18 Dirichlet L-functions, primes in arithmetic progressions

Transcription

1 Number theory I Fll 2017 Lecture #18 11/8/ Dirichlet L-functions, rimes in rithmetic rogressions Hving roved the rime number theorem, we would like to rove n nlogous result for rimes in rithmetic rogressions. We begin with Dirichlet s theorem on rimes in rithmetic rogressions, result tht redtes the rime number theorem by sixty yers. Theorem 18.1 (Dirichlet 1837). For ll corime integers nd m there re infinitely mny rimes mod m. In fct Dirichlet roved more thn this. In sense tht we will mke recise below, he roved tht for every fixed modulus m the rimes re equidistributed mong the residue clsses in (Z/mZ). The equidistribution sttement tht Dirichlet ws ble to rove is bit weker thn one might like, but it is more thn enough to estblish Theorem Remrk Mny of the stndrd tools of comlex nlysis we tke for grnted were not vilble to Dirichlet in Riemnn ws the first to seriously study ζ(s) s function of comlex vrible, some twenty yers fter Dirichlet roved Theorem We will work in more modern setting, but our roch still follows the sirit of Dirichlet s roof Infinitely mny rimes To motivte Dirichlet s method of roof, let us consider the following (dmittedly clumsy) roof tht there re infinitely mny rimes. It is sufficient to show tht the Euler roduct ζ(s) = (1 s ) 1 diverges s s 1 +. Of course we know ζ(s) hs ole t s = 1 (by Theorem 16.3), but let us suose for the moment tht we did not lredy know this. Tking logrithms yields log ζ(s) = log(1 s ) = s + O(1), (1) s s 1 +, where we hve used the symtotic bounds log(1 x) = x + O(x 2 ) (s x 0) nd O( 2s ) = O(1) (Re(s) > 1/2). We cn estimte 1 x vi Mertens second theorem, one of three he roved in [4]. Theorem 18.3 (Mertens 1874). As x we hve (1) x log = log x + R(x), where R(x) < 2. 1 (2) 1 = log log x + B + O( 1 log x), where B = is Mertens constnt; x (3) x log ( 1 1 ) = log log x γ + O ( 1 log x), where γ = is Euler s constnt. 1 In fct, R(x) = B 3 + o(1) where B 3 = is n exlicit constnt. Andrew V. Sutherlnd

2 Proof. See Problem Set 9. Thus not only does s diverge s s 1 +, we cn sy with fir degree of recision how quickly this hens. We should note, however, tht Mertens estimte is not s strong s the rime number theorem. Indeed, s you will rove on Problem Set 9, the Prime Number Theorem is equivlent to the sttement 1 ( ) = log log x + B + o 1, log x x which is (ever so slightly) shrer thn Mertens estimte Infinitely mny rimes congruent to 1 modulo 4 To demonstrte how the rgument bove generlizes to rimes in rithmetic rogressions, let us rove there re infinitely mny rimes congruent to 1 mod 4. We might initilly consider (1 s ) 1 = n s, 1 mod 4 n 1 n 1 mod 4 but the sum on the RHS is bit wkwrd. Let us insted define Dirichlet chrcter 1 if n 1 mod 4, χ(n) := 1 if n 1 mod 4, 0 otherwise, nd consider the Dirichlet L-function L(s, χ) := (1 χ() s ) 1 = χ(n)n s = 1 3 s + 5 s 7 s + 9 s +, which converges bsolutely on Re(s) > 1. As s 1 + we hve log L(s, χ) = log(1 χ() s ) = χ() s + O(1) nd thus log ζ(s) = n 1 = s 1 mod 4 3 mod 4 s + 1 mod 4 3 mod 4 log ζ(s) + log L(s, χ) = 2 s + O(1), s + O(1), s + O(1). 1 mod 4 Provided log L(s, χ) = O(1) s s 1 +, the LHS (nd hence the RHS) must tend to infinity s s 1 +, since ζ(s) s s 1 +. It thus suffices to show tht L(s, χ) hs n nlytic 2 The error term in the PNT ctully imlies 1 x = log log x + B + O( ) 1 x, but n o( 1 ) bound is log x lredy enough to show π(x) x/ log x. Tht the difference between little-o nd big-o is the difference between roving the PNT nd not roving it demonstrtes how criticl it is to understnd error terms Fll 2017, Lecture #18, Pge 2

3 continution to neighborhood of s = 1 with L(1, χ) 0 (in which cse there is brnch of the comlex logrithm holomorhic on neighborhood of L(1, χ)). We will rove this in the next lecture. Assuming this for the moment, we then hve 1 x 1 mod 4 = 1 log log x + O(1). 2 Mertens second theorem imlies tht the sme holds if we insted sum over 3 mod 4. The rimes re thus equidistributed modulo m = 4 in the sense tht for ll integers corime to m we hve log log x. x x mod m We should note tht this sttement is weker thn the rime number theorem for rithmetic rogressions, which sttes tht π(x; m, ) 1 π(x), where π(x; m, ) counts the rimes x for which mod m (see Problem Set 9). Dirichlet did not hve Mertens symtotic bounds so he stted his results in different wy, using wht is now clled the Dirichlet density of set of rimes S, d(s) := lim s 1 + s S, s defined whenever this limit exists (one cn lso define notions of lower nd uer Dirichlet density using lim inf nd lim su tht re lwys defined nd gree whenever d(s) is defined). This definition differs from the more common notion of nturl density #{ x : S} δ(s) := lim. x #{ x} Dirichlet roved tht for ll corime integers nd m the set of rimes mod m hs Dirichlet density 1/, wheres the rime number theorem for rithmetic rogressions sttes tht this set hs nturl density 1/. If set of rimes S hs nturl density then it hs Dirichlet density nd the two re equl, but the converse need not hold: there re sets of rimes tht hve Dirichlet density but no nturl density (see Problem Set 9). In order to comlete our roof tht there re infinitely mny rimes 1 mod 4, we still need to show L(1, χ) 0. We will chieve this in the next lecture, but for now let us show tht this reduces to understnding the behvior of the Dedekind zet function 3 ζ Q (i) (s) t s = 1. In generl the Dedekind zet function of number field K is defined by ζ K (s) := N() s = (1 N() s ) 1, 3 The Dedekind zet function is nmed fter Richrd Dedekind, the lst doctorl student of Guss. He received his Ph.D. in 1854, the sme yer s Riemnn, nother student of Guss. Dedekind nd Riemnn both studied under Dirichlet s well Fll 2017, Lecture #18, Pge 3

4 where the sum rnges over nonzero idels of the ring of integers O K, the roduct rnges over nonzero rime idels of O K (rimes of K), nd N() := [O K : ] is the bsolute norm. Note tht ζ Q (s) = ζ(s), so this is nturl generliztion of the Riemnn zet function. Tht the Euler roduct for ζ K (s) converges for Re(s) > 1 follows esily from the cse ζ Q (s) = ζ(s) roved in Theorem We use unique fctoriztion of idels in the Dedekind domin O K to convert the sum over idels into roduct over rime idels. We then note tht for ech rtionl rime we hve #{ } [K : Q] = n nd N() = f (by Theorems 5.34 nd 6.10), nd it follows tht log (1 N() s ) n log(1 s ). The sum on the RHS converges on Re(s) > 1, so the sum on the LHS must s well. For K = Q(i) we cn rewrite the Euler roduct for ζ K (s) s ζ K (s) = (1 N() s ) 1 = (1 N() s ) 1 = (1 2 s ) 1 (1 s ) 1 (1 s ) 1 (1 2s ) 1 1 mod 4 3 mod 4 = (1 2 s ) 1 (1 s ) 1 (1 s ) 1 (1 s ) 1 (1 + s ) 1 1 mod 4 3 mod 4 = (1 s ) 1 (1 χ() s ) 1 = ζ(s)l(s, χ), where we hve used the fct tht we hve one rime of norm N() = 2 bove the single rime = 2 tht rmifies in Q(i); two rimes, of norm N() = N( ) = bove ech rime tht sits in Q(i), equivlently, the rimes 1 mod 4; one rime of norm N() = 2 bove ech rime tht remins inert in Q(i), equivlently, the rimes 3 mod 4. We know ζ(s) hs simle ole t s = 1. If we cn show ζ K (s) extends to meromorhic function with simle ole t s = 1, then L(s, χ) must extend to function tht is holomorhic nd nonvnishing t s = 1, since ord s=1 L(s, χ) = ord s=1 ζ K (s) ord s=1 ζ ( s) = 1 ( 1) = 0. In fct, ζ K (s) extends to meromorhic function on Re(s) > 1 with simle ole t 2 s = 1; this cn be roved directly, but it follows from much more generl nd striking result, the nlytic clss number formul, which ws lso roved by Dirichlet (t lest for qudrtic fields). We will rove the nlytic clss number formul in the next lecture. For the reminder of this lecture we will focus on generlizing our roch to hndle rbitrry moduli m Fll 2017, Lecture #18, Pge 4

5 18.2 Dirichlet chrcters We now define the notion of Dirichlet chrcter. Historiclly, these receded the notion of grou chrcter; they were introduced by Dirichlet in 1831, well before the notion of n bstrct grou ws in common use. 4 In order to simlify the exosition we will occsionlly invoke some stndrd fcts bout chrcters of finite belin grous tht we recll in Definition A function f : Z C is clled n rithmetic function. 5 The function f is multilictive if f(1) = 1 nd f(mn) = f(m)f(n) for ll corime m, n Z; it is totlly multilictive (or comletely multilictive) if f(1) = 1 nd f(mn) = f(m)f(n) for ll m, n Z. For m Z >0 we sy tht f is m-eriodic if f(n + m) = f(n) for ll n Z, nd we cll m the eriod of f it is the lest m > 0 for which this holds. Definition A Dirichlet chrcter is eriodic totlly multilictive rithmetic function χ: Z C. The imge of Dirichlet chrcter is finite multilictively closed subset of C, hence the union of finite subgrou of U(1) nd subset of {0}. The constnt function 1(n) := 1 is the trivil Dirichlet chrcter; it is the unique Dirichlet chrcter of eriod 1. Ech m-eriodic Dirichlet chrcter χ restricts to grou chrcter χ on (Z/mZ). Conversely, every grou chrcter χ of (Z/mZ) cn be extended to Dirichlet chrcter χ by defining χ(n) = 0 for n (Z/mZ) ; this is clled extension by zero. 6 Definition A Dirichlet chrcter of modulus m is n m-eriodic Dirichlet chrcter χ tht is the extension by zero of grou chrcter on (Z/mZ) ; equivlently, n m-eriodic Dirichlet chrcter for which n (Z/mZ) χ(n) 0. Remrk Some uthors only define Dirichlet chrcters of modulus m, thereby bking m into the definition of Dirichlet chrcter; we simly view Dirichlet chrcters s functions Z C tht stisfy certin roerties. Note tht single Dirichlet chrcter my be Dirichlet chrcter of modulus m for infinitely mny m (for exmle, the unique Dirichlet chrcter of modulus 2 is lso Dirichlet chrcter of modulus 2 k for ll k 1). The Dirichlet chrcters of modulus m form grou under ointwise multiliction tht is cnoniclly isomorhic to the chrcter grou of (Z/mZ). Not every m-eriodic Dirichlet chrcter χ is Dirichlet chrcter of modulus m, since n m-eriodic Dirichlet chrcter need not vnish on n (Z/mZ). More generlly, we hve the following lemm. Lemm Let χ be Dirichlet chrcter of eriod m. Then χ is Dirichlet chrcter of modulus m if nd only if m m m k for some k (which holds in rticulr for m = m). Proof. Suose for the ske of contrdiction tht χ(n) 0 for some n Z tht hs rime fctor in common with m. Then χ() 0, since χ()χ(n/) = χ(n) 0, nd for r Z, χ(r)χ() = χ(r) = χ(r + m) = χ(r + m/)χ(), which imlies χ(r) = χ(r + m/), since χ() 0. Thus χ is (m/)-eriodic, but this contrdicts the minimlity of the eriod m. Therefore χ(n) = 0 for ll n (Z/mZ). 4 Glois seminl er ws rejected tht sme yer; it wsn t ublished until 12 yers fter his deth. 5 Mny uthors restrict the domin of n rithmetic function to Z 1 ; for the eriodic rithmetic functions we re interested in here, this distinction is irrelevnt, nd it is slightly more nturlly to work with Z. 6 When we write n (Z/mZ) we of course refer to the imge of n under the quotient m Z Z/mZ Fll 2017, Lecture #18, Pge 5

6 Conversely, for ny n (Z/mZ) we cn ick n integer = n e 1 mod m so tht χ(1) = χ() = χ(n e ) = χ(n) e 0 nd χ(n) 0. So χ is Dirichlet chrcter of modulus m. If m m m k, then the rime divisors of m coincide with those of m. It follows tht n (Z/m Z) n (Z/mZ) χ(n) 0, nd χ is clerly m -eriodic (since m m ), so χ is Dirichlet chrcter of modulus m. Conversely, if χ is Dirichlet chrcter of modulus m, then χ is m -eriodic, nd therefore m m, since m is the eriod of χ. And since χ is Dirichlet chrcter of modulus m nd of modulus m, for ech rime we hve (Z/mZ) χ() = 0 (Z/m Z), thus the rime divisors of m nd m coincide nd m must divide some ower m k of m Primitive Dirichlet chrcters Given Dirichlet chrcter χ 1 of modulus m 1 dividing m 2, we cn lwys crete Dirichlet chrcter χ 2 of modulus m 2 by tking the extension by zero of the restriction of χ 1 to (Z/m 2 Z) ; in other words, let χ 2 (n) := χ 1 (n) for n (Z/m 2 Z) nd χ 2 (n) := 0 otherwise. If m 2 is divisible by rime tht does not divide m 1, the Dirichlet chrcters χ 1 nd χ 2 will not be the sme (χ 2 () = 0 χ 1 (), for exmle), they will gree on n (Z/m 2 Z) but not n (Z/m 1 Z). 7 We cn crete infinitely mny new Dirichlet chrcters from χ 1 in this wy, but they will differ from χ 1 only in rther trivil sense. We would like to distinguish the Dirichlet chrcters tht rise in this wy from those tht do not. Definition Let χ 1 nd χ 2 be Dirichlet chrcters of modulus m 1 nd m 2, resectively, with m 1 m 2. If χ 2 (n) = χ 1 (n) for n (Z/m 2 Z) then χ 2 is induced by χ 1. A Dirichlet chrcter tht is not induced by ny chrcter other thn itself is rimitive. Lemm A Dirichlet chrcter χ 2 of modulus m 2 is induced by Dirichlet chrcter of modulus m 1 m 2 if nd only if χ 2 is constnt on residue clsses in (Z/m 2 Z) tht re congruent modulo m 1. When this holds, the Dirichlet chrcter χ 1 of modulus m 1 tht induces χ 2 is uniquely determined. Proof. If χ 2 is induced by χ 1 then it must be constnt on residue clsses in (Z/m 2 Z) tht re congruent modulo m 1, since χ 1 is. To rove the converse we first show tht the surjective ring homomorhism Z/m 2 Z Z/m 1 Z given by reduction modulo m 1 induces surjective homomorhism π : (Z/m 2 Z) (Z/m 1 Z) of unit grous, 8 Suose u 1 Z is unit modulo m 1. Let be the roduct of ll rimes dividing m 2 /m 1 but not u 1. Then u 2 = u 1 + m 1 is not divisible by ny rime m 1 (since u 1 isn t), nor is it divisible by ny rime (m 2 /m 1 ): by construction, such divides exctly one of u 1 nd m 1. Thus u 2 is unit modulo m 2 tht reduces to u 1 modulo m 1 nd π is surjective. If χ 2 is Dirichlet chrcter of modulus m 2 constnt on fibers of π we cn define Dirichlet chrcter χ 1 of modulus m 1 vi χ 1 (n 1 ) := χ 2 (n 2 ) for n 1 (Z/m 1 Z) with n 2 π 1 (n 1 ) (ny such n 2 will do). This χ 1 induces χ 2, nd if χ 1 lso induces χ 2 it must stisfy the sme condition χ 1 (n 1 ) = χ 2 (n 2 ) tht uniquely determines χ 1. 7 Note tht while #(Z/m 1Z) #(Z/m 2Z), the set of integers n (Z/m 1Z) (the n corime to m 1) contins the set of integers n (Z/m 2Z) (the n corime to m 2), nd for m 1 m 2 is lrger. 8 In fct, one cn show tht every surjective homomorhism of finite rings induces surjective homomorhism of unit grous, but this does not hold in generl (consider Z Z/5Z, for exmle) Fll 2017, Lecture #18, Pge 6

7 Definition A Dirichlet chrcter χ induced by 1 is clled rincil (nd is rimitive if only if χ = 1). For m Z >0 we use 1 m to denote the rincil Dirichlet chrcter of modulus m; it corresonds to the trivil chrcter of (Z/mZ). Lemm Let χ be Dirichlet chrcter of modulus m. Then χ(n) 0 χ = 1 m. n Z/mZ Proof. We hve χ(n) = 0 for n (Z/mZ), nd the sum over (Z/mZ) is nonzero if nd only if χ restricts to the trivil chrcter on (Z/mZ), by the orthogonlity of chrcters; see Corollry Note tht the rincil Dirichlet chrcters 1 m nd 1 m necessrily coincide when m m m k ; for exmle the rincil Dirichlet chrcter of modulus 2 (the rity function) is the sme s the rincil Dirichlet chrcter of modulus 4 (nd every ower of 2). Theorem Every Dirichlet chrcter χ is induced by rimitive Dirichlet chrcter χ tht is uniquely determined by χ. Proof. Let us define rtil ordering on the set of ll Dirichlet chrcters by defining χ 1 χ 2 if χ 1 induces χ 2. The reltion is clerly reflexive, nd it follows from Lemm tht it is trnsitive. Let χ be Dirichlet chrcter of eriod m nd consider the set X = {χ : χ χ}. Ech χ X necessrily hs eriod m dividing m nd there is t most one χ of eriod m for ech divisor m of m, by Lemm Thus X is finite, nd nonemty (since χ X). Suose χ 1, χ 2 X hve eriods m 1 nd m 2, resectively. Then m 1 nd m 2 both divide m, s does m 3 = gcd(m 1, m 2 ). We hve commuttive squre of surjective unit grou homomorhisms induced by reduction ms: (Z/mZ) (Z/m 1 Z) (Z/m 2 Z) (Z/m 3 Z) From Lemm we know tht χ is constnt on residue clsses in (Z/mZ) tht re congruent modulo either m 1 or m 2, nd therefore χ is constnt on residue clsses in (Z/mZ) tht re congruent modulo m 3, s re χ 1 nd χ 2 (which re determined by χ). It follows tht there is unique Dirichlet chrcter χ 3 of modulus m 3 tht induces χ, χ 1, nd χ 2. Thus every ir χ 1, χ 2 X hs lower bound χ 3 under the rtil ordering tht is comtible with the totl ordering of X by eriod. This imlies tht X contins unique element χ tht is miniml, both with resect to the rtil ordering nd with resect to the totl ordering by eriod; it must be rimitive, by the trnsitivity of. Definition The conductor of Dirichlet chrcter χ is the eriod of the unique rimitive Dirichlet chrcter χ tht induces χ. Corollry For Dirichlet chrcter χ of modulus m we hve n Z/mZ χ(n) 0 if nd only if χ hs conductor 1. Proof. This follows immeditely from Lemm Fll 2017, Lecture #18, Pge 7

8 Corollry Let M(m) denote the set of Dirichlet chrcters of modulus m, let X(m) denote the set of rimitive Dirichlet chrcters of conductor dividing m, nd let G(m) denote the chrcter grou of (Z/mZ). We hve cnonicl bijections M(m) X(m) G (m) χ χ ( n χ (n)). Proof. By Theorem 18.13, the m χ χ is injective, nd it is lso surjective: ech χ X(m) induces the chrcter χ M(m) by setting χ(n) := χ(n) for n (Z/mZ) nd extending by zero. As reviously noted, the m χ (m χ(m)) defines bijection M Ĝ(m) ( grou isomorhism, in fct), nd this bijection fctors through the m χ χ, since χ (n) = χ(n) for n (Z/mZ). Remrk Corollry imlies tht we cn mke X(m) grou by defining χ 1χ 2 := χ 1χ 2. Note tht χ 1χ 2 is not the ointwise roduct of χ 1 nd χ 2 (which is tyiclly not rimitive), it is the unique rimitive chrcter tht induces the ointwise roduct. Exmle eriodic Dirichlet chrcters, ordered by eriod m nd conductor c. m c mod-12 rincil rimitive no yes yes no yes no no yes no no no yes no no yes yes yes no yes no no yes no no yes no yes The fct tht χ(n) {0, ±1} for ll 12-eriodic Dirichlet chrcters χ follows from the fct tht the exonent of (Z/mZ) is 2; thus (im χ) U(1) µ 2 = {±1} Dirichlet L-functions Definition The Dirichlet L-function ssocited to Dirichlet chrcter χ is L(s, χ) := (1 χ() s ) 1 = χ(n)n s. The sum nd roduct converge bsolutely for Re s > 1, since χ(n) 1, thus L(s, χ) is holomorhic on Re(s) > 1. For the trivil Dirichlet chrcter 1 hve L(s, 1) = ζ(s). For the rincil chrcter 1 m of modulus m induced by 1 we hve n 1 ζ(s) = L(s, 1 m ) (1 s ) 1. The roduct on the RHS is finite, hence bounded nd nonzero s s 1 +, so the L-function L(s, 1 m ) hs simle ole t s = 1 with residue res s=1 L(s, 1 m ) = lim (s 1)ζ(s) (1 s ) = (1 1 ) =. s 1 m m m m The L-functions of non-rincil Dirichlet chrcters do not hve ole t s = Fll 2017, Lecture #18, Pge 8

9 Proosition Let χ be non-rincil Dirichlet chrcter of modulus m. L(s, χ) extends to holomorhic function on Re s > 0. Then Proof. Define the function T : R 0 C by For ny x R 0 Lemm imlies T (x + m) T (x) = T (x) := χ(n). 0<n x χ(n) = χ(n) = 0, x<n x+m n Z/mZ since χ is non-rincil. Thus T (x) is eriodic modulo m nd therefore bounded. Writing L(s, χ) s Stieltjes integrl (see 18.5) nd integrting by rts yields L(s, χ) = χ(n)n s n 1 = x s dt (x) 0 = x s T (x) 0 0 T (x)d(x s ) = 0 T (x)( sx s 1 )dx 0 = s T (x)x s 1 dx. 0 The RHS is holomorhic on Re s > 0, since it is the limit of the uniformly converging sen quence of functions φ n (s) := s T (x)x 0 s 1 dx (here we use the fct tht T (x) is bounded), nd is thus the nlytic continution of L(x, χ) to Re(s) > 0. Remrk In fct, L(s, χ) extends to holomorhic function on C whenever χ is non-rincil Primes in rithmetic rogressions We now return to our gol of roving Dirichlet s theorem on rimes in rithmetic rogressions. Let nd m be corime integers. We wnt to show tht the sum mod m is unbounded s s 1 +. To convert this to sum over ll rimes we use Proosition to construct the indictor function { 1 1 if mod m, χ(/) = 0 otherwise χ X(m) where / is comuted modulo m nd χ rnges over rimitive Dirichlet chrcters of conductor dividing m (which we identify with the chrcter grou of (Z/mZ) vi Corollry 18.16). s Fll 2017, Lecture #18, Pge 9

10 As s 1 + we hve s = s 1 χ mod m = = = χ X(m) χ X(m) log ζ(s) X(m) χ (/) χ(1/) χ () s χ(1/) ( ) log L(s, χ) + O(1) + χ X(m) χ 1 χ(1/) log L(s, χ) + O(1). We now mke the key clim tht so long s χ is not rincil, we hve L(1, χ) 0. This imlies tht log L(1, χ) = O(1) s s 1 + nd therefore ) s log ζ(s = + O(1) mod m is unbounded s s 1 +, since ζ(s) is. Moreover, Mertens second theorem imlies 1 x mod m log log x, nd we cn comute the Dirichlet density of S := { mod m}: d(s) = lim s 1 + s S s = 1. We will rove the clim tht L(1, χ) 0 whenever χ is not rincil in the next lecture Stieltjes integrls For the benefit of those who hve not seen them before, we recll few fcts bout Stieltjes integrls (lso clled Riemnn-Stieltjes integrls), tken from [1, Ch. 7]. These generlize the Riemnn integrl but re less generl thn the Lebesgue integrl; they rovide hndy wy for converting sums to integrls tht is often used in nlytic number theory. Definition Let f nd g be (rel or comlex vlued) functions defined on nonemty rel intervl [, b]. For ny rtition P = (x 0,..., x n ) of [, b] nd sequence T = (t 1,..., t k ) with t k [x k 1, x k ], we define the Riemnn-Stieltjes sum n S(P, T, f, g) := f(t k ) ( ) g(x k ) g(x k 1 ) k=1 We sy tht f is Riemnn-Stieltjes integrble with resect to g nd write f S(g) if there is (rel or comlex) number S such tht for every ɛ > 0 there is rtition P ɛ of [, b] Fll 2017, Lecture #18, Pge 10

11 such tht for every refinement P = (x 0,..., x n ) of P ɛ nd every sequence T = (t 1,..., t n ) with t k [x k 1, x k ] we hve S(P, T, f, g) S < ɛ. 9 When such n S exists it is necessrily unique nd we denote it by b f dg, the Riemnn- Stieltjes integrl of f with resect to g. Imroer Riemnn-Stieltjes integrls re then defined s limits b f dg := lim f dg b (nd similrly for the lower limit), nd we define f dg = f dg nd f dg = 0. b b Tking g(x) = x yields the Riemnn integrl. The Riemnn-Stieltjes integrl stisfies the usul roerties of linerity, summbility, nd integrtion by rts. Proosition Let f, g, nd h be functions on [, b] nd let c 1 nd c 2 be constnts. The following hold: b b b If f, g S(h) then (c 1 f + c 2 g) dh = c 1 f dh + c 2 g dh. b b b If f S(g), S(h) then f d(c 1 g + c 2 h) = c i f dg + c 2 f dh. b c b If f S(g) then for ny c [, b] we hve f dg = f dg + f dg. b If f S(g) then g S(f) nd f dg + g df = f(b)g(b) f()g(). If f = f 1 + if 2 nd g = g 1 + ig 2 with f 1, f 2 S(g 1 ), S(g 2 ) then b b ( b b ) ( b b ) f dg = f 1 dg 1 f 2 dg 2 + i f 2 dg 1 + f 1 dg 2. Proof. See [1, Thm ,7.50]. The lst identity llows us to reduce comlex-vlued integrls to rel-vlued integrls. The following roosition llows us to reduce Stieltjes integrls to Riemnn integrls. Proosition Let f nd g be rel-vlued functions on [, b] nd suose g hs continuous derivtive g on [, b]. Then Proof. See [1, Thm. 7.8]. b f dg = b f(x)g (x)dx. A key dvntge of the Stieltjes integrl f dg is tht neither the integrnd f nor the integrtor g is required to be continuous. It suffices for f nd g to be of bounded vrition nd not shre ny discontinuities (nd they cn even shre certin discontinuities, see Theorem 18.26). Definition Let f be (rel or comlex vlued) function defined on nonemty rel intervl [, b]. Then f is of bounded vrition if there exists (rel or comlex) number M such tht n 1 f(x i+1 ) f(x i ) < M b c i=0 9 This definition (due to Pollrd) is more generl thn tht originlly given by Stieltjes but is now stndrd Fll 2017, Lecture #18, Pge 11

12 for every rtition P = (x 0,..., x n ) of [, b]. If f hs continuous derivtive f on [, b] b this is equivlent to requiring f (x) dx <. Every iecewise monotone function is of bounded vrition. In rticulr, ny ste function with finitely mny discontinuities on [, b] is of bounded vrition. Theorem Let f nd g be functions on [, b] of bounded vrition such tht for every c [, b] the function f is continuous from the left t c nd the function g is continuous b b from the right t c. Then f dg nd g df both exist. Proof. See [2, Thm. 3.7]. Corollry Let f nd g be functions on [, b] such tht f nd g re not both discon<n x g(n). tinuous from the left or from the right t integers n [, b], nd let G(x) = Then b f(n)g(n) = f(x) dg(x). <n b In rticulr, the integrl on the RHS lwys exists. Proof. See [1, Thm. 7.11]. As n exmle of using Stieltjes integrls, let us derive n symtotic estimte for the the hrmonic sum 1 H(x) := n. 1 n x Theorem For x R 1, s x we hve H(x) = log x + γ + O ( ) 1 x where γ = lim x (H(x) log x) = is Euler s constnt. Proof. Let [t ] denote the gretest integer function. Alying Corollry with g(t) = 1 nd G(t) = 1 n t 1 = [t], we hve 1 x H(x) = n = 1 1 n x 1 t d[t] = [t] x x 1 t [t] d 1 1 t = [x] x x + [t] 1 t 2 dt = [x] x + x x 1 t [t] dt dt 1 t 1 t2 = [x] x t [t] + log x dt, x 1 t 2 where we used integrtion by rts in the second line nd lied Proosition to get the third line. Now let γ = 1 1 (t [t])/t 2 dt. Then [x] H(x) = + log x 1 + γ + x ( [x] x = log x + γ + + x t [t] dt x t 2 x t [t] dt t 2 ). (2) Fll 2017, Lecture #18, Pge 12

13 Both summnds in the renthesized quntity in (2) re clerly O( 1 ); thus x nd the theorem follows. γ = lim (H(x) log x), x Remrk We cn refine this estimte by lying similr nlysis to the renthesized quntity in (2); the key oint is tht the error term is n exct exression, not n symtotic estimte, nd we cn continue this rocess until we obtin n symtotic exnsion to whtever recision we require. For exmle, one finds tht 1 H(x) = log x + γ + 2x 1 2x x 4 + O ( ) 1. x A quick rec of the chrcter theory of finite belin grous In this section we recll some stndrd results on chrcters of finite belin grous. Definition A chrcter of grou G is homomorhism χ: G U(1). 10 chrcter grou (or dul grou) of G is the belin grou The Ĝ := Hom(G, U(1)) under ointwise multiliction: (χ 1 χ 2 )(g) := χ 1 (g)χ 2 (g). The inverse of χ is given by comlex conjugtion: χ 1 (g) = χ(g) := χ(g). The identity element of G is the trivil chrcter g 1. Remrk This definition generlizes to loclly comct belin grous G, in which cse ech chrcter χ: G U(1) is homomorhism of toologicl grous nd the dul grou Ĝ is loclly comct under the comct-oen toology which hs bsis of neighborhoods of the identity the sets U(C, V ) := {χ Ĝ : χ(c) V }, where C rnges over comct subsets of G nd V rnges over oen neighborhoods of the identity in U(1). The loclly comct grou G is clled the Pontrygin dul of G. 11 When G is finite it necessrily hs the discrete toology (since it must be Husdorff), every homomorhism G U(1) is utomticlly continuous, nd the comct-oen toology on G is lso discrete. Proosition Let G be finite belin grou with chrcter grou G. Then G G. Proof. As finite belin grou we cn write G s direct roduct of cyclic grous G = g 1 g n Z/n 1 Z Z/n r Z, e with n i = g i, nd ech g G cn be uniquely written s g = i gi i with 0 e i < n 1. Now fix (not necessrily distinct) rimitive n i -th roots of unity α i U(1) nd for 1 i r define χ i G vi { α i if i = j, χ i (g j ) := 1 if i j. 10 Some uthors cll these unitry chrcters, llowing chrcters to hve imge in C. When G is finite every chrcter is unitry chrcter, so this distinction won t concern us. 11 Some uthors define the toology on the Pontrygin dulity using uniform convergence on comct sets; for toologicl grous this is equivlent to the comct-oen toology. The unitry grou U(1) R/Z is lso referred to s the 1-torus or circle grou nd my be denoted T or S 1 nd viewed s n dditive grou Fll 2017, Lecture #18, Pge 13

14 e Then χ = α = n, nd ech χ G i i i cn be written uniquely s i χi i with 0 e i < n i, ei where χ(g i ) = α i (becuse chrcter is comletely determined by its vlues on genertors nd the χ i re clerly orthogonl). Therefore Ĝ = χ 1 χ n Z/n 1 Z Z/n r Z. Corollry Let G be finite belin grou. Then g G is the identity if nd only if χ(g) = 1 for ll χ G nd χ G is the identity if nd only if χ(g) = 1 for ll g G. The isomorhism in Proosition is not cnonicl. Indeed, there re #Aut(G) distinct wys to choose the α i used to construct the isomorhism G G. But there is cnonicl isomorhism from G to the chrcter grou of G, the double dul of G. Corollry Let G be finite belin grou. The evlution m g (χ χ(g)) is cnonicl isomorhism from G to its double dul. Proof. It is cler tht the m bove is homomorhism, nd Proosition imlies tht G is isomorhic to its dul grou Ĝ, which is in turn isomorhic to its dul grou, the double dul of G). So it suffices to show the m is injective, which follows from Corollry 18.33: if g lies in the kernel then χ(g) = 1 for ll χ Ĝ nd g = 1 G, by Corollry 18.33, Corollry llows us to view G s the chrcter grou of Ĝ by defining g(χ) := χ(g). Remrk Corollry is secil cse of Pontrygin dulity, which lies to ny loclly comct belin grou G. For infinite grous, G nd G need not be isomorhic; for exmle, the chrcter grou of Z is isomorhic to U(1) (but in some cses they re, s when G is R or Q, or ny locl field, see [3, XV, Lemm 2.2.1]). But the cnonicl isomorhism between G nd its double dul lwys holds. This is nlogous to the sitution with vector sces: finite dimensionl vector sce (which my be n infinite belin grou) is non-cnoniclly isomorhic to its dul sce but cnoniclly isomorhic to its double dul vi the evlution m. We should note tht for loclly comct toologicl vector sce V over field k, the Pontrygin dul is not the sme thing s the vector sce dul: the Pontrygin dul corresonds to Hom(V, U(1)) (morhisms of loclly comct grous) while the vector sce dul corresonds to Hom k (V, k) (morhisms of toologicl k-vector sces). For exmle, the vector sce dul of Q is isomorhic to Q, s is its double dul, but the Pontrygin dul of Q is uncountble (thus not isomorhic to Q), even though the Pontrygin double dul is isomorhic to Q. Proosition Let G be finite belin grou. For ll g 1, g 2 G we hve { 1 1 if g 1 = g 2, g 1, g 2 := χ(g 1 ) χ(g 2 ) = #G 0 if g 1 g 2, χ Ĝ nd for ll χ 1, χ 2 Ĝ we hve { 1 1 if χ 1 = χ 2, χ 1, χ 2 := χ 1 (g) χ 2 (g) = #G 0 if χ 1 χ 2. g G Fll 2017, Lecture #18, Pge 14

15 Proof. By dulity it suffices to consider g 1, g 1. If g 1 = g 2 then χ(g 1 )χ(g 2 ) = 1 for ll χ Ĝ nd g 1, g 2 = #G/ #G = 1. If g1 g 2 then by Corollry there exists λ G for which α := λ(g 1 ) λ 1 (g 2 ) = λ(g 1 g 1 ) 1. We then hve χ Ĝ 2 α g 1, g 2 = 1 ( λχ)(g1 )(λχ)(g 2 ) = 1 χ(g 1 )χ(g 2 ) = g 1, g 2, #G #G which imlies g 1, g 2 = 0, since α 1. Corollry For χ G we hve g G χ λĝ χ(g) 0 if nd only χ is the trivil chrcter. Remrk The orthogonlity of chrcters given by Proosition is secil cse of the orthogonlity of chrcters one encounters in Fourier nlysis on comct grous; since G is finite, the weighted sum over G mounts to integrting ginst its Hr mesure (the counting mesure µ normlized so tht µ(g) = 1). We conclude our discussion of chrcter grous with theorem nlogous to the fundmentl theorem of Glois theory. Proosition Let G be finite belin grou. There is n inclusion reversing bijection ϕ between subgrous H of G nd subgrous K of G defined by The inverse bijection φ is given by nd Ĥ G/ϕ ϕ(h) := {χ Ĝ : χ(h) = 1 for ll h H}. φ(k) := {g G : χ(g) = 1 for ll χ K}, (H) nd K G/φ(K); in rticulr, #H = [G:ϕ(H)] nd #K = [G:φ(K)]. Proof. Its cler from the definitions tht ϕ nd φ re inclusion reversing. Let H be subof G whose kernel contins H. grou of G. The grou K = ϕ(h) consists of the chrcters It is cler tht H := φ(k) contins H, since it is equl to the intersection of these kernels, nd by dulity it is similrly cler tht K := ϕ(h ) contins K. We then hve H H nd ϕ(h) ϕ(h ), but ϕ is inclusion reversing so H = H ; thus φ ϕ is the identity m, nd by dulity, so is ϕ φ. The restriction m Ĝ Ĥ defined by χ χ H is grou homomorhism with kernel K = ϕ(h). It is surjective becuse if we let χ 1 := 1 then we hve G #H#K = χ(h) = χ(h ) χ 1 (h) = χ(g)χ 1 (g) = #G, χ K g G χ K h H χ K h H by Proosition 18.36, nd therefore #Ĥ#K = #Ĝ (by Proosition 18.32). It follows tht Ĥ G/ϕ (H), nd by dulity, K G/φ(K). References [1] Tom Aostol, Mthemticl nlysis, 2nd edition, Addison-Wesley, [2] Pul Btemn nd Hrold Dimond, Anlytic number theory: An introductory course, World Scientific, Fll 2017, Lecture #18, Pge 15

16 [3] J.W.S. Cssels nd A. Fröhlich, Algebric number theory, 2nd edition, London Mthemticl Society, [4] Frnz Mertenz, Ein Beitrg zur nlytischen Zhlentheorie, J. reine gnew. Mth., 78 (1874), Fll 2017, Lecture #18, Pge 16

17 MIT OenCourseWre htts://ocw.mit.edu Number Theory I Fll 2017 For informtion bout citing these mterils or our Terms of Use, visit: htts://ocw.mit.edu/terms.