j ∈ P . We already know that α ∈ H ( L, ρ ) with support of α being P ζ and α ( ζ ) = 1 W ∨ .Let w ∈ W and we have seen in section 9.2 that the way H ( L, ρ ) acts on Hom P ( ρ , πν ) isgiven by: ( α.f )( w ) = Z L ( πν )( l ) f ( α ∨ ( l − ) w ) dl = Z L ( πν )( l ) f (( α ( l )) ∨ w ) dl = Z P ( πν )( pζ ) f (( α ( pζ )) ∨ w ) dp = Z P ( πν )( pζ ) f (( ρ ∨ ( p ) α ( ζ )) ∨ w ) dp = Z P ( πν )( pζ ) f (( ρ ∨ ( p )1 W ∨ ) ∨ w ) dp = Z P ( πν )( pζ ) f (( ρ ∨ ( p )) ∨ w ) dp = Z P π ( pζ ) ν ( pζ ) f (( ρ ∨ ( p )) ∨ w ) dp = Z P π ( pζ ) ν ( ζ ) f (( ρ ∨ ( p )) ∨ w ) dp. Now h , i : W × W ∨ −→ C is given by: h w, ρ ∨ ( p ) w ∨ i = h ρ ( p − ) w, w ∨ i for p ∈ P , w ∈ W .So we have ( ρ ∨ ( p )) ∨ = ρ ( p − ) for p ∈ P . Hence ( α.f )( w ) = Z P π ( pζ ) ν ( ζ ) f ( ρ ( p − ) w ) dp. As f ∈ Hom P ( ρ , πν ), so ( πν )( p ) f ( w ) = f ( ρ ( p ) w ) for p ∈ P , w ∈ W . Hence( α.f )( w ) = ν ( ζ ) Z P π ( pζ )( πν )( p − ) f ( w ) dp = ν ( ζ ) Z P π ( pζ ) π ( p − ) ν ( p − ) f ( w ) dp = ν ( ζ ) Z P π ( pζ ) π ( p − ) f ( w ) dp. Now as π = c - Ind L f P e ρ and › P = h ζ i P , f ρ ( ζ k j ) = ρ ( j ) for j ∈ P , k ∈ Z , so π ( pζ ) = π ( p ) f ρ ( ζ ) = π ( p ) ρ (1) = π ( p )1 W ∨ . Therefore( α.f )( w ) = ν ( ζ ) Z P π ( p ) π ( p − ) f ( w ) dp = ν ( ζ ) f ( w )Vol( P )= ν ( ζ ) f ( w )So ( α.f )( w ) = ν ( ζ ) f ( w ) for w ∈ W . So α acts on f by multiplication by ν ( ζ ). Recallfor λ ∈ C × , we write C λ for the H ( L, ρ )-module with underlying abelian group C such that α.z = λz for z ∈ C λ . Therefore m L ( πν ) ∼ = C ν ( ζ ) .10. 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
Observe that πν lies in R [ L,π ] L ( L ). 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 [7], we have γ = γ = q n/ for unramified case when n isodd and γ = γ = q n/ for ramified case when n is even. As in Propn. 1.6 of [7], let Γ = { γ γ , − γ γ − , − γ − γ , ( γ γ ) − } . So by Propn. 1.6 in [7], ( T P ) ∗ ( C ν ( ζ ) ) is a simple H ( G, ρ )-module ⇐⇒ ν ( ζ ) / ∈ Γ. Recall π = Ind LZ ( L ) P f ρ where f 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 we can 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 REDUCIBILITY PROBLEM FOR EVEN UNITARY GROUPS: THE DEPTH ZERO CASE 43 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 [8] 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. Sofrom the commutative diagram (CD) on the previous page, we can conclude that ι GP ( πν ) isirreducible for any unramified character ν of L . So we conclude with the following theorem. Theorem 2.
Let G = U( n, n ) . Let P be the Siegel parabolic subgroup of G and L be the SiegelLevi component of P . Let π = c - Ind LZ ( L ) P f ρ be a smooth irreducible supercuspidal depth zerorepresentation of L ∼ = GL n ( E ) where f ρ ( ζ k j ) = ρ ( j ) for j ∈ P , k ∈ Z and ρ = τ θ for someregular character θ of l × with [ l : k E ] = n and | k F | = q . Consider the family 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/ , − } . References [1] M. F. Atiyah and I. G. Macdonald.
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