Gross–Prasad periods for reducible representations

Theorem 3.1 (Prasad, Harris–Scholl).

Let π1, π2, π3 be representations of GL2⁡(F) of Whittaker type, with central characters ωi such that ω1⁢ω2⁢ω3=1. Then we have

and

Theorem 4.1 (a).

Let π be an irreducible generic representation of GL2⁡(E) such that ωπ|F×=1. Then we have dim⁡HomH⁡(π⊠ΣF,1)=1.

Remark 4.2.

Note that the case when E/F is unramified, and π is unramified and tempered, is part of [6, Theorem 4.1.1]. However, the proof of this statement given in [6] has a minor error which means the argument does not work when π is the normalised induction of the trivial character of BE. So the argument below fixes this small gap.

Proof.

We first observe that HomH⁡(π⊠ΣF,𝟙) is non-zero. Since π is generic, it has a Whittaker model 𝒲⁢(π) with respect to any non-trivial additive character of E. We may suppose that this additive character is trivial on F, so that we may define the Asai zeta-integral

for W∈𝒲⁢(π) and Φ∈𝒮⁢(F2) (the space of Schwartz functions on F). Here NH is the upper-triangular unipotent subgroup of H.

It is well known that this integral converges for ℜ⁡(s)≫0 and has meromorphic continuation to the whole complex plane; and the values of Z⁢(-,-,s) span a non-zero fractional ideal of 𝐂⁢[qs,q-s], generated by an L-factor independent of Φ and W, which is the Asai L-factor L⁢(As⁡(π),s). Thus the map

defines a non-zero, H-invariant bilinear form W⁢(π)⊗𝒮⁢(F2)→𝐂. Since the maximal quotient of 𝒮⁢(F2) on which F× acts trivially is isomorphic to ΣF (see for example [10, Proposition 3.3 (b)]), this shows that HomH⁡(π⊠ΣF,𝟙)≠0 as claimed.

So, to prove Theorem 4.1 (a), it suffices to show that dim⁡HomH⁡(π⊠ΣF,𝟙)⩽1. As π has unitary central character, it is either a discrete-series representation, in which case it is automatically tempered, or an irreducible principal series, which may or may not be tempered. We shall consider these cases separately.

Note that [1, Theorem 1.1] states that if π is an irreducible tempered representation of GL2⁡(E), then we have dim⁡HomM⁢(F)⁡(π,𝟙)=1, where M⁢(F)={(⋆⋆01)} is the mirabolic subgroup of GL2⁡(F). If we assume ωπ|F×=1, then since F×⋅M⁢(F)=B⁢(F) is the Borel subgroup of GL2⁡(F), we have

As IndB⁢(F)H⁡(𝟙)=ΣF∨, this proves Theorem 4.1 (a) for tempered π.

We now consider the principal-series case. For α,β smooth characters of E×, we write IE⁢(α,β) for the normalised induction to GL2⁡(E) of the character α⊠β of B⁢(E). Note that this representation is tempered if and only if α and β are unitary. We suppose α/β≠|⋅|E±1 and α⁢β|F×=1. Then we have the following results:

1. HomH⁡(π⊠StF,𝟙) is zero if α⁢βc=1, and 1-dimensional otherwise, where βc denotes the character x↦β⁢(xc). See [17, Remark 4.1.1].

2. HomH⁡(π⊠𝟙,𝟙) is 1-dimensional if α⁢βc=1, or if α|F×=β|F×=1; otherwise it is 0. See [13, Theorem 5.2].

We conclude that exactly one of HomH⁡(π⊠StF,𝟙) and HomH⁡(π⊠𝟙,𝟙) is non-zero (and Theorem 4.1 (a) therefore follows), unless π is of the form IE⁢(α,β) with α|F×=β|F×=1 and α⁢βc≠1. However, in this exceptional case α and β are unitary, and thus π is tempered, so Theorem 4.1 (a) has already been established for π above. This completes the proof of Theorem 4.1 (a). ∎

Remark 4.3.

It follows, in particular, that for a generic irreducible representation π of GL2⁡(E), we have HomH⁡(π,𝟙)≠0 (i.e. π is “F-distinguished”) if and only if the zeta-integral ((${\dagger}$)) factors through the 1-dimensional quotient of ΣF, and thus vanishes on all Φ with Φ⁢(0,0)=0; that is, s=0 is an exceptional pole of the Asai L-factor. This is the n=2 case of a theorem due to Matringe [12, Theorem 3.1] applying to GLn⁡(E)-representations. See [10] for analogous results and conjectures regarding poles of zeta-integrals for GSp4 and GSp4×GL2.

Lemma 4.4.

Let σ be an irreducible generic representation of GL2⁡(F) with ωσ=1. Then

Moreover, if σ≠StF, then we have

while for σ=StF we have

Proof.

If σ is not a twist of Steinberg, then its Weil–Deligne representation has trivial monodromy action, so we compute that

Since σ has trivial central character, ε⁢(σ)=±1. If σ is supercuspidal we are done, since in this case ρσIF=0. If σ is principal series, then ρσIF must be either 0, or all of ρσ, since ρσ has determinant 1. Thus det(-Frob:ρσIF)=1, so ε⁢(As⁡(StE)×σ)=ε⁢(σ)⁢ε⁢(σ×ωE/F), proving the claim in this case. The case when σ is a twist of the Steinberg by a non-trivial (necessarily quadratic) character can be computed similarly. ∎

Theorem 4.1 (b).

Let σ be an irreducible generic representation of GL2⁡(F) with ωσ=1. Then:

1. If ε⁢(σ)⁢ε⁢(σ×ωE/F)=ωE/F⁢(-1), then dim⁡HomH⁡(ΣE⊠σ,𝟙)=1 and HomH′⁡(ΣE⊠σ′,𝟙)=0.

2. If ε⁢(σ)⁢ε⁢(σ×ωE/F)=-ωE/F⁢(-1), then HomH⁡(ΣE⊠σ,𝟙)=0 and dim⁡HomH′⁡(ΣE⊠σ′,𝟙)=1.

Proof.

We first consider the situation for H′. This case is relatively simple, since H′ is compact modulo centre, and hence the functor of H′-invariants is exact on the category of H′-representations trivial on F×. So we have

Using Prasad’s results for HomH′⁡(StE⊗σ′,𝟙) and the preceding lemma, we see that dim⁡HomH′⁡(ΣE⊗σ′,𝟙) has dimension 1 if ε⁢(σ)⁢ε⁢(σ×ωE/F)=-ωE/F⁢(-1) and is zero otherwise, as required.

For the group H, the situation is a little more complicated: since σ is generic, we have HomH⁡(σ,𝟙) is zero, and hence there is an exact sequence

This gives the desired formula for dim⁡HomH⁡(ΣE⊗σ,𝟙) in all cases except when σ=StF, in which case we must show that the non-trivial H-invariant period of StE⊗StF does not lift to ΣE⊗StF. This can be done directly: we can compute ΣE|GL2⁡(F) via Mackey theory, using the two orbits of H on 𝐏1⁢(E) to obtain the exact sequence

The latter representation is irreducible and has no homomorphisms to StF; and we saw in the proof of Theorem 4.1 (a) that

This shows that HomH⁡(ΣE⊠StF,𝟙)=0, completing the proof. ∎

Remark 4.5.

We are grateful to the anonymous referee for pointing out the significance of the vanishing of ExtPGL2⁡(F)1⁡(σ,𝟙); the original version of this paper used a different and rather more complicated argument.

Theorem 4.1 (c).

Let η be a quadratic character of F× (possibly trivial). Then we have

Proof.

The computation of the ε-factor is immediate; and by a zeta-integral argument as before, we can show that HomH⁡(ΣE⊠ΣF,η)≠0 (since the representation ΣE, despite being reducible, has a well-defined Whittaker model). So it suffices to show that the hom-space has dimension ⩽1.

If η is not the trivial character, then

so the desired Hom-space injects into HomH⁡(StE⊠ΣF,η), which is 1-dimensional by Theorem 4.1 (a).

If η is trivial, then we have seen above that HomH⁡(ΣE⊠StF,𝟙) is zero. So

From the Mackey decomposition of ΣE|GL2⁡(F) above, one sees easily that this space is 1-dimensional. ∎

Theorem 5.1.

Let π be a Whittaker-type representation of GL2⁡(E). Then the space HomH⁡(π,1) has dimension 1 if ε⁢(As⁡(π))⁢ωA⁢(-1)=1 and is zero otherwise.

Proof.

The case of irreducible generic π is proved in [17] assuming π non-supercuspidal, and the supercuspidal case is filled in by [18]. In this case, the only example of a reducible Whittaker-type representation of G is ΣE⊗η, where η is a character of E×; and the central-character condition implies that λ=η|F× must be trivial or quadratic.

The ε-factors ε⁢(As⁡(StE)×λ) are computed in [17, Section 8]. We find that ε⁢(As⁡(ΣE)×λ)⁢ωE/F⁢(-1) is always +1. On the other hand, ε⁢(As⁡(StE)×λ)⁢ωE/F⁢(-1) is +1 if λ is non-trivial quadratic, and -1 if λ=1. So it follows that exactly one of HomH⁡(𝟙,λ) and HomH⁡(StE,λ) is non-zero, implying that dim⁡HomH⁡(ΣE⊗η,𝟙)⩽1.

To complete the proof, we must show that when λ≠1, the H-invariant homomorphism HomH⁡(StE,λ) extends to ΣE. However, this is clear since the obstruction lies in ExtH1⁡(𝟙,λ), which is zero. ∎