Number Theory

## Towards a Modulo p Langlands Correspondence for GL2 by Chistophe Breuil, Vytautas Paskunas

Posted On March 23, 2017 at 12:48 pm by / Comments Off on Towards a Modulo p Langlands Correspondence for GL2 by Chistophe Breuil, Vytautas Paskunas

The authors build new households of soft admissible $\overline{\mathbb{F}}_p$-representations of $\mathrm{GL}_2(F)$, the place $F$ is a finite extension of $\mathbb{Q}_p$. whilst $F$ is unramified, those representations have the $\mathrm{GL}_2({\mathcal O}_F)$-socle anticipated through the new generalizations of Serre's modularity conjecture. The authors' motivation is a hypothetical mod $p$ Langlands correspondence

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Extra resources for Towards a Modulo p Langlands Correspondence for GL2

Example text

L (u) := [f (u)](ns ), Then the map f → (ψ u , ψ l ) induces an isomorphism: ∼ s H 1 (I1 /Z1 , IndG P χ) −→ Hom((I1 ∩ P )/Z1 , Fp ) ⊕ Hom((I1 ∩ P )/Z1 , Fp ). (16) Proof. Since G = P I1 IndG P P ns I1 , we have an isomorphism: χ|I1 ∼ = IndII11 ∩P 1 ⊕ IndII11 ∩P s 1. G/Z As Z1 acts trivially on both sides, we may rewrite this as IndP/Z11 χ|I1 /Z1 ∼ = I /Z I /Z Ind(I11 ∩P1 )/Z1 1 ⊕ Ind(I11 ∩P1 s )/Z1 1. 5] that: 1 s ∼ 1 H 1 (I1 /Z1 , IndG P χ) = H ((I1 ∩ P )/Z1 , 1) ⊕ H ((I1 ∩ P )/Z1 , 1) which implies the assertion.

1 that ε ≺ δ implies |ε| + 1 = |δ|. 7 gives then the assertion. 9. Let τ be a subrepresentation of V2p−2−r . Assume that there exists an integer k ∈ {0, · · · , f } such that, if Vr(ε) ⊗ dete(ε) occurs in cosocΓ τ , then |ε| = k. Then the graded pieces of the cosocle ﬁltration of τ are given by: τi ∼ = Vr(ε) ⊗ dete(ε) . ε∈Σr (τ ) |ε|=k−i Proof. 7 together with ε ≺ δ ⇒ |ε| = |δ| − 1. 10. Let λ, λ ∈ I(x0 , · · · , xf −1 ) (see Chapter 3). We say λ and λ are compatible if, whenever λi (xi ) ∈ {p − 2 − xi − ±1, xi ± 1} and λi (xi ) ∈ {p − 2 − xi − ±1, xi ± 1} for the same i, then the signs of the ±1 are the same in λi (xi ) and λi (xi ).

5. Let c ∈ [ϕTns ] ns −1 1 )ns . −[μ] OF then: 1 0 c 1 − [ϕTns ](ns ) = μr ψ l μ∈F× q 1 [μ2 ]c 0 1 . 7. COMPUTATION OF R1 I FOR PRINCIPAL SERIES 41 1 0 Proof. Note that the left hand side is equal to μ∈Fq ϕ ( 01 −c 1 )( −[μ] 1 ) − 1 0 1 pF ϕ ( −[μ] 1 ) . Since ( 0 1 ) is contained in the derived subgroup of I1 ∩ P , the term corresponding to μ = 0 is: 1 −c 0 1 ϕ 1 −c 0 1 − ϕ(1) = ψ u = 0. 4 we get: −1 1 0 1 0 1 −c −[μ] ϕ ( 01 −c 1 )( −[μ] 1 ) −ϕ ( −[μ] 1 ) = ϕ ( 0 1 )( 0 = ϕ (−[μ] 0 = χ (−[μ] 0 −1 −1 1 )n −[μ] s − ϕ (−[μ] 0 1 )n ( 12 0 ) −[μ] s [μ ]c 1 0 ) −[μ] −1 −ϕ (−[μ] 0 1 )n −[μ] s −1 1 )n −[μ] s ψ l ( [μ12 ]c 01 ) .