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The twin prime conjecture is one of the oldest unsolved problems in analytic number theory. There are several reasons why this conjecture remains out of reach of current techniques, but the most important obstacle is the parity problem which prevents purely sieve-theoretic methods (or many other popular methods in analytic number theory, such as the circle method) from detecting pairs of prime twins in a way that can distinguish them from other twins of almost primes. The parity problem is discussed in these previous blog posts; this obstruction is ultimately powered by the Möbius pseudorandomness principle that asserts that the Möbius function ${\mu}$ is asymptotically orthogonal to all “structured” functions (and in particular, to the weight functions constructed from sieve theory methods).

However, there is an intriguing “alternate universe” in which the Möbius function is strongly correlated with some structured functions, and specifically with some Dirichlet characters, leading to the existence of the infamous “Siegel zero“. In this scenario, the parity problem obstruction disappears, and it becomes possible, in principle, to attack problems such as the twin prime conjecture. In particular, we have the following result of Heath-Brown:

Theorem 1 At least one of the following two statements are true:

• (Twin prime conjecture) There are infinitely many primes ${p}$ such that ${p+2}$ is also prime.
• (No Siegel zeroes) There exists a constant ${c>0}$ such that for every real Dirichlet character ${\chi}$ of conductor ${q > 1}$, the associated Dirichlet ${L}$-function ${s \mapsto L(s,\chi)}$ has no zeroes in the interval ${[1-\frac{c}{\log q}, 1]}$.

Informally, this result asserts that if one had an infinite sequence of Siegel zeroes, one could use this to generate infinitely many twin primes. See this survey of Friedlander and Iwaniec for more on this “illusory” or “ghostly” parallel universe in analytic number theory that should not actually exist, but is surprisingly self-consistent and to date proven to be impossible to banish from the realm of possibility.

The strategy of Heath-Brown’s proof is fairly straightforward to describe. The usual starting point is to try to lower bound

$\displaystyle \sum_{x \leq n \leq 2x} \Lambda(n) \Lambda(n+2) \ \ \ \ \ (1)$

for some large value of ${x}$, where ${\Lambda}$ is the von Mangoldt function. Actually, in this post we will work with the slight variant

$\displaystyle \sum_{x \leq n \leq 2x} \Lambda_2(n(n+2)) \nu(n(n+2))$

where

$\displaystyle \Lambda_2(n) = (\mu * L^2)(n) = \sum_{d|n} \mu(d) \log^2 \frac{n}{d}$

is the second von Mangoldt function, and ${*}$ denotes Dirichlet convolution, and ${\nu}$ is an (unsquared) Selberg sieve that damps out small prime factors. This sum also detects twin primes, but will lead to slightly simpler computations. For technical reasons we will also smooth out the interval ${x \leq n \leq 2x}$ and remove very small primes from ${n}$, but we will skip over these steps for the purpose of this informal discussion. (In Heath-Brown’s original paper, the Selberg sieve ${\nu}$ is essentially replaced by the more combinatorial restriction ${1_{(n(n+2),q^{1/C}\#)=1}}$ for some large ${C}$, where ${q^{1/C}\#}$ is the primorial of ${q^{1/C}}$, but I found the computations to be slightly easier if one works with a Selberg sieve, particularly if the sieve is not squared to make it nonnegative.)

If there is a Siegel zero ${L(\beta,\chi)=0}$ with ${\beta}$ close to ${1}$ and ${\chi}$ a Dirichlet character of conductor ${q}$, then multiplicative number theory methods can be used to show that the Möbius function ${\mu}$ “pretends” to be like the character ${\chi}$ in the sense that ${\mu(p) \approx \chi(p)}$ for “most” primes ${p}$ near ${q}$ (e.g. in the range ${q^\varepsilon \leq p \leq q^C}$ for some small ${\varepsilon>0}$ and large ${C>0}$). Traditionally, one uses complex-analytic methods to demonstrate this, but one can also use elementary multiplicative number theory methods to establish these results (qualitatively at least), as will be shown below the fold.

The fact that ${\mu}$ pretends to be like ${\chi}$ can be used to construct a tractable approximation (after inserting the sieve weight ${\nu}$) in the range ${[x,2x]}$ (where ${x = q^C}$ for some large ${C}$) for the second von Mangoldt function ${\Lambda_2}$, namely the function

$\displaystyle \tilde \Lambda_2(n) := (\chi * L)(n) = \sum_{d|n} \chi(d) \log^2 \frac{n}{d}.$

Roughly speaking, we think of the periodic function ${\chi}$ and the slowly varying function ${\log^2}$ as being of about the same “complexity” as the constant function ${1}$, so that ${\tilde \Lambda_2}$ is roughly of the same “complexity” as the divisor function

$\displaystyle \tau(n) := (1*1)(n) = \sum_{d|n} 1,$

which is considerably simpler to obtain asymptotics for than the von Mangoldt function as the Möbius function is no longer present. (For instance, note from the Dirichlet hyperbola method that one can estimate ${\sum_{x \leq n \leq 2x} \tau(n)}$ to accuracy ${O(\sqrt{x})}$ with little difficulty, whereas to obtain a comparable level of accuracy for ${\sum_{x \leq n \leq 2x} \Lambda(n)}$ or ${\sum_{x \leq n \leq 2x} \Lambda_2(n)}$ is essentially the Riemann hypothesis.)

One expects ${\tilde \Lambda_2(n)}$ to be a good approximant to ${\Lambda_2(n)}$ if ${n}$ is of size ${O(x)}$ and has no prime factors less than ${q^{1/C}}$ for some large constant ${C}$. The Selberg sieve ${\nu}$ will be mostly supported on numbers with no prime factor less than ${q^{1/C}}$. As such, one can hope to approximate (1) by the expression

$\displaystyle \sum_{x \leq n \leq 2x} \tilde \Lambda_2(n(n+2)) \nu(n(n+2)); \ \ \ \ \ (2)$

as it turns out, the error between this expression and (1) is easily controlled by sieve-theoretic techniques. Let us ignore the Selberg sieve for now and focus on the slightly simpler sum

$\displaystyle \sum_{x \leq n \leq 2x} \tilde \Lambda_2(n(n+2)).$

As discussed above, this sum should be thought of as a slightly more complicated version of the sum

$\displaystyle \sum_{x \leq n \leq 2x} \tau(n(n+2)). \ \ \ \ \ (3)$

Accordingly, let us look (somewhat informally) at the task of estimating the model sum (3). One can think of this problem as basically that of counting solutions to the equation ${ab+2=cd}$ with ${a,b,c,d}$ in various ranges; this is clearly related to understanding the equidistribution of the hyperbola ${\{ (a,b) \in {\bf Z}/d{\bf Z}: ab + 2 = 0 \hbox{ mod } d \}}$ in ${({\bf Z}/d{\bf Z})^2}$. Taking Fourier transforms, the latter problem is closely related to estimation of the Kloosterman sums

$\displaystyle \sum_{m \in ({\bf Z}/r{\bf Z})^\times} e( \frac{a_1 m + a_2 \overline{m}}{r} )$

where ${\overline{m}}$ denotes the inverse of ${m}$ in ${({\bf Z}/r{\bf Z})^\times}$. One can then use the Weil bound

$\displaystyle \sum_{m \in ({\bf Z}/r{\bf Z})^\times} e( \frac{am+b\overline{m}}{r} ) \ll r^{1/2 + o(1)} (a,b,r)^{1/2} \ \ \ \ \ (4)$

where ${(a,b,r)}$ is the greatest common divisor of ${a,b,r}$ (with the convention that this is equal to ${r}$ if ${a,b}$ vanish), and the ${o(1)}$ decays to zero as ${r \rightarrow \infty}$. The Weil bound yields good enough control on error terms to estimate (3), and as it turns out the same method also works to estimate (2) (provided that ${x=q^C}$ with ${C}$ large enough).

Actually one does not need the full strength of the Weil bound here; any power savings over the trivial bound of ${r}$ will do. In particular, it will suffice to use the weaker, but easier to prove, bounds of Kloosterman:

Lemma 2 (Kloosterman bound) One has

$\displaystyle \sum_{m \in ({\bf Z}/r{\bf Z})^\times} e( \frac{am+b\overline{m}}{r} ) \ll r^{3/4 + o(1)} (a,b,r)^{1/4} \ \ \ \ \ (5)$

whenever ${r \geq 1}$ and ${a,b}$ are coprime to ${r}$, where the ${o(1)}$ is with respect to the limit ${r \rightarrow \infty}$ (and is uniform in ${a,b}$).

Proof: Observe from change of variables that the Kloosterman sum ${\sum_{m \in ({\bf Z}/r{\bf Z})^\times} e( \frac{am+b\overline{m}}{r} )}$ is unchanged if one replaces ${(a,b)}$ with ${(\lambda a, \lambda^{-1} b)}$ for ${\lambda \in ({\bf Z}/d{\bf Z})^\times}$. For fixed ${a,b}$, the number of such pairs ${(\lambda a, \lambda^{-1} b)}$ is at least ${r^{1-o(1)} / (a,b,r)}$, thanks to the divisor bound. Thus it will suffice to establish the fourth moment bound

$\displaystyle \sum_{a,b \in {\bf Z}/r{\bf Z}} |\sum_{m \in ({\bf Z}/r{\bf Z})^\times} e\left( \frac{am+b\overline{m}}{r} \right)|^4 \ll d^{4+o(1)}.$

The left-hand side can be rearranged as

$\displaystyle \sum_{m_1,m_2,m_3,m_4 \in ({\bf Z}/r{\bf Z})^\times} \sum_{a,b \in {\bf Z}/d{\bf Z}}$

$\displaystyle e\left( \frac{a(m_1+m_2-m_3-m_4) + b(\overline{m_1}+\overline{m_2}-\overline{m_3}-\overline{m_4})}{r} \right)$

which by Fourier summation is equal to

$\displaystyle d^2 \# \{ (m_1,m_2,m_3,m_4) \in (({\bf Z}/r{\bf Z})^\times)^4:$

$\displaystyle m_1+m_2-m_3-m_4 = \frac{1}{m_1} + \frac{1}{m_2} - \frac{1}{m_3} - \frac{1}{m_4} = 0 \hbox{ mod } r \}.$

Observe from the quadratic formula and the divisor bound that each pair ${(x,y)\in ({\bf Z}/r{\bf Z})^2}$ has at most ${O(r^{o(1)})}$ solutions ${(m_1,m_2)}$ to the system of equations ${m_1+m_2=x; \frac{1}{m_1} + \frac{1}{m_2} = y}$. Hence the number of quadruples ${(m_1,m_2,m_3,m_4)}$ of the desired form is ${r^{2+o(1)}}$, and the claim follows. $\Box$

We will also need another easy case of the Weil bound to handle some other portions of (2):

Lemma 3 (Easy Weil bound) Let ${\chi}$ be a primitive real Dirichlet character of conductor ${q}$, and let ${a,b,c,d \in{\bf Z}/q{\bf Z}}$. Then

$\displaystyle \sum_{n \in {\bf Z}/q{\bf Z}} \chi(an+b) \chi(cn+d) \ll q^{o(1)} (ad-bc, q).$

Proof: As ${q}$ is the conductor of a primitive real Dirichlet character, ${q}$ is equal to ${2^j}$ times a squarefree odd number for some ${j \leq 3}$. By the Chinese remainder theorem, it thus suffices to establish the claim when ${q}$ is an odd prime. We may assume that ${ad-bc}$ is not divisible by this prime ${q}$, as the claim is trivial otherwise. If ${a}$ vanishes then ${c}$ does not vanish, and the claim follows from the mean zero nature of ${\chi}$; similarly if ${c}$ vanishes. Hence we may assume that ${a,c}$ do not vanish, and then we can normalise them to equal ${1}$. By completing the square it now suffices to show that

$\displaystyle \sum_{n \in {\bf Z}/p{\bf Z}} \chi( n^2 - b ) \ll 1$

whenever ${b \neq 0 \hbox{ mod } p}$. As ${\chi}$ is ${+1}$ on the quadratic residues and ${-1}$ on the non-residues, it now suffices to show that

$\displaystyle \# \{ (m,n) \in ({\bf Z}/p{\bf Z})^2: n^2 - b = m^2 \} = p + O(1).$

But by making the change of variables ${(x,y) = (n+m,n-m)}$, the left-hand side becomes ${\# \{ (x,y) \in ({\bf Z}/p{\bf Z})^2: xy=b\}}$, and the claim follows. $\Box$

While the basic strategy of Heath-Brown’s argument is relatively straightforward, implementing it requires a large amount of computation to control both main terms and error terms. I experimented for a while with rearranging the argument to try to reduce the amount of computation; I did not fully succeed in arriving at a satisfactorily minimal amount of superfluous calculation, but I was able to at least reduce this amount a bit, mostly by replacing a combinatorial sieve with a Selberg-type sieve (which was not needed to be positive, so I dispensed with the squaring aspect of the Selberg sieve to simplify the calculations a little further; also for minor reasons it was convenient to retain a tiny portion of the combinatorial sieve to eliminate extremely small primes). Also some modest reductions in complexity can be obtained by using the second von Mangoldt function ${\Lambda_2(n(n+2))}$ in place of ${\Lambda(n) \Lambda(n+2)}$. These exercises were primarily for my own benefit, but I am placing them here in case they are of interest to some other readers.