Create Matthew Koschel The Riemann hypothesis
Riemann, in 1859, in a paper [R] written on the occasion of his admission to the Berlin Academy of Sciences and read to the Academy by none other than Encke, devised an analytic way to understand the error term in Gauss approximation, via the zeros of the zeta-function (s) = X n=1 1 ns . The connection of (s) with prime numbers was found by Euler via his product formula X n=1 1 ns = Y p (1 p s ) 1 where there is one factor for each prime number p. This formula encodes the fundamental theorem of arithmetic that every integer is a product of primes in a unique way. Riemann saw that the zeros of what we now call the Riemann zeta-function were the key to an analytic expression for (x). Riemann observes that s/2 (s/2) (s) = Z 0 (x)x s 1 dx where (x) = X n=1 e n2x and then uses 2 (x) + 1 = x 1 2 2 1 x + 1 , which follows from a formula of Jacobi, to transform the part of the integral on 0 x 1. In this way he finds that (s) is a meromorphic function of s with its only a pole a simple pole at s = 1 and that (s) = 1 2 s(s 1) s/2 (s/2) (s) is an entire function of order 1 which satisfies the functional equation (s) = (1 s). As a consequence, (s) has zeros at s = 2n for n Z+, these are the so called trivial zeros, as well as a denser infinite sequence of zeros in the critical strip 0 <s 1. The Euler product precludes any zeros with real part larger than 1. Also, for any non-trivial zero = +i there is a dual zero 1 by the functional equation. Also, and 1 are zeros since (s) is real for real s but note that coincides with 1 whenever = 1/2. Riemann 4 BRIAN CONREY used Stirling s formula and the functional equation to evaluate the number of non-trivial zeros in the critical strip as N(T) := #{ = + i : 0 < T} = T 2 log T 2 e + 7/8 + S(T) + O(1/T) where S(T) = 1 arg (1/2 + iT) where the argument is determined by beginning with arg (2) = 1 and continuous variation along line segments from 2 to 2 + iT and then to 1/2 + iT, taking appropriate action if a zero is on the path. Riemann implied that S(T) = O(log T) a fact that was later proven rigorously by Backlund [Bac18]. Thus, the zeros get denser as one moves up the critical strip. The functional equation together with Riemann s formula for the number of zeros of (s) up to a height T help give us a picture of (1/2 + it). In particular Hardy defined a function Z(t) which is a real function of a real variable having the property that | (1/2+it)| = |Z(t)|. It may be defined by Z(t) = (1/2 it) 1/2 (1/2 + it) where (s) is the factor from the functional equation which may be written in asymmetric form as (1 s) = 2(2 ) s (s) cos s 2 . Here are some graphs of Z(t): 10 20 30 40 50 -3 -2 -1 1 2 3 ZHtL for 0<t<50 1010 1020 1030 1040 1050 -8 -6 -4 -2 2 4 6 ZHtL for 1000<t<1050 1.1. Riemann s formula for primes. Riemann found an exact formula for (x). If we invert Euler s formula we find 1 (s) = Y p (1 p s ) = 1 2 s 3 s 5 s + 6 s + = X n=1 (n) ns where is known as the M obius mu-function. A simple way to explain the value of (n) is that it is 0 if n is divisible by the square of any prime, while if n is squarefree then it is +1 if n has an even number of prime divisors and 1 if n has an odd number of prime divisors. Riemann s formula is (x) = X n=1 (n) n f(x 1/n) RIEMANN S HYPOTHESIS 5 where f(x) = li(x) X li(x ) ln 2 + Z x dt t(t 2 1) log t . Here the = +i are the zeros of (s) and the sum over the is to be taken symmetrically, i.e. to pair the zero with its dual 1 as the sum is performed. Thus the difference between Riemann s formula and Gauss conjecture is, to a first estimation, about li(x 0 ) where 0 is the largest or the supremum of the real parts of the zeros. Riemann conjectured that all of the zeros have real part = 1/2 so that the error term is of size x 1/2 log x. This assertion of the perfect balance of the zeros, and so of the primes, is Riemann s Hypothesis. In 1896 Hadamard and de la Vall ee Poussin independently proved that (1 + it) 6= 0 and concluded that (N) li(N) a theorem which is known as the prime number theorem. 2. Riemann and the zeros After his evaluation of N(T) T 2 log T he asserted that we find about this many real zeros of (t) := (1/2 + it) in 0 < t T. This is an assertion which is still unproven and is the subject of speculation. His memoir Ueber die Anzahl der Primzahlen unter einer gegebenen Gr osse is only 8 pages. But in the early 1930s his Nachlass was delivered from the library at G ottingen to Princeton where C. L. Siegel [?] looked over Riemann s notes at the Institute for Advanced Study. In the notes were found an approximate functional equation, which had been independently found by Hardy and Littlewood [HL29]: (s) = X n t 2 1 ns + (s) X n t 2 1 n1 s + O(t /2 ) for s = + it. Here (s) is the factor from the asymmetric form of the functional equation (s) = (s) (1 s) with (s) = (1 s)/2 ((1 s)/2) s/2 (s/2) = 2(2 ) s 1 (1 s) sin s 2 . Now | (1/2 + it)| = 1; in fact (1/2 + it) = e itlog t/2 (1 + O(1/t)). One might be led to believe that 1 + (s) is a reasonable approximation to (s), i.e. that the contributions from the oscillatory terms 2 s etc. might be small overall. This approximation has zeros on s = 1/2 + it at a rate sufficient to produce asymptotically all of the zeros of (s), so it seems reasonable to conclude that almost all of the zeros are on this line, and to go on and conjecture that ALL of the zeros are on the one-line. But we have found it hard to make this reasoning precise. 6 BRIAN CONREY Riemann computed the first few zeros: 1/2 + i14.13 . . . , 1/2 + i21.02 . . . , 1/2 + i25.01 . . . , . . . A good way to be convinced that these are indeed zeros is to use the easily proven formula (1 2 1 s ) (s) = 1 1 2 s + 1 3 s 1 4 s . . . . The alternating series on the right converges for 0 and so, for example, s = 1/2 + i14.1347251417346937904572519835624 . . . can be substituted into a truncation of this series (using a computer algebra system) to see that it is very close to 0. (See www.lmfdb.org to find a list of high precision zeros of (s) as well as a wealth of information about (s) and similar functions called L-functions.) 3. Elementary equivalents of the Riemann Hypothesis We ve mentioned that the Riemann Hypothesis implies a good error bound for the prime number theorem. The converse is also true: the Riemann Hypothesis is equivalent to (x) := X p x 1 = Z x 2 du log u + O(x 1/2 log x), and to (x) := X p k x log p = x + O(x 1/2 log2 x) Equivalences may also be phrased in terms of the M obius function (n) where 1 (s) = X n=1 (n) ns . It is not difficult to show that the Riemann Hypothesis is equivalent to the assertion that this series is (conditionally) convergent for any s with 1/2 < < 1. The Riemann Hypothesis is also equivalent to each of M(x) := X n x (n) = O(x 1/2+ ) and Z X 1 ( (x) x) 2 dx x 2 C log X. The assertion that Z X 1 M(x) 2 dx x 2 C log X implies the Riemann Hypothesis and that all of the zeros are simple. A question is whether the converse is true. RIEMANN S HYPOTHESIS 7 Stieltjes thought that he had found a proof that M(x) = O(x 1/2 ) and so of the Riemann Hypothesis. His claim appeared in Comptes Rendus Mathematique. Consequently de la Hadamard was somewhat apologetic about his inconsequential offering in his own paper [Had96] which proves the prime number theorem! Figure 1. A plot of M(x) versus x 4. The general distribution of the zeros An immediate consequence of Euler s product formula (s) = Y p 1 p s 1 is that (s) 6= 0 if 1. A subsequent consequence of Riemann s functional equation is that (s) 6= 0 if <s < 0 except at s = 2, 4, 6, . . . , the so-called trivial zeros. The prime number theorem (x) x log x is equivalent to the assertion that (1 +it) 6= 0; equivalently (it) 6= 0. In order to be precise about the error term in the prime number theorem it is necessary to prove that there is a 8 BRIAN CONREY region near the line = 1 in which there are no zeros. It was shown by de la Vall ee Poussin in 1899 [Val96] that ( + it) 6= 0 for > c log(2+|t|) for a specific c. This is known as a zero-free region. The best known shape of the zero-free region is due to Korobov [Kor58] and Vinogradov [Vin58] in 1958: ( + it) is free of zeros when > 1 C (log t) 2/3 (log log t) 1/3 . The best explicit value of C is due to Kevin Ford [For00] who showed that C = 1/54.57 is admissible. 4.1. Density results. Bounds for the quantity N( , T) := #{ = + i : and 0 < T} are known as density estimates. Near to = 1 we have [For00] N( , T) T 58.05(1 ) 3/2 (log T) 15 . As we move away from the line = 1 our estimates get weaker but are still pretty good. Bounds often take the shape N( , T) T k( )+ ; there are many forms of admissible k( ). A strong classical one due to Ingham in 1940 [Ing40] is that N( , T) = O(T 3(1 )/(2 ) log5 T); this is still the best bound when 1/2 < < 3/4. It is known that k( ) = 3/2 is also admissible. The unproven Density Hypothesis is that the above holds with k( ) = 2(1 ). It is known that an estimate of the sort (1/2 + it) t c (log t) c 0 implies that N( , T) T 2(1+2c)(1 ) log5 T; see [Tit86]. Thus, the Density Hypothesis is a consequence of the Lindel of Hypothesis (for which see below). A consequence of the Density Hypothesis is that for any > 0 there is a C( ) such that pn+1 pn C( )n 1/2+ where pn denotes the nth prime. This estimate is not quite strong enough to conclude that there is always a prime between consecutive squares. Here is a plot of the exponent in density theorems (the minimum of the two graphs is an admissible density exponent): RIEMANN S HYPOTHESIS 9 0.5 0.6 0.7 0.8 0.9 1.5 2.5 3 3.5 4 4.5 Density exponent kHsigmaL 4.2. Zeros near the 1/2-line. It has been known for quite some time that almost all of the zeros are near the 1/2-line. For example at least 99% of the zeros = + i satisfy | 1/2| < 8 log . and almost all are within ( )/ log of the critical line where is any function which goes to infinity. Thus, we know that the zeros cluster around the critical line. 4.3. Zeros on the critical line. Many people have worked on verifying the Riemann Hypothesis. Today it is known that the first ten trillion zeros are all on the critical line <s = 1/2! Hardy was the first one to show that there are infinitely many zeros on the 1/2-line. He and Littlewood [HL18] later gave proofs that the number of zeros on the 1/2-line up to a height T is more than a positive constant times T. In 1942 Selberg [Sel42] proved that a positive proportion of the zeros are on the critical line. In 1973 N. Levinson [Lev74] proved that at least 1/3 of the zeros are on the half-line. This was improved in 1989 to at least 2/5 of the zeros are on the line. The current record is Feng [Fen12] with 0.412; for simple zeros the record proportion is due to Bui, Conrey, and Young [BCY11] who show that at least 0.405 of the zeros of (s) are on the critical line and simple. It follows from the Riemann Hypothesis that all of the zeros of all of the derivatives (k) (s) are on the critical line. Along these lines it can be shown, for example, that more than 4/5 of the zeros of 0 (s) are on the critical line and more than 99% of the zeros of (5)(s) are on the critical line, see [Con83]. 5. The Lindelof Hypothesis The assertion that for any > 0, (1/2 + it) t is known as the Lindel of Hypothesis and is a consequence of the Riemann Hypothesis. It is a consequence of the functional equation, trivial bounds for (it) and (1 + it), and general 10 BRIAN CONREY principles of the growth of analytic functions that (1/2 + it) t 1 4 + ; this is known as the convexity bound. Weyl, using exponential sums, improved the bound to (1/2 + it) t 1 6 + . Bombieri and Iwaniec [BI86] used some novel ideas to show (1/2 + it) t 89/560+ . Huxley [Hux05] obtained (1/2 + it) t 32/205 logc t. Recently Bourgain [Bou14] announced that (1/2 + it) t 53/342+ 5.1. Estimates for (s) near the 1-line. Richert [Ric67] proved the important estimate that for an explicit c > 0, | ( + it)| < ct100(1 ) 3/2 log2/3 t for 1/2 1, t 2. Such a bound is useful for zero-free regions, the error term in the prime number theorem, and zero density results near 1. K. Ford [For02] has improved these made the constants explicit: | ( + it)| < 76.2t 4.45(1 ) 3/2 log2/3 t. 5.2. 1 versus 2. RH implies that | (1/2 + it)| exp log 2 2 log t log log t + O log tlog log | log t (log log t) 2 see [ChS11] . It can be proven, see [Sou08] that every interval [T, 2T] contains a t for which | (1/2 + it)| exp (1 + o(1)) (log t) 1/2 (log log t) 1/2 . Which of these is closer to the true largest order of magnitude of on the 1/2-line? It is difficult to say, though most people (not the author!) think that the lower bound ( -result) is closer to the truth. Farmer, Gonek, and Hughes [FGH07] conjecture that | (1/2 + it)| exp s 1 2 + o(1) (log t)(log log t) ! . RIEMANN S HYPOTHESIS 11 6. Computations Turing was the first to use a computer to calculate the zeros of (s). He proposed an efficient rigorous method to verify RH up to a given height, or indeed within an interval. It involves using a precise version of the approximate functional equation, known as the Riemann - Siegel formula, to evaluate Z(t) and detect sign changes, together with an explicit bound for the average of S(t) namely if t2 > t1 > 168 then Z t2 t1 S(t) dt = 2.30 + 0.128 log t2 2 to verify that all of the zeros are accounted for. (Here represents a number that is at most 1 in absolute value.) Goldfeld has pointed out that if (s) had a double zero somewhere up the line, the computational verification of RH would come to a halt because it would be impossible to distinguish a double zero from two very close zeros either on or off the line. Here is one of Turing s versions of the Riemann-Siegel formula: Theorem 1. Let m and be respectively the integral and non-integral parts of 1/2 and 64, ( ) = 1 4 i log ( 1 4 + i ) ( 1 4 i ) 1 4 log , Z( ) = (1/2 + 2 i )e 2 ( ) , 1( ) = 1 2 ( log 1 2 ), h( ) = cos 2 ( 2 1 16 ) cos 2 . Then Z( ) is real and Z( ) = 2Xm n=1 n 1 2 cos 2 { log n ( )} + ( 1)m+1 1 4 h( ) + (1.09 3 4 , ( ) = 1( ) + (0.006 1 ). In 1988, Andrew Odlyzko and Sch onhage [OS88] invented an algorithm which allowed for the very speedy calculation of many values of (s) at once. The Riemann-Siegel allows for a single computation of (1/2 + it) with T < t < T + T 1/2 in time T 1/2+ . The Odlyzko - Sch onhage algorithm allows for a single computation in time T after a pre-computation of time T 1/2+ . This led Odlyzko to compile extensive statistics about the zeros at enormous heights - up to 1023 and higher. His famous graphs showed an incredible match between data for zeros of (s) and for the proven statistical distributions for random matrices. Here is a list of contributors to verifying RH in an initial segment of the 1/2-line and the year they did the work. 12 BRIAN CONREY G. H. B. Riemann 3 1859 J. P. Gram 15 1903 R. J. Backlund 79 1914 E. C. Titchmarsh 1041 1935 A. M. Turing 1104 1953 D. H. Lehmer 15000 1956 D. H. Lehmer 25000 1956 N. A. Meller 35337 1958 R S. Lehman 250000 1966 J. B. Rosser , J. M. Yohe, L. Schoenfeld 3500000 1968 R. P. Brent 40000000 1977 R. P. Brent 81000000 1979 R. P. Brent, J. van de Lune, H. J. J. te Riele, D. T. Winter 200000001 1982 J. van de Lune, H. J. J. te Riele 300000001 1983 J. van de Lune, H. J. J. te Riele, D. T. Winter 1500000001 1986 J. van de Lune 10000000000 2001 S. Wedeniwski 900000000000 2004 X. Gourdon, P. Demichel 10000000000000 2004 Ghaith Hiary [Hia11] has improved these algorithms. He can compute one value of (1/2+ it) in time T 1/3+ using an algorithm that has been implemented by Jonathan Bober and Hiary; he has a more complicated algorithm that will work in time T 4 13 + . They have verified RH in some small ranges around the 1033 zero! Bober s website [Bob14] has some great pictures of large values of Z(t). 7. Why do we think RH is true? The main reason is because of the beauty of the conjecture. It strikes our sensibilities as appropriate that something so incredibly symmetric should be true in mathematics. The second reason is that the first 10 trillion zeros are all on the line. If there were a counterexample it should have shown itself by now. A third reason is that the numerical evidence for all L-functions ever computed lead to this conclusion; some have thought that a counterexample to RH might show itself when computing zeros of L-functions associated with Maass forms because these have no arithmetic-geometry interpretation (eg. their coefficients are generally believed to be transcendental); however the computations reveal that the zeros are still on the 1/2-line. A fourth reason is probabilistic. RH is known to be equivalent to the assertion that M(x) := P n x (n) x 1/2 log2 x. This sum represents the difference between the number of squarefree integers up to x with an even number of prime factors and the number with an odd number of prime factors. It is similar to the difference in the number of heads and tails when one flips x coins, and so should be around x 1/2 . Here is another more elaborate reason. Suppose that a Dirichlet series F(s) = P n=1 ann s converges for > 0, and suppose that it has a zero with real part > 1/2. We might reasonably expect it then to have T zeros in > , 0 < t < T for any large T by RIEMANN S HYPOTHESIS 13 almost periodicity. But zero density results tell us that there are T 1 zeros in 0 and t < T. 7.1. Almost periodicity. As just mentioned a possible strategy is to try to prove that if (s) has one zero off the line then it has infinitely many off the line. Bombieri [Bom00] has come closest to achieving this. Here is a conjecture that attempts to encapsulate this idea: Conjecture 1. Suppose that the Dirichlet series F(s) = X n=1 ann s converges for > 0 and has a zero in the half-plane > 1/2. Then there is a number CF > 0 such that F(s) has > CF T zeros in > 1/2, |t| T. This seemingly innocent conjecture implies the Riemann Hypothesis for virtually any primitive L-function (except curiously possibly the Riemann zeta-function itself!). And it seems that the Euler product condition has already been used (in the density result above); i.e. the hard part is already done. Note that the 1/2 in the conjecture needs to be there as the example X n=1 (n)/n1/2 ns demonstrates. Assuming RH, this series converges for > 0 and its lone zero is at s = 1/2. This example is possibly at the boundary of what is possible. 8. A spectral interpretation Hilbert and P olya are reputed to have suggested that the zeros of (s) should be interpreted as eigenvalues of an appropriate operator. Odlyzko wrote to P olya to ask about this. Here is the text of Odlyzko s letter, dated Dec. 8, 1981. Dear Professor P olya: I have heard on several occasions that you and Hilbert had independently conjectured that the zeros of the Riemann zeta function correspond to the eigenvalues of a self-adjoint hermitian operator. Could you provide me with any references? Could you also tell me when this conjecture was made, and what was your reasoning behind this conjecture at that time? The reason for my questions is that I am planning to write a survey paper on the distribution of zeros of the zeta function. In addition to some theoretical results, I have performed extensive computations of the zeros of the zeta function, comparing their distribution to that of random hermitian matrices, which have been studied very seriously by physicists. If a hermitian operator associated to the zeta function exists, then in some respects we might expect it to behave like a random hermitian operator, which in turn ught to resemble a random hermitian matrix. I have discovered that the distribution of zeros of 14 BRIAN CONREY the zeta function does indeed resemble the distribution of eigenvalues of random hermitian matrices of unitary type. Any information or comments you might care to provide would be greatly appreciated. Sincerely yours, Andrew Odlyzko and P olya s response, dated January 3, 1982. Dear Mr. Odlyzko, Many thanks for your letter of Dec. 8. I can only tell you what happened to me. I spent two years in G ottingen ending around the beginning of 1914. I tried to learn analytic number theory from Landau. He asked me one day: You know some physics. Do you know a physical reason that the Riemann Hypothesis should be true? This would be the case, I answered, if the non-trivial zeros of the function were so connected with the physical problem that the Riemann Hypothesis would be equivalent to the fact that all the eigenvalues of the physical problem are real. I never published this remark, but somehow it became known and it is still remembered. With best regards. Yours sincerely, George, P olya 9. The vertical spacing of zeros In the 1950s physicists predicted that excited nuclear particles emit energy at levels which are distributed like the eigenvalues of random matrices. This was verified experimentally in the 1970s and 1980s; Oriol Bohigas was the first to put this data together in a way that demonstrated this law. Figure 2 shows 96 zeros of (s) starting at a height T = 1200 wrapped once around a circle for the purposes of comparing with the eigenvalues of a randomly chosen 96 96 unitary matrix, and with 96 points chosen randomly independently on a circle (Poisson). It should be clear that the zeros of (s) do not have a Poisson distribution (and would have been clear to anyone looking carefully at them, say in the mid 1900 s!). In 1972 Hugh Montgomery, then a graduate student at Cambridge, delivered a lecture at a symposium on analytic number theory in St. Louis, outlining his work on the spacings between zeros of the Riemann zeta-function. This was the first time anyone had considered such a question. On his flight back to Cambridge he stopped over in Princeton to show his work to Selberg. At afternoon tea at the Institute for Advanced Study, Chowla insisted that Montgomery meet the famous physicist - and former number theorist - Freeman Dyson. When Montgomery explained to Dyson the kernel he had found that seemed to govern the spacings of pairs of zeros, Dyson immediately responded that it was the same kernel that RIEMANN S HYPOTHESIS 15 (a) Unitary (b) Poisson (c) -zeros Figure 2. 96 points of three different types of spacings governs pairs of eigenvalues of random matrices. Montgomery [Mon73] proved that X 1, 2 [0,T] w( 1 2)f log T 2 ( 1 2) = T log T 2 f(0) + Z f(u) " 1 sin( u) u 2 # du + o(1)! assuming the Riemann Hypothesis and that the Fourier transform f of f vanishes outside of [ 1, 1] and w(x) = 4/(4 + x 2 ). The sum here is over pairs of zeros 1/2 + i 1 and 1/2 + i 2. The conjecture is that the assumption on the support of f is not necessary. Odlyzko did extensive numerical calculations to test this conjecture; the numerics are stunning! The pair-correlation function in Figure 3 is 1 sin x x 2 . The nearest-neighbor density function is more complicated. It may be given as 1 4 d 2 dt2 exp Z t 0 (2u) u du where = (s) is a solution of a Painlev e equation: (s 00) 2 + 4(s 0 ) ( 0 ) 2 + s 0 = 0 with a boundary condition (s) s s 2 2 as s 0, as discovered by Jimbo, Miwa, Mori, and Sato, see [JMMS80]. 16 BRIAN CONREY (a) Pair correlation (b) Nearest neighbor Figure 3. Odlyzko s graphics Now we have the challenge of not only explaining why all of the zeros are on a straight line, but also why they are distributed on this line the way they are! The connections with Random Matrix theory first discovered by Montgomery and Dyson have received a great deal of support from seminal papers of Katz and Sarnak [KaSa99] and Keating and Snaith [KS00]. The last 15 years have seen an explosion of work around these ideas. In particular, it definitely seems like there should be a spectral interpretation of the zeros `a la Hilbert and P olya. 10. Some initial thoughts about proving RH 10.1. Fourier integrals with all real zeros. Riemann proved that (t) := (1/2 + it) = Z (u)e iut du where (u) = X n=1 (4 2n 4 e 9u/2 6n 2 e5u/2 ) exp( n 2 e2u ) It is known that (u) is even, is positive for real u and is (rapidly!) decreasing for u > 0. Consequently, we can write (t) = 2 Z 0 (u) cos ut du = X n=0 ( 1)n bn (2n)! t 2n RIEMANN S HYPOTHESIS 17 where bn := Z (u)u 2n du. The Riemann Hypothesis is the assertion that all of the zeros of (t) are real. This has prompted investigations into Fourier integrals with all real zeros. Polya [Pol27] and deBruijn [deB50] spent a lot of time with such investigations. A sample theorem is Theorem 2. Let f(u) be an even nonconstant entire function of u, f(u) 0 for real u, and such that f 0 (u) = exp ( u2 )g(u), where 0 and g(u) is an entire function of genus 1 with purely imaginary zeros only. Then (z) = R exp { f(u)}e izudt has real zeros only. Now (u) > 0 for all real u and 0 (u) < 0 for u 0. Thus, we can write (u) = e f(u) . The functional equation for (s) is equivalent to the assertion that (u) is even. In particular, it was shown by P olya [Pol26] that all of the zeros of the Fourier transform of a first approximation (u) to (u) (u) = (2 cosh(9u/2) 3 cosh 5u/2) exp( 2 cosh 2u) are real. These ideas have been further explored by deBruijn, Newman [New76], Hejhal, Haseo Ki [KK02], [KK03] and others. Hejhal [Hej90] has shown that almost all of the zeros of the Fourier transform of any partial sum of (u) are real. A goal of this approach is to determine necessary and sufficient conditions that describe the Fourier transform of a function all of whose zeros are real. 10.2. Jensen s inequalities. In section 14.32 of [Tit86], we find the assertion that RH is equivalent to Z Z ( ) ( )e i( + )x e ( )y ( ) 2 d d 0 for all real x and y where (u) is as in the last section. We quote a passage from P olya s collected works, volume I, page 427, written by M. Marden commenting on the paper of P olya. In this paper Professor P olya reports his findings on examining the Nachlass of the Danish mathematician J. L. W. V. Jensen who died in 1925. Fourteen years earlier Jensen had announced that he would publish a paper regarding his algebraic-function theoretic research on the Riemann -function. In view of Jensen s well-known interest in the zeros of polynomials and entire functions, expectations were high that Jensen would contribute to the solution of the Riemann hypothesis problem regading the zeros of the -function. However, this paper was never published, and so on Jensen s death it was a matter of paramount importance to have his papers examined by an expert in this area. Professor P olya undertook this task, but after an arduous examination he found no clue to any progress that Jensen may have made towards the Riemann hypothesis. 18 BRIAN CONREY Professor P olya does sketch Jensen s algebraic-function-theoretic investigations, many of which were advanced considerably by P olya s own work. In this paper, P olya gives two more necessary and sufficient conditions for RH. RH is equivalent to Z Z ( ) ( )e i( + )x ( ) 2n d d 0 for all real values of x and n = 0, 1, 2, . . . ; and finally RH is equivalent to Z Z ( ) ( )(x + i ) n (x + i ) n ( ) 2 d d 0 for all real values of x and n = 0, 1, 2, . . . . P olya points out that the first equivalence to RH follows immediately from the more general theorem that all of the zeros of a real entire function F(z) of genus at most 1 are real if and only if 2 y2 |F(z)| 2 0 for all z = x + iy. To see that this condition is necessary for polynomials suppose that F(z) = QJ j=1(z rj ) and let f(x, y) = |F(z)| 2 . Then log f = PJ j=1 log(z rj ) + log(z rj ) so that fy f = X J j=1 i z rj i z rj = 2X J j=1 =(z rj ) |z rj | 2 . Taking another partial with respect to y leads to fyy (fy) 2 f 2 = X J j=1 1 (z rj ) 2 + 1 (z rj ) 2 = 2X J j=1 <(z rj ) 2 |z rj | 4 . If all of the rj are real we have fyy f = 4y 2 X J j=1 1 |z rj | 2 !2 + 2X J j=1 (x rj ) 2 |z rj | 4 2y 2X J j=1 1 |z rj | 4 . The middle term is clearly positive and the first term is clearly greater than 4y 2 PJ j=1 1 |z rj | 4 which is twice the third term in absolute value. Thus the condition is necessary. The second equivalent to RH is a consequence of the fact that if for each real x the function f(x, y) = |F(x + iy)| 2 is expanded into a power series in y then all of the coefficients should be non-negative. To see this, again for polynomials, let the notation be as above. We have y n fy f y=0 = i n 1n! X J j=1 1 (z rj ) n+1 + ( 1)n+1 (z rj ) n+1 . RIEMANN S HYPOTHESIS 19 Now f is even in y so fy/f is odd in y. Thus (fy/f) (n) |y=0 is 0 when n is even. For odd n we have (fy/f) (n) |y=0 = 2( 1)(n 1)/2n! X J j=1 (x rj ) n 1 ; (we have used the fact that each rj has a conjugate that is also a root). Suppose that all of the rj are real. Letting k = k(x) = PJ j=1(x rj ) k , we are led to fyy f = 2 2 = 2!E1 f (4) f = 12( 2 2 4) = 4!E2 f (6) f = 120(2 6 + 3 2 3 2 4) = 6!E3 f (8) f = 1680( 6 8 + 4 2 6 2 2 4 + 3 2 4 + 8 2 6) = 8!E4 where En = En(x) is the nth elementary symmetric function of the (x rj ) 2 . Thus we see that n yn f(x, y) 0 for each n and all x in the case of all real roots rj . The final equivalence is a consequence of the assertion that if F(z) = a0 + a1 1! z + a2 2! z 2 + . . . and if Fn(z) := a0z n + n 1 a1z n 1 + n 2 a2z n 2 + + an, then for all real x and n = 1, 2, . . . the inequality Fn 2 (x) Fn 1(x)Fn+1(x) > 0 holds. The application of these to RH comes about because of the formulae | (z)| 2 = Z Z ( ) ( )e i( + )x e ( )y d d = X n=0 y 2n (2n)! Z Z ( ) ( )e i( + )x ( ) 2n d d and n(z) = Z (u)(z + iu) n du. Note, for example, the third equivalence with n = 2 implies that if RH is true then it must be the case that b0b1X 4 + (3b 2 1 b0b2)X 2 + b1b2 > 0 20 BRIAN CONREY for all real X where we are using the notation bn = Z (u)u 2n du from above. This inequality holds in turn if the discriminant of the quadratic in X2 is negative: 9b 4 1 10b 2 0 b1b2 + b 2 0 b 2 2 < 0 i.e. (9b 2 1 b0b2)(b 2 1 b0b2) < 0. A consequence is that b0b2 < 9b 2 1 . (The Turan inequalities, see below, imply that 3b 2 1 > b0b2, that 5b 2 2 > 3b1b3, that 7b 2 3 > 5b2b4 etc. and Cauchy s inequality implies that b 2 n bn abn+a for a = 1, 2, . . . , n so in particular 3b 2 1 > b0b2 > b2 1 . In fact it is easily calculated that the ratio b0b2 b 2 1 = 2.79 . . . . Note that the Karlin-Nuttall inequality below would have this ratio smaller than 6. ) For n = 3 the Jensen inequality implies that b0b1X 6 + (6b 2 1 3b0b2)X 4 + 3b1b2X 2 + b 2 2 > 0 for all X. The discriminant of this cubic in X2 is 746496b0b 6 2 b 3 0 b 3 2 7b 2 0 b 2 1 b 2 2 + 11b0b 4 1 b2 5b 6 1 2 < 0 so that the cubic has only one real root. Since the value at x = 0 is positive, the real root is negative and so the third Jensen inequality is always true. For n = 4 the condition is b0b1X 8 + (10b 2 1 6b0b2)X 6 + (5b1b2 + b0b3)X 4 + (10b 2 2 6b1b3)X 2 + b2b3 > 0 for all X. 11. Grommer inequalities In 1914 Grommer [Gro14] found a necessary and sufficient condition for the reality of the zeros of an entire function. We describe how it applies to the Riemann Hypothesis. Let (t) = (1/2 +it) so that RH is the assertion that all zeros of are real. Now the functional equation for is equivalent to the fact that (t) is even. Let Y (t) = ( t) and let Y 0 Y (t) = s1 + s2t + s3t 2 + . . . . Then RH is equivalent to the assertion that for each n, Dn = det s2 s3 . . . sn+1 s3 s4 . . . sn+2 . . . . . . . . . sn+1 sn+2 . . . s2n > 0. RIEMANN S HYPOTHESIS 21 The collection of inequalities for n = 1 applied to Y (t) = ( t) and all of its derivatives are sometimes known as the Turan inequalities. Here is a proof of the necessity of Grommer s criterion. First off we consider the polynomial case. Assume that P is a polynomial with P(0) 6= 0. Let P(z) = Yn r=1 (z 1/zr) be a polynomial with real coefficients.We have P 0 P (z) = Xn r=1 1 z 1/zr = Xn r=1 zr 1 zzr = Xn r=1 X m=0 z mzr m+1 so that P 0 P (z) = s1 + s2z + s3z 2 + . . . where sm = Xn r=1 z m r is the sum of the mth powers of the reciprocal roots. Let Dm be the m m Grommer determinant as above. The key observation is that Dn = (z1, . . . , zn) 2Yn r=1 z 2 r where is the Vandermonde determinant for which we have the formula (z1, . . . , zn) = Y 1 i<j n (zj zi). More generally, if m n, then Dm = X Z {z1,...,zn} |Z|=m Y zr Z z 2 r (Z) 2 and Dm = 0 if m > n. (Note that whereas (Z) has an ambiguous sign, the notation (Z) 2 makes sense.) Thus, it is clear that if all of the zr are real, then all of the Dm 0, so Grommer s condition is a necessary condition for the reality of the zeros of P. We can show that the condition is sufficient if there are an odd number of conjugate pairs of non-real zeros. If only one pair, say z1, z2 with z2 = z1 is complex, and all of the rest are distinct reals, then Dn = |z1| 2Yn r=3 z 2 r (z3, . . . , zn) 2 (z1 z2) 2Yn r=3 |zr z1| 2 . 22 BRIAN CONREY All of the factors here are positive with the exception of (z1 z2) 2 = 4(=z1) 2 < 0. Thus, Dn < 0. The same argument works anytime there are an odd number of pairs of complex zeros. If there are an even number of pairs of non-real complex conjugate pairs of zeros, say m of them, then it seems that Dn m < 0 but we don t see how to prove this. Grommer s argument proceeds via the Euler-Stieltjes theory of continued fractions, which study contains the genesis of the theory of orthogonal polynomials. The second set of Grommer inequalities asserts that 10b 2 0 b2b4 21b 2 0 b 2 3 30b0b 2 1 b4 + 350b0b1b2b3 350b0b 3 2 420b 3 1 b3 + 525b 2 1 b 2 2 > 0. 12. Turan inequalities The entire function (t) can be expanded into an everywhere convergent power series: (t) = X n=0 ( 1)n bnt 2n (2n)! where bn = Z (u)u 2n du. Let Y (t) = ( t) = X n=0 ( 1)n bnt n (2n)! . Then Y is entire of order 1/2 and the Riemann Hypothesis implies that all of its zeros are real, and in addition, that all of the zeros of all derivatives Y (m) (t) are real. From the Grommer inequalities, a necessary condition for all of the zeros of Y (t) to be real is that s2 > 0 where Y 0 Y (t) = s1 + s2t 2 + s3t 3 + . . . ; in other words Y 0 Y 0 (0) < 0. Thus, RH implies that Y (m+1) Y (m) 0 (0) < 0 for m = 0, 2, 4, . . . . It is easy to check that this condition translates to b 2 m > 2m 1 2m + 1 bm 1bm+1 (m = 1, 2, . . .); these are known as the Turan inequalities and give a necessary but not sufficient condition for the reality of all of the zeros of (t). Matiyasevich [Mat82] and Csordas, Norfolk, and RIEMANN S HYPOTHESIS 23 Varga [CNV86] proved the Turan inequalities for . Conrey and Ghosh [CoGh94] considered these for the function associated with the Ramanujan -function. In conjunction with this, they show Theorem 3. Let F C 3 (R). Let F(u) be positive, even, and decreasing for positive u, and suppose that F 0/F is decreasing and concave for u > 0. Suppose that F is rapidly decreasing so that X(t) = Z F(u)e itu du is an entire function of t. Then X(t) satisfies the Tur an inequalities. 12.1. Karlin and Nuttall. We let (u) be Riemann s function as earlier. We let (t) = X n=0 ( 1)n bn (2n)!t 2n where bn = Z (u)u 2n du as before. Define B(i, j) = bj i (2j 2i)! if i j 0 if i > j Then RH is equivalent to D(n, r) > 0 for all positive r and non-negative n where D(n, r) = det r r B(i, j + n)|i,j=1,r (see [Kar68] chapter 8). The case r = 1 here is clear since F(u) > 0. The case r = 2 is slightly weaker than the Turan inequalities; it asserts that b 2 m > m m + 1 (2m 1) (2m + 1) bm 1bm+1. Nuttall [Nut13] has established the case r = 3 which asserts that b 3 m ((2m)!)3 2bm 1bm+1bm (2m)!(2m 2)!(2m + 2)! bm 2bm+2bm (2m)!(2m 4)!(2m + 4)! + b 2 m 1 bm+2 ((2m 2)!)2 (2m + 4)! + bm 2b 2 m+1 (2m 4)!((2m + 2)!)2 > 0 for all m 2. For m = 2 this is b 3 2 4 5 b1b3b2 1 70 b0b4b2 + 2 75 b0b 2 3 + 3 35 b 2 1 b4 > 0. 24 BRIAN CONREY 13. Turan inequalities, 2 Ramanujan s tau-function may be defined by equating coefficients of the power series on both sides of X n=1 (n)x n = x Y n=1 (1 x n ) 24 The associated Dirichlet series is L(s) = L (s) = X n=1 (n)n s This series is absolutely convergent for = 13/2. The xi-function for is given by (s) = (2 ) s (s)L(s) and it satisfies the functional equation (s) = (12 s). This functional equation is equivalent to the fact that (z) = X n=1 (n)e(nz) is a holomorphic cusp form of weight 12 for the full modular group which in turn is equivalent to: (i) (z) is expressible in terms of a Fourier series in z in which coefficients of e(nz) with n 0 vanish and (ii) satisfies the transformation formula ( 1/z) = z 12 (z). It is believed that all of the zeros of (s) are on the line <s = 6; this is the Riemann Hypothesis for L . See Hardy [Har78], Chapter X for introductory information about . Now (s) = Z 0 (iy)y s dy y so that (t) = (6 + it) = Z (ieu )e 6u e iut du is an entire even function of t. We define (u) = (ieu )e 6u We see that (u) is an even function of u by the functional equation for . The fact that (u) > 0 for real u is immediately obvious from the product formula for : (u) = e 6u e 2 eu Y n=1 1 e 2 neu 24 RIEMANN S HYPOTHESIS 25 We can also see that (u) is decreasing for positive u by calculating the logarithmic derivative. We first observe that y d dy X n=1 log(1 y n ) = y d dy X m,n=1 y mn m = X m,n=1 nymn = X n=1 (n)y n where (n) = X d|n d is the sum of divisors of n. Let x = 2 eu and y = e x . Then (u) = e 6u y Y n=1 (1 y n ) 24 so that 0 (u) = 6 + 1 y 24X n=1 (n)y n ! dy du = 6 x(1 0(x)) where k(x) = 24X n=1 n k (n)y n (The expansion of 0 / above is related to the Fourier expansion of the Eisenstein series E2: E2(z) = 1 24X n=1 (n)e(nz) E2 is not a modular form of weight 2; it transforms according to the formulae E2( 1/z) = z 2E2(z) + 12z 2 i and E2(z + 1) = E2(z). Note also that P(y) = E2(e 2y ) satisfies the Chazy equation P 000 2P P00 + 3(P 0 ) 2 = 0; the Chazy equation is related to a Painlev e equation.) Now 0 / is an odd function of u so that 0 / (0) = 0. Thus, to show that 0 (u) < 0 for u > 0 it suffices to prove that 0 0 (u) < 0 26 BRIAN CONREY for u > 0. But 0 0 (u) = ( 1 + 0(x) + x 0 0 (x)) dx du = x + x 0(x) x 2 1(x) since 0 k (x) = k+1(x) and this then is = x 1 24X n=1 (n)y n (1 nx) ! Since u 0 corresponds to x 2 each of the terms 1 nx < 0 so that the whole expression is negative. Arguing in the same way we see that 0 00 (u) is odd and 0 000 (u) = x(x 3 3(x) 6x 2 2(x) + 7x 1(x) 0(x) + 1) = x(1 24X n=1 (n)y nP3(nx)) where P3(x) = 1 7x + 6x 2 x 3 < 0 for x > 6. Thus we conclude that 0 00 (u) < 0 for u > 0, i.e. that 0 is concave for u > 0; see [CoGh94] for more details. 13.1. A difficulty with classifying functions whose Fourier transforms have real zeros. Let g(u) = u 4 + eu 12 + X n=1 1(n)e 2 neu . Here 1(n) = P d|n d 1 is the sum of the reciprocals of the positive divisors of n. Then g(u) is positive, even, decreasing, and its logarithmic derivative is decreasing and concave for u > 0. So k(t) = Z e kg(u) e iut du might seem to be a good candidate for a function to have only real zeros. In fact k = 24 is the case we ve just been discussing about the Ramanujan tau-function. And k = 1 corresponds to the Xi-function associated with the Dirichlet L-function associated to the unique primitive character of modulus 12, and so all of its zeros should be rea // Extend manually.
Riemann, in 1859, in a paper [R] written on the occasion of his admission to the Berlin Academy of Sciences and read to the Academy by none other than Encke, devised an analytic way to understand the error term in Gauss approximation, via the zeros of the zeta-function (s) = X n=1 1 ns . The connection of (s) with prime numbers was found by Euler via his product formula X n=1 1 ns = Y p (1 p s ) 1 where there is one factor for each prime number p. This formula encodes the fundamental theorem of arithmetic that every integer is a product of primes in a unique way. Riemann saw that the zeros of what we now call the Riemann zeta-function were the key to an analytic expression for (x). Riemann observes that s/2 (s/2) (s) = Z 0 (x)x s 1 dx where (x) = X n=1 e n2x and then uses 2 (x) + 1 = x 1 2 2 1 x + 1 , which follows from a formula of Jacobi, to transform the part of the integral on 0 x 1. In this way he finds that (s) is a meromorphic function of s with its only a pole a simple pole at s = 1 and that (s) = 1 2 s(s 1) s/2 (s/2) (s) is an entire function of order 1 which satisfies the functional equation (s) = (1 s). As a consequence, (s) has zeros at s = 2n for n Z+, these are the so called trivial zeros, as well as a denser infinite sequence of zeros in the critical strip 0 <s 1. The Euler product precludes any zeros with real part larger than 1. Also, for any non-trivial zero = +i there is a dual zero 1 by the functional equation. Also, and 1 are zeros since (s) is real for real s but note that coincides with 1 whenever = 1/2. Riemann 4 BRIAN CONREY used Stirling s formula and the functional equation to evaluate the number of non-trivial zeros in the critical strip as N(T) := #{ = + i : 0 < T} = T 2 log T 2 e + 7/8 + S(T) + O(1/T) where S(T) = 1 arg (1/2 + iT) where the argument is determined by beginning with arg (2) = 1 and continuous variation along line segments from 2 to 2 + iT and then to 1/2 + iT, taking appropriate action if a zero is on the path. Riemann implied that S(T) = O(log T) a fact that was later proven rigorously by Backlund [Bac18]. Thus, the zeros get denser as one moves up the critical strip. The functional equation together with Riemann s formula for the number of zeros of (s) up to a height T help give us a picture of (1/2 + it). In particular Hardy defined a function Z(t) which is a real function of a real variable having the property that | (1/2+it)| = |Z(t)|. It may be defined by Z(t) = (1/2 it) 1/2 (1/2 + it) where (s) is the factor from the functional equation which may be written in asymmetric form as (1 s) = 2(2 ) s (s) cos s 2 . Here are some graphs of Z(t): 10 20 30 40 50 -3 -2 -1 1 2 3 ZHtL for 0<t<50 1010 1020 1030 1040 1050 -8 -6 -4 -2 2 4 6 ZHtL for 1000<t<1050 1.1. Riemann s formula for primes. Riemann found an exact formula for (x). If we invert Euler s formula we find 1 (s) = Y p (1 p s ) = 1 2 s 3 s 5 s + 6 s + = X n=1 (n) ns where is known as the M obius mu-function. A simple way to explain the value of (n) is that it is 0 if n is divisible by the square of any prime, while if n is squarefree then it is +1 if n has an even number of prime divisors and 1 if n has an odd number of prime divisors. Riemann s formula is (x) = X n=1 (n) n f(x 1/n) RIEMANN S HYPOTHESIS 5 where f(x) = li(x) X li(x ) ln 2 + Z x dt t(t 2 1) log t . Here the = +i are the zeros of (s) and the sum over the is to be taken symmetrically, i.e. to pair the zero with its dual 1 as the sum is performed. Thus the difference between Riemann s formula and Gauss conjecture is, to a first estimation, about li(x 0 ) where 0 is the largest or the supremum of the real parts of the zeros. Riemann conjectured that all of the zeros have real part = 1/2 so that the error term is of size x 1/2 log x. This assertion of the perfect balance of the zeros, and so of the primes, is Riemann s Hypothesis. In 1896 Hadamard and de la Vall ee Poussin independently proved that (1 + it) 6= 0 and concluded that (N) li(N) a theorem which is known as the prime number theorem. 2. Riemann and the zeros After his evaluation of N(T) T 2 log T he asserted that we find about this many real zeros of (t) := (1/2 + it) in 0 < t T. This is an assertion which is still unproven and is the subject of speculation. His memoir Ueber die Anzahl der Primzahlen unter einer gegebenen Gr osse is only 8 pages. But in the early 1930s his Nachlass was delivered from the library at G ottingen to Princeton where C. L. Siegel [?] looked over Riemann s notes at the Institute for Advanced Study. In the notes were found an approximate functional equation, which had been independently found by Hardy and Littlewood [HL29]: (s) = X n t 2 1 ns + (s) X n t 2 1 n1 s + O(t /2 ) for s = + it. Here (s) is the factor from the asymmetric form of the functional equation (s) = (s) (1 s) with (s) = (1 s)/2 ((1 s)/2) s/2 (s/2) = 2(2 ) s 1 (1 s) sin s 2 . Now | (1/2 + it)| = 1; in fact (1/2 + it) = e itlog t/2 (1 + O(1/t)). One might be led to believe that 1 + (s) is a reasonable approximation to (s), i.e. that the contributions from the oscillatory terms 2 s etc. might be small overall. This approximation has zeros on s = 1/2 + it at a rate sufficient to produce asymptotically all of the zeros of (s), so it seems reasonable to conclude that almost all of the zeros are on this line, and to go on and conjecture that ALL of the zeros are on the one-line. But we have found it hard to make this reasoning precise. 6 BRIAN CONREY Riemann computed the first few zeros: 1/2 + i14.13 . . . , 1/2 + i21.02 . . . , 1/2 + i25.01 . . . , . . . A good way to be convinced that these are indeed zeros is to use the easily proven formula (1 2 1 s ) (s) = 1 1 2 s + 1 3 s 1 4 s . . . . The alternating series on the right converges for 0 and so, for example, s = 1/2 + i14.1347251417346937904572519835624 . . . can be substituted into a truncation of this series (using a computer algebra system) to see that it is very close to 0. (See www.lmfdb.org to find a list of high precision zeros of (s) as well as a wealth of information about (s) and similar functions called L-functions.) 3. Elementary equivalents of the Riemann Hypothesis We ve mentioned that the Riemann Hypothesis implies a good error bound for the prime number theorem. The converse is also true: the Riemann Hypothesis is equivalent to (x) := X p x 1 = Z x 2 du log u + O(x 1/2 log x), and to (x) := X p k x log p = x + O(x 1/2 log2 x) Equivalences may also be phrased in terms of the M obius function (n) where 1 (s) = X n=1 (n) ns . It is not difficult to show that the Riemann Hypothesis is equivalent to the assertion that this series is (conditionally) convergent for any s with 1/2 < < 1. The Riemann Hypothesis is also equivalent to each of M(x) := X n x (n) = O(x 1/2+ ) and Z X 1 ( (x) x) 2 dx x 2 C log X. The assertion that Z X 1 M(x) 2 dx x 2 C log X implies the Riemann Hypothesis and that all of the zeros are simple. A question is whether the converse is true. RIEMANN S HYPOTHESIS 7 Stieltjes thought that he had found a proof that M(x) = O(x 1/2 ) and so of the Riemann Hypothesis. His claim appeared in Comptes Rendus Mathematique. Consequently de la Hadamard was somewhat apologetic about his inconsequential offering in his own paper [Had96] which proves the prime number theorem! Figure 1. A plot of M(x) versus x 4. The general distribution of the zeros An immediate consequence of Euler s product formula (s) = Y p 1 p s 1 is that (s) 6= 0 if 1. A subsequent consequence of Riemann s functional equation is that (s) 6= 0 if <s < 0 except at s = 2, 4, 6, . . . , the so-called trivial zeros. The prime number theorem (x) x log x is equivalent to the assertion that (1 +it) 6= 0; equivalently (it) 6= 0. In order to be precise about the error term in the prime number theorem it is necessary to prove that there is a 8 BRIAN CONREY region near the line = 1 in which there are no zeros. It was shown by de la Vall ee Poussin in 1899 [Val96] that ( + it) 6= 0 for > c log(2+|t|) for a specific c. This is known as a zero-free region. The best known shape of the zero-free region is due to Korobov [Kor58] and Vinogradov [Vin58] in 1958: ( + it) is free of zeros when > 1 C (log t) 2/3 (log log t) 1/3 . The best explicit value of C is due to Kevin Ford [For00] who showed that C = 1/54.57 is admissible. 4.1. Density results. Bounds for the quantity N( , T) := #{ = + i : and 0 < T} are known as density estimates. Near to = 1 we have [For00] N( , T) T 58.05(1 ) 3/2 (log T) 15 . As we move away from the line = 1 our estimates get weaker but are still pretty good. Bounds often take the shape N( , T) T k( )+ ; there are many forms of admissible k( ). A strong classical one due to Ingham in 1940 [Ing40] is that N( , T) = O(T 3(1 )/(2 ) log5 T); this is still the best bound when 1/2 < < 3/4. It is known that k( ) = 3/2 is also admissible. The unproven Density Hypothesis is that the above holds with k( ) = 2(1 ). It is known that an estimate of the sort (1/2 + it) t c (log t) c 0 implies that N( , T) T 2(1+2c)(1 ) log5 T; see [Tit86]. Thus, the Density Hypothesis is a consequence of the Lindel of Hypothesis (for which see below). A consequence of the Density Hypothesis is that for any > 0 there is a C( ) such that pn+1 pn C( )n 1/2+ where pn denotes the nth prime. This estimate is not quite strong enough to conclude that there is always a prime between consecutive squares. Here is a plot of the exponent in density theorems (the minimum of the two graphs is an admissible density exponent): RIEMANN S HYPOTHESIS 9 0.5 0.6 0.7 0.8 0.9 1.5 2.5 3 3.5 4 4.5 Density exponent kHsigmaL 4.2. Zeros near the 1/2-line. It has been known for quite some time that almost all of the zeros are near the 1/2-line. For example at least 99% of the zeros = + i satisfy | 1/2| < 8 log . and almost all are within ( )/ log of the critical line where is any function which goes to infinity. Thus, we know that the zeros cluster around the critical line. 4.3. Zeros on the critical line. Many people have worked on verifying the Riemann Hypothesis. Today it is known that the first ten trillion zeros are all on the critical line <s = 1/2! Hardy was the first one to show that there are infinitely many zeros on the 1/2-line. He and Littlewood [HL18] later gave proofs that the number of zeros on the 1/2-line up to a height T is more than a positive constant times T. In 1942 Selberg [Sel42] proved that a positive proportion of the zeros are on the critical line. In 1973 N. Levinson [Lev74] proved that at least 1/3 of the zeros are on the half-line. This was improved in 1989 to at least 2/5 of the zeros are on the line. The current record is Feng [Fen12] with 0.412; for simple zeros the record proportion is due to Bui, Conrey, and Young [BCY11] who show that at least 0.405 of the zeros of (s) are on the critical line and simple. It follows from the Riemann Hypothesis that all of the zeros of all of the derivatives (k) (s) are on the critical line. Along these lines it can be shown, for example, that more than 4/5 of the zeros of 0 (s) are on the critical line and more than 99% of the zeros of (5)(s) are on the critical line, see [Con83]. 5. The Lindelof Hypothesis The assertion that for any > 0, (1/2 + it) t is known as the Lindel of Hypothesis and is a consequence of the Riemann Hypothesis. It is a consequence of the functional equation, trivial bounds for (it) and (1 + it), and general 10 BRIAN CONREY principles of the growth of analytic functions that (1/2 + it) t 1 4 + ; this is known as the convexity bound. Weyl, using exponential sums, improved the bound to (1/2 + it) t 1 6 + . Bombieri and Iwaniec [BI86] used some novel ideas to show (1/2 + it) t 89/560+ . Huxley [Hux05] obtained (1/2 + it) t 32/205 logc t. Recently Bourgain [Bou14] announced that (1/2 + it) t 53/342+ 5.1. Estimates for (s) near the 1-line. Richert [Ric67] proved the important estimate that for an explicit c > 0, | ( + it)| < ct100(1 ) 3/2 log2/3 t for 1/2 1, t 2. Such a bound is useful for zero-free regions, the error term in the prime number theorem, and zero density results near 1. K. Ford [For02] has improved these made the constants explicit: | ( + it)| < 76.2t 4.45(1 ) 3/2 log2/3 t. 5.2. 1 versus 2. RH implies that | (1/2 + it)| exp log 2 2 log t log log t + O log tlog log | log t (log log t) 2 see [ChS11] . It can be proven, see [Sou08] that every interval [T, 2T] contains a t for which | (1/2 + it)| exp (1 + o(1)) (log t) 1/2 (log log t) 1/2 . Which of these is closer to the true largest order of magnitude of on the 1/2-line? It is difficult to say, though most people (not the author!) think that the lower bound ( -result) is closer to the truth. Farmer, Gonek, and Hughes [FGH07] conjecture that | (1/2 + it)| exp s 1 2 + o(1) (log t)(log log t) ! . RIEMANN S HYPOTHESIS 11 6. Computations Turing was the first to use a computer to calculate the zeros of (s). He proposed an efficient rigorous method to verify RH up to a given height, or indeed within an interval. It involves using a precise version of the approximate functional equation, known as the Riemann - Siegel formula, to evaluate Z(t) and detect sign changes, together with an explicit bound for the average of S(t) namely if t2 > t1 > 168 then Z t2 t1 S(t) dt = 2.30 + 0.128 log t2 2 to verify that all of the zeros are accounted for. (Here represents a number that is at most 1 in absolute value.) Goldfeld has pointed out that if (s) had a double zero somewhere up the line, the computational verification of RH would come to a halt because it would be impossible to distinguish a double zero from two very close zeros either on or off the line. Here is one of Turing s versions of the Riemann-Siegel formula: Theorem 1. Let m and be respectively the integral and non-integral parts of 1/2 and 64, ( ) = 1 4 i log ( 1 4 + i ) ( 1 4 i ) 1 4 log , Z( ) = (1/2 + 2 i )e 2 ( ) , 1( ) = 1 2 ( log 1 2 ), h( ) = cos 2 ( 2 1 16 ) cos 2 . Then Z( ) is real and Z( ) = 2Xm n=1 n 1 2 cos 2 { log n ( )} + ( 1)m+1 1 4 h( ) + (1.09 3 4 , ( ) = 1( ) + (0.006 1 ). In 1988, Andrew Odlyzko and Sch onhage [OS88] invented an algorithm which allowed for the very speedy calculation of many values of (s) at once. The Riemann-Siegel allows for a single computation of (1/2 + it) with T < t < T + T 1/2 in time T 1/2+ . The Odlyzko - Sch onhage algorithm allows for a single computation in time T after a pre-computation of time T 1/2+ . This led Odlyzko to compile extensive statistics about the zeros at enormous heights - up to 1023 and higher. His famous graphs showed an incredible match between data for zeros of (s) and for the proven statistical distributions for random matrices. Here is a list of contributors to verifying RH in an initial segment of the 1/2-line and the year they did the work. 12 BRIAN CONREY G. H. B. Riemann 3 1859 J. P. Gram 15 1903 R. J. Backlund 79 1914 E. C. Titchmarsh 1041 1935 A. M. Turing 1104 1953 D. H. Lehmer 15000 1956 D. H. Lehmer 25000 1956 N. A. Meller 35337 1958 R S. Lehman 250000 1966 J. B. Rosser , J. M. Yohe, L. Schoenfeld 3500000 1968 R. P. Brent 40000000 1977 R. P. Brent 81000000 1979 R. P. Brent, J. van de Lune, H. J. J. te Riele, D. T. Winter 200000001 1982 J. van de Lune, H. J. J. te Riele 300000001 1983 J. van de Lune, H. J. J. te Riele, D. T. Winter 1500000001 1986 J. van de Lune 10000000000 2001 S. Wedeniwski 900000000000 2004 X. Gourdon, P. Demichel 10000000000000 2004 Ghaith Hiary [Hia11] has improved these algorithms. He can compute one value of (1/2+ it) in time T 1/3+ using an algorithm that has been implemented by Jonathan Bober and Hiary; he has a more complicated algorithm that will work in time T 4 13 + . They have verified RH in some small ranges around the 1033 zero! Bober s website [Bob14] has some great pictures of large values of Z(t). 7. Why do we think RH is true? The main reason is because of the beauty of the conjecture. It strikes our sensibilities as appropriate that something so incredibly symmetric should be true in mathematics. The second reason is that the first 10 trillion zeros are all on the line. If there were a counterexample it should have shown itself by now. A third reason is that the numerical evidence for all L-functions ever computed lead to this conclusion; some have thought that a counterexample to RH might show itself when computing zeros of L-functions associated with Maass forms because these have no arithmetic-geometry interpretation (eg. their coefficients are generally believed to be transcendental); however the computations reveal that the zeros are still on the 1/2-line. A fourth reason is probabilistic. RH is known to be equivalent to the assertion that M(x) := P n x (n) x 1/2 log2 x. This sum represents the difference between the number of squarefree integers up to x with an even number of prime factors and the number with an odd number of prime factors. It is similar to the difference in the number of heads and tails when one flips x coins, and so should be around x 1/2 . Here is another more elaborate reason. Suppose that a Dirichlet series F(s) = P n=1 ann s converges for > 0, and suppose that it has a zero with real part > 1/2. We might reasonably expect it then to have T zeros in > , 0 < t < T for any large T by RIEMANN S HYPOTHESIS 13 almost periodicity. But zero density results tell us that there are T 1 zeros in 0 and t < T. 7.1. Almost periodicity. As just mentioned a possible strategy is to try to prove that if (s) has one zero off the line then it has infinitely many off the line. Bombieri [Bom00] has come closest to achieving this. Here is a conjecture that attempts to encapsulate this idea: Conjecture 1. Suppose that the Dirichlet series F(s) = X n=1 ann s converges for > 0 and has a zero in the half-plane > 1/2. Then there is a number CF > 0 such that F(s) has > CF T zeros in > 1/2, |t| T. This seemingly innocent conjecture implies the Riemann Hypothesis for virtually any primitive L-function (except curiously possibly the Riemann zeta-function itself!). And it seems that the Euler product condition has already been used (in the density result above); i.e. the hard part is already done. Note that the 1/2 in the conjecture needs to be there as the example X n=1 (n)/n1/2 ns demonstrates. Assuming RH, this series converges for > 0 and its lone zero is at s = 1/2. This example is possibly at the boundary of what is possible. 8. A spectral interpretation Hilbert and P olya are reputed to have suggested that the zeros of (s) should be interpreted as eigenvalues of an appropriate operator. Odlyzko wrote to P olya to ask about this. Here is the text of Odlyzko s letter, dated Dec. 8, 1981. Dear Professor P olya: I have heard on several occasions that you and Hilbert had independently conjectured that the zeros of the Riemann zeta function correspond to the eigenvalues of a self-adjoint hermitian operator. Could you provide me with any references? Could you also tell me when this conjecture was made, and what was your reasoning behind this conjecture at that time? The reason for my questions is that I am planning to write a survey paper on the distribution of zeros of the zeta function. In addition to some theoretical results, I have performed extensive computations of the zeros of the zeta function, comparing their distribution to that of random hermitian matrices, which have been studied very seriously by physicists. If a hermitian operator associated to the zeta function exists, then in some respects we might expect it to behave like a random hermitian operator, which in turn ught to resemble a random hermitian matrix. I have discovered that the distribution of zeros of 14 BRIAN CONREY the zeta function does indeed resemble the distribution of eigenvalues of random hermitian matrices of unitary type. Any information or comments you might care to provide would be greatly appreciated. Sincerely yours, Andrew Odlyzko and P olya s response, dated January 3, 1982. Dear Mr. Odlyzko, Many thanks for your letter of Dec. 8. I can only tell you what happened to me. I spent two years in G ottingen ending around the beginning of 1914. I tried to learn analytic number theory from Landau. He asked me one day: You know some physics. Do you know a physical reason that the Riemann Hypothesis should be true? This would be the case, I answered, if the non-trivial zeros of the function were so connected with the physical problem that the Riemann Hypothesis would be equivalent to the fact that all the eigenvalues of the physical problem are real. I never published this remark, but somehow it became known and it is still remembered. With best regards. Yours sincerely, George, P olya 9. The vertical spacing of zeros In the 1950s physicists predicted that excited nuclear particles emit energy at levels which are distributed like the eigenvalues of random matrices. This was verified experimentally in the 1970s and 1980s; Oriol Bohigas was the first to put this data together in a way that demonstrated this law. Figure 2 shows 96 zeros of (s) starting at a height T = 1200 wrapped once around a circle for the purposes of comparing with the eigenvalues of a randomly chosen 96 96 unitary matrix, and with 96 points chosen randomly independently on a circle (Poisson). It should be clear that the zeros of (s) do not have a Poisson distribution (and would have been clear to anyone looking carefully at them, say in the mid 1900 s!). In 1972 Hugh Montgomery, then a graduate student at Cambridge, delivered a lecture at a symposium on analytic number theory in St. Louis, outlining his work on the spacings between zeros of the Riemann zeta-function. This was the first time anyone had considered such a question. On his flight back to Cambridge he stopped over in Princeton to show his work to Selberg. At afternoon tea at the Institute for Advanced Study, Chowla insisted that Montgomery meet the famous physicist - and former number theorist - Freeman Dyson. When Montgomery explained to Dyson the kernel he had found that seemed to govern the spacings of pairs of zeros, Dyson immediately responded that it was the same kernel that RIEMANN S HYPOTHESIS 15 (a) Unitary (b) Poisson (c) -zeros Figure 2. 96 points of three different types of spacings governs pairs of eigenvalues of random matrices. Montgomery [Mon73] proved that X 1, 2 [0,T] w( 1 2)f log T 2 ( 1 2) = T log T 2 f(0) + Z f(u) " 1 sin( u) u 2 # du + o(1)! assuming the Riemann Hypothesis and that the Fourier transform f of f vanishes outside of [ 1, 1] and w(x) = 4/(4 + x 2 ). The sum here is over pairs of zeros 1/2 + i 1 and 1/2 + i 2. The conjecture is that the assumption on the support of f is not necessary. Odlyzko did extensive numerical calculations to test this conjecture; the numerics are stunning! The pair-correlation function in Figure 3 is 1 sin x x 2 . The nearest-neighbor density function is more complicated. It may be given as 1 4 d 2 dt2 exp Z t 0 (2u) u du where = (s) is a solution of a Painlev e equation: (s 00) 2 + 4(s 0 ) ( 0 ) 2 + s 0 = 0 with a boundary condition (s) s s 2 2 as s 0, as discovered by Jimbo, Miwa, Mori, and Sato, see [JMMS80]. 16 BRIAN CONREY (a) Pair correlation (b) Nearest neighbor Figure 3. Odlyzko s graphics Now we have the challenge of not only explaining why all of the zeros are on a straight line, but also why they are distributed on this line the way they are! The connections with Random Matrix theory first discovered by Montgomery and Dyson have received a great deal of support from seminal papers of Katz and Sarnak [KaSa99] and Keating and Snaith [KS00]. The last 15 years have seen an explosion of work around these ideas. In particular, it definitely seems like there should be a spectral interpretation of the zeros `a la Hilbert and P olya. 10. Some initial thoughts about proving RH 10.1. Fourier integrals with all real zeros. Riemann proved that (t) := (1/2 + it) = Z (u)e iut du where (u) = X n=1 (4 2n 4 e 9u/2 6n 2 e5u/2 ) exp( n 2 e2u ) It is known that (u) is even, is positive for real u and is (rapidly!) decreasing for u > 0. Consequently, we can write (t) = 2 Z 0 (u) cos ut du = X n=0 ( 1)n bn (2n)! t 2n RIEMANN S HYPOTHESIS 17 where bn := Z (u)u 2n du. The Riemann Hypothesis is the assertion that all of the zeros of (t) are real. This has prompted investigations into Fourier integrals with all real zeros. Polya [Pol27] and deBruijn [deB50] spent a lot of time with such investigations. A sample theorem is Theorem 2. Let f(u) be an even nonconstant entire function of u, f(u) 0 for real u, and such that f 0 (u) = exp ( u2 )g(u), where 0 and g(u) is an entire function of genus 1 with purely imaginary zeros only. Then (z) = R exp { f(u)}e izudt has real zeros only. Now (u) > 0 for all real u and 0 (u) < 0 for u 0. Thus, we can write (u) = e f(u) . The functional equation for (s) is equivalent to the assertion that (u) is even. In particular, it was shown by P olya [Pol26] that all of the zeros of the Fourier transform of a first approximation (u) to (u) (u) = (2 cosh(9u/2) 3 cosh 5u/2) exp( 2 cosh 2u) are real. These ideas have been further explored by deBruijn, Newman [New76], Hejhal, Haseo Ki [KK02], [KK03] and others. Hejhal [Hej90] has shown that almost all of the zeros of the Fourier transform of any partial sum of (u) are real. A goal of this approach is to determine necessary and sufficient conditions that describe the Fourier transform of a function all of whose zeros are real. 10.2. Jensen s inequalities. In section 14.32 of [Tit86], we find the assertion that RH is equivalent to Z Z ( ) ( )e i( + )x e ( )y ( ) 2 d d 0 for all real x and y where (u) is as in the last section. We quote a passage from P olya s collected works, volume I, page 427, written by M. Marden commenting on the paper of P olya. In this paper Professor P olya reports his findings on examining the Nachlass of the Danish mathematician J. L. W. V. Jensen who died in 1925. Fourteen years earlier Jensen had announced that he would publish a paper regarding his algebraic-function theoretic research on the Riemann -function. In view of Jensen s well-known interest in the zeros of polynomials and entire functions, expectations were high that Jensen would contribute to the solution of the Riemann hypothesis problem regading the zeros of the -function. However, this paper was never published, and so on Jensen s death it was a matter of paramount importance to have his papers examined by an expert in this area. Professor P olya undertook this task, but after an arduous examination he found no clue to any progress that Jensen may have made towards the Riemann hypothesis. 18 BRIAN CONREY Professor P olya does sketch Jensen s algebraic-function-theoretic investigations, many of which were advanced considerably by P olya s own work. In this paper, P olya gives two more necessary and sufficient conditions for RH. RH is equivalent to Z Z ( ) ( )e i( + )x ( ) 2n d d 0 for all real values of x and n = 0, 1, 2, . . . ; and finally RH is equivalent to Z Z ( ) ( )(x + i ) n (x + i ) n ( ) 2 d d 0 for all real values of x and n = 0, 1, 2, . . . . P olya points out that the first equivalence to RH follows immediately from the more general theorem that all of the zeros of a real entire function F(z) of genus at most 1 are real if and only if 2 y2 |F(z)| 2 0 for all z = x + iy. To see that this condition is necessary for polynomials suppose that F(z) = QJ j=1(z rj ) and let f(x, y) = |F(z)| 2 . Then log f = PJ j=1 log(z rj ) + log(z rj ) so that fy f = X J j=1 i z rj i z rj = 2X J j=1 =(z rj ) |z rj | 2 . Taking another partial with respect to y leads to fyy (fy) 2 f 2 = X J j=1 1 (z rj ) 2 + 1 (z rj ) 2 = 2X J j=1 <(z rj ) 2 |z rj | 4 . If all of the rj are real we have fyy f = 4y 2 X J j=1 1 |z rj | 2 !2 + 2X J j=1 (x rj ) 2 |z rj | 4 2y 2X J j=1 1 |z rj | 4 . The middle term is clearly positive and the first term is clearly greater than 4y 2 PJ j=1 1 |z rj | 4 which is twice the third term in absolute value. Thus the condition is necessary. The second equivalent to RH is a consequence of the fact that if for each real x the function f(x, y) = |F(x + iy)| 2 is expanded into a power series in y then all of the coefficients should be non-negative. To see this, again for polynomials, let the notation be as above. We have y n fy f y=0 = i n 1n! X J j=1 1 (z rj ) n+1 + ( 1)n+1 (z rj ) n+1 . RIEMANN S HYPOTHESIS 19 Now f is even in y so fy/f is odd in y. Thus (fy/f) (n) |y=0 is 0 when n is even. For odd n we have (fy/f) (n) |y=0 = 2( 1)(n 1)/2n! X J j=1 (x rj ) n 1 ; (we have used the fact that each rj has a conjugate that is also a root). Suppose that all of the rj are real. Letting k = k(x) = PJ j=1(x rj ) k , we are led to fyy f = 2 2 = 2!E1 f (4) f = 12( 2 2 4) = 4!E2 f (6) f = 120(2 6 + 3 2 3 2 4) = 6!E3 f (8) f = 1680( 6 8 + 4 2 6 2 2 4 + 3 2 4 + 8 2 6) = 8!E4 where En = En(x) is the nth elementary symmetric function of the (x rj ) 2 . Thus we see that n yn f(x, y) 0 for each n and all x in the case of all real roots rj . The final equivalence is a consequence of the assertion that if F(z) = a0 + a1 1! z + a2 2! z 2 + . . . and if Fn(z) := a0z n + n 1 a1z n 1 + n 2 a2z n 2 + + an, then for all real x and n = 1, 2, . . . the inequality Fn 2 (x) Fn 1(x)Fn+1(x) > 0 holds. The application of these to RH comes about because of the formulae | (z)| 2 = Z Z ( ) ( )e i( + )x e ( )y d d = X n=0 y 2n (2n)! Z Z ( ) ( )e i( + )x ( ) 2n d d and n(z) = Z (u)(z + iu) n du. Note, for example, the third equivalence with n = 2 implies that if RH is true then it must be the case that b0b1X 4 + (3b 2 1 b0b2)X 2 + b1b2 > 0 20 BRIAN CONREY for all real X where we are using the notation bn = Z (u)u 2n du from above. This inequality holds in turn if the discriminant of the quadratic in X2 is negative: 9b 4 1 10b 2 0 b1b2 + b 2 0 b 2 2 < 0 i.e. (9b 2 1 b0b2)(b 2 1 b0b2) < 0. A consequence is that b0b2 < 9b 2 1 . (The Turan inequalities, see below, imply that 3b 2 1 > b0b2, that 5b 2 2 > 3b1b3, that 7b 2 3 > 5b2b4 etc. and Cauchy s inequality implies that b 2 n bn abn+a for a = 1, 2, . . . , n so in particular 3b 2 1 > b0b2 > b2 1 . In fact it is easily calculated that the ratio b0b2 b 2 1 = 2.79 . . . . Note that the Karlin-Nuttall inequality below would have this ratio smaller than 6. ) For n = 3 the Jensen inequality implies that b0b1X 6 + (6b 2 1 3b0b2)X 4 + 3b1b2X 2 + b 2 2 > 0 for all X. The discriminant of this cubic in X2 is 746496b0b 6 2 b 3 0 b 3 2 7b 2 0 b 2 1 b 2 2 + 11b0b 4 1 b2 5b 6 1 2 < 0 so that the cubic has only one real root. Since the value at x = 0 is positive, the real root is negative and so the third Jensen inequality is always true. For n = 4 the condition is b0b1X 8 + (10b 2 1 6b0b2)X 6 + (5b1b2 + b0b3)X 4 + (10b 2 2 6b1b3)X 2 + b2b3 > 0 for all X. 11. Grommer inequalities In 1914 Grommer [Gro14] found a necessary and sufficient condition for the reality of the zeros of an entire function. We describe how it applies to the Riemann Hypothesis. Let (t) = (1/2 +it) so that RH is the assertion that all zeros of are real. Now the functional equation for is equivalent to the fact that (t) is even. Let Y (t) = ( t) and let Y 0 Y (t) = s1 + s2t + s3t 2 + . . . . Then RH is equivalent to the assertion that for each n, Dn = det s2 s3 . . . sn+1 s3 s4 . . . sn+2 . . . . . . . . . sn+1 sn+2 . . . s2n > 0. RIEMANN S HYPOTHESIS 21 The collection of inequalities for n = 1 applied to Y (t) = ( t) and all of its derivatives are sometimes known as the Turan inequalities. Here is a proof of the necessity of Grommer s criterion. First off we consider the polynomial case. Assume that P is a polynomial with P(0) 6= 0. Let P(z) = Yn r=1 (z 1/zr) be a polynomial with real coefficients.We have P 0 P (z) = Xn r=1 1 z 1/zr = Xn r=1 zr 1 zzr = Xn r=1 X m=0 z mzr m+1 so that P 0 P (z) = s1 + s2z + s3z 2 + . . . where sm = Xn r=1 z m r is the sum of the mth powers of the reciprocal roots. Let Dm be the m m Grommer determinant as above. The key observation is that Dn = (z1, . . . , zn) 2Yn r=1 z 2 r where is the Vandermonde determinant for which we have the formula (z1, . . . , zn) = Y 1 i<j n (zj zi). More generally, if m n, then Dm = X Z {z1,...,zn} |Z|=m Y zr Z z 2 r (Z) 2 and Dm = 0 if m > n. (Note that whereas (Z) has an ambiguous sign, the notation (Z) 2 makes sense.) Thus, it is clear that if all of the zr are real, then all of the Dm 0, so Grommer s condition is a necessary condition for the reality of the zeros of P. We can show that the condition is sufficient if there are an odd number of conjugate pairs of non-real zeros. If only one pair, say z1, z2 with z2 = z1 is complex, and all of the rest are distinct reals, then Dn = |z1| 2Yn r=3 z 2 r (z3, . . . , zn) 2 (z1 z2) 2Yn r=3 |zr z1| 2 . 22 BRIAN CONREY All of the factors here are positive with the exception of (z1 z2) 2 = 4(=z1) 2 < 0. Thus, Dn < 0. The same argument works anytime there are an odd number of pairs of complex zeros. If there are an even number of pairs of non-real complex conjugate pairs of zeros, say m of them, then it seems that Dn m < 0 but we don t see how to prove this. Grommer s argument proceeds via the Euler-Stieltjes theory of continued fractions, which study contains the genesis of the theory of orthogonal polynomials. The second set of Grommer inequalities asserts that 10b 2 0 b2b4 21b 2 0 b 2 3 30b0b 2 1 b4 + 350b0b1b2b3 350b0b 3 2 420b 3 1 b3 + 525b 2 1 b 2 2 > 0. 12. Turan inequalities The entire function (t) can be expanded into an everywhere convergent power series: (t) = X n=0 ( 1)n bnt 2n (2n)! where bn = Z (u)u 2n du. Let Y (t) = ( t) = X n=0 ( 1)n bnt n (2n)! . Then Y is entire of order 1/2 and the Riemann Hypothesis implies that all of its zeros are real, and in addition, that all of the zeros of all derivatives Y (m) (t) are real. From the Grommer inequalities, a necessary condition for all of the zeros of Y (t) to be real is that s2 > 0 where Y 0 Y (t) = s1 + s2t 2 + s3t 3 + . . . ; in other words Y 0 Y 0 (0) < 0. Thus, RH implies that Y (m+1) Y (m) 0 (0) < 0 for m = 0, 2, 4, . . . . It is easy to check that this condition translates to b 2 m > 2m 1 2m + 1 bm 1bm+1 (m = 1, 2, . . .); these are known as the Turan inequalities and give a necessary but not sufficient condition for the reality of all of the zeros of (t). Matiyasevich [Mat82] and Csordas, Norfolk, and RIEMANN S HYPOTHESIS 23 Varga [CNV86] proved the Turan inequalities for . Conrey and Ghosh [CoGh94] considered these for the function associated with the Ramanujan -function. In conjunction with this, they show Theorem 3. Let F C 3 (R). Let F(u) be positive, even, and decreasing for positive u, and suppose that F 0/F is decreasing and concave for u > 0. Suppose that F is rapidly decreasing so that X(t) = Z F(u)e itu du is an entire function of t. Then X(t) satisfies the Tur an inequalities. 12.1. Karlin and Nuttall. We let (u) be Riemann s function as earlier. We let (t) = X n=0 ( 1)n bn (2n)!t 2n where bn = Z (u)u 2n du as before. Define B(i, j) = bj i (2j 2i)! if i j 0 if i > j Then RH is equivalent to D(n, r) > 0 for all positive r and non-negative n where D(n, r) = det r r B(i, j + n)|i,j=1,r (see [Kar68] chapter 8). The case r = 1 here is clear since F(u) > 0. The case r = 2 is slightly weaker than the Turan inequalities; it asserts that b 2 m > m m + 1 (2m 1) (2m + 1) bm 1bm+1. Nuttall [Nut13] has established the case r = 3 which asserts that b 3 m ((2m)!)3 2bm 1bm+1bm (2m)!(2m 2)!(2m + 2)! bm 2bm+2bm (2m)!(2m 4)!(2m + 4)! + b 2 m 1 bm+2 ((2m 2)!)2 (2m + 4)! + bm 2b 2 m+1 (2m 4)!((2m + 2)!)2 > 0 for all m 2. For m = 2 this is b 3 2 4 5 b1b3b2 1 70 b0b4b2 + 2 75 b0b 2 3 + 3 35 b 2 1 b4 > 0. 24 BRIAN CONREY 13. Turan inequalities, 2 Ramanujan s tau-function may be defined by equating coefficients of the power series on both sides of X n=1 (n)x n = x Y n=1 (1 x n ) 24 The associated Dirichlet series is L(s) = L (s) = X n=1 (n)n s This series is absolutely convergent for = 13/2. The xi-function for is given by (s) = (2 ) s (s)L(s) and it satisfies the functional equation (s) = (12 s). This functional equation is equivalent to the fact that (z) = X n=1 (n)e(nz) is a holomorphic cusp form of weight 12 for the full modular group which in turn is equivalent to: (i) (z) is expressible in terms of a Fourier series in z in which coefficients of e(nz) with n 0 vanish and (ii) satisfies the transformation formula ( 1/z) = z 12 (z). It is believed that all of the zeros of (s) are on the line <s = 6; this is the Riemann Hypothesis for L . See Hardy [Har78], Chapter X for introductory information about . Now (s) = Z 0 (iy)y s dy y so that (t) = (6 + it) = Z (ieu )e 6u e iut du is an entire even function of t. We define (u) = (ieu )e 6u We see that (u) is an even function of u by the functional equation for . The fact that (u) > 0 for real u is immediately obvious from the product formula for : (u) = e 6u e 2 eu Y n=1 1 e 2 neu 24 RIEMANN S HYPOTHESIS 25 We can also see that (u) is decreasing for positive u by calculating the logarithmic derivative. We first observe that y d dy X n=1 log(1 y n ) = y d dy X m,n=1 y mn m = X m,n=1 nymn = X n=1 (n)y n where (n) = X d|n d is the sum of divisors of n. Let x = 2 eu and y = e x . Then (u) = e 6u y Y n=1 (1 y n ) 24 so that 0 (u) = 6 + 1 y 24X n=1 (n)y n ! dy du = 6 x(1 0(x)) where k(x) = 24X n=1 n k (n)y n (The expansion of 0 / above is related to the Fourier expansion of the Eisenstein series E2: E2(z) = 1 24X n=1 (n)e(nz) E2 is not a modular form of weight 2; it transforms according to the formulae E2( 1/z) = z 2E2(z) + 12z 2 i and E2(z + 1) = E2(z). Note also that P(y) = E2(e 2y ) satisfies the Chazy equation P 000 2P P00 + 3(P 0 ) 2 = 0; the Chazy equation is related to a Painlev e equation.) Now 0 / is an odd function of u so that 0 / (0) = 0. Thus, to show that 0 (u) < 0 for u > 0 it suffices to prove that 0 0 (u) < 0 26 BRIAN CONREY for u > 0. But 0 0 (u) = ( 1 + 0(x) + x 0 0 (x)) dx du = x + x 0(x) x 2 1(x) since 0 k (x) = k+1(x) and this then is = x 1 24X n=1 (n)y n (1 nx) ! Since u 0 corresponds to x 2 each of the terms 1 nx < 0 so that the whole expression is negative. Arguing in the same way we see that 0 00 (u) is odd and 0 000 (u) = x(x 3 3(x) 6x 2 2(x) + 7x 1(x) 0(x) + 1) = x(1 24X n=1 (n)y nP3(nx)) where P3(x) = 1 7x + 6x 2 x 3 < 0 for x > 6. Thus we conclude that 0 00 (u) < 0 for u > 0, i.e. that 0 is concave for u > 0; see [CoGh94] for more details. 13.1. A difficulty with classifying functions whose Fourier transforms have real zeros. Let g(u) = u 4 + eu 12 + X n=1 1(n)e 2 neu . Here 1(n) = P d|n d 1 is the sum of the reciprocals of the positive divisors of n. Then g(u) is positive, even, decreasing, and its logarithmic derivative is decreasing and concave for u > 0. So k(t) = Z e kg(u) e iut du might seem to be a good candidate for a function to have only real zeros. In fact k = 24 is the case we ve just been discussing about the Ramanujan tau-function. And k = 1 corresponds to the Xi-function associated with the Dirichlet L-function associated to the unique primitive character of modulus 12, and so all of its zeros should be rea // Extend manually.