Quadratic reciprocity (after Weil) 1. Standard set-up and Poisson summation

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1 (December 19, 010 Quadratic reciprocity (after Weil Paul Garrett garrett/ I show that over global fields k (characteristic not the quadratic norm residue symbol is a Hecke character, that is, a k -invariant continuous character on the ideles of k. From this reciprocity law one directly obtains the traditional reciprocity laws for quadratic Hilbert symbols, and for quadratic symbols. Striking is the role of certain quadratic exponential functions, as tempered distributions. The archimedean prototype is S x (z = e πix z (x R and z C This argument is suggested by, and essential to, a careful treatment of Weil representations, and proof that theta series are automorphic forms. 1. Standard set-up and Poisson summation. Weil s quadratic exponential distributions 3. Quadratic norm residue symbols and local integrals 4. The reciprocity law for quadratic norm residue symbols 5. Quadratic Hilbert-symbol reciprocity 6. Quadratic reciprocity 7. The simplest examples 1. Standard set-up and Poisson summation This section reviews standard items: measures, convolutions, characters, bilinear forms. We are concerned not with L Fourier inversion, but rather with a Schwartz-Bruhat Fourier inversion, which is easier. Let k be a global field of characteristic not. On each completion k v of k fix a non-trivial additive unitary character ψ v. For global applications, make these choices so that, for all but finitely-many v, ψ v (xy = 1 for all y o v x o v (o v the local ring of integers Further, choose the family of characters so that the global character ψ = v ψ v is trivial on k k A. For a character and (additive Haar measure on k v there is a local Fourier transform F f(x = ˆf(x = Take the Haar measure so that Fourier inversion holds: k v f (x = f( x f(y ψ v (xy dy The aggregate of local measures should make the total measure of the quotient A k /k be 1. Let K be either a quadratic field extension of k or isomorphic to k k. In either case let σ be the non-trivial k-algebra automorphism of K. Define a k-valued k-bilinear form, on K by α, β = tr K/k (αβ σ (with tr K/k (α = α + α σ 1

2 Paul Garrett: Quadratic reciprocity (after Weil (December 19, 010 Extend this k v -linearly to a k v -valued k v -bilinear form, on = K k k v Give the spaces additive compatible Haar measures, such that Fourier Inversion holds locally everywhere, with respect to the pairing a b ψ a, b Locally and globally the convolution on Schwartz-Bruhat functions is (f ϕ(a = f(a b ϕ(b db The groups are abelian, so convolution is commutative: f ϕ = ϕ f For a Schwartz-Bruhat function f on K A, Poisson summation is f(x = f(x x K x K Poisson summation is equivalent to the assertion that the tempered distribution u defined by u(f = f(x x K is its own Fourier transform. This may be proven by proving the one-dimensionality of the space of distributions supported on K and annihilated by multiplication by all the functions x ψ ξ, x (for ξ K Fourier transform exchanges these conditions, from which follows the Poisson formula up to a constant. Our normalization of measure makes the constant be 1. [1.0.1] Lemma: (p-adic version For Schwartz-Bruhat f on and smooth ϕ on, F (ϕf = F ϕ F f where F ϕ is the Fourier transform of the tempered distribution ϕ. [1.0.] Note: In the lemma, smooth means locally constant. The analogue of the lemma in the archimedean case is more delicate, since in that case Schwartz-Bruhat and test functions differ, and the notion of moderate growth is needed in addition to smoothness. Proof: (p-adic case The proof is a reduction to the corresponding property for Schwartz-Bruhat functions. Let L b be the (left regular representation operator on Schwartz-Bruhat functions by L b f(a = f(a b For a function h on, let θh(x = h( x

3 Paul Garrett: Quadratic reciprocity (after Weil (December 19, 010 The convolution of a distribution u and a test function f is [1] (u f(a = u(l a θf When the distribution u is integration against a function S, convolution is the usual convolution of functions. WIth ch be the characteristic function of a (large compact open subgroup of ch ϕ ϕ (in the topology of tempered distributions Indeed, for large enough depending upon the Schwartz-Bruhat function f, ϕ(a f(a da = ϕ(a f(a da since in the p-adic case f has compact support. Likewise, since the Fourier transform is a topological automorphism of Schwartz-Bruhat functions, lim F (ch u = F u Thus, (F ϕ F f(a = (F ϕ(l a θf f = lim (F (ch ϕ(l a θf f = lim (F (ch ϕ F f(a = lim F (ch ϕf(a = F (ϕf(a using the identity for Schwartz-Bruhat functions. F (α β = F α F β. Weil s quadratic exponential distributions Weil s quadratic exponential functions are introduced, as tempered distributions. contain the germ of the reciprocity law. The first two lemmas For x k v define S x (a = ψ v ( x a, a View this as a tempered distribution, as usual, by identifying it with the integration-against functional f S x (a f(a da [.0.1] Lemma: (p-adic case where F S x γ(x = lim = γ(x S x 1 ψ v ( x a, a da as ranges over larger and larger compact open subgroups of. In fact, there is a large-enough compact open subgroup Y of K so that the limit is reached for any Y. [1] Expression of convolution in terms of left and right regular representations proceeds similarly on any unimodular topological group, by changing variables in the integrals, using the invariance of the measure. For that matter, the precise relationship is often less significant than the fact that there is a relationship. 3

4 Paul Garrett: Quadratic reciprocity (after Weil (December 19, 010 [.0.] Lemma: Let f be a Schwartz-Bruhat function on. For x k v and a (S x f(a = S x (a F (S x f(xa Proof: (of first lemma In the course of the proof, we show that the limit exists in the stronger sense indicated. By the usual definition of Fourier transform of a tempered distribution, F S x (f = S x (F f = S x (a F f(a da v Since F f is also a Schwartz-Bruhat function, the characteristic function ch of any sufficiently large compact open set can be inserted into the integral without affecting its value. Thus, F S x (f = S x (a ch (a F f(a da Then S x ch is itself a Schwartz-Bruhat function, so apply the identity K f 1 (a F f (a da = F f 1 (a f (a da Thus, Since generally given a, = F (S x ch (a = F S x (f = F (S x ch (a f(a da ψ v a, b = ψ v ( a, b σ = ψ v b, a S x (b ψ v a, b db = ψ v ( x b, b 1 a, b 1 b, a db = ψ v ( x b, b a, b db ψ v ( x b x 1 a, b x 1 a x 1 a, a db For large enough (depending upon a, replace b by b + x 1 a to obtain F (S x ch (a = S x 1 a, a ψ v ( x b, b db Thus, as claimed. F S x = S x 1 lim ψ v ( x b, b db Proof: (of second lemma From the definitions, and from we have (f S x (a = (S x f(a = = K This is all we want here. K ψ a, b = ψ b, a S x (a b f(b db = S x (a ψ xa, b S x (b f(b db = S x (a K K ψ ( x a, a x a, b x b, a + x b, b f(b db ψ xa, b S x (b f(b db = S x (a F (S x f(xa 4

5 Paul Garrett: Quadratic reciprocity (after Weil (December 19, Quadratic norm residue symbols and local integrals One more bit of preparation is required. We define the local norm residue symbol ν v : k v {±1} attached to the separable quadratic extension /k v as follows. If = K k k v is not a field, then just put ν v (x = 1 for all x k v. If is a field, put ν v (x = 1 if x is a norm from, otherwise ν v (x = 1. It is standard but non-trivial that the norms from are of index two in k v for a separable quadratic field extension of k v. We invoke this know that ν v is a group homomorphism. As in the previous section, let γ(x = γ v (x = lim S x (a da where ranges over larger and larger compact open subgroups of, and x k v. [3.0.1] Lemma: (p-adic case γ v (x = ν v (x x 1 k v γ v (1 (for x k v Proof: From the definition, γ(x = lim ψ(xaa σ da and the limit is reached for sufficiently large. For x k v of the form x = bb σ, replacing a by ab 1 in the integral gives γ(x = b 1 lim ψ(aa σ da We are using the local norms making the product formula hold, so b x kv = bb σ kv = b Kv Thus, we have the desired formula when x is a local norm. When x is not a local norm, then it must be that is a field, otherwise the local norm map is onto. Let Θ be the subgroup of of elements of norm 1; it is compact. Letting vary over Θ-stable compact open subgroups, γ(x = lim ψ(xaa σ da = lim ψ(xaθθ σ a σ dθ da = lim ψ(xaa σ da Θ\ Θ Θ\ where we give Θ total measure 1. Taking the quotient of K v by the kernel Θ of the norm, is an isomorphism to its image. Note that ϕ : Θ\K v k v (by α αα σ d α = α 1 dα d y = y 1 k v dy are multiplicative Haar measures on and k v, respectively. The (topological isomorphism just above yields an identity γ(x = lim ψ(xaa σ da = lim ψ(xαα σ α Kv d α = lim ψ(xy y kv d y Y Θ\ Θ\ 5 Y

6 Paul Garrett: Quadratic reciprocity (after Weil (December 19, 010 where y = αα σ, = 0, and Y is the image of under the norm map. (Here we choose some compatible normalizations of the measures: it doesn t matter which. Since in this quadratic field extension the norms are of index, lim Y Y ψ(xy d y = lim ψ(xy 1 (1 + ord v(y y kv d y where runs over larger and larger compact open additive subgroups of k v (ignoring the point 0 k v. A typical elementary cancellation argument shows Then lim ψ(xy y kv d y = 0 (for x 0 γ(x = lim ψ(xy 1 ord v(y y kv d y Replace y by yx 1 to obtain the desired identity. 4. Reciprocity law for quadratic norm residue symbols The first reciprocity law is that (quadratic global norm residue symbols x ν K/k (x = Π v ν v (x (with x an idele of k are Hecke characters, that is, are trivial on k. Continuity is clear. Proof: This global assertion needs a global source: Poisson summation. For f an adelic Schwartz-Bruhat function, x k, and an adele a = {a v }, write S x (a = Π v S v x(a v where now S v x(a = ψ v ( x a va σ v Since S x is 1 on K, By Poisson summation, F (S x f(a = f(a = S x (a f(a (F S x F f(a = γ(x (S x 1 F f(a by the first lemma, which computed the Fourier transform of S x as tempered distribution. By the second lemma, which computed S x f, this is γ(x S x 1(a F (S x 1F f(xa = γ(x F (S x 1F f(xa since S x 1 = 1 on K. Change variables in the sum, replacing a by ax 1, to obtain (so far f(a = γ(x F (S x 1F f(a = γ(x S x 1(a F f(a 6

7 Paul Garrett: Quadratic reciprocity (after Weil (December 19, 010 the latter by Poisson summation, and this is = γ(x F f(a = γ(x f(a since S x 1(a = 1, and again applying Poisson summation. Taking any f so that f(a 0 necessarily γ(x = 1 (for all x k Then 1 = γ(x = Π v γ v (x = Π v x 1 k v ν v (xγ v (x = Π v ν v (xγ v (1 = ν(xγ(1 from the product formula and from the earlier result that γ v (x = ν v (xγ v (1 Thus, ν is a Hecke character. 5. Quadratic Hilbert-symbol reciprocity Recall the definition of quadratic Hilbert symbols, and obtain the their reciprocity law from the fact that the norm residue symbol is a Hecke character. For a, b k v the (quadratic Hilbert symbol is (a, b v = 1 1 (when ax + by = z has non-trivial solution in k v (otherwise [5.0.1] Theorem: For a, b k Π v (a, b v = 1 Proof: We prove this from the fact that the quadratic norm residue symbol is a Hecke character. When b (or a is a square in k, the equation ax + by = z has a solution over the global field k, with x = 0. Then there is a solution over k v for all v, so all the Hilbert symbols are all 1. Thus, the reciprocity assertion holds in this case. Suppose that b is not a square in k. Rewrite the equation as where K = k( b is a quadratic field extension of k. ax = z by = Norm K/k (z + y b At a prime v of k split in K, the local extension K k k v is not a field, and the norm is a surjection, so ν v 1 in that case. 7

8 Paul Garrett: Quadratic reciprocity (after Weil (December 19, 010 At a prime v of k not split in K, the local extension K k k v is a field, so ax = z by can have no (non-trivial solution x, y, z even in k v unless x 0. In that case, divide by x and find that a is a norm if and only if this equation has a solution. In summary, the values (a, b v of the Hilbert symbol coincide with the values of the local norm residue symbol ν v (a attached to the local field extension k( b/k. Thus, the reciprocity law for the norm residue symbol gives the corresponding result for the Hilbert symbol. In the proof, we proved a local statement stronger than that required to obtain the reciprocity law: we showed that the quadratic Hilbert symbol (a, b v is the quadratic local norm residue symbol ν v (a attached to the local extension k v ( b, for any non-zero a, b k v. When b is a square in k we must interpret this extension as k v [] mod b 6. Quadratic reciprocity Now we obtain the most traditional quadratic reciprocity law from the reciprocity law for the quadratic Hilbert symbol. We only derive what is often called the main part, refering to non-archimedean odd primes. Inspection of the relation (indicated in the proof of the quadratic symbols to Hilbert symbols will make clear how to obtain the auxiliary parts of quadratic reciprocity. Fix a ring of integers o inside k: for number fields k take the integral closure of in k, and for a function field which is a separable extension of F q ( take the integral closure of F q [] in k. For a (non-archimedean prime v of o, and for x o define the quadratic symbol ( x v = 1 (forx a non-zero square mod v 0 (for x = 0 mod v 1 (for x a non-square mod v For π o generating a prime ideal v, also write ( x π = ( x v A prime v is odd when the cardinality of its residue class field is odd. In the number field case, a prime is infinite when it lies over the real prime of Q. In the function field case, the prime at infinity in F q (T is given by the valuation : P q deg P A prime (i.e., valuation v of a finite separable extension k of F q (T, lying over, is an infinite prime of k. The reciprocity law here is: Quadratic Reciprocity ( main part : Let π and ϖ be two elements of o generating distinct odd prime ideals. Then ( ϖ ( π = Π v (π, ϖ v π ϖ where v runs over all even or infinite primes, and (, v is the (quadratic Hilbert symbol. 8

9 Paul Garrett: Quadratic reciprocity (after Weil (December 19, 010 Quadratic Reciprocity ( supplementary part : Let π o generate an odd prime ideal. For any other element α of o which is a πo-unit, ( α = Π v (π, α v π where v runs over all even or infinite primes and over all primes at which α is not a local unit. The proof of the main part illustrates well-enough the connection, so we omit explicit proof of the supplementary part. Proof: (of main part We claim that, since πo and ϖo are odd primes, (π, ϖ v = ( ϖ π for v = πo ( π ϖ for v = ϖo 1 for v odd and v πo, ϖo Let v = πo. Suppose that there is a solution x, y, z in k v to πx + ϖy = z Then (via the ultrametric property ord v y and ord v z must be identical, and less than ord v x, since ϖ is a v-unit and ord v πx is odd. Then multiply through by π n so that π n y and π n z are v-units. Then we see that ϖ must be a square modulo v. On the other hand, if ϖ is a square modulo v, then we can use Hensel s lemma to infer that ϖ is a square in k v. Then ϖy = z certainly has a non-trivial solution. Further, for v is an odd prime distinct from both πo and ϖo, then both π and ϖ are v-units. If ϖ is a square in k v, then ϖ = z certainly has a solution, so the Hilbert symbol is 1. Suppose ϖ is not a square in k v. Then, k v ( ϖ is an unramified field extension of k v, since v is odd. Thus, the norm map is surjective to units in k v. Thus, there are y, z k v so that π = Norm(z + y ϖ = z ϖy Thus, all but the even prime and infinite prime quadratic Hilbert symbols have interpretations in terms of quadratic symbols. 7. The simplest examples First, let s recover quadratic reciprocity for two (positive odd prime numbers p, q in. We wish to recover the assertion ( ( q p = ( 1 (p 1(q 1/4 p q We have proven so far that ( q p ( p q = (p, q (p, q where (p, q is the -adic Hilbert symbol and (p, q is the Q R Hilbert symbol. 9

10 Paul Garrett: Quadratic reciprocity (after Weil (December 19, 010 Since both p, q are positive, the equation px + qy = z has non-trivial real solutions x, y, z. That is, the real Hilbert symbol (p, q for the archimedean completion of Q has the value 1. Therefore, only the -adic Hilbert symbol contributes to the right-hand side of Gauss formula: so far ( ( q p = (p, q p q Hensel s lemma shows that the solvability of the equation above (for p, q both -adic units depends only upon their residue classes mod 8. The usual formula is but one way of interpolating the -adic Hilbert symbol by elementary-looking formulas. For contrast, let us derive the analogue for F q [T ] with q odd: for distinct monic irreducible polynomials π, ϖ in F q [T ], we have ( ϖ ( π ( (deg π(deg ϖ 1 = π ϖ F q ( 1 where F q is ± depending upon whether 1 is a square in F q or not. So far, from the general assertion of the previous section, where is the prime (valuation ( ϖ π ( π ϖ P q deg P = (π, ϖ This valuation has valuation ring consisting of all rational functions in T which can be written as power series in the local parameter t = T 1. Then π = t deg π (1 + t (... where (1 + t (... is some power series in t. A similar assertion holds for ϖ. Thus, if either degree is even, then one of π, ϖ is a local square, so the Hilbert symbol is +1. If t deg π (1 + t (... is a non-square, then deg π is odd. Nevertheless, any expression of the form is a local square (by Hensel s lemma. contemplating the equation 1 + t (... Thus, without loss of generality for local purposes, we are t (x + y = z The t -order of the right-hand side is even. If there is no 1 in F q, then the left-hand side is t -times a norm from the unramified extension F q ( 1(T = F q (T ( 1 so has odd order. This is impossible. On the other hand if there is a 1 in F q then the equation has non-trivial solutions. Thus, if neither π nor ϖ is a local square (i.e., both are of odd degree, then the Hilbert symbol is 1 if and only if there is a 1 in F q. The formula given above is an elementary interpolation of this assertion (much as was done for the case k = Q. 10