arXiv:2111.07591v1 [math.NT] 15 Nov 2021

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ARTHUR PACKETS FOR QUASISPLIT GSp(2n) AND GO(2n) OVER A p-ADIC

FIELD

BIN XU

Abstract. We construct the Arthur packets for symplectic and even orthogonal similitude groups over ap-adic field and show that they are stable and satisfy the twisted endoscopic character relations.

1. Introduction

Let F be a number field and G a quasisplit symplectic or special even orthogonal group over F . LetAF be the adele ring of F . We fix an automorphism θ0 of G preserving an F -splitting. It induces a dual

automorphism θ0 on the dual group G. When G is symplectic, θ0 is trivial. When G is special evenorthogonal, we require θ0 to be the unique nontrivial outer automorphism induced from the conjugationof the full orthogonal group. We say two irreducible admissible representations of G(AF ) are θ0-conjugateif they are θ0-conjugate at every place. In [Art13] Arthur proved that the discrete automorphic spectrumof G(AF ) can be decomposed as follows

L2disc(G(F )\G(AF )) =

⊕

[ψ]∈Ψ2(G)

⊕

[π]∈Πψ

mψ(π)π(1.1)

modulo θ0-conjugation. Here Ψ2(G) is the set of θ0-conjugacy classes of the discrete Arthur parametersof G, which can be understood in terms of automorphic representations of general linear groups. For ψ ∈Ψ2(G), Arthur associated a multi-set Πψ of isomorphism classes of irreducible admissible representationsof G(AF ) modulo θ0-conjugation and mψ(π) is the multiplicity of π contributed by [ψ] in the discretespectrum modulo θ0-conjugation. More precisely,

Πψ = ⊗′vΠψv .

Here Πψv is a finite multi-set of θ0-conjugacy classes of isomorphism classes of irreducible admissiblerepresentations of G(Fv). It is equipped with a map

Πψv → Sψv , [πv] 7→ 〈·, πv〉

where Sψv = π0(ZG(ψv)/Z(Gv)Γv ) and Sψv is the set of irreducible characters of Sψv . If πv is unramified,

then 〈·, πv〉 is trivial. For [π] ∈ Πψ, let

〈s, π〉 :=∏

v

〈sv, πv〉

through a natural homomorphism

Sψ → Sψv , s 7→ sv

Then

mψ(π) = mψ|Sψ|−1∑

s∈Sψ

〈s, π〉ǫψ(s)

where εψ is some linear character of Sψ and mψ = 1 or 2.

2010 Mathematics Subject Classification. 22E50 (primary); 11F70 (secondary).Key words and phrases. similitude group, twisted endoscopic transfer, Arthur packet.Supported by NSFC No. 20191300979 and Tsinghua University Initiative Scientific Research Program No. 2019Z07L02016.

1

http://arxiv.org/abs/2111.07591v1

2 BIN XU

Let G be the group of symplectic or orthogonal similitudes over F , whose derived group is G. Then θ0extends to an automorphism of G. We have a short exact sequence

1 // G // GλG

// Gm// 1,

where λG is the similitude character. On the dual side, we have

1 // C× //G

p// G // 1.

Let ζ be a character of ZG(AF )/ZG(F ) and ζ its restriction to ZG(AF ). We will parametrize the θ0-

conjugacy classes of the discrete Arthur parameters of G associated with central character ζ by pairs

(ψ, ω), where ψ ∈ Ψ2(G, ζ) and ω ∈ Y := Hom(G(AF )/G(F )ZG(AF )G(AF ),C×). There are exact

sequences

1 // Sψι

//

��

Sψα

//

��

Hom(G(AF )/G(F )G(AF ),C×)

��

1 // Sψvιv

// Sψvαv

// Hom(G(Fv)/G(Fv),C×).

(cf. [Xu21a, 2.18]). Then Arthur’s conjectural description of the discrete automorphic spectrum of G(AF )(cf. (1.1)) can be reformulated as follows.

Conjecture 1.1. For any ψ ∈ Ψ2(G), there exists a multi-set Πψ of irreducible admissible representations

of G(AF ) modulo θ0-conjugation, which is unique up to twist by Y , such that

L2disc(G(F )\G(AF ), ζ) =

⊕

ψ∈Ψ2(G,ζ)

⊕

ω∈Y/α(Sψ)

⊕

[π]∈Πψ⊗ω

mψ(π) π.

Here Πψ = ⊗′vΠψv . There is a correspondence

Πψv −→ Πψv , [πv] 7→ [πv]

such that πv ⊆ πv|G(Fv) and it fits into a commutative diagram

Πψv//

��

Sψv , [πv] 7→ 〈s, πv〉

Πψv // Sψv .

OO

At last,

mψ(π) = mψ|Sψ|−1∑

s∈Sψ

〈s, π〉εψ(s)

where εψ = εψ|Sψ .

This conjecture has been proved in the tempered case (cf. [Xu21a, Theorem 1.1]). The main goal ofthis paper is to construct Πψv at the nonarchimedean places.

Let F be a p-adic field, ΓF the absolute Galois group of F and WF the Weil group of F . Let LG :=

G⋊WF be the Langlands dual group of G. An Arthur parameter of G is a G-conjugacy class of admissiblehomomorphisms

ψ :WF × SL(2,C)× SL(2,C) −→ LG

such that ψ(WF ) is bounded. We denote the set of θ0-conjugacy classes of Arthur parameters of G (resp.

G) by Ψ(G) (resp. Ψ(G)). For [ψ] ∈ Ψ(G), Mœglin [Mœg11] showed that Πψ is multiplicity free. Let ζ be

a character of ZG(F ) extending the central character of Πψ and ˜Πψ,ζ the subset of θ0-conjugacy classes of

ARTHUR PACKETS FOR QUASISPLIT GSp(2n) AND GO(2n) OVER A p-ADIC FIELD 3

isomorphism classes of irreducible admissible representations of G(F ) with central character ζ, such that

their restrictions to G(F ) have irreducible constituents in Πψ. Suppose ψ = p ◦ ψ for ψ ∈ Ψ(G), we can

define a map ˜Πψ,ζ → Sψ such that it satisfies the following diagram

˜Πψ,ζ//

��

Sψ

Πψ // Sψ

OO

(cf. Corollary 5.2). Let H(G(F )) to be the space θ0-invariant smooth compactly supported functions on

G(F ). Let X = Hom(G(F )/G(F )ZG(F ),C×). For ψ = φ trivial on the second SL(2,C), we have shown

in [Xu18, Theorem 4.6] that there exists a subset Πφ of ˜Πφ,ζ unique up to twists by X such that

(1) ⊕

[π]∈Πφ

[π|G(F )] =⊕

[π]∈Πφ

[π],

(2)

f(φ) :=∑

[π]∈Πφ

fG(π), f ∈ H(G(F ))

is stable,(3) for any semisimple s ∈ Sφ and (G′, φ′) → (φ, s) (see (2.4)), we have

f ′(φ′) =∑

[π]∈Πφ

〈s, π〉fG(π), f ∈ H(G(F )),

where f ′ is the Langlands-Shelstad transfer of f and f ′(φ′) is defined as in (2) with respect tosome Πφ′ .

Our main result generalize this to Ψ(G).

Theorem 1.2. For [ψ] ∈ Ψ(G), there exists a subset Πψ of ˜Πψ,ζ such that

(1) ⊕

[π]∈Πψ

[π|G(F )] =⊕

[π]∈Πψ

[π],

(2)

f(ψ) :=∑

[π]∈Πψ

〈sψ, π〉 fG(π), f ∈ H(G(F ))

is stable, where sψ is the image of ψ(1, 1,−1) in Sψ,

(3) for semisimple s ∈ Sψ and (G′, ψ′) → (ψ, s) (see (2.4)), we have

f ′(ψ′) =∑

[π]∈Πψ

〈ssψ, π〉 fG(π), f ∈ H(G(F )).

where f ′ is the Langlands-Shelstad transfer of f and f ′(ψ′) is defined as in (2) with respect to Πψ′ .

This result is contained in Theorem 5.4, where we have also shown the twisted character relations underwhich Πψ is unique up to twist by X. One of our principle for constructing Πψ is that its elements should

have the same infinitesimal character, which is inspired by Conjecture 2.1. We follow the notion of infini-tesimal character introduced by Vogan in [Vog93], which depends on the local Langlands correspondence.In Section 2, we review this notion and study how the lift of infinitesimal character would determinethe lifts of Langlands parameters of elements in Πψ, through the geometry of Langlands parameters, cf.

4 BIN XU

[CFM+21]. In Section 4, we construct a candidate for the local Langlands correspondence of G, through

which we could define the infinitesimal characters for representations of G(F ) modulo θ0-conjugation. InSection 5, we formulate the main result (cf. Theorem 5.4) and treat the special case when the lift ofinfinitesimal character determine the lift of Langlands parameters of elements in Πψ. The key input ofthe proof is that the parabolic induction preserves the infinitesimal character (cf. Proposition 4.1). InSection 6, we treat the general case following Mœglin’s strategy [Mœg09] for constructing Πψ. It reducesto the case of discrete L-packets, which has already been treated in our earlier work [Xu18]. This sectionis the most difficult part of the paper. In the last section, we address the uniqueness part of the mainresult.

2. Infinitesimal character

Let G be a quasisplit connected reductive algebraic group over a p-adic field F . Let Π(G(F )) be the setof isomorphism classes of irreducible admissible representations of G(F ) and Φ(G) the set of Langlands

parameters of G, i.e., G-conjugacy classes of admissible homomorphisms from WF × SL(2,C) to LG. Weassume the local Langlands correspondence for G, i.e., there is a surjection

Π(G(F )) → Φ(G), π 7→ φπ

whose fibers are called L-packets. Then the infinitesimal character of π ∈ Π(G(F )) is defined to be that

of φπ, namely the G-conjugacy class of

λ : WF → LG, w 7→ φπ

(w,

(|w|1/2 0

0 |w|−1/2

)).

Fix an infinitesimal character λ. Let ZG(λ) be the centralizer of λ in G. The set Φ(G)λ of Langlandsparameters with infinitesimal character λ is in bijection with ZG(λ)-orbits in

Vλ :={x ∈ g

λ(IF )|Ad(λ(Fr))x = qx}

through the map

φ 7→ dφ|SL(2,C)

((0 10 0

)),

where g is the Lie algebra of G and q is the order of the residue field of F (cf. [CFM+21, Proposition4.2.2]). The number of orbits is finite and there is a unique open orbit (cf. [CFM+21, Proposition5.6.1]). For φ ∈ Φ(G)λ, let us denote the corresponding orbit by Cφ. We can also view λ as a Langlandsparameter, which is trivial on SL(2,C). It corresponds to the origin in Vλ. For φ ∈ Φ(G), we say it isbounded (resp. discrete) if φ(WF ) is bounded (resp. Z

G(φ) is finite). Denote the set of bounded (resp.

discrete) parameters by Φ(G) (resp. Φ2(G)). We have Φ2(G) ⊆ Φbdd(G).

Let Ψ(G) be the set of Arthur parameters of G, i.e., G-conjugacy classes of admissible homomorphismsfrom WF ×SL(2,C)×SL(2,C) to LG such that ψ(WF ) is bounded. Then we can view Φbdd(G) ⊆ Ψ(G).There is an inclusion

Ψ(G) → Φ(G), ψ 7→ φψ(w, x) := ψ(w, x,

(|w|

12

|w|−12

))

(cf. [CFM+21, Lemma 3.6.1]). We define the infinitesimal character of ψ to be that of φψ, denoted by λψ.Let Ψ(G)λ be the subset of Ψ(G) with infinitesimal character λ. For ψ ∈ Ψ(G), let ψd := ψ◦∆ ∈ Φbdd(G),where

∆ : WF × SL(2,C) → WF × SL(2,C)× SL(2,C), (w, x) 7→ (w, x, x).

We can also view ψd as an Arthur parameter, which is trivial on the second SL(2,C). Then ψd correspondsto the unique open orbit in Vλ (cf. [CFM+21, Proposition 6.1.1]). Let Π(G(F ))λ be the subset of Π(G(F ))with infinitesimal character λ. The following conjecture is suggested by [ABV92] [Vog93] [CFM+21].

Conjecture 2.1. For ψ ∈ Ψ(G), Πψ ⊆{π ∈ Π(G(F ))λ : Cφπ ⊇ Cφψ

}.


SupposeG is a quasisplit symplectic or special even orthogonal group over F and G is the corresponding

similitude group. We have G ∼= (Gm × G)/(Z/2Z), where Z/2Z is embedded diagonally into the centreof each factor. The similitude character λG is square on Gm and trivial on the other factor. We fix anautomorphism θ0 of G preserving an F -splitting. When G is symplectic, we require θ0 to be trivial. WhenG is special even orthogonal, we require θ0 to be the unique nontrivial outer automorphism induced from

the conjugation of the full orthogonal group. Clearly, θ20 = 1, θ0 extends to G by acting trivially on ZG,

and λG is θ0-invariant. It induces a dual automorphism θ0. Let Σ0 = 〈θ0〉 and Σ0 = 〈θ0〉. More generally,we can consider

G = G1 × · · · ×Gq(2.1)

where Gi is a quasisplit symplectic or special even orthogonal group. We define

G = (Gm ×G1 ×G2 × · · · ×Gq)/(Z/2Z),

where Z/2Z is embedded diagonally into the center of each factor. We also define a character λG

of G,which is square on Gm and trivial on the other factors. Then there is an exact sequence

1 // G // GλG

// Gm// 1.

On the dual side, we have

1 // C× //G

p// G // 1.

We can also view G as a subgroup of G1 × · · · × Gq by

G ∼={(gi) ∈

∏

i

Gi |λG1(g1) = · · · = λGq(gq)

}.

We define a group of automorphisms of G by taking the product of Σ0 on each factor, and we denote

this group again by Σ0. We can extend Σ0 to G by the trivial action on ZG. It induces a group Σ0 of

automorphisms ofG. Let

GΣ0 = G⋊ Σ0, GΣ0 = G⋊ Σ0,

GΣ0 = G⋊ Σ0,G

Σ0

=G⋊ Σ0.

For admissible representations πi of Gi(F ), we define the restriction of ⊗i πi to G(F ) by ⊗i πi. Let

Π(G(F )) be the set of Σ0-orbits in Π(G(F )) and Φ(G) the set of Σ0-orbits in Φ(G). We will also considerΠ(GΣ0(F )) the set of isomorphism classes of irreducible admissible representation of GΣ0(F ). There is anatural surjection from Π(GΣ0(F )) to Π(G(F )).

For G in (2.1), the local Langlands correspondence is known modulo Σ0-conjugation (cf. [Art13,Theorem 1.5.1]), namely we have a surjection with finite fibers

Π(G(F )) → Φ(G), [π] 7→ [φπ].

So we can define the Σ0-conjugacy classes of infinitesimal characters for Π(G(F )), which will be calledΣ0-infinitesimal characters. For πΣ0 ∈ Π(GΣ0(F )), we define its Σ0-infinitesimal character to that of its

restriction to G(F ). We fix an infinitesimal character λ of G and let λ = p ◦ λ. There is a surjection

Φ(G)λ → Φ(G)λ by the composition with p (cf. [Xu18, Theorem 2.9]). This can also be seen from the

natural isomorphism Vλ∼=−→ Vλ under p, which is compatible with the actions by Z

G(λ) → ZG(λ). Let

Φ(G)λ (resp. Φ(G)λ) be the subset of Φ(G) (resp. Φ(G)) with Σ0-infinitesimal character [λ] (resp. [λ]).

6 BIN XU

Then we have the following commutative diagram.

Φ(G)λ //

≃

��

Φ(G)λ

≃

��

Vλ/ZG(λ)// Vλ/ZGΣ0

(λ)

Φ(G)λ//

AA✄✄✄✄✄✄✄✄✄✄✄✄✄✄✄✄✄✄

≃

��

Φ(G)λ

??⑧⑧⑧⑧⑧⑧⑧⑧⑧⑧⑧⑧⑧⑧⑧⑧⑧⑧⑧

≃

��

Vλ/Z G(λ) //

AA☎☎☎☎☎☎☎☎☎☎☎☎☎☎☎☎☎☎☎

Vλ/Z G

Σ0 (λ)

@@��

.

Let H1(WF ,C×) := Im{H1(WF ,C×) → H1(WF , Z(G))}. For φ ∈ Φ(G)λ, let

X(φ) :={ω ∈ H1(WF ,C×) : φ⊗ ω = φ

}, XΣ0(φ) :=

{ω ∈ H1(WF ,C×) : [φ⊗ ω] = [φ]

},

which only depend on the image of φ in Φ(G)λ. Then

Φ(G)λ/X(λ) ∼= Φ(G)λ, Φ(G)λ/XΣ0(λ) ∼= Φ(G)λ.

So the fiber of

Φ(G)λ → Φ(G)λ

(resp. Φ(G)λ → Φ(G)λ

)

over φ (resp. [φ]) is in bijection with

X(λ)/X(φ)(resp. XΣ0(λ)/XΣ0(φ)

).

Let us identify Vλ with Vλ. Suppose φo ∈ Φ(G)λ corresponds to the unique open Z G(λ)-orbit in Vλ,

then this orbit must also be the unique open ZG(λ)-orbit in Vλ. Let φo ∈ Φ(G)λ be the corresponding

parameter. So λ determines a unique lift φo of φo, hence X(λ) = X(φo). By the uniqueness, we can also

conclude that this orbit is invariant under ZGΣ0

(λ). So [λ] determines a unique lift [φo] of [φo], hence

XΣ0(λ) = XΣ0(φo).

By the same argument as for Langlands parameters, one can show that there is a surjection Ψ(G)λ →

Ψ(G)λ through the composition with p. For ψ ∈ Ψ(G)λ, let

X(ψ) :={ω ∈ H1(WF ,C

×) : ψ ⊗ ω = ψ}, XΣ0(ψ) :=

{ω ∈ H1(WF ,C

×) : [ψ ⊗ ω] = [ψ]},

which only depend on the image of ψ in Ψ(G)λ. The fiber of

Ψ(G)λ → Ψ(G)λ

(resp. Ψ(G)λ → Ψ(G)λ

)

over ψ (resp. [ψ]) is in bijection with

X(λ)/X(ψ)(resp. XΣ0(λ)/XΣ0(ψ)

).

From the inclusion Ψ(G) → Φ(G), we have X(ψ) = X(φψ) and XΣ0(ψ) = XΣ0(φψ).

For ψ ∈ Ψ(G) and ψ = p ◦ ψ ∈ Ψ(G), we define

SΣ0ψ = ZGΣ0

(ψ), SΣ0

ψ= Z

G

Σ0 (ψ).

The short exact sequence

1 // C× //G

Σ0// GΣ0 // 1


induces a long exact sequence

1 // C× // SΣ0

ψ// SΣ0ψ

δ// H1(WF ,C×),

and hence

1 // SΣ0

ψ/C× ι

// SΣ0ψ

δ// H1(WF ,C×).

Taking quotient by Z(G)ΓF , we get

1 // SΣ0

ψ

ι// SΣ0ψ

δ// H1(WF ,C×),

where

SΣ0

ψ= SΣ0

ψ/Z(

G)ΓF , SΣ0

ψ = SΣ0ψ /Z(G)ΓF

and H1(WF ,C×) = H1(WF ,C×)/δ(Z(G)Γ). One can show that Im δ is finite as in [Xu18, Lemma 2.1],then S0

ψ= S0

ψ. After taking the quotients by the identity components, we get

1 // SΣ0

ψ

ι// SΣ0ψ

δ// H1(WF ,C×),(2.2)

where

SΣ0

ψ= SΣ0

ψ/S0

ψ, SΣ0

ψ = SΣ0ψ /S0

ψ.

There are natural maps from SΣ0ψ , SΣ0

ψ , and SΣ0ψ to Σ0, and for θ ∈ Σ0, we denote the preimages of θ ∈ Σ0

by Sθψ, Sθψ and Sθψ respectively. The above discussion also applies to φ ∈ Φ(G) and φ = p ◦ φ ∈ Φ(G).

To understand H1(WF ,C×), we take

1 // C× // Z(G) // Z(G) // 1

and it induces

Z(G)ΓFδ

// H1(WF ,C×) // H1(WF , Z(G)) // H1(WF , Z(G)) // H2(WF ,C×) = 1.

So

H1(WF ,C×) ∼= Im{H1(WF ,C

×) → H1(WF , Z(G))}

and

1 // H1(WF ,C×) // H1(WF , Z(G)) // H1(WF , Z(G)) // 1.(2.3)

We will identify H1(WF , D) ∼= Hom(G(F )/G(F ),C×) (cf. [Xu18, Section 2.2]) and denote δ by α. Indeed

the image of α lies in X = Hom(G(F )/G(F )ZG(F ),C×). Moreover, one can show as in [Xu16, Lemma

3.3] that

Lemma 2.2.

X(ψ) = α(Sψ), XΣ0(ψ) = α(SΣ0ψ ),

X(φ) = α(Sφ), XΣ0(φ) = α(SΣ0φ ).

For θ ∈ Σ0, any semisimple s ∈ Sθψ and ω = α(s), let H := ZG(s)0 and it can be equipped with a

Galois action given by ψ. This determines a quasisplit connected reductive group H, and ψ will factorthrough LH for some θ-twisted endoscopic datum (H, s, ξ) of G. Hence we get a parameter ψH ∈ Ψ(H).In this way, we call (H,ψH) corresponds to (ψ, s), and denote this relation by

(H,ψH) → (ψ, s).(2.4)

8 BIN XU

LetH be the preimage of H in

G and it can be equipped with a Galois action given by ψ. This determines

a quasisplit connected reductive group H, and ψ will factor through LH for some (θ, ω)-twisted endoscopic

datum (H, s, ξ) of G. Hence we get a parameter ψH ∈ Ψ(H).From [Xu21a, Section 2.1], we have a factorization

H ∼=∏

i

GL(ni)×G′

where G′ is of the form (2.1), and

H ∼=∏

i

GL(ni)× G′

We extend the action of Σ0 on G′ trivially to H. When H = G′, it is called elliptic. If ψH =∏i ψi × ψ′

for ψi ∈ Ψ(GL(ni)) and ψ′ ∈ Ψ(G′), then SΣ0

ψH∼= SΣ0

ψ′ and SΣ0

ψH∼= SΣ0

ψ′.

3. Parabolic induction and Jacquet module

From now on, let G be a quasisplit symplectic or special even orthogonal group over F and G thecorresponding similitude group. We define the split symplectic group Sp(2n) (resp. split special evenorthogonal group SO(2n)) in GL(2n) with respect to

(0 −JnJn 0

) (resp.

(0 JnJn 0

))for Jn =

11

. ..

1

We denote the outer twist of SO(2n) with respect to a quadratic extension E/F by SO(2n, ηE/F ), where

ηE/F is the quadratic character associated to E/F by the local class field theory. The Levi subgroup M

of G is isomorphic to

GL(n1)× · · · ×GL(nr)× G−,(3.1)

where G− is of the same type as G with semisimple rank n− > 0 and n =∑r

i=1 ni + n−. Throughout

this paper we fix a Borel subgroup B of G consisting of upper-triangular matrices and choose M to becontained in the group

GL(n1) 0. . .

GL(nr)

G−GL(nr)

. . .

0 GL(n1)

.

This gives all the standard Levi subgroups if G is GSp(2n) or GSO(2n, η) (η 6= 1), and GO(2n)-conjugacy

classes of standard Levi subgroups if G is GSO(2n). We fix an isomorphism from (3.1) to M as follows

(g1, · · · gr, g) −→ diag{g1, · · · , gr, g, λ(g)tg−1r , · · · , λ(g)tg

−11 }

if n− > 0, and

(g1, · · · gr, g) −→ diag{g1, · · · , gr, λ(g)tg−1r , · · · , λ(g)tg

−11 }

if n− = 0. Here tgi = JnitgiJ

−1ni for 1 6 i 6 r. If M is θ0-stable, we write MΣ0 := M ⋊ Σ0. Otherwise,

we let MΣ0 = M . Suppose σΣ0 ∈ Rep(MΣ0(F )), πΣ0 ∈ Rep(GΣ0(F )), we define the parabolic inductionand Jacquet module as follows.


• If Mθ0 = M , we define the normalized parabolic induction IndGΣ0 (F )

PΣ0 (F )σΣ0 to be the extension of

the representation IndG(F )

P (F )(σΣ0 |

M(F )) by an induced action of Σ0, and we define the normalized

Jacquet module JacPΣ0 (F )πΣ0 to be the extension of the representation JacP (F )(π

Σ0 |G(F )) by an

induced action of Σ0.

• If Mθ0 6= M , we define the normalized parabolic induction IndGΣ0 (F )

PΣ0 (F )σΣ0 to be Ind

GΣ0 (F )

G(F )Ind

G(F )

P (F )(σΣ0 |

M(F )),

and we define the normalized Jacquet module JacPΣ0 (F )

πΣ0 to be JacP (F )

(πΣ0 |G(F )

).

For σ = τ1 ⊗ · · · ⊗ τr ⊗ πΣ0− , where τi ∈ Rep(GL(ni, F )) for 1 6 i 6 r and πΣ0

− ∈ Rep(GΣ0− (F )),

we denote the normalized parabolic induction IndGΣ0 (F )

PΣ0(F )(σΣ0) by τ1 × · · · × τr ⋊ πΣ0

− and its socle by

〈τ1×· · ·×τr⋊ πΣ0− 〉. We also denote by τ1×· · ·×τr the normalized parabolic induction in GL(

∑ri=1 ni, F )

and its socle by 〈τ1 × · · · × τr〉.Suppose ρ is a unitary irreducible supercuspidal representation of GL(dρ, F ). For an increasing (resp.

decreasing) sequence {x, · · · , y} of real numbers of common distance 1, we denote

〈x, · · · , y〉ρ := 〈ρ||x × · · · × ρ||y〉.

Let St(ρ, a) := 〈a−12 , · · · ,−a−12 〉ρ. More generally, we denotex11 · · · x1n...

...xm1 · · · xmn

ρ

:= 〈×i∈[1,m]〈xi1, · · · , xin〉ρ〉

where each row is a decreasing (resp. increasing) and each column is an increasing (resp. decreasing) ofcommon distance 1. This includes the case that

Sp(St(ρ, a), b) := 〈St(ρ, a)||−(b−1)/2 × St(ρ, a)||−(b−3)/2 × · · · × St(ρ, a)||(b−1)/2〉.

Suppose M = GL(dρ)× G−. For πΣ0 ∈ Rep(GΣ0(F )), we can decompose the semisimplification of the

Jacquet module

s.s.JacPΣ0 (F )(πΣ0) =

⊕

i

τi ⊗ σΣ0i ,

where τi ∈ Rep(GL(dρ, F )) and σΣ0i ∈ Rep(GΣ0

− (F )), both of which are irreducible. We define Jacρx(πΣ0)

for any real number x to be

Jacρx(πΣ0) :=

⊕

τi=ρ||x

σΣ0i .

If we have an ordered sequence of real numbers {x1, · · · , xs}, we can define

Jacρx1,··· ,xs πΣ0 := Jacρxs ◦ · · · ◦ Jac

ρx1 π

Σ0 .

Lemma 3.1. For A,B ∈ R such that A−B ∈ Z\{0} and ζ a sign, let

πΣ0 = 〈ζB, · · · ,−ζA〉ρ ⋊ πΣ0− .

Then

JacρζB πΣ0 = 〈ζ(B − 1), · · · ,−ζA〉ρ ⋊ πΣ0

− + 〈ζB, · · · ,−ζA〉ρ ⋊ JacρζB πΣ0− ,

JacρζA πΣ0 = 〈ζB, · · · ,−ζ(A− 1)〉ρ ⋊ (πΣ0

− ⊗ ηρω−ζAρ ) + 〈ζB, · · · ,−ζA〉ρ ⋊ JacρζA π

Σ0− ,

where ωρ = | · |dρ and ηρ is the central character of ρ.

Proof. It follows from [Tad95, Theorem 5.2]. �

10 BIN XU

For

XT(ρ,A,B,ζ) :=

ζ(B + T ) · · · ζ(B + 1)

......

ζ(A+ T ) · · · ζ(A+ 1)

let |XT(ρ,A,B,ζ)| be the number of entries in XT

(ρ,A,B,ζ). We define

Jac(ρ,A+T,B+T,ζ)7→(ρ,A,B,ζ) := ◦x∈XT(ρ,A,B,ζ)

Jacρx,

where x ranges over XT(ρ,A,B,ζ) from top to bottom and left to right.

Let Rep(G(F )) be the category of finite-length smooth representations of G(F ) viewed as H(G(F ))-

modules. We denote the elements in Rep(G(F )) by [π] for π ∈ Rep(G(F )), and we call [π] is irreducibleif π is irreducible. Let

JacP (F )

=

{JacP (F ) + JacP (F ) ◦ θ0, if G = GSO(2n) and Mθ0 6= M ,

JacP (F )

, otherwise.

We can define parabolic induction and Jacquet module on Rep(G(F )) as follows

IndG(F )

P (F )[σ] := [Ind

G(F )

P (F )σ] and JacP (F )[π] := [JacP (F )π].

We can define Jacρx similarly. The above discussion also applies to G(F ) (cf. [Xu17a]).

4. Local Langlands correspondence for similitude groups

In order to define the Σ0-infinitesimal characters of Π(G(F )), we need the local Langlands correspon-

dence for Gmodulo θ0-conjugation. From [Xu21a, Theorem 8.12], we can construct such a correspondence,which depends on the choices for simple parameters of G. Although this correspondence is not canonical,it is compatible with parabolic induction and twisted endoscopic transfer. For our application, we wouldfurther require that the Σ0-infinitesimal characters are preserved under parabolic inductions, cf. Propo-sition 4.1. So we will make the following choices for simple parameters. Let φa := ρ ⊗ νa ∈ Φsim(Ga),where ρ is a self-dual irreducible unitary dρ-dimensional representation of WF and νa is the irreducible a-dimensional representation of SL(2,C). By the local Langlands correspondence for general linear groups[HT01] [Hen00] [Sch13], we can associate ρ with a self-dual irreducible unitary supercuspidal representa-tion of GL(dρ, F ), denoted again by ρ. Let ηρ be the central character of ρ. When ρ is of orthogonal type,it parametrizes an L-packet Πρ of Gρ(F ), which is a singleton. Then we fix a correspondence ρ 7→ Πρsuch that

(1) Πρ⊗ω = Πρ ⊗ ω for any ω ∈ H1(WF , Z(Gρ)) ∼= Hom(Gρ(F ),C×),

(2) the central character χρ of Πρ is given by the composition of ρ withGρ → Z

Gρ.

Next we construct the correspondence for φa by induction on a. Note Πφa is also a singleton and we have

Πφa → ρ||a−12 ⋊ Πφa−2

induced from M := GL(dρ)×Ga−2 (cf. [Xu17b, Proposition 8.1]). Suppose we have associated φa−2 with

a packet Πφa−2. Let φa be the lift of φa such that the Σ0-infinitesimal character is the same as that of

ρ||a−12 ⋊ φa−2. Then we associate φa with Πφa such that

Πφa → ρ||a−12 ⋊ Πφa−2

.

Proposition 4.1. For any Levi subgroup M of G and Σ0-infinitesimal character [λM] for M , we have

IndG(F )

P (F )Π(M(F ))λ

M⊆ Π(G(F ))λ

for [λ] = [ιM

◦ λM], where ι

M: LM → LG.


This kind of statement is due to Haines [Hai14]. In the case of classical groups, this has been provedby Moussaoui [Mou17]. In our case, the critical step is the following lemma.

Lemma 4.2. For [π] ∈ Π(G(F ))λ with cuspidal support (L, πcusp), if [λL] is the Σ0-infinitesimal character

of πcusp, then [λ] = [ιL◦ λ

L].

Proof. Let [φ] ∈ Φbdd(G) be the Langlands parameter of [π], then φ factors through φM ⊗ ξ ∈ Φ(M) for

some parabolic subgroup P of G with Levi factor M and π is the Langlands quotient of IndG(F )

P (F )(σ ⊗ ξ)

for some tempered representation [σ] ∈ ΠφM . It suffices to prove the lemma for [σ]. So we can assume π

is tempered. In this case, φ factors through φM ∈ Φ2(M) and π is a direct summand of IndG(F )

P (F )(σ) for

some discrete series representation [σ] ∈ ΠφM . Therefore we can further reduce to the discrete series case.

Suppose π is a non-cuspidal discrete series representation. Let [φ] := [p ◦ φ] ∈ Φ2(G). Then we can

find (ρ, a) ∈ Jord(φ) such that π− := Jacρ

a−12

π is an irreducible tempered representation of G−(F ) (cf.

Proposition 6.3). If φ = φa, then the lemma follows from our construction of the correspondence for φa.Otherwise, we can factor φ through φH := φa×φ

′ ∈ Φbdd(H) for a twisted endoscopic group H := Ga×G′

of G, where φa = ρ ⊗ νa if dρ is even (resp. φa = (ρ ⊗ ηρ) ⊗ νa if dρ is odd). Correspondingly, φ factors

through φH = p ◦ (φa × φ′) ∈ Φbdd(H) for a twisted endoscopic group H of G. Let φ− ∈ Φbdd(G−) factor

through φH−= φa−2 × φ′ for a twisted endoscopic group H− := Ga−2 × G′ of G−. Let φ− ∈ Φbdd(G−)

factor through φH−= p◦ (φa−2 × φ

′) for the corresponding twisted endoscopic group H−. Here φa−2 is so

chosen that φa has the same Σ0-infinitesimal character as that of ρ||a−12 ⋊ φa−2 (resp. ρ⊗ ηρ||

a−12 ⋊ φa−2).

The twisted endoscopic embeddings are so chosen that

LH // LG

L(GL(dρ)× H−)

OO

// L(GL(dρ)× G−)

OO

commutes. Then φ has the same Σ0-infinitesimal character as that of ρ||a−12 ⋊ φ−. We claim that [φ−]

is the Langlands parameter of [π−]. First of all, Πφa−2= Jac

ρa−12Πφa (resp. Πφa−2

= Jacρ⊗ηρa−12

Πφa) from

our construction. Secondly, Πφ = Tran (Πφa ⊗ Πφ′) and Πφ− = Tran (Πφa−2⊗ Πφ′) (cf. [Xu21a, Theorem

8.12]). At last, it follows from the compatibility of the twisted endoscopic transfer with Jacquet module[Xu17b, Appendix C] that Πφ− = Jac

ρa−12Πφ. Hence [π−] ∈ Πφ− . We may assume that the lemma holds

for π− by induction on the rank of groups, then the rest is clear. �

Remark 4.3. In [Xu17b, Appendix C], we have only considered the twist by automorphism. Our case can

be deduced from there by taking M+ := G×GL(1) with the automorphism θ+0 (g, x) := (θ0(g), λ(g)x−1).

Note a representation π ⊗ ω of M+ is θ+0 -invariant if and only if πθ0 ⊗ ω ∼= π.

Now we can complete the proof of Proposition 4.1.

Proof. Suppose the cuspidal support of [σ] ∈ Π(M(F ))λM

is (L, σcusp). Then this is also the cuspidal

support of π. Let [λL] be the Σ0-infinitesimal character of σcusp. Then by this lemma, the Σ0-infinitesimal

character of π is [λ] = [ιL◦ λ

L] = [ι

M◦ ιM

L◦ λ

L] = [ι

M◦ λ

M]. �

Corollary 4.4. For any Levi subgroup M of G and Σ0-infinitesimal character [λM] for M , we have

IndGΣ0 (F )

PΣ0 (F )Π(MΣ0(F ))λ

M⊆ Π(GΣ0(F ))λ

for [λ] = [ιM

◦ λM], where ι

M: LM → LG.

12 BIN XU

At last, for πΣ0 ∈ Π(GΣ0(F )) and π ∈ Π(G(F )) such that π ⊆ πΣ0 |G(F )

, let

X(πΣ0) = {ω ∈ X | πΣ0 ⊗ ω ∼= πΣ0},

X(π) = {ω ∈ X | π ⊗ ω ∼= π},

X([π]) = {ω ∈ X | [π ⊗ ω] = [π]}.

We have X(πΣ0) = X([π]).

Lemma 4.5. Suppose φ ∈ Φ(G) and πΣ0 ∈ ΠΣ0

φ, [π] ∈ Πφ such that π ⊆ πΣ0 |

G(F ), then

X(πΣ0) = XΣ0(φ), X(π) = X(φ).

Proof. It follows from Lemma 2.2 and [Xu18, Corollary 4.2]. �

5. Statement of main results

For [ψ] ∈ Ψ(G) and λ = λψ, Arthur [Art13, Theorem 1.5.1] associated it with a multiset Πψ overΠ(G(F )). It follows from Mœglin’s multiplicity one result [Mœg11] and [Art13, (2.2.12)] that Πψ is asubset of Π(G(F ))λ. It is equipped with a map

Πψ −→ Sψ, [π] 7→ 〈·, π〉W(5.1)

such that

(1)

fW (ψ) :=∑

[π]∈Πψ

〈sψ, π〉W fG(π), f ∈ H(G)

is stable.(2) For θ ∈ Σ0, s ∈ Sθψ and (H,ψH ) → (ψ, s), we have

fHW (ψH) =∑

[π]∈Πψ

〈ssψ, π〉W fG(π), f ∈ H(G),(5.2)

where fH is the transfer of f and fHW (ψ) is defined with respect to ΠψH := ⊗iΠψi⊗Πψ′ for ψH =∏i ψi×ψ

′

with ψi ∈ Ψ(GL(ni)) and ψ′ ∈ Ψ(G′).

For ε ∈ Sψ, let πW (ψ, ε) be the direct sum of the preimages of ε under (5.1). We define ΠΣ0ψ to be the

set of irreducible representations of GΣ0(F ), whose restriction to G(F ) belong to Πψ. If SΣ0ψ 6= Sψ, then

πθ0 ∼= π for any irreducible constituent [π] in πW (ψ, ε) (cf. [Xu17a, Section 8]). So we can define

ΠΣ0ψ −→ SΣ0

ψ , πΣ0 7→= 〈·, πΣ0〉W(5.3)

and for ε ∈ SΣ0ψ , let πΣ0

W (ψ, ε) be the direct sum of the preimages of ε under (5.3) such that

• [πΣ0W (ψ, ε)|G(F )] = 2πW (ψ, ε) if G is special even orthogonal and SΣ0

ψ = Sψ, or πW (ψ, ε) otherwise.

• For any s ∈ SΣ0ψ but not in Sψ and (H,ψH) → (ψ, s), the following identity holds

fHW (ψH) =∑

[π]∈Πψ

〈ssψ, πΣ0〉W fG(π

Σ0), f ∈ C∞c (G(F )⋊ θ0).(5.4)

where πΣ0 is any preimage of [π].

Recall that there is an exact sequence

1 // SΣ0

ψ

ι// SΣ0ψ

α// Hom(G(F )/G(F ),C×) .(5.5)

Lemma 5.1.


(1) πW (ψ, ε1), πW (ψ, ε2) are conjugate under G(F ) if and only if ε1/ε2 is trivial on Sψ.

(2) πΣ0W (ψ, ε1), π

Σ0W (ψ, ε2) are conjugate under GΣ0(F ) if and only if ε1/ε2 is trivial on SΣ0

ψ.

Proof. Since fW (ψ) is stable, then Πψ is stable under G(F )-conjugation. Moreover, the character relation

(5.2) implies that for any g ∈ G(F ) and x ∈ Sψ,∑

ε∈Sψ

ε(sψx)πW (ψ, ε)g−1

=∑

ε∈Sψ

ωx(g)ε(sψx)πW (ψ, ε),

where ωx(g) = α(x)(g) (cf. [Xu16, Lemma 3.13]). Note sψ = sψ, so ωsψ(·) = 1. It follows that

πW (ψ, ε)g = πW (ψ, ω(·)(g) ε).(5.6)

Since α induces an injection Sψ/Sψ → Hom(G(F )/G(F ),C×), then

G(F )/G(F ) → Hom(Sψ/Sψ, Z/2Z), g 7→ ω(·)(g)

is surjective. Part (1) follows from this. Part (2) also follows when SΣ0ψ = Sψ.

Now let us suppose SΣ0ψ 6= Sψ, then πθ0 ∼= π for any irreducible constituent [π] in πW (ψ, ε). The

character relation (5.4) implies that for any g ∈ G(F ) and x ∈ Sθ0ψ ,∑

ε∈Sψ

ε(sψx) fG(πΣ0W (ψ, ε)g

−1) =

∑

ε∈Sψ

ωx(g)ε(sψx) fG(πΣ0W (ψ, ε)), f ∈ C∞c (G(F ) ⋊ θ0).

Combining with (5.6), we get

πΣ0W (ψ, ε)g = πΣ0

W (ψ, ω(·)(g) ε).(5.7)

Since α induces an injection SΣ0ψ /SΣ0

ψ→ Hom(G(F )/G(F ),C×), then

G(F )/G(F ) → Hom(SΣ0ψ /SΣ0

ψ, Z/2Z), g 7→ ω(·)(g)

is surjective. Part (2) follows from this. �

Let ζ be a character of ZG(F ) whose restriction to ZG(F ) is the central character of Πψ. Let˜Πψ,ζ (resp.

ΠΣ0

ψ,ζ) be the subset of Π(G(F )) (resp. Π(GΣ0(F ))) with central character ζ, such that the restrictions to

G(F ) (resp, GΣ0(F )) belong to Πψ (resp. ΠΣ0ψ ).

Corollary 5.2. There exist unique maps

˜Πψ,ζ → Sψ and ΠΣ0

ψ,ζ→ SΣ0

ψ(5.8)

such that

˜Πψ,ζ//

��

Sψ

Πψ // Sψ

OO

and ΠΣ0

ψ,ζ//

��

SΣ0

ψ

ΠΣ0ψ

// SΣ0ψ

OO

commute respectively.

Proof. For πΣ0 ∈ ΠΣ0

ψ,ζ, we choose πΣ0 in the restriction of πΣ0 . Then we define

〈·, πΣ0〉W := 〈·, πΣ0〉W |SΣ0ψ

.

By Lemma 5.1, we see that 〈·, πΣ0〉W is independent of the choice of πΣ0 . The commutativity of thediagram is clear from our definition. The uniqueness is also clear. By restriction, we can deduce the othercase. �

14 BIN XU

Lemma 5.3. For [π] ∈ Πψ, X(ψ) ⊆ X(φπ) and XΣ0(ψ) ⊆ XΣ0(φπ). Moreover for any ω ∈ H1(WF ,C×),

ψθ0 = ψ ⊗ ω ⇒ φθ0π = φπ ⊗ ω.(5.9)

Proof. Let G(π) (resp. G([π])) be the stabilizer of π (resp. [π]) in G(F ) under the conjugation action.By the refinement of L-packets in the case of even orthogonal groups [Art13, Theorem 8.4.1], we have

G(π) = G([π]). Since

G([π]) ⊆ G(πW (ψ, ε)) =⋂

ω∈α(Sψ)

Kerω,

then

X(φπ) = X(π) = (G/G(π))∗ ⊇ (G/G(πW (ψ, ε)))∗ = α(Sψ) = X(ψ).

This also settles (5.9) in the case when θ0 = id. For the remaining cases, it suffices to assume SΣ0ψ 6= Sψ.

Then πθ0 ∼= π. Suppose ψθ0 = ψ ⊗ ω, there exists x0 ∈ Sθ0ψ such that ω = α(x0) = ωx0 . From (5.7), we

can deduce that for g ∈ G(π) and f ∈ C∞c (G(F ) ⋊ θ0),

〈x0, (πΣ0)g〉W fG(π

Σ0) = ωx0(g)〈x0, πΣ0〉W fG(π

Σ0) = ωx0(g)〈x0, (πΣ0)g〉W fG((π

Σ0)g).

Hence

fG((πΣ0)g) = ωx0(g)

−1fG(πΣ0).

From this we can conclude that πθ0 ∼= π⊗ωx0 as in the proof of [Xu16, Proposition 6.16]. This completes

the proof of (5.9). As a consequence, we get XΣ0(ψ) ⊆ XΣ0(φπ). �

Let ZF be a closed subgroup of ZG(F ) such that λ(ZF ) has finite index in F× and ZF = ZF ∩G(F ).

Let χ = ζ|ZF and χ = χ|ZF . Define H(G(F ), χ) (resp. H(G(F ), χ)) to be the space χ−1 (resp. χ−1)-

equivariant θ0-invariant smooth compactly supported functions on G(F ) (resp. G(F )).

Theorem 5.4. For [ψ] ∈ Ψ(G), there exists a subset Πψ of ˜Πψ,ζ unique up to twisting by X, such that

the following properties are satisfied.

(1) For ˜ε ∈ Sψ, let πW (ψ, ˜ε) be the direct sum of the preimages of ˜ε in Πψ under (5.8), then

πW (ψ, ˜ε)|G(F ) =⊕

ε∈Sψ: ε|Sψ=˜ε

πW (ψ, ε).

(2)

fW (ψ) :=∑

[π]∈Πψ

〈sψ, π〉W fG(π), f ∈ H(G(F ), χ)

is stable.(3) Suppose θ ∈ Σ0, s ∈ Sθψ with ω = α(s) and (H,ψH) → (ψ, s), fix ΠψH := ⊗iΠψi ⊗ Πψ′ for

ψH =∏i ψi × ψ′ with ψi ∈ Ψ(GL(ni)) and ψ′ ∈ Ψ(G′), then we can choose Πψ such that

f HW (ψH) =∑

[π]∈Πψ

fGθ,W

(π, ω), f ∈ H(G(F ), χ)(5.10)

where fGθ,W

(π, ω) = tr(π(f) ◦ Aπ(θ, ω)), and Aπ(θ, ω) is an intertwining operator between π ⊗ ω

and πθ, which is normalized in a way so that if f is the restriction of f on G(F ), then

(f |ZFG(F )

)Gθ,W

(π, ω) =∑

π⊆π|G(F )

〈sψs, π+〉W fGθ(π)(5.11)

where π+ is an extension of π to G+(F ) = G(F )⋊ 〈θ〉 and fGθ(π) = tr(π(f) ◦ π+(θ)).


Let πW (ψ, ˜ε) := πW (ψ, ε) for any ε ∈ Sψ such that ε|Sψ= ˜ε. Then πW (ψ, ε1) = πW (ψ, ε2) if and only

if ε1/ε2 is trivial on Sψ. Moreover,

πW (ψ, ε)|G(F ) =⊕

ε′∈(Sψ/Sψ)∧

πW (ψ, εε′).

We define ΠΣ0

ψto be the set of irreducible representations of GΣ0(F ), whose restrictions to G(F ) belong

to Πψ. If SΣ0

ψ6= Sψ, i.e., ψ

θ0 = ψ, then πθ0 ∼= π for any [π] ∈ ˜Πψ,ζ by (5.9). So we can define

ΠΣ0

ψ−→ SΣ0

ψ, πΣ0 7→ 〈·, πΣ0〉W(5.12)

and for ε ∈ SΣ0

ψ, let πΣ0

W (ψ, ε) be the direct sum of the preimages of ε under (5.12) such that

• [πΣ0W (ψ, ε)|

G] = 2πW (ψ, ˜ε) if G is special even orthogonal and SΣ0

ψ= Sψ, or πW (ψ, ˜ε) otherwise.

• For any s ∈ SΣ0

ψbut not in Sψ and (H,ψH) → (ψ, s), the following identity holds

f HW (ψH) =∑

[π]∈Πψ

〈ssψ, πΣ0〉W fG(π

Σ0), f ∈ C∞c (G(F )⋊ θ0).

Similarly we will define πΣ0W (ψ, ε) := πΣ0

W (ψ, ε) for ε = ε|SΣ0ψ

. Then πΣ0W (ψ, ε1) = πΣ0

W (ψ, ε2) if and only if

ε1/ε2 is trivial on SΣ0

ψ. Moreover,

πΣ0W (ψ, ε)|GΣ0 =

⊕

ε′∈(SΣ0ψ /S

Σ0ψ

)∧

πΣ0W (ψ, εε′).

5.1. A special case. Suppose [ψ] ∈ Ψ(G) satisifies XΣ0(λ) = XΣ0(ψ). Then for any [π] ∈ Πψ, we have

XΣ0(λ) = XΣ0(φπ) by Lemma 5.3, i.e., there exists a unique [π] ∈ Π(G(F ))λ such that [π] is in therestriction of [π] (cf. Section 2). So we can define

Πψ :={[π] ∈ Π(G(F ))λ : [π]|G(F ) ⊆ Πψ

}.

Theorem 5.5. For [ψ] ∈ Ψ(G) such that XΣ0(λ) = XΣ0(ψ), Πψ satisfies (1), (2) in Theorem 5.4.

Proof. Since XΣ0(λ) = XΣ0(ψ), the elements in Πψ can be distinguished from Π(G(F ))λ by their restric-

tions to G(F ). Let us write

fW (ψ) =∑

[φ]∈Φ(G)λ

m(ψ, φ)f(φ), f ∈ H(G(F ), χ),

where φ factors through φM ⊗ ξ for φM ∈ Φbdd(M),

f(φ) =∑

[σ]∈ΠφM

fG(IndG(F )P (F ) σ ⊗ ξ)

(cf. [Art13, (2.2.12)]). We form a stable distribution on G(F ) by∑

φ∈Φ(G)λ

m(ψ, φ)f(φ), f ∈ H(G(F ), χ)(5.13)

where m(ψ, φ) = |X(λ)/X(φ)|−1m(ψ, φ) and φ factors through φM ⊗ ξ for φM ∈ Φbdd(M ),

f(φ) =∑

[σ]∈ΠφM

fG(IndG(F )

P (F )σ ⊗ ξ).

16 BIN XU

It is clear that the restriction of (5.13) to ZFG(F ) is fW (ψ). We would like to show that it is equal to

fW (ψ). By Proposition 4.1, the irreducible constituents of IndG(F )

P (F )σ⊗ ξ for σ ∈ Πφ

Mand [φ] ∈ Φ(G)λ all

belong to Π(G(F ))λ. Since (5.13) is invariant underXΣ0(λ), then for any [π] ∈ Π(G(F ))λ that contributes

to (5.13), [π ⊗ ω] must have the same coefficients in (5.13) for all ω ∈ XΣ0(λ). So their restrictions toG(F ) do not cancel. This means that [π] ∈ Πψ. The rest is clear by considering the restriction of (5.13)

to ZFG(F ) again.�

Theorem 5.6. Suppose [ψ] ∈ Ψ(G) satisifies XΣ0(λ) = XΣ0(ψ). For θ ∈ Σ0, s ∈ Sθψ with ω = α(s) and

(H,ψH ) → (ψ, s), let λH = λψH . If XΣ0(ψH) = XΣ0(λH), then (5.10) holds.

Proof. Let us write

fHW (ψH) =∑

φH∈Φ(H)λH

m(φH , ψH)fH(φH), f ∈ H(G(F ), χ),

where fH is the transfer of f , then we can show as in Theorem 5.5 that

f HW (ψH) =∑

φH∈Φ(H)λH

m(ψH , φH)fH(φH), f ∈ H(G(F ), χ),

where m(ψH , φH) = |X(λH)/X(φH )|−1m(ψH , φH) and f H is the transfer of f . Note the restriction of

f HW (ψH) to ZFG(F ) is fHW (ψH). Since f

HW (ψH) is invariant under X

Σ0(λ), we can further show as before

that it is supported on Πψ. Then the rest is clear by considering the restriction of f HW (ψH) to ZFG(F )again. �

Remark 5.7. In case that XΣ0(λ) = 1, the condition in Theorem 5.6 is always satisfied.

6. Construction

Mœglin [Mœg06b] [Mœg09] give a construction of Arthur packets for classical groups, which reduces

to the tempered case. Since we have already constructed the tempered packets of GΣ0(F ) in [Xu18], the

idea is to extend Moeglin’s construction to GΣ0(F ).

6.1. Combinatorial description of Arthur parameters. Let N = 2n+1 if G = Sp(2n) and N = 2n

if G = SO(2n, ηE/F ). Let ω0 be the character of GΣ0(F )/G(F ) ∼= GΣ0(F )/G(F ). For [ψ] ∈ Ψ(G), we getan equivalence class of N -dimensional self-dual representation of WF ×SL(2,C)×SL(2,C) by composingwith a GL(N,C)-conjugacy class of twisted endoscopic embedding ξG : LG −→ GL(N,C) such that

• ξG|G is the standard representation of G,• ξG|WF

is trivial if N is odd, and factors through ΓE/F with the nontrivial element sent to areflection if N is even.

Then we can decompose ψ into a direct sum of irreducible subrepresentations

ψ =

r⊕

i=1

liψi =

r⊕

i=1

li(ρi ⊗ νai ⊗ νbi).(6.1)

Here ρi are equivalence classes of irreducible unitary representations of WF and νai (resp. νbi) are the(ai − 1)-th (resp. (bi − 1)-th) symmetric power representations of SL(2,C). The irreducible constituentρi ⊗ νai ⊗ νbi has dimension ni = n(ρi,ai,bi) and multiplicity li. Let ηi = ηaibiρi . We define the multi-set ofJordan blocks for ψ as follows,

Jord(ψ) := {(ρi, ai, bi) with multiplicity li : 1 6 i 6 r}.

Moreover, for any ρ let us define

Jordρ(ψ) := {(ρ′, a′, b′) ∈ Jord(ψ) : ρ′ = ρ}.


One can define the parity for self-dual irreducible unitary representations ρ of WF as in ([Xu17b], Section3). Then we say (ρi, ai, bi) is of orthogonal type if ai + bi is even when ρi is of orthogonal type, andai + bi is odd when ρi is of symplectic type. Similarly we say (ρi, ai, bi) is of symplectic type if ai + biis odd when ρi is of orthogonal type, and ai + bi is even when ρi is of symplectic type. Let ψp be the

parameter whose Jordan blocks consist of those in Jord(ψ) with the same parity as G, and let ψnp beany parameter of general linear groups such that

ψ = ψnp ⊕ ψp ⊕ ψ∨np,

where ψ∨np is the dual of ψnp. We also denote by Jord(ψ)p the set of Jordan blocks in Jord(ψp) without

multiplicity. Let s0 = (s0,i) ∈ Z/2ZJord(ψ)p be defined as s0,i = 1 if li is even and s0,i = −1 if li is odd.Then

SΣ0ψ

∼= {s = (si) ∈ Z/2ZJord(ψ)p}/〈s0〉,

andSψ ∼= {s = (si) ∈ Z/2ZJord(ψ)p :

∏

i

(si)ni = 1}/〈s0〉

if G is special even orthogonal. Under this interpretation, one can show

α : SΣ0ψ → Hom(G(F )/G(F ),C×), s 7→ (

∏

i:si=−1

ηi) ◦ λG(6.2)

as in [Xu18, Lemma 6.9]).

There is a natural inner product on Z/2ZJord(ψ)p which identify its dual with itself. Let ε = (εi) and

s = (si) be two elements in Z/2ZJord(ψ)p , then their inner product is defined by ε(s) =∏i(εi ∗ si), where

εi ∗ si =

{−1, if εi = si = −1,

1, otherwise.

So on the dual side,

SΣ0ψ

∼= {ε = (εi) ∈ Z/2ZJord(ψ)p :∏

i

εlii = 1}.

When G is special even orthogonal, let ε0 = (ε0,i) ∈ Z/2ZJord(ψ)p be defined as ε0,i = 1 if ni is even, or

ε0,i = −1 if ni is odd, then ε0 ∈ SΣ0ψ is always trivial when restricted to Sψ, and

Sψ ∼= {ε = (εi) ∈ Z/2ZJord(ψ)p :∏

i

εlii = 1}/〈ε0〉.

In general, we can let ε0 = 1 if G is not special even orthogonal.There is a natural projection

Z/2ZJord(ψp) Cont// Z/2ZJord(ψ)p

s ✤// s′

(6.3)

such thats′(ρ, a, b) =

∏

(ρ′,a′,b′)∈Jord(ψp)(ρ′,a′,b′)=(ρ,a,b) in Jord(ψ)p

s(ρ′, a′, b′)

for (ρ, a, b) ∈ Jord(ψ)p. In particular, s0 has a natural representative s>0 in Z/2ZJord(ψp) given bys>0 (ρ, a, b) = −1 for all (ρ, a, b) ∈ Jord(ψp). We define

SΣ0

ψ> = {s(·) ∈ Z/2ZJord(ψp)}/〈s>0 〉,

andSψ> = {s(·) ∈ Z/2ZJord(ψp) :

∏

(ρ,a,b)∈Jord(ψp)

s(ρ, a, b)n(ρ,a,b) = 1}/〈s>0 〉

18 BIN XU

if G is special even orthogonal. Then there are surjections SΣ0

ψ> → SΣ0ψ and Sψ> → Sψ.

On the dual side, we have a natural inclusion

Z/2ZJord(ψ)p �� Ext

// Z/2ZJord(ψp)

ε ✤ // ε′

such thatε′(ρ, a, b) = ε(ρ, a, b)

for (ρ, a, b) ∈ Jord(ψp). We define an inner product on Z/2ZJord(ψp) as for Z/2ZJord(ψ)p . Then

SΣ0

ψ>∼= {ε(·) ∈ Z/2ZJord(ψp) :

∏


ε(ρ, a, b) = 1},

andSψ> ∼= {ε(·) ∈ Z/2ZJord(ψp) :

∏


ε(ρ, a, b) = 1}/〈ε0〉

if G is special even orthogonal. There are inclusions SΣ0ψ → SΣ0

ψ> and Sψ → Sψ> . For ε ∈ SΣ0

ψ> , we denote

its image in Sψ> by ε.Suppose ψ = ψp. For (ρ, a, b) ∈ Jord(ψ), let us write A = (a + b)/2 − 1, B = |a − b|/2, and set

ζ = ζa,b = Sign(a− b) if a 6= b and arbitrary otherwise. Then we can denote (ρ, a, b) also by (ρ,A,B, ζ).We would like to impose some total order >ψ on Jordρ(ψ) for each ρ. We say >ψ is “admissible” if itsatisfies

(P) : ∀(ρ,A,B, ζ), (ρ,A′, B′, ζ ′) ∈ Jord(ψ) with A > A′, B > B′ and ζ = ζ ′,

then (ρ,A,B, ζ) >ψ (ρ,A′, B′, ζ ′).

For a fixed admissible order >ψ, we have introduced εMW/Wψ , ε

M/MWψ ∈ SΣ0

ψ> in [Xu17a]. For [π] ∈ Πψ

and s ∈ SΣ0

ψ> , we define

〈s, π〉MW := εMW/Wψ (s)〈s, π〉W , 〈s, π〉M := ε

M/MWψ (s)〈s, π〉MW .

For s ∈ SΣ0

ψ> , it gives a partition of

Jord(ψ) = Jord+ ⊔ Jord−.

depending on s(ρ, a, b) = 1 or −1. Let

NI =∑

(ρ,a,b)∈Jord+

n(ρ,a,b), NII =∑

(ρ,a,b)∈Jord−

n(ρ,a,b)

andηI =

∏

(ρ,a,b)∈Jord+

η(ρ,a,b), ηII =∏

(ρ,a,b)∈Jord−

η(ρ,a,b).

Then ψ factors through ψH := ψI × ψII ∈ Ψ(H) for a twisted elliptic endoscopic group H = GI ×GII ofG, where

GI = SO(NI), GII = SO(NII)

and ψI ∈ Ψ(GI), ψII ∈ Ψ(GII) such that

Jord(ψI) =

{{(ρ⊗ ηI , a, b) : (ρ, a, b) ∈ JordI} if NI is odd

JordI if NI is even

Jord(ψII) =

{{(ρ⊗ ηII , a, b) : (ρ, a, b) ∈ JordII} if NII is odd

JordII if NII is even


Comparing with (2.4), this amounts to choosing a special representative of s in SΣ0ψ . So we will say that

(H,ψH ) → (ψ, s).

6.2. Elementary case. Suppose [ψ] ∈ Ψ(G) is elementary, i.e., [ψd] ∈ Φ2(G) and A = B for all(ρ,A,B, ζ) ∈ Jord(ψ) (cf. [Xu17a, Section 6]). For simplicity, we denote by Jordρ(ψd) the set of in-tegers α such that (ρ, α, 1) ∈ Jord(ψd), and we write (ρ, α, δα) for (ρ, (α− 1)/2, (α− 1)/2, δα) ∈ Jord(ψ).We always impose the natural order >ψ, i.e., (ρ, α, δα) >ψ (ρ, α′, δα′) if α > α′. Let λ = λψ. In this case,we have

X(ψ) = X(ψd) = X(λ) and XΣ0(ψ) = XΣ0(ψd) = XΣ0(λ).

by Lemma 2.2 and (6.2). So this case has been covered in Theorem 5.5. Theorerm 5.6 also applies to thiscase for ψH = ψI × ψII , where ψI , ψII are both elementary. Moreover,

ΠΣ0ψ −→ SΣ0

ψ

is a bijection, cf. [Xu17a, Section 6.4]. Let us define

πΣ0MW (ψ, ε) := πΣ0

W (ψ, εεMW/Wψ ), πΣ0

M (ψ, ε) := πΣ0MW (ψ, εε

M/MWψ )


W (ψ, εεMW/W ), πΣ0M (ψ, ε) := πΣ0

MW (ψ, εεM/MW ).

Let ρ be a unitary irreducible supercuspidal representation of GL(dρ, F ) and PΣ0dρ

the set of Σ0-

conjugacy classes of standard parabolic subgroups P of G, whose Levi component M is isomorphicto

GL(a1dρ)× · · · ×GL(aldρ)×G−.(6.4)

If dρ 6= 1, we require G− 6= SO(2). For P ∈ PΣ0dρ

and σΣ0 ∈ Rep(MΣ0(F )), we denote by (σΣ0)ρ<x0 the

direct sum of irreducible constitutes of σΣ0 whose cuspidal support on the general linear factors consistonly of ρ||x with |x| < x0. In particular, when G− = SO(2) ∼= GL(1), we also impose this condition onG−. Define

JacPΣ0(F )(πΣ0) =

{JacPΣ0 (F )(π

Σ0)⊗ ω0, if G− = SO(2),

JacPΣ0 (F )(πΣ0), otherwise.

For X0 ∈ N, x0 = (X0 − 1)/2, define

invρ<X0(πΣ0) :=

∑

P∈PΣ0dρ

(−1)dimAM IndGΣ0 (F )

PΣ0 (F )(JacPΣ0(F )(π

Σ0)ρ<x0).

Define invρ6X0

(πΣ0) similarly. Let ψ♯ be obtained from ψ by changing δα to −δα for all α ∈ Jordρ(ψ)

such that α < X0. Mœglin [Mœg06b, Theorem 5] showed that

ε(sψ)β(ψ, ρ,< X0) invρ<X0

πΣ0M (ψ, ε) = ε(sψ♯)π

Σ0M (ψ♯, ε)

where β(ψ, ρ,< X0) is a sign, cf. [Xu17a, Section 6.2]. By taking |invρ<X0| (resp. |invρ

6X0|), we forget the

sign. Then

πΣ0M (ψ, ε) = ◦(ρ,a,δa)∈Jord(ψ)

δa=−1

(|invρ<a| ◦ |invρ6a|)π

Σ0W (ψd, ε).(6.5)

We would like to extend this result to GΣ0(F ). Define

invρ

<X0(πΣ0) :=

∑

P∈PΣ0dρ

(−1)dimAM IndGΣ0 (F )

PΣ0 (F )(JacPΣ0 (F )(π

Σ0)ρ<x0 )

and invρ

6X0(πΣ0) similarly. We denote inv

ρ

∞ if X0 is taken to infinity.

Lemma 6.1.

ε(sψ)β(ψ, ρ,< X0) invρ

<X0πΣ0M (ψ, ε) = ε(sψ♯)π

Σ0M (ψ♯, ε).

20 BIN XU

Proof. It suffices to show any πΣ0 in IndGΣ0 (F )

PΣ0 (F )(Jac

PΣ0(F )πΣ0M (ψ, ε))ρ<x0 such that X(πΣ0) ( XΣ0(λ) is

cancelled in invρ

<X0πΣ0M (ψ, ε). If not, then it follows from{ω ∈ X | inv

ρ

<X0πΣ0M (ψ, ε)⊗ ω = inv

ρ

<X0πΣ0M (ψ, ε)

}= XΣ0(λ)

that the restriction of πΣ0 to GΣ0(F ) is not cancelled in

(ε(sψ)β(ψ, ρ,< X0) invρ

<X0πΣ0M (ψ, ε))|GΣ0 (F ) = ε(sψ)β(ψ, ρ,< X0) inv

ρ<X0

(πΣ0M (ψ, ε)|GΣ0 (F ))

=∑

ε′∈(SΣ0ψ /S

Σ0ψ

)∧

ε(sψ)β(ψ, ρ,< X0) invρ<X0

πΣ0M (ψ, εε′) =

∑

ε′∈(SΣ0ψ♯/S

Σ0ψ♯

)∧

ε(sψ♯)πΣ0M (ψ♯, εε′).

Here we have used the fact that sψ = sψ and sψ♯ = sψ♯ . Since XΣ0(ψ♯) = XΣ0(λ), then the irreducible

constituents of πΣ0 |GΣ0 (F ) can not belong to πΣ0M (ψ♯, εε′). So we get a contradiction. �

As a consequence, we have

Corollary 6.2.

πΣ0M (ψ, ε) = ◦(ρ,a,δa)∈Jord(ψ)

δa=−1

(|invρ

<a| ◦ |invρ

6a|) πΣ0W (ψd, ε).(6.6)

6.3. Case of discrete diagonal restriction.

6.3.1. Construction. Suppose [ψ] ∈ Ψ(G) has discrete diagonal restriction, i.e., [ψd] ∈ Φ2(G) (cf. [Xu17a,Section 7]). We always impose the natural order i.e., (ρ,A,B, ζ) >ψ (ρ,A′, B′, ζ ′) if A > A′. Let


W (ψ, εεMW/Wψ ), πΣ0

M (ψ, ε) := πΣ0MW (ψ, εε

M/MWψ ).

Suppose there exists (ρ,A,B, ζ) ∈ Jord(ψ) such that A > B. For ε ∈ SΣ0ψ , let η0 = ε(ρ,A,B, ζ) and

Mœglin [Mœg09] (also see [Xu17a, Theorem 7.14]) showed that

πΣ0M (ψ, ε) =⊕C∈]B,A] (−1)A−C〈ζB, · · · ,−ζC〉ρ ⋊ Jacρζ(B+2),··· ,ζC π

Σ0M

(ψ′, ε′, (ρ,A,B + 2, ζ; η0)

)

⊕η=± (−1)[(A−B+1)/2]ηA−B+1ηA−B0 πΣ0M

(ψ′, ε′, (ρ,A,B + 1, ζ; η), (ρ,B,B, ζ; ηη0)

).

Let J(ψ) be the set of ρ appearing in Jord(ψ). Our idea is to construct a family of Arthur packets forthe similitude groups of various ranks, so that they satisfy similar recursive formulas. Let

XC := 〈ζB, · · · ,−ζC〉ρ ⋊ Jacρζ(B+2),··· ,ζC πΣ0M

(ψ′, ε′, (ρ,A,B + 2, ζ; η0)

)⊗ ηCρ ω

ζγB(C)ρ χρ,

Xη := πΣ0M

(ψ′, ε′, (ρ,A,B + 1, ζ; η), (ρ,B,B, ζ; ηη0)

),

where ωρ = | · |dρ , γB(C) = B+12 +

((B + 2) + · · ·+C

), and χρ is the central character of Πρ for dρ even

(resp. Πρ⊗ηρ

for dρ odd) when ρ is of orthogonal type and trivial otherwise. If X+ 6= X−, we define

πΣ0M (ψ, ε) = ⊕C∈]B,A](−1)A−CXC ⊕η=± (−1)[(A−B+1)/2]ηA−B+1ηA−B0 Xη.(6.7)

If X+ = X−, then it is necessary that ρ is of orthogonal type with ηρ 6= 1 and a, b are even, so A−B +1is also even, and we define

πΣ0M (ψ, ε) = ⊕C∈]B,A](−1)A−CXC ⊕ (−1)(A−B+1)/2η0X+.(6.8)

The construction reduces to the elementary case, which further reduces to the case of discrete L-packetsby (6.6). So it suffices to specify a family of discrete L-packets. If ρ ∈ J(ψ) is of orthogonal type, we fix


Πρ for dρ even (resp. Πρ⊗ηρ

for dρ odd). In the case that ρ ∈ J(ψ) and φa = ρ ⊗ νa for dρ even (resp.

φa = (ρ⊗ ηρ)⊗ νa for dρ odd), we define Πφa inductively by requiring

Πφa → ρ||a−12 ⋊ (Πφa−2

⊗ ω− a−1

4ρ ) for dρ even

(resp. Πφa → ρ⊗ ηρ||

a−12 ⋊ (Πφa−2

⊗ ω− a−1

4ρ ) for dρ odd

).

For φ ∈ Φ2(G) such that J(φ) ⊆ J(ψ), we will define Πφ through a sequence of twisted endoscopic

transfers. According to [Xu21a, Section 8.3], we need to specify the factorization of φ and liftings ofthe twisted endoscopic embeddings, which depend on some choices of 1-cochains of WF in C×. WhenJ(φ) = {ρ}, we choose 1-cochains of ΓE/F in C× for E/F associated with ηρ. By [Xu21a, Corollary 8.5

and Theorem 8.12], the resulting packet Πφ is independent of the choice of factorizations of φ. In general,

we fix an order on J(ψ) and decompose φ = ⊕ρ∈J(ψ)φρ such that J(φρ) = {ρ}. Then we will factorφ by factoring out φρ one by one according to this order. To lift the twisted endoscopic embeddings,we will choose 1-cochains of ΓE/F in C× for E/F associated with ηφρ whenever possible, and fix somearbitrary choices depending on ρ otherwise. We can further make these choices independent of the discreteparameters φ relevant in the reductions of Πψ. Denote the product of 1-cochains for defining Πψd by cψ.

At last, Πφ determines ΠΣ0

φ. Under these constructions we have the analogue of [Xu17b, Proposition 9.3]

for GΣ0(F ).

Proposition 6.3. Let φ ∈ Φ2(G) such that J(φ) ⊆ J(ψ) and ε ∈ SΣ0φ . For any (ρ, a, 1) ∈ Jord(φ), we

denote by a− the biggest positive integer smaller than a in Jordρ(φ), denote by amin the minimum ofJordρ(φ). If a = amin, we let a− = 0 if a is even, and −1 otherwise. In this case, we always assumeε(ρ, a, 1)ε(ρ, a− , 1) = −1.

(1) If ε(ρ, a, 1)ε(ρ, a−, 1) = −1 and a− < a− 2, then

πΣ0(φ, ε) → 〈(a− 1)/2, · · · , (a− + 3)/2〉ρ ⋊ πΣ0(φ′, ε′)⊗ ω−((a−1)+···+(a−+3))/4ρ(6.9)

as the unique irreducible subrepresentation, where

Jord(φ′) = Jord(φ) ∪ {(ρ, a− + 2, 1)}\{(ρ, a, 1)},

and

ε′(·) = ε(·) over Jord(φ)\{(ρ, a, 1)}, ε′(ρ, a− + 2, 1) = ε(ρ, a, 1).

(2) If ε(ρ, a, 1)ε(ρ, a−, 1) = 1, then

πΣ0(φ, ε) → 〈(a− 1)/2, · · · ,−(a− − 1)/2〉ρ ⋊ πΣ0(φ′, ε′)⊗ ω−((a−1)+···+(a−+1))/4ρ χρ,(6.10)

where

Jord(φ′) = Jord(φ)\{(ρ, a, 1), (ρ, a− , 1)},

and ε′(·) is the restriction of ε(·).(3) If ε(ρ, amin, 1) = 1 and amin is even, then

πΣ0(φ, ε) → 〈(amin − 1)/2, · · · , 1/2〉ρ ⋊ πΣ0(φ′, ε′)⊗ ω−((amin−1)+···+1)/4ρ(6.11)

as the unique irreducible subrepresentation, where

Jord(φ′) = Jord(φ)\{(ρ, amin, 1)},

and ε′(·) is the restriction of ε(·).

Proof. Comparing with [Xu17b, Proposition 9.3], it suffices to show

Jacρ(a−1)/2,··· ,(a−+3)/2 Πφ = Πφ′ ⊗ ω−((a−1)+···+(a−+3))/4

ρ

for (1), and

Jacρ(a−1)/2,··· ,(a−+1)/2 Πφ = 〈(a− − 1)/2, · · · ,−(a− − 1)/2〉ρ ⋊ Πφ′ ⊗ ω−((a−1)+···+(a−+1))/4

ρ

22 BIN XU

for (2), and

Jacρ(amin−1)/2,··· ,1/2 Πφ = Πφ′ ⊗ ω−((amin−1)+···+1)/4

ρ

for (3). Here (1) and (3) follow directly from our construction of the L-packets and the compatibility oftwisted endoscopic transfer with Jacquet module. By the same argument, we know in case (2) that

Jacρ(a−1)/2,··· ,(a−+1)/2 Πφ = Tran ΠφH ⊗ ω−((a−1)+···+(a−+1))/4

ρ

where φH = p ◦ (φa− × φ′′) and Jord(φ′′) = Jord(φ)\{(ρ, a)}. By [Xu21a, Lemma 8.2], Tran ΠφH differs

from〈(a− − 1)/2, · · · ,−(a− − 1)/2〉ρ ⋊ Πφ′

by the central character χφa−of Πφa−

. Since χφa = ηρχφa−2for any a ∈ Z, then χφa−

= η(a−−1)/2ρ χρ. In

view of πΣ0(φ, ε) ⊗ ηρ ∼= πΣ0(φ, ε), we can drop η(a−−1)/2ρ . �

We could also define πM (ψ, ε) in a similar way. Let

Jord(ψ1) = Jord(ψ′) ∪ {(ρ,A,B + 2, ζ)},

andJord(ψ2) = Jord(ψ′) ∪ {(ρ,A,B + 1, ζ), (ρ,B,B, ζ)}.

We can identify Sψ ∼= Sψ1 by sending (ρ,A,B, ζ) to (ρ,A,B + 2, ζ) if A > B + 1. It induces Sψ∼= Sψ1 .

We also map s ∈ Sψ into Sψ2 by letting

s(ρ,A,B + 1, ζ) = s(ρ,B,B, ζ) := s(ρ,A,B, ζ).

Then Sψ → Sψ2 is of index 1 or 2. It also induces Sψ → Sψ2 of index 1 or 2. We denote the image of ε

in Sψ1 by ε1. Let us define

πM (ψ, ε) := ⊕C∈]B,A] (−1)A−C〈ζB, · · · ,−ζC〉ρ ⋊ Jacρζ(B+2),··· ,ζC πM (ψ1, ε1)⊗ ηCρ ω

ζγB(C)ρ χρ

⊕ ˜ε← ˜ε2∈Sψ2(−1)[(A−B+1)/2]ε2(ρ,A,B + 1, ζ)A−B+1ε(ρ,A,B, ζ)A−B πM (ψ2, ε2).

By induction, one can show [πΣ0M (ψ, ε)|

G] = 2πM (ψ, ε) if G is special even orthogonal and SΣ0

ψ= Sψ, or

πM (ψ, ε) otherwise.

Lemma 6.4. The definitions of πΣ0M (ψ, ε) and πM (ψ, ε) are independent of the choice of (ρ,A,B, ζ) such

that A > B.

Proof. Suppose there exists another Jordan block (ρ′, A′, B′, ζ ′) ∈ Jord(ψ) such that A′ > B′. We can

substitute in πΣ0M (ψ, ε) the recursive formulas for πΣ0

M (ψ′, ε′, (ρ,A,B + 2, ζ; η0)) and πΣ0M (ψ′, ε′, (ρ,A,B +

1, ζ; η), (ρ,B,B, ζ; ηη0)) with respect to (ρ′, A′, B′, ζ ′). To simplify the result, we can use the facts that

Jacρζ(B+2),··· ,ζC

(〈ζ ′B, · · · ,−ζ ′C ′〉ρ′ ⋊ Jacρ

′

ζ′(B′+2),··· ,ζ′C′ πΣ0M

(ψ′′, ε′′, (ρ,A,B + 2, ζ; η0), (ρ

′, A′, B′ + 2, ζ ′; η′0)))

=〈ζ ′B, · · · ,−ζ ′C ′〉ρ′ ⋊ Jacρζ(B+2),··· ,ζC ◦ Jacρ′

ζ′(B′+2),··· ,ζ′C′ πΣ0M

(ψ′′, ε′′, (ρ,A,B + 2, ζ; η0), (ρ

′, A′, B′ + 2, ζ ′; η′0))

and〈ζB, · · · ,−ζC〉ρ × 〈ζ ′B′, · · · ,−ζ ′C ′〉ρ′ ∼= 〈ζ ′B′, · · · ,−ζ ′C ′〉ρ′ × 〈ζB, · · · ,−ζC〉ρ

and Jacρζ(B+2),··· ,ζC commutes with Jacρ′

ζ′(B′+2),··· ,ζ′C′ . Then the result would be the same if we first define

πΣ0M (ψ, ε) with respect to (ρ′, A′, B′, ζ ′) and expand further with respect to (ρ,A,B, ζ). The case of

πM (ψ, ε) follows by restricting πΣ0M (ψ, ε) to G(F ).

�

Lemma 6.5.

(1) For ε′ ∈ (SΣ0ψ /SΣ0

ψ)∧ (resp. ε′ ∈ (Sψ/Sψ)

∧),

πΣ0M (ψ, ε) = πΣ0

M (ψ, εε′) (resp. πM (ψ, ε) = πM (ψ, εε′) ).


(2)

πΣ0M (ψ, ε)|GΣ0 (F ) =

⊕

ε′∈(SΣ0ψ /S

Σ0ψ

)∧

πΣ0M (ψ, εε′).(6.12)

πM (ψ, ε)|G(F ) =⊕

ε′∈(Sψ/Sψ)∧

πM(ψ, εε′).(6.13)

Proof. It follows from the recursive formulas and the results in the discrete case. �

So far we have only defined πΣ0M (ψ, ε) in the Grothendieck group, and it is by no means clear that this

defines a representation. Indeed, Mœglin first defined πΣ0M (ψ, ε) by the recursive formula and then showed

that it is a representation by direct computation (cf. [Mœg09, Theorem 4.1]). We will follow the samestrategy below.

Theorem 6.6. Let

πΣ0M (ψ, ε)main =

⟨〈ζB, · · · ,−ζA〉ρ ⋊ πΣ0

M

(ψ′, ε′, (ρ,A − 1, B + 1, ζ; η0)

)⊗ ηAρ ω

ζ((B+1)+···+A)/2ρ χρ

⟩

and

πΣ0M (ψ, ε)com,η = πΣ0

M

(ψ′, ε′,∪C∈[B,A](ρ,C,C, (−1)Cη)

)

for

η ∈ Sη0 :={η = ± | η0 =

∏

C∈[B,A]

(−1)Cη}.

If πΣ0M (ψ, ε)com,η 6= πΣ0

M (ψ, ε)com,−η , then


M (ψ, ε)main ⊕η∈Sη0πΣ0M (ψ, ε)com,η .(6.14)

Otherwise,


M (ψ, ε)main ⊕ πΣ0M (ψ, ε)com,η(6.15)

for any η ∈ Sη0 .

Corollary 6.7. πΣ0M (ψ, ε) is a representation of GΣ0(F ).

We define ΠΣ0

ψto be the set of irreducible constituents of

⊕

ε∈Sψ

πΣ0M (ψ, ε).

From [Mœg09, Theorem 4.2] (also see [Xu17a, Theorem 7.8]), the irreducible constitutes of πΣ0M (ψ, ε) can

be parametrized by pairs of integer-valued functions (l, η) over Jord(ψ), such that

l(ρ,A,B, ζ) ∈ [0, [(A −B + 1)/2]] and η(ρ,A,B, ζ) ∈ {±1},(6.16)

and

ε(ρ,A,B, ζ) = εl,η(ρ,A,B, ζ) := η(ρ,A,B, ζ)A−B+1(−1)[(A−B+1)/2]+l(ρ,A,B,ζ).(6.17)

Moreover,

πΣ0M (ψ, l, η) → ×(ρ,A,B,ζ)∈Jord(ψ)

ζB · · · −ζA...

...ζ(B + l(ρ,A,B, ζ) − 1) · · · −ζ(A− l(ρ,A,B, ζ) + 1)

ρ

× πΣ0M

(∪(ρ,A,B,ζ)∈Jord(ψ) ∪C∈[B+l(ρ,A,B,ζ),A−l(ρ,A,B,ζ)](ρ,C,C, ζ; η(ρ,A,B, ζ)(−1)C−B−l(ρ,A,B,ζ))

)

24 BIN XU

as the unique irreducible subrepresentation. There is an obvious equivalence relation to be made here onpairs (l, η), namely

(l, η) ∼Σ0 (l′, η′)

if and only if l = l′ and (η/η′)(ρ,A,B, ζ) = 1 unless l(ρ,A,B, ζ) = (A−B + 1)/2. Then

πΣ0M (ψ, ε) =

⊕

{(l,η): ε=εl,η}/∼Σ0

πΣ0M (ψ, l, η).(6.18)

We define πΣ0M (ψ, l, η) to be the element in ΠΣ0

ψcontaining πΣ0

M (ψ, l, η) in its restriction to GΣ0(F ). Then

πΣ0M (ψ, l, η) → ×(ρ,A,B,ζ)∈Jord(ψ)

ζB · · · −ζA...

...ζ(B + l(ρ,A,B, ζ) − 1) · · · −ζ(A− l(ρ,A,B, ζ) + 1)

ρ

⋊ πΣ0M


)

⊗(ρ,A,B,ζ)∈Jord(ψ) η(A−(l(ρ,A,B,ζ)−1)/2)l(ρ,A,B,ζ)ρ ω(A+B+1)(A−B−l(ρ,A,B,ζ)+1)l(ρ,A,B,ζ)/2

ρ χl(ρ,A,B,ζ)ρ

as the unique irreducible subrepresentation. Moreover, πΣ0M (ψ, l, η) = πΣ0

M (ψ, l′, η′) if and only if l = l′ and

πΣ0M


)

= πΣ0M

(∪(ρ,A,B,ζ)∈Jord(ψ) ∪C∈[B+l′(ρ,A,B,ζ),A−l′(ρ,A,B,ζ)](ρ,C,C, ζ; η

′(ρ,A,B, ζ)(−1)C−B−l′(ρ,A,B,ζ))

).

This defines an equivalence relation ∼GΣ0on (l, η). Hence

πΣ0M (ψ, ε) =

⊕

{(l,η): ε=εl,η}/∼GΣ0

πΣ0M (ψ, l, η).(6.19)

In the rest of this section, we will prove Theorem 6.6. Let us assume the theorem holds for∑

(ρ,A,B,ζ)∈Jord(ψ)(A−

B) < K. In our discussions below, we consider∑

(ρ,A,B,ζ)∈Jord(ψ)(A−B) 6 K. Before starting the proof,

we make the following preparations.

Lemma 6.8. The representations on the right hand sides of (6.14), (6.15) are independent of the choiceof (ρ,A,B, ζ) for A > B.

Proof. By the induction assumption, we can conclude that the right hand sides of (6.14), (6.15) are equalto that of (6.19), which is independent of the choice of (ρ,A,B, ζ) ∈ Jord(ψ) for A > B. �

Lemma 6.9. Suppose ψT 0 is obtained from ψ by shifting (ρ,A,B, ζ) to (ρ,A+ T0, B+ T0, ζ) such that ithas discrete diagonal restriction and any Jordan block in Jordρ(ψ) between the two has sign −ζ. Then

Jac(ρ,A+T0,B+T0,ζ)7→(ρ,A,B,ζ) πΣ0M (ψT0 , εT0) = πΣ0

M (ψ, ε)⊗ ω−ζ(|X

T0(ρ,A,B,ζ)

|)/2ρ .(6.20)

where εT0 is related to ε by the change of order formula (cf. [Xu19, Theorem 6.3]).

Proof. We will prove this by induction on∑

(ρ,A,B,ζ)∈Jord(ψ)(A − B). If∑

(ρ,A,B,ζ)∈Jord(ψ)(A − B) = 0,

i.e., ψ is elementary, then πΣ0M (ψT0 , εT0) is a representation by definition (cf. Theorem 5.5). Since

Jac(ρ,A+T0,B+T0,ζ)7→(ρ,A,B,ζ) πΣ0M (ψT0 , εT0) = πΣ0

M (ψ, ε),

it suffices to know

Jac(ρ,A+T0,B+T0,ζ)7→(ρ,A,B,ζ) ΠψT0= Πψ ⊗ ω

−ζ(|X(ρ,A,B,ζ)|)/2ρ ,

which follows from the compatibility of twisted endoscopic transfer with Jacquet modules. If there exists(ρ′, A′, B′, ζ ′) ∈ Jord(ψ) distinct from (ρ,A,B, ζ) such that A′ > B′. Then we can expand πΣ0

M (ψT0 , εT0)


according to the recursive formula with respect to (ρ′, A′, B′, ζ ′). Since we can take Jac(ρ,A+T0,B+T0,ζ)7→(ρ,A,B,ζ)

in front of

πΣ0M

(ψ′T0 , ε

′T0 , (ρ,A

′, B′ + 2, ζ ′; η′T0,0)), πΣ0

M

(ψ′T0 , ε

′T0 , (ρ,A

′, B′ + 1, ζ ′; η′), (ρ,B′, B′, ζ ′; η′η′T0,0)),

then the result follows from the induction assumption.Now we may assume (ρ,A,B, ζ) is the only Jordan block such that A > B. For 0 6 T 6 T0, let us

define

πΣ0M (ψT , εT ) := Jac(ρ,A+T0,B+T0,ζ)7→(ρ,A+T,B+T,ζ) π

Σ0M (ψT0 , εT0)⊗ ω

−ζ(|XT0−T

(ρ,A+T,B+T,ζ)|)/2

ρ .(6.21)

Note the restriction of πΣ0M (ψT , εT ) to G

Σ0(F ) is⊕

ε′∈(SΣ0ψT

/SΣ0ψT

)∧

πΣ0M (ψT , εT ε

′).

We claim that πΣ0M (ψT , εT ) can be expressed as (6.7) or (6.8), where

XT,C := 〈ζ(B + T ), · · · ,−ζ(C + T )〉ρ ⋊ Jacρζ(B+T+2),··· ,ζ(C+T )

πΣ0M

(ψ′T , ε

′T , (ρ,A + T,B + T + 2, ζ; ηT,0)

)⊗ ηC+T

ρ ωζγB+T (C+T )ρ χρ

for B + 1 6 C 6 A and

πΣ0M

(ψ′T , ε

′T , (ρ,A+ T,B + T + 2, ζ; ηT,0)

)= Jac(ρ,A+T0,B+T0+2,ζ)7→(ρ,A+T,B+T+2,ζ)

πΣ0M

(ψ′T , ε

′T , (ρ,A+ T0, B + T0 + 2, ζ; ηT,0)

)⊗ ω

−ζ(|XT−T0(ρ,A+T,B+T+2,ζ)

|)/2ρ

and

XT,η :=Jac(ρ,A+T0,B+T0,ζ)7→(ρ,A+T,B+T,ζ) XT0,η ⊗ ω−ζ(|X

T−T0(ρ,A+T,B+T,ζ)

|)/2ρ

=Jac(ρ,A+T0,B+T0+1,ζ)7→(ρ,A+T,B+T+1,ζ) ◦ Jac(ρ,B+T0,B+T0,ζ)7→(ρ,B+T,B+T,ζ)

XT0,η ⊗ ω−ζ(|X

T−T0(ρ,B+T,B+T,ζ)

|)/2ρ ω

−ζ(|XT−T0(ρ,A+T,B+T+1,ζ)

|)/2ρ .

Then the lemma follows from the case T = 0 and the induction assumption.At last, we shall prove our claim by induction on T0 − T . It suffices to show that

Jacρζ(B+T ),··· ,ζ(A+T ) XT,C = XT−1,C ⊗ ω−ζ((B+T )+···+(A+T ))/2ρ(6.22)

and

Jacρζ(B+T ),··· ,ζ(A+T ) XT,η = XT−1,η ⊗ ω−ζ((B+T )+···+(A+T ))/2ρ .(6.23)

The equality (6.23) is clear from the definition. For (6.22), we write

Jacρζ(B+T ),··· ,ζ(A+T )XT,C = Jacρζ(C+T+1),··· ,ζ(A+T ) ◦ Jacρζ(B+T+1),··· ,ζ(C+T ) ◦ Jac

ρζ(B+T )XT,C .

First we have

Jacρζ(B+T )XT,C = 〈ζ(B + T − 1), · · · ,−ζ(C + T )〉ρ ⋊ Jacρζ(B+T+2),··· ,ζ(C+T )

πΣ0M

(ψ′T , ε

′T , (ρ,A + T,B + T + 2, ζ; ηT,0)

)⊗ ηC+T

ρ ωζγB+T (C+T )ρ χρ.

Next we claim that

Jacρζ(B+T+1),··· ,ζ(C+T ) ◦ Jacρζ(B+T+2),··· ,ζ(C+T )π

Σ0M

(ψ′T , ε

′T , (ρ,A + T,B + T + 2, ζ; ηT,0)

)= 0.(6.24)

Since πΣ0M

(ψ′T , ε

′T , (ρ,A + T,B + T + 2, ζ; ηT,0)

)is a representation, it suffices to show

Jacρζ(B+T+1),··· ,ζ(C+T ) ◦ Jacρζ(B+T+2),··· ,ζ(C+T )π

Σ0M

(ψ′T , ε

′T , (ρ,A+ T,B + T + 2, ζ; ηT,0)

)= 0.

26 BIN XU

For any irreducible constituent πΣ0 in πΣ0M

(ψ′T , ε

′T , (ρ,A + T,B + T + 2, ζ; ηT,0)

), we have

πΣ0 →

ζ(B + T + 2) ζ(B + T + 1)

......

ζ(C + T ) ζ(C + T − 1)

ρ

⋊ σΣ0 .

If JacρζxσΣ0 = 0 for all x ∈ [B + T + 1, C + T ], the vanishing statement is clear. Suppose Jacρζxσ

Σ0 6=

0 for some x ∈ [B + T + 1, C + T ], then JacρζxπΣ0 6= 0. It is necessary that x = B + T + 2. So

Jacρζ(B+T+2),ζ(B+T+2)πΣ0 6= 0. But this is impossible by [Xu17a, Proposition 8.3]. At last, it follows from

(6.24) that

Jacρζ(B+T ),··· ,ζ(A+T )XT,C = 〈ζ(B + T − 1), · · · ,−ζ(C + T − 1)〉ρ ⋊ Jacρζ(C+T+1),··· ,ζ(A+T ) ◦ Jacρζ(B+T+1),··· ,ζ(C+T−1)

◦ Jacρζ(B+T+2),··· ,ζ(C+T )πΣ0M

(ψ′T , ε

′T , (ρ, (A + T ), (B + T + 2), ζ; ηT,0)

)⊗ ηC+T

ρ ωζγB+T (C+T )ρ χρ · ηρω

−ζ(C+T )ρ

= 〈ζ(B + T − 1), · · · ,−ζ(C + T − 1)〉ρ ⋊ Jacζ(B+T+1),··· ,ζ(C+T−1)

◦ Jacρζ(B+T+2),··· ,ζ(A+T )πΣ0M

(ψ′T , ε

′T , (ρ,A + T,B + T + 2, ζ; ηT,0)

)⊗ ηC+T

ρ ωζγB+T (C+T )ρ χρ · ηρω

−ζ(C+T )ρ

= 〈ζ(B + T − 1), · · · ,−ζ(C + T − 1)〉ρ ⋊ Jacζ(B+T+1),··· ,ζ(C+T−1)

πΣ0M

(ψ′T , ε

′T , (ρ,A+ T − 1, B + T + 1, ζ; ηT,0)

)⊗ ηC+T−1

ρ ωζγB+T (C+T−1)ρ χρ · ω

−ζ((B+T+2)+···+(A+T ))/2ρ

NoteγB+T (C + T − 1) = γB+T−1(C + T − 1)− (B + T + 1)/2 − (B + T )/2.

This finishes the proof of our claim.�

Remark 6.10. We will only need the special case that there are no Jordan blocks between (ρ,A,B, ζ) and(ρ,A+T0, B+T0, ζ) until the end of the proof. The idea of shifting the Jordan block is motivated by theconstruction in the general case, cf. Section 6.4.

Now we can begin the process of the proof. First of all, let us determine the Σ0-infinitesimal charactersof irreducible representations in πΣ0

M (ψ, ε).

Proposition 6.11. The Σ0-infinitesimal characters of irreducible constituents in XC , Xη are the same

as that of ΠΣ0

ψd.

Proof. We will prove this by induction on∑

(ρ,A,B,ζ)∈Jord(ψ)(A − B). The case of Xη follows from the

induction assumption directly. For XC , it is the same to determine that of

〈B, · · · ,−C〉ρ ⋊ JacρB+2,··· ,C πΣ0M (ψ1

d , 1)⊗ ηCρ ωγB(C)ρ χρ,

which is the same as

〈C, · · · ,−B〉ρ ⋊ JacρB+2,··· ,C πΣ0M (ψ1

d, 1) ⊗ ηCρ ωγB(C)ρ χρ · ω

−((B+1)+···+C)ρ ηC−Bρ .

Note

JacρB+2,··· ,C πΣ0M (ψ1

d, 1) = πΣ0M

(ψ′d, 1,∪C′∈[B+1,A]\{C}(ρ,C

′, C ′,+;+1))⊗ ω−((B+2)+···+C)/2

ρ .

On the other hand, one can show

πΣ0M

(ψ′d, 1,∪C′∈[C+1,A](ρ,C

′, C ′,−; +1), (ρ,C,C,+; (−1)C−B−1),∪C′∈[B+1,C−1](ρ,C′, C ′,−;−1), (ρ,B,B,+;+1)

)

→ 〈C, · · · ,−B〉ρ ⋊ πΣ0M

(ψ′d, 1,∪C′∈[B+1,A]\{C}(ρ,C

′, C ′,−; +1))⊗ ω−(C+···+(B+1))/2

ρ χρ,

by the compatibility of twisted endoscopic transfer with Jacquet module. So the Σ0-infinitesimal characteris the same as that of ΠΣ0

ψd.

�


This proposition explains the appearance of twists by ωρ and χρ in XC . The twist by ηρ is more subtle,

since it does not change the Σ0-infinitesimal character. Next we would like to distinguish πΣ0M (ψ, ε)main

and πΣ0M (ψ, ε)com,η from XC , Xη .

Proposition 6.12. The irreducible constituents of πΣ0M (ψ, ε)main (resp. πΣ0

M (ψ, ε)com,η) only appear in

XA (resp. Xη) with multiplicity one.

Proof. By definition, πΣ0M (ψ, ε)main is the socle of XA. To show multiplicity one, we can apply JacρζB,··· ,−ζA

to XA. By (6.20),

JacρζB,··· ,−ζAXA = Jacρζ(B+2),··· ,ζAπM

(ψ′, ε′, (ρ,A,B + 2, ζ, η0)

)⊗ ηAρ ω

ζγB(A)ρ χρ

= πΣ0M

(ψ′, ε′, (ρ,A− 1, B + 1, ζ, η0)

)⊗ ηAρ ω

ζγB(A)ρ χρ · ω

−ζ((B+2)+···+A)/2ρ

= πΣ0M

(ψ′, ε′, (ρ,A− 1, B + 1, ζ, η0)

)⊗ ηAρ ω

ζ((B+1)+···+A)/2ρ χρ

which is multiplicity free. This shows multiplicity one. Moreover,

JacρζB,··· ,−ζAXC , Xη = 0

for C 6= A. So the irreducible constituents of πΣ0M (ψ, ε)main can not appear in XC , Xη for C 6= A. Next

we consider πΣ0M (ψ, ε)com,η , where

η0 =∏

C∈[B,A]

(−1)Cη = ηA−B+1(−1)(A+B)(A−B+1)/2 .

If A−B is odd, then we have η0 = (−1)(A−B+1)/2 and the sign in front of Xη becomes

(−1)[(A−B+1)/2]ηA−B+1ηA−B0 = (−1)(A−B+1)/2η0 = 1.

Moreover, X+ = X− if and only if πΣ0M (ψ, ε)com,η = πΣ0

M (ψ, ε)com,−η. By induction, we can assume

(Xη)com,η′ = πΣ0M

(ψ′, ε′,∪C∈[B+1,A](ρ,C,C, ζ, (−1)Cη′), (ρ,B,B, ζ, ηη0)

)

whereη =

∏

C∈[B+1,A]

(−1)Cη′ = η′(−1)(A+B+1)/2.

One checks (−1)B+1η′ = −ηη0. So

(Xη)com,η′ = πΣ0M (ψ, ε)com,η′ .

If A−B is even, then we have η0 = (−1)(A+B)/2η and the sign in front of Xη becomes

(−1)[(A−B+1)/2]ηA−B+1εA−B0 = (−1)(A−B)/2η.

By induction, we can assume

(Xη)com,η′ = πΣ0M

(ψ′, ε′,∪C∈[B+1,A](ρ,C,C, ζ, (−1)Cη′), (ρ,B,B, ζ, ηη0)

)

whereη =

∏

C∈[B+1,A]

(−1)Cη′ = (−1)(A−B)/2.

One checks (−1)B+1η′ = −ηη0 if and only if η = η′. So

(Xη)com,η = πM(ψ, ε)com,η .

So far we have shown πΣ0M (ψ, ε)com,η (η ∈ Sη0) appears in{

⊕η=±(−1)[(A−B+1)/2]ηA−B+1ηA−B0 Xη if X+ 6= X−

(−1)(A−B+1)/2η0X+ if X+ = X−

28 BIN XU

with coefficient +1. At last, we still need to show the irreducible constituents of πΣ0M (ψ, ε)com,η do not

appear in XC . This follows from Mœglin’s proof of [Mœg09, Theorem 4.1] that the irreducible constituents

of πΣ0M (ψ, ε)com,η|GΣ0 (F ) do not appear in XC |GΣ0 (F ).

�

Corollary 6.13. For any irreducible constituent πΣ0 in XC , Xη, it is in πΣ0M (ψ, ε)main if and only if

JacρζB,··· ,−ζAπΣ0 6= 0.

Next, we would like to show any irreducible constituent πΣ0 in XC , Xη excluded from πΣ0M (ψ, ε)main, π

Σ0M (ψ, ε)com,η

must be cancelled.

Proposition 6.14. For irreducible constituent πΣ0 in XC , Xη, if there exists x ∈ [B + 1, A] such that

JacρζxπΣ0 6= 0, then πΣ0 is cancelled in πΣ0

M (ψ, ǫ).

The idea is to characterize πΣ0 by JacρζxπΣ0 and show the cancellation of the corresponding Jacquet

modules.

Lemma 6.15. For irreducible constituent πΣ0 in XC , Xη and x ∈ [B,A], we have Jacρζx,ζxπΣ0 = 0.

Proof. It suffices to show Jacρζx,ζxXC , Xη = 0 for all x ∈ [B,A]. �

Corollary 6.16. For irreducible constituent πΣ01 , πΣ0

2 in XC , Xη such that JacρζxπΣ0i 6= 0 for some x ∈

[B,A], we have πΣ01 = πΣ0

2 if and only if JacρζxπΣ01 = Jacρζxπ

Σ02 .

Proof. Since JacρζxπΣ0i 6= 0, there exists an irreducible representation σΣ0

i such that

πΣ0i → ρ||ζx ⋊ σΣ0

i

Since Jacρζx,ζxπΣ0i = 0, we must have Jacρζxσ

Σ0i = 0. So

Jacρζx (ρ||ζx ⋊ σΣ0

i ) = σΣ0i .

This is means JacρζxπΣ0i = σΣ0

i and πΣ0i is the unique irreducible subrepresentation of ρ||ζx ⋊ σΣ0

i . Therest is clear. �

For the proof of Proposition 6.14, it remains to show

Lemma 6.17. For x ∈ [B + 1, A], JacρζxπΣ0M (ψ, ε) = 0.

Proof. First let us consider x ∈ [B + 2, A]. We will compute JacρζxXC as follows. If C = x,

JacρxXC = 〈ζB, · · · ,−ζ(C − 1)〉ρ ⋊ JacρB+2,··· ,C πΣ0M

(ψ′, ε′, (ρ,A,B + 2, ζ, η0)

)⊗ ηCρ ω

ζγB(C)ρ χρ · ηρω

−ζCρ

If C = x− 1,

JacρxXC = 〈ζB, · · · ,−ζC〉ρ ⋊ JacρB+2,··· ,C,C+1 πΣ0M

(ψ′, ε′, (ρ,A,B + 2, ζ, η0)

)⊗ ηCρ ω

ζγB(C)ρ χρ

If C 6= x, x− 1, JacρxXC = 0. One checks that

JacρζCXC = JacρζCXC−1.

Moreover, JacρζxXη = 0. So JacρζxπΣ0M (ψ, ε) = 0 for x ∈ [B + 2, A].

Next let us consider the case that x = B + 1. For C > B + 1,

Jacρζ(B+1)XC = 〈ζB, · · · ,−ζC〉ρ ⋊ Jacρζ(B+1)Jacρζ(B+2),··· ,ζC π

Σ0M

(ψ′, ε′, (ρ,A,B + 2, ζ, η0)

)⊗ ηCρ ω

ζγB(C)ρ χρ

To simplify further, we need to expand πΣ0M

(ψ′, ε′, (ρ,A,B + 2, ζ, η0)

). Let

X ′C′ := 〈ζ(B + 2), · · · ,−ζC ′〉ρ ⋊ Jacζ(B+4),··· ,ζC′ πΣ0M

(ψ′, ε′, (ρ,A,B + 4, ζ, η0)

)⊗ ηC

′

ρ ωζγB+2(C

′)ρ χρ


X ′η := πΣ0M

(ψ′, ε′, (ρ,A,B + 3, ζ, η), (ρ,B + 2, B + 2, ζ, ηη0)

)

If X ′+ 6= X ′−, then

πΣ0M

(ψ′, ε′, (ρ,A,B + 2, ζ, η0)

)= ⊕C′∈]B+2,A](−1)A−C

′

X ′C′ ⊕η=± (−1)[(A−B+1)/2]+1ηA−B+1ηA−B0 Xη

If X ′+ = X ′−, then

πΣ0M

(ψ′, ε′, (ρ,A,B + 2, ζ, η0)

)= ⊕C′∈]B+2,A](−1)A−C

′

X ′C′ ⊕ (−1)(A−B+1)/2+1η0X′+

Consider X ′C′ and the corresponding contribution to πΣ0M (ψ, ε). If C < C ′, we get

(−1)C+C′

〈ζB, · · · ,−ζC〉ρ × 〈ζB, · · · ,−ζC ′〉ρ ⋊ Jacρζ(B+3),··· ,ζCJacρζ(B+4),··· ,ζC′

πΣ0M

(ψ′, ε′, (ρ,A,B + 4, ζ, η0)

)⊗ ηC+C′

ρ ωζ(γB(C)+γB+2(C

′))ρ χ2

ρ

If C > C ′, we get

(−1)C+C′

〈ζB, · · · ,−ζC〉ρ × 〈ζB, · · · ,−ζ(C ′ − 1)〉ρ ⋊ Jacρζ(B+3),··· ,ζ(C′−1)Jacρζ(B+4),··· ,ζC

πΣ0M

(ψ′, ε′, (ρ,A,B + 4, ζ, η0)

)⊗ ηC+C′−1

ρ ωζ(γB(C)+γB+2(C

′)−C′)ρ χ2

ρ

Note C > B + 1, C ′ > B + 2. We can pair (C,C ′) with (C ′, C + 1) for C < C ′, and their contributions

cancel each other. The contribution of X ′η to πΣ0M (ψ, ε) is

(−1)[(A−B+1)/2]+1ηA−B+1ηA−B0 · (−1)A−C〈ζB, · · · ,−ζC〉ρ ⋊ Jacρζ(B+3),··· ,ζC

(6.25)

πΣ0M

(ψ′, ε′, (ρ,A,B + 3, ζ, η), (ρ,B,B, ζ, ηη0)

)⊗ ηCρ ω

ζ(γB(C)−(B+2)/2−(B+1)/2)ρ χρ(6.26)

For C = B + 1, we also get contribution from

Jacρζ(B+1)XB+1 = 〈ζB, · · · ,−ζB〉ρ ⋊ πΣ0M

(ψ′, ε′, (ρ,A,B + 2, ζ, η0)

)⊗ ηB+1

ρ ωζγB(B+1)ρ χρ · ηρω

−ζ(B+1)ρ

(6.27)

with sign (−1)A−B−1. To cancel these terms, we need to further expand

Xη := πΣ0M

(ψ′, ε′, (ρ,A,B + 1, ζ, η), (ρ,B,B, ζ, ηη0)

)

Let

X ′′C := 〈ζ(B+1), · · · ,−ζC〉ρ⋊Jacρζ(B+3),··· ,ζC πΣ0M

(ψ′, ε′, (ρ,A,B+3, ζ, η), (ρ,B,B, ζ, ηη0)

)⊗ηCρ ω

ζγB+1(C)ρ χρ

X ′′η′ := πΣ0M

(ψ′, ε′, (ρ,A,B + 2, ζ, η′), (ρ,B + 1, B + 1, ζ, η′η), (ρ,B,B, ζ, ηη0)

)

If X ′′+ 6= X ′′−, then

Xη = ⊕C∈]B+1,A](−1)A−CX ′′C ⊕η′=± (−1)[(A−B)/2]+1η′A−BηA−B−1X ′′η′

If X ′′+ = X ′′−, then

Xη = ⊕C′∈]B+1,A](−1)A−C′

X ′′C ⊕ (−1)(A−B)/2+1ηX ′′+

One checks the contributions from X ′′C for C > B + 1 cancel (6.25). Also note

Jacρζ(B+1)X′′η′ 6= 0

if and only if η′ = η0, in which case

Jacρζ(B+1)X′′η′ = 〈ζB, · · · ,−ζB〉ρ ⋊ πΣ0

M

(ψ′, ε′, (ρ,A,B + 2, ζ, η0)

)⊗ ηBρ ω

−ζ(B+1)/2ρ χρ

which is exactly (6.27). To see the cancellation, we still need to check the sign in front of Jacρζ(B+1)X′′η0

(−1)[(A−B+1)/2]ηA−B+1εA−B0 · (−1)[(A−B)/2]+1ηA−B0 ηA−B−1 = (−1)A−B

30 BIN XU

which is opposite to that of (6.27). �

Lemma 6.18. Assume all (ρ,A′, B′, ζ ′) ∈ Jordρ(ψ) such that B′ > B satisfy ζ ′ = ζ. If πΣ0 is in XC or

Xη such that πΣ0 ⊗ ηρ ≇ πΣ0 and JacρζxπΣ0 = 0 for x ∈ [B + 1, A], then JacρζB,··· ,−ζAπ

Σ0 6= 0.

Proof. Since πΣ0 ⊗ ηρ ≇ πΣ0 , we can write πΣ0 as a submodule of

πΣ0 → ×x′>y′ 〈ζx′, · · · , ζy′〉ρ ⋊ σΣ0(6.28)

where x′ + y′ < 0 and is increasing, and

2∑

x′>y′

(x′ − y′ + 1) =∑

(ρ,a,b)∈Jordρ(ψ)

ab

and σΣ0 is an irreducible representation of GΣ0− (F ). This can be achieved by first considering the standard

representation containing πΣ0 in case ζ = + (resp. Aubert dual of πΣ0 in case ζ = −) as the uniquesubrepresentation and using the fact that the inducing representation can not be invariant under twistby ηρ. We would like to consider those 〈ζx′, · · · , ζy′〉ρ containing ρ||±(A+1) in the cuspidal support. If it

contains ρ||ζ(A+1), then

x′ > A+ 1, y′ < −(A+ 1).

So it also contains ρ||−ζ(A+1). If it contains ρ||−ζ(A+1) but not ρ||ζ(A+1), then

x′ < A+ 1, y′ 6 −(A+ 1).

We will first show the second case never occurs. The representation 〈ζx′′, · · · , ζy′′〉ρ in front of 〈ζx′, · · · , ζy′〉ρin (6.28) satisfies x′′ 6 x′ or y′′ 6 y′. In particular, if the conditions x′′ 6 x′ and y′′ 6 y′ are not bothsatisfied, then we can interchange 〈ζx′, · · · , ζy′〉ρ with 〈ζx′′, · · · , ζy′′〉ρ. As a result, if the second caseoccurs, then

Jacρζx′′,··· ,ζy′′ πΣ0 6= 0(6.29)

for some x′′ < A+ 1 and y′′ 6 −(A+ 1). Since JacρζxπΣ0 = 0 for x ∈ [B + 1, A], then x′′ < B + 1. Since

all (ρ,A′, B′, ζ ′) ∈ Jord(ψ) such that B′ > B satisfy ζ ′ = ζ, we can further show

Jacρζx′′,··· ,ζy′′ XC = 0

Jacρζx′′,··· ,ζy′′ Xη = 0

for any x′′ 6 B and y′′ 6 −(A+ 1), which contradicts to (6.29).

Since only the first case occurs, then the number of ρ||±(A+1) is even in the cuspidal support of πΣ0 ,hence ∑

(ρ,A′,B′,ζ′)∈Jordρ(ψ)B′>A

A′ −B′ + 1

is even. It follows that the number of ρ||±A is odd in the cuspidal support. By the same argument, we seeif 〈ζx′, · · · , ζy′〉ρ in (6.28) contains ρ||ζA, then it must also contain ρ||−ζA. So there exists 〈ζx′, · · · , ζy′〉ρsuch that it only contains ρ||−ζA, but not ρ||A, i.e., x′ < A, y′ 6 −A. In the same way, one can show

Jacρζx′′,··· ,ζy′′ πΣ0 6= 0

for some x′′ < B + 1 and y′′ 6 −A. This is only possible when x′′ = B, y′′ = −A. �

Proposition 6.19. Theorem 6.6 holds if for any ρ′ ∈ J(ψ), either A′ = B′ for all (ρ′, A′, B′, ζ ′) ∈Jordρ′(ψ) or there exists (ρ′, A′, B′, ζ ′) ∈ Jordρ′(ψ) with A

′ > B′ such that all (ρ′, A′′, B′′, ζ ′′) ∈ Jordρ′(ψ)with B′′ > B′ satisfy ζ ′′ = ζ ′.


Proof. Let πΣ0 be an irreducible constituent in XC , Xη excluded from πΣ0M (ψ, ε)main, π

Σ0M (ψ, ε)com,η . If

X(πΣ0) = XΣ0(λ), then πΣ0 is determined by its restriction to GΣ0(F ) together with its Σ0-infinitesimalcharacter. In this case, the cancellation of πΣ0 follows from that of πΣ0 |GΣ0 (F ) (cf. [Mœg09, Theorem

4.1]). Suppose X(πΣ0) 6= XΣ0(λ). Since XΣ0(λ) = XΣ0(ψd), then πΣ0 ⊗ ηρ′ ≇ πΣ0 for some ρ′ ∈ J(ψ). It

implies that Jordρ′(ψ) can not be elementary. By Lemma 6.4, it is enough to consider the case that πΣ0

is also contained in X ′C′ or X ′η′ in the recursive formula with respect to ρ′. Then the result follows fromLemma 6.18 and Lemma 6.8. �

To complete the proof of Theorem 6.6, we still need to remove the assumption in Proposition 6.19.Let us choose an admissible order >ψ such that for any ρ′ ∈ J(ψ), either A′ = B′ for all (ρ′, A′, B′, ζ ′) ∈Jordρ′(ψ) or there exists (ρ

′, A′, B′, ζ ′) ∈ Jordρ′(ψ) with A′ > B′ and all (ρ′, A′′, B′′, ζ ′′) ∈ Jordρ′(ψ) with

(ρ′, A′′, B′′, ζ ′′) >ψ (ρ′, A′, B′, ζ ′) satisfy ζ ′′ = ζ ′. Let ψ≫ be obtained from ψ by shifting certain Jordanblocks such that it has discrete diagonal restriction and >ψ induces the natural order, then Theorem 6.6

holds for ψ≫. In particular, πΣ0M (ψ≫, ε≫) is a representation, where ε≫ is related to ε by the change of

order formula, cf. [Xu19, Theorem 6.3]. By Lemma 6.9,

◦(ρ′,A′,B′,ζ)Jac(ρ′,A′≫,B′

≫,ζ′)7→(ρ,A′,B′,ζ′) πM (ψ≫, ε≫) = πM(ψ, ε) ⊗∏

(ρ′,A′,B′,ζ′)

ω−ζ′|X

A′≫−A′

(ρ′,A′,B′,ζ′)|/2

ρ′ ,(6.30)

where the composition is taken in the decreasing order. As a consequence, πΣ0M (ψ, ε) is a representation.

Then Theorem 6.6 follows from Proposition 6.12 and (6.12).

6.3.2. Stability and character relation. For [ψ] ∈ Ψ(G) with discrete diagonal restriction, we define ΠΣ0

ψ

to be the set of irreducible constituents of ⊕

ε∈Sψ

πΣ0M (ψ, ε).

Let πW (ψ, ε) := πW (ψ, ε) be the direct sum of the preimages of ǫ in ΠΣ0

ψunder (5.8) and πΣ0

MW (ψ, ε) :=

πΣ0W (ψ, εε

MW/Wψ ). It follows from πΣ0

M (ψ, ε) = πΣ0MW (ψ, εε

M/MWψ ) (cf. [Xu17a, Theorem 7.5]) that


MW (ψ, εεM/MWψ ).

For s ∈ Sθψ with ω = α(s), let

ΠM,s(ψ) :=∑

π∈Πψ

(π, Aπ(θ, ω)M ),

where Aπ(θ, ω)M is an intertwining operator between π⊗ω and πθ, which is normalized in a way so that

if f is the restriction of f to G(F ), then

(f |ZFG)Gθ,M (π, ω) =∑

π⊆π|G(F )

〈sψs, π+〉M fGθ(π), f ∈ H(G, χ).

We also define

ΠΣ0M,s(ψ) :=

∑

πΣ0∈ΠΣ0ψ

(πΣ0 , AπΣ0 (ω)M ),

where AπΣ0 (ω)M is an intertwining operator between πΣ0 ⊗ω and πΣ0 given by Aπ(θ, ω)M . If G is specialeven orthogonal, then

(fθ)GΣ0(ΠΣ0

M,s(ψ)) = 2fGθ

(ΠM,s(ψ)), f ∈ H(G, χ),

where fθ(g ⋊ θ) := f(g) is supported on G(F ) ⋊ θ. Similarly we can define ΠMW,s(ψ), ΠΣ0MW,s(ψ). One

can check that

ΠΣ0M,s(ψ) = ε

M/MWψ (s)ΠΣ0

MW,s(ψ), ΠM,s(ψ) = εM/MWψ (s) ΠMW,s(ψ).

32 BIN XU

We denote ΠMW (ψ) = ΠMW,1(ψ) and ΠM (ψ) = ΠM,1(ψ). Note ΠMW (ψ) = ΠM (ψ). It follows from the

construction of πΣ0M (ψ, ε) (reps. πM (ψ, ε)) that ΠΣ0

M,s(ψ) (resp. ΠM,s(ψ)) also satisfies a recursive formula.

Proposition 6.20. Suppose [ψ] ∈ Ψ(G) has discrete diagonal restriction and we fix (ρ,A,B, ζ) ∈ Jord(ψ)

such that A > B, then for any s ∈ SΣ0ψ

ΠΣ0M,s(ψ) =⊕C∈]B,A] (−1)A−C〈ζB, · · · ,−ζC〉ρ ⋊ Jacρζ(B+2),··· ,ζCΠ

Σ0M,s(ψ

1)⊗ ηCρ ωζγB(C)ρ χρ

⊕ (−1)[(A−B+1)/2]ΠΣ0M,s(ψ

2)

and

ΠM,s(ψ) =⊕C∈]B,A] (−1)A−C〈ζB, · · · ,−ζC〉ρ ⋊ Jacρζ(B+2),··· ,ζCΠM,s(ψ

1)⊗ ηCρ ωζγB(C)ρ χρ

⊕ (−1)[(A−B+1)/2]ΠM,s(ψ2).

Proof. The proof is the same as that of [Xu17a, Lemma 7.6]. �

Let (H,ψH) → (ψ, s) and suppose ψH = ψI ×ψII . Let ΠψH = ΠψI ⊗ ΠψII and cI , cII be the 1-cochains

for defining ΠψI,d , ΠψII,d respectively. We choose 1-cochain cs to get the twisted endoscopic embedding

for H. Let ′cψ = cIcIIcs, then we can define a homomorphism χcψ,′cψ : WF → C× (cf. [Xu21a, Lemma

8.6]).

Theorem 6.21. (1)

fMW (ψ) :=∑

[π]∈Πψ

〈sψ, π〉MW fG(π), f ∈ H(G(F ), χ)

is stable.(2) For s ∈ Sθψ with ω = α(s) and (H,ψH ) → (ψ, s), ΠMW,s(ψ) ⊗ χcψ,

′cψ is the twisted endoscopic

transfer ΠMW (ψH).

Proof. Part (1) follows from the recursive formula of ΠMW (ψ) and the fact that parabolic induction andJacquet module preserve stability. The proof of part (2) follows exactly the same line of that of [Xu17a,Theorem 7.5]. Let us suppose ψs := ψH = ψI × ψII . We can assume (ρ,A,B, ζ) ∈ Jord(ψII) for theother case is similar. Let ψ1

s = ψ1I × ψ1

II ∈ Ψ(H1) and ψ2s = ψ2

I × ψ2II ∈ Ψ(H2). In particular, ψI = ψ1

I =

ψ2I . The 1-cochain cs gives rise to twisted endoscopic embeddings for both H1 and H2. By induction

and [Xu21a, Proposition 8.7], we can assume that ΠMW,s(ψ1) ⊗ χcψ ,

′cψ (resp. ΠMW,s(ψ2) ⊗ χcψ,

′cψ)

is the twisted endoscopic transfer of ΠMW (ψ1I ) ⊗ ΠMW (ψ1

II) (resp. ΠMW (ψ2I ) ⊗ ΠMW (ψ2

II)). By thecompatibility of twisted endoscopic transfer with Jacquet module and parabolic induction, we can concludethat ΠM,s(ψ)⊗ χcψ,

′cψ is the twisted endoscopic transfer of

⊕C∈]B,A] (−1)A−CεM/MWψ1 (s)〈ζB, · · · ,−ζC〉ρ ⋊ Jac

ρζ(B+2),··· ,ζC (ΠMW (ψ1

I ) ⊗ ΠMW (ψ1II))⊗ ηCρ ω

ζγB(C)ρ χρ

⊕ (−1)[(A−B+1)/2]εM/MWψ2 (s) ΠMW (ψ2

I ) ⊗ ΠMW (ψ2II).

Note JacρζDΠMW (ψ1I ) = 0 for any B + 2 6 D 6 A. Then we can rewrite it as

⊕C∈]B,A] (−1)A−CεM/MWψ1 (s) ΠMW (ψ1

I ) ⊗(〈ζB, · · · ,−ζC〉ρ ⋊ Jac

ρζ(B+2),··· ,ζCΠMW (ψ1

II)⊗ ηCρ ωζγB(C)ρ χρ

)

⊕ (−1)[(A−B+1)/2]εM/MWψ2 (s) ΠMW (ψ2

I ) ⊗ ΠMW (ψ2II).

Since

εM/MWψ (s) = ε

M/MWψ1 (s) = ε

M/MWψ2 (s),(6.31)

then ΠM,s(ψ)⊗ χcψ ,′cψ is the twisted endoscopic transfer of ε

M/MWψ (s) ΠMW (ψI) ⊗ ΠMW (ψII). �


6.4. General case. We first assume ψ = ψp and fix an admissible order >ψ on Jord(ψ). We say ψ≫ withorder >ψ≫

dominates ψ with respect to >ψ, if there is an order preserving bijection between Jord(ψ≫)and Jord(ψ), which sends (ρ,A≫, B≫, ζ≫) to (ρ,A,B, ζ) such that A≫ −A = B≫ −B > 0 and ζ≫ = ζ.Let us choose a dominating parameter ψ≫ of ψ with discrete diagonal restriction and the natural order.

Identify SΣ0

ψ>∼= SΣ0

ψ≫. For ε ∈ SΣ0

ψ> , we define

πΣ0M,>ψ

(ψ, ε) := ◦(ρ,A,B,ζ)∈Jord(ψ) Jac(ρ,A≫,B≫,ζ)7→(ρ,A,B,ζ)πΣ0M (ψ≫, ε)

⊗(ρ,A,B,ζ)∈Jord(ψ) ω−ζ(|X

A≫−A

(ρ,A,B,ζ)|)/2

ρ ,

and

πM,>ψ(ψ, ε) := ◦(ρ,A,B,ζ)∈Jord(ψ) Jac(ρ,A≫,B≫,ζ)7→(ρ,A,B,ζ)πM (ψ≫, ε)


A≫−A

(ρ,A,B,ζ)|)/2

ρ ,

where the compositions are taken in the decreasing order respectively.

Lemma 6.22. πΣ0M,>ψ

(ψ, ε) (resp. πM,>ψ(ψ, ε)) is independent of the choice of ψ≫.

Proof. Let πΣ0M,>ψ

(ψ, ε)i be defined with respect to ψi≫ for i = 1, 2. We can choose ψ≫ that dominates

both ψi≫ with discrete diagonal restriction and the natural order. By Lemma 6.9,

πΣ0M (ψi≫, ε) = ◦(ρ,A,B,ζ)∈Jord(ψ) Jac(ρ,A≫,B≫,ζ)7→(ρ,Ai≫,Bi≫,ζ)π

Σ0M (ψ≫, ε)


A≫−Ai≫

(ρ,Ai≫,Bi

≫,ζ)|)/2

ρ .

Since

◦(ρ,A,B,ζ)∈Jord(ψ) Jac(ρ,Ai≫,Bi≫,ζ)7→(ρ,A,B,ζ) ◦(ρ,A,B,ζ)∈Jord(ψ) Jac(ρ,A≫,B≫,ζ)7→(ρ,Ai≫,Bi≫,ζ)

= ◦(ρ,A,B,ζ)∈Jord(ψ) Jac(ρ,A≫,B≫,ζ)7→(ρ,A,B,ζ),

then

πΣ0M,>ψ

(ψ, ε)i = ◦(ρ,A,B,ζ)∈Jord(ψ) Jac(ρ,Ai≫,Bi≫,ζ)7→(ρ,A,B,ζ)πΣ0M (ψi≫, ε)


Ai≫−A

(ρ,A,B,ζ)|)/2

ρ

= ◦(ρ,A,B,ζ)∈Jord(ψ) Jac(ρ,A≫,B≫,ζ)7→(ρ,A,B,ζ)πΣ0M (ψ≫, ε)


A≫−A

(ρ,A,B,ζ)|)/2

ρ .

So πΣ0M,>ψ

(ψ, ε)1 = πΣ0M,>ψ

(ψ, ε)2. The case of πM,>ψ(ψ, ε) also follows from this. �

For functions l(ρ,A,B, ζ) ∈ [0, [(A −B + 1)/2]] and η(ρ,A,B, ζ) ∈ Z/2Z on Jord(ψ) such that

εl,η(ρ,A,B, ζ) := η(ρ,A,B, ζ)A−B+1(−1)[(A−B+1)/2]+l(ρ,A,B,ζ)

defines a character εl,η of SΣ0

ψ> , we define

πΣ0M,>ψ

(ψ, l, η) := ◦(ρ,A,B,ζ)∈Jord(ψp) Jac(ρ,A≫,B≫,ζ)7→(ρ,A,B,ζ)πΣ0M (ψ≫, l, η)

⊗(ρ,A,B,ζ)∈Jord(ψp) ω−ζ(|X

A≫−A

(ρ,A,B,ζ)|)/2

ρ ,

where the composition is taken in the decreasing order,

l(ρ,A,B, ζ) = l(ρ,A≫, B≫, ζ) and η(ρ,A,B, ζ) = η(ρ,A≫, B≫, ζ).

34 BIN XU

Note

πΣ0M,>ψ

(ψ, l, η)|GΣ0 (F ) =⊕

(l′,η′)∼GΣ0

(l,η)/∼Σ0

πΣ0M,>ψ

(ψ, l′, η′).(6.32)

So πΣ0M,>ψ

(ψ, l, η) is irreducible or zero.

In general, we define

πΣ0M,>ψ

(ψ, ε) :=(×(ρ,a,b)∈Jord(ψnp) Sp(St(ρ, a), b)

)⋊ πΣ0

M (ψp, ε)

and

πM,>ψ(ψ, ε) :=(×(ρ,a,b)∈Jord(ψnp) Sp(St(ρ, a), b)

)⋊ πM (ψp, ε).

From the definitions it is clear that Lemma 6.5 extends to the general case. We define ΠΣ0

ψto be the set

of irreducible constituents of ⊕

ε∈Sψ

πΣ0M,>ψ

(ψ, ε).

Let πΣ0W (ψ, ε) := πΣ0

W (ψ, ε) be the the direct sum of the preimages of ǫ in ΠΣ0

ψunder (5.8) and πΣ0

MW,>ψ(ψ, ε) :=

πΣ0W (ψ, εε

MW/Wψ ). It follows from πΣ0

M,>ψ(ψ, ε) = πΣ0

MW,>ψ(ψ, εε

M/MWψ ) (cf. [Xu17a, Proposition 8.2]) that

πΣ0M,>ψ

(ψ, ε) = πΣ0MW,>ψ

(ψ, εεM/MWψ ). Hence,

πΣ0M,>ψ

(ψ, ε) =

{πΣ0W (ψ, εε

M/MWψ ε

MW/Wψ ), if εε

M/MWψ ε

MW/Wψ ∈ SΣ0

ψ ,

0, otherwise.

For l(ρ,A,B, ζ) ∈ [0, [(A−B +1)/2]] and η(ρ,A,B, ζ) ∈ Z/2Z on Jord(ψp) such that εl,η ∈ SΣ0

ψ> , we also

define

πΣ0M (ψ, l, η) =

(×(ρ,a,b)∈Jord(ψnp) Sp(St(ρ, a), b)

)⋊ πΣ0

M (ψp, l, η)

and

πM (ψ, l, η) =(×(ρ,a,b)∈Jord(ψnp) Sp(St(ρ, a), b)

)⋊ πM (ψp, l, η).

We still have (6.32) and πΣ0M,>ψ

(ψ, l, η) is irreducible or zero by the similar result for GΣ0(F ), cf. [Mœg06a,

Theorem 6].

For s ∈ Sθψ> with ω = α(s), we can define ΠM,s(ψ),ΠΣ0M,s(ψ), ΠMW,s(ψ),Π

Σ0MW,s(ψ) as in the case of

discrete diagonal restriction. In particular,

ΠΣ0M,s(ψ) =

(×(ρ,a,b)∈Jord(ψnp) Sp(St(ρ, a), b)

)⋊ΠΣ0

M,s(ψp)

where

ΠΣ0M,s(ψp) = ◦(ρ,A,B,ζ)∈Jord(ψp) Jac(ρ,A≫,B≫,ζ)7→(ρ,A,B,ζ)Π

Σ0M,s(ψp,≫)


A≫−A

(ρ,A,B,ζ)|)/2

ρ .

Theorem 6.23.

(1)

fMW (ψ) :=∑

[π]∈Πψ

〈sψ, π〉MW fG(π), f ∈ H(G(F ), χ)

is stable.(2) For s ∈ Sθψ> with ω = α(s) and (H,ψH) → (ψ, s), ΠMW,s(ψ) ⊗ χcψp,≫ ,′cψp,≫ is the twisted

endoscopic transfer of ΠMW (ψH).


Proof. Part (1) follows from the stability of ΠMW (ψp,≫) and the fact that parabolic induction and Jacquet

module preserve stability. Part (2) follows from the result for ψp,≫ and the compatibility of twistedendoscopic transfer with parabolic induction and Jacquet module. �

So far we have constructed Πψ and shown that it satisfies the properties (1), (2) in Theorem 5.4. We

have also proved (3) when H is elliptic (cf. the end of Section 6.1). To treat the remaining cases, we needto show

Proposition 6.24. For ψ = ψp ∈ Ψ(G), if ψ factors through

ψM := (∏

i

ρi ⊗ νai ⊗ νbi)× ψ− ∈ Ψ(M)

for M =∏iGL(ni)×G− and ρi ⊗ νai ⊗ νbi ∈ Ψ(GL(ni)), ψ− ∈ Ψ(G−), then

ΠW (ψ) =∏

i

Sp(St(ρi, ai), bi)⋊ ΠW (ψ−) ⊗i ηAiρi χ

Ai−Bi+1ρi

(6.33)

This will be proved in the next section.

Corollary 6.25. Part (3) of Theorem 5.4 holds for Πψ.

Proof. For θ ∈ Σ0, s ∈ Sθψ with ω = α(s) and (H,ψH) → (ψ, s), the result follows from Theorem 6.23

when H is elliptic. If H is not elliptic, then it is a Levi subgroup of some (θ, ω)-twisted elliptic endoscopic

group G′ of G. By Proposition 6.24, f H(ψH) = f G′

(ψ′). So it reduces to the elliptic case. �

7. Uniqueness

To complete the proof of Theorem 5.4, we still need to show Proposition 6.24 and the uniqueness partof the theorem. First, we would like to show in many cases Πψ can be uniquely characterized by properties

(1), (2) of Theorem 5.4 up to twists by X.

Theorem 7.1. Suppose ψ = ψp ∈ Ψ(G) and Jordρ(ψ) is indexed for each ρ satisfying

ζρ := ζi+1 = ζi, Ai+1 > Ai, Bi+1 > Bi.(7.1)

We fix the order >ψ on Jordρ(ψ) so that

(ρ,Ai, Bi, ζi) >ψ (ρ,Ai−1, Bi−1, ζi−1).

(1) There exists a subset Πψ of Πψ,ζ unique up to twists by X such that the properties (1), (2) of

Theorem 5.4 are satisfied.

(2) If there exists a stable distribution supported on ˜Πψ,ζ , i.e.,

S(f) =∑

π∈ ˜Πψ,ζ

cπ fG(π), f ∈ H(G(F ), χ),

then it must be of the form

S(f) =∑

ω∈X/α(SΣ0ψ )

cω fW (ψ ⊗ ω), f ∈ H(G(F ), χ).

Corollary 7.2. Suppose [ψ] ∈ Ψ(G) satisfies (7.1). Then any stable distribution supported on Πψ (resp.

Πψ) is a scalar multiple of fW (ψ) (resp. fW (ψ)).

Proof. For Πψ, it follows directly from part (2) of Theorem 7.1. The case of Πψ follows from that for Πψby restricting H(G(F ), χ) to ZFG(F ). �

Corollary 7.3. The uniqueness part of Theorem 5.4 holds.

36 BIN XU

Proof. By property (3) of Theorem 5.4, we can reduce to the case that ψ = ρ ⊗ νa ⊗ νb, to whichTheorem 7.1 applies. �

We will deduce Proposition 6.24 from Theorem 7.1. The critical step is to show the following specialcase.

Lemma 7.4. For ψ = ψp = 2ρ⊗ νa ⊗ νb ∈ Ψ(G),

ΠW (ψ) = Sp(St(ρ, a), b) ⋊ ηAρ χA−B+1ρ .

Proof. It is known that ΠW (ψ) = Sp(St(ρ, a), b) ⋊ 1, so the right hand side is supported on ˜Πψ,ζ , where

ζ is the central character of ΠW (ψ). Since the right hand side is also stable, then by Theorem 7.1 it isequal to ∑

ω∈X/α(SΣ0ψ

)

cω ΠW (ψ)⊗ ω.

By comparing the Σ0-infinitesimal characters, it suffices to sum over XΣ0(λ)/α(SΣ0ψ ). NoteXΣ0(λ) = 〈ηρ〉.

Then it is equal to

ΠW (ψ) or ΠW (ψ)⊗ ηρ.

Suppose ΠW (ψ) 6= ΠW (ψ)⊗ ηρ, then it is necessary that a, b are even. For l(ρ,A,B, ζ) = (A−B + 1)/2,we have

πM,>ψ(ψ, l, 1) →

ζB · · · −ζA...

...ζ(A+B − 1)/2 · · · −ζ(A+B + 1)/2

×

ζB · · · −ζA...

...ζ(A+B − 1)/2 · · · −ζ(A+B + 1)/2

⋊ η(A+···+(A+B+1)/2)ρ ωζ(A+B+1)(A−B+1)(A−B)/8

ρ χ(A−B+1)/2ρ

as the unique irreducible subrepresentation. Note

πM,>ψ(ψ, l, 1) 6= πM,>ψ(ψ, l, 1)⊗ ηρ,

and πM,>ψ(ψ, l, 1)⊗ηρ /∈ Πψ. It suffices to show that πM,>ψ(ψ, l, 1)⊗ηρ is not contained in Sp(St(ρ, a), b)⋊

ηAρ χA−B+1ρ . This can be checked by computing JacX ◦ JacX for

X =

ζB · · · −ζA...

...ζ(A+B − 1)/2 · · · −ζ(A+B + 1)/2

.

�

Now we can give the proof of Proposition 6.24.

Proof. By induction on the number of general linear factors in M , it reduces to show the case thatM = GL(n1)×G−. From the above lemma,

ΠW (ψ1) = Sp(St(ρ1, a1), b1)⋊ ηA1ρ1 χ

A1−B1+1ρ1

for ψ1 := 2ρ1 ⊗ νa1 ⊗ νb1 ∈ Ψ(G1). Since M is also a Levi subgroup of G1 ×G−, then

Sp(St(ρ1, a1), b1)⋊ ΠW (ψ−)⊗ ηA1ρ1 χ

A1−B1+1ρ1

is equal to the endoscopic transfer of ΠW (ψ1) ⊗ ΠW (ψ−), which is also equal to ΠW (ψ) by Theorem 6.23.�

Before starting the proof of Theorem 7.1, let us look at some of the implications of the condition (7.1).

Lemma 7.5. Suppose ψ = ψp ∈ Ψ(G) satisfies (7.1) and ζρ = + for all ρ. Then

Πφψ={πM,>ψ(ψ, l, η) ∈ Πψ | l(ρ,A,B, ζ) = [(A−B + 1)/2]

}.


Proof. It suffices to show that

Πφψ ={πM,>ψ(ψ, l, η) ∈ Πψ | l(ρ,A,B, ζ) = [(A−B + 1)/2]

}.

Note the Langlands parameters of elements in Πφψ have been given in [Xu21b, Theorem 4.1]. So theresult follows directly from there. �

Corollary 7.6. Suppose ψ = ψp ∈ Ψ(G) satisfies (7.1) and ζρ = + for all ρ.

(1){πM,>ψ(ψ, l, η) ∈ Πψ | l = 0

}= Πψ if and only if ψ is trivial on the second SL(2,C).

(2) If ψ is not trivial on the second SL(2,C), then there are no nontrivial stable distributions supportedon ⋃

ω∈X/α(SΣ0ψ

)

{πM,>ψ(ψ, l, η) ∈ Πψ | l = 0

}⊗ ω.

Proof. If ψ is not trivial on the second SL(2,C), then Πφψis nonempty and disjoint from

{πM,>ψ(ψ, l, η) ∈

Πψ | l = 0}. So (1) is clear. (2) follows from Theorem 7.1 and Lemma 5.3. �

Now we can start the proof of Theorem 7.1.

Proof. It suffices to show part (2). Let ψ♯ be obtained from ψ by changing ζρ to −ζρ for all ρ such thatζρ = −. Then

◦ρ:ζρ=− |invρ

∞| πM,>ψ(ψ, l, η) = πM,>ψ(ψ♯, l, η).

So it suffices to consider the case that ζρ = + for all ρ. We will prove it by induction on∑

(ρ,A,B,ζ)∈Jord(ψ)(A−

B). If∑

(ρ,A,B,ζ)∈Jord(ψ)(A − B) = 0, then it follows from the tempered case, cf. [Xu18, Corollary 4.8].

So let us assume∑

(ρ,A,B,ζ)∈Jord(ψ)(A−B) > 0. Suppose

R :=∑

ω∈X/α(SΣ0ψ

)

∑

(l,η)/∼GΣ0

cωl,η πM,>ψ(ψ, l, η)⊗ ω

is stable. By subtracting linear combinations of fW (ψ ⊗ ω) for ω ∈ X/α(SΣ0ψ ), we can assume cωl,η = 0 if

l(ρ,A,B, ζ) = [(A−B + 1)/2] and η = 1. Then it is enough to show R = 0.Suppose R 6= 0, then there exists (l, η) such that

∑

ω∈X/α(SΣ0ψ )

cωl,η πM,>ψ(ψ, l, η)⊗ ω 6= 0.

If l(ρ,A,B,+) 6= 0 for some (ρ,A,B,+). Then over Jordρ(ψ) = {(ρ,Ai, Bi,+)}ni=1, we can choose t 6 ssuch that A = As = · · · = At, B = Bs = · · · = Bt and Bs+1 > Bs or s = n, At > At−1 or t = 1. Considerthe map

1

(s− t+ 1)!◦si=t (Jac

ρBi,··· ,−Ai(·)⊗ η−Aiρ ω−((Bi+1)+···+Ai)/2

ρ χ−1ρ ) : ˜Πψ,ζ →˜Πψ′,ζ′ ∪ {0},

where ψ′ is obtained from ψ by replacing (ρ,Ai, Bi,+) by (ρ,Ai − 1, Bi + 1,+) for t 6 i 6 s. It is a

bijection on the preimage of ˜Πψ′,ζ′ (cf. [Xu21b, Lemma 4.3]). Since the image of R is stable and can

not be a nontrivial linear combination of fW (ψ′ ⊗ ω) for ω ∈ X, then it must be zero by the induction

assumption. But the image of πM,>ψ(ψ, l, η) is nonzero, so we get a contradiction.It remains to consider the case that

R =∑

ω∈X/α(SΣ0ψ )

∑

(0,η)/∼GΣ0

cω0,η πM,>ψ(ψ, 0, η)⊗ ω.

Let us choose ρ and minimal t over the index set of Jordρ(ψ) such that At − Bt 6= 0. Then we divide itinto the following two cases.

38 BIN XU

• Bt 6= Bt+1: We can break (ρ,At, Bt,+) into (ρ,At, Bt + 1,+) and (ρ,Bt, Bt,+), and denote thenew parameter by ψ′. It also satisfies (7.1). Note

{πM,>ψ(ψ, l, η) ∈ Πψ | l = 0

}⊆{πM,>ψ′ (ψ

′, l′, η′) ∈ Πψ′ | l′ = 0

}.

If ψ′ is not trivial on the second SL(2,C), then R = 0 by the induction assumption, cf. Corol-lary 7.6. If ψ′ is trivial on the second SL(2,C), then it is necessary that At = Bt + 1. In thiscase, {

πM,>ψ(ψ, l, η) ∈ Πψ | l = 0}( Πψ′ .

Hence R = 0.• Bt = Bt+1 = · · · = Bs < Bs+1: We break (ρ,As, Bs,+) into (ρ,As, Bs + 1,+) and (ρ,Bs, Bs,+),and move (ρ,Bs, Bs,+) after (ρ,At, Bt,+) in the order >ψ. Denote the new parameter by ψ′. Italso satisfies (7.1). Since we have assumed R 6= 0 stable, then we must have

Πφψ′

⊆{πM,>ψ(ψ, l, η) ∈ Πψ | l = 0

}⊆ Πψ′

by the induction assumption. So we can assume Ai −Bi = 1 for t 6 i 6 s and it follows from thechange of order formula (cf. [Xu19, Theorem 6.1]) and Lemma 7.5 that

Πφψ′

⊇{πM,>ψ(ψ, l, η) ∈ Πψ | l = 0

}.

Take1

(s− t)!◦s−1i=t (Jac

ρBi,··· ,−Ai(·)⊗ η−Aiρ ω−((Bi+1)+···+Ai)/2

ρ χ−1ρ ) : Πψ′ → Πψ′′ ∪ {0},

where ψ′′ is obtained from ψ′ by removing (ρ,Ai, Bi,+) for t 6 i 6 s − 1. It induces a bijectionfrom Πφ

ψ′to Πψ′′ . So the stability of R 6= 0 implies that the image of

{πM,>ψ(ψ, l, η) ∈ Πψ | l = 0

}

should be all of Πψ′′ . On the other hand, the image consists of πM,>ψ(ψ′′, 0, η′′) with

η′′(ρ,Bs, Bs,+) = (−1)s−t+1η′′(ρ,As, As,+)

by direct computation.(1) If s− t+ 1 is odd, then the image can not be all of Πψ′′ . So we get a contradiction.

(2) If s − t + 1 is even, then we can further assume that (ρ,Bs, Bs,+) and (ρ,As, As,+) bothhave even multiplicities in Jord(ψ)\{(ρ,Ai, Bi,+)}si=t, otherwise the image can not be all ofΠψ′′ . As a consequence,

{πM,>ψ′ (ψ

′, l′, η′) ∈ Πψ′ | l′ = 0

}6= ∅.

Note α(SΣ0ψ′ ) = α(SΣ0

ψ′d). Then R can not be stable by the induction assumption. So we get a

contradiction again.

�

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Yau Mathematical Sciences Center and Department of Mathematics, Tsinghua University, Beijing, China

Email address: [email protected]

arXiv:2111.07591v1 [math.NT] 15 Nov 2021

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