ANALYTIC RTF: GEOMETRIC SIDE

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1 ANALYTIC RTF: GEOMETRIC SIDE JINGWEI XIAO 1. The big picture Yesterday we defined a certain geometric quantity I r (f). Today we will define an analytic quantity J r (f). Both of these have two expansion: u P 1 (F ) 0 I γ(u, f) I r (f) π I r(π, f). (1.1) and u P 1 (F ) 0 J γ(u, f) J r (f) π J r(π, f). (1.2) The left sides of (1.1), (1.2) are expansions in terms of orbital integrals. The right side are the quantities that we want to compare: J r (π, f) L (r) (π F, 1/2), and I r (π, f) = [Sht T ] π, f [Sht T ] π. The functions f are test functions which provide the flexibility to isolate the terms of interest. 2. The relative trace formula Let F be a global field, the function field of X for X/F q. Let G/F be a reductive group, and H 1, H 2 G subgroups over F. We ll write [G] := G(F )\G(A), and similarly for H i The kernel. Let A = A F, [G] = G(F )\G(A). For f Cc (G(A)), we define the kernel function K f (g 1, g 2 ) := f(g1 1 γg 2). γ G(F ) The point is that G(A) acts on C (G(F )\G(A)), and for φ C (G(F )\G(A)) we have π(f) φ = K f (g 1, g 2 )φ(g 2 ) dg 2. (2.1) G(F )\G(A) The relative trace formula involves the quantity K f (h 1, h 2 ) dh 1 dh 2 (2.2) [H 1 ] [H 2 ] 1

2 2 JINGWEI XIAO 2.2. The geometric expansion. The geometric expansion of (2.2) is 1 γh 2) dh 1 dh 2 [H 1 ] [H 2 ] γ G(F ) = γ H 1 (F )\G(F )/H 2 (F ) Rearranging, one rewrites this as = γ H 1 (F )\G(F )/H 2 (F ) Here γ denotes the stabilizer of γ: [H 1 ] [H 2 ] δ H 1 (F )γh 2 (F ) (H 1 H 2 ) γ(f )\H 1 (A) H 2 (A) (H 1 H 2 ) γ := {(h 1, h 2 ): h 1 1 γh 2 = γ}. 1 δh 2) dh 1 dh 2 1 γh 2) The spectral expansion. Now we rewrite (2.1) in a different way. The idea is to decompose K f π(f)φ φ π cuspidal where φ runs over orthogonal basis of π. This is a bit of lie, as one also needs to consider the residual and Eisenstein parts, but it roughly works. Using this, we can rewrite (2.1) = π(f)φ dh 1 φ dh 2 π [H 1 ] [H 2 ] Here the term [H 1 ] π(f)φ is a period P H 1 (π(f)φ). φ 2.4. The split subtorus. For G = PGL 2, set H 1 = H 2 = A to be the diagonal torus of G. For f Cc (G(A)) we get a kernel function K f. We consider the integral K f (h 1, h 2 ) h 1 h 2 s η(h 2 ) [A] [A] ( x where if h = then h = x/y, and η : A y) F {±1} is the character corresponding by class field theory to F /F. There s an issue with this integral. Since A = G m /F, [A] := F \A F is not compact, since [A]/ O x = Pic(X). x This has infinitely many connected components, which are finite since they are isomorphic to Pic 0 (X). To regularize the integral, define {( } x [A] n = : v(x/y) = n. y) φ We have a map v : A F / O x Z

3 sending v(π x ) = log q (q x ) for any x X. The [A n ] are compact, so we can talk about This is actually a polynomial in q s. ANALYTIC RTF: GEOMETRIC SIDE 3 [A] n1 [A] n2 K(h 1, h 2 ) h 1 h 2 s η(s) ds. Proposition 2.1. For each f, there exists N such that n 1 + n 2 N implies [A] n1 [A] n2 K f (h 1, h 2 )η(h 2 ) h 1 h 2 s = 0. Assuming this claim, we can define the regularized integral reg := [A 1 ] [A 2 ] n 1,n 2 [A] n1 [A] n2 3. Spectral expansion The goal of this section is to establish the identity J r (f) = J γ (u, f). u P 1 (F ) The invariant map. We have seen that in the RTF, we care about the double coset space H 1 (F )\G(F )/H 2 (F ). For G = PGL 2, H 1 = H 2 = A we can define an invariant map A(F )\ PGL 2 (F )/A(F ) P 1 {1} by ( ) a b c d bc ad. Proposition 3.1. For x P 1 (F ) {1}, we have single ( ) orbit ( ) ( inv 1,, (x) = ( ) ( ) ( 1 1 1,, Also, γ is regular semisimple iff inv(γ) 0,. ) 1 ) 1 x 0, x = 0 x =

4 4 JINGWEI XIAO 3.2. Expansion. This lets us write where and J(f, s) = J(γ, f, s) = γ A(F )\G(F )/A(F ) [A] [A] K f,γ (h 1, h 2 ) = For u P 1 (F ) {1}, we define J(u, f, s) = J(γ, f, s) K f,γ (h 1, h 2 ) h 1 h 2 s η(h 2 ) dh 2 δ A(F )γa(f ) γ A(F )\G(F )/A(F ) inv(γ)=u 1 δh 1 2 ) Higher derivatives. Now we define ( ) d r J r (f) = J(f, s) s=0. ds Similarly, we have a decomposition Also J r (u, f) = J r (f) = The goal is to establish the identity Yesterday we defined J(γ, f, s). ( ) d r J(u, f, s) s=0. ds u P 1 (F ) {1} 4. The case r = 0 u P 1 (F ) 0 J r (u, f). I γ (u, f) = I r (f). I 0 (f) := Sht 0 T, f Sht 0 T. Let F /F be a quadratic extension. Define T = Res F /F G m,f /G m,f. It turns out that we also have an equality I 0 (f) = K f (h 1, h 2 ). [T ] [T ] The Waldspurger formula can be reinterpreted in these terms: J 0 (γ, f) = J 0 (f) L(π, 0). γ A(F )\G(F )/A(F )

5 ANALYTIC RTF: GEOMETRIC SIDE 5 While I 0 (γ, f) = I 0 (f) γ T (F )\G(F )/T (F ) [T ] φ π. 5. The equality I 0 (f) = J 0 (f) The strategy to relate the things is to relate the orbital integrals. So we first need to relate the orbits Matching double cosets. Let G = PGL 2 or D /F, where D is a quaternion algebra over F (with an embedding F D). Theorem 5.1. We have A(F )\ PGL 2 (F ) rss /A(F ) = G=PGL 2 or D /F T (F )\G(F ) rss /T (F ). Proof. We consider G = PGL 2 or D. Let H = M 2 (F ) or D, so G = H. We have an embedding F H. There exists ɛ H(F ) such that ɛxɛ 1 = x for x F. The choice of ɛ is unique up to multiplication by (F ). By computation, ɛ 2 Z(H) = F, so [ɛ 2 ] F / Nm(F ) is well-defined. Proposition 5.2. The element [ɛ 2 ] F / Nm(F ) determines H. We have an invariant map sending T (F )\H /T (F ) inv P 1 (F ) {1} h 1 + ɛh 2 h 2h 2 ɛ 2. h 1 h 1 The image lands in ɛ 2 Nm((F ) ), and is equal in the regular semisimple case since this is equivalent to h 1 h Matching orbital integrals. Now that we ve matched up the double cosets, we turn to showing that I 0 (f) = J 0 (f) for f = v X f v H G a bi-k-invariant function. Writing out the expansion, this comes down to I 0 (u, f) = J 0 (u, f) u u This becomes a fundamental lemma type statement. Consider F v/f v a quadratic extension. Let f C (PGL 2 (F v )) be bi-k-invariant. We have A(F v )\ PGL 2 (F )/A(F v ) = T (F v )\ PGL 2 (F v )/T (F v ) T (F v )\D /T (F v ) By Theorem 5.1, each γ on the left side matches up with a γ 1 T (F v )\ PGL 2 (F v )/T (F v ) or γ 2 T (F v )\D /T (F v ). One can then compute by hand that the corresponding orbital integrals are equal.

6 6 JINGWEI XIAO If γ γ 1 T (F v )\ PGL 2 (F v )/T (F v ), then ± 1 γh 2)η(h 2 ) = A(F ) A(F ) T (F ) T (F ) so I 0 (u, f) = J 0 (u, f). If γ γ 2 T (F v )\D /T (F v ), then 1 γh 2)η(h 2 ) dh 2 = 0 A(F ) A(F ) so the extra double cosets do not contribute. This shows that I 0 (f) = J 0 (f). 1 γh 2)