User gh - MathOverflow

Yitang Zhang's preprint on Landau-Siegel zeros

2013-05-20T13:23:27Z

The recent sensational news on bounded gaps between primes made me wonder: what is the status of Yitang Zhang's earlier arXiv preprint on Landau-Siegel zeros? If this result is correct, then (in my opinion) it is even bigger news for analytic number theory. Has anyone checked this paper carefully?

Answer by GH for Power series whose partial sums attain only finitely many values

2013-05-16T00:55:03Z

The phenomenon you observe is a special case of a theorem of Frobenius (1880):

If a series is Cesaro summable then it is also Abel summable, and the Cesaro limit is the same as the Abel limit.

In your case the series is $(1,\zeta,\zeta^2,\dots)$ which ensures Cesaro summability as the sequence of partial sums is periodic.

For more details see Page 4 of Korevaar: Tauberian theory - A century of developments (Springer, 2004).

Answer by GH for Possible ratios of Pythagorean fractions

2013-05-15T00:05:47Z

This is a sketch how to decide the question for $\frac{4}{9}$.

The question is if there are positive integers $a,b,d,e$ such that $\frac{4}{9}=\frac{a/b}{d/e}$ with $a^2+b^2$ and $d^2+e^2$ squares. Denoting $p:=\frac{9a}{b}=\frac{4d}{e}$, the question is if there is $p\in\mathbb{Q}^\times$ such that both $p^2+4^2$ and $p^2+9^2$ are rational squares. That is, the question is if the quadrics $x^2+4y^2=t^2$ and $x^2+9y^2=z^2$ over $\mathbb{Q}$ intersect in a point with $x,y\neq 0$. The intersection of the two quadrics is isomorphic to the elliptic curve $$ Y^2=8(X-1)(X+1)(9X-1) $$ according to a 1997 preprint by R.G.E Pinch: Square values of quadratic polynomials (which used to be here but is no longer available, unfortunately). If this elliptic curve has finitely many rational points (something that I cannot check at the moment for lack of time) then finding them explicitly will list all points $(x,y,t,z)$ lying on both quadrics, so the question if there is a point with $x,y\neq 0$ can be settled. Otherwise there surely will be a point with $x,y\neq 0$.

Added. It seems that François Brunault filled in the details (see his comments to this post), and $\frac{4}{9}$ is indeed not a ratio of two Pythagorean fractions.

Least prime in an arithmetic progression and the Selberg sieve

2013-05-09T19:39:52Z

My question concerns a technical step in the proof of Linnik's theorem on the least prime in an arithmetic progression, as presented in Chapter 18 of Iwaniec-Kowalski: Analytic number theory.

The proof uses certain weights $\theta_b$ coming from the theory of the Selberg sieve. The sequence is supported on square-free numbers up to $y$ coprime with $q$ (the modulus of the arithmetic progression), and its main feature is, cf. (18.19)-(18.23) in the book, $$ |\theta_b|\leq 1,\qquad \theta_1=1, $$ $$ 0<\sum_{b_1,b_2}\frac{\theta_{b_1}\theta_{b_2}}{[b_1,b_2]}\leq\frac{q}{\varphi(q)\log y}. $$ The corresponding Selberg upper sieve coefficients are defined by $$ \sigma_m:=\sum_{[b_1,b_2]=m}\theta_{b_1}\theta_{b_2}, $$ so that with the notation $$ \nu(n):=\sum_{b\mid n}\theta_b $$ we have for any integer $n\geq 1$ $$ \sum_{m\mid n}\sigma_m=\nu(n)^2\geq\sum_{m\mid n}\mu(m). $$

By (18.70) of the book we have, "applying a sieve of dimension 8 (see the Fundamental Lemma 6.3)", $$ \sum_{n\leq x}\nu^2(n)\frac{\tau^3(n)}{n}\ll\left(\frac{\log x}{\log y}\right)^8. $$ Here $\tau(n)$ is the number of divisors of $n$. Can anyone help me understand why this is true? The quoted lemma is about Brun's sieve (where $\sigma_m$ would be $\mu(m)$ restricted to certain integers), but even if I accept it for the Selberg sieve, I do not see the stated bound. Similarly, I do not understand why (18.75) in the book is true.

Added. Based on the expert response of Dimitris Koukoulopoulos I think that the proof of Linnik's theorem, as presented in Iwaniec-Kowalski's book, works fine if we replace the Selberg sieve $\sigma_m$ with an upper $\beta$-sieve $\beta_m$. More precisely, we redefine $\nu(n)$ so that $$\nu(n)^2=\sum_{m\mid n}\beta_m,$$ which makes sense as the right hand side is nonnegative.

Answer by GH for Asymptotics of a function

2013-05-09T07:59:07Z

From the comments so far (including mine above) it follows that $$ f(n) = \sum_{i=1}^{n} \frac{i^n}{n^{4i}}=\frac{n!}{(4\ln n)^{n+1}}\left(1+O(n^{-1/2}\ln n)\right). $$

Answer by GH for zeta(2k+1) is a rational multiple of pi^{2k} zeta'(-2 k) ?

2013-05-07T11:47:12Z

I am also sure that this is all known, but here is a quick proof that

$$ \zeta(2k+1)=\frac{(-1)^k2^{2k+1}}{(2k)!}\pi^{2k}\zeta'(-2k). $$

Following Carl Dettmann's suggestion, let us start from the functional equation $$ \pi^{-\frac{s}{2}}\Gamma\left(\frac{s}{2}\right)\zeta(s)=\pi^{-\frac{1-s}{2}}\Gamma\left(\frac{1-s}{2}\right)\zeta(1-s). $$ For $s=-2k+\epsilon$ this gives $$ \zeta(2k+1-\epsilon)=\pi^{2k+\frac{1}{2}-\epsilon}\frac{\Gamma(-k+\epsilon/2)}{\Gamma(k+1/2-\epsilon/2)}\zeta(-2k+\epsilon). $$ On the right hand side, for $\epsilon\to 0$, $$ \Gamma(-k+\epsilon/2)\sim \frac{2(-1)^k}{k!\epsilon}$$ and $$ \zeta(-2k+\epsilon)\sim\epsilon\zeta'(-2k), $$ hence we have at $\epsilon=0$ i.e. at $s=-2k$ $$ \zeta(2k+1)=\pi^{2k+\frac{1}{2}}\frac{2(-1)^k}{k!\Gamma(k+1/2)}\zeta'(-2k). $$ On the right hand side $$ \Gamma\left(k+\frac{1}{2}\right)=\left(k-\frac{1}{2}\right)\left(k-\frac{3}{2}\right)\cdots\frac{1}{2}\Gamma\left(\frac{1}{2}\right)=\frac{(2k-1)(2k-3)\cdots 1}{2^k}\pi^{\frac{1}{2}}, $$ whence the stated identity follows.

Answer by GH for Weyl law for arithmetic Fuchsian groups known?

2013-05-04T17:43:17Z

The Weyl law has been proven in great generality, e.g. for congruence subgroups of $PGL(n,\mathbb{R})$, and beyond. See the introduction in Lindenstrauss-Venkatesh: Existence and Weyl's law for spherical cusp forms, GAFA 17 (2007), 220-251 (manuscript available here). See also Marc Palm's recent thesis.

Answer by GH for No Exceptional Eigenvalues of Weight 1/2 Maass Forms on $\Gamma_0(4)$?

2013-05-02T09:17:38Z

A weight $1/2$ Hecke-Maass form of eigenvalue $(1-s^2)/4$ on $\Gamma_0(4)$ has a Shimura lift to a weight $0$ even Hecke-Maass form of eigenvalue $1/4-s^2$ on $\Gamma_0(1)$, see e.g. the proof of Theorem 1.5 in Baruch-Mao: A generalized Kohnen-Zagier formula for Maass forms (manuscript here). So the database for weight $0$ forms furnishes the required information for weight $1/2$ forms.

Answer by GH for The divisor bound for binary quadratic forms

2013-04-25T19:50:18Z

Let $r_Q^*(n)$ be the number of primitive representations of $n$ by $Q$. Let $\mathcal{C}$ be a set of representatives of the classes of binary quadratic forms of discriminant $\Delta$. We can assume that $Q\in\mathcal{C}$. It is known that $\sum_{Q'\in\mathcal{C}}r_{Q'}^*(n)$ equals the number of residue classes $b\pmod{2n}$ such that $b^2\equiv\Delta\pmod{4n}$, see e.g. Page 66 in Zagier: Zetafunktionen und quadratische Körper (Springer 1981). By the Chinese Remainder Theorem, we can calculate this as a product of local densities, in fact this is an instance of Siegel's mass formula. Estimating crudely we get $$ \sum_{Q'\in\mathcal{C}}r_{Q'}^*(n) \ll_\Delta 2^{\omega(n)},$$ where $\omega(n)$ is the number of prime factors of $n$, whence $r_Q^*(n)\ll_{\Delta,\epsilon} n^\epsilon$ readily follows. Finally, $$ r_Q(n)=\sum_{d^2\mid n}r_Q^*(n/d^2) \ll_{\Delta,\epsilon} n^\epsilon. $$

Answer by GH for q-series related to d(n)=# of divisors of n

2013-04-23T20:37:21Z

The coefficient $d(n)$ is the limit of the $n$-th Hecke eigenvalue of the (nonholomorphic) Eisenstein series $E(z,s)$ as $s\to 1/2$. The limit of the Eisenstein series itself is zero, hence it is more natural to consider the $(\partial/\partial s)E(z,s)$ at $s=1/2$, which is precisely $$ \sqrt{y}\log y+4\sqrt{y}\sum_{n=1}^\infty d(n)K_0(2\pi ny)\cos(2\pi nx). $$

In short, you have to leave the realm of holomorphic modular forms to find the object you are looking for: $d(n)$ is "morally" the $n$-th Hecke eigenvalue of the Eisenstein series with Laplace eigenvalue $1/4$ (which is zero eventually, so we pass to the derivative). This is another reason why Maass forms are so natural, something that not every arithmetician appreciates!

You can read more about this fascinating story in Iwaniec: Introduction to the spectral theory of automorphic forms. The limit Eisenstein series above is discussed at the end of Chapter 3.

Answer by GH for Computing certain class numbers modulo 4

2013-03-29T10:28:46Z

Your conjecture is true, it follows from Corollary 1 to Theorem 39 (Page 181) combined with Theorem 41 (Page 190) in Fröhlich-Taylor: Algebraic number theory (Cambridge University Press, 1991).

Answer by GH for How can we understand Baker's theorem about transcendence ?

2013-03-11T02:49:32Z

The heart of all transcendence proofs is the analytic fact that there is no integer between $0$ and $1$. This is one of the things I learned from Baker's book "Transcendental Number Theory".

Answer by GH for Estimating $\prod_{p\mid n}(1+1/p)$ in terms of n

2013-03-08T20:09:08Z

It is easy to verify that for sufficiently large $n$ $$ \sum_{p\mid n}\frac{1}{p}<\sum_{p<2\log n}\frac{1}{p}<\log\log\log n+O(1) $$ whence $$ \prod_{p\mid n} \left( 1+\frac{1}{p} \right) \ll \log\log n.$$ In other words, you can choose any $b>0$. Also, one can further refine the above bound.

Answer by GH for Counterexamples for (weak) multiplicity one

2013-03-06T16:20:37Z

Very concretely: if $f\in S_k^{\text{new}}(\Gamma_0(N), \chi)$, and $d>1$ is an integer, then $f(z)$ and $f(dz)$ are elements of $S_k(\Gamma_0(dN), \chi)$ with the same Hecke eigenvalues at $m$ coprime with $dN$, yet they are not multiples of each other. Of course this is the same what Marc was saying, but in more classical terms.

In particular, if $dN=p$ (i.e. $d=p$, $N=1$, $\chi$ is trivial), then you get a counterexample for $S_k(\Gamma_0(p))$.

I suggest that you read this famous paper by Atkin and Lehner.

Answer by GH for Hecke eigenvalue at p and at p^k

2013-03-06T11:34:22Z

For each $k$ there is such a polynomial and can be determined recursively from the usual relation $\lambda(m)\lambda(n)=\sum_{d\mid (m,n)}\chi(d)\lambda(mn/d^2)$, where $\chi$ is the nebentypus (for weight zero Maass newforms). Am I missing something?

Actually, for unramified primes $P_k(x)=U_k(x/2)$, where $U_k$ is the $k$-th Chebyshev polynomial of the second kind, while for ramified primes $P_k(x)=x^k$. See Section 2.3 in Young's paper.

Answer by GH for Divisibility in a set

2013-03-04T16:07:01Z

The set $A$ has positive lower density, hence the result follows from the main theorem in this paper of Paul Erdős (the proof takes one page).

Answer by GH for Equivalence of two well-known forms of (RH): reference-request.

2013-02-18T05:05:45Z

To complement Greg Martin's response, here is a proof that (1) implies (2).

Write $\pi(t)=\mathrm{Li}(t)+f(t)$, so that $f(t)=O(t^{1/2}\log t)$ by assumption. Then $$ \theta(x) = \int_{2-}^x\log t\ d\pi(t)=\int_{2-}^x\log t\ d\mathrm{Li}(t)+\int_{2-}^x\log t\ d f(t). $$ Here the first integral is $$ \int_{2-}^x\log t\ d\mathrm{Li}(t) = \int_2^x\log t\frac{dt}{\log t} = x-2 $$ and the second integral is $$ \int_{2-}^x\log t\ df(t) = f(x)\log x - \int_2^x \frac{f(t)}{t}dt = O(x^{1/2}\log^2 x) + \int_2^x O(t^{-1/2}\log t)\ dt. $$ The last integral is clearly $$ O(\log x)\int_2^x t^{-1/2}\ dt = O(x^{1/2}\log x),$$ hence altogether $$ \theta(x) = x-2 + O(x^{1/2}\log^2 x)+O(x^{1/2}\log x) = x+O(x^{1/2}\log^2 x). $$

Answer by GH for Sums of two squares: What is known about the distribution of r(n)?

2013-02-08T20:57:49Z

The topic is too general, and your question is too vague.

At any rate, for a large $x$, the average value of $r(n)$ for $1\leq n\leq x$ is about $\pi$, while the standard deviation is about $2\sqrt{\log x}$. There are many ways to prove this, perhaps the most instructive is to look at the Dirichlet series $\sum r(n)n^{-s}$ and $\sum r(n)^2n^{-s}$ which have poles of order $1$ and $2$, respectively (as follows from their Euler product decomposition). The following recent paper discusses the analogous problem in short intervals: Garaev-Kühleitner-Luca-Nowak, Asymptotic formulas for certain arithmetic functions, Math. Slovaca 58 (2008), 301–308.

Added. One can turn Greg Martin's response into a rigorous disproof of the added conjecture as follows. Assume that the conjecture is true, then in any interval $(a+k\sqrt{a},a+(k+1)\sqrt{a})$ with $0\leq k\leq\sqrt{a}$ the number of $n$'s with $r(n)>0$ is $\Omega(\sqrt{a})$. This implies that in $(a,2a)$ the number of $n$'s with $r(n)>0$ is $\Omega(a)$, contradicting Landau's result $O(a/\sqrt{\log a})$ for the number of such $n$'s.

Answer by GH for Trichotomies in mathematics

2013-02-04T08:03:30Z

Each set in a topological space partitions the space into three parts: interior, boundary, exterior.

Answer by GH for Hecke Relations on Fourier Coeficients for GL(n), n>2

2013-01-31T20:24:06Z

Have you checked Sections 9.3 and 9.4 in Goldfeld: Automorphic forms and L-functions for the group GL(n,R)? It might have what you need.

Note also that the Euler product translates into the Hecke relations, even over GL(n). Here it is good to know that at an unramified prime $p$ the Euler factor is of the form $H_p(p^{-s})^{-1}$, where $H_p(x)$ is a polynomial of degree $n$ and constant term $1$ (while for a ramified prime $p$ it is of degree less than $n$).

You might also find useful Lemma 5.2 in Qu: Linnik-type problems for automorphic L-functions (Journal of Number Theory 130 (2010), 786-802), which is a variant of the Newton-Girard formulae for power sums. Lemma 5.3 is a nice consequence, which might also serve you well.

Answer by GH for Mersenne primes problem

2013-01-29T14:09:44Z

It is an easy exercise to prove your statement for all Mersenne primes, not just the known ones, using that $2^6\equiv 1\pmod{9}$. Indeed, $p$ must be prime for $M_p=2^p-1$ to be prime. Hence either $p\equiv 1\pmod{6}$ in which case $M_p\equiv 2^1-1\equiv 1\pmod{9}$ and so $M_p\equiv 1\pmod{18}$, or $p\equiv 5\pmod{6}$ in which case $M_p\equiv 2^5-1\equiv 4\pmod{9}$ and so $M_p\equiv 13\pmod{18}$. I told you earlier that your questions are not of research level. Try MathStackExchange next time.

Answer by GH for Numbers of a certain form not expressible as squares

2013-01-28T18:12:32Z

I recommend this survey about the solution of Catalan's conjecture. We learn from here that the special case you are considering, namely $x^p-y^q=1$ for $p=2$, was solved by Chao Ko in 1964. In 1976 Chein published a simpler proof, see here.

Answer by GH for Understanding the "idea" behind Langlands

2013-01-19T18:13:11Z

I would disagree with your last two points just as wccanard does in his comment: automorphicity of $L$-functions is part of global Langlands functoriality, not the local conjectures (although the two are related).

I would also disagree with your third point: instead of nonabelian harmonic analysis, it is automorphicity that translates into functional equations and vice versa. For example, by the automorphicity of certain Eisenstein series we can see that certain $L$-functions coming from automorphic forms satisfy a functional equation, and from this we can sometimes deduce that the $L$-function is itself automorphic (not just pretends to be).

Regarding nonabelian harmonic analysis I would say that it naturally leads to the notion of automorphic forms and representations (via the spectral decomposition), and it also provides a good framework to study them (including establishing certain cases of functoriality without using $L$-functions).

Answer by GH for The Riemann Hypothesis and the Langlands program

2013-01-19T17:53:55Z

One can use Langlands functoriality to eliminate the so-called Siegel zeros of an automorphic $L$-function. For example, Hoffstein-Ramakrishnan (IMRN 1995) proved that the $L$-function of a $GL(n)$ cusp form for $n>1$ has no Siegel zero if all $GL(m)\times GL(n)$ $L$-functions are $GL(mn)$ $L$-functions. There are several unconditional results along this line, e.g. in the same paper it is shown that the $L$-function of a $GL(2)$ cusp form has no Siegel zero.

Answer by GH for Permanent of a matrix of odd integers

2013-01-16T20:24:08Z

Let us consider $s$ with $2^s\leq n<2^{s+1}$. First we prove the conjecture when all the entries of $A$ are $1$'s. Then $\mathrm{perm}(A)=n!$, hence by Legendre's formula the exponent of $2$ in it equals $n-t$, where $t$ is the number of $1$'s in the binary expansion of $n$. For $n=2^{s+1}-1$ we have $t=s+1$, hence the exponent equals $n-s-1$. For $2^s\leq n< 2^{s+1}-1$ we have $t\leq s$, hence the exponent is at least $n-s$.

For the general case we can assume that the statement holds for $n-1$. By the special case above, it suffices to show that if we increase by $2$ a single entry $a_{ij}$ of $A$, then the resulting matrix $B$ satisfies $$ \mathrm{perm}(B) \equiv \mathrm{perm}(A) \pmod{2^{n-s}}. $$ Clearly, $$ \mathrm{perm}(B) = \mathrm{perm}(A) + 2\mathrm{perm}(C),$$ where $C$ is the $(n-1)\times(n-1)$ matrix that results from $A$ by deleting the $i$-th row and the $j$-th column. Note that $2^s-1\leq n-1<2^{s+1}-1$, hence by the induction hypothesis we have $$ \mathrm{perm}(C) \equiv 0 \pmod{2^{n-1-s}}. $$ The claim follows, and we are done.

Answer by GH for Simultaneous diophantine approximation

2012-09-12T14:08:03Z

Let me share a simple proof I found during a childbirth class 8 years ago:

Let $x_1,\dots,x_d\in\mathbb{R}$ such that $1,x_1,...,x_d$ are linearly independent over $\mathbb{Q}$. Let $\epsilon>0$ and $a_1,\dots,a_d\in\mathbb{R}$ be arbitrary. We want to show that there are $n\in\mathbb{Z}$ and $y_1,\dots,y_d\in\mathbb{Z}$ such that $$|nx_i-y_i-a_i|<\epsilon,\quad 1\leq i\leq d.$$ We proceed by induction on $d$, the case of $d=0$ being trivial. The hypothesis is invariant under replacing $x_i$ with $nx_i-y_i$ for any nonzero $n\in\mathbb{Z}$ and any $y_1,\dots,y_d\in\mathbb{Z}$, while the conclusion only becomes stronger. Hence by Dirichlet's theorem on simultaneous diophantine approximation we can assume from the beginning that $$|x_i|<\epsilon,\quad 1\leq i\leq d.$$ By the induction hypothesis applied for $x_1/x_d,\dots,x_{d-1}/x_d$, there are $m\in\mathbb{Z}$ and $y_1,\dots,y_d\in\mathbb{Z}$ such that such that $r:=(m+a_d)/x_d$ satisfies $$|rx_i-y_i-a_i|<\epsilon/2,\quad 1\leq i\leq d.$$ Note that for $i=d$ this inequality is automatic with $y_d:=m$. Let $n$ be the closest integer to $r$, then $$|nx_i-y_i-a_i|\leq |rx_i-y_i-a_i|+|(n-r)x_i|<\epsilon/2+\epsilon/2=\epsilon,\quad 1\leq i\leq d.$$ The proof is complete.

Remark 1. I clarified the proof in response to some criticism.

Remark 2. Using Dirichlet's theorem again, there are infinitely many $n$'s with the required properties.

Connectedness of the complement of small subsets (extended question)

2013-01-03T03:08:14Z

The following questions occurred to me while browsing this site and looking at Exercise 20 here.

Question 1. Let $n>1$. Does there exist a countable dense subset $A\subset\mathbb{R}^n$ for which the complement $\mathbb{R}^n\setminus A$ is disconnected?

EDIT. I deleted an erroneous paragraph. Let me add two more questions, the first one being Gerald Edgar's comment here, the second one correcting the erroneous paragraph.

Question 2. Let $n>1$. Is it true that for any subset $A\subset\mathbb{R}^n$ of Hausdorff dimension less than $n-1$ the complement $\mathbb{R}^n\setminus A$ is connected?

It seems that Joel's argument answers this in the affirmative as well.

Question 3. Let $n>1$. Are there two countable dense subsets $A,B\subset\mathbb{R}^n$ whose complements are not homeomorphic?

Answer by GH for Need reference to an assertion re: Riemann

2013-01-08T07:39:23Z

We have $$ \log\text{lcm}(1,\dots,n)=\sum_p\log p\left[\frac{\log n}{\log p}\right] =\sum_p \log p\sum_{k\log p\leq \log n} 1 = \sum_{p^k\leq n}\log p.$$ The right hand side is the summatory function of the von Mangoldt function, which is known as the second Chebyshev function: $$ \psi(n):=\sum_{m\leq n}\Lambda(m). $$ That is, the bound you are asking about is $$ |\psi(n)-n|<\sqrt{n}\log^2 n.$$ The fact that the Riemann Hypothesis implies this bound for $n>73$ (hence probably also for $n>2$) follows from this paper. The other direction, that such a bound with any constant in front of $\sqrt{n}\log ^2 n$ implies the Riemann Hypothesis, is more classical and can be found in many textbooks (see e.g. the bottom of p.463 in Montgomery-Vaughan: Multiplicative Number Theory I).

Answer by GH for What are conjectures that are true for primes but then turned out to be false for some composite number?

2013-01-03T07:25:00Z

Once it was conjectured (for a short time) that $2^p-2$ cannot be divisible by $p^2$ when $p$ is prime. The two known counterexamples are $1093$ and $3511$. For more detail and context read here.

Answer by GH for Question about ideals in quadratic extensions.

2013-01-02T09:25:02Z

Your $O_\mathbb{K}$ is a PID (principal ideal domain) because its class number is $1$. Over a PID any torsion free module is free, hence your $I$ is indeed of the form $I=O_{\mathbb{K}}x+O_{\mathbb{K}}y$.

More generally, without the class number condition, $I=O_{\mathbb{K}}x+Jy$ for some $x,y\in\mathbb{L}$ and an ideal $J\subseteq O_{\mathbb{K}}$ whose ideal class is uniquely determined by $I$. This is a special case of the structure theorem of torsion free modules over Dedekind domains.

Comment by GH

2013-05-21T03:49:14Z

@Terry: Thanks for your insight!

Comment by GH

2013-05-20T19:28:59Z

Perhaps I should add that back in 2007 I did look at this preprint briefly. What I was missing is an essential and transparent idea that is customary with big advances like the current bounded gaps theorem (or the earlier result by Goldston, Pintz, Yildirim). But who knows, maybe the result is correct, but noone checked it carefully?

Comment by GH

2013-05-20T19:25:04Z

@unknown: My question was about the status of this preprint, i.e. if it had been checked and if it is correct.

Comment by GH

2013-05-20T16:56:41Z

@zy and @quid: I agree with both of you.

Comment by GH

2013-05-20T01:16:44Z

@Kálmán: Thanks for the clarification.

Comment by GH

2013-05-19T23:01:01Z

Also, I think your opening line "What is known to follow from the existence of Siegel zeros?" does not reflect what you want (in the light of your comments). You really want: What is known to follow from both the existence and the absence of Siegel zeros. That is, what can we prove unconditionally by using the notion of Siegel zeros on the way.

Comment by GH

2013-05-19T23:00:48Z

@Kálmán: The notion of exceptional zero does not make sense for a single zero (this is also what Terry Tao tried to say). At any rate, HB's result can be formulated as: if the twin prime conjecture is false, then there is no exceptional zero (with an appropriate constant in the definition of the exceptional zero). In other words, there is $c>0$ such that if there exist a quadratic $χ$ modulo q and a real zero $\sigma>1−c/\log q$ of $L(s,\chi)$, then the twin prime conjecture is true. I continue in next comment.

Comment by GH

2013-05-19T22:06:56Z

@Joël: I disagree. I believe analytic number theory was a constantly busy subject throughout the 20th century, just think of the circle method, sieves, zero density results, automorphic forms and automorphic L-fuctions, and several really big heros in the subject, some of who are very much alive and still working. On the other hand, I do agree that the subject got more fashionable lately, perhaps because of the constantly growing number of connections to other subjects, and the recent fascinating easy-to-state but hard-to-prove developments.

Comment by GH

2013-05-19T21:44:41Z

@Kálmán: I am not sure if you understood what kiskis said below. Heath-Brown proved that the existence of an exceptional zero implies the twin prime conjecture: Prime twins and Siegel zeros, Proc. London Math. Soc. (3) 47 (1983), no. 2, 193-224.

Comment by GH

2013-05-17T18:34:02Z

Joël: According to several experts the result seems to be true. Very astonishing indeed.

Comment by GH

2013-05-17T18:31:48Z

I don't understand the question. According to the twin prime conjecture, gap 2 occurs infinitely often, hence one cannot give a better lower bound than 2. Perhaps you meant: how large are the gaps between P2 numbers, provably?

Comment by GH

2013-05-16T16:59:02Z

I am glad I could help!

Comment by GH

2013-05-15T11:52:37Z

@François: Thank you for filling in the details!

Comment by GH

2013-05-15T01:21:42Z

@Vanchinathan: I misunderstood the question, and updated my response.

Comment by GH

2013-05-15T01:20:24Z

I updated my response.