David Goldberg* - Purdue Universitygoldberg/papers/crelle.pdf · 2006. 6. 22. · 2 DAVID GOLDBERG* Section 2, we prove that if ˇw’ˇ; then I(ˇ) is irreducible if and only if

SOME RESULTS ON REDUCIBILITY FOR UNITARY

GROUPS AND LOCAL ASAI L-FUNCTIONS

David Goldberg*

Introduction. Let F be a p -adic field of characteristic zero and let F be the

algebraic closure of F. In [20], Shahidi describes the relationship of the poles

of certain Langlands L-functions attached to representations of GL(n, F ), to the

theory of twisted endoscopy. Here we carry out this program for the generalized Asai

L-function attached to a representation of GL(n,E), where E/F is a quadratic

extension. The problem is equivalent to certain reducibility questions for unitary

groups.

More precisely, let G = U(n, n) be the quasi-split unitary group in 2n vari-

ables defined with respect to E/F. Let G = G(F ). Then there is a maximal

parabolic subgroup P = MN of G, with M isomorphic to ResE/F (GLn).

Thus, the F -points are given by M = M(F ) ' GL(n,E). The L-group LM of

M is isomorphic to

(GL(n,C)×GL(n,C))oWF ,

where WF

is the Weil group of F /F. Let Ψ be the adjoint representation of

LM acting on the Lie algebra Ln of LN [17, §4]. This is the generalization to

rank n of the situation studied by Asai [1]. Let w be the non-trivial element of

the Weyl group of G with respect to M. Suppose π is an irreducible unitary

supercuspidal representation of M. Then πw ' π if and only if π is invariant

under the automorphism ε of GL(n,E) which takes g to tg−1 (here g is the

Galois conjugate of g ). If πw ' π, then the unitarily induced representation

IndGP (π ⊗ 1N ) is irreducible if and only if L(s, π,Ψ) has a pole at s = 0 [18].

We compute the poles of L(s, π,Ψ) by computing the residue of the standard

intertwining operator that determines the reducibility of I(π) = IndGP (π⊗ 1N ). In

*Partially supported by NSF Postdoctoral Fellowship DMS9206246

Typeset by AMS-TEX

2 DAVID GOLDBERG*

Section 2, we prove that if πw ' π, then I(π) is irreducible if and only if there

is some matrix coefficient ϕ of π for which a certain sum of ε-twisted orbital

integrals is non-zero (cf. Theorems 2.7 and 2.8). More precisely, if πw ' π, then

I(π) is irreducible if and only if there is some matrix coefficient ϕ of π such that

Φκε (δ, ϕ) 6= 0 if n is even

Φstε (δ, ϕ) 6= 0 if n is odd,

where δ is a hermitian form in GL(n,E), Φκε (δ, ϕ) is the non-stable sum of

twisted orbital integrals, and Φstε (δ, ϕ) is the stable sum (cf. Section 1). The

theory of twisted endoscopy, as developed by Kottwitz and Shelstad [12,13], says

that the non-vanishing of such sums points to the representation π being a lift

from the group U(n) (cf. Section 4). In Section 4 we show that, for n = 3, the

pole of L(s, π,Ψ) at s = 0 determines whether or not π is a standard base

change lift from U(3) [14]. This is similar to our results for n = 2 [6].

In Section 5, we explicitly determine the generalized Asai L-function L(s, σ,Ψ)

for any irreducible admissible representation σ of GL(n,E). We use the results

of Section 2 to compute L(s, π,Ψ) for any irreducible supercuspidal representation

π of GL(n,E). We then use Theorem 3.5 of [18] to compute L(s, σ,Ψ) for σ

in the discrete series. This determines L(s, σ,Ψ) in general. One consequence of

these computations is the following identity (cf. Corollary 5.5);

(1) L(s, σ × σ) = L(s, σ,Ψ) L(s, σ ⊗ µ det,Ψ),

where σ(g) = σ(g), µ is an extension to E× of the local class field theory

character attached to E/F, and L(s, σ × σ) is the Rankin-Selberg product L-

function attached to σ and σ [10].

In Section 6 we consider a second group where reducibility of induced representa-

tions is also determined by Asai L-functions (by means of (1)). Let G′ = U(n, n+

1) be the quasi-split unitary group in 2n+1 variables defined with respect to E/F.

Then there is a maximal parabolic P′ = M′N′, with M′ ' ResE/F (GLn)×U(1).

Thus, M ′ = M′(F ) ' GL(n,E)× U(1). If π is an irreducible unitary represen-

tation of GL(n,E), and ν is a character of U(1), then we denote by (π, ν) the

representation of M ′ given by (π, ν)(g, y) = π(g)ν(y det(gε(g)). Suppose that

SOME RESULTS ON REDUCIBILITY FOR UNITARY GROUPS AND LOCAL ASAI L-FUNCTIONS3

π is an irreducible unitary supercuspidal representation of GL(n,E). Then, un-

less n = 1 and π = 1, the reducibility of I(π, ν)′ = IndG′

P ′(π, ν) is determined

by whether or not L(s, π ⊗ µ det,Ψ) has a pole at s = 0. If πw ' π, then

L(s, π × π) has a simple pole at s = 0 [10], and therefore, by (1), exactly one

of L(s, π,Ψ) and L(s, π ⊗ µ det,Ψ) has a pole at s = 0. Thus, exactly one of

I(π) and I(π, ν)′ is reducible (cf. Proposition 6.2). The results of Section 2 allow

us to explicitly describe the reducibility criteria for I(π, ν)′ in terms of twisted

endoscopy (cf. Theorem 6.3). It is interesting to note how these results parallel

those of Shahidi [20], Theorem 6.3.

In Section 7, we use the explicit formulas for L(s, σ,Ψ), with σ a discrete se-

ries representation to determine the reducibility for IndGP (σ) and IndG′

P ′(σ, ν) (cf.

Theorem 7.1). These results are similar to the reducibility criteria for Sp(2n, F )

and SO(2n+ 1, F ) determined by Shahidi [20].

I would like to thank Freydoon Shahidi for encouraging me to study these ques-

tions, and for many discussions and insights. Among other things, Lemmas 5.3,

5.4, and Corollary 5.5 are due to him. I would like to thank Robert Kottwitz for

pointing out a mistake in an earlier proof of some of these results.

§1 Twisted Orbital Integrals for GL(n) .

Let F be a nonarchimedean local field of characteristic 0. Let RF

be its ring of

integers and pF

the unique maximal ideal in RF. Let $

Fbe a uniformizer in

F, that is, pF

= $FRF. Let q

F= |R

F/p

F| be the residual characteristic of F.

Let F be a (separable) algebraic closure of F.

Let E be a quadratic extension of F. Suppose E = F (β), with β2 ∈ F× \

(F×)2. Let RE, p

E, $

E, and q

Ebe the appropriate objects in E. Let τ :

E −→ E be the non-trivial Galois automorphism of E/F. We also denote the

action of τ by τ(x) = x. Let N : E× −→ F× be the norm map, N(x) = xx.

We assume that β = −β.

4 DAVID GOLDBERG*

Let H = U(n) as an algebraic group over F. H is defined as follows. Let

Φn =

1

−11

..

, and δn =

Φn n odd

βΦn n even.

Then we let H = g ∈ GL(n)|gδntg = δn . Let H = ResE/F (H). Then H is an

algebraic group such that H(F ) = H(E) = GL(n,E) [3]. Over E, H ' H×H.

Let H = H(F ) be the F -rational points of H. Let H = H(F ) be the F -rational

points of H. We define the automorphism ε : H → H by g 7→ tg−1.

Definition 1.1. An element δ of H is said to be ε-semisimple if (δ, ε) is

semisimple in the non-connected group Ho < ε > .

Definition 1.2. Two elements δ and δ′ of H are said to be ε-conjugate if

there is a g ∈ H such that δ′ = g−1δε(g).

Let δ be an ε-semisimple element of H. Let Hδε =g ∈ H | g−1δε(g) = δ

,

and let H′

δε =g ∈ H | g−1δε(g)δ−1 ∈ ZF

, where ZF is the set of F scalar

matrices. Similarly, for γ a semisimple element of H, we define Hγ =g ∈ H | g−1γg = γ

.

definition 1.3. An element δ ∈ H is stably ε-conjugate to δ′ if there is a

g ∈ H(F ) so that δ = g−1δ′ε(g). In this case τ(g)g−1 ∈ Hδε [14]. Two elements

of H are stably conjugate if they are conjugate by some g in H(F ). This

implies that τ(g)g−1 ∈ Hγ .

Lemma 1.4.

(1) A stable conjugacy class in H is a union of conjugacy classes.

(2) A stable ε-conjugacy class in H is the union of ε-conjugacy classes.

For γ ∈ H we let O(γ) be the conjugacy class of γ and Ost(γ) the stable

conjugacy class of γ in H. For δ ∈ H we let Oε(δ) be the ε-conjugacy class

of δ and Oε−st(δ) the stable ε-conjugacy class of δ in H.


Definition 1.5. We define the norm map for δ ∈ H by N(δ) = δε(δ). Note

that N(g−1δε(g)) = g−1N(δ)g. Thus, N defines an injection N : [δ] 7→ N([δ])

from ε-stable conjugacy classes of H, to the set of stable conjugacy classes of H.

Proposition 1.6 (Rogawski [14, proposition 3.11.1(c)]). The norm map de-

fines a bijection between stable ε-conjugacy classes in H and stable conjugacy

classes in H.

Corollary 1.7. For any n, Oε−st(δn) consists of all hermitian matrices in H.

Proof. δn is hermitian, i.e. δn = tδn. Therefore, N(δn) = In. Since In is

a stable conjugacy class in H, proposition 1.6 implies δ is stably ε-conjugate to

δn if and only if N(δ) = In. This holds if and only if δ is hermitian.

Lemma 1.8. For any n,

Oε−st(δn) = Oε(δn) ∪Oε(δ′n),

where δ′n is any nondegenerate hermitian form which is inequivalent with δn.

Proof. Note that if δ is hermitian, then δ′ is ε-conjuagte to δ if and only if

δ and δ′ define equivalent hermitian forms. For each n there are two classes of

hermitian forms in GL(n,E), indexed by F×/NE×. That is, if δ and δ′ are

hermitian, then δ is equivalent to δ′ if and only if det (δ′δ−1) ∈ NE× [9]. The

result then follows from Corollary 1.7.

Remark. Suppose z ∈ F×. Then δ and zδ define the same unitary group.

Thus, the isomorphism class of Hδε only depends on F×/NE×(F×)n. Therefore,

when n is odd, Hδnε ' Hδ′nε.

Let E1 be the norm 1 elements in E, i.e. E1 = z ∈ E×|zz = 1 . Note that

Z = Z(H) ' E1, and Z = Z(H) ' E×. Let ω be a character of E1. Let

C(H,ω) be the space of locally constant functions, compactly supported modulo

Z, such that f(zg) = ω−1(z)f(g) for all z ∈ Z, and g ∈ H. Let ω be the

character of E× given by ω(z) = ω(z/z). Then we let C(H, ω) be the space

of locally constant functions, compactly supported modulo Z, such that ϕ(zg) =

ω−1(z)ϕ(g) for all z ∈ Z, and g ∈ H.

6 DAVID GOLDBERG*

Definition 1.9.

(a) For γ a semisimple element of H and f ∈ C(H,ω), we define

Φ(γ, f) =

∫Hγ\H

f(g−1γg) dg×,

where dg× is the right invariant measure on the quotient coming from Haar mea-

sure dg on H. This is referred to as the orbital integral of f at γ.

(b) Similarly, for δ an ε-semisimple element of H and ϕ ∈ C(H, ω), we

define

Φε(δ, ϕ) =

∫H′δε\H

ϕ(g−1δε(g)) dg×,

where again the measure dg× is the right invariant one coming from Haar measure.

This is called the ε-twisted orbital integral of ϕ at δ.

definition 1.10.

(a) Let G be a reductive F group. Let q(G) be the F rank of the derived

group of G. Let e(G) = (−1)q(G)−q(G′), where G′ is the quasi split form of G.

(b) Let γ be a semisimple element of H. Then we let e(γ) = e(Hγ).

(c) Let δ be an ε-semisimple element of H. Then we let e(δ) = e(Hδε).

definition 1.11. Let γ ∈ H be a semisimple element and let γ′ be a collection

of representatives of the conjugacy classes in Ost(γ). Let ω be a character of E1

and let f ∈ C(H,ω). Then we let

Φst(γ, f) =∑γ′

e(γ′)Φ(γ′, f).

Definition 1.12. Let ∆n = δn, δ′n be the chosen representatives of the ε-

classes in Oε−st(δn). Consider the bijection ∆n ↔ F×/NE× given by δ 7−→det(δ) (mod NE×). Let κ : F×/NE× −→ ±1 be a character. Let ϕ ∈

C(H, ω). Define

Φκε (δn, ϕ) =∑∆n

κ(δ)e(δ)Φε(δ, ϕ).


When κ = 1 we write Φstε (δn, ϕ). By the remark following Lemma 1.8 we know

that, when n is odd, e(δn) = e(δ′n) = 1. Therefore, in this case Φstε (δn, ϕ) =

Φε(δn, ϕ) + Φε(δ′n, ϕ). However, when n is even, e(δn) = 1 and e(δ′n) = −1.

Therefore, in the even case, the sum of the two twisted orbital integrals is Φκε (δn, ϕ),

where κ is the non-trivial character.

§2 Reducibility Criteria. In this Section we repeat the argument of Section 2 of

[6] in the wider context of the group U(n, n). Let E/F be as in §1. Recall that

E = F (β). In this section we use the hermitian form J ′ =

(0 βIn−βIn 0

). Let

G = U(2n) be defined with respect to J ′, and G = U(n, n) = G(F ). Let T be

the maximal torus of diagonal elements in G. Then

T = T(F ) =

x1

. . .

xnx−1

1

. . .

x−1n

∣∣∣∣∣ xi ∈ E×

.

Let Td be the maximal F -split sub-torus of T. Then

Td = Td(F ) =

x1

. . .

xnx−1

1

. . .

x−1n

∣∣∣∣∣ xi ∈ F×

.

The restricted root system Φ(G,Td) is of type Cn. Let A be the subtorus

of Td given by θ = ei − ei+1n−1i=1 . Then

A = A(F ) =

(xIn 00 x−1In

) ∣∣∣∣ x ∈ F× .Let M be the centralizer of A in G. Then

M =

(g 00 tg−1

) ∣∣∣∣g ∈ ResE/F

(GLn)

' H.

8 DAVID GOLDBERG*

The Weyl group W (A) is of order two, with the non-trivial element w represented

by

(0 −InIn 0

). Let P = MN with

N =

(In X0 In

) ∣∣∣∣tX = X

.

Then P is a maximal parabolic subgroup of G. Let P = P(F ) = MN, with

M = M(F ) ' GL(n,E), and

N = N(F ) =

(In X0 In

) ∣∣∣∣X ∈M(n,E); tX = X

.

Let X(M)F

denote the F -rational characters of M. Let a be the real Lie

algebra of A. Then a = Hom(X(M)F,R) [7]. Let a∗ = X(M)

F⊗Z R be its

dual, and let a∗C = a∗ ⊗R C. There is a homomorphism [7] HP : M → a defined

by

q<χ,HP (m)>F

= |χ(m)|F, ∀ χ ∈ X(M)

F.

Let ρ be half the sum of the positive roots in N. Let α = 2en be the unique

simple root in N. Let α = < ρ, α >−1 ρ, where < γ1, γ2 > is defined as follows.

Choose non-restricted roots γ′1 and γ′2 restricting to γ1 and γ2, respectively.

Then

< γ1, γ2 > =2(γ′1, γ

′2)

(γ′2, γ′2),

where (·, ·) is the standard Euclidean inner product on Φ(G,T) [17]. Clearly <

ρ, α > is well defined. Let z be the complexified lie algebra of the split component

of G. Then we identify a∗C/z with C via the map s 7→ sα. Note that the non-

restricted roots Φ(G,T) are of type A2n−1. Let ∆ = β1, . . . , β2n−1 be the set

of simple roots of Φ(G,T), with βi = ei − ei+1. Then θ = ∆ \ βn. Thus,

ρθ =n

2

n∑j=1

j(βj + β2n−j).

Therefore, < ρ, α >= (ρθ, βn) = n. This implies that, in terms of non-restricted

roots,

α =1

2

n∑j=1

j(βj + β2n−j).


Thus, if m ∈ GL(n,E), we have

q<α,HP (m)>F

= | detm|1/2E.

Let E(M) be the set of equivalence classes of irreducible unitary supercuspidal

representations of M. Let (π, V ) ∈E(M). Let ω be the central character of π.

Let s ∈ C and let V (s, π) =

f ∈ C∞(G, V )

∣∣f(mng) = π(m)q<sα,HP (m)>F

δ1/2P (m)f(g)∀ g ∈ G,m ∈M,n ∈ N

.

Then G acts on V (s, π) by right translations. We denote this action by

I(s, π) = IndGP

(π ⊗ q<sα,HP ()>

F

)= IndGP

(π ⊗ | det()|s/2

E

).

We write I(π), or IndGP (π) for I(0, π). Note that by Bruhat theory [7] IndGP (π)

is irreducible if πw 6' π. If m =

(g 00 tg−1

)with g ∈ GL(n,E), then wmw−1 =(

tg−1 00 g

). Therefore, πw = πε, where πε(g) = π(ε(g)).

We formally define an operator A(s, π) on V (s, π) by

[A (s, π)f ] (g) =

∫N

f(w−1ng) dn

for f ∈ V (s, π), g ∈ G. If A(s, π) converges, then it defines an intertwining

operator between I(s, π) and I(−s, πw). It is a Theorem of Harish-Chandra [15]

that, for π supercuspidal, A(s, π) converges for Re s > 0. Moreover, s 7→A(s, π) is meromorphic as an operator valued function, and has a meromorphic

continuation to the whole plane. This means that there is some fixed polynomial

P (t) so that s 7→ P (q−sF

) < A(s, π)f(g), v > is holomorphic for each g ∈ G, v ∈

V , f ∈ V (s, π), and the operator P (q−sF

)A(s, π) is non-zero.

Harish-Chandra’s completeness Theorem, [21], implies IndGP (π) is reducible if

and only if π ' πw and 0 is not a pole of s 7→ A(s, π).

10 DAVID GOLDBERG*

Lemma 2.1 (Rallis, Shahidi [20]). Let

V (s, π)0 =

f ∈ V (s, π)

∣∣∣∣ supp f ⊂ PN and is compact mod P

.

Then every pole of s 7→ A(s, π) is a pole of s 7→ A(s, π)f(e) for some f ∈

V (s, π)0.

Thus, we study poles of s 7→ A(s, π)f(e) for f ∈ V (s, π)0 and π ' πε. Let

L = M(n, pmE

) for some m ∈ Z+. Let L′ ⊂ N be given by

L′ =

(I 0x I

) ∣∣∣∣x ∈ L .Let f ∈ V (s, π)0. We assume that there is a v ∈ V so that, for y ∈ N,

f(y) =

v y ∈ L′

0 otherwise.

Lemma 2.2. If y ∈ N, then w−1y ∈ PN if and only if y =

(I a0 I

)with

a ∈ GL(n,E) and a = ta.

Proof. Suppose a, b, c ∈M(n,E), and g ∈ GL(n,E). If(0 −II 0

)(I a0 I

)=

(g b0 tg−1

)(I 0c I

),

then

(0 −II a

)=

(g + bc btg−1c tg−1

).

Therefore, a = tg−1, which implies a ∈ GL(n,E). Moreover, if y ∈ N then

a = ta. Conversely, if a ∈ GL(n,E) with ta = a, then the above calculation

shows (0 −II a

)=

(a−1 −I0 a

)(I 0a−1 I

).


Let v ∈ V . By Lemma 2.2

〈A(s, π)f(e), v〉 =

∫N

< f(w−1y), v > dy

=

∫det a 6=0

a= ta

⟨f

((ε(a) −I

0 a

)(I 0a−1 I

)), v

⟩da

=

∫det a 6=0

a= ta

⟨π(ε(a))f

((I 0a−1 I

)), v

⟩| det a|−s/2−<ρ,α>

Eda

=

∫det a 6=0

a=taa−1∈L

〈π(ε(a))v, v〉 | det a|−s/2−<ρ,α>E

da.(2.1)

Lemma 2.3. If d×a = | det a|−<ρ,α>E

da then d×a−1 = d×a.

Proof. Let M act on N by conjugation. If tX = X ∈ M(n,E) and g ∈GL(n,E) then

(g−1 00 tg

)(I X0 I

)(g 00 tg−1

)=

(I g−1Xtg−1

0 I

).

Therefore, on N, d(g−1Xtg−1)/dX = | detX |−2<ρ,α>E

.

Now suppose g = X. Then

| detX−1|−<ρ,α>E

dX−1 = | detX |−<ρ,α>E

dX,

and therefore, d×a−1 = d×a.

Let ϕ(g) = < π(g)v, v > . For x ∈M(n,E) let

ξL(x) =

1 x ∈ L0 x 6∈ L.

Then using the relation a = ta and Lemma 2.3, we rewrite (2.1) as

< A(s, π)f(e), v > =

∫det a 6=0

a= ta

ϕ(a−1)| det a|−s/2E

ξL(a−1) d×a

=

∫det a 6=0

a= ta

ϕ(a)| deta|s/2EξL(a) d×a.(2.2)

12 DAVID GOLDBERG*

Remark. Since v and v were arbitrary, any matrix coefficient ϕ of π can

appear in (2.2).

Recall that ∆n = δn, δ′n are the representatives of the ε-classes in Oε−st(δn).

If a ∈ GL(n,E) is hermitian, then a = g−1δε(g) for some g and a unique

δ ∈ ∆n. Thus, using the notation in Section 1, we can rewrite (2.2) as

(2.3)

< A(s, π)f(e), v > =∑∆n

∫Hδε\H

ϕ(g−1δε(g))| det(g−1δε(g))|s/2EξL(g−1δε(g)) d×g,

where d×g is the invariant measure on the quotient coming from d×a.

Notice that the integrals in (2.3) are not of the form given in Section 1. The

next lemma allows us to decompose the integrals in (2.3) into an iterated integral

involving twisted orbital integrals, as defined in Section 1.

Lemma 2.4.

(a) If n is even then Hδε\H ′δε ' F×, for any δ ∈ ∆n.

(b) [14, proposition 3.11.2(c)] If n is odd then Hδε\H ′δε ' NE×, for any

δ ∈ ∆n.

Proof. (a) Note that ψ : H ′δε −→ F× given by ψ(g) = g−1δε(g)δ−1 is a

homomorphism. By its definition Hδε is the kernel of ψ. Let z ∈ F×. Note that

Hδε = H(zδ)ε, and therefore, since n is even, the two hermitian forms δ and zδ

are equivalent. Since ε-conjugacy is equivalence of hermitian forms, there is some

g ∈ H so that g−1δε(g) = zδ. Therefore, g is in H ′δε. Since ψ(g) = z, we see

ψ is surjective.

The proof of (b) is similar, once one notes that if δ and zδ are equivalent,

then z ∈ NE× [14].

Suppose g ∈ H is a representative of a coset of H ′δε\H. If n is even and

z0 ∈ F×, then there is some g0 ∈ H ′δε so that

(g0g)−1δε(g0g) = (z0In)g−1δε(g).

If n is odd, and z0 ∈ E×, then there is some g0 ∈ H ′δε so that

(g0g)−1δε(g0g) = (Nz0In)g−1δε(g).


Therefore, we choose representatives for H ′δε\H so that, if z ∈F× n odd

NE× n even,

then

(2.4) g−1zδε(g) ∈ L if and only if |z|E≤ 1.

If n is even, we now can rewrite (2.3) as < A(s, π)f(e), v >=

∑∆n

∫F×

∫H′δε\H

ϕ(g−1(zδ)ε(g))| det(g−1(zδ)ε(g))|s/2EξL(g−1(zδ)ε(g))d×g d×z

=

∫RF

ω(z)|z|ns/2E

d×z∑∆n

∫H′δε\H

ϕ(g−1δε(g))| det(g−1δε(g))|s/2E

d×g.(2.5a)

If n is odd, we rewrite (2.3) as < A(s, π)f(e), v >=

∑∆n

∫NE×

∫H′δε\H

ϕ(g−1(zδ)ε(g))| det(g−1(zδ)ε(g))|s/2EξL(g−1(zδ)ε(g))d×g d×z

=

∫RF∩NE×

ω(z)|z|ns/2E

d×z∑∆n

∫H′δε\H


d×g.

(2.5b)

Lemma 2.5. g ∈ H ′δε\H∣∣ g−1δε(g) ∈ suppϕ is compact.

Proof. Let || ||∞ denote the supremum norm on M(n,E). Let X = q−mE

, q−m+1E

.

Let L0 = x∣∣ ||x||∞ ∈ X ⊂ M(n,E). We have already chosen representatives g

with g−1δε(g) ∈ L0. Let S ⊂ M(n,E) be the set of hermitian matrices. Since

transposition and Galois conjugation are continuous, S is closed in M(n,E).

Define ψ : H −→ S by ψ(g) = g−1δε(g). Then

Im(ψ) = s ∈ S∣∣det s ≡ det δ mod NE×.

Since both NE× and its complement are closed in F×, and the determinant is

continuous, Im(ψ) is closed in S, thus is closed in H.

If y ∈ S and ψ(g) = y, then ψ−1(y) = Hδεg. Note that ψ−1(ZFy) = H ′δεg.

Let C be a compact subset of H such that suppϕ ⊂ CZ. Since x 7→ ||x||∞

14 DAVID GOLDBERG*

and x 7→ | det x|E

are continuous we can choose integers j, k, l, t so that qkE≤

||c||∞ ≤ qjE

and qlE≤ | det c|

E≤ qt

Efor all c ∈ C. Therefore, if c ∈ C and

cz ∈ CZ ∩ L0 then q−m−jE

≤ |z|E≤ q−m+1−k

E. We let

Ω = z∣∣ q−m−j

E≤ |z|

E≤ q−m+1−k

E.

Then CZ∩L0 ⊂ CΩ, which is compact. Therefore, Imψ∩suppϕ∩L0 is compact

in H. Hence the lemma holds.

By Lemma 2.5,

lims→0

∑∆n

∫H′δε\H


dg× =

∑∆n

∫H′δε\H

ϕ(g−1δε(g)) d×g =

Φκε (δn, ϕ) n even

Φstε (δn, ϕ) n odd.(2.6)

Consequently, if n is even, the residue at 0 of A(s, π)f(e) is given by

(2.7a) lims→0

s < A(s, π)f(e), v > =

(lims→0

s

∫RF

ω(z)|z|ns/2E

d×z

)Φκε (δn, ϕ).

Similarly, when n is odd, the residue is given by

(2.7b) lims→0

s < A(s, π)f(e), v > =

(lims→0

s

∫RF∩NE×

ω(z)|z|ns/2E

d×z

)Φstε (δn, ϕ).

Consider the case where n is even. If A(s, π)f(e) has a pole at s = 0 then

s 7→∫RF

ω(z)|z|ns/2E

d×z

has a pole at s = 0.

∫RF

ω(z)|z|ns/2E

d×z =

∫RF

ω(z)|z|nsFd×z = L(ns, ω|F× ),(2.8)


where L(s, χ) is the local Hecke-Tate L-function attached to a character χ of

F×. Thus, (2.7a) is zero, unless ω|F× is unramified, in which case it is proportional

to

(2.9)

( ∞∑m=0

(ω($

F)q−nsF

)m)Φκε (δn, ϕ) =

(1

1− ω($F

)q−nsF

)Φκε (δn, ϕ).

Therefore, there is no pole at s = 0 unless ω($F

) = 1. Since ω|F× is unram-

ified and ω($F

) = 1, ω|F× ≡ 1.

Lemma 2.6. Let η : E× −→ C× be a character.

(a) Suppose n is even and, for some ψ ∈ C(H, η), Φκε (δn, ψ) 6= 0. Then

η(z) = 1 for all z ∈ F×.

(b) Suppose n is odd and ,for some ψ ∈ C(H, η), Φstε (δn, ψ) 6= 0. Then

η(z) = 1 for any z ∈ NE×.

Proof. (a) Since Φstε (δn, ψ) 6= 0, we know that Φε(δ,ψ) 6= 0, for some δ ∈ ∆n.

Let z ∈ F×. By Lemma 2.4 we can choose some g0 ∈ H ′δε with g−10 δε(g0) = zδ.

Therefore, changing g to g0g in Φε(δ, ψ), we have

0 6= Φε(δ, ψ) =

∫H′δε\H

ψ(g−1δε(g)) dg×

=

∫H′δε\H

ψ((g0g)−1δε(g0g)) dg× = η(z−1)Φε(δ, ψ).

The proof of (b) is similar.

Therefore, the condition ω|F× ≡ 1 is guaranteed by the non-vanishing of

Φκε (δn, ψ). We can now state the result precisely.

Theorem 2.7. Let n be even. Let π ∈ E(M). The intertwining operator

A(s, π) has a pole at s = 0 if and only if π ' πε, and there is a matrix co-

efficient ϕ of π such that Φκε (δn, ϕ) 6= 0.

Now consider the case where n is odd. Since we are assuming that π ' πε, we

know that ω(zz) = 1. Therefore, the integral appearing in (2.7b) can be rewritten

16 DAVID GOLDBERG*

as

(2.10)

∫RE

|Nz|ns/2E

d×z =

∫RE

|z|nsEd×z = (1− 1/q

E)

1

1− q−nsE

.

Therefore, we have the following Theorem.

Theorem 2.8. Let n be odd and π ∈ E(M). Then the intertwining operator

A(s, π) has a pole at s = 0 if and only if π ' πε, and there is a matrix coefficient

ϕ of π such that Φstε (δn, ϕ) 6= 0.

Theorem 2.9. Let G = U(n, n) and M = GL(n,E) as above. Let π ∈E(M).

Let κ denote the trivial character of F×/NE× if n is odd, and the non-trivial

character if n is even. Then I(π) is reducible if and only if π ' πε and

Φκε (δn, ϕ) = 0 for every matrix coefficient ϕ of π.

§3 Complementary series. In this Section we use the results of Shahidi [18] to

determine when the representation I(s, π) is reducible, for s 6∈ iR. Let F be

a separable algebraic closure of F, and let ΓF

= Gal(F /F ). We denote by WF

the Weil group of F over F [23].

Recall that, for a connected, reductive, algebraic group G, defined over F, the

L-group is given by LG = LG0 oWF, where LG0 is the complex group whose

canonical root datum is dual to that of G, and WF

acts on LG0 through the

action of ΓF

on root data [3].

For H = U(n) we know that H(E) ' GL(n,E), and therefore LH0 =

GL(n,C). Since H(E) ' H(E)×H(E), we have LH0 ' GL(n,C)×GL(n,C).

Let τ be the generator of Gal(E/F ). Then the action of τ on LH0 is given by

τ(x) = Φn(tx−1)Φ−1n . The Galois group Gal(E/F ) acts on LH0 by τ(x, y) =

(τ(y), τ(x)). The action of τ determines the action of WF

in each case [14, pg.

47].

We compute the constituents of the adjoint representation of LM on Ln, where

LM is the L-group of M, Ln is the Lie algebra of LN, and LP = LMLN [3].

Recall that G = U(2n) arises from an action of Gal(E/F ) on the root system

A2n−1. Let ∆ = αi2n−1i=1 be the simple roots in A2n−1, with αi = ei − ei+1.

The action is given by τ(αi) = α2n−i. This is case 2A2n−1 − 2 of [17].


By the above discussion, LG0 = GL(2n,C) is a group of type A2n−1, while

the restricted root system Φ(G,Td) is of type Cn. The Levi subgroup M is gen-

erated by the subset θ = αin−1i=1 . As described above, each αi is the restriction

of two roots from A2n−1, namely αi and α2n−i. Thus, θ = αii6=n.

Therefore, LM0 =

(g 00 h

) ∣∣∣∣g, h ∈ GL(n,C)

' GL(n,C)×GL(n,C). Note

that the action of Gal(E/F ) on LM0 is given by

τ(g, h) = Φ2n

(tg 00 th

)−1

Φ−12n =

(Φn

th−1Φ−1n 0

0 Φntg−1Φ−1

n

)= (τ(h), τ(g)),

which is consistent with the description given above. LM =L M0 oWF

with this

action. The unipotent radical LN = LN0 is given by LN =

(I2 X0 I2

) ∣∣∣∣X ∈M(n,E)

.

Thus, Ln =

(0 X0 0

) ∣∣∣∣X ∈M(n,E)

. Let LM act on Ln by the adjoint rep-

resentation. We denote this representation by Ψ. Then Ψ(m)Y = mYm−1. Let

(g, h) ∈ LM0. Then

Ψ((g, h, 1))

(0 X0 0

)=

(0 gXh−1

0 0

).

Therefore, r|LM0' ρn⊗ρn, where ρn is the standard representation of GL(n,C).

Thus, Ψ|LM0is irreducible, so Ψ must also be irreducible. Note that the action

of τ on ∆ shows that

Ψ((1, 1, τ))

(0 (xij)0 0

)=

(0 (yij)0 0

)

where (xij) ∈ M(n,C) and yij = x(n+1−j)(n+1−i). We will use this in Section 5

in computing L(s, σ,Ψ) for σ in the discrete series of M.

Theorem 3.1. Let G = U(n, n) and suppose P = MN, with M ' GL(n,E).

Let π ∈E(M). Suppose that π ' πε and I(π) is irreducible (see Theorem 2.9).

(a) For 0 < s < 1 the representation I(s, π) = IndGP(π ⊗ | det()|s/2

E

)is irre-

ducible and unitarizable (i.e. in the complementary series).

18 DAVID GOLDBERG*

(b) The representation I(1, π) is reducible. It has a unique generic non-supercuspidal

discrete series subrepresentation. Its Langlands quotient is degenerate (non-generic)

pre-unitary, and nontempered.

(c) If s > 1 then I(s, π) is irreducible and never unitarizable.

Proof. Fix a non-trivial additive character ψF

of F. Theorem 3.5 of [18] de-

scribes a function γ(s, π,Ψ, ψF

), which is rational in q−sF. If Pπ(t) is the polyno-

mial such that Pπ(0) = 1 and Pπ(q−sF

) is the numerator of γ(s, π,Ψ, ψF

), then

L(s, π,Ψ) = P (q−sF

)−1 [18, §7]. Since Ψ is irreducible, Pπ = Pπ,1 in the lan-

guage of [18]. Therefore, Corollary 7.6 of [18] implies that the polynomial Pπ(t)

has a zero at t = 1. Since π ' πε and I(π) is irreducible, (a), (b) and (c) follow

immediately from Theorem 8.1 of [18].

Theorem 3.2. Suppose π ' πε and I(π) is reducible (see Theorem 2.9). Then

for all s > 0, the representation I(s, π) is irreducible, but not unitarizable.

Proof. This follows from Theorem 2.9 and Theorem 8.1 (d) of [18].

§4 Reducibility and Base Change.

In this Section we discuss the relation of the reducibility criteria derived in

Section 2 to the theory of base change from U(n) to GL(n,E). We will obtain

specific results when n = 3. In [6] we obtained similar results when n = 2.

Let ωE/F be the local class field theory character attached to E/F. That is

ωE/F (x) =

1 x ∈ NE×

−1 x 6∈ NE×.

Let µ : E× −→ C be any character whose restriction to F× is ωE/F .

There are two base change maps ψH , ψ′H : LH −→ LH. These homomorphisms

of L-groups are described by Rogawski [14, pg. 50]. The maps ψH and ψ′H are

related by the cocycle α : WF−→ LH which defines χµ = µ det over F. We

will describe some properties of these base change maps.

Definition 4.1. A distribution on C(H,ω) is called stable if it vanishes on every

f such that Φst(γ, f) = 0 for every regular semisimple γ ∈ H.


Let E(H) be the set of equivalence classes irreducible admissible representa-

tions of H, E2(H) ⊂ E(H) the collection of square integrable equivalence classes,

and E(H) the unitary supercuspidal equivalence classes. We make no notational

distinction between a class [π] and its representative π. For any π ∈ E(H) with

central character ω, we define the distribution character of π on C(H,ω) by

χπ(f) = Tr(π(f)), where

π(f) =

∫Z(H)\H

f(g)π(g) dg .

We assume that, for every reductive group G = G(F ), E2(G) can be parti-

tioned into finite subsets, called L-packets, with the following property: for every

L-packet Π we can choose non-zero integers, m(π) for each π ∈ Π, so that

χΠ =∑π∈Π

m(π)χπ is a stable distribution. This distribution is referred to as the

stable character of the L-packet Π. We let E2(G) be the collection of discrete

series L-packets. One can then define tempered L-packets by parabolic induction

[18, §9]. We let E(G) be the collection of tempered L-packets.

For each Π ∈ E(H) there should be two base change lifts, ψH(Π) and ψ′H(Π),

given by the Langlands correspondence. Namely, if ξ : WF−→ LH is an admissi-

ble homomorphism defining Π then ψH ξ : WF−→ LH should define ψH(Π).

Similarly ψ′H ξ : WF−→ LH should define ψ′H(Π). Since L-packets of GL(n)

are singletons, these lifts will actually define representations. Since ψH and ψ′H

differ by the cocycle α, Theorem 10.3(2) of [3] implies the two lifts are related by

(4.1) ψH(Π) = ψ′H(Π)⊗ χµ.

Let π = ψH(Π). One of the properties of base change lifting is that the central

characters of Π and ψH(Π) should be related by ωπ = ωΠ N, where N is

the norm map. Recall that Z(H) ' E× and Z(H) ' E1. If z ∈ F× then

ωπ(z) = ωΠ(zz−1) = 1. Therefore, ψH(Π) always has central character whose

restriction to F× is trivial. Note that if n is even, then χµ is trivial on F×

as well, while if n is odd, then χµ|F× = ωE/F . Therefore, if π′ = ψ′H(Π), then

ωπ′|F× = ωE/F if n is odd, and ωπ′|F× = 1 if n is even.

20 DAVID GOLDBERG*

In [6] we discussed the relation of the reducibility criteria given in Theorem 2.7

to the theory of base change when n = 2. For the remainder of this section we

consider the case where n = 3. Recall that we must determine when Φstε (δ3, ϕ) 6= 0

for some matrix coefficient ϕ of π. Let ω be a character of E1, and let ω be

the character of E defined by ω(z) = ω(z/z).

Theorem 4.2 (Rogawski). Let n = 3 and ϕ ∈ C(H, ω).

(a) There exists an f ∈ C(H,ω) so that Φstε (δ, ϕ) = Φst(γ, f) whenever δ is a

regular ε-semisimple element of H and γ = N(δ) ∈ H [14, Proposition 4.10.1].

(b) If γ = N(δ) is central then Φstε (δ, ϕ) = f(γ), where f is given by (a)

[14, Proposition 8.4.1].

Let E ′(H) be the collection of irreducible admissible representations of H such

that πε ' π, and the central character of π has trivial restriction to F×. Let

π(ε) be an equivalence between π and πε. Let ω be the central character of π.

We define a distribution on C(H, ω) by χπε(ϕ) = Tr(π(ϕ)π(ε)).

definition 4.3. If π ∈ E ′(H) is tempered, then π = ψH(Π) if and only if there

is a choice of π(ε) so that the character identity χπε(ϕ) = χΠ(f) holds whenever

ϕ→ f as in Theorem 4.2 [14, pg. 200].

Let H0 = U(2)×U(1). Then H0 → H, and we consider H0 as a subgroup of

H. Moreover, H0 is an endoscopic group for H [14]. The L-packets for H0 are

understood, and some L-packets for H should arise from the map LH0 −→ LH

[14]. Such L-packets are said to come from endoscopic transfer in the sense of

character identities. Suppose ρ0 is an L-packet of H0. Denote by ξH0(ρ0) the

corresponding L-packet of H.

Theorem 4.4 (Rogawski [14, Theorem 13.1.1,proposition 13.1.3]).

(1) An L-packet Π of H has more than one element if and only if Π =

ξH0(ρ0) for some L-packet ρ0 of H0.

(2) If Π is a tempered L-packet of H with more than one element, then Π

is either a discrete series L-packet, or Π consists of the constituents of

IndHB (θ), for a character θ of the torus U(1)× U(1)× U(1). Here B is

the standard Borel subgroup of H.


(3) If Π is a tempered L-packet of H with more than one element, then

χΠ =∑ρ∈Π

χρ,

i.e. m(ρ) = 1, ∀ρ ∈ Π.

(4) If Π ∈ E2(H) and Π = ξH0(ρ0) then every element of Π has the same

formal degree.

Theorem 4.5 (Rogawski [14, propositions 13.2.1, 13.2.2]).

(1) For every tempered L-packet Π of H, there is a unique standard base

change lift π = ψH(Π) ∈ E ′(H). The map ψH : E(H) → E ′(H) is injec-

tive.

(2) The representation ψH(Π) is square integrable if and only if Π is a square

integrable L-packet which consists of one element. Moreover, if Es(H) is

the collection of singleton square integrable L-packets, then

ψH : Es(H) −→ E ′(H) ∩ E2(H)

is bijective. Furthermore, ψH(Π) is supercuspidal if and only if Π ∈ Es(H)

is supercuspidal.

Therefore, if π is a supercuspidal representation of H such that πε ' π, then

π = ψH(Π), or π = ψ′H(Π) for a unique L-packet Π of H. Which map π lifts

through is determined by the restriction of the central character of π to F×.

Corollary 4.6. Suppose G = U(3, 3) and M = GL(3, E). If π ∈ E(M) and

A(s, π) has a pole at s = 0, then π is a base change lift from U(3).

Proof. This follows immediately from Theorem 2.8.

We will now proceed to show that we can find a matrix coefficient ϕ of π

such that Φstε (δ3, ϕ) 6= 0 if and only if π is a standard base change lift. Since

N(δ3) = I3, Theorem 4.2(b) implies Φstε (δ3, ϕ) = f(I3), where ϕ → f is as in

Theorem 4.2(a).

Let e = I3. We use the Plancherel formula [8] to expand f(e). The descrip-

tion of the tempered L-packets, given in Theorem 4.4, allows us to rewrite the

22 DAVID GOLDBERG*

Plancherel formula in terms of stable tempered characters. Using the character re-

lation which defines base change, and orthogonality relations of twisted characters,

we show that f(e) 6= 0 for some choice of ϕ if and only if π = ψH(Π) for some

Π. Note that if π has central character ω then ϕ ∈ C(H, ω−1).

Let M be the collection of standard Levi components of H. Suppose ρ is

a discrete series representation of L for some L ∈ M. Then we write ρ for

IndHLN (ρ), where N is the standard unipotent radical associated to L. For each

L ∈ M, let E2(L)ω−1 be the collection of discrete series representations of L with

central character ω−1. We denote by E2(L)ω−1 the collection of discrete series

L-packets of L with central character ω−1. Then, by the Plancherel formula

(4.2) f(e) =∑L∈M

C(L)

∫ρ∈E2(L)ω−1

d(ρ)µ(ρ)χρ(f) dρ,

where C(L) > 0 is a constant, d(ρ) is the formal degree of ρ, µ(ρ) is its

Plancherel measure, and dρ is the Euclidean measure given in [8].

Suppose Π ∈ E2(L)ω−1 . Let Π be the tempered L-packet of H obtained

by induction. If Π is not a singleton L-packet, then Π is either a discrete

series L-packet, or the collection of constituents of IndHB (θ). In the first case the

representations in Π appear in (4.2) with L = H. So ρ = ρ and µ(ρ) = 1 for all

ρ ∈ Π = Π. By Theorem 4.4(4), d(ρ) = d(ρ′) for all ρ, ρ′ ∈ Π. We let λH(Π) be

this common formal degree. Then, collecting terms for this L-packet, and applying

Theorem 4.4(3), the stable character χΠ appears with coefficient λH(Π) in (4.2).

In the second case d(θ) = 1, and part (3) of Theorem 4.4 implies χΠ appears in

(4.2) with coefficient λH(Π) = µ(θ). If Π = ρ then we let λH(Π) = d(ρ)µ(ρ).

Collecting terms according to L-packets we rewrite (4.2) as

(4.3) f(e) =∑L∈M

C(L)

∫Π∈E2(L)ω−1

λH(Π) χΠ(f) dΠ.

The character identity defining base change (Definition 4.3) shows that we can

replace χΠ(f) by χπ′ε(ϕ), where π′ = ψH(Π). We need the following orthogo-

nality relation whose proof is identical to that of Lemma 2.7 of [6].


Lemma 4.7. Let π, π′ be irreducible admissible representations of M. Suppose

their central characters are related by ω′ ' ω−1. Further suppose that π is

supercuspidal and π′ 6' π. Then, for any matrix coefficient ϕ of π, π′(ϕ) = 0.

In particular, if π and π′ are both ε-invariant, then χπ′ε(ϕ) = 0.

Theorem 4.8. Let π ∈E(M). Then A(s, π) has a pole at s = 0 if and only if

π is a standard base change lift from U(3), i.e. π = ψH(Π) for some Π ∈ E(H).

Proof. We already noted, in Corollary 4.6, that if A(s, π) has a pole at s = 0,

then π is a base change lift from U(3). Now suppose that π is a base change lift.

By Theorem 2.8, A(s, π) will have a pole at s = 0 if and only if Φstε (δ3, ϕ) 6= 0,

for some matrix coefficient ϕ of π. Let f ∈ C(H,ω−1) be the function associated

to ϕ by Theorem 4.2(a). Then f(e) is given by (4.3). By Theorem 4.5(2), π is

also a base change lift from H. Moreover, by considering central characters, π is

a standard base change lift if and only if π is a standard base change lift.

Then Lemma 4.7 and Definition 4.3 show that every term of (4.3) vanishes,

unless π = ψH(Π) for some Π. Thus, if π = ψ′H(Π), then Theorem 4.5(2)

implies f(e) = 0. Therefore, if π is a non-standard base change lift, then there

is no pole of A(s, π) at s = 0. On the other hand, if π = ψH(Π), then

Φstε (δ3, ϕ) = f(e) = cχπε(ϕ),

for some c > 0. Thus, it would be enough to know that, for some matrix coefficient

ϕ of π, χπε(ϕ) 6= 0. Such a matrix coefficient is called an ε- pseudo coefficient,

and their existence is guaranteed by [14, pg. 188]. Therefore, we can find a matrix

coefficient ϕ of π for which Φstε (δ3, ϕ) 6= 0 and hence A(s, π) has a pole at

s = π.

§5 Computation of Local Asai L-functions.

We wish to compute the local L-function L(s, π,Ψ) referred to in Section 3.

Once we have computed this L-function when π is supercuspidal, we will compute

L(s, σ,Ψ) in general.

Recall that G is of type A2n−1. Let ∆ = βj be the set of simple roots,

where

24 DAVID GOLDBERG*

βj = ej − ej+1. Then, as described in Section 3, the Galois action identifies βj

and β2n−j for j = 1, 2, . . . , n − 1. Therefore, the L-function in question is the

generalization of the Asai L-function [1] as described in Section 4 of [17].

Let π be a supercuspidal representation of M. Then L(s, π,Ψ) = Pπ(q−sF

)−1,

where Pπ is the polynomial defined in Section 7 of [18]. Let πs = π ⊗ | det()|s/2E.

Then, for any s1 ∈ C, Theorem 3.5(2) of [18] implies,

Pπs1 (q−sF

) = Pπ(q−s−s1F

).

Suppose that, for every s ∈ C, we have πs 6' (πs)ε. Then 1 = Pπs(1) = Pπ(q−s

F).

Thus, under this assumption, L(s, π,Ψ) = 1. Therefore, we may suppose that π '

πε. We need to find the normalized polynomial Pπ(q−sF

) such that Pπ(q−sF

)A(s, π)

is holomorphic and non-zero.

Let f ∈ V (s, π)0 be as in Section 2, and let v ∈ V . Let ϕ be the matrix

coefficient associated to f and v. For s ∈ C we let ϕs(g) = ϕ(g)| det g|s/2E.

Then ϕs is a matrix coefficient of πs.

Suppose n is odd. The poles of A(s, π) are among those of(1− q−ns

E

)−1.

Let s C. Suppose q−ns0E

= 1. Then (2.7b) shows that

Ress=s0

< A(s, π)f(e), v >=

(Ress=s0

(1

1− q−nsE

))Φstε (δn, ϕs0).

Therefore, A(s, π) has a pole at s = s0 if and only if Φstε (δn, ϕs0) 6= 0, for some

matrix coefficient ϕ of π.

Since πε ' π, the central character ω of π satisfies ω(z) = 1 for all z ∈

NE×. By Lemma 2.6(b), the non-vanishing of Φstε (δn, ϕs0) implies that ω| |−ns0/2E

is trivial on NE×. Therefore, the unramified character | |−ns0/2E

is trivial on

NE×. Thus, evaluating at N$E, we see that s = s0 is a root of 1− q−ns

E= 0.

So, if Φstε (δn, ϕs0) 6= 0 for some ϕ, then (1−qs0−sE

) divides Pπ(q−sF

). Note that

1− qs0−sE

=

1− qs0−s

FE/F ramified

1− q2s0−2sF

E/F unramified.

Finally notice that if η = | |s0/2E

, then (π⊗ηdet)ε ' πε⊗η−1det . Therefore,

the above discussion shows that π⊗ηdet is ε-invariant at the points in question.

We summarize our results below.


Theorem 5.1. Let n be odd. Suppose that π is an irreducible supercuspidal

representation of M such that π ' πε. Let Λ be the set of all unramified

characters η ∈ E×, no two of which have equal squares, such that Φstε (δn, ψ) 6= 0

for some matrix coefficient ψ of π ⊗ η det .

(a) Suppose E/F is ramified. Then

L(s, π,Ψ) =∏η∈Λ

(1− η2($E

)q−sE

)−1 =∏η∈Λ

(1− η($F

)q−sF

)−1.

(b) If E/F is unramified then


(1− η2($E

)q−sE

)−1 =∏η∈Λ

(1− η2($F

)q−2sF

)−1.

Note than when n = 3, then Lemma 4.7 and the argument of Theorem 4.8 show

that η ∈ Λ if and only if π ⊗ η det = ψH(Π) for some supercuspidal L-packet

Π of H = U(3).

Now suppose that n is even, and π ' πε. Then (2.9) shows that the poles

of A(s, π) are among those of(1− ω($

F)q−nsF

)−1. Moreover, if s0 is a zero of

(1− ω($F

)q−nsF

), then s0 is a pole of A(s, π) if and only if Φκε (δn, ψ) 6= 0 for

some matrix coefficient ψ of π⊗| det()|s0/2E

. Since ω is trivial on NE×, Lemma

2.6(a) implies | |s0/2E

= ω on F×. Therefore, if η = | |s0/2E

, then π ⊗ η det is

ε-invariant.

Theorem 5.2. Let n be even. Suppose that π is an irreducible supercuspidal

representation of M such that π ' πε. Let Λ be the collection of unrami-

fied characters η ∈ E×, no two of which have equal value at $F, such that

Φκε (δn, ψ) 6= 0 for some matrix coefficient ψ of π ⊗ η. Then


(1− η($F

)q−sF

)−1.

If n = 2, then the results of [6] show that Λ is the set of η such that

π ⊗ η det = ψ′H(Π) for some supercuspidal L-packet Π of U(2).

We now compute the L-function L(s, σ,Ψ) for any irreducible admissible repre-

sentation σ of M. By the discussion in Section 7 of [18], it is enough to compute

26 DAVID GOLDBERG*

L(s, σ,Ψ) where σ is in the discrete series of M. From now on we write Ψn for

Ψ, and denote by α the character | det()|E.

Let σ be an irreducible admissible discrete series representation of M. Let

ψF

be a non-trivial additive character of F. Let γ(s, σ,Ψn, ψF ) be the rational

function of q−sF

attached to σ and Ψn by Theorem 3.5 of [18]. Then L(s, σ,Ψn)

is the inverse of the normalized numerator of γ(s, σ,Ψn, ψF ). That is, there is a

monomial ε(s, σ,Ψn, ψF ) in q−sF, such that

(5.1) γ(s, σ,Ψn, ψF ) = ε(s, σ,Ψn, ψF )L(1− s, σ,Ψn)/L(s, σ,Ψn).

Let ∆ be the set of simple roots of M. Suppose σ ∈ E2(M). By Jacquet’s

theorem [7], we can find a parabolic M ′N ′ of M, and an irreducible supercuspidal

representation σ1 of M ′ so that σ is a subquotient of IndMM ′N ′(σ1 ⊗ 1N ′). We

can assume that there is a subset θ of ∆ so that M ′N ′ = MθNθ. By [2,24]

there are integers a and b, with ab = n, so that

Mθ ' GL(a, E)× · · · ×GL(a, E).

Moreover, we can assume that there is an irreducible unitary supercuspidal repre-

sentation π0 of GL(a, E) so that σ1 = π1 ⊗ · · · ⊗ πb, where

πi = π0 ⊗ α(b+1−2i)/2.

Note that

LM0θ = GL(a,C)× · · · ×GL(a,C) ⊂ GL(2n,C).

Suppose that

(5.2) g = (g1, . . . , gb, h1, . . . , hb) ∈ LM0θ .

Then, computing directly, we see that

τ(g) = (τ(hb), . . . , τ(h1), τ(gb), . . . , τ(g1)) ,

where τ(gi) is, up to an inner automorphism, the element described in Section 3.


Consider the restriction Ψθ of Ψn to LMθ. If X ∈M(n,C), then we write

X =

X11 . . . X1b...

. . ....

Xb1 . . . Xbb

with each Xij ∈M(a,C). Then if g is given by (5.2), we have

Ψθ((g, 1)) · (Xij) = (giXijh−1j ) and(5.3)

Ψθ((1, τ)) · (Xij) = (X(b+1−j)(b+1−i)).(5.4)

We look at the irreducible constituents of Ψθ. Let Vkl be the subspace of Ln

given by (Xij)

∣∣Xij = 0, (i, j) 6= (k, l).

Let G1 = U(a, a), M1 ' GL(a, E), and P1 = M1N1. Then, we see that

LM1 = (GL(a,C)×GL(a,C))nWF. Let Ln1 be the lie algebra of LN1. Then, for

1 ≤ i ≤ b, Vi(b+1−i) is irreducible, and Ψθ restricted to Vi(b+1−i) is isomorphic

to the representation Ψa of LM1 on Ln1.

If j 6= b+ 1− i, then Wij = Vij ∪V(b+1−j)(b+1−i) is irreducible, and the action

of Ψθ restricted to Wij is given by IndLM1LM0

1(ρa ⊗ ρa), where ρa is the standard

representation of GL(a,C).

The following two lemmas and their corollary were pointed out to me by Frey-

doon Shahidi. I would like to thank him again for his time and effort in this matter.

Lemma 5.3. Consider ρ = ρa ⊗ ρa as a representation of LM01 . Let η be

the non-trivial character of Gal(E/F ). Let I(ρ) = IndLM1LM0

1(ρ). Then I(ρ) =

Ψa ⊕ (Ψa ⊗ η).

Proof. Let ei be a basis for Ca, and let e∗i be the dual basis for (Ca)∗.

Then

ρ(g1, g2, 1)(ei ⊗ e∗j ) = g1ei ⊗ e∗jg−12 .

Let τ(ρ) be the representation of LM01 given by

τ(ρ)(g1, g2, 1) = ρ(

(1, 1, τ) (g1, g2, 1) (1, 1, τ)−1)

= ρ(tg−12 , tg−1

1 , 1).

28 DAVID GOLDBERG*

Note that I(ρ)((1, 1, τ))(ei⊗ e∗j ) is a basis for τ(ρ) as a subspace of I(ρ)|LM01

.

Let vij = ei ⊗ e∗j + I(ρ)((1, 1, τ))(ej ⊗ e∗i ). Let V0 be the subspace generated

by vij |1 ≤ i, j ≤ a. Then we will show that V0 is invariant and isomorphic to

Ψa. First note that

I(ρ)((g1, g2, 1))vij = I(ρ)((g1, g2, 1))ei ⊗ e∗j + I(ρ)((g1, g2, 1)I(ρ)((1, 1, τ))ej ⊗ e∗i

= g1ei ⊗ e∗jg−12 + I(ρ)((1, 1, τ))τ(ρ)((g1, g2, 1))ej ⊗ ei∗

= g1ei ⊗ e∗jg−12 + I(ρ)((1, 1, τ))

(tg−1

2 ej ⊗ e∗i tg1

).

Let

gei =a∑l=1

clel and e∗jg−12 =

a∑k=1

dke∗k.

Then

e∗itg1 =

a∑l=1

cle∗l and tg−1

2 ej =a∑k=1

dkek.

Thus,

g1ei ⊗ e∗jg−12 =

a∑l,k=1

cldkel ⊗ e∗k,

while

tg−12 ej ⊗ e∗i tg1 =

a∑l,k=1

cldkek ⊗ e∗l .

Therefore,

I(ρ)((g1, g2, 1))vij =∑l,k

cldkvlk ∈ V0.

Furthermore,

I(ρ)((1, 1, τ))vij = I(ρ)((1, 1, τ))ei⊗ e∗j + ej ⊗ e∗i = vji.

Thus, V0 is invariant, and the description of I(ρ) acting on V0 given above

clearly shows that V0 is isomorphic to Ψa (see Section 3).


Note that the complement of V0 in I(ρ) is generated by

wij = ei ⊗ e∗j − I(ρ)((1, 1, τ))ej ⊗ e∗i .

Moreover, I(ρ)((1, 1, τ))wij = −wji. Therefore, I(ρ) acts on this subspace as

Ψa ⊗ η acts on Ln1. This proves the lemma.

Remark. For the following lemma, we allow the possibility that F = R (and

E = C or R⊕ R ).

Lemma 5.4. Let ρ and I(ρ) be as in Lemma 5.3. Suppose that π is any

irreducible admissible representation of GL(a, E) which can be parameterized, i.e.

F = R, or π is unramified in the sense of [3]. Then L(s, σ, I(ρ)) = L(s, σ × σ).

Here σ(g) = σ(g), and the L-function on the right is the Rankin-Selberg product

L-function attached to σ and σ. [10].

Proof.

Since ρa is the standard representation, ρa is isomorphic to the representation

ρτ , given by ρτ (g) = ρ(τ(g)). The action of Gal(E/F ) on ResE/F (GLn) sends

σ to σ. Taking the viewpoint of E-groups, we have L(s, σ, I(ρ)) = L(s, σ, ρ).

However, by definition L(s, σ, ρ) = L(s, σ × σ).

Corollary 5.5. Let π be an irreducible admissible supercuspidal representation

of GL(a, E). Then

L(s, π × π) = L(s, π,Ψa) L(s, π ⊗ χµ,Ψa).

Proof. By proposition 5.1 of [18] we can choose a number field K, with Kv0= F

for some place v0 of K, a quadratic extension K ′/K, a place w0 lying over

v0 so that K ′wo = E, and a cusp form Π =⊗

w Πw of GL(a,AK′), such that

Πw is unramified for every finite place w 6= w0, and Πw0' π. For each non-split

place w of K ′ we let ηw be the non-trivial character of Gal(K ′w/Kv), where

w lies over v. By local class field theory, ηw corresponds to the character χµw

of GL(a,K ′w). Let ηK′ be the non-trivial character of Gal(K ′/K). By Lemmas

5.3 and 5.4 we have

(5.5) L(s,Πw × Πw) = L(s,Πw,Ψa) L(s,Πw ⊗ χµw ,Ψa),

30 DAVID GOLDBERG*

for every w 6= w0. We have a global functional equation

L(s,Π× Π) = ε(s,Π× Π)L(1− s, Π× ˜Π),

as well as one for L(s,Π,Ψa) and L(s,Π,Ψa ⊗ η′K

). At each place of K ′, we

have

L(s, Πw × ˜Πw) = L(s,Πw, ρ) and

L(s, Πw,Ψa) = L(s,Πw, Ψa).

Using global functional equations for each of these L-functions and equation (5.5)

for every w 6= w0, we conclude

L(s, π × π) = L(s, π,Ψa)L(s, π⊗ χµ,Ψa),

as desired.

We return to the situation in which σ is a discrete series subquotient of IndMMθNθ(σ1),

with σ1 =⊗

i πi. In what follows we use the symbol ≡ to denote two rational

functions which differ by a monomial. Applying part 3 of Theorem 3.5 of [18] along

with Lemmas 5.3 and 5.4 to (5.3), and (5.4), we find

γ(s, σ,Ψn, ψF ) =

b∏i=1

γ(s, πi,Ψa, ψF )∏

1≤i<j≤bγ(s, πi × πj, ψF ),

where γ(s, πi× πj , ψF ) is the Rankin-Selberg factor attached to (πi, πj) [10,16].

By part (2) of Theorem 3.5 of [18] and page 409 of [10], we have

γ(s, πi,Ψa, ψF ) = γ(s+ b+ 1− 2i, π0,Ψa, ψF ),(5.6)

and

γ(s, πi × πj , ψF ) = γ(s+ b+ 1− (i+ j), π0 × π0, ψF ).(5.7)

Using this, and (5.1) we have

γ(s, σ,Ψn, ψF ) ≡

b∏i=1

L(1− (s+ b+ 1− 2i), π0,Ψa)

L(s+ b+ 1− 2i, π0,Ψa)

∏i<j

L(1− (s+ b+ 1− (i+ j)), π0 × ˜π0)

L(s+ b+ 1− (i+ j), π0 × π0).


Since π0 is unitary and supercuspidal, [19] implies that, up to a monomial, the

right hand side is equal to

b∏i=1

L(s+ b− 2i, π0,Ψa)

L(s+ b+ 1− 2i, π0,Ψa)

∏i<j

L(s+ b− (i+ j), π0 × π0)

L(s+ b+ 1− (i+ j), π0 × π0)

=

(b−1)/2∏ν=−(b−1)/2

L(s+ 2ν − 1, π0,Ψa)

L(s+ 2ν, π0,Ψa)

b−1∏i=1

L(s− i, π0 × π0)

L(s+ b− 2i, π0 × π0).

By [10], L(s, π0 × π0) is identically one unless some unramified twist π0 ⊗ ηsatisfies π0 ⊗ η ' (π0⊗η)˜. By [2, §7], the contragredient of an irreducible admis-

sible representation of GL(a, E) is given by composition with the automorphism

g 7→ tg−1. Therefore, L(s, π0× π0) is identically one unless there is an unramified

character η such that π0 ⊗ η ' (π0 ⊗ η)ε. By Theorems 5.1 and 5.2, this is the

same as the condition for L(s, π0,Ψa) to be non-trivial. Note that

IndMMθNθ(σε1 ⊗ 1Nθ) = IndMMθNθ

(˜σ1 ⊗ 1Nθ)

'(

IndMMθNθ(σ1 ⊗ 1Nθ)

) ˜ ' (IndMMθNθ(σ1 ⊗ 1Nθ)

)ε(see [4]). Therefore, if L(s, σ,Ψn) 6≡ 1, then σ1 ' σε1, which implies that π0 '

πε0.

Suppose that b is even. Then

b−1∏i=1

L(s− i, π0 × π0)

L(s+ b− 2i, π0 × π0)=

(b−2)/2∏k=0

L(s− (2k + 1), π0 × π0)

L(s+ 2k, π0 × π0),

and

(b−1)/2∏ν=−(b−1)/2

L(s+ 2ν − 1, π0,Ψa)

L(s+ 2ν, π0,Ψa)=

(b−2)/2∏k=0

L(s+ 2k, π0,Ψa)L(s− 2(k + 1), π0,Ψa)

L(s− (2k + 1), π0,Ψa)L(s+ (2k + 1), π0,Ψa).

Therefore, using Corollary 5.5, we have

γ(s, σ,Ψn, ψF ) =

(b−2)/2∏k=0

L(s+ (2k + 1), π0,Ψa)−1L(s+ 2k, π0 ⊗ χµ,Ψa)−1

L(s− 2(k + 1), π0,Ψa)−1L(s− (2k + 1), π0 ⊗ χµ,Ψa)−1.

32 DAVID GOLDBERG*

Note that both the numerator and denominator of the above expression are

polynomials in q−sF, and there are no further cancellations. For each 0 ≤ k ≤

(b− 2)/2, we let i = (b/2)− k. Then 1 ≤ i ≤ b/2. Moreover,

L(s+ (2k + 1), π0,Ψa) = L(s, πi,Ψa), and

L(s+ 2k, π0 ⊗ χµ,Ψa) = L(s, πi ⊗ χµ ⊗ α−1/2,Ψa).

Let πi = π0 ⊗ αb+1−2i/2. Then

L(s− 2(k + 1), π0,Ψa) ≡ L(1− s, πi,Ψa) and

L(s− (2k + 1), π0 ⊗ χµ,Ψa) ≡ L(1− s, πi ⊗ χ−1µ ⊗ α−1/2,Ψa).

Now suppose that b is odd. Then

b−1∏i=1

L(s− i, π0 × π0)

L(s+ b− 2i, π0 × π0)=

(b−1)/2∏k=1

L(s− 2k, π0 × π0)

L(s+ (2k − 1), π0 × π0),

and

(b−1)/2∏ν=−(b−1)/2

L(s+ 2ν − 1, π0,Ψa)

L(s+ 2ν, π0,Ψa)=

(b−1)/2∏k=1

L(s+ (2k − 1), π0,Ψa)

L(s− 2k, π0,Ψa)

(b−1)/2∏k=0

L(s− (2k + 1), π0,Ψa)

L(s+ 2k, π0,Ψa).

Thus,

γ(s, σ,Ψn, ψF ) ≡

(b−1)/2∏k=0

L(s+ 2k, π0,Ψa)−1

L(s− (2k + 1), π0,Ψa)−1

(b−1)/2∏k=1

L(s+ (2k − 1), π0 ⊗ χµ,Ψa)−1

L(s− 2k, π0 ⊗ χµ,Ψa)−1.

Rewriting the product above in terms of the πi we get part (b) of the following

theorem.


Theorem 5.6. Let σ be a discrete series representation of GL(n,E). Choose an

irreducible unitary supercuspidal representation π0 of GL(a, E), n = ab, such

that σ is the unique discrete series component of the representation of GL(n,E)

induced from π1 ⊗ · · · ⊗ πb, πi = π0 ⊗ α(b+1−2i)/2. Let πi = π0 ⊗ α(b+1−2i)/2.

(a) Suppose b is even. Then

L(s, σ,Ψn) =

b/2∏i=1

L(s, πi,Ψa)L(s, πi ⊗ χµ ⊗ α−1/2,Ψa),

and

L(s, σ,Ψn) =

b/2∏i=1

L(s, πi,Ψa)L(s, πi ⊗ χ−1µ ⊗ α−1/2,Ψa).

(b) Suppose b is odd. Then

L(s, σ,Ψn) =

(b+1)/2∏i=1

L(s, πi,Ψa)

(b−1)/2∏i=1

L(s, πi ⊗ χµ ⊗ α−1/2,Ψa),

and

L(s, σ,Ψn) =

(b+1)/2∏i=1

L(s, πi,Ψa)

(b−1)/2∏i=1

L(s, πi ⊗ χ−1µ ⊗ α−1/2,Ψa).

corollary 5.7. Let σ be an irreducible admissible representation of GL(n,E).

Then

L(s, σ × σ) = L(s, σ,Ψn)L(s, σ ⊗ χµ,Ψn).

Proof. It is enough to prove the claim for the case when σ is in the discrete series.

Suppose that σ is of the form described in Theorem 5.6. Then, by Theorem 8.2

of [10],

L(s, σ × σ) =b∏i=1

L(s, π1 × πi).

34 DAVID GOLDBERG*

We rewrite this as

(5.8)b∏i=1

L(s+ b− i, π0 × π0) =b∏i=1

L(s+ b− i, π0,Ψa)L(s, π0 ⊗ χµ,Ψa).

Examining the formulas in Theorem 5.6 we see that L(s, σ,Ψa)L(s, σ ⊗ χµ,Ψa)

also has the form (5.8).

§6 Reducibility for U(n, n+ 1) .

We now use the results of Section 5 to determine reducibility criteria for the

group U(n, n + 1). Let E/F be a quadratic extension of local nonarchimedean

fields of characteristic 0. Let n ≥ 1. Let J =

βIn1

−βIn

, where β ∈

E \ F satisfies β = −β. Let G = U(2n + 1), defined with respect to J. Let

G = G(F ) = U(n, n + 1). Then G = g ∈ GL(2n + 1, E) | gJ tg = J. Let

T = T(F ) be the maximal torus of diagonal elements. Then

T =

x1

. . .

xny

x−11

. . .

x−1n

∣∣∣∣∣ xi ∈ E×y ∈ E1

.

Let Td be the maximal F -split subtorus. Then

Td = Td(F ) = (x1, . . . , xn, y) ∈ T |xi ∈ F×, y = 1.

The restricted root system Φ(G,Td) is of type BCn. We choose the set of simple

roots ∆ = ei − ei+1n−1i=1 ∪ en. Let θ = ∆ \ en. Let P = Pθ, be the

maximal parabolic subgroup of G associated to θ. Let P = MN be the Levi

decomposition of P, and P = P(F ) = MN. Then

M = M(F ) =

g

ytg−1

∣∣∣∣∣ g ∈ GL(n,E)y ∈ E1

' GL(n,E)× U(1).


Let Π be an irreducible admissible representation of M. Then we write Π =

(π, ν), in the following way. Π(g, y) = π(g)ν(det(gε(g))y), i.e., (π, ν) = (π ⊗ ν

det)⊗ ν, with π an irreducible admissible representation of GL(n,E), ν ∈ E1,

and ν(z) = ν(z/z). We choose this normalization so that our notation is consistent

with that of Keys [11] and Rogawski [14].

Let A be the split component of M. Since

0 0 In0 1 0−In 0 0

represents the

non-trivial element w of W (A), we see that Π ' Πw if and only if π ' πε.

We now consider the L-groups. Note that LG0 = GL(2n + 1,C). If g ∈GL(2n+ 1,C), then τ(g) = J(tg−1)J−1. We also note that

LM0 =

g

ah

∣∣∣∣∣ g, h ∈ GL(n,C)a ∈ C×

' GL(n,C)×GL(1,C)×GL(n,C).

The action of Gal(E/F ) restricted to LM0 is given by τ((g, a, h)) = (th−1, a−1, tg−1).

The Lie algebra Ln of LN is given by

0 y X

0 0 tz0 0 0

∣∣∣∣∣ z, y ∈ CnX ∈M(n,C)

.

The adjoint action r of LM on Ln has two constituents. We order these

constituents as in [18] and write r = r1 ⊕ r2. Thus,

r1| LM0' ρn ⊗ ρ1 ⊕ ρ1 ⊗ ρn.

Moreover, [18, pp. 297-298], shows that L(s,Π, r2) = L(s, π,Ψn⊗ η), where η is

the nontrivial character of Gal(E/F ). Thus, L(s,Π, r2) = L(s, π ⊗ χµ,Ψn).

Lemma 6.1. Let π be an irreducible admissible representation of GL(n,E), and

ν ∈ E1. Suppose Π = (π, ν). Then L(s,Π, r1) = L(s, π), where L(s, π) is the

Godement-Jacquet L-function attached to π [5].

Proof. Let ϕ : WF −→ GL(1,C) o WF be an admissible homomorphism at-

tached to ν, and ϕ0 : WF−→ GL(1,C), the associated 1-cocycle. Then ϕ0 :

36 DAVID GOLDBERG*

WF −→ GL(1,C) × GL(1,C), given by ϕ0(w) = (ϕ0(w), ϕ0(w)), is attached

to the character ν of E× [14, pg. 50]. Let M1 = ResE/F (GLn). Then

LM1 = (GL(n,C)× GL(n, C)) oWF. We consider a representation r′1 of LM1

on Cn ⊕Cn. Namely,

r′1(g, h, 1) · (y, tz) = (gy, tzh−1),

and

r1(1, 1, τ) · (y, tz) = (z′, ty′),

where (y1, . . . , yn)′ = (yn, . . . , y1). Suppose π is a class one representation, and

ξ : WE−→ GL(n,C) parameterizes π. Then there is an admissible homomor-

phism ξ′ : WF−→ LM1 which parameterizes π as a representation of an F -

group [14, pg. 48]. For simplicity, we write ξ′0(w) = (ξ1(w), ξ2(w)). Then, by our

choice of normalization, Π is parameterized by the map ψ : WF−→ LM, given

by

ψ(w) =

ξ1(w)ϕ0(w)Inϕ0(w)

ξ2(w)ϕ0(w)In

, w

.

Let ψ0 be the map from WF

to LM0. Thus, if (y, tz) ∈ Cn ⊕Cn, then

r1 ψ0(w) · (y,t z) = (ξ1(w)y, tzξ2(w)−1) = r′1 ξ′0(w) · (y, tz).

Similarly, r1(1, 1, 1, τ) = r′1(1, 1, τ). Therefore, L(s,Π, r1) = L(s, π, r′1) = L(s, π).

The lemma now follows from global considerations, as in the proof of Corollary

5.5.

By Lemma 6.1 and Proposition 5.11 of [5], L(s,Π, r1) = 1, for n 6= 1. If

n = 1, then, by [22], L(s, π) is holomorphic and non-zero at s = 0, unless

π = 1. Assume that if n = 1, then π 6= 1. By Corollary 7.6 of [18], we know that

I(Π) = IndGP (Π) is reducible if and only if Πw ' Π and L(s,Π, ri) has no pole at

s = 0 for i = 1 or 2. Since L(s,Π, r1) is holomorphic and non-zero at s = 0,

we know that I(Π) is reducible if and only if πε ' π, and L(s, π⊗ χµ,Ψn) has

no pole at s = 0.


Proposition 6.2. Let π be an irreducible unitary supercuspidal representation of

GL(n,E), with πε ' π, and let ν be a unitary character of E1. If n = 1, then

we assume that π 6= 1. Let G = U(n, n), P = MN, with M ' GL(n,E). Let

G′ = U(n, n+ 1), P ′ = M ′N ′, with M ′ ' GL(n,E)× U(1). Let Π = (π, ν) as

above. We denote by I(π) the representation of G induced from π, and I(Π)′

the representation of G′ induced from Π. Then I(π) is reducible if and only if

I(Π)′ is irreducible.

Proof. Since π ' πε, [10] implies L(s, π × π) has a simple pole at s = 0. By

Corollary 5.5, exactly one of L(s, π,Ψn) and L(s, π⊗χµ,Ψn) has a pole at s = 0.

The first L-function determines reducibility of I(π), while the second determines

the reducibility of I(Π)′. Therefore, the proposition holds.

For the remainder of this section we let G, G′, P, and P ′ be as in Proposition

6.2. Let ρ = ρθ be the half sum of the positive roots in N ′. Let α = en be the

unique simple root in N ′, and let α =< ρ, α >−1 ρ, where < ρ, α > is defined

as in Section 2. Then ∆ = β1, . . . , β2n, and θ = ∆ \ βn, βn+1. Here, as is

Section 2, βi = ei − ei+1 is the i-th simple, non-restricted root. Therefore,

ρθ =n+ 1

2

n∑j=1

j(βj + β2n+1−j),

and thus

< ρ, α >= (ρθ, βn) =n+ 1

2.

So, we have

q<α,HP ′ (m)>F

= |det(m)|E.

Therefore, I(s,Π)′ = IndG′

P ′(Π⊗ | det()|sE

).

Theorem 6.3. Let G′, and P ′ be as in Proposition 6.2. Let π be an irreducible

unitary supercuspidal representation of GL(n,E), and let Π = (π, ν) ∈ E(M ′).

Let κ be the trivial character of F×/NE× if n is odd, and the non-trivial

character if n is even.

(1) I(Π)′ is reducible if and only if π ' πε and there is some matrix coefficient

ϕ of π so that Φκε (δn, ϕ) 6= 0.

38 DAVID GOLDBERG*

(2) Suppose that if n = 1, then π 6= 1. If π ' πε and I(Π)′ is irreducible,

then the following hold.

(a) For 0 < s < 1/2, I(s,Π)′ is irreducible and unitarizable.

(b)The representation I(1/2,Π)′ is reducible. It has a unique generic

non-supercuspidal discrete series subrepresentation. It’s Langlands quotient

is degenerate, pre-unitary and non-tempered.

(c) If s > 1/2 then I(s,Π)′ is irreducible and never unitarizable.

(3) If n = 1 and π = 1, then I(Π)′ is irreducible, and the complementary

series is of length 1. That is, we replace “1/2” by “1” in (a)-(c) above.

(4) If I(Π)′ is reducible, then I(s,Π)′ is irreducible and never unitarizable

for s > 0.

Proof. (1) follows from Theorem 2.9 and Proposition 6.2. Parts (2) and (4)

follow from Theorem 8.1 of [18]. Part (3) follows from Corollary 5.5, part (2), and

Theorem 8.1 of [18]. Part (3) also follows from [11].

definition 6.4. Let π be an irreducible unitary supercuspidal representation of

GL(n,E) satisfying π ' πε. Let κ be the non-trivial character of F×/NE×.

Then we say that π is a standard lift from U(n) if there is a matrix coefficient

ϕ of π satisfying Φstε (δn, ϕ) 6= 0. We say that π is a κ-lift from U(n) if there

is a matrix coefficient ϕ of π so that Φκε (δn, ϕ) 6= 0.

In this language, Theorems 2.7, 2.8, 6.3(1), and Proposition 6.2 can be restated

as follows.

Theorem 6.5. Let G = U(n, n) and M = GL(n,E). Let G′ = U(n, n+1) and

M ′ = GL(n,E)×U(1). Let π ∈E(M) with π ' πε. If n = 1, then we suppose

that π 6= 1. Suppose n is odd. Then the following are equivalent:

(1) π is a standard lift from U(n),

(2) L(s, π,Ψn) has a pole at s = 0.

(3) π ⊗ χµ is not a standard lift from U(n).

(4) I(π) is irreducible and I(π, ν)′ is reducible for any character ν of U(1).

Suppose that n is even. Then (1)-(4) are valid if we replace “standard lift” by

“ κ-lift” in (1) and (3).


§7 Reducibility from discrete series.

We finally compute the reducibility of I(σ) for discrete series representations of

M. We first consider the group G = U(n, n) and M = GL(n,E). Suppose that

σ ∈ E2(M). If σ 6' σε, then I(σ) is irreducible [21]. So we assume that σ ' σε.Let a and b be integers with ab = n, such that σ is the unique discrete series

constituent of the representation induced from π1⊗· · ·⊗πb. Then σ ' σε implies

π0 ' πε0.

By [21], Corollary 5.4.2.3, we have to check whether µ(s, σ) has a zero at s = 0.

Here µ(s, σ) is the Plancherel measure attached to σ and s. By Corollary 3.6 of

[18], we need to check whether L(1 + s, σ,Ψn)L(s, σ,Ψn)−1 has a zero at s = 0.

Suppose that b is even. Then Theorem 5.6 and equations (5.6) and (5.7) imply

L(1, σ,Ψn)L(0, σ,Ψn)−1 =

(7.1)

b/2∏i=1

L(b+ 2− 2i, π0,Ψa)L(b+ 1− 2i, π0 ⊗ χµ,Ψa)

L(b+ 1− 2i, π0,Ψa)L(b− 2i, π0 ⊗ χµ,Ψa).

Since both L(s, π0,Ψa) and L(s, π0⊗χµ,Ψa) are holomorphic for Re s > 1, we

see that (7.1) is zero if and only if L(s, π0 ⊗ χµ,Ψa) has a pole at s = 0. By

Theorem 6.5 this holds if and only if π0 ⊗ χµ is a standard lift from U(a) if a

is odd, and a κ-lift from U(a) if a is even. Now suppose b is odd. Then using

Theorem 5.6, and making a computation similar to (7.1) we find that µ(s, σ) has

a zero at s = 0 if and only if L(s, π0,Ψa) has a pole at s = 0. Therefore I(σ)

is irreducible if and only if π0 is a standard lift from U(a) if a is odd, and a

κ-lift from U(a), otherwise.

Now suppose that G′ = U(n, n + 1) and M ′ = GL(n,E) × U(1). Let σ1 ∈

E2(M ′), and suppose that σ1 ' (σ, ν) = (σ⊗νdet)⊗ν. If σ 6' σε then I(σ1)′ is

irreducible. Let σ ∈ E2(GL(n,E)) be the discrete series constituent of an induced

representation as above. We again wish to determine whether µ(0, σ) = 0. By

Corollary 3.6 of [18] we need to examine the behavior of

(7.2) L(1 + s, σ1, r1)L(1 + 2s, σ1 ⊗ χµ,Ψn)L(s, σ1, r1)−1L(2s, σ ⊗ χµ,Ψn)−1

at s = 0. Here r1, is as in Section 6. If a > 1, then L(s, σ1, r1) ≡ 1. In this

case, our computations for U(n, n) above determine the zeros of (7.2). If b is

40 DAVID GOLDBERG*

even then (7.2) is zero if and only if π0 is a standard lift from U(a) if a is odd,

and a κ-lift from U(a) if a is even. If b is odd, then (7.2) is zero if and only if

π0 ⊗ χµ is a standard lift if a is odd, and a κ-lift if a is even.

Now suppose a = 1. If n = 1, then π0 = χ is a character of E×. Let

H = U(1, 1) and B be it’s Borel subgroup. Then IndHB (χ) is reducible if and

only if χ|F× = ωE/F . Therefore L(s, 1,Ψ1) has a pole at s=0, and L(s, χµ,Ψ1)

does not. By [11] IndGP (χ, ν) is reducible if and only if χ 6= 1 and χ|F× = 1.

Therefore, if χ = 1, then L(s, (χ, ν), r1) must have a pole at s = 0. Notice

that this is consistent with L(s, (χ, ν), r1) = L(s, χ), which has a pole at s = 0

if and only if χ = 1 [22]. If n > 1, then Proposition 7.11 of [5] shows that

L(s, σ1, r1) = L(s, π1), which cannot have a pole at s = 0.

Theorem 7.1. Let G = U(n, n), M = GL(n,E), G′ = U(n, n + 1), and

M ′ = GL(n,E) × U(1). Let σ, σ1, a, b, ν, and π0 be as above. Suppose

that σ ' σε. We first assume that if n = 1, then σ 6= 1. If both a and b are

odd, then I(σ) is irreducible if and only if π0 is a standard lift from U(a). If a

is even and b is odd then I(σ) is irreducible if and only if π0 is a κ-lift from

U(a). If a is odd and b is even, then I(σ) is irreducible if and only if π0 ⊗ χµis a standard lift from U(a). If a and b are both even then I(σ) is irreducible

if and only if π0 ⊗χµ is a κ− lift from U(a). Furthermore, I(σ) is irreducible

if and only if I(σ1)′ is reducible. Finally, if n = 1 and σ = 1, then both I(σ)

and I(σ1)′ are irreducible.

Proof. Only the next to last statement remains to be proved. But this follows

immediately from the above computations and Theorem 6.5.

1991 AMS Subject Classification: 11S40, 22E35, 22E50

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Department of Mathematics

Purdue University

West Lafayette, IN 47907

Documents

David Goldberg* - Purdue Universitygoldberg/papers/crelle.pdf · 2006. 6. 22. · 2 DAVID GOLDBERG* Section 2, we prove that if ˇw’ˇ; then I(ˇ) is irreducible if and only if