Character identities in the twisted endoscopy of real by Paul Mezo

By Paul Mezo

Feel G is a true reductive algebraic workforce, ? is an automorphism of G, and ? is a quasicharacter of the crowd of actual issues G(R). less than a few extra assumptions, the speculation of twisted endoscopy affiliates to this triple actual reductive teams H. The neighborhood Langlands Correspondence walls the admissible representations of H(R) and G(R) into L-packets. the writer proves twisted personality identities among L-packets of H(R) and G(R) made out of crucial discrete sequence or limits of discrete sequence

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In particular, the subset of representations π ∈ Πϕ satisfying π ∼ = ω ⊗ (π )θ is {πw−1 Λ : w ∈ (Ω(G, S)/ΩR (G, S))δθ }. 42 6. SPECTRAL TRANSFER FOR SQUARE-INTEGRABLE REPRESENTATIONS Proof. It is a simple exercise to verify that if U is an intertwining operator satisfying U ◦ (ω −1 ⊗ π ) = (π )θ ◦ U , for some π ∈ Πϕ , then π (δ)U is an intertwining operator satisfying π (δ)U ◦ (ω −1 ⊗ π ) = (π )δθ ◦ π (δ)U . Similarly if U intertwines ω −1 ⊗ π with (π )δθ then π (δ)−1 U intertwines ω −1 ⊗ π with (π )θ .

1. Some work of Harish-Chandra and Duflo. Suppose that 1 is an irreducible square-integrable representation of L0 . In [Duf82], Duflo parametrizes the set of representations of L (up to equivalence), whose restriction to L0 is equal to 0 1 . 4, we specialize to L = Gder (R) , δθ , and Duflo’s parametrization will be conjoined with the intertwining operators occurring in Lemma 3. A discussion similar to this one may be found in §18 [Ren97]. Let us recall some of the work of Harish-Chandra ([HC66], IX and XII in [Kna86]).

Proof. By hypothesis, we have sδθ = (g −1 g )s δθ(g −1 g )−1 . (53) The centralizer in gder of the left-hand side of this equation is equal to sδθ der . The centralizer of the right-hand side is equal to Ad(g −1 g )sδθ . After exponentiating, der this identity of centralizers translates to δθ δθ (R)0 = (g −1 g ) Sder (R)0 (g −1 g )−1 . Sder δθ This identity shows that g −1 g normalizes Sder (R)0 . We may also rearrange equation (53) to obtain δθ(g −1 g )−1 δθ −1 (g −1 g ) = (s )−1 (g −1 g )−1 s(g −1 g ).

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