Theta correspondence for p-adic dual pairs of type I

Theorem A.1.

Let l denote the largest integer which satisfies the following criteria:

1. ϕ contains χV⁢Sl,χV⁢Sl-2,…,χV⁢Sκ,

2. the multiplicity of χV⁢Si in ϕ is odd, for i=κ,κ+2,…,l-2,

3. η⁢(χV⁢Si)=-η⁢(χV⁢Si+2), i=κ,κ+2,…,l-2,

4. if κ=2, then

   η⁢(χV⁢S2)={ϵδ(χV=𝟙)if ⁢E=F⁢ and ⁢m≢n⁢(mod⁡2),-1if ⁢E≠F⁢ and ⁢m≡n⁢(mod⁡2).

   Here we use

   δ(χV=𝟙)={1if ⁢χV=𝟙,-1if ⁢χV≠𝟙.

Then l⁢(τ)=l. (We set l=-1 if no positive integer meets the above criteria.)

Theorem A.2.

The following statements hold.

1. (Low rank) Let l>0 and let (ϕ′,η′) denote the L-parameter of θl⁢(τ). Then θl⁢(τ) is tempered and

   ϕ′=(ϕ⊗χV-1⁢χW)-χW⁢Sl.

   Furthermore, η′ can be deduced from η.

2. (Almost equal rank) Let

   l0={0if ⁢κ=2,1if ⁢κ=1.

   Let (ϕ′,η′) denote the L-parameter of θ-l0⁢(τ). Then

   ϕ′={(ϕ⊗χV-1⁢χW)+χW⁢S1if ⁢l0=1,ϕ⊗χV-1⁢χWif ⁢l0=0,

   and η′ can be deduced from η . In particular, θ-l0⁢(τ) is tempered.

3. (high rank) Let l>0. Then

   θ-l(τ)=L(χW|⋅|l-12,χW|⋅|l-32,…,χW|⋅|1-l02;θ-l0(τ)).

Theorem A.3.

Let l0=l⁢(τ). Recall that the first non-zero lift of τ on the going-up tower is θ-2-l0⁢(τ).

1. (First lifts, odd multiplicity) Assume that l0=0 or that Sl0 appears in ϕ with odd multiplicity. Then θ-2-l0⁢(τ) is tempered. If (ϕ′,η′) is its parameter, we have

   ϕ′=(ϕ⊗χV-1⁢χW)+χW⁢Sl0+2

   and η′ can be deduced from η.

2. (First lifts, even multiplicity) Assume that l0>0 and that Sl0 appears in ϕ with even multiplicity, say 2⁢h>0. Then there exists an irreducible tempered representation τ0 such that τ is a direct summand of

   χV⁢Stl0×…×χV⁢Stl0⏟h⁢ times⋊τ0;

   the parameter of τ0 is obtained by removing all the occurrences of χV⁢Sl0 from ϕ . In particular, we have l⁢(τ0)=l0-2; the first lift of τ0 on the going-up tower is θ-l0⁢(τ0) and is described by part a) of this theorem. The representation

   τ′=χW⁢Stl0×…×χW⁢Stl0⏟h-1⁢ times⋊θ-l0⁢(τ0)

   is irreducible and tempered, and we have

   θ-2-l0(τ)=L(χWδ(|⋅|1-l02,|⋅|l0+12);τ′).

3. (Higher lifts) For any l>l0+2, θ-l⁢(τ) is the unique irreducible quotient of

   χW|⋅|l-12×χW|⋅|l-32×…×χW|⋅|3+l02⋊θ-l0-2(τ).

Lemma B.1.

Let σ∈Irr⁢(Gn) be an irreducible discrete series representation. Assume that Θ-l⁢(σ) possesses a non-tempered irreducible subquotient τ. Then l>0 and

where τ1 is an irreducible subquotient of Θ2-l⁢(σ). In particular, Θ2-l⁢(σ)≠0.

Proof.

This corresponds directly to [30, Theorem 4.1]. Since τ is non-tempered, there is a segment [ρ⁢να,ρ⁢νβ] such that β-α∈ℤ, α+β<0 and an irreducible representation τ1 such that

Since τ is a subquotient of Θ-l⁢(σ) and all the representations of the form χW⁢ρ⁢νγ are cuspidal, it follows immediately from the above that there is a quotient of RP⁢(Θ-l⁢(σ)) of the form χW⁢ρ⁢νβ⊗…⊗χW⁢ρ⁢να⊗τ2 for some irreducible representation τ2. Therefore, we have a non-zero intertwining

Now let k≥1 be minimal such that there exists a sequence β=γ0>γ1>…>γk=α-1 such that the image of ψ is contained in

In that case, because of the minimality of k, we can permute the segments δ⁢([ρ⁢νγj+1,ρ⁢νγj]). In other words, for any j=0,…,k-1 we get an intertwining

for some irreducible τ3 (which depends on j). We know that σ⊗Θ-l⁢(σ) is a quotient of the Weil representation ωm,n with m=n+ϵ0+l. Therefore we have a non-zero Gn-intertwining

Frobenius reciprocity and Proposition 4.5 now imply that we have only two options: either

or

in which case ρνγj+1+1=|⋅|1-l2. Now let j=k-1. We show that (B.1) is not possible. Indeed, for j=k-1 we have

Therefore (B.1) contradicts the square integrability criterion for σ. Thus (B.2) holds, and the same argument now shows that γk-1=γk+1. Indeed, we look at the segment appearing in (B.2). If it is non-trivial, then setting j=k-1 we get

which again contradicts the fact that σ is a discrete series representation. Notice that the last inequality holds because (B.2) necessarily implies that α=1-l2∈12⁢ℤ. Since β-α∈ℤ, this means that α+β<0 implies the stronger inequality α+β≤-1.

This proves that (B.2) must hold with

The next step is to show k=1. If not, we can look at j=k-2. We first show that (B.1) is not possible. If it did hold, we would have

which again contradicts the square integrability of σ. But that means that (B.2) must hold, which implies γk-1+1=1-l2=γk+1. This contradicts γk-1>γk, so is impossible. Therefore k=1.

We now know that β=α=1-l2. In particular, since α+β<0, we have l>0. In short, if τ is non-tempered we have shown that τ↪χW|⋅|1-l2⋊τ1, i.e.

with l>0. Lemma B.1 now follows from Lemma 6.5. ∎

Lemma B.2.

Let σ∈Irr⁢(Gn) be a tempered representation. Let l>0 and let τ be an irreducible subquotient of Θ-l⁢(σ), and at the same time a quotient of χW|⋅|l-12⋊τ1 for some irreducible representation τ1. Then τ1 is a subquotient of Θ2-l⁢(σ); in particular, Θ2-l⁢(σ)≠0.

Proof of Lemma B.2.

Since τ↪χW|⋅|1-l2⋊τ1, Frobenius reciprocity gives

As, by assumption, τ is a subquotient of Θ-l⁢(σ), this implies that χW|⋅|1-l2⊗τ1 is a subquotient of RQ1⁢(Θ-l⁢(σ)). We now use the fact that Θ-l⁢(σ) – and therefore also RQ1⁢(Θ-l⁢(σ)) – is admissible. We can thus decompose

where μ runs over a finite set of characters of GL1⁡(E)=E×, and RQ1⁢(Θ-l⁢(σ))μ is the maximal subrepresentation of RQ1⁢(Θ-l⁢(σ)) on which GL1⁡(E) acts by μ. The above discussion now shows that RQ1⁢(Θ-l⁢(σ))χW|⋅|1-l2 is non-zero. Note that σ⊗RQ1⁢(Θ-l⁢(σ))χW|⋅|1-l2 is a quotient of RQ1⁢(ωm,n). Using Theorem 4.4, we see that RQ1⁢(ωm,n) has the following filtration:

We claim that J1 cannot participate in the map RQ1⁢(ωm,n)→σ⊗RQ1⁢(Θ-l⁢(σ))χW|⋅|1-l2. Assume the contrary, i.e. that

Then Bernstein’s Frobenius reciprocity shows that

here P¯1 denotes the parabolic opposite to P1. This would imply that RP¯1⁢(σ) has a subrepresentation of the form χV|⋅|l-12⊗σ′ for some σ′≠0. But this contradicts the square integrability criterion applied to σ. Therefore σ⊗RQ1⁢(Θ-l⁢(σ))χW|⋅|1-l2 is a quotient of J0, which immediately implies that τ1 is a subquotient of Θ2-l⁢(σ). ∎