The Gelfand–Graev representation of classical groups in terms of Hecke algebras

Abstract Let G be a p-adic classical group. The representations in a given Bernstein component can be viewed as modules for the corresponding Hecke algebra—the endomorphism algebra of a pro-generator of the given component. Using Heiermann’s construction of these algebras, we describe the Bernstein components of the Gelfand–Graev representation for 
$G=\mathrm {SO}(2n+1)$
 , 
$\mathrm {Sp}(2n)$
 , and 
$\mathrm {O}(2n)$
 .


Introduction
Let F be a non-Archimedean local field of residue characteristic q. Let G be the group of F-points of a connected, split reductive algebraic group defined over F; in particular, the group G contains a Borel subgroup. Let U be the unipotent radical of the Borel subgroup, and fix a nondegenerate (Whittaker) character ψ ∶ U → C × . The Gelfand-Graev representation of G is c − ind G U (ψ), where c − ind stands for induction with compact support. The goal of this paper is to give an explicit description of the Bernstein components of the Gelfand-Graev representation.
Let us briefly describe what is known. Let K be a special maximal compact subgroup of G, and let I be an Iwahori subgroup contained in K. Let H be the Iwahori-Hecke algebra of I-biinvariant functions on G, and let H K be the subalgebra consisting of functions supported on K. Then H K is isomorphic to the group algebra of the Weyl group W of G, and thus it has a one-dimensional representation ε (the sign character). As an H-module, (c − ind G U ψ) I is isomorphic to the projective H-module [10] H ⊗ HK ε.
If G = GL n , then a similar statement holds for all Bernstein components with appropriate Hecke algebras arising from Bushnell-Kutzko types [11]. We build on methods of that paper. We finish this paragraph by mentioning a recent article of Mishra and Pattanayak [20] that considers Bernstein components of c − ind G U (ψ) corresponding to representations induced from the Borel subgroup. Their result is formulated in terms of Hecke algebras arising from types constructed by Roche.

Notation
Throughout the paper, F will denote a non-Archimedean local field of residue characteristic q and uniformizer , equipped with the absolute value | ⋅ | normalized in the usual way.
We let G denote the special odd orthogonal group, the symplectic group, or the (full) even orthogonal group. If we want to emphasize the rank, we use G n to denote SO(2n + 1, F), Sp(2n, F), or O(2n, F). By Rep(G), we denote the category of smooth complex representations of G.
For an arbitrary group H, we let X(H) denote the group of complex characters of H.

Unramified characters
If M is a Levi subgroup of G, we let M ○ = ⋂ χ ker |χ|, the intersection taken over the set of all rational characters χ ∶ M → F × . We say that a (complex) character χ of M is unramified if it is trivial on M ○ ; we let X nr (M) denote the group of all unramified characters on M. Then M/M ○ is a free Z-module of finite rank, and the group X nr (M) = X(M/M ○ ) has a natural structure of a complex affine variety. For any element m ∈ M, we denote by b m the evaluation χ ↦ χ(m). Now, let σ be an irreducible cuspidal representation of M, and set M σ = {m ∈ M ∶ m σ ≅ σ}. Then M/M σ is a finite abelian group and, abusing notation, we let A denote the ring of regular functions on the quotient variety X(M/M ○ )/X(M/M σ ). Since M σ /M ○ is once again a free Z-module (of the same rank as M/M ○ ), we have A ≅ C[M σ /M ○ ], by m ↦ b m . Furthermore, letting σ 0 denote an arbitrary irreducible constituent of σ| M ○ , we have a canonical isomorphism A ≅ End M (c − ind M M ○ σ 0 ). Indeed, this follows from a simple application of Mackey theory. We refer the reader to [17,Sections 1.17 and 4] for additional details.

The Hecke algebra of a Bernstein component
If π is an irreducible representation of G, there is a Levi subgroup M of G and an irreducible cuspidal representation σ of M such that π is (isomorphic to) a subquotient of i G P (σ). Here, P is a parabolic subgroup of G with a Levi component M. The pair (M, σ) is determined by π up to conjugacy; we call (M, σ) the cuspidal support of π.
We say that the two pairs (M 1 , σ 1 ) and (M 2 , σ 2 ) as above are inertially equivalent if there exist an element g ∈ G and an unramified character χ of M 2 such that taken over the set of all inertial equivalence classes. We refer to Rep (M,σ) (G) as the Bernstein component attached to the pair (M, σ). For a detailed discussion of the above results, see [3] or [4].
The fact that Γ (M,σ) is a projective generator implies that this is an equivalence of categories. This is [4,Lemma 22]; a detailed proof of this fact is also given in [22,Theorem 1.5.3.1]. Given a Bernstein component attached to s = (M, σ), we use H s to denote End G (Γ s ) and refer to it as the Hecke algebra attached to the component s. Furthermore, for any π ∈ Rep(G), we let π s denote the corresponding H s -module Hom(Γ s , π).
Although we do not use it here, we point out that there is another highly useful approach to analyzing Bernstein components, based on the theory of types developed by Bushnell and Kutzko [8]. One can show that the Hecke algebra used by Bushnell and Kutzko is in fact isomorphic to the algebra H s introduced above; we prove this fact in Appendix A. Therefore-for the purposes of this paper-the two approaches are equivalent.
The Gelfand-Graev representation of classical groups in terms of Hecke algebras 1347

Cuspidal representations
Here, we briefly recall some facts and introduce notation related to cuspidal representations of classical groups.
Let ρ and τ be irreducible unitarizable cuspidal representations of GL k (F) and G n0 , respectively. We consider the representation ν α ρ ⋊ τ, where α ∈ R. Here, and throughout the paper, we use ν to denote the unramified character |det| of the general linear group. If ρ is not self-dual, the above representation never reduces. If ρ is self-dual, then there exists a unique α ≥ 0 such that ν α ρ ⋊ τ is reducible; we denote it by α ρ .
The number α ρ has a natural description in terms of Langlands parameters. Let ϕ be the L-parameter of τ. Then ϕ decomposes into a direct sum of irreducible representations of W F × SL 2 (C). We view ρ as a representation of W F ; we say that it is of the same type as ϕ if the corresponding W F -representation factors through a group of the same type (orthogonal/symplectic) as ϕ. Letting S a denote the (unique) irreducible algebraic a-dimensional representation of SL 2 (C), we now set a ρ = max{a ∶ ρ ⊗ S a appears in ϕ}.
If the above set is empty, we let With this description of a ρ , we have α ρ = aρ +1 2 .

The structure of the Hecke algebra
We retain the notation ρ, τ, and G n from the previous subsection, and consider the cuspidal component s attached to the representation In the rest of the paper, we restrict our attention to cuspidal components of the above form. This does not present a significant loss of generality, since the Hecke algebra of a general cuspidal component is the product of algebras corresponding to the components described above. To simplify notation, we set H = H s . The structure of the Hecke algebra H has been completely described by Heiermann [16,17]. In his work, Heiermann shows that H is a Hecke algebra with parameters (the type of the algebra and the parameters depending on the specifics of the given case). When the component in question is of the form described above, we have three distinct cases, which we now summarize. For basic definitions and results on Hecke algebras with parameters, we refer to the work of Lusztig [19].
In what follows, we let t denote the order of the (finite) group {χ ∈ X nr (GL k (F)) ∶ ρ ⊗ χ ≅ ρ}. In all three cases, the commutative algebra A (see Section 2.3) is a subalgebra of H. In the present setting, the rank of the free module M σ /M ○ is equal to n. We can thus identify A ≅ C[M σ /M ○ ] with the algebra of Laurent polynomials C[X ± 1 , . . . , X ± n ]. We fix this isomorphism explicitly: For i = 1, . . . , n, let h i be the element of M which is equal to diag( , 1, . . . , 1) on the ith GL factor, and equal to the identity elsewhere. Then X i = b t h i . The three cases are: (i) No representation of the form ρ ⊗ χ with χ ∈ X nr (GL k (F)) is self-dual.
In this case, the algebra H is described by an affine Coxeter diagram of typẽ A n−1 with equal parameters t. In other words, it is isomorphic to the algebra H n described in [11]: There are elements T 1 , . . . , T n−1 which satisfy the quadratic relation and commutation relations where f s i is obtained from f ∈ A by swapping X i and X i+1 .
In the two remaining cases, there is an unramified character χ such that ρ ⊗ χ is self-dual. Without loss of generality, we may assume that ρ is self-dual. Then, up to isomorphism, there is a unique representation of the form ρ ⊗ χ ≇ ρ which is also self-dual; we denote it by ρ − . We set α = α ρ and β = α ρ − (see Section 2.5 for notation).
Since the situation is symmetric, we may (and will) assume that α ≥ β. The description of H now involves two additional operators T 0 and T n (see Remark 2.1). We have the following two cases: (ii) α = β = 0. In this case, H is described by an affine Coxeter diagram of typeC n : The nodes correspond to operators T 0 , . . . , T n which satisfy the quadratic relations and the braid relations as prescribed by the diagram. The commutation relations for T i , i = 1, . . . , n − 1, are the same as in Case (i), whereas T n satisfies In this case, H is described by an affine Coxeter diagram of typeC n : Here, s = t(α − β) and r = t(α + β). Again, the nodes correspond to operators T 0 , . . . , T n which satisfy quadratic relations analogous to those in Case (ii), along with the braid relations. The commutation relations for T i , i = 1, . . . , n − 1, are the same as in Case (i), whereas T n satisfies Cases (i)-(iii) correspond to Cases (I)-(III) listed in [16, Section 3.1]. The above results are collected in Section 3.4 of [16]. We take a moment to explain the situation in the even orthogonal case. Papers [16,17] do not treat the full orthogonal group; rather, they contain results about the special orthogonal group SO(2n). In the special orthogonal case, there is a nontrivial R-group (see [14]) which complicates the structure of the Hecke algebra; this was ultimately worked out by Heiermann in [18]. Because of this, we choose to work with O(2n) instead. This is indeed justified: Annex A of [18] shows that the results of [16,17] generalize to the full orthogonal case.
A detailed construction of the operators T i (starting from standard intertwining operators) is the subject matter of [17]; we do not need the details here, except in a special case discussed in the final part of Section 3.2. To facilitate the comparison of the above summary to the works of Heiermann [16][17][18], we point out the ways in which our summary deviates from them.
we use is different than the one used in [16]; there, Heiermann sets The operator T 0 which appears in Cases (ii) and (iii) above is not needed to describe H, and is therefore not used in [16,17]. To be precise, the Hecke algebra is generated over A by the operators T 1 , . . . , T n and determined by the quadratic and braid relations they satisfy, along with the commutation relations listed above. Each of the operators T 1 , . . . , T n corresponds to a simple reflection in the Weyl group, whereas the operator T 0 corresponds to the reflection given by the (in this case, unique) minimal element of the root system (see [19,Section 1.4]). In fact, we define T 0 by setting [19,Sections 2.8 and 3.3]). We use T 0 out of convenience, as it allows a more symmetric description of certain H-modules. (c) The description of H in Case (ii) differs from the one given in [16], which views T n as the nontrivial element of the R-group. However, one can verify that the description we use is equivalent. With our description, (ii) can be viewed as a special case of (iii) (with r = s = 0); however, since our results in (ii) require additional analysis, we still state the two cases separately.

Generic representations
We recall only the most basic facts here; a general reference is, e.g., [24]. Assume that G is split, and let U is be a maximal unipotent subgroup of G. Fix a nondegenerate character ψ of U. Recall that a character of U is said to be nondegenerate if it is nontrivial on every root subgroup corresponding to a simple 1350 P. Bakić and G. Savin root. We say that a representation (π, V ) of G is ψ-generic if there exists a so-called Whittaker functional-that is, a linear functional L ∶ V → C such that The key fact we use throughout is that the space of Whittaker functionals is at most one-dimensional. However, this fact does not hold for the disconnected O(2n, F), and we need to adjust the definition of Whittaker character as follows. In this case, the Levi factor of the normalizer of Observe that the order of α is 2. We extend ψ to a characterψ ofŨ = U ⋊ ⟨α⟩ byψ(α) = 1. With this extension, the space of Whittaker functionals for any irreducible representation of O(2n, F) is at most one-dimensional. Now, let P = MN be a parabolic subgroup of G. If σ is an irreducible generic representation of M, then one can construct a Whittaker functional on i G P σ (see [24, Proposition 3.1] and equation (3.11)); in other words, the induced representation is ψ-generic as well. We use this fact later, in Section 3.2.

The Gelfand-Graev representation
Continuing with split G, let U be a maximal unipotent subgroup of G and fix a nondegenerate character ψ) is replaced by the pair (Ũ ,ψ) in this definition. With this modification for O(2n, F) in mind, the Gelfand-Graev representation is the "universal" ψ-generic representation: Every ψ-generic representation of G appears as a quotient (with multiplicity one).
From this point on, we assume that the cuspidal representation τ-used to define the Bernstein component s in Section 2.6-is generic. We let Π denote (c − ind G U (ψ)) viewed as an H-module. Our goal is to determine the structure of Π.
We begin by investigating the structure of Π as an A-module. We point out that the proof of the following proposition applies, without modification, to any split reductive p-adic group.

Proposition 3.1 As
Recall that σ 0 was taken to be an arbitrary irreducible constituent of σ| M ○ . However, having now fixed the Whittaker datum for M (and thus for M ○ ), there exists a unique irreducible summand of σ| M ○ which is ψ-generic. Thus, from now on, we assume that σ 0 is this unique ψ-generic constituent of σ| M ○ .
To view Π as an A = End M (c − ind M M ○ (σ 0 ))-module, we use the Bernstein version of Frobenius reciprocity: The Gelfand-Graev representation of classical groups in terms of Hecke algebras 1351 here, r N denotes the Jacquet functor with respect to P = MN, the parabolic opposite to P. We now use the fact that Since σ 0 appears with multiplicity one, and no other m-conjugate of σ 0 is generic, , which proves the above claim about isotypic components. Thus, viewed as an A-module, Π is isomorphic to 2 We point out that the above differs from the proof of the analogous statement in [11]. It is shown there that any H-module Π which is (i) projective; (ii) finitely generated; and which satisfies (iii) dim Hom H (Π, π) ≤ 1 for any principal series representation π is isomorphic to A when viewed as an A-module (see [11, Lemmas 2.2 and 2.3]). The Gelfand-Graev representation can be shown to satisfy properties (i)-(iii): Property (i) is provided by Corollary 8.6 of [11]; (ii) is proved in [6], and (iii) follows from the multiplicity one property of generic representations. In Section 4, we present another useful application of the above approach to proving that an H-module is isomorphic to A.
Proposition 3.1 suggests the following approach to determine the H-module structure of Π: First, we find all possible H-module structures on A. After that, we need to only determine which one of those structures describes Π. In the following subsection, we compute the possible H-structures on A.

H-module structures on A
In order to treat the case of general Bernstein components-and not just those described in Section 2.6-we work in a slightly more general setting in this section. We thus investigate the possible H-module structures on A (where H is generated by T 0 , . . . , T n over A), but we assume that A = A ′ [X ± 1 , . . . , X ± n ], where A ′ is an integral domain containing C as a subring. For Bernstein components described in Section 2.6, we have A ′ = C; in general, A ′ itself is a (Laurent) polynomial ring over C.
First, assume that we are in Case (i) (see Section 2.6). Then the situation is precisely the one treated in [11], and the possible H-module structures on A are determined in Section 2.2 there. We have the following.

Proposition 3.3 (Case (i)) Let Π be an H-module which is isomorphic to A as an
Here, H Sn denotes the finite-dimensional algebra generated by T 1 , . . . , T n−1 ; we have H = A ⊗ C H Sn . Furthermore, H Sn has precisely two one-dimensional representations: We now treat Cases (ii) and (iii) simultaneously. Recall that, in these cases, the algebra H is described by an affine Coxeter diagram of typeC n . We let H 0 and H n denote the algebras obtained by removing the vertices which correspond to T 0 and T n , respectively. In other words, H 0 is generated by T 1 , . . . , T n as an A-algebra, whereas H n is generated by T 0 , . . . , T n−1 . Note that we have H = A ⊗ C H n = A ⊗ C H 0 . We now prove the following result.

Proposition 3.4 (Cases (ii) and (iii)) Let Π be an H-module which is isomorphic to
Here, ε 0 (resp. ε n ) is a one-dimensional representation of H 0 (resp. H n ).
Proof We first restrict our attention to the subalgebra generated by T 1 , . . . , T n−1 , which is contained in both H 0 and H n . This is precisely the algebra H Sn discussed in [11]. The possible H Sn -structures on A are determined in Section 2.2 there. To summarize the relevant results, there exists an invertible element g 0 ∈ A on which the operators T 1 , . . . , T n act by the same scalar, either q t or −1.
We now determine how T 0 and T n act on g 0 . Since g 0 is invertible, we have T n g 0 = f g 0 for some f ∈ A. Recall that T n satisfies the quadratic relation T 2 n = (q r − 1)T n + q r as well as the commutation relation Here, and throughout the proof, we let r = s = 0 if we are considering Case (ii). Recall that f ∨ denotes the function f ∨ (X 1 , . . . , X n ) = f (X 1 , . . . , X n−1 , 1 Xn ). Using the above and comparing the two sides of T 2 n g 0 = (q r − 1)T n g 0 + q r g 0 , we get To simplify notation, we now set b = q r − 1 and c = ( √ q r+s − √ q r−s ). We also temporarily drop the index n, writing X instead of X n . Clearing out the denominators, we rearrange the above equation into Our first goal is to find the possible solutions f ∈ A of this equation.

The Gelfand-Graev representation of classical groups in terms of Hecke algebras 1353
Lemma 3. 5 The above equation has the following solutions: along with the constant solutions f = q t and f = −1.
∎ Proof Each f ∈ A can be written as We write maxdeg( f ) for k and mindeg( f ) for −l. Now, let f be a solution of (*). We begin our analysis of (*) by solving some special cases. We claim the following: t h e n a 0 = q r or a 0 = −1. If f = a 1 X, t h e n a 1 = ∓ √ q r±s .
If f = a 0 + a −1 X −1 and a −1 ≠ 0, then a 0 = b and a −1 = ± √ q r±s . (3.1) To verify this, we first look at solutions f = a 0 . In this case, the equation (*) reduces to a 2 0 = ba 0 + q r . This equation has two constant solutions, a 0 = −1 and a 0 = q r . These are also the only solutions, since A has no zero divisors. When f (X) = a 1 X, the equation becomes a 2 1 + a 1 c − q r = 0. Again, the only two solutions of this equation are the constant ones: a 1 = ∓ √ q r±s . Finally, when f = a 0 + a −1 X −1 , the equation reduces to the following system: Since we are assuming that a 1 ≠ 0, the first equation gives us a 0 = b, and then the second becomes a 2 −1 − ca −1 − q r = 0. Again, we have two solutions: a −1 = ± √ q r±s .
Next, when f is a solution of (*) given by ( †), we observe k and l cannot both be positive. Indeed, let LHS and RHS denote the left-hand side and the right-hand side of (*), respectively. We then have maxdeg(LHS) = k + l + 2, whereas maxdeg(RHS) ≤ max{l + 2, k + 1, 2}. Therefore, equality of degrees cannot be achieved unless k ≤ 0 or l ≤ 0. In fact, the same argument gives us a slightly stronger statement in one case: If k > 0, then a 0 = 0.

P. Bakić and G. Savin
Finally, we make use of the following fact, which is readily verified by direct computation:

For any positive integer d, f is a solution of (*) if and only if
We are now ready to find all the solutions. By (3.2), any solution of f contains either only positive powers of X, or only nonpositive. We therefore consider two separate cases.
Let d = ⌊l/2⌋. We use (3.4) and look at another solution, g = X 2d f − R d . We first assume that l = 2d is even. In this case, g only has nonnegative powers of X, but it has a nonzero constant term, a −l . Therefore, (3.3) shows that the coefficients next to the positive powers must be zero: shows that there are only two possibilities for the constant term: a −l = q t or a −l = −1. We thus get two solutions: Next, assume that l = 2d + 1 is odd. Now, g has a nonzero coefficient (i.e., a −l ) next to X −1 , so by (3.2) the coefficients next to positive powers must be equal to 0. This gives us a 0 = b, a −1 = c, . . . , a 2−l = c. Furthermore, g is thus of the form a 1−l + a −l X −1 , so we can read off the coefficients a 1−l and a −l from (3.1). We thus arrive at two more solutions: Case B: f only has positive powers, i.e., f = a k X k + ⋅ ⋅ ⋅ + a 1 X.
This time, we set d = ⌊k/2⌋ and use (3.4) to obtain the solution g = 1 X 2d ( f + R d ). First, assume that k = 2d + 1 is odd. Then g has a nonzero coefficient (i.e., a k ) next to X, so (3.2) and (3.3) imply that all the lower coefficients are zero. This immediately gives us a 1 = −c, a 2 = −b, . . . , a 2d = −b. Furthermore, we have g = a k X, so (3.1) shows that we have two possibilities for a k . We therefore get two solutions: Finally, assume that k = 2d is even. First, if k > 2, consider another solution g ′ = X 2−2d ( f + R 2d−2 ). Now, g ′ has a nonzero coefficient (i.e., a k ) next to X 2 , so the coefficient next to nonpositive powers of X have to be 0 by (3.2) and (3.3). This gives us a 1 = −c, a 2 = −b, . . . , a 2d−2 = −b. In particular, this shows that g = (a k + b) + (a k−1 + c)X −1 . Since a k + b ≠ b (i.e., a k ≠ 0), (3.1) shows that we have only two possibilities:

The Gelfand-Graev representation of classical groups in terms of Hecke algebras 1355
In other words, a k−1 = −c and a k ∈ {−q r , 1}. We thus get the remaining solutions: We continue the proof of Proposition 3.4. We have just proved that T n g 0 = f g 0 where f ∈ A is one of the elements listed in Lemma 3.5. First, assume that f is one of the constant solutions, i.e., f = −1 or f = q r . Then g 0 is an invertible element of A on which T 1 , . . . , T n−1 , T n all act as scalars. In other words, we have a onedimensional representation ε 0 of the algebra H 0 . Since H = A ⊗ C H 0 , it follows that the corresponding H-module structure on A is isomorphic to Now, if f is of type (i) or (ii) listed in the statement of Lemma 3.5, set Since (X 1 X 2 ⋅ ⋯ ⋅ X n ) −d commutes with T 1 , . . . , T n−1 , g 1 is still an eigenvector for each of these operators. We claim that g 1 is an eigenvector for T n as well. Indeed, using the appropriate commutation relation and the fact that T n commutes with X 1 , . . . , X n−1 , we get Simplifying the expression in the parentheses, we obtain λX −d n , so that T n g 1 = λg 1 , where λ = q t (resp. −1) when f is of type (i) (resp. (ii)). We have thus once more found a common eigenvector for T 1 , . . . , T n−1 , T n . Again, we deduce that the corresponding H-module structure is isomorphic to H ⊗ H0 ε 0 , where ε 0 is a one-dimensional representation of H 0 .
When f is of type (v) or (vi), we use the same argument and arrive at the same conclusion. The only difference in this case is that we have to set g 1 = (X 1 X 2 ⋅ ⋯ ⋅ X n ) d g 0 in order to obtain a common eigenvector for T 1 , . . . , T n−1 , T n .
In the remaining cases-that is, when f is of type (iii) or (iv)-we cannot find such an eigenvector, but we claim that we can find an invertible g 1 ∈ A which is a common eigenvector for T 0 , T 1 , . . . , T n−1 . Just like in the previous cases, this will imply that the H-structure on A is isomorphic to H ⊗ Hn ε n for some one-dimensional representation ε n of H n .
If T n g 0 = f g 0 with f of type (iii), we set g 1 = (X 1 X 2 ⋅ ⋯ ⋅ X n ) −d g 0 . If f is of type (iv), let g 1 = (X 1 X 2 ⋅ ⋯ ⋅ X n ) d+1 g 0 . In both cases, g 1 is an eigenvector for T 1 , . . . , T n−1 and a computation analogous to the one we carried out in for Cases (i) and (ii) shows P. Bakić and G. Savin that we have The following lemma then shows that g 1 is also an eigenvector for T 0 and thus concludes the proof of Proposition 3.4.

Lemma 3.6
Let g be an invertible element of A, which is an eigenvector for T 1 , . . . , T n−1 , such that T n g = (b ± √ q r±s X −1 n )g. Then g is also an eigenvector for T 0 .
In both cases, all the operators T 1 , . . . , T n−1 act by the same scalar λ ∈ {−1, q t }. We therefore have We now recall that T −1 n = 1 q r (T n − b); this follows from the quadratic relation for T n . Therefore, by the assumption in the statement of the lemma, T −1 n g = ± √ q ±s−r X −1 n . Thus, with μ ∈ {−1, q s }. Finally, it remains to notice that, for every i = 1, . . . , n − 1, we have Indeed, from the quadratic relation, we have T −1 i = 1 q t (T i − (q t − 1)). Combining this with the commutation relation for T i , we get (3.6). Successively applying (3.6) to (3.5) (and taking into account that each T i acts on g by λ), we get which we needed to prove. Notice that the possible eigenvalues are precisely the zeros of (x − q s )(x + 1) = 0, the quadratic equation satisfied by T 0 . ∎ The above lemma shows that, in Cases (iii) and (iv), we have an invertible element g 1 ∈ A which is a common eigenvector for T 0 , T 1 , . . . , T n . Consequently, the Hmodule structure on A is given by H ⊗ Hn ε n for some one-dimensional representation ε n of H n . This concludes the proof of Proposition 3.4.
In view of Proposition 3.4, there are eight candidates for the H-structure (four, if n = 1): First, we may take the tensor product over H 0 or H n ; after that, there are four one-dimensional representations of H 0 (resp. H n ) to choose from. To verify this, note that the braid relations imply that-in any one-dimensional representation-the operators T 1 , . . . , T n−1 act by the same scalar, which has to be a zero of the quadratic relation satisfied by T i : (x − q t )(x + 1) = 0. We therefore have two possibilities for the action of the operators T i , and two additional possibilities (again, the zeros of the quadratic relation) for T n (resp. T 0 ). For example, the one-dimensional representations of H 0 The Gelfand-Graev representation of classical groups in terms of Hecke algebras 1357 are given by where ε is a one-dimensional representation of H W .
Proof Recall that H is a tensor product of Hecke algebras each of which is isomorphic to the Iwahori Hecke algebra of GL n or an algebra of typeC n with unequal parameters. Propositions 3.3 and 3.4 deal with these two cases, with additional flexibility that Thus, the corollary follows by repeated application of these two propositions. ∎

The Gelfand-Graev module
To complete the analysis of the Gelfand-Graev representation, we need to determine which of the H-module structures from the previous section is isomorphic to Π = (c − ind G U (ψ)) s . We consider Cases (i)-(iii) separately. Case (i). Let δ be the unique irreducible subrepresentation of ρν Since ρ ∨ is not an unramified twist of ρ in this case, the above Hom-space is only onedimensional. By Proposition 3.3, To determine which, we need only look at the action of H on the one-dimensional module π. We now need to examine the definition of the operators T i , i = 1, . . . , n − 1. In [17], T i is defined in Section 5.2 by the formula The intertwining operator R i has a pole at 0, and a zero at the point of reducibility (see [17,Section 1.8]). Since ν 3−n 2 −i ρ × ν 3−n 2 −i+1 ρ reduces, the operator R i acts by 0 in this case. It therefore remains to determine the action of X i /X i+1 . Equation (3.7) shows that it suffices to determine the action of X i /X i+1 on

P. Bakić and G. Savin
Recalling the definition of X i (Section 2.6), we immediately see that X i /X i+1 acts by This implies that T i also acts by (q t − 1) q t q t − 1 = q t . Since π is a quotient of Π, we conclude that we must have Π ≅ H ⊗ H Sn ε q t .
Case (iii). In this situation, the s-component of the Gelfand-Graev representation has two irreducible generic representations whose H-module is one-dimensional. These are the two (generalized) Steinberg representations: π and π ′ , which are the unique irreducible subrepresentations of respectively. Recall that α (resp. β) is the unique positive real number such that ν α ρ ⋊ τ (resp. ν β ρ − ⋊ τ) reduces (see Section 2.6). We now compare the action of the operators T 0 , . . . , T n on these two representations-that is, on Hom G (Γ s , π) and Hom G (Γ s , π − ), where Γ s is the projective generator defined in Section 2.4. We start by analyzing the action on π. We first focus on T i , i = 1, . . . , n − 1. Again, T i is defined by (3.8), and once more, the operator R i acts by 0. By the Bernstein version of Frobenius reciprocity, we have . We immediately see that X i /X i+1 acts by Again, this shows that T i acts by (q t − 1) q t q t − 1 = q t . For T n , we have a similar formula: Once more, R n acts by 0, and X n acts by (| | −α ) t = q tα . Recalling that r = t(α + β), we see that T n acts by q r . Finally, since and since X 1 acts by q (α+n−1)t , we see that T 0 acts by √ q r+2t(n−1)+s q 2t(n−1) ⋅ q r q (α+n−1)t = q s . We do the same with π − . Again, X i /X i+1 acts by q t , which shows that T i acts by q t as well. This time X n acts by −q tβ : Recall that ρ − = χ 0 ⊗ ρ with X n (χ 0 ) = −1, so X n (χ 0 ν −β ) = −q tβ . Repeating the above calculations, we now see that T n acts by q r , whereas T 0 acts by −1.

The Gelfand-Graev representation of classical groups in terms of Hecke algebras 1359
The above analysis allows us to single out the H-module structure on Π. Since T 0 does not act by the same scalar on π and π − , we deduce that Π = H ⊗ H0 ε for some one-dimensional representation ε of H 0 . Now, since every T i (i = 1, . . . , n − 1) acts by q t and T n acts by q r , we deduce that Π = H ⊗ H0 ε q r ,q t (see the end of Section 3.1 for notation).
Case (ii) The first part of our analysis remains the same as in Case (iii). The representation has two irreducible subrepresentations (both of which are in discrete series when n > 1, and tempered when n = 1), only one of which is generic. Denote the generic subrepresentation by π. Let π − denote the generic representation resulting from an analogous construction, when ρ is replaced by ρ − . Again, the H-modules corresponding to π and π − are one-dimensional, and the same calculations we used in Case (iii) show that the operators T i , i = 1, . . . , n − 1, act by q t . This leaves us four possible H structures to consider H ⊗ H0 ε 0 , with ε 0 (T n ) = ±1 (and ε 0 (T i ) = q t , i = 1, . . . , n − 1); and H ⊗ Hn ε n , with ε n (T 0 ) = ±1 (and ε n (T i ) = q t , i = 1, . . . , n − 1). (3.10) So far, we have been able to view Case (ii) as a special instance of Case (iii) which occurs when r = s = 0. However, to obtain an explicit description of the Gelfand-Graev module, we need more information than we used above in Case (iii). The reason is that the standard intertwining operator χρ ⋊ τ → χ −1 ρ ∨ ⋊ τ no longer has a pole when X n (χ) = ±1. In Case (iii), the operator R n (see formula (3.9))-which is constructed from the standard intertwining operator-vanishes at the point of reducibility, and the action of T n is determined by the action of the function (q r − 1) X n (X n − q tβ − q tα q r − 1 ) X 2 n − 1 used to remove the poles of R n . In this case, however, R n no longer vanishes and is regular at the point of reducibility; consequently, the above function does not appear in the construction and we have T n = R n . We know that this operator acts by 1 or −1 on the H-modules π and π − , but we still have a certain amount of freedom in our choices. Indeed, as one verifies easily, the operator T ′ n = (−1) e X f n T n (where e ∈ {0, 1} and f ∈ Z) satisfies the same relations as T n . Therefore, we obtain the same Hecke algebra if we replace T n by T ′ n , but the action of T ′ n on π obviously differs from the action of T n .
In fact, we know that X n acts on π by 1, and on π − by −1. Therefore, X 2 n acts by 1 on both, so replacing T n by X 2 n T n does not affect our description of the Gelfand-Graev module. We thus have four choices that affect the description (e = 0 or 1; f even or odd), and as we vary the four choices, the description of the Gelfand-Graev module varies through all the four possibilities described in (3.10).
This discussion shows that-to determine the action explicitly-we need to specify the choices appearing in the construction of the operator R n . We now explain one possible normalization using Whittaker models. To be concrete, we now focus on G = SO(2N + 1); the same approach is possible when G is symplectic or even orthogonal. We also specialize our discussion to the case n = 1 to simplify notation (thus, the cuspidal representation which defines the component is ρ ⊗ τ); the general case is analogous and follows from this one. We thus drop the subscripts and write T, X instead of T n , X n .
We fix a nondegenerate character ψ of the unipotent radical U of G = SO(2N + 1). Let V ρ denote the space of the representation ρ, and let λ be a ψ-Whittaker functional Notice that λ is then also a ψ-Whittaker functional for ρ ⊗ χ for any unramified character χ ∈ GL k (F): We have since det u = 1 and thus u ∈ ker χ. Abusing notation, we also let λ denote the ψ-Whittaker functional of ρ ⊗ τ (or χρ ⊗ τ for any unramified χ, as we have just shown). Following Proposition 3.1 of [24], we now form a ψ-Whittaker functional Λ χ on the space of i G P (χρ ⊗ τ) by setting where w is a representative of the nontrivial element of the Weyl group; in our case, we take w to be the block antidiagonal matrix Since π and π − are generic, it suffices to determine the action of T on their respective Whittaker functionals if we want to determine how T acts on the H-modules Hom G (Γ s , π) and Hom G (Γ s , π − ).

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Let Λ ∨ χ denote the Whittaker functional on i G P (χρ ∨ ⊗ τ) obtained using (3.11) from a fixed Whittaker functional λ ∨ for ρ ∨ . By the uniqueness of Whittaker functionals, Λ χ ○ φ = c ⋅ Λ ∨ χ for some constant c. Furthermore, since φ is induced from an isomorphism φ ∶ ρ ∨ ↦ ρ, it follows immediately that c does not depend on χ. Therefore, we have Note that there is a natural way to normalize φ in such a way that c = 1. We denote by g τ the transpose of an element g ∈ GL k (F) with respect to the antidiagonal (and with g −τ its inverse). One can then define a new representation ρ 1 by ρ 1 (g) = ρ(g −τ ). This representation is isomorphic to the contragredient of ρ; the advantage is that it acts on V ρ , the space of ρ. Furthermore, for any diagonal matrix (i.e., an element of the maximal torus) t ∈ GL k (F), we may conjugate ρ 1 to get ρ 2 (g) = t ρ 1 (g) = ρ 1 (t −1 gt). Then ρ 2 ≅ ρ 1 , and with a suitable choice of t, ρ 2 becomes ψ-generic with the same Whittaker functional λ. For example, assume that ψ is given by where ψ 0 is a nontrivial additive character of F, and u is an upper-triangular unipotent matrix with entries u 1,2 , . . . , u k−1,k above the main diagonal. Then one checks immediately that t = diag(1, −1, . . . , (−1) k−1 ) gives for any v ∈ V ρ . In short, we may assume that Λ χ ○ sp χ ○ T = Λ ∨ χ ○ J(χ −1 ) ○ sp χ −1 . This leads to the second choice we have to make in the construction of T: that of the normalization of the intertwining operator J. Here, we choose the standard normalization introduced by Shahidi (cf. [24,Theorem 3.1]). Under this assumption, we have Λ ∨ χ ○ J(χ −1 ) = Λ χ −1 for every unramified character χ. Thus, With this, we are ready to compare the action of T on π and π − . For π, we specialize at χ = 1; this gives us i.e., T acts trivially.