By Lisa Jacobsen
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Extra info for Analytic theory of continued fractions III
Let τ be a smooth admissible irreducible non-supersingular representation of G over Fp with central character χ with χ as in ( 17). Then there exists a short exact sequence: (18) 0 → Ext1H =1 (I(τ ), M ) → Ext1G,χ (τ, π) → HomH (I(τ ), R1I(π)) → 0 where Ext1G,χ (τ, π) denotes the Fp -vector space of G-extensions with central character χ. Proof. Let E be the class of an exact sequence in RepG,χ : 0 /π / /τ /0. Taking I1 -invariants we obtain an exact sequence of H-modules: 0 / I(π) / I( ) / I(τ ) φE / R1 I(π) .
Assume r = 0 or r = q − 1. Set M := M (r, λ) and M := M (q − 1 − r, λ), then dimFp Ext1H =1 (M, M ) = dimFp Ext1H =1 (M, M ) = 1. Proof. 3 are projective resolution for M . If r = q − 1 then for all H =1 -modules N we have ExtiH =1 (M, N ) = 0 for i > 1 and an exact sequence: 0 −→ HomH Zγ −λ =1 (M, N ) −→ N e1 Tns −→ N e1 Tns −→ Ext1H =1 (M, N ) −→ 0. The assertion for r = q − 1 follows from this exact sequence. The case r = 0 is analogous. 8. Assume λ = 0 then we have: T (M (r, λ, η)) ∼ = π(r, λ, η), I(π(r, λ, η)) ∼ = M (r, λ, η).
Since eχs is an idempotent, this is a projective resolution of E. Applying T , we s obtain an exact sequence using T (eχs H =1 ) ∼ = c-IndG IZ χ : 0 / L1 T (E) s / c-IndG IZ χ Zγ −λ s / c-IndG IZ χ / T (E) / 0. s ∼ Now Zγ − λ is an injection since it is an injection on I(c-IndG IZ χ ) = eχs H =1 . G 1 s ∼ Hence L T (E) = 0 and T (E) = c-IndIZ χ /(Zγ − λ). Set π := π(q − 1 − r, λ(−1)r , ω r ), π := π(r, λ) and M := M (q − 1 − r, λ(−1)r , ω r ). Applying T /M /E /M / 0 we obtain an exact to the exact sequence 0 sequence: 0 / L1 T (M ) /π / T (E) /π / 0.
Analytic theory of continued fractions III by Lisa Jacobsen