LANGLANDS FOR GL(2): GALOIS TO AUTOMORPHIC, III (D APRÈS DRINFELD)

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1 LANGLANDS FOR GL(2): GALOIS TO AUTOMORPHIC, III (D APRÈS DRINFELD) TONY FENG 1. Recollections Let ω be a meromorphic differential on X and ψ 0 : F q Q l be an additive character. Last time we produced f C (GL 2 (O)\ GL 2 (A K )/B(K), Q l ) by the formula (( ) ( a 1 z f b 1)) r(div ω + Div(b/a))ψ 0 ( z, ω ) (1.1) ω Ω K where Ω K = Ω K 0. We know that f would be a cuspidal eigenform for GL 2 if we can show that f is actually right-invariant by the full GL 2 (K), and that is what we aim to show. 2. A geometric formula We are going to give a geometric interpretation of the function f defined in (1.1). We interpret GL 2 (O)\ GL 2 (A K )/ GL 2 (K) = Bun GL2 (k) =: Bun 2. and GL 2 (O)\ GL 2 (A K )/B(K) = Bun B (k) =: Flag 2. Here Flag 2 parametrizes flags (A, L) where L is a rank 2 vector bundle on X and A is a line-subbundle of L such that L/A is a invertible (i.e. A is a maximal line sub-bundle of L). We want to reinterpret f in these terms. First, what flag corresponds to ( ) ( a 1 z? b 1) We can view a and b as GL 1 -adeles, whose correspondence we understand well: it is the correspondence between divisors and line bundles: a A and b B := L/A. Then L should be determined by an extension class 0 A L B 0 in Ext 1 (B, A) = H 1 (B 1 A). This is the image of z under the natural isomorphism A K /(K + b a O) = H 1 (B 1 A) (2.1) Date: March 4,

2 2 TONY FENG Proof of (2.1). Consider the short exact sequence 0 O(D) K K/O(D) 0 We have an identification H 0 (K/O(D)) A K /D 1 O described by taking polar parts. (For clarity, if D = n v ϖ v then D 1 O means v ϖ 1 v O v.) The desired isomorphism then follows from the long exact sequence, noting that the higher cohomology of K vanishes since it is constant: A K /(K + b a O) H 1 (B 1 A) H 0 (K/O(Div(a/b))) H 1 (O(Div(a/b))). Let s first rewrite (1.1) slightly. Notice that changing ω by an F q -multiple doesn t affect the divisor. Therefore, (( ) ( a 1 z f q r(div ω+div(b/a)) b 1)) r(div ω+div(b/a)). ω Ω K /F q ω,z =0 ω Ω K /F q ω,z =0 (There is a little shuffling around of terms here, and using the fact that the sum over F q of a non-trivial character is 1.) Let P (L, A) = PH 0 (A 1 B Ω) and H(L, A) P (L, A) be the hyperplane cut out by z = 0. Then our geometric reformulation is f q r(d) r(d). (2.2) H(L,A) P (L,A) In these geometric terms the task is to show that the above formula is independent of the sub-bundle A, hence descends to a function of L. 3. Geometric interpretation of Hecke operators Definition 3.1. A lower modification of L at v X is a rank two sub-bundle L such that L( v) L L An upper modification of L is an L such that L is a lower modification of L at v.

3 LANGLANDS FOR GL(2): GALOIS TO AUTOMORPHIC, III (D APRÈS DRINFELD) 3 The space of lower modifications of L at v can be pictured as a projective space P 1 k v, since it is determined by a choice of line in L/L( v) = k 2 v. The space of modifications of L at v up to scalar is the affine Grassmannian for PGL 2, which can be pictured as the Cayley graph for a free group on q v + 1 generators. In these terms, the Hecke operators on Bun 2 have the following interpretation: (U v f)(l) = f(l( v)) (T v f)(l) = f(l ). The Hecke operators on Flag 2 have a completely analogous interpretation: (U v f)(l, A) = f(l( v), A( v)) (T v f)(l, A) = f(l, A L ). To first order, we can imagine that there are two transverse direction in the affine Grassmannian for GL 2 : one modifying lattices by scalar and one which looks like the affine Grassmannian for PGL 2. The Hecke operator U v translates in the first direction, and the Hecke operator T v averages in the second direction. Example 3.2. For fun (and later use) let s think about what A L looks like as L ranges over Lower v (L). Looking just in the formal neighborhood of v, we have that lower modifications are represented by choices of lines in L/L( v). Also A represents a line in L/L( v), and so exactly one lower modification L contains A, namely the one corresponding to the line of A. What about upper modifications? There is exactly one upper modification of L in which A is not maximal, namely the pushout of L along A A(v). This can also be described as the pullback of B B(v) in the short exact sequence 0 A(v) L B 0 0 A(v) L(v) B(v) 0

4 4 TONY FENG 4. Geometric formulation of invariance Definition 4.1. For L Bun 2 let h(l) denote the least degree of invertible quotient sheaves of L F q on X F q. Note that h(l) >, because H 1 (X, L) is finite-dimensional. If we had a quotient L B with deg B 0, then the long exact sequence associated to 0 A L B 0 tells us that H 1 (L) H 1 (B); then Riemann-Roch gives a lower bound on deg B. We aim to prove the following result. Theorem 4.2. Let f : Flag 2 Q l be a Hecke eigenfunction. Suppose that for some N Z the following condition is satisfied: f(l, A) = f(l, A ) for all L, A, A such that deg A, A < h(l) N. Then f(l, A) is independent of A. In other words, if f(l, A) is independent of A whenever deg A is sufficiently negative, then f(l, A) is completely independent of A. Define the function g(l) := f(l, A) for any A with degree < h(l) N. We want to show that f(l, A) = g(l) for all A. First, we require a little Lemma guaranteeing that A is actually welldefined. Lemma 4.3. For any rank 2 vector bundle L on X, there exist maximal line subbundles A L of arbitrarily negative degree. Proof. The set of line sub-bundles of any fixed degree is finite, since Pic 0 (X) is finite (the rational poitns of a finite type variety over a finite field) and for every A the space Hom(A, L) is finite (being a finite-dimensional vector space over a finite field). However, the set of maximal rank one sub-bundles is infinite, being in bijection with lines in the generic fiber of L, which can be identified with K 2. Indeed, any such line defines a section of PL on some open subset, which can be completed by the valuative criterion to some section of the whole projective bundle, thus picking out a line sub-bundle of L with the right generic fiber. Now we ll use the Hecke condition to propagate the equality f(l, A) = g(l) through the Bun 2. Suppose f has eigenvalues t v, u v for T v, U v. Then g is also an eigenfunction with eigenvalues t v, u v. Indeed, U v g(l) = g(l( v), A( v)) and since deg A < h(l) 1 and h(l( v)) = h(l) 1 this is still in the range to be f(l( v), A( v)) = u v f(l, A). Similarly, T v g(l) = g(l ) = f(l, A L )

5 LANGLANDS FOR GL(2): GALOIS TO AUTOMORPHIC, III (D APRÈS DRINFELD) 5 for deg A h(l ), and by the Hecke eigenfunction property of f this is t v g(l). Notice that there is something funny going on here: if we choose deg A to be right at the limit h(l) N then one lower modification L has L A = A. Thus the Hecke eigensheaf property lets us propagate the invariance to higher degree relative to h(l). That is the idea exploited in the following proposition. Proposition 4.4. Let (L, A) Flag 2 and L be an upper modification of L at v such that A is maximal as a subsheaf of L. Suppose that for every L L we have f( L, A) = g( L). Then f(l, A) = g(l). Recall that the hypothesis of Theorem 4.2 is that f is independent of A once A has small enough degree with respect to h(l); since A is smaller with respect to L than L, this is progress. Proof. Consider applying T v to f(l;, A): we get t v f(l, A) = f(l, A) + and also t v g(l ) = g(l) + L Lower v(l ) L Lower v(l ) f(l, A L ) (4.1) g(l ) (4.2) but also by hypothesis f(l, A) = g(l ). Recall from Example 3.2 that L A = A( v), so f(l, A L ) = f(l, A( v)) = u v f(l (v), A) where now L (v) is an upper modification of L with A as a maximal subbundle. L (v) L L L By assumption f(l (v), A) = g(l (v)) and by the Hecke eigensheaf property for U v we have g(l (v)) = u 1 v g(l ), so the conclusion is that f(l, A L ) = g(l ). But then comparing (4.1) and (4.2) gives the result. Lemma 4.5. Let (L, A) Flag 2 with deg L > 2h(L). Then there exists an upper modification L of L at v such that h(l ) > h(l) and A is maximal as a subsheaf of L. Proof. Let A L be a line sub-bundle of maximal degree, so h(l) = deg(l/a ). There is one upper modification of L such that A is not maximal and one such that A is not maximal, so of the q v + 1 upper modifications there is at least one L in which both remain maximal. We claim that this L does the trick. Indeed, if it maps

6 6 TONY FENG to a bundle Q of degree h(l) then the kernel is of larger degree than A. Since deg A > deg h(l) the composite map A L Q is zero, so the kernel strictly contains A, a contradiction. Now we conclude the proof of Theorem 4.2. Consider the statement P (m, n): For every L containing A as a maximal invertible subsheaf and such that h(l) m, deg L n we have f(l, A) = g(l). The hypothesis is that P (m, n) holds for m > deg A + N, which is everything sufficiently far to the right in the (m, n) plane. Next, Lemma 4.5 implies that we can find an upper modification L preserving the maximality of A if n > 2m, but increasing m m+1. A further upper modification can only increase h, so Proposition 4.4 then implies that P (m + 1, n) = P (m, n) if n > 2m. This gives the truth for P (m, n) for all sufficiently large n. Finally, Proposition 4.4 implies that P (m, n + 1) = P (m, n). This completes the truth of P (m, n) for all m, n. References [1] Drinfeld, Two-dimensional l-adic representations of the fundamental group of a curve over a finite field and automorphic forms on GL(2). American Journal of Mathematics 105 (1983).