Beurling Integers: Part 2
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1 Beurlig Itegers: Part 2 Isomorphisms Devi Platt July 11, Prime Factorizatio Sequeces I the last article we itroduced the Beurlig geeralized itegers, which ca be represeted as a sequece of real umbers or as a multiplicatio fuctio over the atural umbers. These two represetatios are isomorphic, though we really oly showed how to get a multiplicatio fuctio from a give sequece of real umbers. To go i the opposite directio, we actually eed to itroduce a third represetatio: a sequece of prime factorizatios. To discuss prime factorizatios, we eed to have a way of represetig the factorizatios as objects themselves. Oe represetatio we ca use is expoet vectors, which are best explaied with examples (see Table 1). This otatio ivites aalogies to a vector space, although the atural umbers do ot form a field so we do ot have a vector space. Noetheless, some typical operatios have atural correspodece. Multiplicatio correspods to vector additio ad expoetiatio correspods to scalar multiplicatio. The prime umbers correspod to uit (basis) vectors i a ifiite dimesioal space. Sice the vectors live i a ifiite dimesioal space, we require a fiite represetatio. I the table above the vectors have their tail sequeces of zeros cut off, but geerally this ist the most compact represetatio. We also write factorizatios as sets of (prime, expoet) pairs eg. 2, 0, 5 correspods to {(1, 2), (3, 5)} This pair represetatio is aki to writig a vector as a liear combiatio of basis vectors. The set of pairs represetatio teds to be more coveiet whe a vector is sparse (ie. it has may zeros Table 1: Comparig represetatios of the sequece of atural umbers Natural Number Prime Factorizatio Sequece of Prime Expoets Expoet Vector (Trucated Zeros) 1 oe 0,0,0,0,0,0, ,0,0,0,0,0, ,1,0,0,0,0,... 0, ,0,0,0,0,0, ,0,1,0,0,0,... 0, 0, ,1,0,0,0,0,... 1, ,0,0,1,0,0,... 0, 0, 0, ,0,0,0,0,0, ,2,0,0,0,0,... 0, ,0,1,0,0,0,... 1, 0, ,0,0,0,1,0,... 0, 0, 0, 0, ,1,0,0,0,0,... 2, 1 1
2 Table 2: Comparig factorizatio orderigs of the atural umbers ad the odds Natural Number Natural Vector Odd Vector Odd Number , 1 0, , 0, , 0, , 1 0, 0, 0, , 0, 0, 1 0, 0, 0, 0, , , 2 {(6,1)} , 0, 1 {(7,1)} , 0, 0, 0, 1 1, 0, , 1 {(8,1)} 23 betwee o-zero elemets it cotais a large prime, but is a relatively small umber). Moreover, the pair represetatio makes sese whe discussig powers of primes because the each umber has oly oe pair (i this case represetatio is really as cocise as is possible). Sice multiplicatio ca be chaied, we ca cosider it as a map from multisets (sets where multiplicity is allowed) of atural umbers to a atural umber, istead of a map just from pairs of atural umbers to a atural umber. If we restrict the domai to multisets of prime umbers, the the uiqueess of prime factorizatio guaratees that this map is ijective oly oe item i the domai maps to some umber i the rage. We could also defie the map from the sigleto multiset (of a prime umber) as a idetity mappig, which would allow the map to be surjective. This ew form of the multiplicatio fuctio is thus a bijectio with the atural umbers as its rage, which meas that it is equivalet to a sequece: a orderig of its domai. So every multiplicatio fuctio correspods to a orderig of prime factorizatios. For example, we ca compare the orderigs of the atural umbers ad the odd umbers as i Table 2. We ca cosider a triagle of isomorphic represetatios of the Beurlig geeralized itegers as see i Figure 1. We have show that the multiplicatio fuctios ad sequeces of Factorizatios are isomorphic. We have also show that sequeces of real umbers determie multiplicatio fuctios. Figure 1: Isomorphism Triagle Now quite obviously, a sequece of all factorizatios determies the orderigs of prime powers, sice these are just subsequeces. I sectio 4 of the paper, Beurlig Zeta Fuctios, Geeralised Primes, ad Fractal Membraes, Lapidus ad Hilberdik prove the isomorphism betwee orderigs of prime powers ad the sequece of real umbers represetatio, which completes our isomorphism 2
3 triagle. I shall reproduce their proof here (usig some of their otatios, which hopefully ca be iferred if I eglect to defie them), which takes the form of two theorems ad a lemma. Just provig oe directio, that a orderig of prime powers determies a sequece of real umbers, should be sufficiet to complete the isomorphism triagle. (As a aside, the basics of the proof rely o usig the orderig to pi values of ay power of ay prime p betwee some power of p 1 ad the ext power +1. I had always suspected such a approach could be used i this proof, but my attempts failed i vai. For this reaso the relative simplicity ad the elegace of the followig proofs always impressed me.) 2 The Lapidus-Hilberdik Proof Notes: This proof uses some elemetary real aalysis. If youre ot iterested, you ca skim or skip the proof ad move o to the other posts i the series. For a give system of geeralized itegers, the authors use P to refer to the set of associated primes (as real umbers). They use N to refer to the set of itegers geerated by P. This proof requires the ifiitude of both primes ad composites. This is largely a verbatim copy of their proof, with some small chages ad my ow otes throw i. For the proofs of each idividual theorem/lemma I provide a digest i brackets [], mostly for my ow coveiece as this article serves as a persoal referece. The first part of the proof comes from that the recogitio that if a sequece of umbers is defied as aother sequece with each elemet take to some fixed power greater tha zero, the the multiplicatio fuctio remais the same. For example, with the squares of the atural umbers, four times four equals 16, so m(2, 2) = 4. I geeral, if b = a λ, the b 1 (x) = a 1 (x 1/λ ) ad multiplicatio for b ca be defied as: m(x, y) = b 1 (b x b y ) = a 1 ((a λ x a λ y) 1/λ ) = a 1 (a x a y ) It turs out the coverse is true, if the multiplicatio fuctio (ad hece the factorizatio orderig) is the same for two sequeces of real umbers, the the sequeces must be some power of each other. (Note that the coverse is oly true assumig that we have i ifiite umber of composite umbers.) Here is the exact theorem due to Lapidus ad Hilberdik (with a early verbatim copy of the proof): 2.1 Theorem: Let P 1 ad P 2 be two geeralized prime systems with geeralized itegers N 1 ad N 2 respectively. If the orderigs of N 1 ad N 2 coicide, the P 1 = P λ 2 for some λ > 0 ad hece N 1 = N λ 2. [Proof otes: boud p m k, qm k by powers of p 1, q 1 to, + 1, take log. Now forms of p m k ad qm k are bouded by the same umbers, + 1 ad we ca use substitutio to get ps ad qs i same iequality. Takig limit as m gives us our result.] 3
4 2.2 Proof: Deote the primes i P 1 by p1, p 2,..., ad those i P 2 by q 1, q 2,.... Let k, m N. The p m k [p 1, p 1 + 1) for some N. Sice N 2 has the same orderig as N 1 we also have qk m [q 1, q 1 + 1). Takig logs gives: mlog(p k) log(p 1 ) < + 1 ad mlog(q k) log(q 1 ) < + 1 ie. = mlog(p k) log(p 1 ) = mlog(q k) log(q 1 ). So we ca relate the sequeces q ad p mlog(p k ) log(p 1 ) 1 < mlog(q k) log(q 1 ) < + 1 mlog(p k) logp ad hece This holds for all m, so lettig m gives log(p k ) log(p 1 ) 1 m < log(q k) log(q 1 ) < log(p k) log(p 1 ) + 1 m log(p k ) log(p 1 ) = log(q k) log(q 1 ) i.e. p k = q λ k, where λ = log(p 1) log(q 1 ), ad hece P 1 = P λ Axioms Lapidus ad Hilberdik ote the subsequece of N as of powers of real umbers, Q = {p : p P, N} ad remark that Q is isomorphic to N 2 via the isomorphism p m (m, ). They defie three axioms regardig the order Q imposes o N 2, which correspod to the properties we defied for multiplicatio earlier: A1: (m, ) (m, ) wheever both m m ad, with strict iequality if < A2: (m, ) (m) implies (m, k) (m, k) for every k N, with strict iequality if (m, ) < (m, ). A3: Fiiteess: (i) For all N, there exists k N such that (1, ) < (k, 1). (ii) For all m N, there exists l N such that (m, 1) < (1, l). A3 correspods to havig ifiitely may composites ad ifiitely may primes, so that we have a order o all of N 2. Oe ca view A1 ad A2 as correspodig to icreasigess ad traslatio ivariace, respectively. The fial theorem requires a lemma. The purpose of the lemma is as follows. We kow that sequeces of umbers geerate the same system whe they are powers of each other, ad we just showed that this is actually the oly case. Thus we should be able to geerate uique values of the primes give a order of prime powers if we choose a value of the first prime (the values of all the itegers are i tur determied by the values of the primes, hece why we oly eed to care about the order of prime powers.) We could the write the value of ay k th prime as a real (ad o-itegral) power of the first prime, power(k). We ca boud ay power of a k th prime by powers of the first prime, ad if we choose some power f k () to give the 4
5 maximum lower boud the f k () + 1 will give the miimum upper boud (if oe of these bouds eed be ostrict the other could be strict.) Now for ay such we have a iequality: f k () < power(k) < f k() + 1 For example, with the atural umbers , ie. power(2) = (power(2) refers to 3 as the secod prime). Our first iequalities imply 2 < 3 < 4 = 1 < power(2) < 2 8 < 9 < 16 = 1.5 < power(2) < 2 16 < 27 < 32 = < power(2) < Eve after the first few iequalities, we are able to boud 1.5 < power(2) < , which give p 1 = 2 puts our guess for 3s value at < 3 < We would hope that sup f k() f = power() or eve that lim k () = power(). (This basically is what I had hoped before fidig the Lapidus-Hilberdik paper, but I was ever able to arrive at a proof.) The followig lemma aids i our proof of the theorem by showig a coditio for which the limit of f k() exists. 2.4 Lemma Let f : N R be such that for all m, N f(m) m f() the lim exists. [Proof otes: show that f() f() 1 is bouded. The show that f(m) m lim if f() (usig the fact that there exists a sequece k for which f( k) k The use this to show that f() approaches this limit.] 2.5 Proof By puttig = 1 ito the coditio (1) it follows that f() f() α = lim if is bouded. Let By defiitio of α, there exists sequece k tedig to ifiity with k such that f( k ) k α as k Hece, for fixed m N, we have f(m k) m k α as k, sice f(m k ) f( k) m k k 1 k (ad f( k) (1) approaches the same limit as approaches this limit). k α while 1 k 0) Now fix N, put m = k i the iitial coditio ad let k. The f( k ) f() k k 1 5
6 Sice f( k) k α as m we have f() α 1 ad the result follows. 2.6 Theorem Give a order o N 2 satisfyig axioms A1-A3, there exists a geeralized prime system P which iduces this order. [Proof otes: Fix the k th prime. Boud the th power of this prime (k, ) from below by some maximal power f k () of p 1 (so f k ()+1 bouds it above). Note that this iequality is preserved if we replace by some m (because it is simply a substitutio), or if we multiply all the powers by some m (due to traslatio ivariace). We mix these these derived iequalities sice the middle term (k, m) is the same. For this proof Lapidus specifically gives that the lefthad side of oe iequality is less tha the righthad side of the other, sice the iequalities have the same middle term, but ote that a eve stroger result is quite obvious: sice f k (m) gives a maximal lower boud (ad hece f k (m) gives a miimal upper boud) we have that the lower boud mf k () f k (m) ad the upper boud m(f k () + 1) f k (m). We ca write this i oe lie: (1, mf k ()) (1, f k (m)) (k, m) < (1, f k (m) + 1) (1, m(f k () + 1)) ad it is eve more apparet that the first term is less tha the fourth term ad the secod term is less tha the fifth. Takig those two iequalities, we ca create correspodig iequalities for the expoets, sice the iequalities are o powers of the same prime. These ew iequalities ca be combied ad adjusted so that the coditio (1) of the Lemma is applicable. We defie our sequece of real umbers by lettig p k = p α k 1, where α k is the limit of f k(). At this poit, lettig m leads to our result, sice iequalities p m < p m o our sequece of real umbers imply iequalities o f k (), which i tur implies iequalities (m, ) < (m, ) o our orderig of N 2.] 2.7 Proof Fix k N. For N, let f k () be the uique positive iteger for which (1, f k ()) (k, ) < (1, f k () + 1) (2) (*Im ot sure why ostrict iequality is ecessary here (for k=1?), but it doest really matter.) This exists o accout of A2 ad A3, ad is uique by A1. Replacig by m, we have (1, f k (m)) (k, m) < (1, f k (m) + 1) (3) O the other had, A2 implies that (give the first equatio, (2)) (3) ad (4) give (1, mf k ()) (k, m) < (1, m(f k () + 1)) (4) (1, f k (m)) < (1, m(f k () + 1)), ad (1, mf k ()) < (1, f k (m) + 1) 6
7 These iequalities ivolve the same prime base ad thus we ca apply the iequality to their expoets Combiig these ad dividig through by m gives f k () f k (m) < mf k () + m ad (5) mf k () < f k (m) + 1 (6) 1 m < f(m) m < f k() + 1 Thus f k () satisfies the coditio of the Lemma (1), that its distace from f(m) m is less tha 1. This meas that f k() α k for some α k. Now choose some p 1 > 1 ad defie a system of geeralized primes such that p k = p α k 1. We shall prove that this system iduces the give orderig o prime powers. Lettig m i (6) we have that If p m < p m (i.e. α m < α m ) the, f k () < α k < f k () + 1 f m () α m < α m f m () + 1 Sice f m () ad f m () are itegers, this implies that f m () f m (), i.e. (m, ) (m, ). [The equality case ca be illustrated by example. Imagie the umbers 25 ad 27 i the case of the atural umbers. We have that p 2 3 < p3 2, but that f 3(2) = f 2 (3). However, log 2 (25 3 ) < log 2 (27 3 ), so f 3 (3 2) < f 2 (3 3). So we ca amed the argumet with a call to traslatio ivariace.] Now if (m, ) = (m, ), the (m, k) = (m, k) for all k N by A2. Hece fm(k) k = fm(k) k ad, lettig k, we have α m = α m, i.e. p m = p m. This shows that p m < p m implies that (m, ) < (m, ). This proof cocludes Part 2 of the series ad our itroductio to the Beurlig geeralized itegers. The followig articles i the series will explore the partial order that the geeralized itegers impose o the set of factorizatios. (7) 7
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