Prime Gaps

With conditional information on the level of distribution of the primes, the author shows that there are infinitely often two primes in a given \(k\)-tuple for sufficiently large \(k\). With the same approach unconditionally, the author establishes the existence of very small gaps between primes which go slowly compare to the size of primes.

Let \(n \in \mathbb{N}\), \(\mathcal{H}=\{h_1,\ldots,h_k\}\), \(h_i\in \mathbb{N}\cup\{0\}\), \(h_i < h_j \le h\), \(i < j\), the \(k\)-tuple is

\[ (n+h_1,\, n+h_2,\, \ldots,\, n+h_k). \]

Let \(\nu_p(\mathcal{H})\) denote the number of distinct residue classes modulo \(p\) occupied by the integers \(h_i\), and called \(\mathcal{H}\) is admissible if \(\nu_p(\mathcal{H}) < p\) for all \(p\) primes. For square free \(d\), we extend the definition to \(\nu_d(\mathcal{H})\) by multiplicity. Take \(N > h_k\), the goal is to construct a suitable \(w(n) \ge 0\) for which

\[ \sum_{N < n \le 2N}\left(\sum_{i=1}^k \theta(n+h_i)-\rho\log 3N\right)w(n) > 0. \]

where \(\rho > 0\). This yields that there exist \(n\in (N,2N]\), s.t. there are at least \(\lfloor \rho \rfloor +1\) prime components in the \(k\)-tuple. Therefore, \(w(n)\) functions to detect the prime components of this tuple and it is natural to use the Selberg-style weight

\[ w(n)=\left(\sum_{\substack{d\mid P(n)\\ d\le R}} \lambda(d)\right)^2 \]

where \(P(n)=(n+h_1)\cdots(n+h_k)\), \(R < N^{\frac15}\). In their arguement, they relied on the \(k\)-th generalized Von Mangoldt function

\[ \Lambda_k(n)=\frac{1}{k!}\sum_{d\mid n}\mu(d)\left(\log \frac{n}{d}\right)^k \]

which vanishes if \(n\) has more than \(k\) distinct prime factors. Hence, they select

\[ w(n):=\left(\sum_{\substack{d\mid P(n)\\ d\le R}} \mu(d)\,\frac{1}{m!}\left(\log \frac{R}{d}\right)^m\right)^2, \]

\[ \lambda(d):=\mu(d)\,\frac{1}{m!}\left(\log \frac{R}{d}\right)^m \]

for some \(m=k+\ell\). Note if a component is highly composite then the whole weight would be dragged down to \(0\).

\[ \mathfrak{S}(\mathcal{H}):=\prod_p \left(1-\frac{1}{p}\right)^{-k}\left(1-\frac{\nu_p(\mathcal{H})}{p}\right). \]

since \(\nu_p(\mathcal{H})=k\) for \(p > h\) if admissible, then such singular series is convergent. We expand the sum above and obtain

\[ \sum_{d_1,d_2\le R}\lambda(d_1)\lambda(d_2) \left( \sum_{i=1}^k \sum_{\substack{N < n \le 2N\\ [d_1,d_2]\mid P(n)}} \theta(n+h_i) -\rho\log 3N \sum_{\substack{N < n \le 2N\\ [d_1,d_2]\mid P(n)}} 1 \right). \]

Notice that for \(p\mid P(n)\), then \(n\equiv -h_i\pmod p\) for some \(h_i\)'s, and \(\nu_p(\mathcal{H})\) counts the number of solutions of \(n\) modulo \(p\). Then by the Chinese Remainder Theorem, we have the parentheses in the above equal to

\[ k\left(\nu_q(\mathcal{H})^*\right)\left(\frac{X_N}{\varphi(q)}+O(E(N,q))\right) -\rho\log 3N\, \nu_q(\mathcal{H})\left(\frac{N}{q}+O(1)\right) \]

\[ X_N=\sum_{N < n \le 2N}\theta(n), \]

\[ E(N,q)=\sup_{(a,q)=1}\left|\sum_{\substack{N < n \le 2N\\ n\equiv a\, (\mathrm{mod}\ q)}}\theta(n)-\frac{X_N}{\varphi(q)}\right|, \]

\[ \nu_q(\mathcal{H})^*=(\nu_{p_{{\alpha}_1}}(\mathcal{H})-1)\cdots(\nu_{p_{{\alpha}_t}}(\mathcal{H})-1), \qquad q=p_{{\alpha}_1}\cdots p_{{\alpha}_t}, \]

since for \(p\mid P(n)\), \(n\equiv -h_j\pmod p\), \(n+h_i\equiv h_i-h_j\pmod p\), and we need \((h_i-h_j,p)=1\) for \(\theta(n+h_i)\ne 0\), thus \(\nu_p(\mathcal{H})-1\) residue classes modulo \(p\).

We first deal with the error term. Let \(\omega(q)\) be the number of prime divisors of \(q\), then \(\nu_q(\mathcal{H})\le k^{\omega(q)}=d_k(q)\). The error is then bounded by

\[ (\log R)^{2m}k\sum_{q\le R^2}\mu(q)^2\, d_k(q)\, d_3(q)\, E(N,q) \]

where we have the number of possible \(q=[d_1,d_2]\) with \(d_1,d_2\le R\) is bounded by the number of representations of \(q\) written as \(3\) factors, i.e. \(d_3(q)\). Give the bound

\[ \sum_{q\le x}\mu(q)^2\frac{d_k(q)}{q}\le (k+1+\log x)^{k+1}, \qquad \sum_{q\le x}\mu(q)^2 d_k(q)\le x(k+1+\log x)^{k+1}, \]

\[ kR^2(3\log R)^{3k+2m+1}+k(\log R)^{2m} \sqrt{\sum_{q\le R^2}\frac{\mu(q)^2d_{9k^2}(q)}{q}} \sqrt{\sum_{q\le R^2} q(E(N,q)-1)^2}. \]

by the trivial bound for \(E(N,q)-1\le \dfrac{2N}{q}\log N\), we have the bound for the latter term

\[ \ll k(\log R)^{2m}\sqrt{\log(R)^{k+2\ell}}\sqrt{2N\log N} \sqrt{\sum_{q\le R^2}(E(N,q)-1)} \]

\[ \ll kN(\log N)^{\frac{9k^2+1-A}{2}} \]

where we applied the Bombieri-Vinogradov Theorem by letting \(R\ll N^{\frac14}/\log N^B, B=B(m,A) \text{ where } m,A,k\le \sqrt{\log N}/18 \text{ are constants}.\) Hence, the error is \(N^{1-\varepsilon}\).

The author uses Perron's Formula to calculate the main term.
Let \((c)\) denote the contour \(s=c+it\), \(-\infty < t < \infty\). For \(c > 0\), we have

\[ \frac{1}{2\pi i}\int_{(c)} \frac{x^s}{s^{k+1}}\,ds= \begin{cases} 0 & \text{if } 0 \le x \le 1,\\[4pt] \dfrac{1}{k!}(\log x)^k & \text{if } x \ge 1. \end{cases} \]

\[ \sum_{d\le R}\frac{\lambda(d)\nu_d(\mathcal{H})}{d} =\frac{1}{2\pi i}\int_{(1)} F(s)\frac{R^s}{s^{k+1}}\,ds \]

\[ F(s)=\sum_{d=1}^{\infty}\frac{\mu(d)\nu_d(\mathcal{H})}{d^{1+s}} =\prod_p\left(1-\frac{\nu_p(\mathcal{H})}{p^{1+s}}\right). \]

\[ G_{\mathcal{H}}(s)=\prod_p\left(1-\frac{\nu_p(\mathcal{H})}{p^{1+s}}\right) \left(1-\frac{1}{p^{1+s}}\right)^{-k} \]

\[ =\prod_p\left(1+\frac{k-\nu_p(\mathcal{H})}{p^{1+s}}+O_h\!\left(\frac{k^2}{p^{2+2\sigma}}\right)\right) \]

since \(\nu_p(\mathcal{H})=k\) for \(p > h\), then it is analytic and uniformly bounded for \(\sigma > -\frac12+\delta\) for any \(\delta > 0\). Also, we see that

\[ F(s)=\frac{G_{\mathcal{H}}(s)}{s(1+s)^k}, \qquad G_{\mathcal{H}}(0)=\mathfrak{S}(\mathcal{H}). \]

Note, if \(\mathcal{H}\) not admissible, then the integrand has a simple pole at \(0\) in the region bounded by \(\mathcal{L}: s=-\dfrac{c}{\log(|t|+2)}+it\) and \((1)\) with residue \(\mathfrak{S}(\mathcal{H})\); if \(\mathcal{H}\) is not admissible, then it is analytic everywhere. Therefore, we can calculate the integral by moving to the contour \(\mathcal{L}\), plus an extra term of \(\mathfrak{S}(\mathcal{H})\).
As for the main term, we write \(d_1=a_1a_{12}\) and \(d_2=a_2a_{12}\), \(a_1,a_2,a_{12}\) are relative prime to each other, then \(q=a_1a_2a_{12}\). Therefore, after some expansion of terms, we may apply the Perron's Formula twice and obtain new \(F(s_1,s_2)\), \(G_{\mathcal{H}}(s_1,s_2)\) with \(G_{\mathcal{H}}(0,0)=\mathfrak{S}(\mathcal{H})\), Then we are done by evaluating the integral.