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Base change for GL(2)
Base change for GL(2)
Robert P. Langlands
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The problem of base change or of lifting for automorphic representations can be introduced in several ways. It emerges very quickly when one pursues the formal principles expounded in the article 20 which can in fact be reduced to one, viz., the functoriality of automorphic forms with respect to what is now referred as the Lgroup.
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Year:
1980
Edition:
Princeton
Publisher:
Princeton University Press
Language:
english
Pages:
158
ISBN 10:
0691082723
ISBN 13:
9780691082639
Series:
Annals of mathematics studies 96
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BASE CHANGE FOR GL(2)† by R. P. Langlands † Appeared originally as Annals of Mathematics Study 96 (1980). Base change i FOREWORD These are the notes from a course of lectures given at The Institute for Advanced Study in the fall of 1975. Following a suggestion of A. Borel, I have added a section (§2) with an outline of the material and have discussed the applications to Artin Lfunctions in more detail (§3), including some which were discovered only after the course was completed. I have also made corrections and other improvements suggested to me by him, and by T. Callahan, A. Knapp, and R. Kottwitz. But on the whole I have preferred to leave the notes in their original, rude form, on the principle that bad ideas are best allowed to languish, and that a good idea will make its own way in the world, eventually discovering that it had so many fathers it could dispense with a mother. R. P. Langlands Base change ii Table of Contents 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. Foreword Introduction Properties of Base Change Applications to Artin Lfunctions σ Conjugacy Spherical Functions Orbital Integrals Characters and Local Lifting Convolution The Primnitive State of our Subject Revealed The Trace formula The Comparison References 1. INTRODUCTION The problem of base change or of lifting for automorphic representation can be introduced in several ways. It emerges very quickly when one pursues the formal principles expounded in the article [20] which can in fact be reduced to one, viz., the functoriality of automorphic forms with respect to what is now referred to as the Lgroup. This is not the place to rehearse in any generality the considerations which led to the principle, or its theoretical background, for which it is best to consult [4]; but it is useful to review them briefly in the form which is here pertinent. Suppose that F is a nonarchimedean local field and G is GL(2). If O is the ring of integers of F the Hecke algebra H of compactly supported functions on the double cosets of G(F )//; G(O) has a known structure. It is in particular isomorphic to the algebra of functions on GL(2, C) obtained by taking linear combinations of characters of finitedimensional analytic representations. According to the definitions of [20], the Lgroup of G over F is the direct product L G =LGo × G(K/F ). Here L Go , the connected component of L G, is GL(2, C), and K is simply a finite Galois extension of F , large enough for all purposes at hand. If K/F is unramified the Frobenius element Φ in G(K/F ) is defined and the Hecke algebra H is also isomorphic to the algebra of functions on L Go × Φ ⊆LG obtained by restriction of linear combinations of characters of analytic representations of the complex Lie group L G. Suppose E is a finite separable extension of F . The group G obtained from G by restriction of scalars from E to F is so defined that G(F ) = G(E). As a group over F it has an associated Lgroup, whose connected o component L G is Y G(K/E)\G(K/F ) o GL(2, C). The group G(K/F ) operates on L G via its action on coordinates. The Lgroup G is a semidirect product L o G =L G × G(K/F ). If E/F and K/F are unramified the Hecke algebra HE of G(E) with respect to G(OE ) is isomorphic to the algebra of functions on L o G × Φ ⊆L G Base change 2 obtained by the restriction of linear combinations of characters of finitedimensional analytic representations of L G. At first this is a little baffling for Hecke algebras on G(E) and G(F ) are the same, while the first is isomorphic ∨ o to the representation ring of GL(2, C) and the second to an algebra of functions on L G ×Φ. If f ∨ and f represent the same element of the Hecke algebra then (1.1) ∨ f ((g1 , · · · , g` ) × Φ) = f ∨ (g` · · · g2 g1 ) ` = [E : F ]. The homomorphism ϕ : (g × τ ) → (g, · · · , g) × τ o o of L G to L G takes L Go × Φ to L G × Φ. It allows us to pull back functions from L G × Φ to L Go × Φ, and yields especially a homomorphism ϕ∗ : HE → H. To give an irreducible admissible representation π of G(F ) which contains the trivial representation of G(O) is tantamount to giving a homomorphism λ of H onto C, and to give an irreducible representation Π of G(E) which contains the trivial representation of G(OE ) is tantamount to giving a homomorphism λ0 of HE onto C. We say that Π is a lifting of π if λ0 = λ ◦ ϕ∗ . The notion of lifting may also be introduced when E is simply a direct sum of finite separable extensions. For example if E = F ⊕ · · · ⊕ F then G(F ) = G(E) = G(F ) × · · · × G(F ) and L G is the direct product GL(2, C) × · · · × GL(2, C) × G(K/F ). We may define ϕ as before. The algebra HE is H ⊗ · · · ⊗ H. It is easily verified that if f1 ⊗ · · · ⊗ f` lies in HE then ϕ∗ (f1 ⊗ · · · ⊗ f` ) is the convolution f1 ∗ · · · ∗ f` , and so the lifting of π , defined by the same formal properties as before, turns out to be nothing but π ⊗ · · · ⊗ π . Thus when E is a direct sum of several copies of F , the concept of a lifting is very simple, and can be extended immediately to all irreducible, admissible representations. However when E is a field, it is not at all clear how to extend the notion to cover ramified π . Nonetheless class field theory suggests not only that this might be possible but also that it might be possible to introduce the notion of a lifting over a global field. The principal constraint on these notions will be the compatibility between the local and the global liftings. If F is a global field and E is a finite separable extension of F then for each place v of F we define Ev to be E ⊗F Fv . If π = ⊗v πv is an automorphic representation of G(A) ([3]), where A is the adèle ring of F , then the automorphic representation Π of G(AE ) will be a lifting of π if and only if Πv is a lifting of πv for all v . Since πv Base change 3 is unramified for almost all v and the strong form of the multiplicity one theorem implies that, in general, Π is determined when almost all Πv are given, this is a strong constraint. Proceeding more formally, we may define the Lgroups L G and L G over a global field F too. L and L G= We also introduce Y G = GL(2, A) × G(K/F ) G(K/E)\G(K/F ) GL(2, A) o G(K/F ). ϕ : (g, τ ) → (g, · · · , g) × τ once again. If v is a place of F we may extend it to a place of K . The imbedding G(Kv /Fv ) ,→ G(K/F ) yields imbeddings of the local Lgroups L Gv = GL(2, C) × G(Kv /Fv ) ,→L G, L Gv ,→L G. The restriction ϕv of ϕ to L Gv carries it to L Gv and is the homomorphism we met before. If Π(G/F ) is the set of automorphic representations of G(A) and Π(G/F ) is the same set for G(A) = G(AE ), the global form of the principle of functoriality in the associate group should associate to ϕ a map Π(ϕ) : Π(G/F ) → Π(G/F ). The lifting Π of π would be the image of π under Π(ϕ). Since the principle, although unproved, is supported by all available evidence we expect Π to exist. The local form of the principle should associate to ϕv a map Π(ϕv ) from Π(G/Fv ), the set of classes of irreducible admissible representations of G(Fv ), to Π(G/Fv ) and hence should give a local lifting. Whatever other properties this local lifting may have it should be compatible with that defined above when E v /Fv is unramified and πv contains the trivial representation of G(O). Moreover, as observed already, local and global lifting should be compatible so that if π = ⊗πv lifts to Π = ⊗Πv then Πv should be a lifting of π for each v . The main purpose of these notes is to establish the existence of a lifting when E/F is cyclic of prime degree. It is worthwhile, before stating the results, to describe some other paths to the lifting problem. If H is the group consisting of a single element then the associate group L H is just G(K/F ) and a homomorphism ϕ :LH →LG compatible with the projections of the two groups on G(K/F ) is simply a twodimensional representation ρ of G(K/F ). Since Π(H/F ) consists of a single element, all Π(ϕ) should do now is select a particular automorphic representation π = π(ρ) in Π(G/F ). Base change 4 The local functoriality should associate to ϕv a representation πv = π(ρv ), where ρv is the restriction of ρ to the decomposition group G(Kv /Fv ). We let Φv be the Frobenius at a place v at which Kv is unramified and suppose ϕv : Φv → tv × Φv . The associated homomorphism ϕ∗v of the Hecke algebra Hv of G at v into that of H at v , namely, to C is obtained by identifying Hv with the representation ring of GL(2, C) and evaluating at tv . For such a v, πv = π(ρv ) is defined as the representation corresponding to this homomorphism. We may define π = π(ρ) globally by demanding that π = ⊗πv with πv = π(ρv ) for almost all v . This of course does not prove that it exists. It is also possible to characterize π(ρv ) for all v (§12 of [14]), although not in a truly satisfactory manner. Nonetheless π(ρv ) can now be shown to exist ([17]), but by purely local methods quite different from those of these notes, where the emphasis is on the existence of π(ρ) globally. These considerations can be generalized. If ρ is a continuous twodimensional representation of the Weil group WK/F by semisimple matrices we may define ρv as the restriction of ρ to WKv /Fv . For almost all v, ρv factors through WKv /Fv → Z → GL(2, C). If tv is the image of a Frobenius element, that is, of 1 ∈ Z, and πv the representation of G(Fv ) which contains the trivial representation of G(Ov ) and yields the homomorphism of Hv into C defined by evaluation at tv , we say that π(ρv ) = πv . We may, at least for irreducible ρ, define π = π(ρ) globally by the demand that π = ⊗π v and πv = π(ρv ) for almost all v . If π(ρ) exists its local factors πv can be characterized in terms of ρv . If ρ is reducible the existence of π(ρ) is proved in the theory of Eisenstein series. If ρ is dihedral, by which I shall mean, in spite of justified reproofs, induced from a quasicharacter of the Weil group of a quadratic extension, the existence of π(ρ) is implicit in the work of Hecke and of Maass. But nothing more was known when, late in 1966 or early in 1967, the principle of functoriality, and hence the existence of π(ρ), was first suggested by the general theory of Eisenstein series. It was desirable to test a principle with so many consequences – for example, the existence of π(ρ) implies the Artin conjecture for the Artin Lfunction L(s, ρ) – as thoroughly as possible. Weil’s elaboration of the Hecke theory, which had been completed not long before, together with a careful analysis ([21]) of the factors appearing in the functional equation of the Artin Lfunctions, enabled one to show that the existence of π(ρ) was implied by Weil’s form of the Artin conjecture ([14]), and to obtain at the same time a much better understanding of the local maps ρv → π(ρv ). In retrospect it was clear that one could argue for the existence of π(ρ) by comparing the form of the functional equation for the Artin Lfunctions on one hand and of the Euler products associated by Hecke and Maass to automorphic forms on the other. This is especially so when F = Q and ρ∞ factors through ρo ∞ GL(2, C) WC/R −→ G(C/R) −→ Base change 5 with the second homomorphism taking complex conjugation to 1 0 0 −1 . This argument is simple, can be formulated in classical terms, and resembles closely the argument which led Weil to his conjecture relating elliptic curves and automorphic forms, and thus has the sanction of both tradition and authority, and that is a comfort to many. The emphasis on holomorphic forms of weight one is misleading, but the connection with elliptic curves is not, for, as Weil himself has pointed out ([33]), the consequent pursuit of his conjecture leads ineluctably to the supposition that π(ρ) exists, at least when F is a function field. Once the conjecture that π(ρ) existed began to be accepted, the question of characterizing those automorphic representations π which equal π(ρ) for some twodimensional representation of the Galois group arose. It seems to have been generally suspected, for reasons which are no longer clear to me, that if F is a number field then π is a π(ρ) if and only if, for each archimedean place v, πv = π(ρv ), where ρv is a representation of G(F v /Fv ); but there was no cogent argument for giving any credence to this suspicion before the work of Deligne and Serre ([6]) who established that it is correct if F = Q and π∞ = π(ρo∞ ). This aside, it was clear that one of the impediments to proving the existence of π(ρ) was the absence of a process analogous to composition with the norm, which in class field theory enables one to pass from a field to an extension, that is, to effect a lifting or a base change. The expectation that there will be a close relation between automorphic Lfunctions on one hand and motivic Lfunctions on the other entails the existence of such a process, for it implies that to any operation on motives must correspond an analogous operation on automorphic representations, and one of the simplest operations on motives is to pass to a larger field of definition, or, as one says, to change the base. For motives defined by a representation of a Galois group or a Weil group over F , base change is simply restriction to the Galois or Weil group over E . If F is a local field and ρ : WK/F → GL(2, C) an unramified twodimensional representation of the local Weil group we have already defined π(ρ). It must be observed that if E/F is unramified and σ the restriction of ρ to WK/E then π(σ) is the lifting of π . Otherwise, base change for automorphic forms would be incompatible with base change for motives. That π(σ) is the lifting of π follows from formula (1.1) and the definition of π(σ) and π(ρ). Although the lifting problem emerges from the general principle of functoriality in the Lgroup, some of its historical roots and most of the sources of progress lie elsewhere. The initial steps were taken for F = Q and E quadratic by Doi and Naganuma. It is instructive to review their early work ([7],[8]). We first recall the relevant facts about Lfunctions associated to automorphic forms. If ρ is any analytic representation of L G and π an automorphic representation it is possible ([20]) to introduce an Euler product L(s, π, ρ) = Y v L(s, πv , ρ). Base change 6 To be frank it is at the moment only possible to define almost all of the factors on the right. For a few ρ it is possible to define them all; for example, if ρ is the projection ρo of L G on its first factor GL(2, C) then L(s, π, ρ) is the Hecke function L(s, π) studied in [14]. One basic property of these Euler products is that L(s, π, ρ1 ⊕ ρ2 ) = L(s, π, ρ1 ) L(s, π, ρ2 ). If ρ is a representation of L G and Π an automorphic representation of G(A) we may also introduce L(s, Π, ρ) ([20]). These functions are so defined that if ϕ :L G →L G is defined as above and Π is the lifting of π then L(s, Π, ρ) = L(s, π, ρ ◦ ϕ). If ρ ◦ ϕ is reducible the function on the right is a product. An automorphic representation for G(A) is also one for G(AE ), because the two groups are the same. However, L GE , the associate group of GL(2) over E , is GL(2, C) × G(K/E). Given a representation ρE of L GE we may define a representation ρ of L G so that L(s, Π, ρ) = L(s, Π, ρE ). Choose a set of representations τ1 , · · · , τ` for G(K/E)\G(K/F ) and let τi τ = σi (τ )τj(i) , with σi (τ ) ∈ G(K/E). Set ρ(g1 , · · · , g` ) = ρE (g1 ) ⊕ · · · ⊕ ρE (g` ) and let ρ(τ ) : ⊕vi → ⊕ρE (σi (τ ))vj(i) τ ∈ G(K/F ). The role played by passage from ρ to ρ is analogous to, and in fact an amplification of, that played by induction in the study of Artin Lfunctions. Suppose for example that ρ = ρoE is the standard twodimensional representation of L GE , obtained by projection on the first factor, and Π is the lifting of π . If E/F is cyclic of prime degree, let ω be a nontrivial character of G(E/F ) and hence of G(K/F ) and let ρi be the representation of L G defined by ρi (g × τ ) = ω i (τ )ρo (g). Then ρ◦ϕ= and M`−1 i=0 ρi L(s, Π, ρoE ) = L(s, Π, ρ) = L(s, π, ρ ◦ ϕ) = However, ω may also be regarded as a character of F × \IF and L(s, π, ρi ) = L(s, ω i ⊗ π). Y L(s, π, ρi ). Base change 7 Take F to be Q and E to be a real quadratic field. Suppose G1 is the multiplicative group of a quaternion algebra over E which splits at only one of its two infinite places. The Lgroups of G1 and G over E are the same. There is also associated to G1 a family of algebraic curves S which are defined over E and called Shimura curves. The HasseWeil zeta function of S can be written as a quotient of products of the Lfunctions L(s, Π1 ) = L(s, Π1 , ρoE ) corresponding to automorphic representations of G1 (AE ). It can happen that S not only is connected and elliptic, so that the nontrivial part of its zetafunction is exactly L(s, Π1 ) for a certain Π1 , but also has a model defined over Q ([7]). Then the conjecture of Taniyama as refined by Shimura and Weil ([32]) affirms that there is an automorphic representation π of G(AQ ) such that the interesting part of the zetafunction of the model is L(s, π). Hence L(s, Π1 ) = L(s, π) L(s, ω ⊗ π) if ω is the character of Q× \IQ defined by E . This equation is tantamount to the assertion that Π1 is a lifting of π ; and the problem of lifting as posed by Doi and Naganuma was not from π to Π but from π to Π1 , where G1 was some quaternion algebra over E . However, if Π1 is any automorphic representation of G1 (AE ) there is always (cf. [14], and especially the references therein to the work of Shimizu) an automorphic representation Π of G(AE ) such that L(s, Π) = L(s, Π1 ) and the problem of lifting from π to Π1 becomes the problem of lifting from π to Π. Following a suggestion of Shimura they were able to establish the existence of Π for a large number of π by combining an idea of Rankin with the theory of Hecke ([8]), at least when F = Q and E is a real quadratic field. Their idea was pursued by Jacquet ([13]) who removed the restriction on F as well as the restrictions on π which are inevitable when working in the context of holomorphic automorphic forms. However, the method was limited to quadratic extensions, and could establish the existence of a lifting, but could not characterize those Π which were liftings. The next step was taken by Saito ([27]), who applied what one can refer to as the twisted trace formula to establish the existence of a lifting and to characterize them when E/F is cyclic of prime degree. This is in fact not what he did, for he worked with holomorphic forms in the customary sense, without any knowledge of representation theory; and the language of holomorphic forms seems to be inadequate to the statement of a theorem of any generality much less to its proof. It is not simply that one can only deal with π = ⊗πv for which πv belongs to the discrete series at each infinite place, although this alone precludes the applications of these lectures, but rather that one is in addition confined to forms of low level. But Saito certainly does establish the usefulness of the twisted trace formula, the application of which may have been suggested by some computations of Busam and Hirzebruch. Base change 8 To carry over an idea in the theory of automorphic forms from a functiontheoretic to a presentationtheoretic context is seldom straightforward and usually demand new insight. What was needed to give suppleness and power to the idea of Saito was the correct notion of a local lifting. This was supplied by Shintani, who sketched his ideas during the U.S.Japan seminar on number theory held at Ann Arbor in 1975, and has now published them in more detail in [30]. It was Shintani who fired my interest in the twisted trace formula. It soon became clear* that his ideas, coupled with those of Saito, could, when pursued along lines which he had perhaps foreseen, be applied in a striking, but after this lengthy introduction no longer surprising, fashion to the study of Artin Lfunctions. Before giving the applications, I describe the results on lifting yielded by a fully developed – but only for GL(2) and only for cyclic extensions of prime degree! – theory. Moreover, only fields of characteristic zero will be considered. This is largely a result of indolence. * when reflecting upon these matters not long after the seminar at our cabin in the Laurentians Base change 9 2. PROPERTIES OF BASE CHANGE So far all applications of the trace formula to the comparison of automorphic representations of two different groups have been accompanied by local comparison theorems for characters, the typical example being provided by twisted forms of GL(2) ([14]). Base change for cyclic extensions is no exception, and, following Shintani, local liftings can be defined by character relations. Suppose F is a local field, and E a cyclic extension of prime degree `. The Galois group G = G(E/F ) acts on G(E), and we introduce the semidirect product G0 (E) = G(E) × G. The group G operates on irreducible admissible representations of G(E), or rather on their classes, Πτ : g → Π(τ (g)) and Π can be extended to a representation Π0 of G0 (E) on the same space if and only if Πτ ∼ Π for all τ . Fix a generator σ of G. Then Πτ ∼ Π for all τ if and only if Πσ ∼ Π. The representation Π0 is not unique, but any other extension is of the form ω ⊗ Π0 , where ω is a character of G. There are ` choices for ω . It will be shown in §7 that the character of Π0 exists as a locally integrable function. If g lies in G(E), we form N g = gσ(g) · · · σ `−1 (g). This operation, introduced by Saito, is easy to study. Its properties are described in §4. It is not the element N g which is important, but rather its conjugacy class in G(E), and indeed the intersection of that conjugacy class with G(F ), which is then a conjugacy class in G(F ). We also denote an element of that class by N g . The class of N g in G(F ) depends only on the class of g × σ in G0 (E). The representation Π of G(E) is said to be a lifting of the representation π of G(F ) if one of the following two conditions is satisfied: (i) Π is π(µ0 , ν 0 ), π is π(µ, ν), and µ0 (x) = µ(NE/F x), ν 0 (x) = ν(NE/F x) for x ∈ E × . (ii) Π is fixed by G and for some choice of Π0 the equality χΠ0 (g × σ) = χπ (h) is valid whenever h = N g has distinct eigenvalues. The representation π(µ, ν) associated to two characters of F × is defined on p. 103 of [14]. The characters χπ0 , χΠ0 of π and Π0 are welldefined functions where N g has distinct eigenvalues, so that the equality of (ii) is meaningful. It should perhaps be underlined that it is understood that π and Π are irreducible and admissible, Base change 10 and that they are sometimes representations, and sometimes classes of equivalent representations. It is at first sight dismaying the liftings cannot be universally characterized by character identities, but it is so, and we are meeting here a particular manifestation of a widespread phenomenon. We shall prove the following results on local lifting for fields of characteristic zero. a) Every π has a unique lifting. b) Π is a lifting if and only if Πτ ' Π for all τ ∈ G. c) Suppose Π is a lifting of π and of π0 . If π = π(µ, ν) then π 0 = π(µ0 , ν 0 ) where µ−1 µ0 and ν −1 ν 0 are characters of N E × \F × . Otherwise π 0 ' ω ⊗ π where ω is a character of N E × \F × . If ω is nontrivial then π ' ω ⊗ π if and only if ` is 2 and there is a quasicharacter θ of E× such that π = π(τ ) with τ = Ind(WE/F , WE/E , θ). d) If k ⊂ F ⊂ E and E/k, F/k are Galois and τ ∈ G(E/k) then the lifting of πτ is Πτ if the lifting of π is Π. e) If ρ is reducible or dihedral and π = π(ρ) then the lifting of π is π(P ) if P is the restriction of ρ to WK/E . f) If Π is the lifting of π and Π and π have central characters ωΠ and ωπ respectively then ωΠ (z) = ωπ (NE/F z). g) The notion of local lifting is independent of the choice of σ . The assertion (e) cries out for improvement. One can, without difficulty, use the results of §3 to extend it to tetrahedral ρ, but it is not clear that the methods of these notes can, unaided, establish it for octahedral ρ. I have not pursued the question. Many of the properties of local liftings will be proved by global means, namely, the trace formula. For this it is important that the map on characters χπ → χΠ0 which appears in the definition of local liftings is dual to a map φ → f of functions. It is only the values of χΠ0 on G(E) × σ which matter, and thus φ will be a function on G(E) × σ , or, more simply, a function on G(E). Since the χπ are class functions, it is not necessary – or possible – to specify f uniquely. It is only its orbital integrals which are relevant, and these must be specified by the orbital integrals of φ. But these will be integrals over conjugacy classes on G(E) × σ , a subset of G0 (E). As a step preliminary to the introduction of the trace formula, the map φ → f will be defined and introduced in §6. Objections can be made to the arrow, because the map is in fact only a correspondence, but the notation is convenient, and not lightly to be abandoned. There are other local problems to be treated before broaching the trace formula, but before describing them it will be best to recall the function of the trace formula. Let F be for now a global field and E a cyclic extension of prime degree `. Let Z be the group of scalar matrices, and set ZE (A) = Z(F )NE/F Z(AE ). Base change 11 Let ξ be a unitary character of ZE (A) trivial on Z(F ). We introduce the space Ls (ξ) of measurable functions ϕ on G(F )\G(A) which satisfy (a) ϕ(zg) = ξ(z)ϕ(g) for all z ∈ ZE (A) (b) Z ϕ(g)2 dg < ∞. ZE (A)G(F )\G(A) G(A) acts on Ls (ξ) by left translations. The space Ls (ξ) is the direct sum of two mutually orthogonal invariant subspaces: Lsp (ξ), the space of squareintegrable cusp forms; and Lse (ξ), its orthogonal complement. The theory of Eisenstein series decomposes Lse (ξ) further, into the sum of L0se (ξ), the span of the onedimensional invariant subspaces of Ls (ξ), and L1se (ξ). We denote by r the representation of G(A) on the sum of Lsp (ξ) and L0se (ξ). Suppose we have a collection of functions fv , one for each place v of F , satisfying the following conditions. i) fv is a function on G(Fv ), smooth and compactly supported modulo Z(Fv ). ii) fv (zg) = ξ −1 (z)fv (g) for z ∈ NEv /Fv Z(Ev ). iii) For almost all v , fv is invariant under G(OFv ), is supported on the product G(OFv )NEv /Fv Z(Ev ), and satisfies Z fv (g)dg = 1. NEv /Fv Z(Ev )\G(OFv )NEv /Fv Z(Ev ) Then we may define a function f on G(A) by f (g) = where g = (gv ). The operator r(f ) = Z Y v fv (gv ), f (g)r(g)dg NE/F Z(AE )\G(A) is defined and of trace class. Let ξE be the character z → ξ(NE/F z) of Z(AE ). We may also introduce the space Ls (ξE ) of measurable functions ϕ on G(E)\G(AE ) satisfying (a) ϕ(zg) = ξE (z)ϕ(g) for all z ∈ Z(AE ) (b) Z Z(AE )\G(AE ) ϕ(g)2 dg < ∞. Base change 12 Once again we have a representation r of G(AE ) on the sum of Lsp (ξE ) and L0se (ξE ). But r now extends to a representation of the semidirect product G0 (AE ) = G(AE ) × G. An element τ of G sends ϕ to ϕ0 with ϕ0 (h) = ϕ(τ −1 (h)). We will consider functions φ on G(AE ) defined by a collection φv , one for each place of F , satisfying i) φv is a function on G(Ev ), smooth and compactly supported modulo Z(Ev ). −1 ii) φV (zg) = ξE (z)fv (g) for z ∈ Z(Ev ). iii) For almost all v, φv is invariant under G(OEv ), is supported on Z(Ev )G(OEv ), and satisfies Z φ(g)dg = 1. Z(Ev )\Z(Ev )G(OEv ) Then φ(g) = Q v φv (gv ), and r(φ) = Z φ(g)r(g)dg Z(AE )\G(AE ) is defined and of trace class. We now introduce another representation, R, of G0 (AE ). If ` is odd, then R is the direct sum of ` copies of r. The definition of R for ` even is best postponed to §11. The function of the trace formula is to show that, for compatible choices of φ and f , (2.1) trace R(φ)R(σ) = trace r(f ). Here σ is the fixed generator of G(E/F ). The trace formula for the left side is somewhat different than the usual trace formula, and is usually referred to as the twisted trace formula. It will be reviewed in §10. The condition of compatibility means that φv → fv for all v . As we observed, the meaning of the arrow will be explained in §6 for those v which remain prime in E . Its meaning for v which split will be explained later, in the very brief §8. It is very important that when v does not ramify in E , φv lies in HEv , and fv is its image in HFv under the homomorphism introduced in §1, then the relation φ v → fv is satisfied. This was verified by Saito [27], who had no occasion to mention that the homomorphism from HEv to HFv was just one of many provided by the general theory of spherical functions and the formalism of the Lgroup. In §5 another verification is given; it exploits the simplest of the buildings introduced by Bruhat–Tits. The definition of the arrow φv → fv and the structure of the trace formula together imply immediately that the two sides of (2.1) are almost equal. The difference is made up of terms contributed to the trace formula by the cusps. There is a place for insight and elegance in the proof that it is indeed zero, but in these notes the proof is Base change 13 regarded as a technical difficulty to be bashed through somehow or other. The local information accumulated in §5 and in §9, which is primarily technical and of interest only to specialists, allows us to put the difference of the two sides of (2.1) in a form sufficiently tractable that we can exploit the fact that we are dealing with a difference of two traces to establish equality. This is the first step taken in §11. The equality (2.1) available, one chooses a finite set of places, V , including the infinite places and the places ramifying in E , and for each v ∈ / V an unramified representation Πv of G(Ev ) such that Πσv ∼ Πv . Let A be the set of irreducible constituents Π of R, counted with multiplicity, such that Π v is the given Πv outside of V . By the strong form of multiplicity one, A is either empty or consists of a single repeated element, and if Π ∈ A then Πσ ∼ Π. If Πσ ∼ Π then G0 (AE ) leaves the space of Π invariant, and so we obtain a representation Π0 of G0 (AE ), as well as local representations Π0v . Set A= X Π∈A Y trace Πv (φv )Π0v (σ). v∈V Let B be the set of constituents π = ⊗v πv of r such that Πv is a lifting of πv for each v outside of V . Set B= X π∈B Y v∈V trace πv (fv ). Elementary functional analysis enables us to deduce from (2.1) that A = B . This equality is local, although the set V may contain more than one element, and we have no control on the size of B . Nonetheless, when combined with some local harmonic analysis, it will yield the asserted results on local lifting. The necessary harmonic analysis is carried out in §7. Some of the results are simple; none can surprise a specialist. They are proved because they are needed. The last part of §7, from Lemma 7.17 on, contains material that was originally intended for inclusion in [18], and found its way into these notes only because they were written first. It is joint work with J.P. Labesse, and it was he who observed Lemma 7.17. Although [18] was written later, the work was carried out earlier, and the methods are less developed than those of these notes. At the time, one hesitated to strike out on a global expedition without providing in advance for all foreseeable local needs. One could probably, reworking [18], dispense with some of the computations of §7. But little would be gained. A word might be in order to explain why the last part of §7 and the more elaborate definition of R when ` = 2 are called for. When [E : F ] = 2 there are twodimensional representations ρ of the Weil group WF induced from characters of the Weil group WE . These representations have several distinctive properties, which we must expect to be mirrored by the π(ρ). For example, ρ can be irreducible but its restriction to W E will be reducible. If F is global this means that the cuspidal representation π(ρ) becomes Eisensteinian upon lifting, and this complicates the proofs. In the course of proving the results on local lifting, we also obtain the existence of global liftings, at least for a cyclic extension of prime degree `. If Π is an automorphic representation of G(AE ) then, for each place v of F , Base change 14 Π determines a representation Πv of G(Ev ), and Π is said to be a lifting of π if Πv is a lifting of πv for each v . The first properties of global liftings are: A) Every π has a unique lifting. B) If Π is isobaric in the sense of [24], in particular cuspidal, then Π is a lifting if and only if Πτ ∼ Π for all τ ∈ G(E/F ). C) Suppose π lifts to Π. If π = π(µ, ν) with two characters of the idèle class group ([14]), then the only other automorphic representations lifting to Π are π(µ1 µ, ν1 ν), where µ1 , ν1 are characters of F × NE/F IE \IF . If π is cuspidal then π 0 lifts to Π if and only if π0 = ω ⊗ π where ω is again a character of F × NE/F IE . The number of such π 0 is ` unless ` = 2 and π = π(τ ) where τ is a twodimensional representation of WE/F induced by a character of E × \IE , when it is one, for π ∼ ω ⊗ π in this case. D) Suppose k ⊂ F ⊂ E and F/k, E/k are Galois. If τ ∈ G(E/k) and Π is a lifting of π then Πτ is a lifting of πτ . E) The central character ωπ of π is defined by π(z) = ωπ (z)I, z ∈ Z(A) = IF , and ωΠ is defined in a similar fashion. If Π is a lifting of π then ωΠ (z) = ωπ (NE/F z). If Π is cuspidal then Π is said to be a quasilifting of π if Πv is a lifting of πv for almost all v . A property of global liftings that has considerable influence on the structure of the proofs is: F) A quasilifting is a lifting. It is worthwhile to remark, and easy to verify, that the first five of these properties have analogues for twodimensional representations of the Weil group WF of F if lifting is replaced by restriction to WE . The central character is replaced by the determinant. Base change 15 3. APPLICATIONS TO ARTIN LFUNCTIONS Suppose F is a global field and ρ is a twodimensional representation of the Weil group WK/F , K being some large Galois extension. There are two possible definitions of π(ρ). If π(ρv ) is characterized as in §12 of [14], we could say that π = π(ρ) if π = ⊗πv and πv = π(ρv ) for all v . On the other hand we could say that π = π(ρ) if π is isobaric, in the sense of [24], and πv = π(ρv ) for almost all v . The second definition is easier to work with, for it does not presuppose any elaborate local theory, while for the first the relation L(s, π) = L(s, ρ) is clear. It will be useful to know that they are equivalent. The first condition is easily seen to imply the second. To show that the second implies the first, we use improved forms of results of [5] and [14] which were communicated to me by T. Callahan. He also provided a proof of the following strong form of the multiplicity one theorem. Lemma 3.1 Suppose π and π 0 are two isobaric automorphic representations of GL(2, A). If πv ∼ πv0 for almost all v then π ∼ π 0 . If π is isobaric and not cuspidal then π = π(µ, ν), where µ, ν are two idèle class characters. An examination of the associated Lfunctions L(s, ω ⊗ π) and L(s, ω ⊗ π 0 ) shows easily that if πv ∼ πv0 for almost all v and π = π(µ, ν) then π0 = π(µ0 , ν 0 ). Thus the lemma is quickly reduced to the case that π and π 0 are cuspidal. It is stronger than the theorem of Casselman ([5]) because it does not assume that πv ∼ πv0 for archimedean v , but the proof is similar. One has to observe that if v is archimedean and if for every character ωv of Fv× and some fixed nontrivial character ψv of Fv the function of s given by ε0 (s, ωv ⊗ πv , ψv ) = L(1 − s, ωv−1 ⊗ πv )ε(s, ωv ⊗ πv , ψv ) L(s, ωv ⊗ πv ) is a constant multiple of ε0 (s, ωv ⊗ πv0 , ψv ) then πv ∼ πv0 . This is an archimedean analogue of Corollary 2.19 of [14], and is a result of the formulae for ε0 (s, ωv ⊗ πv , ψv ) given in the proofs of Lemma 5.18 and Corollary 6.6 of [14]. One needs in addition the following variant of Lemma 12.5 of [14]. Lemma 3.2 Suppose that we are given at each infinite place v of F a character χv of Fv× and, in addition, an integral ideal A of F . Then there exists an idèleclass character ω which is such that ωv is close to χv for each archimedean v and whose conductor is divisible by A. Base change 16 This lemma, whose proof will not be given, can also be used, in conjunction with the methods of §12 of [14], to show that the second definition of π(ρ) implies the first. Again, if π(ρ) is not cuspidal then it is π(µ, ν) and ρ must be the direct sum of the onedimensional representations µ and ν . There is one further property of liftings which is now clear. G) If E/F is cyclic of prime degree, if π = π(ρ), Π is the lifting of π , and P the restriction of ρ to the Weil group over E , then Π = π(P ). Of course the definition of π(ρ) does not imply that it always exists. If ρ is irreducible and π = π(ρ) then π is necessarily cuspidal and, since L(s, π) = L(s, ρ), the Artin Lfunction attached to ρ is entire, as it should be. In this paragraph we take the results of global liftings announced in the previous paragraph for granted, and see what can be deduced about the existence of π(ρ). The representation ρ is of course to be twodimensional, and we may as well assume that it is neither reducible nor dihedral. If ρ is a representation of WK/F the image of K × \IK will then consist of scalar matrices, and passing to P GL(2, C) ' SO(3, C) we obtain a finite group which will be tetrahedral, octahedral, or icosahedral. About the last I can say nothing. I consider the other two in turn. i) Tetrahedral type There are three pairs of opposite edges so that we obtain a map of G(K/F ) into S3 . Since we only obtain proper motions of the tetrahedron the image must in fact be A 3 ' Z3 . The kernel defines a cyclic extension E of degree 3. The restriction P of ρ to WK/E must be dihedral, and so Π = π(P ) exists as an automorphic representation of G(AE ). If τ ∈ G(E/F ) has a representative u in WK/F then Πτ is clearly π(P τ ) if P τ is the representation, or rather the class of representations, defined by P τ (w) = P (uwu−1 ). However, P and P τ are equivalent so Πτ ' Π and Π is a lifting of an automorphic representation π of G(A). If ωρ = det ρ then ωρ and ωπ pull back to the same quasicharacter of E × \IE . Thus there is a character ω of Base change 17 F × N IE \ IF such that ωρ = ω 2 ωπ . Replacing π by ω ⊗ π , we can arrange that ωρ = ωπ and that π lifts to π(P ). This determines π uniquely. We write π = πps (ρ), which is to be read πpseudo (ρ). P GL(2, C) (D.1) ..... ..... ..... ..... ..... .... ..... .... ..... ... ..... . . ..... ... . ..... . ..... ... . . ..... .. . . ..... . ....... ... . . ... .... . ................................................................ GL(2,.. C) ........ ........ ........ ........ ........ ........ ........ ........ ........ .......... ..... SL(3, C) ϕ ...... . GL(3, C) It follows from (C) and (G) that if π(ρ) exists then it must be πps (ρ), but at the moment all we have in our hands is πps (ρ), and the problem is to show that it is in fact π(ρ). This will be deduced from results of Gelbart, Jacquet, PiatetskiiShapiro, and Shalika (cf. [11]). Consider the commutative diagram (D.1) in which the skewed arrow on the right is given by the adjoint representation. Taking the product with G(K/F ), we obtain a diagram of Lgroups with G1 = SL(2), H1 = P GL(3), H = GL(3). The representation σ = ϕ ◦ ρ is a representation of G(K/F ). Each of the onedimensional subspaces defined by an axis passing through opposite edges of the tetrahedron is fixed by G(K/E) and thus defines a character θ of G(K/E). It is easy to see that σ = Ind(G(K/F ), G(K/E), θ). L G1 ... ....... ...... ...... ...... . . . . ..... ..... ...... ..... (D.2) LG ..... ..... ..... ..... ..... ..... ..... ........ ........ LH1 ........ ......... ........ ......... ......... ........ ......... ........ ..... ......... . ......... .......... ϕ LH For each finite place v at which σv is unramified one attaches a conjugacy class in GL(3, C) to σv , namely that of σv (Φ) if Φ is the Frobenius at v . Moreover one also attaches a conjugacy class {A(πv1 )} in GL(3, C) to each unramified representation πv1 of GL(3, Fv ) (cf. [3], [20], [26]). The representation is determined by the conjugacy class, and one says that πv1 = π(σv ) if {A(πv1 )} = {σv (Φ)}. The following instance of the principle of functoriality is due to PiatetskiiShapiro ([16]): 1) There is a cuspidal representation π1 of GL(3, A) such that πv1 = π(σv ) for almost all v . There is another instance of the principle due to GelbartJacquet ([12]): 2) Let π = πps (ρ). Then there is a cuspidal representation π2 of GL(3, A) such that {A(πv2 )} = {ϕ(A(πv ))}. Recall that evaluation at the class {A(πv )} defines the homomorphism of the Hecke algebra into C associated to πv . Base change 18 It is to be expected that π 1 and π2 are equivalent, and this can indeed be established, using a criterion of Jacquet–Shalika ([15]). Let π−1 be the contragredient of π 1 . All that need be verified is, in the notation of [15], that L(s, πv1 × π̃v1 ) = L(s, πv2 × π̃v−1 ) for almost all v . The left side is det−1 (1 − $v s A(πv1 ) ⊗ t A−1 (πv1 )), (3.1) and the right side is det−1 (1 − $v s A(πv2 ) ⊗ t A−1 (πv1 )). (3.2) In general, if πv1 = π(σv ) and σv is unramified then det(1 − $v s B ⊗ t A−1 (πv1 )) = Y wv det(1 − $v n(w)sB n(w) ) if n(w) is the degree [Ew : Fv ]. If v splits completely in E then ρv (Φ) is conjugate to A(πv ). Since {A(πv1 )} = {ϕ(ρv (Φ))} and {A(πv2 )} = {ϕ(A(πv ))} the equality of (3.1) and (3.2) is clear. If v does not split in E then n(w) = 3, and, by definition, {A3 (πv )} = {ρ3v (Φ)}. The equality is again clear. To show that πps (ρ) is π(ρ) we have to show that {A(πv )} = {ρv (Φ)} even when v does not split in E . We have so chosen π that both sides have the same determinant. Thus we may suppose that {ρv (Φ)} = and that {A(πv )} = ( ( a 0 0 b ) ξa 0 0 ξ2b ) Base change 19 with ξ 3 = 1. We need to show that ξ may be taken to be 1. Since π 1 and π2 are equivalent, ( ) ( ) a 0 ξa 0 ϕ( ) = ϕ( ) 0 b 0 ξ2b in GL(3, C). This implies either that ξ = 1, and then we are finished, or that a2 = ξb2 . From this equation we conclude that either a = ξ 2 b, which also leads to the desired conclusion, or a = −ξ 2 b, which implies that ϕ(ρv (Φ)) has order 6 if ξ is not 1. Since the tetrahedral group contains no element of order 6, the last possibility is precluded. We have proved the following theorem. Theorem 3.3 If F is a number field and ρ a twodimensional representation of the Weil group of F of tetrahedral type then the Lfunction L(s, ρ) is entire. ii) Octahedral type. Rather than an octahedron I draw a cube in which I inscribe a tetrahedron. The subgroup of G(K/F ) which takes the tetrahedron to itself defines a quadratic extension E of F . The restriction P of ρ to WK/E is of tetrahedral type; so Π = π(P ) exists. If τ ∈ G(E/F ) then Πτ = π(P τ ). Since P τ ' P we conclude that Πτ ' Π. Hence Π is the lifting of exactly two automorphic representations π, π 0 of G(A), one of which can be obtained from the other by tensoring with the nontrivial character ω of F × N IE \IF . We are no longer able to define πps (ρ) uniquely; we take it to be either of the two representations π, π 0 . We are only able to show that one of the πps (ρ) is in fact π(ρ) in very special cases. We will exploit a result of DeligneSerre. There is a general observation to be made first. Suppose E is a cyclic extension of arbitrary prime degree ` and ρ a twodimensional representation of the Weil group of F . Suppose in addition that the restriction P of ρ to the Weil group of E is irreducible and that Π = π(P ) exists. Let π lift to Π, and suppose that π is π(ρ0 ) for some ρ0 , perhaps different from ρ. Base change 20 If P 0 is the restriction of ρ0 to the Weil group of E then Pw0 = Pw for almost all places of E and thus (see, for example, Lemma 12.3 of [14]) P 0 = P . Consequently, ρ = ω ⊗ ρ 0 and π(ρ) = ω ⊗ π exists, so that L(s, ρ) is entire. Here ω is a characer of F × NE/F IE \IF . Thus, for ρ of tetrahedral type and E the associated quadratic extension, we can conclude that one of π or π0 is π(ρ) if we can show that π is π(ρ0 ) for some ρ0 . By the result of DeligneSerre ([6]), this will be so if F = Q and the infinite component π∞ of π is π(ρ0∞ ) where ρ0∞ = µ ⊕ ν, µ, ν being two characters of R× with µ(x) = ν(x)sgn x. This is the condition that guarantees that the tensor product of π with some idèle class character is the automorphic representation defined by a holomorphic form of weight one. We will not be able to show that π ∞ has this form unless we assume that ρ∞ , the infinite component of ρ, has the same form as ρ0∞ . Interpreted concretely this means that the image of complex conjugation in the octahedral group is rotation through an angle of 180◦ about some axis. This axis passes either through the center of a face of the cube or through the center of an edge. If it passes through the center of a face then complex conjugation fixes E , which is therefore a real quadratic field. If v is either of the infinite places of E , then π∞ is equivalent to Πv and Πv = π(Pv ). Since Pv = ρ∞ , the representation π∞ satisfies the condition which allows us to apply the theorem of Deligne–Serre. Theorem 3.4 Suppose ρ is a twodimensional representation of the Weil group of Q which is of octahedral type. If the image of complex conjugation is rotation through an angle of 180◦ about an axis passing through a vertex of the octahedron or, what is the same, the center of a face of the dual cube, then L(s, ρ) is entire. There is one other condition which allows us to conclude that π∞ is of the desired type. We continue to suppose that the image of complex conjugation is rotation through an angle of 180◦ . If ωπ is the central character of π and ωρ the determinant of ρ then η = ω π ωρ−1 is of order two. Since its local component is trivial at all places which split in E , it is either trivial itself or the quadratic character associated to the extension E . π∞ has the desired form if and only if η∞ is trivial. If E is a real quadratic field then η∞ is necessarily trivial, and so we obtain the previous theorem. If E is an imaginary quadratic field then η∞ is trivial if and only if η is; and η is trivial if and only if ηv is trivial for some place of F which does not split in E . Theorem 3.5 Suppose ρ is a twodimensional representation of the Weil group of Q which is of octahedral type. Suppose the image of complex conjugation is rotation through 180◦ about an axis passing through the center of an edge. If for some place v which does not split in E , the quadratic field defined by the tetrahedral subgroup, the local representation ρv is dihedral then L(s, ρ) is entire. Base change 21 −1 It is clear that ηv = ωπv ωπ(ρ . However πv and π(ρv ) have the same lifting to G(Ev ). Thus, by property (c) v) of local liftings, π(ρv ) = ω ⊗ πv with ω of order two. We conclude that ω π(ρv ) = ωπv . Base change 22 4. σ –CONJUGACY Suppose F is a field and E is a cyclic extension of prime degree `. Fix a generator σ of G = G(E/F ). If x and y belong to G(E) we say that they are σ conjugate if for some h ∈ G(E) y = h−1 xσ(h). Then yσ(y) · · · σ`−1 (y) = h−1 xσ(x) · · · σ`−1 (x)h. We set N x = xσ(x) · · · σ`−1 (x). If u = N x and v = h−1 uh then v = N y . Lemma 4.1 If u = N x then u is conjugate in G(E) to an element of G(F ). Let F be an algebraic closure of F containing E . It is sufficient to verify that the set of eigenvalues of N x, with multiplicities, is invariant under G(F /F ), or even under those σ 0 ∈ G(F /F ) with image σ in G. Acting on the set with σ 0 we obtain the eigenvalues of σ(u), and σ(u) = x−1 ux. The invariance follows. Suppose u = N x lies in G(F ). Let Gu be the centralizer of u and let Gσx (E) be the set of all g in G(E) for which x = g −1 xσ(g). The matrix x belongs to Gu (E) and y → xσ(y)x−1 is an automorphism of Gu (E) of order `. It therefore defines a twisted form Gσu of Gu . Clearly Gσu (F ) = Gσx (E). If M is the algebra of 2 × 2 matrices and Mu the centralizer of u, we may also introduce the twisted form M uσ of Mu . Then Gσu is the group of invertible elements in M uσ , and it follows readily from the exercise on p. 160 of [28] that H 1 (G, Gσu (E)) = {1}. Lemma 4.2 If N x and N y are conjugate then x and y are σ conjugate. We reduce ourselves immediately to the case that u = N x lies in G(F ) and N x = N y . If τ = σ r belongs to G set cτ = yσ(y) · · · σr−1 (y)σr−1 (x)−1 · · · σ(x)−1 x−1 . Base change 23 Since N x = N y , cτ is well defined and cσ xσ(cτ )x−1 = cστ . In other words τ → cτ defines a cocycle of G with values in Gσu (E). Therefore there is an h satisfying yx−1 = cσ = h−1 xσ(h)x−1 and y = h−1 xσ(h). Occasionally in later paragraphs N x will simply stand for an element of G(F ) which is conjugate to xσ(x) · · · σ`−1 (x), but for now it is best to retain the convention that N x = xσ(x) · · · σ`−1 (x). Lemma 4.3 Suppose u= a av 0 a with v 6= 0. Then u = N x for some x in G(E) if and only if a ∈ N E × . If u = N x and h ∈ G(E) then h−1 xσ(h) is uppertriangular if and only if h itself is. If u = N x then x ∈ Gu (E) and has the form b by 0 b Consequently Nx = Nb . 1 tr y 0 1 . The first assertion follows. To obtain the second we observe that if h−1 xσ(h) is uppertriangular, then h−1 uh is also. Lemma 4.4 Suppose u= a1 0 0 a2 with a1 = a2 . Then u = N x if and only if a1 and a2 lie in N E × . If y is uppertriangular then N y is of the form a1 v 0 a2 if and only if y = h−1 xσ(h) with an uppertriangular h. If h−1 xσ(h) is diagonal then h is of one of the two forms α 0 0 β or 0 β α 0 . Base change 24 Since Gu is the group of diagonal matrices the first and last assertions are clear. Suppose y is uppertriangular and Ny = a1 v 0 a2 . Replacing y by g −1 yσ(g) with g diagonal we may suppose that y=x perhaps with a different v . If then 1 −w 0 1 x 1 v 0 1 b1 0 0 b2 =x x= 1 σ(w) 0 1 1 σ(w) − b−1 1 b2 w 0 1 . To complete the proof we need only verify the following supplementary lemma. Lemma 4.5 If b ∈ E regard w → σ(w) − bw as a linear transformation of the vector space E over F . The determinant of this linear transformation is (−1)` (N b − 1). To compute the determinant we may extend scalars to E . Since E ⊗F E ' (E ⊕ · · · ⊕ E) and the linear transformation becomes (x1 , · · · , x` ) → (x2 − bx1 , x3 − σ(b) x2 , · · · , x1 − σ `−1 (b)x` ), we are reduced to calculating −b 1 −σ(b) 1 · · . · −σ `−2 (b) 1 1 −σ `−1 (b) Elementary row and column operations yield the desired result. Lemma 4.6 Suppose u ∈ G(F ) has distinct eigenvalues which do not lie in F . Let E 0 = F (u) be the centralizer of u in M (F ). (a) If E 0 is isomorphic to E over F then u = N x has a solution. (b) If E 0 is not isomorphic to E over F and L = E 0 ⊗F E then u = N x has a solution if and only if u ∈ NL/E 0 L× . Base change 25 If T is the Cartan subgroup in which u lies and u = N x then x ∈ T (E) = L× . The second statement is therefore clear. If E 0 is isomorphic to E then E 0 ⊗F E is isomorphic to E 0 ⊕ E 0 with E 0 imbedded as {(y, y)}. Since σ acts as (y1 , y2 ) → (y2 , y1 ) every invariant is a norm. Corollary 4.7 If F is a local field and u ∈ G(F ) has distinct eigenvalues which do not lie in F then u = N x has a solution if and only if det u ∈ NE/F E × . This follows from the previous lemma and local class field theory. Lemma 4.8 Suppose u ∈ Z(F ) = F × . If ` is odd then u = N x, x ∈ G(E), if and only if u ∈ N E × . If ` = 2 then u = N x always has a solution. If u = N x then u2 = det u ∈ N E × . This makes the first statement clear. If ` = 2 we may imbed E in M (F ); so the second statement follows from the proof of Lemma 4.6. If F is a global field and v a place of it we set Ev = E ⊗F Fv . G acts on Ev . Either Ev is a field and G = G(Ev /Fv ) or Ev is isomorphic to a direct sum of ` copies of Fv and σ acts as (x1 , · · · , x` ) → (x2 , · · · , x` , x1 ). Then G(Ev ) = G(Fv ) × · · · × G(Fv ). If u = (u, · · · , u) lies in G(Fv ) ⊆ G(Ev ) and x = (u, 1, · · · , 1) then u = N X ; so at a place which splits in E every element is a norm. Lemma 4.9 Suppose F is a global field and u ∈ G(F ). Then u = N x has a solution in G(E) if and only if it has a solution in G(Ev ) for each place v . It is enough to show that the equation u = N x can be solved globally if it can be solved locally. We know that a ∈ F × lies in N E × if and only if it lies in N Ev× for all v . If u is conjugate to an uppertriangular matrix the desired result follows from this and Lemmas 4.3, 4.4 and 4.8. Otherwise we apply Lemma 4.6. Observe that if u ∈ F × then the number of places v for which u ∈ / N Ev× is finite and even. We close this paragraph with a simple lemma which will be used frequently below. Suppose S is an abelian algebraic group over F , either a torus or the additive group Ga , and ω an invariant form of maximum degree on it. Let T be the group over F obtained from S over E by restriction of scalars and let ν be an invariant form of maximal degree on T . The two forms ω and ν and the exact sequences N (4.1) 1 −→ T 1−σ −→ T −→ S −→ 1 (4.2) 1 −→ S −→ T −→ T 1−σ −→ 1 1−σ Base change 26 yield forms µ1 and µ2 on T 1−σ . Lemma 4.10 The forms µ1 and µ2 are equal, except perhaps for sign. The lemma need only be verified over the algebraic closure F of F . So we may assume S is either G a or Gm and T is either Ga × · · · × Ga or Gm × · · · × Gm . Suppose first that S is Ga . Then T 1−σ = (x1 , · · · , x` ) X xi = 0 . We may suppose ω is dx and ν is dx1 ∧ · · · ∧ dx` . The pullback of dx to T is T to S is dx. We may take x1 , · · · , x`−1 as coordinates on T 1−σ . Then P dxi and the restriction of dx1 from µ1 = dx1 ∧ · · · ∧ dx`−1 . Pulling back µ1 from T 1−σ to T we obtain d(x1 − x2 ) ∧ d(x2 − x3 ) ∧ · · · ∧ d(x`−1 − x` ). Multiplying by dx1 we obtain (−1)`−1 dx1 ∧ · · · ∧ dx` so µ2 = (−1)`−1 µ1 . A similar computation can be made for Gm . Base change 27 5. SPHERICAL FUNCTIONS In this paragraph F is a nonarchimedean local field and O = OF is the ring of integers in F . We want to study the algebra H of compactly supported functions on G(F ) spherical with respect to G(O). A is the group of diagonal matrices and X ∗ , which is isomorphic to Z2 , its lattice of rational characters. Set X∗ = Hom(X ∗ , Z). If $ is a generator of the prime ideal of O then the map γ → λ(γ), where λ(γ) ∈ X∗ is defined by λ(γ) = $<λ,λ(γ)> establishes an isomorphism of A(O)\A(F ) with X∗ . If γ= set ∆(γ) = and, if ∆(γ) = 0, let Ff (γ) = ∆(γ) a 0 0 b (a − b)2 ab Z 1/2 f (g −1 γg)dg. A(F )\G(F ) If f ∈ H then Ff (γ) depends only on λ(γ); so we write Ff (λ). This function is invariant under permutation of the two coordinates of λ, and f →meas A(O)Ff (λ) defines an isomorphism of H with the subalgebra of the group ring of X∗ over C, formed by the invariant elements. We may look at this in a slightly different way. X ∗ may also be regarded as the lattice of rational characters of the diagonal matrices A(C) in GL(2, C), and every element P a(λ)λ of the group ring defines a function t→ X a(λ)λ(t) on A(C). The symmetric elements are precisely the functions obtained by restricting the elements of the representation ring of GL(2, C) to A(C). Thus H is isomorphic to an algebra of functions on A(C). Let f ∨ be the function corresponding to f . There are a number of distributions, which will arise in the trace formula, whose value on f we shall have to be able to express in terms of f ∨ . We begin this paragraph by verifying the necessary formulae. Our method of verification will be simply to check that both sides are equal for f = fλ , the characteristic function of a double coset G(O)γG(O) with λ(γ) = λ. It is easy to verify that m(λ), the measure of G(O)γG(O), is meas G(O) if λ = (k, k) and is q <α,λ> 1 meas G(O) 1+ q Base change 28 if λ = (k0 , k), k0 > k . q is the number of elements in the residue field of O and α is the root for which < α, λ > > 0, that is < α, λ >= k 0 − k . Lemma 5.1 If < α, λ > ≥ 0, then f λ∨ (t) is given by <α,λ> m(λ) · q− 2 1 + 1q 1 − q −1 α−1 (t) 1 − q −1 α(t) λ(t) + λ̃(t) . 1 − α−1 (t) 1 − α(t) Here λ̃ is obtained from λ by permuting its two coordinates. Taking α(t) < 1 and expanding the denominators in a Laurent expansion we find that this expression is equal to meas G(O)λ(t) meas G(O)q meas G(O)q X <α,λ> 2 < α, λ >= 0 <α,λ> 2 <α,λ> j=0 (λ(t) + λ̃(t)) λ(t)α−j (t) − 1 X<α,λ>−1 j=1 q < α, λ >= 1 −j λ(t)α (t) < α, λ > ≥ 2. To verify the lemma we have only to calculate Ffλ (µ) explicitly. Let λ(γ) = µ and choose δ in A(F ) with λ(δ) = λ. To make the calculation we use the building associated by Bruhat and Tits to SL(2, F ). This building is a tree X, the vertices of which are equivalence classes of lattices in F 2 , two lattices being equivalent if one is a scalar multiple of the other. The vertices defined by lattices M 1 , M2 are joined by an edge if there are scalars α and β such that αM1 ⊃ βM2 ⊃ $αM1 . 6= 6= If M0 is the lattice of integral vectors let p0 be the corresponding vertex. The action of G(F ) on lattices induces an action on X. Every vertex of X lies on q + 1 edges. We associate to A an apartment A. This is a subtree whose vertices are the points tp0 , t ∈ A(F ), and whose edges are the edges joining two such points. The apartment A is a line; every vertex lies on two edges. If p 1 , p2 are two points in X there is a g in G(F ) and a t in A(F ) such that p1 = gtp0 , p2 = gp0 . If λ(t) = (k0 , k) then k 0 − k is uniquely determined, and is just the distance from p2 to p1 . We may also associate a simplicial complex X0 to GL(2, F ) = G(F ). The points are lattices, two lattices M1 and M2 being joined by an edge if M1 ⊃ M2 ⊃ $M1 or M2 ⊃ M1 ⊃ $M2 . We may define an apartment A0 6= 6= 6= 6= and the type of an ordered pair (p01 , p02 ). It is a λ = (k0 , k), the pair, not the ordered pair, (k0 , k) being uniquely determined, so that the type is in fact a double coset. There is an obvious map p0 → p of X0 to X. The type of (γp0 , p0 ) depends only on the orbit under A(F ) to which p0 belongs. If τ (p1 , p2 ) denotes the type of (p1 , p2 ) the integral Z A(F )\G(F ) fλ (g −1 γg)dg Base change 29 is a sum over representatives of the orbits of A(F ) in X 0 X τ (γp0 ,p0 )=λ meas Gp0 ∩ A(F )\Gp0 . Here Gp0 is the stablizer of p0 . We may choose the representatives p0 so that the closest point to p in A is p0 . If p00 is the vertex of X0 determined by the lattice of integral vectors and p 0 = gp00 , let d(p0 ) be defined by 0 det g = $d(p ) . Any point p lifts uniquely to a p0 with d(p0 ) = dist(p, p0 ). We may also demand that the representatives p0 be chosen so that d(p0 ) = dist(p, p0 ). Then A(F ) ∩ Gp0 will lie in A(O). The number of choices for representatives satisfying the two conditions is [A(O) : A(O) ∩ Gp0 ]. Since meas Gp = meas G(O) the integral is equal to meas G(O) X 1. τ (γp0 ,p)=λ meas A(O) (5.1) The sum is over all p0 for which, in addition to the condition τ (γp0 , p0 ) = λ on the type of γp0 , p0 , d(p0 ) = dist(p, p0 ) = dist(p, A). This type of reduction will be used repeatedly, but without further comment, in the present paragraph. We may suppose < α, µ > ≥ 0 and < α, λ > ≥ 0. We have to show that ∆(γ) times the sum appearing in (5.1) is q <α,λ> 2 if λ = µ, q <α,λ> 2 1 − 1q if λ = µ + nα, n > 0, and 0 otherwise. There are two possibilities which have to be treated in different fashions. Suppose the following picture 6= 1. Then we have γp p .............. ... ... ... ........... ... ... ... ... ......... ... ... ... ... . .......... .......... ......... b a p0 .......... .......... .......... .......... . .. ... .. . . . . ........ .. .. .. .... . . ........ . ... ... .. . ......... ......... .. γ p0 .......... A The distance between p0 and γp0 is m0 − m if µ = (m0 , m). If the distance of p from p0 is k then the type of γp, p is 2k + m0 − m, provided d(p, p0 ) = d(p, A), and the type of (γp0 , p0 ) is (m0 + k, m − k). If k = 0 there is one choice for p and if k > 0 there are q k 1 − ∆(γ) = 1 q . Since b a 1/2 1− a b = b a 1/2 =q m0 −m 2 Base change 30 and q <α,λ> 2 = q k+ m0 −m 2 if λ = (m0 + k, m − k) the required equality follows. Before treating the second possibility we establish another lemma. Lemma 5.2 Suppose a b = 1 and 1 − a b = q −r . Then the points of X fixed by γ are precisely those at a distance less than or equal to r from A. Since G(F ) = A(F )N (F )K , with K = G(O) and N (F ) = 1 x 0 1 x ∈ F any point of X is of the form tnp0 , t ∈ A(F ), n ∈ N (F ). Moreover γ fixes tnp0 if and only if it fixes np0 ; and dist(tnp0 , A) = dist(np0 , A). We may index the vertices of A by Z, the integer z corresponding to the vertex pz = 1 x This vertex is fixed by n = 0 1 np0 = p0 for z ≤ 0. Otherwise 1 0 0 $z p0 . if and only if $z x ∈ O. If z is the smallest integer for which $ z x ∈ O then dist(np0 , A) = dist(np0 , pz ) = dist(np0 , npz ) = z. Pictorially, np0 .............. ........ ........ p0 ......... ... ... ..... ...... ... ... ... ...... ...... ... ... ... ...... ..... ... ... ... ..... ......... ......... .... ......... pz Certainly γ fixes np0 if and only if n−1 γn or γ −1 n−1 γn belongs to K . Since γ −1 n−1 γn = the lemma follows. 1 x 1− 0 1 b a Base change 31 To complete the proof of the first lemma, we have still to treat the case that ∆(γ) = q −r p ............ ........ ........ ......... ..... .... ....... .... .... ... ...... . . . . . . .. ......... ...... ... ... ....... ......... ...... . .... .. .... ......... ... ... ... . ......... ........ ......... b a = 1. Let 1 − a b = r so that γp ........ p0 ......... A If the distance of p from p0 is k + r with k > 0 then the type of (γp0 , p0 ) is (m0 + k, m − k), with m0 now equal to m. There are q k+r 1 − q1 possible such points. If the distance of p from p0 is less than or equal to r the type of (γp0 , p0 ) is (m0 , m). There are 1+ Xr j=1 qj 1 1− = qr q such points. This gives the desired equality once again. The group A0 (C) of elements in A(C) whose eigenvalues have absolute value 1 is compact. We introduce an inner product in the group ring of X∗ by setting < f1 , f2 >= Z A 0 ( C) f1 (t)f 2 (t). The total measure of the group is taken to be one. Lemma 5.3 Suppose γ lies in Z(F ) and µ = λ(γ). If f belongs to H and ϕγ (t) = 1+ 1 q 1 − α(t) 1 − α−1 (t) ∨ λ (t) 2 meas G(O) 1 − q −1 α(t) 1 − q −1 α−1 (t) then f (γ) =< f ∨ , ϕγ > . We verify this for f = fλ . If λ = (k 0 , k), k0 ≥ k and µ = (m, m) then both sides are 0 unless k0 + k = 2m. If this condition is satisfied f ∨ (t)ϕγ (t) is constant with respect to elements u 0 0 u z 0 0 1 so that the integration may be taken with respect to . This gives k−k0 m(λ) q 2 meas G(O) 2 Base change 32 times 1 2πi Z z=1 1−z 1 − z −1 k0 −m + z z k−m 1 − q −1 z 1 − q −1 z −1 dz . z Since k 0 ≥ m ≥ k this integral is seen by inspection to be 0 unless k 0 = m = k when it is 2. These are the required values. Corollary 5.4 If f 1 and f2 belong to H then Z G(F ) f1 (g)f 2 (g)dg = 1 q 1+ 2 meas G(O) Z ∨ A∨ 0 ( C) f1∨ (t)f 2 (t) 1 − α(t) 1 − α−1 (t) · . 1 − q −1 α(t) 1 − q −1 α−1 (t) Apply the previous formula for f = f1 ∗ f2∗ and γ = 1 with f2∗ (g) = f 2 (g −1 ). Let ν(t) = 1 q 1+ · 2 1 − α(t) 1 − α−1 (t) · . 1 − q −1 α(t) 1 − q −1 α−1 (t) Then the family of function fλ∨ is orthogonal with respect to ν(t) and Z fλ∨ (t)2 ν(t) = m(λ)meas G(O). A 0 ( C) Let n0 = Lemma 5.5 If a ∈ F × , n = an0 , and µ = λ Z a 0 0 a 1 1 0 1 . , then f (g −1 ng)dg Gn (F )\G(F ) is equal to 1 < f ∨, µ > . meas Gn (O) 1 − 1q Since {p0 ∈ A0 d(p0 ) = dist(p, p0 )} is a set of representatives for the orbits of Gn (F ) in X0 , the integral is equal to meas G(O) · X p0 ∈A0 d(p0 )=dist(p,p0 ) τ (np0 ,p0 )=λ 1 meas Gn (F ) ∩ Gp0 when f = fλ . If µ = (m, m) this expression is 0 unless λ = (m + k, m − k), k ≥ 0. If k = 0 the sum is X∞ 1 1 1 1 = · . z z=0 q meas Gn (O) meas Gn (O) 1 − 1q If k > 0 there is only one term in the sum and it equals qk . meas Gn (O) Base change 33 Comparing with the explicit expansion of fλ∨ we obtain the lemma. For these calculations we of course rely on the diagram n p ................ ......... ........ .... ....... ....... ... ... ..... ........ ....... ......... p p0 ........ ......... ......... If γ is any semisimple element in G(F ) with eigenvalues a and b we may set (a − b)2 ab ∆(γ) = and Ff (γ) = ∆(γ) Z 1/2 f (g −1 γg)dg T (F )\G(F ) if T is the Cartan subgroup containing γ . Lemma 5.6 If f belongs to H and T splits over the unramified quadratic extension F 0 then Z ∆(γ) meas G(O) 1 meas Gn (O) f (g −1 ng)dg − 2 Ff (γ) = 1 + · f (z). q meas T (O) Gn (F )\G(F ) q − 1 meas T (O) Here z ∈ F is determined by z = a = b, and n = zn0 . The group T (O) consists of all matrices in T (F ) whose eigenvalues are units. The Bruhat–Tits buildings X and X0 over F may be regarded as subtrees of the buildings X(F 0 ) and X0 (F 0 ) over F 0 . The torus T splits over F 0 and we may introduce the associated apartments AT (F 0 ) and A0T (F 0 ). They consist of all vertices fixed by T (O 0 ) and the edges joining them. G(F 0 /F ) operates on these buildings and, because H 1 (G(F 0 /F ), G(O0 )) is trivial, X and X0 are formed by the fixed points of (F 0 /F ). The intersection X ∩ AT (F 0 ) consists of p0 alone and X0 ∩ A0T (F 0 ) is formed by the points lying over p0 . p ......... ......... ......... ...... ...... ... .... ....... ... .... ........ .... .. .. .. ... ........ .. ... ... ........ .......... p0 γp ......... ......... AΥ (F ) The integral defining Ffλ (γ) is equal to meas G(O) X 1 τ (γp0 ,p0 )=λ meas T (O) Base change 34 where p0 runs over those points for which not only τ (γp0 , p0 ) = λ but also d(p0 ) = dist(p, p0 ). Since the closest point to p in AT (F 0 ) is p0 and since the shortest path joining p to p0 must lie completely in X there are q r+m 1 + such points if ξ = λ(z) + (m, −m), m > 0, and ∆(γ) = q −r , and there are Xr−1 1 + (q + 1) k=0 qk = qr · 1 q q+1 2 − q−1 q−1 if λ = λ(z), but none otherwise. The lemma follows upon comparison with the calculations for the proof of Lemma 5.5. Suppose the torus T splits over a ramified quadratic extension F 0 . It is no longer X and X0 but their first barycentric subdivisions X1 and X01 which are subcomplexes of X(F 0 ) and X(F 0 ). We may again introduce AT (F 0 ) and A0T (F 0 ) as well as the action of G(F 0 /F ). There is exactly one point pT of AT (F 0 ) fixed by G(F 0 /F ) and it is a vertex. If p is a vertex of X the closest point to it on A T (F 0 ) is pT . There can be at most two points on X at a minimal distance from AT (F 0 ) and these two points must be a distance 1 apart in X, for every second point on the path of shortest length joining them lies in X p1 .......... ......... ......... ......... ........ p ........ 2 .... ... ... ...... ......... ...... . .... .. .. ......... ... .. ... . ......... ......... pΥ ......... ......... AΥ (F 0 ) There must be at least two such points p1 , p2 , for the set of them is fixed by T (F ), and T (F ) contains an element whose determinant has order 1, and which, as a consequence, fixes no point of X. Let δ be the distance of p3 from pT in X(F 0 ). Lemma 5.7 Suppose det γ = $2m+1 and set µ = (m + 1, m). If −δ−1 q 2 ϕγ (t) = 2 meas T (O) µ e(t) µ(t) + 1 − q −1 α(t) 1 − q −1 α−1 (t) then Ff (γ) =< f ∨ , ϕγ > for f ∨ in H. Here T (O) is the stabilizer of p0T in T (F ). Observe that by Lemma 5.2, ∆(γ) = q −δ/2 Base change 35 for γ must certainly interchange p1 and p2 and therefore p3 is a fixed point of γ at maximal distance from AT (F 0 ). Arguing from a diagram p ............ .... ....... .... .... .... .... ..... . . ........ . . ...... ...... .... .... .... .... ........ ..... ....... ........ . . . .... .... ...... . . 1 .............. 2 ... . ... ... ....... ... .... .. . ........ ......... ........ ......... ......... p ......... ......... γp p PT ......... as usual we see that Ffλ (γ) is 0 unless λ = (m + 1 + r, m − r), r ≥ 0 when it is ∆(γ) meas G(O) r q . meas T (O) Moreover < fλ∨ , ϕγ > is 0 unless λ = (m + 1 + r, m − r) when it is meas G(O) r− δ q 2 2 meas T (O) times 1 2πi Z z=1 zr (1 − q −1 z −1 ) (1 − q −1 z) z −r r+1 −r−1 + z + z + 1 − z −1 (1 − z −1 )(1 − q −1 z) (1 − z)(1 − q −1 z −1 ) 1−z dz . z This contour integral can be evaluated by shrinking the path a little and then integrating term by term. The first two terms have no poles inside the contour of integration and yield 0; the last two integrals are evaluated by moving the path to ∞, and each yields the residue 1 at z = 1. The lemma follows. If det γ = $2m we may choose z ∈ F so that zF 0 = aF 0 = bF 0 . Lemma 5.8 If det γ = $2m then Ff (γ) = q −δ−1 2 meas Gn (O) meas T (O) Z f (g −1 ng)dg − Gn (F )\G(F ) ∆(γ) meas G(O) f (z) q − 1 meas T(O) with n = zn0 . If ∆(γ) = q −α then α ≥ 0, 2α − δ − 1 is even, and γ fixes all points in X(F 0 ) at a distance less than or equal to 2α from AT (F 0 ). If j − δ − 1 is even and nonnegative there are 2q j−δ−1 2 points in X whose distance from AT (F 0 ) is j . Certainly Ffλ (γ) is 0 unless λ = (m + r, m − r), r ≥ 0. If r > 0 it is equal to meas G(O) r+ −δ−1 2 q meas T (O) Base change 36 and if r = 0 it equals −δ+1 meas G(O) q 2 − q −α meas G(O) X 2α−δ−1 2 qj = . j=0 meas T (O) meas T (O) q−1 γp ............ .... .... .... .............................. ..... .... ........ .... ........ .... . . .... .. . . . . . 1 .................... 2 ....... ... .... .... ........ ... .. ... . . ......... ......... ......... ......... p ......... p p ........ ......... ........ PT AT (F 0 ) The factor 2 disappears because the orbits under T (F ) are twice as large as the orbits under T (O). The lemma follows upon comparison with the proof of Lemma 5.5. If g=t 1 x 0 1 k with t in A(F ) and k in G(O), we set λ(g) = 1 if x ∈ O and λ(g) = x−2 otherwise. Then `nλ(g) is 2`n$ times the distance of gp0 from A. If ∆(γ) 6= 0 set A1 (γ, f ) = ∆(γ) Z f (g −1 γg)`nλ(g)dg. A(F )\G(F ) If t= t1 0 0 t2 lies in A(C) and f belongs to H we write f ∨ (t) = X 0 af (j 0 , j)tj1 tj2 . j 0 ,j Lemma 5.9 Let γ= a 0 0 b and let λ(γ) = (m0 , m) or (m, m0 ), m0 ≥ m. If m0 > m then X `n$ 1 0 0 A1 (γ, f ) = 1 − s af (j 0 , j) j +j=m +m meas A(O) q j 0 −j>m0 −m with 2s = j 0 − j − (m0 − m). If m0 = m then A1 (γ, f ) is equal to the sum of three terms: X `n$ 1 0 0 1 − af (j 0 , j) j +j=m +m s meas A(O) q 0 j −j>0 and `n∆(γ) X 1 0 0 0 a (j , j) 2af (m0 , m) + f j +j=m +m meas A(O) q s+1 j 0 −j>0 1− 1 q Base change 37 and, if ∆(γ) = q−α and z ∈ F × satisfies z = $m , 1 meas G(O) Xα−1 j−2 1− jq 2`n$ f (z). j=0 meas A(O) q It is enough to verify these formulae for f = f λ . Suppose first that m0 > m. The integral appearing in the definition of A1 (γ, f ) is equal to meas G(O) meas A(O) times the sum over all p0 for which τ (γp0 , p0 ) = λ and d(p0 ) = dist(p, p0 ) = dist(p, A) of 2`n$dist(p, A). p γ p ............. ... ... . ......... ... ... ... . ......... γ p0 ......... ........ ........ .......... ... .. .. . . ........ ... ... .. . ........ ........ ......... p0 The sum is empty unless λ = (m0 + r, m − r), r ≥ 0. However if this condition is satisfied it equals 1 q r `nω. 2r 1 − q The sum appearing in the formula claimed for A1 (γ, f ) is also 0 unless λ has this form when, by the explicit expansion of f ∨ (t), it equals 2 1 1 1 1 1 1 1 1 1− + 1− 1+ +· · ·+ 1 − 1+ +· · ·+ r−2 + 1+ +· · ·+ r−1 q r meas G(O) q q q q q q q q which is easily shown by induction to be 2rq r meas G(O). If m0 = m we base our calculation on the diagram ........ .......... ..... ... ... .... ..... .... . . . . . ....... ..... ........ .... ... .... .... ........ ... ...... .......... . . . .... .... .... .... ....... ...... ... .... .. ... ......... .. .... ... . . ........ ....... ......... ......... ......... α ......... ........ ......... ......... A1 (γ, fλ ) is certainly 0 unless λ = (m + r, m − r), r ≥ 0. If this condition is satisfied it equals 1 meas G(O) r 1− 2`n$(r + α)q , r > 0, q meas A(O) or 2q −α `n$ The contribution 1 meas G(O) Xα jq j 1 − , j=0 meas A(O) q meas G(O) r rq 2`n$ meas A(O) 1 1− q r = 0. Base change 38 is accounted for by the first of the three summands in the lemma. The contribution meas G(O) r 2`n$ αq meas A(O) 1 1− q by the second, and the remainder, which is 0 for r > 0 and 1 meas G(O) Xα−1 j−α 1− 2`n$ jq j=0 meas A(O) q for r = 0, by the third. The purpose of this paragraph is not simply to consider the algebra H by itself, but rather to compare it with the algebra HE of spherical functions on G(E), where E is an unramified extension of F of degree `. The comparison can be motivated by the point of view exposed in [20]. We have already seen that H is isomorphic to the representation ring of GL(2, C). With G = G(E/F ) we form the direct product L G = GL(2, C) × G which is the Lgroup of G. Let Φ be the Frobenius element in G. The representation ring of GL(2, C) is isomorphic, by means of the map g → g × Φ from GL(2, C) to GL(2, C) × Φ ⊆LG, to the algebra H obtained by restricting to GL(2, C) × Φ the representation ring of LG, which is the algebra of functions on LG formed by linear combinations of characters of finitedimensional complex analytic representations of LG. It is the isomorphism of H with H which is now important. We may regard G(E) as GE (F ) where GE is obtained from G by restriction of scalars. Its Lgroup is formed by setting L o GE = Y G GL(2, C), on which we let G act by right translations on the coordinates, and then taking the semidirect product L GE =LGoE × G. For simplicity index the coordinate g ∈LGoE corresponding to Φj by j . Then (h1 , · · · , h` )−1 · (g1 , · · · , g` ) × Φ · (h1 , · · · , h` ) is equal to −1 −1 (h−1 1 g1 h2 , h2 g2 h3 , · · · , h` g` h1 ) × Φ. Taking h2 = g2 h3 , h3 = g3 h4 , · · · , h` = g1 h1 , and h1 = h we obtain (h−1 g1 g2 · · · g` , h, 1, · · · , 1) × Φ. Base change 39 Thus conjugacy classes in LGE which project to Φ stand in a bijective correspondence with conjugacy classes in GL(2, C). It follows easily that HE is isomorphic to the algebra of functions HE obtained by restricting the representation ring of LGE to LGoE × Φ. The map of LG to LGE given by g × τ → (g, · · · , g) × τ yields a homomorphism HE → H and hence a homomorphism HE → H. It is this homomorphism which must be studied. If φ in HE has Fourier transform φ∨ , then maps to f , which is defined by f ∨ (t) = φ∨ (t` ). Fix σ ∈ G, σ 6= 1. We have observed that if γ ∈ G(F ), δ ∈ G(E), and γ = N δ then Gσδ (E) equals Gσγ (F ), where Gσγ is a twisted form of Gγ . We may therefore use the convention of [14] to transport Tamagawa measures from Gγ (F ) to Gσγ (E). Lemma 5.10 Suppose φ in HE maps to f in H. If γ = N δ then Z φ(g −1 δσ(g))dg = ξ(γ) Gσ (E)\G(E) δ Z f (g −1 γg)dg. Gγ (F )\G(F ) Here ξ(γ) is 1 unless γ is central and δ is not σ conjugate to a central element when it is 1. Moreover if γ in G(F ) is the norm of no element in G(E) then Z f (g −1 γg)dg = 0. Gγ (F )\G(F ) We check this when φ = φλ , the characteristic function of the double coset G(OE )tG(OE ) = KE tKE , where λ(t) = λ. X(E) and X0 (E) are the Bruhat–Tits buildings over E . To prove the lemma we are unfortunately, but probably inevitably, reduced to considering cases. Suppose first that δ is a scalar so that Gσδ (E) = G(F ). Then Z φ(g −1 δσ(g))dg Gσ (E)\G(E) δ is equal to the sum over representatives p0 = gp00 of the orbits of G(F ) in X(E) for which the type of the pair (δσ(p0 ), p0 ) is λ of meas G(OE ) . meas G(F ) ∩ gG(OE )g −1 We choose representatives p0 so that d(p0 ) = dist(p, p0 ) and so that dist(p, p0 ) = dist(p, X). The reduction used repeatedly before shows that the integral is equal to meas G(OE ) X d(p0 )=dist(p,p0 )=dist(p,X) 1. meas G(O) τ (δσ(p0 ),p0 )=λ Base change 40 If λ(δ) = (m, m), this is 0 unless λ = (m+r, m−r), r ≥ 0. Since δp = p, the type τ (δσ(p0 ), p0 ) is λ = (m+r, m−r) if and only if dist(σ(p), p) = 2r. p ............ ......... ........ ... ... .. ... .. . . .. . ........... ...... ..... ... .. ... .. . ... . .. . ........ .......... ... .... ... .. ... ... ........ .......... ........ ......... σ(p) p0 ........ X Since X is the set of fixed points of σ in X(E), the paths from p0 to p and from p0 to σ(p) must start off in different directions. In other words the initial edge of the path from p0 to p does not lie in X . This shows that there are q `r (1 − q 1−` ) possibilities for the p0 or, what is the same, the p occurring in the above sum if r > 0 and just 1 if r = 0. To complete the verification in this case we have to evaluate Z A 0 ( C) φλ (t` )ϕγ (t) with ϕγ defined as in Lemma 5.3. Since λ(γ) = `λ(δ), the integral is certainly 0 unless λ = (m + r, m − r). If this condition is satisfied it equals mE (λ) 1 + q −1 −r` q 2 meas G(O) 1 + q −` times 1 2πi Z z=1 1 − z −1 1−z 1 − q −` z ` 1 − z −1 1−z 1 − q −` z −` `r · · z + · · z −`r −` −1 −1 −1 ` −1 −1 1−z 1−q z 1−q z 1−z 1−q z 1 − q −1 z We have to show that this integral is 2 1 − q −` 1 + q −1 2 1 + q −` 1 + q −1 if r > 0 and dz . z if r = 0. Once we shrink the circle of integration a little, we may integrate term by term. The first term will have only one pole inside the new circle, that at 0, where the residue is 0 if r > 0 and q 1−` if r = 0. The second term we write as 1 1−z 1 − z −1 1 · · ` −1 1 − z 1 − q z 1 − q −1 z −1 z 1− z` q` 1 1 − q −` 1−z 1 − z −1 1 z −`r − 1 − ` + · · · . ` −1 q 1−z 1 − q z 1 − q −1 z −1 z Base change 41 The first summand is integratd by moving the path out. The residues are at q and ∞ and yield 1 1 − q −1 · q − 1 · 1 − q` 1 − q −2 0 1 1− ` + 1 q q`−1 r>0 r=0. The second is integrated by moving in; the residues are at 0 and 1q . They yield 1− 1 1 1 − q −1 q −1 − 1 1 · · · + 1 − q. q ` 1 − q −` 1 − q −2 q −1 q` If everything is put together the result follows. Now suppose that γ is central but δ is not σ conjugate to a central element. Then as we observed in the previous paragraph ` = 2. Moreover since E is unramified det δ = $2m+1 for some integer m. Let Σ be the map p → δσ(p) of X(E) to itself. Then Σ has no fixed points, for Σ : gp0 → (δσ(g)g −1 )gp0 and det(δσ(g)g−1 ) = $2m+1 . Suppose p1 is a point for which dist(p1 , Σp1 ) is a minimum. Since Σ2 is the identity, Σ defines an inversion of the path of shortest length joining p1 to Σp1 . It follows immediately that dist(p1 , Σp1 ) = 1. I claim that if dist(p2 , Σp2 ) = 1 then p2 ∈ {p1 , Σp1 }. If not take the path of shortest length joining p2 to this set. Replacing p2 by Σp2 if necessary we may suppose the path runs from p2 to p1 . Then Σ applied to the path joins Σp2 to Σp1 , and we obtain a nontrivial cycle p2 . ..... ..... ..... .... . . . . . ........ .... ... .... .... .... . .... .... ... . . ... . . . ........ .... .... .... .... ... p1 This is a contradiction. The integral Z . .... .... ... .... . . . .. ....... .... .... .... ... .... ... .... . ... . . . .. . . . .. ...... ..... ..... ..... ..... ..... . P P p2 p1 φ(g −1 δσ(g))dg Gσ (E)\G(E) δ is equal to meas G(OE ) meas Gσδ (E) ∩ Gp01 X d(p0 )−d(p01 )=dist(p,p1 )<dist(p,Σp1 ) τ (p0 ,Σp0 )=λ 1 . Base change 42 Here p01 is any fixed lifting of p1 to X(E). The sum is empty unless λ = (m + 1 + r, m − r), r ≥ 0, when it is equal p ........ ... ... ... . ......... ... ... ... ..... ...... ...... ..... ... .. . ...... ...... .. .. . . . .......... P p1 P p p1 to q `r . Since Gσδ (E) ∩ Gp01 is a maximal compact subgroup of Gσδ (E), we can easily verify that (cf. p. 475 of [14]) meas Gσδ (E) ∩ Gp01 = meas G(O) . q−1 Since λ(γ) must be (2m + 1, 2m + 1), the integral Z A 0 ( C) φλ (t` )ϕγ (t) (` = 2) is 0 unless λ = (m + 1 + r, m − r). If this condition is satisfied this inner product is mE (λ) 1 + q −1 −(r+1/2)` meas G(OE ) 1 q −(r+1/2)` · q = · 1+ 2 meas G(O) 1 + q −` 2 meas (G(O) q times 1 2πi Z 1 − q −` z −` 1 − z −1 1 − z `r+` 1 − q −` z ` 1 − z −1 1 − z −`r−` · · z + · · z −` −1 −1 −1 ` −1 −1 1−z 1−q z 1−q z 1−z 1−q z 1 − q −1 z z=1 dz (` = 2). z This integral has to be shown to equal −2 q−1 . q+1 This can be done much as before. Once the contour of integration is shrunk a little, the integral of the first term becomes 0. The second term is written as 1 1 − z −1 1−z · · ` 1 − z 1 − q −1 z −1 1 − q −1 z z` 1−z 1 1 − q −` 1 − z −1 1 − ` z −`(r+1) − 1 − ` · . + q q 1 − z ` 1 − q −1 z −1 1 − q −1 z To integrate the first summand we move the path out. There is a residue at q which yields q−1 1 1 − q −1 1 · · q(q − 1) · 1 − ` = − 1 − q ` 1 − q −2 q q+1 For the second we move the path in; the residue at 1 q (` = 2). is 1 − q −1 q−1 1 −1 · =− . −2 q 1−q q+1 If γ is central but is not a norm then ` is odd and λ(γ) = (m, m) with m prime to `. It follows immediately that λ(t` ) is always orthogonal to ϕγ , so that f ∨ (γ) = 0 if f is the image of φ. Base change 43 We next suppose that γ = an0 = a Then 1 1 0 1 . a 0 µ = λ( )0 = (m, m) 0 a and γ is a norm if and only if ` divides m. It is clear that < f ∨ , µ >= 0 if ` does not divide m. Suppose then γ = N δ . We may write δ=b 1 v 0 1 . Then Gσδ (E) = Gγ (F ). We may choose as a set of representatives for the orbits of Gγ (F ) in X(E) the collection {p0 = np0z }, where z ∈ Z, where p 0z is defined to be that element of A 0 which projects to pz in A and satisfies d(p0z ) = dist(pz , p0 ), and where n= with x running over E/F + $ −zOE . Observe that 1 x 0 1 meas G(OE ) meas Gp0 = q −z . 0 meas Gp ∩ Gγ (F ) meas Gγ (O) m ` . Moreover τ (Σp0 , p0 ) is (k + r, k − r), k = with r equal to 0 if order(σ(x) − x − v) ≥ −z and equal to −order(σ(x) − x − v) − z otherwise. Thus Z Gσ (E)\G(E) δ φλ (g −1 δσ(g))dg is equal to 0 unless λ = (k + r, k − r). Since trace(σ(x) − x − v) = 1, the order of σ(x) − x − v is always less than or equal to 0. If we assume, as we may, that order v = 0, then ord(σ(x) − x − v) is ord(σ(x) − x) if this is negative and is 0 otherwise. If λ = (k, k) then the integral equals meas G(OE ) 1 meas G(OE ) X∞ −z q = · . z=0 meas Gγ (O) meas Gγ (O) 1 − q −1 If λ = (k + r, k − r) with r > 0 then the integral is meas G(OE ) meas Gγ (O) which equals X∞ q (`−1)r · q r + z=−r+1 q −z q (`−1)r (1 − q 1−` ) meas G(OE ) 1 − q −` `r ·q . meas Gγ (O) 1 − q −1 Base change 44 Since we can easily compute Z A 0 ( C) ` φ∨ λ (t )µ(t) by using the explicit expansion of φ∨ λ , the required equality follows from Lemma 5.5. We have still to treat the case that γ is regular and semisimple. Let T be the Cartan subgroup containing γ . If γ = N δ then δ also belongs to T . If T is A then γ is a norm if and only if µ = λ(γ) = (m0 , m) with both m0 and m divisible by `. Since Z f (g −1 γg)dg = ∆(γ)−1 Ff (γ) A(F )\G(F ) this integral is certainly 0 if m0 and m are not both divisible by `. However if ` divides m 0 and m and φ = φλ , with λ = (k 0 , k), the integral equals 0 ` meas G(O) ∆(γ)−1 q 2 (k −k) meas A(O) m0 = `k 0 , m = `k and 0 ` meas G(O) ∆(γ)−1 q 2 (k −k) (1 − q −` ) m0 = `k 0 − `r ≥ m = `k + `r, r > 0 meas A(O) but 0 otherwise. The integral Z (5.2) is equal to If m0 6= m then ∆(γ) = q φλ (g −1 δσ(g))dg A(F )\G(E) meas G(O) X d(p0 )=dist(p,p0 )=d(p,A) 1. meas A(O) τ (Σp0 ,p0 )=λ m0 −m 2 and the relevant diagram is p ........... ......... .. ... ... ..... . ... . ...... ..... ... ... ... ......... ......... ......... σ(p) ......... P .......... ... ... ... ..... ...... ... ... ... ... ......... ........ (p) ......... ......... If d(p, p0 ) = r then the type of (Σp0 , p0 ) is (m0 + r, m − r). For a given r, there are q `r (1 − q −` ) r>0 1 r=0 or possibilities for p or p0 . The equality follows. Base change 45 Suppose m0 = m and ∆(γ) = q −α , α ≥ 0. Then γ fixes all points in X(E) which are at a distance at most α from A, but no other points. We so choose δ that N δ lies in G(F ) and take γ = N δ . If, as usual, Σ : p → δσ(p) then Σ` p = γp so that Σ fixes p only if γ does. Suppose Σ fixes p and dist(p, A) < α. Then γ fixes all points which can be joined by p by an edge. If p = gp 0 these are the points gkp1 , k ∈ KE = G(OE ), p1 being one of the points in the apartment A adjacent to p0 . Thus g −1 γg ∈ G(OE ) and has trivial image in G(κE ), if κE is the residue field of OE . Moreover p0 = g −1 p = g −1 Σp = g −1 δσ(g)p0 ; so g −1 δσ(g) ∈ KE . Since (g −1 δσ(g))σ(g−1 δσ(g)) · · · σ`−1 (g −1 δσ(g)) = g −1 γg, we conclude that g −1 δσ(g) defines a cocycle of G in G(κE ). But all such cocycles are trivial; so we may suppose, upon replacing g by gk , that the image of g −1 δσ(g) in G(κE ) is 1. Then Σ(gkp1 ) = g(g −1 δσ(g)σ(k))p1 = gσ(k)p1 . This is equal to gkp1 if and only if k−1 σ(k)p1 = p1 . It follows that the number of points in X(E) which can be joined to p by an edge and are fixed by Σ is the same as the number of points in X which can be joined to p 0 by an edge, namely q + 1. The relevant diagram is now p ......... ........ The integral is certainly 0 unless λ = q α−1 `r P p0 m ` integral is meas G(OE ) meas A(O) ......... ........... ... ... ... .... ... .. ...... ...... ... ... ... ..... ........ ..... q (1 − q p ......... ........ ......... ......... A + r, m ` − r , r ≥ 0. If λ has this form and r > 0 the value of the 1−` 1 `r meas G(OE ) α+`r q (1 − q −`r ) )+q q 1− = q meas A(O) α and if r = 0 it is meas G(OE ) α q . meas A(O) This again yields the correct result. We suppose next that T is not split over F but that it splits over E . Then ` = 2 and the equation γ = N δ can always be solved. If the eigenvalues of δ are a, b those of γ are aσ(b), bσ(a). Σ has exactly one fixed point in AT (E) and this point is a vertex or not according as the order of the eigenvalues of γ is even or odd. If it is odd, say 2m + 1, then the diagram to be used is Base change 46 p ........... ........ ......... ... ... .. ......... ... ... .. .. ......... p1 P ........ .. ... .. . . ......... . .. .. . . .. ........ ........ P p ........ ......... p1 AT (E) The integral (5.2) (with A replaced by T ), is 0, unless λ = (m + 1 + r, m − r), r ≥ 0 when it is meas G(OE ) `r 2q meas T (O) (5.3) for the only forbidden initial direction for the path from p1 to p is the edge joining p1 and Σp1 . Define z ∈ F × = Z(F ) by z = aσ(b) = bσ(a) the set n = zn0 . If we appeal, as we shall now constantly have occasion to do, to the calculations made for γ a scalar or a scalar times a unipotent we see that (5.3) equals ∆(γ)−1 times Z 2∆(γ) meas G(O) 1 meas Gn (O) f (g −1 ng)dg − f (z) 1+ q meas T (O) Gn (F )\G(F ) q − 1 meas T (O) if f is the image of φλ . Observe in particular that the integral appearing here is 0 because the order of z is odd. In any case the desired equality follows from Lemma 5.6. If the order of the eigenvalues is even, say 2m, then the diagram to be brought into play is: ......... ......... ......... P .... ...... ... .... ....... ....... ..... .... .. .... ......... ... .. .... ...... ......... ..... p ................ p ........ ......... ......... ......... AT (E) Thus (5.2) is 0 unless λ = (m + r, m − r), r ≥ 0. If ∆(γ) = q−α and λ is of this form then it equals meas G(OE ) meas T (O) Xα−1 (1 − q 1−` )q `r 1 + (q + 1) q j + q `r (q + 1)q α−` j=0 or (5.5) meas G(OE ) meas T (O) 2q `r (1 − q 1−` ) α+`r q + 1 −` q · · (1 − q ) − q−1 q−1 if r > 0, and (5.6) meas G(OE ) meas T (O) Xα−1 meas G(OE ) −2 q+1 1 + (q + 1) qj = + qα · j=0 meas T (O) q−1 q−1 if r = 0. Our previous calculations show once again that this is equal to (5.4), so that we have only to appeal to Lemma 5.6. Base change 47 Suppose that T does not split over E but that it does split over an unramified extension. Then ` is odd. If the order of the eigenvalues of γ is m then γ is a norm if and only if ` divides m. It is clear from Lemma 5.6 and the cases previously discussed that Z (5.7) f (g −1 γg)dg = 0 T (F )\G(F ) when f is the image of φλ , if ` does not divide m. Suppose ` divides m. Let E 0 be the quadratic extension over which T splits and let ∆(γ) = q −α . There is one point, denoted p1 , in AT (EE 0 ) ∩ X(E) P p ............... ......... ......... ......... ......... ... ... .... ... .. ...... ...... ..... . ... ......... ......... p ......... p1 ......... ......... AT (EE 0 ) We can analyze the fixed points in Σ in X(E) and evaluate (5.2) as before. It is 0 unless λ = m ` when it is given by (5.5) and (5.6). + r, m ` − r , r ≥ 0, It remains to treat the case that T splits over a ramified quadratic extension E 0 . We shall appeal to Lemmas 5.7 and 5.8 as well as to some of our previous calculations. We know that γ is a norm if and only if det γ ∈ NE/F E × , that is, if and only if the order of det γ is divisible by `. The apartments AT (E 0 ) and AT (EE 0 ) are the same. Since this apartment is fixed by G(EE 0 /F ), the vertices in X(E) closest to it lie in X. Let them be p 1 , p2 as before p1................ ......... ......... ......... ... ....... 2 ... ... ... ...... ......... .... .. ... ........ ......... p pT ......... ......... ........ AT (E 0 ) We have all the information needed to calculate (5.2) (with T replacing A) at our disposal. If the order of det γ is odd, say 2m + 1, then Σ interchanges p1 and p2 , and (5.2) is 0 unless λ = (m + 1 + r, m − r), r ≥ 0, when it is meas G(OE ) `r q . meas T (O) If ` = 2 this is −1 meas G(O) f (z) q − 1 meas T (O) if z ∈ F × and order z = 2m + 1. Since the order of z is odd Z Gn (F )\G(F ) f (g −1 ng)dg = 0 Base change 48 for n = zn0 ; so we may conclude by an appeal to Lemma 5.8. If ` is odd we have to appeal to Lemma 5.7. This forces us to evaluate < f ∨ , ϕγ > . ∆(γ) (5.8) If the order of det γ is 2m0 + 1 the inner product is certainly 0 unless λ = (k 0 , k) with `(k 0 + k) = 2m0 + 1, and this implies in particular that it is always 0 unless ` divides 2m0 + 1. If ` divides 2m0 + 1, so that we can solve (`−1) 2 . γ = N δ , then `(2m + 1) = 2m 0 + 1 and m0 = `m + When these necessary relations between k 0 , k and m0 are satisfied the expression (5.8) is equal to q `r+`−1 2 meas G(OE ) meas T (O) 2 times 1 2πi Z `−1 `+1 `+1 (`−1) 1 − q −` z −` z `r+ 2 1 − q `z −` z `r+ 2 1 − q −`z ` z −`r− 2 1 − q −`z ` z −`r− 2 + + + · −` −1 −1 −` −1 ` −1 −1 1−z 1−q z 1−z 1−q z 1−z 1−q z 1 − z` 1 − q −1 z z=1 dz . z We again shrink the contour a little and then integrate term by term. The first two integrals are 0. For the last two we push the contours out to infinity. The only residues are at z ` = 1 and they are independent of r ≥ 0. We may therefore evaluate the last two integrals by setting r = 0 and shrinking the contour to 0. The third integral then has residues at 1 q and 0, which yield altogether `+1 `−1 1 − q −2` `+1 q 2 − q 2 = q− 2 . −` 1−q The residue at 0 is easy to calculate because ` − `+1 2 = `−1 2 is positive, so that it is the same as the residue of `+1 z− 2 . z − q −1 The fourth has a residue only at 0 and there it is q−( p `−1 2 ) . The desired equality follows. P p ........ ......... p ......... . ... .. . .................................. ..... .... ... ...... ....... ........ . ... .. 2 1 ......... ....... ....... .... ..... .. . .......... ......... ......... ........ p ......... ........ pT If the order of detγ is even, 2m0 , but not divisible by ` then Lemma 5.8 together with some of the previous calculations show that (5.7) is 0. Suppose γ = N δ and order (det δ) = 2m. Then Σ fixes p1 and p2 and the Base change 49 integral (5.2) is 0 unless λ = (m + r, m − r), r ≥ 0. Let ∆(γ) = q− λ = (m + r, m − r) with r > 0 the integral (5.2) equals q `r 1 − If r = 0 it equals 1 q `−1 meas G(OE ) meas T (O) X α=1 j=0 qj meas G(OE ) meas T (O) · + q `r · q α = q `r · q α · δ−1 2 −α . Here α is necessarily integral. If times 1 − q −` q `r − · (1 − q 1−` ). 1 − q −1 q−1 times Xα qα 1 − . −1 1−q q−1 qj = j=0 Our previous calculations show that these expressions equal Z −δ−1 meas Gn (O) q 2 ∆(γ) meas T (O) f (g −1 ng)dg − Gn (F )\G(F ) meas G(O) 1 f (z) q − 1 meas T (O) if z ∈ F × and order z = m0 . We have now merely to appeal to Lemma 5.8. Lemma 5.10 is now completely proved but the tedious sequence of calculations is not quite finished. There is one more lemma to be proved, but its proof will be briefer. The function λ(g) was defined in the preamble to Lemma 5.9. If δ ∈ A(E) and Gσδ (E) ⊆ A(E) we set A1 (δ, φ) = ∆(γ) Z φ(g −1 δσ(g)λ(g))dg A(F )\G(E) with γ = N δ . Lemma 5.11 Suppose φ maps to f . Then `A1 (γ, f ) = A1 (δ, φ). Let δ= a 0 0 b 0 and let a = $m , b = $m . Suppose first that m0 > m. The relevant diagram is P p ........... ........ ......... ... ... . .......... ... .. ... .......... p0 ......... ......... ......... ... .. .. . . .......... .. .. .. . . ........ .......... P p ......... p0 If φ = φλ then A1 (δ, φ) is 0 unless λ = (m0 + r, m − r), r ≥ 0, when it is 2r∆(γ)`n$` meas G(OE ) `r q meas A(O) 1 1− ` . q A1 (γ, f ) may be computed by combining the formula of Lemma 5.9 with the explicit expansion of φ ∨ λ . This yields meas G(OE ) `r X 1 0 0 2∆(γ)`n$ aφ (j, j 0 ) q 1 − j+j =m +m `s meas A(O) q 0 0 j −j>m −m Base change 50 if meas G(OE )aφ (j, j 0 ) = aφ (j, j 0 ), for ∆(γ) = q ` m0 −m 2 . The above sum is 1 1 1 1 1 1 1− ` +···+ 1− ` 1 − `(r−1) + 1 − `r = r 1 − ` 1− ` q q q q q q as required. Now take m0 = m. Let ∆(γ) = q −a p ......... ......... ......... .......... .... .... .... .... .... ... . . . . . ........ . ...... ...... ... ... ... ....... ....... .... ... .. ... . ......... ........ ......... P p ......... ......... A1 (δ, φλ ) is 0 unless λ = (m + r, m − r), r ≥ 0, when it is (5.9) 2q −α `n$ ` meas G(OE ) meas A(O) `r rq 1 − α−1 X `r + (j +r)q 1 − `−1 1 q j=1 1 1 `r+α j 1− q 1 − +(α+r)q q `−1 q q 1 if r > 0 and 2q −α `n$` (5.10) 1 meas G(OE ) Xα jq j 1 − j=0 meas A(O) q if r = 0. We sort this out and compare with the formula for A1 (γ, f ) given by Lemma 5.9 rq `r 1− 1 q `−1 + 1− 1 q `−1 X α−1 j=1 1 1 1 α `r+α +q = rq 1− 1− ` . q 1− q q q j This yields the part corresponding to the first summand of the lemma. The second summand of the lemma equals meas G(OE ) 1 α`n$ 1− q meas A(O) times 2q `r and is therefore given by the term X r−1 1 1 1 1 − + = 2q `r s=0 q `s q` q `r αq `r+α 1 1− q in the parentheses of (5.9) or by the last term of (5.10). This leaves from (5.9) X α−1 1 meas G(OE ) `r 1 j−α q 2`n$ 1 − `−1 jq 1− j=0 meas A(O) q q ` Base change 51 and from (5.10) meas G(OE ) Xα−1 j−α 1 . 2`n$ jq 1− j=1 meas A(O) q ` We know from the calculations made in the proof of Lemma 5.10 that these two expressions are equal to the last summand of Lemma 5.9. We now have all the formulae for spherical functions that we need, but unfortunately for the wrong spherical functions. Suppose ξ is an unramified character of NE/F Z(E) and H0 is the algebra of functions f 0 on G(F ) which are biinvariant with respect to G(O), of compact support modulo NE/F Z(E), and satisfy f 0 (zg) = ξ −1 (z)f 0 (g) Multiplication is defined by Z NE/F Z(E)\G(F ) 0 The map f → f with 0 f (g) = z ∈ NE/F Z(E). f10 (gh−1 )f20 (h)dh. Z f (zg) ξ(z)dz NE/F Z(E) is a surjective homomorphism from H to H0 . There is a simple and obvious relation between the orbital integrals of f and f 0 as well as between A1 (γ, f ) and A1 (γ, f 0 ). For example A1 (γ, f 0 ) = Z A1 (zγ, f ) ξ(z)dz. NE/F Z(E) If ξE is the composite of ξ with the norm we may define H 0 E in a similar manner. If φ0 ∈ H0 E then −1 φ0 (zg) = ξE (z)φ0 (g) for z ∈ Z(E). Moreover φ → φ0 with φ0 (g) = Z φ(zg) ξE (z)dz. Z(E) There is also a commutative diagram H0 y H 0 −−−−→ HE y −−−−→ HE If φ0 → f 0 then an analogue of Lemma 5.10 is valid. Z 0 φ (g Z(E)Gσ (E)\G(E) δ −1 δσ(g))dg = Z f 0 (g −1 γg)dg. Gγ (F )\G(F ) To verify this we begin with Z Z(E)Gσ (E)\G(E) δ Z Z(F )\Z(E) φ(z −1 g −1 δσ(g)σ(z))dzdg = Z f (g −1 γg)dg. Gγ (F )\G(F ) Base change 52 Replace δ by δv, v ∈ Z(E) and hence γ by δN v . Both sides are then functions on Z 1−σ (E)\Z(E). Multiply by ξ −1 (z) and integrate. The right side becomes Z f 0 (g −1 γg)dg. Gγ (F )\G(F ) Because of Lemma 5.10, the left side is Z φ0 (g −1 δσ(g))dg. Z(E)Gσ (E)\G(E) δ In order that the analogue of Lemma 5.11 be valid, we must set A1 (δ, φ0 ) = ∆(γ) Z φ(g −1 δσ(g))λ(g)dg. Z(E)A(F )\G(E) It is not difficult to see that H 0 is isomorphic to the algebra of functions on α 0 0 β ` ` ∈ A(C)  (αβ) = ξ($) obtained by restriction from some f ∨ , f ∈ H. This enables us to speak of (f 0 )∨ . Every homomorphism H0 → C is of the form α 0 f → (f ) ( ) 0 β 0 We may also speak of (φ0 )∨ . 0 ∨ (αβ)` = ξ($` ). Base change 53 6. ORBITAL INTEGRALS The study of orbital integrals was initialed by HarishChandra in his papers on harmonic analysis on semisimple Lie groups; the same integrals on padic groups were afterwards studied by Shalika. Some basic questions remain, however, unanswered. If they had been answered, much of this paragraph, which provides the information about orbital integrals, and twisted orbital integrals, to be used later, would be superfluous. But they are not and stopgaps must be provided. No elegance will be attempted here; I shall simply knock together proofs out of the material nearest at hand. Let F be a local field of characteristic 0. If f is a smooth function with compact support on G(F ), T is a Cartan subgroup of G over F , and γ is a regular element in T (F ) then we set Z Φf (γ, T ) = f (g −1 γg)dg. T (F )\G(F ) The integral depends on the choice of measures on G(F ) and T (F ), measures which we always take to be defined by invariant forms ωT and ωG . When it is useful to be explicit we write Φ f (γ, T ; ωT , ωG ). Since they complicate the formulae we do not use the local Tamagawa measures associated to forms ω as on p. 70 of [23] but simply the measures ω, which could be termed the unnormalized Tamagawa measures. It was observed on p. 77 of [23] that the map γ → Ch(γ) = (trace γ, det γ) of G to the affine plane X is smooth except at the scalar matrices. If a ∈ X is given then a two form µ on X which is regular and does not vanish in some neighborhood of a may be used to define an invariant form µ0 on Gγ \G if γ is regular and Ch(γ) is close to a. Set Φf (γ, µ) = Z f (g −1 γg)dµ0 . Gγ (F )\G(F ) If γ ∈ T is regular there is a form ωT (µ) such that Φf (γ, µ) = Φf (T, γ; ωT (µ), ωG ). ωT (µ) depends on γ . We shall call a function γa, T → Φ(γ, T ) = Φ(γ, T ; ωT , ωG ) an HCS family if it satisfies the following conditions. 0 (i) If ωT0 = αωT and ωG = βωG with α, β ∈ F × then 0 Φ(γ, T ; ωT0 , ωG )= β Φ(γ, T ; ωT , ωG ). α (ii) If h ∈ G(F ), T 0 = h−1 T h, γ 0 = h−1 γh, and if ωT 0 is obtained from ωT by transport of structure then Φ(γ 0 , T 0 ; ωT 0 , ωG ) = Φ(γ, T ; ωT , ωG ). Base change 54 (iii) For each T, γ → Φ(γ, T ) is a smooth function on the set of regular elements in T (F ) and its support is relatively compact in T (F ). (iv) Suppose z ∈ Z(F ) and a = Ch(z). Suppose µ is a twoform on X which is regular and nonzero in a neighborhood of a. There is a neighborhood U of a and for each T two smooth functions Φ0 (γ, T ; µ) and Φ00 (γ, T ; µ) on TU (F ) = {γ ∈ T (F )Ch(γ) ∈ U } such that Φ(γ, T ; ωT (µ), ωG ) = Φ0 (γ, T, µ) − meas(T (F )\G0 (F ))Φ00 (γ, T ; µ). Here G0 is the multiplicative group of the quaternion algebra over F . In the exceptional case that T is split, when G0 may not exist, the function Φ 00 (γ, T ; µ) is not defined and we take meas(T (F )\G0 (F )) = 0. Otherwise we regard T as a subgroup of G0 . The measure on T is to be ωT (µ) and that on G0 is given by the conventions on pp. 475–478 of [14]. If F is archimedean, X belongs to the center of the universal enveloping algebra and XT is its image under the canonical isomorphism of HarishChandra [25] then the restriction of XT Φ0 (γ, T, µ) to Z(F ) must be independent of T . Lemma 6.1. The collection {Φ(γ, T } is an HCS family if and only if there is a smooth function f with compact support such that Φ(γ, T ) = Φf (γ, T ) for all T and γ . Then for z ∈ F × = Z(F ) Φ0 (z, T, µ) = Φf (n, µ) with n=z 1 1 0 1 and, if T is not split, Φ00 (z, T, µ) = f (z). If F is archimedean, X belongs to the center of the universal enveloping algebra of the Lie algebra of T and X T is its image under the canonical isomorphism of HarishChandra then XT Φ0 (z, T ; µ) = ΦXf (n, µ) and XT Φ00 (z, T ; µ) = Xf (z). Base change 55 If F is nonarchimedean this is simply Lemma 6.2 of [23]. I observe however that in the formula for aT (γ) on p. 81 of [23] the function ξ(z) should be replaced by ξ(z) 1/2 de