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19 the Analytic Class Number Formula 18.785 Number theory I Fall 2017 Lecture #19 11/13/2017 19 The analytic class number formula In the previous lecture we proved Dirichlet's theorem on primes in arithmetic progressions modulo the claim that the L-function L(s; χ) is holomorphic and nonvanishing at s = 1 for all non-principal Dirichlet characters χ. To establish this claim we will prove a more general result that has many other applications. Recall that the Dedekind zeta function of a number field K is defined by X −s Y −s −1 ζK (s) := N(a) = (1 − N(p) ) ; a p where a ranges over nonzero ideals of OK and p ranges over nonzero prime ideals of OK ; as we showed in the previous lecture the sum and product converge absolutely on Re(s) > 1. The following theorem is often attributed to Dirichlet, although he originally proved it only for quadratic fields (this is all he needed to prove his theorem on primes in arithmetic progressions, but we will use it in a stronger form). The formula for the limit in the theorem was proved by Dedekind [2, Supplement XI] (as a limit from the right, without an analytic continuation to a punctured neighborhood of z = 1), and analytic continuation was proved by Landau [3]. Hecke later showed that, like the Riemann zeta function, the Dedekind zeta function has an analytic continuation to all of C and satisfies a functional equation [1], but we won't take the time to prove this; see Remark x19.13 for details. Theorem (Analytic Class Number Formula). Let K be a number field of degree n. 1 The Dedekind zeta function ζK (z) extends to a meromorphic function on Re(z) > 1 − n that is holomorphic except for a simple pole at z = 1 with residue 2r(2π)sh R lim (z − 1)ζ (z) = K K ; K 1=2 z!1+ wK jDK j where r and s are the number of real and complex places of K, respectively, hK := # cl OK is the class number, RK is the regulator, wK := #µK is the number of roots of unity, and DK := disc OK is the absolute discriminant. 1=2 r s Recall that jDj is the covolume of OK as a lattice in KR := K ⊗Q R ' R × C × (Proposition 14.12), and RK is the covolume of ΛK := Log(OK ) as a lattice in the trace- r+s zero hyperplane R0 (see Definition 15.14) The residue of ζK (z) at z = 1 thus reflects both the additive and multiplicative structure of the ring of integers OK . Remark 19.1. In practice the class number hK is usually the most difficult quantity in the analytic class number formula to compute. We can approximate the limit on the LHS to any desired precision using a finite truncation of either the sum or product defining ζK (s). Provided we can compute the other quantities to similar precision, this provides a method for computing (or at least bounding) the class number hK ; this explains the origin of the term \analytic class number formula". You will have an opportunity to explore a computational application of this formula on Problem Set 9. Example 19.2. For K = Q we have n = 1, r = 1, s = 0, h = 1, w = #{±1g = 2, D = 1, and the regulator R is the covolume of a lattice in a zero-dimensional vector space, equivalently, the determinant of a 0 × 0 matrix, which is 1. In this case the theorem states 1 that ζQ(z) = ζ(z) is holomorphic on Re z > 1 − 1 = 0 except for a simple pole at z = 1 with residue 21(2π)0 · 1 · 1 lim (z − 1)ζQ(z) = = 1: z!1+ 2 · j1j1=2 Andrew V. Sutherland 19.1 Lipschitz parametrizability In order to prove the analytic class number formula we need an asymptotic estimate for the number of nonzero OK -ideals a with absolute norm N(a) bounded by a parameter t 2 R>0 that we will let tend to infinity; this is necessary for us to understand the behavior of P −z + × ζK (z) = a N(a) as z ! 1 . Our strategy is to count points in Log(OK \ K ) that lie r+s inside a suitably chosen region S of R that we will than scale by t. In order to bound this count as a function of t we need a condition on S that ensures that the count grows smoothly with t; this requires S to have a \reasonable" shape. A sufficient condition for this is Lipschitz parametrizability. Definition 19.3. Let X and Y be metric spaces. A function f : X ! Y is Lipschitz continuous if there exists c > 0 such that for all distinct x1; x2 2 X d(f(x1); f(x2)) ≤ c · d(x1; x2): Every Lipschitz continuous function is uniformly continuous, but the converse need p not hold. For example, the function f(x) = x on [0; 1] is uniformly continuous but not p Lipschitz continuous, since jp1=n − 0j=j1=n − 0j = n is unbounded as 1=n ! 0. Definition 19.4. A set B in a metric space X is d-Lipschitz parametrizable if it is the d union of the images of a finite number of Lipschitz continuous functions fi : [0; 1] ! B. Before stating our next result, we recall the asymptotic notation f(t) = g(t) + O(h(t)) (as t ! a); for real or complex valued functions f; g; h of a real variable t, which means f(t) − g(t) lim sup < 1: t!a h(t) Typically a = 1, and this is assumed if a is not specified. n 0 Lemma 19.5. Let S ⊆ R be a set whose boundary @S := S − S is (n − 1)-Lipschitz parametrizable. Then n n n−1 #(tS \ Z ) = µ(S)t + O(t ); n as t ! 1, where µ is the standard Lebesgue measure on R . n n Proof. It suffices to prove the lemma for positive integers, since #(tS \ Z ) and µ(S)t are both monotonically increasing functions of t and µ(S)(t + 1)n − µ(S)tn = O(tn−1). We can n partition R as the disjoint union of half-open cubes of the form C(a1; : : : ; an) = f(x1; : : : ; xn) 2 Rn : xi 2 [ai; ai + 1)g; with a1; : : : ; an 2 Z. Let C be the set of all such half-open cubes C. For each t > 0 define B0(t) := #fC 2 C : C ⊆ tSg; B1(t) := #fC 2 C : C \ tS 6= ;g: For every t > 0 we have n B0(t) ≤ #(tS \ Z ) ≤ B1(t): 18.785 Fall 2017, Lecture #19, Page 2 We can bound B1(t)−B0(t) by noting that each C(a1; : : : ; an) counted by this difference n p contains a point (a1; : : : ; an) 2 Z within a distance n = O(1) of a point in @tS = t@S. n−1 Let f1; : : : ; fm be Lipschitz functions [0; 1] ! @S whose images cover @S, and let n−1 c1; : : : cm be constants such that d(fi(x1); fi(x2)) ≤ cid(x1; x2) for all x1; x2 2 [0; 1] For any y 2 @S, we have y = fi(x1; : : : ; xn−1) for some i, and if we put rj = btxjc 2 Z so that 0 ≤ xj − rj=t ≤ 1=t, then p r1 rn−1 r1 rn−1 d(y; fi( t ;:::; t )) ≤ ci · d (x1; : : : ; xn−1); ( t ;:::; t ) < ci n=t ≤ c=t; p where c := n maxi ci. Thus every y 2 @S lies within a distance c=t of a point in the set r1 rn−1 P = fi t ;:::; t : 1 ≤ i ≤ m; 0 ≤ r1; : : : ; rn−1 ≤ t : which has cardinality m(t + 1)n−1 = O(tn−1). It follows that every point of @tS is within a distance c of one of the O(tn−1) points in tP. The number of integer lattice points within p a distance n of a point in t@S is thus also O(tn−1), and therefore n−1 B1(t) − B0(t) = O(t ): n We now note that B0(t) ≤ µ(tS) ≤ B1(t) and µ(tS) = t µ(S); the lemma follows. n Corollary 19.6. Let Λ be a lattice in an R-vector space V ' R and let S ⊆ V be a set whose boundary is (n − 1)-Lipschitz parametrizable. Then µ(S) #(tS \ Λ) = tn + O(tn−1): covol(Λ) n Proof. The case Λ ⊆ Z is given by the lemma; note that the normalization of the Haar measure µ is irrelevant, since we are taking a ratio of volumes which is necessarily preserved n under the isomorphism of topological vector spaces V ' R . We now note that if the corollary holds for sΛ, for some s > 0, then it also holds for Λ, since tS \ sΛ = (t=s)S \ Λ. For any lattice Λ, we can choose s > 0 so that sΛ is arbitrarily close to an integer lattice (for example, take s to be the LCM of all denominators appearing in rational approximations of the coordinates of a basis for Λ). The corollary follows. Remark 19.7. Recall that covol(Λ) = µ(F ) for any fundamental region F for Λ, so the ratio µ(S)= covol(Λ) = µ(S)/µ(F ) in Corollary 19.6 does not depend on the normalization of the Haar measure µ. However, we plan to apply the corollary to Λ = OK and want to p 1=2 replace covol(OK ) with j disc(OK )j = jDK j via Proposition 14.12, which requires us to use the normalized Haar measure on KR defined in x14.2.
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