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# ME 6 ME 6

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This 29 page Class Notes was uploaded by Daren Beatty Jr. on Thursday October 22, 2015. The Class Notes belongs to ME 6 at University of California Santa Barbara taught by Staff in Fall. Since its upload, it has received 51 views. For similar materials see /class/227084/me-6-university-of-california-santa-barbara in Mechanical Engineering at University of California Santa Barbara.

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Notes on the Deuring Heilbronn Phenomenon Jeffrey S topple Introduction Analytic number theory studies Lfunctions gen eralizations of the Riemann zeta function s It can be difficult to see why this is number theory In fact the Class Number Formula 6 of Dirichlet gives the number Mid of classes of binary qua dratic forms of discriminant 7d as the value of such an Lfunction at s l The location of the zeros is important since the functions are continuous the value is in uenced by any zero of the function near s 1 Such a zero would of course contradict the Generalized Riemann Hypothesis GRH The DeuringHeilbronn phenomenon says that such a counterexample to the GRH for one L function would influence the horizontal and ver tical distribution of the zeros of otherLfunctions They would be forced to lie on the critical line s 12 it at least up to some height This is the Local GRH More surprisingly the imaginary parts If would be restricted to a set which is very nearly periodic This is a very beautiful result indeed Standard analogies interpret the imaginary parts If as frequencies the DeuringHeilbronn phenom enon means these frequencies are in harmony We give an overview of the proof first in the case Mid 1 before treating Mid gt 1 Even though the class number 1 problem is now solved the es sential features of the general problem are visible there We also look at some examples which indi cate that even in the absence of a contradiction to Jeffrey Stopple is professor of mathematics at the University of California Santa Barbara His email address is stopplemathucsbedu NOTICES or THE AMS GRH near contradictions still cause a tendency towards such a phenomenon The author would like to thank David Farmer for suggesting the calculations in the last section Notations The complex variable s is written 039 it We write fX ltlt hX resp fX aX 0hX if there is some constant C so that lfXl s ChX resp fX 7am ltlt hX usually for X approaching some limiting value which may not be explicitly stated We write fX N hX if fXhX 1 Binary Quadratic Forms Algebraic number theory has its roots in the beau tiful theorem ofFermat that an odd prime p is the sum of two squares 1724ry2 ltgt p21mod4 Euler Lagrange and Gauss developed many gen eralizations of this for example p 7 can be writ ten as px2xy2y2 cgt p2124mod7 The necessity of the congruence is the easy half as it is in Fermat s theorem 4p 2x y2 7y2 now reduce modulo 7 In general one studies pos itive definite binary quadratic forms functions VOLUME 53 NUMBER 8 Qxy ax2 bxy cyz with 7db274aclt0 the discriminant We build the minus sign into the notation to simplify later when we want square root or logarithm of the absolute value TWo forms Q and Q are said to be equivalent if there is an integer matrixM with determinant 1 such that Q XY QXYM Such forms obviously have the same range as func tions ie represent the same integers A calcula tion shows that equivalent forms have the same dis criminant and it is not difficult to show the number of classes h7d is finite In fact a very deep result of Gauss is that they form a finite abelian group C7d Below we will need the Kronecker symbol X4 at tached to a discriminant 7d In the simplest case that d is an odd prime X4 reduces to the Legendre symbol 0 if dln X4101 1 if n E a square 7 1 otherwise Of course this definition works just fine for pos itive discriminants as well For odd composite dis criminants we can define a Jacobi symbol via mul tiplicatiVity This no longer detects squares for exa ple X7159 X732X52 1 fl l but 2 is not a square modulo 15 It does however detect whether a prime is represented by some form of discriminant id just as in the example with discriminant 77 above For primes p with X71507 1 then px2xy4y2 ltgt pl4mod15 p2x2xy2y2 lt2 p228mod15 Primes with X15p 71 are not represented by any form of discriminant 715 The natural slightly complicated extension of this function to even discriminants as well is called the Kronecker sym bol A weaker relation than equivalence is also use ful two forms are in the same genus if they rep resent the same congruence classes in the multi plicative group modulo d For example x2 14y2 and 2x2 7y2 both have discriminant 756 they must be in dif ferent classes since the first represents 1 while the second does not However they are in the same genus since SEPTEMBER 2006 25271257Elmod56 In this case M756 4 the other two classes of forms are 3x2 2xy Syz Both these forms represent 3 while neither of the first two forms can represent an integer congru ent to 3 mod 56 reduce modulo 7 So there are two genera each consisting of two classes For more details on the algebraic theory of binary qua dratic forms see 2 To bring analysis to the study of binary quadratic forms we introduce the classical Riemann zeta function and the Dirichlet Lfunction 1 7 rsrl 713111 M8 s 7 US X4 Z X dsm 711 1 I l 1 1 7 Xdpp 5 1 p We will also want the Epstein zeta function ms 2 QXy s xy where the in the sum means omit the term 0 0 The factor of 12 in the definition accounts for the automorphism Xy a X 7y This is in fact the only automorphism ifwe assume d gt 4 which we do from now on Forms with positive discriminant have infinitely many automorphisms For another way of writing this we group together all the terms in which the form takes on the same value and count them with the representation numbers rQn ran Xy l QXy n This gives Asymptotic Behavior We begin by generalizing the proof that l as S 1 o1 The idea is that the sum 2MB rQn of the repre sentation numbers is one half the number of lat tice points inside the ellipse Qxy B which as B a co is approximately the area The change of variables that converts the ellipse to a circle has Jacobian 2d independent of Q which gives Tr quot2d rQn N WE NOTICES OF THE AMS 865 In fact one can show the error is 0312 From this and a calculus identity for n we can compute a residue Qs Z rQnsJ X s ldx 711 quot SJ rQngt x s 1 dx 1 nltx 5 J00 ix ONE x s 1 dx 1 xH The integral of the error term is finite for 039 gt 12 which gives 1 ms 01 1 1 i 3 s 7 1 The key fact is that the residue is the same for all forms Q of discriminant 7d This same circle ofideas will actually give us a little more information there is a constant C so that for 039 gt 1 we have d 1270 C 7 is lt2 mus a mm s 4 U Here s a sketch ofa proof The expression a s 23 is exactly the contribution to Qs of the terms Qm n with n 0 So we want to bound the re maining terms 2 Z Qmn s mEZ 711 We use the fact that the sum is absolutely convergent for 039 gt 1 and the usual trick of comparing a sum to an integral There are some things to check care fullyhere which is why this is only a sketch We may as well assume the middle coefficient 17 0 if not a real rotation by i arctan ba 7 C2 makes it so without changing the value of the integral or the discriminant So we want to estimate 2000 4000 6000 8000 10000 Figure l Discriminants vs class numbers NOTICES OF THE AMS JIM cyz dxdy where the annular region of integration r gt 1 i s 0 s i 1T contains ourrotated lattice points A change of variables X r cos0 y r sin9 with Jacobian Mr Mr converts this to GYMquot 1M The point of 2 is that for s in the region 039 gt 1 we get 3 ms a a s 25 as vEa a co We will see below that we can extend 3 to the larger region 039 gt 05 a 1 This is in some sense the rea son why the DeuringHeilbronn phenomenon oc curs Class Number Formula The representation numbers rQn of the individ ual forms are mysterious but there is a nice ex pression for rdn the total number of ways 11 is represented by any form of discriminant 7d 4 2 ran Z Xdm Q mlquot The proof uses the Chinese Remainder Theorem The right side above is the Dirichlet convolution of the multiplicative functions X4 and 1 the constant function Together with 4 this implies 2 6115 2 l2 LAW n s Q 111 mln 5 SLS X711 The previous calculation 1 of the residues at s 1 for the Epstein zeta function gives us when we sum over classes Q Dirichlet s Analytic Class Number Formula 1Tl ld From this it is not too hard to prove upper bounds on the class number ma ltlt iogmw 6 Lllerd See 12 for an exposition Figure 1 shows a scat ter plot of discriminants and class numbers for d lt 10000 You can see the square root upper bound in the upper envelope of the points roughly a parabola Lower bounds are much much harder Gauss conjectured in Art 303 of Disquisitiones Arith meticae VOLUME 53 NUMBER 8 the series of discriminants corre sponding to the same given classifica tion ie the given number ofboth gen era and classes always seems to terminate with a finite numberThere seems to be no doubt that the preced ing series does in fact terminate andby analogy it is permissible to extend the same conclusion to any other classifi cationsHowever rigorous proofs of these observations seem to be very dif ficult Getting good answers to these questions are still the main open problems in the theory Fourier Expansion So far we have viewed the forms Qxy ax2 bxycy2 as discrete objects A slightly different point of view lets us view them as sitting in a continuous family parametrized by a variable 2 in the complex upper half plane H We deho mogenize the form to find roots Heegner points azzbzc0 z 7blH 2a Then Qmn aim 7 MP Eds a 2 W 7 11245 mm As a function of z in the complex upper half plane this is invariant under z 7 z 1 since Qmn N Qmn f1 and thus the Epstein zeta function has a Fourier expansion in x Rez Aphorism Hecke A periodic function should always be expanded into its Fourier series In fact we will end up with a Fourier cosine series since our Epstein zeta function is an even function of the X bZa parameter the forms Qmn am2 bmn CH2 and Q 1m n am2 7 bmn CH2 represent the same integers so their Epstein zeta functions are identical The Fourier coefficients will each be a function of the Dirichlet series vari able 5 as well as lmz xHZa To see what the Fourier expansion looks like we will need the K Bessel function Kvy which is the solution to the second order ODE derivatives with respect to y l v 2 u u 71 u0 y y2 SEPTEMBER 2006 which tends to 0 as y 7 00 In fact 12 7 Kvlty explt7ylt1olty1gt Also significant for us is that it has the integral rep resentation Mellin transform 8 KVU J eXp7y2T 1TTv dTT A good reference for this is S The change of vari ables T 7 T 1 in the integral 8 gives 9 Kvy K407 Theorem ChowlaSelberg e d s712 AQs 2quot rams Ts Tl 7 s Us where s712 Ts a sTs 25 and Um ns71Zol28ltn x Kai2 nala COSmea The divisor function o vn is defined by 2mm mV Notice that each term ns lZO39IQSM is invariant under s 7 l 7 s as is the KBessel function by 9 and this gives as a Corollary the analytic continu ation and functional equation AQs AQl 7 s Proof extremely sketchy The appearance of the term Ts is not surprising just as before it is exactly the contribution of pairs m n that have n 0 The remaining terms contribute a sum over m in Z and n in N Each summand can be written as a Mellin transform and the sum pulled through the integral Poisson Summation gives the integral 8 for the KBessel function El We wrote the Fourier expansion this way iso lating the constant term Ts Tl 7 s because this term will be dominant The details are messy because the implicit constant in 7 depends on s but it is shown in 1 that for 0 s 039 s l 2 T039 Us 5 WW exp71TVEa On the critical line 039 12 one can get from 8 an estimate independent of t NOTICES OF THE AMS 867 6 U12 10 5 W examEm An important consequence of this is that we ex tend the asymptotic behavior 3 10 Qs 7 a s 2s as Mda 7 00 to s in the region 039 gt 0 and s 1 This has to fail at s 1 of course Qs has a pole there and 2s does not It is slightly surprising where this pole appears in our Fourier expansion it is in the T1 7 5 term the function HS has a simple pole at s 0 The Siegel Zero and Consequences Hecke s aphorism pays off now by showing us a deep connection between the arithmetic and the analysis in the following three theorems Theorem ChowlaSelberg For 1 a2 gt 200 i The Epstein zeta function has a real zero in l 2 1 ii If h7d 1 then the Generalized Riemann Hypothesis GRH is false Proof We start by computing AQs at s 12 Although the term Q25 has a pole at s 12 a calculation will show that Ts T1 7 s has a re movable singularity at s 12 ms 51122 US by 710g45 7 12 057 122 7 quot1 7 1 Iowa2mm a e W T E y05712 S7 12057122 Here y is Euler s constant and these calculations are classical enough that Mathematica can do them for us So Ts is S1T zz HTa2 y longSrran Os 7 12 Adding the expansion of T1 7 5 kills off all pow ers of S 7 12 with odd exponent including the pole and we find TU T1 Sls12 HTa y longSrran Since U1 2 is exponentially small AQ1 2 gt 0 for da2 gtgt 1 in fact bigger than 200 But recall Tr 1 QS 7 W 39 a is negative for s 7 1 By the Intermediate Value Theorem Qs has a real zero in 121 This proves i O1 NOTICES OF THE AMS Now suppose h7d 1 so by genus theory 1 is a prime congruent to 3 mod 4 an QXy 7 x2 xy 1dy2 a 1 make use of S which now says Qs sLsXd The fact that the Riemann zeta function has no real zeros in 0 1 follows from Euler s identity 17 21 s 5 71quot 1n s M8 711 The series converges for real 3 gt 0 by the Alter nating Series Test now group the terms in pairs to see that the sum is positive So Ls X4 has a real zero in 12 1 a Siegel zero The Argument Principle tells us the number of zeros minus the number of poles inside the circle of say radius 14 around 3 1 is given by L I 425 d 21Ti s7ll4 QS By 10 for sufficiently large 1 this is 1 C25 d 0 21Tl Ls7114 U25 S S because 2s has neither zeros nor poles near 3 1 Of course one has to justify passing the limit through the integral The convergence is not uniform but it is uniform on a compact set con taining the path of integration Since the Epstein zeta function has one simple pole there is only one Siegel zero close to s 1 El In fact you can say even more about the loca tion of this zero Exercise Use Mathematica to compute the Laurent expansion C71 TsT1 s 7 S71 C0Os71 around 3 1 Neglecting the higher order terms and the Us contribution show the zero is at 1 7 B for 6a THH Still under the hypothesis h7d 1 andd gtgt 1 Theorem Deuring Except for the Siegel zero all other zeros ofAQs in0 lt 039 lt 10 5 t lt d have real part 039 12 This is the Local GRH as a7oo El Outline of Proof Write 7 7 T1 7 5 Ts T1 s 7 Ts 1 TS The term Ts is never zero for 039 2 12 HS has no zeros and the fact that s is nonzero for VOLUME 53 NUMBER 8 a 2 l is equivalent to the Prime Number Theo rem 1 Deuring shows T1 STS lt l as long as a gt 12 and S 1 gt 14 This depends on known estimates for TS and C2S but basi cally works because for d large relative to t the xd2139r5 12 term dominates as long as we stay away from the pole of T1 S at S 1 Thus TST1 S 0 gt 0 12 Next he shows these zeros are simple as follows Write fS for T1 STS so the zeros occur at p 12 it such that fp 1 Then N d Erma fS H Tpf p For d d f p gm gm sgt f m fp TS T1 s 5 he can get a lower bound f p gtgt logd from known estimates for the logarithmic derivative of the Riemann zeta function as long as t is small enough relative to d Similarly known estimates give Tp gtgt 1 logd He then shows that TST1 S gt US DJ and so by Rouche s Theorem AQS has the same number of zeros in the box 0 lt a lt l 0 s t lt xd as does TS T1 S Here he needs t lt xd since by Stirling s formula TS also has exponential decay as t increases Around each zero p of TS T1 S Deuring puts a circle of radius Xp 1TH and uses the Cauchy Integral Formula to get upper bounds on the Taylor series coefficients 4 TS T1 S C1S p cnS pquot of the form cn ltlt K 1 By 2 above he already has c1 gtgt 1 The triangle inequality and summing a geometric series gives TS T1 S gtgt Xp 1Ta on the circle He can then apply Rouche s Theorem again to see AQS has one zero in that circle Since any zeros off the line would come in symmetric pairs by the func tional equation S gt 1 S that zero is on the line III Deuring s theorem is quite strong Since the zeros of CS are a subset of the zeros of QS whenever h d l by S as a Corollary we get that either there are only finitely many d with h d 1 or the Riemann hypothesis for CS is true Of course it is now known that the former is true but Deuring s theorem was an essential first step in solving the problem1 SEPTEMBER 2006 10 113 I a0 g 1 31 a2 and 119 8 g 33 e1 42 From B4 and BS 9 X 120 32 1437 262 003 892 375 1 no a 3 410399 where a 46l 786 552 where the 340109 means that lao 3 lt Ado9 From 119 116 and 113 ge see that 18 la lt 10 Hence we can rewrite 117 as 2 121 x2 xlai 10d9 We see experimentally that I2 1 9 122 3999 999 660 e 3 a i 2 10 31 l2 This degac of closeness of 3 3 a to an integer is 1 minus the discriminant of the nineth complex quadratic field with class number one In tact see 114 not completely surprising since x1 3 corresponds to p 165 Figure 2 A page from Stark s thesis Folklore Theorem Deuring Heilbronn In the presence of a Siege zero the lowlying zeros S 12 it ofLS xed are very regularly spaced 11 t n forinteger n N 1T IOgWHZn Idea of proof We make use of the fact that for S 12 it 1 S S The zeros of LSxd are zeros of AQS which we have seen are very near the zeros of TS T1 S TS TS TS m 2ReTS 2 TS cos ltargltd21T TS 2Sgtgt We saw above that the term TS is never zero on the line a l 2 Meanwhile TS is very near to real for t ltlt 1 so does not contribute much to the ar gument And 1 2 2SS12yOS 12 1An editorial comment this result made me have a lot more respect for Rouche39 S Theorem which previouslyl thought existed only to provide problems for analysis qualifying exams NOTICES OF THE AMS 869 719 77743 3 J o w 1 Q Figure3 119 Figure 4d43 7d 767 7d 7163 I 30 3 KJ f Figure 5 d 67 so arg 23 N NnZ for s l2 it N 12 Fi nally argltVH21Tittlogva2Tr and cos tlog ME2n N 1T2 sin tlog ME2n So tlogltVE21TN mT Actually it is not enough above that lTsl is nonzero To make this argument precise we need to estimate a lower bound To the extent that one can bound the tail of the Fourier expansion every NOTICES OF THE AMS Figure 6 1 163 11 above corresponds to a zero by Rouch s Theo rem The other factor s of Qs has no lowlying zeros but the same analysis shows that the zeros p 12 iy of s with y ltlt xH make Tp nearly pure imaginary Experimental Observations Even in the absence of a Siegel zero can one see this effect for a class number extremely small rel ative to its discriminant Stark in his 1964 PhD thesis was the first to investigate this numerically see Figure 2 for a cryptic comment For a graphical interpretation one can use Math ematica to plot VOLUME 53 NUMBER 8 k1 737 k2 72311 f 1 k11 k2 4391027 f 1 is WW5 k Figure 7d 85507 23 37 k1 89 k2 in76217403 f 1 W 4 Figure 8 d 991027 k1 1129 k2 490123 f 1 Figure 9 d 553348867 5507 89 1129 w s71 2 NS 7 Fs 25 on the critical line S 12 it for various values of d that have Mid 1 How close to the imagi nary axis is Tp for p that are known zeros of s Figures 376 show several examples for various 1 with the same five lowest zeros p of s indicated in red on each Since Stirling s Formula for Fs makes lTsl decay exponentially as tincreases one does not see the function wrap around the ori gin so 1 have renormalized the absolute value by taking the logarithm without changing the argu ment Increasing t corresponds to spiraling in coun terclockwise SEPTEMBER 2006 Figure IO 1 553348867 5507 89 1129 Figure 6 shows graphically what Stark was re ferring to the zeros of s can see the extremal discriminant 7163 This is as Stark later showed the largest d with Mid 1 Lfunction Magic Since the class number problem Mid 1 is al ready solved we want to think about lower bounds in general In this case the Dedekind zeta function Z Eds SLS X4 Q is a sum over all classes of Epstein zetas Whenever da2 gt 200 the corresponding Qs has a zero in 12 1 by Chowla Selberg This does not contra dict any GRH since the individual Epstein zetas do NOTICES OF THE AMS k1 75507 kZI 100481 f 1 k117 k2783 f1 15 Figure lld 553348867 5507 89 1129 not have Euler products over primes only the sum oes The reduced representatives of the forms can have a as big as Hi3 which means the exponen tial bound on the tail Us in the Fourier expansion can be as big as exp7Tr z 0044 for each of the Mid terms We still get the benefit of the 114 term in the denominator though so if Mid is very small relative to JR ie much smaller than 114 then the sum of the tails is still small Under this hypothesis Ls X4 has a Siegel zero and the Deur ingHeilbronn Phenomenon reappears as before we get the Local GRH and the lowlying zeros are uni formly spaced just as in l 1 But more is true Given a homomorphism on the class group p C41 a X we can form an Lfunction M w Z wQ QS Q More generally for an auxiliary discriminant f we can twist the Epstein zeta function Q20 Xf Z xfnrqnn s 711 and form Lsw Xf Z wQ QSXr Q For any odd fundamental discriminant k1 we can make the Dirichlet Lfunction LSXk1 appear as a factor of such an expression by careful choice oftp and f Factor 711 D1 D2 as a product of fun damental discriminants in such a way that D1 gcdk1d By a theorem of Kronecker this factorization corresponds to a genus character p with NOTICES OF THE AMS r5 5 10 15 Figure2dl4ll 83 I7 12 US w LSXD1LSXD2 generalizing S In fact the genera of quadratic forms mentioned above are exactly the cosets of the class group C41 modulo the subgroup C412 of squares of classes The genus group is therefore a product of copies of 12 and Gauss showed the number of terms is g 7 l where g is the number of prime factors of d The corresponding genus characters are exactly those taking only the values 1 By a comparison of the corresponding Dirichlet series in 1 Z LIIQ Z rQnn s Z anddxn Wd H s Q ngt0 ngt0 sin and the uniqueness ofDirichlet series coefficients we deduce that for all n 13 Z wQrqn Z XD1CXD2nC Q cln This is a generalization of 4 Theorem Letf k1D1 k2 sz Then 14 LSXk1LSXk2 M w Xf Proof The idea goes back to Heilbronn LUMP 39 Xf ZWQKQUJH Q Z WQ 2 XrnrQnn s Q ngt0 2 mm 2 WQWle n s Q ngt0 By13we get VOLUME 53 NUMBER 8 k1747 k2209 f1 k17167 k2185 f1 10 U a v 710 Figure l3d9823ll 19 47 HS LII Xf Z Xfn ZXD1CXD2nC n s ngt0 cln Z 1 XfCXD1CXfnCXD2nC n s ngt0 cln Z inCXk2nCgt n s ngt0 in LS Xk1LSXk2 From the right side ofl4 we have a Fourier ex pansion srlZ gt rltsLltsxkLltsxk2 Ts TI 7 s Us similar to the previous one with now srlZ Ts rltslt2sPfltsAlts 2n and PAS W1 7 1145 Am Z wQXraa s Plf Q In fact in the examples below we take f I In this case we have a linear combination of the Chowla Selberg Fourier expansions the coefficients are merely the character values tpQ If the class number Mid is too small or LSXd has a Siegel zero the DeuringHeilbronn phenomenon 11 appears for Lsxk1 as well as long as fis not too big Experimental Observations As in the case of Mid I we can plot Ts on the line 3 12 it and see where the zeros ofLsXk1 SEPTEMBER 2006 1 Figure l4d 30895 167 5 37 and LsXk2 end up For extreme values of 7d will they tend towards the imaginary axis Figures 7711 show some examples with small class num ber The lowest five zeros of Ls Xkl are shown in red while the lowest five zeros of Lsxk2 are shown in blue In some cases not all five are visi ble if they lie nearly on top of each other I The discriminant 785507 7231137 has class group isomorphic to 122 so h7dH z 075 There is one nontrivial genus character The discriminant 7991027 has class group iso morphic to 163 there is only the trivial genus character This is the famous example of Shanks h7dH z 063 which minimizes this ratio for all 1 lt 103 Because the zeros of s Lsxk1 are so high up we show instead 10 zeros of Ls X4 Lsxk2 in Figure 8 The discriminant 7553348867 75507 89 1129 has class group isomorphic to 1732 XZZ so hemmu 062 There are three nontrivial genus characters Observe that in each of the examples the para meter f which in some sense measures the corre lation between 711 and the auxiliary discriminant k1 is as small as possible f I or in other words k1 id Even so the correlation between the zeros is by no means trivial There is no obvious relation between the class groups Ckl and C41 one is not a direct factor of the other Of course these examples are hand picked to show off this tendency towards the Deuring Heilbronn phenomenon in general one sees noth ing like this In the next section below we discuss contemporary conjectures about the distribution of zeros It is also interesting to look at some examples of discriminants 7d that are famous for LS X41 having a very lowlying zero Figures 13717 show N w NOTICES OF THE AMS k1 113 k2 71019 f 1 k1 71427 k2 3 123329 f 1 10 1quot 710 Figure15d115147 1019 113 some examples The lowlying zero does not appear in the figures rather it is causing the Deuring Heilbronn phenomenon l The Lfunction Lsx1411 has a zero at s 12 i 0077967 The discriminant 71411 783 17 has one nontrivial genus character The Lfunction Lsxggzg has a zero at s 12 i 0058725 The discriminant 79823 711 719 747 has three nontrivial genus characters The Lfunction Lsx3oggg has a zero at s 12 i 0018494 The discriminant 730895 7167 5 37 also has three nontriv ial genus characters The Lfunction Lsx115147 has a zero at s 12 i 0003158 The discriminant 7115147 71019 113 has one nontrivial genus character LSX1ygggo433 has a zero at s l2 i 0000475 The discriminant 7d 7175990483 719 71427 76491 has three nontrivial genus characters In both sets of examples Tp is very nearly pure imaginary for zeros p of Ls Xkl or Ls m and so Tp Tl 7 p is also very near 0 Neces sarily this means that the tail of the Fourier ex pansion Up is also very near 0 much smaller than our estimate 0 h7dd 14 It would be nice to have examples where the zeros p not only forced Ts to be nearly purely imaginary but also were restricted to near integer multiples of Tr logH21T as in 11 This would require argFs 2s to be very near its limiting value 1T2 and thus k1 very large in order that Lsxk1 have several zeros so low But this may allowf gt 1 as well N w 3 U1 NOTICES OF THE AMS L Figure 16d 175990483 19 1427 6491 Can You Hear the Class Number In 1966 Mark Kac posed the question Can you hear the shape of a drum In fact what one hears are solutions to the wave equation ie eigenvalues of the Laplace operator The mathematical meaning of Kac s question is what does this spectrum de termine about the geometry In the very useful analogy between spectral geometry and number theory eigenvalues of the Laplacian correspond to zeros of Lfunctions while geometric properties correspond to properties of primes It is very in teresting that the DeuringHeilbronn phenomenon l 1 if it occurs corresponds in this analogy to fre quencies in harmony Above I mentioned Stark s PhD thesis in which he used precise values of zeros of the Riemann zeta function to show that a certain range of discrimi nants did not have Mid 1 He later extended this to the problem of Mid 2 Montgomery and Weinberger used lowlying zeros of auxiliary Lsxk1 to attack Mid 2 and 3 in 6 This work led Montgomery to investigate the question If GRH is true and there are no Siegel zeros and no DeuringHeilbronn phenomenon what is the ver tical distribution of the zeros on the critical line Remarkably he discovered 7 that the pair cor relation of the zeros is the same as that for the eigenvalues of random unitary matrices the Gauss ian Unitary Ensemble GUE Montgomery s proof works only for a restricted range of test functions not in general but the GUE hypothesis is also sup ported by the statistics of 10 billion zeros of the zeta function computed by Odlyzko 8 This sug gests the zeros of Lfunctions may indeed have a spectral interpretation as conjectured by Hilbert and Polya VOLUME 53 NUMBER 8 Atmospheric and Oceanic Circulations con nued Chapter 6 Lecture 14 4 February 2005 Figure Credit Earth s 3imate by W Ruddiman Outgoing radiation adiation intensity WmZ 100 Incoming sola 50 radiation 0 O I O I O I O I O I O I 0 I O 0 10 20 30 40 50 60 80N NE 100 Latitude E 50 A 30 8 O E 25 V 20 f g 15 I 393 u g 50 g 10 3 100 2 5 l a 0 g I0 I o I o I o I o I o I o I o I 5 i a 0 1O 20 3O 4O 50 60 SUN I I I 60 5410 2390 0 2390 4390 8390 N Latitude Latitude ED 0 Precipitation P vs evaporation E Figure Credit Earth s Chmate by W Ruddiman V nd simply put wind is the horizontal ow of air in response to differences in air pressure these pressure differences are usually due to uneven solar heating at the surface ill Low Pressure Credit WWW physicalgeugraphy net Wind winds are designated as direction from not direction to oceanographers do itthe opposite 0 wind com ass so a westerly wind would be coming from what angular direction 270quot 180 Four forces that determine winds 1 Gravity pulls gas molecules close to Earth density amp pressure decrease with height 2 Pressure gradient force the difference in air pressure between areas 3 Coriolis force deflects wind from a straight line to the right or left depending on hemisphere 4 Friction force the drag on air flow from the Earth s surface Pressure vs Pressure Gradient The value of pressure itself is NOT important The CHANGE in pressure over DISTANCE is Change over distance is a GRADIENT The GRADIENT in pressure gives winds amp ocean currents their push UIdUIUIII ruruc rur Closer spacing Wider spacing o o n n a a m E E E E E E E E no 1 O D N m 0 m N m V o on e e e e e e 92 Gradual pressure ep pressure gradient 39 E p High EIb Low pressure pressure 29 2 9 Isobar LIGHT STRONG WINDS WINDS Eobar a line of equal pressure gradient is 16 mb analogous to iSOtherm note the closer isobars the PGF acts at right 90 angles to the isobars o 00 400 MILES 00 400 KILOMETEHS I E Wind Direction Hm kilnmeiers Pressure Gradienl 01 rubkilometer 1020 mb 1 39quotb Charge row mi 1020 mb 4quot 39quotb Chmge 990 mb Wind Direction 11m kilometers Pressure Gradienl 04 mblkilurneier Wind speed will be 4 limes greaier Wind speed Const Pressure Gradient Credit WWW pnysmaigeagrapny net Here a 4x increase in PGF corresponds to a 4x increase in wind speed Pressure Gradient Force and Isobars isobar isobar i 1 ssure Top view and h High 5 Low 4 Top gradient side View or a view H force movement in pressure area and High Low iowpressure area I Side pressure pressure on a nonrotating J K View Earth H39High H a Low ab W Earth 5 surface Descending Ascending diverging converging a Pressure gradient lnrce alone if there were no other forces acting on wind it would flow in straight lines perpendicular to isobars from high to low pressure zones Coriolis Force just the facts Rotation of Earth acts to deflect any motion from a straight line Deflection is to right NH to the left SH Coriolis force act on a right angle to the motion Coriolis Force is NOT a real force but is caused by viewing motion on a rotating planet Coriolis Force Show the merrygo round video Polar person W spinning in place mount of rotation about a tical axis spinning is gtltimum at the poles and inimum at the equator 031 9rlealper50n 3i f00enijti4ard Sec A Figure Credit Earth s Climatequot by W Ruddiman Cnrinlis Fnrma N an object initial eas velocity maintair velocity e it passes surfaces different ve as a res appears deflectec that surfac in NH left a Hemisphere I it V Rotation Flotation Flight path on a Flight path deflected quot 39 taquot g Eanh by Coriolis force

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