# Can the exact number of twin primes $\leq n$ be proved using a "twin-prime zeta function"?

Let $$\pi(n)$$ denote the amount of primes $$\leq n$$ and let $$\pi_2(n)$$ the equivalent for twin primes. Properties of $$\pi(n)$$ can be proved using a well-known formula involving the zeros of the Riemann Zeta function, but no such relationship is known for twin primes.

Why not then define a "Twin-Prime Zeta function" with a corresponding twin-prime Euler product, continue it analytically then define the error term between $$\pi_2(n)$$ and $$\text{Li}_2(n)=\int_2^x log^{-2}t \ dt$$ in terms of a sum over all the zeroes within the critical strip in order to obtain an exact value for $$\pi_2(n)$$?

Obviously there are major challenges in doing so, otherwise we would have an exact formula for the amount of twin primes not exceeding $$n$$. What are these challenges? Assuming there are infinitely many twin primes, why is there no plausible conjecture about a Riemann-like exact formula for $$\pi_2(n)$$?

• Let $a(n)=1$ if $n$ is prime, $a(n)=0$ otherwise. You can define $F(s) = \sum_{n=1}^\infty a(n)a(n-2) n^{-s}$ but you won't be able to say anything about $\exp(F(s))$. The situation is very different with $G(s) = \sum_{n=1}^\infty a(n) n^{-s}$ since $\exp(G(s))$ has an expression in term of $\zeta(s)$ which follows from $\log \zeta(s) = \sum_{k=1}^\infty \frac{G(sk)}{k}$ Commented Oct 13, 2018 at 3:35
• $\zeta(s)$ is a very simple function, we can say a lot about it, which is useful to deduce non-trivial properties for the much more complicated function $\log \zeta(s)$. In particular $(s-1)\zeta(s)$ is analytic, thus so is $\frac{\zeta'}{\zeta}(s)+\frac{1}{s-1}$ which is +/- defined entirely by its poles (the zeros of $\zeta(s)$) Commented Oct 13, 2018 at 3:38

The main problem is with the analytic continuation of the twin prime zeta function. No one knows how to do that!

Can you please elaborate? Perhaps give an expression for the twin prime zeta function and show why this is difficult?

At present the best results about the occurrence of twin primes are obtained by sieve methods, which are more combinatorial. Usually sieve methods do not exploit analytic continuation as occurs when working with the zeta function. but it is always interesting to experiment with different possible approaches. I try to say more in what follows.

Your question concerning the possible discovery of an Euler product over twin primes is a natural one. But as also pointed out in the discussion, there are very significant problems that would need to be overcome. A fairly elementary explanation of the difficulties might go something like this---in the half plane$$\{s = \sigma + it : 1 < \sigma\}\tag*{(1)}$$we have the familiar Euler product identity$$\prod_p (1 - p^{-s})^{-1} = \sum_{n = 1}^\infty n^{-s},\tag*{(2)}$$where here $$p$$ always denotes a prime number. When proving the identity $$(2)$$ one makes use of the fundamental theorem of arithmetic---each positive integer has a unique representation as a product of prime numbers, where "unique" has a certain technical meaning that I am sure that the reader understands. As is well known, if $$K$$ is a compact subset of the half plane $$(1)$$, then both the product on the left of $$(2)$$, and the sum on the right of $$(2)$$, converge absolutely and uniformly on $$K$$. This easily leads to the conclusion that both sides of $$(2)$$ define the same analytic function of $$s$$ in the half plane $$(1)$$. The really striking feature of the identity $$(2)$$ is that the prime numbers index the product on the left side, but the positive integers index the sum on the right side.

If we wrote $$\zeta(s)$$ for the function defined in the half plane $$(1)$$ by $$(2)$$, then it appears that we could investigate the properties of $$\zeta(s)$$ using the representation on the right side of $$(2)$$, and so learn all about $$\zeta(s)$$ without needing to know anything about prime numbers. Then we could use our knowledge of $$\zeta(s)$$ to discover results about prime numbers. This seems to be an excellent plan for research, but so far the plan has been only partially successful. For example, using the representation for $$\zeta(s)$$ as the sum---the Dirichlet series---on the right of $$(2)$$, it is possible to discover the functional equation for $$\zeta(s)$$, and to also discover the analytic continuation of $$\zeta(s)$$ to an analytic function $$\mathbb{C}$$ except for the simple pole at $$s = 1$$. Several ways of establishing the functional equation were indicated in Riemann's only paper on the zeta function, which was published in 1859. Riemann also made it clear that when tryinng to establish results about prime numbers, a crucial role would be played by the zeros of the function $$\zeta(s)$$. Riemann produced an explicit formula for a prime counting function in which there is sum over the zeros of the function $$\zeta(s)$$. And of course he conjectured that the nontrivial zeros would have real part equal to $$1/2$$.

As the theory of the zeta function developed, it became clear that to investigate the nontrivial zeros of the zeta function it would be necessary to exploit the Euler product representation of the left of $$(2)$$. I assume that the reader is familiar with the proof that $$\zeta(1 + it) \neq 0$$ using the elementary inequality$$0 \le 2(1 + \cos \theta)^2 = 3 + 4\cos\theta + \cos2\theta.$$You will note that the proof makes use of the Euler product representation on the left of $$(2)$$. Today, the best zero-free regions known for the zeta function all use the Euler product in a nontrivial way. Thus the plan outlined above, to learn all about $$\zeta(s)$$ by using the Dirichlet series on the right side of $$(2)$$, has turned out to be somewhat naive. Today it seems that both the Dirichlet series for the zeta function and the Euler product for the zeta function must be exploited.

In your question you proposed a (roughly) analogous approach to studying the distribution of twin primes. Here is a slight variant of your approach. First define$$T = \{p : p \text{ and }p - 2\text{ are both prime numbers}\} = \{5, 7, 13, 19, \ldots\}.$$Then in the half plane $$(1)$$ we could define a corresponding Euler product. A simple calculation shows that$$\prod_{p \in T} (1 - p^{-s})^{-1} = \prod_{p \in T} (1 + p^{-s} + p^{-2s} + p^{-3s} + \ldots) = \sum_{n = 1}^\infty b(n)n^{-s},\tag*{(3)}$$where the coefficients in the Dirichlet series on the right of $$(3)$$ are given by$$b(n) = \begin{cases} 1 & \text{if }n = 1 \\ 1 & \text{if }n = p_1^{e_1}p_2^{e_2}\ldots p_L^{e_L},\text{ where }\{p_1, \ldots, p_L\} \subseteq T \\ 0 & \text{if }p \mid n,\text{ where }p\text{ is prime and }p \notin T.\end{cases}\tag*{(4)}$$Again the elementary manipulations of the product and series in $$(3)$$ can be justified by absolute and uniform convergence of the relevant partial products and partial sums on compact subset of the half plane $$(1)$$. As with the zeta function, we can conclude that the expressions on both sides of $$(3)$$ define the same analytic function in the right half plane $$(1)$$. Write $$\zeta_T(s)$$ for the function defined in the right half plane $$(1)$$ by both the left and right side of $$(3)$$. Now, however, we encounter a formidable difficulty. The product on the left of $$(3)$$ which defines $$\zeta_T(s)$$ is indexed by the mysterious twin primes. And the sum on the right of $$(3)$$ which also defines $$\zeta_T(s)$$ contains the equally mysterious function $$b(n)$$. Thus neither side of $$(3)$$ can be investigated in a simple way. Alternatively, if we define$$U = \{n : 1 \le n \text{ and }b(n) = 1\},$$then unlike the situation with the primes and the positive integers, both $$T$$ and $$U$$ are mysterious subsets of the positive integers.

A basic unsolved problem is to decide if $$T$$ is a finite set or an infinite set. Plainly we would not need to use the Riemann zeta function to decide of the set of (ordinary) prime numbers is finite or infinite. So let us consider the possibility that $$\zeta_T(s)$$, as defined by $$(3)$$, might shed some light on the question---is $$T$$ a finite set or an infinite set? Let us start by assuming that $$T$$ is a finite set. Under this assumption the Euler product on the left of $$(3)$$ is finite, and so it defines an analytic function at every point $$s$$ in $$\mathbb{C}$$ except those points $$s$$ that satisfy $$(1 - p^{-s}) = 0$$ for some prime $$p$$ in the finite set $$T$$. It is easy to see that if $$p$$ is a prime number in $$T$$, then$$\{s \in \mathbb{C} : (1 - p^{-s}) = 0\} = \left\{ {{2\pi i m}\over{\log p}} : m \in \mathbb{Z}\right\}.\tag*{(5)}$$Also, if $$p_1$$ and $$p_2$$ are distinct prime numbres in $$T$$, if $$m_1$$ and $$m_2$$ are integers, we may ask if$${{2\pi im_1}\over{\log p_1}} = {{2\pi im_2}\over{\log p_2}}.\tag*{(6)}$$But $$(6)$$ implies that $$p_1^{m_2} = p_2^{m_1}$$, and by the fundamental theorem of arithmetic this happens if and only if $$m_1 = m_2 = 0$$. Thus we conclude that $$\zeta_T(s)$$ is analytic at all points of $$\mathbb{C}$$ except at each point$$s = {{2\pi im}\over{\log p}}, \text{ where }p \in T,\text{ and }m \in \mathbb{Z}, \text{ and }m \neq 0,$$where it has a simple pole---that is a pole of order $$1$$. And at $$s = 0$$, where $$\zeta_T(s)$$ has a pole of order $$|T|$$, where $$|T|$$ is the number of primes in $$T$$. To summarize, if $$T$$ is a finite set we know a great deal about the analytic function $$\zeta_T(s)$$. We know that $$\zeta_T(s)$$ has countably many poles, each pole occurs on the imaginary axis, there is a pole of order $$|T|$$ at $$s = 0$$, and all the remaining poles are simple poles at the nonzero points in the sets $$(5)$$. We also know that $$\zeta_T(s)$$ never takes the value $$0$$ in $$\mathbb{C}$$. This follows from the observation that$${1\over{\zeta_T(s)}} = \prod_{p \in T} ( 1 - p^{-s})\tag*{(7)}$$is analytic everywhere in $$\mathbb{C}$$. If $$\zeta_T(s)$$ has a zero at $$s = \alpha$$, then its reciprocal $$(7)$$ would have a pole at $$s = \alpha$$.

We have drawn these conclusions from the assumption that $$T$$ is a finite set, and this may well be the actual state of affairs. However, if we want to prove that $$T$$ is an infinite set, then we might try to derive a contradiction to some fact we have derived about $$\zeta_T(s)$$ under the assumption that $$T$$ is finite. We might try to derive a contradiction by exploiting the fact that we have another representation for $$\zeta_T(s)$$ given by the Dirichlet series$$\zeta_T(s) = \sum_{n = 1}^\infty b(n)n^{-s}.\tag*{(8)}$$At present the representation $$(8)$$ has not been useful because it contains the mysterious coefficients $$b(n)$$. The coefficients $$b(n)$$ encode twin prime information in a manner that differs significantly from the way that $$T$$ encodes twin prime information. Thus we can say quite a bit about the function $$\zeta_T(s)$$ if $$T$$ is finite, but the basic identity $$(3)$$ does not really supply us with a new tool, or with the right tool, to go further.

Update. Let me take another stab at the question. The point that the Euler product converges at $$s = 1$$ is quite good. One way of responding to the question at hand is simply that the Dirichlet series is much harder to understand. We want to define$$F(s) = \prod_{p \text{ twin prime}} \left(1 - \frac{1}{p^s} \right)^{-1} = \sum_n \frac{b(n)}{n^s}$$for some arithmetic functions $$b(n)$$---essentially recording whether $$n$$ is a product of twin primes or not. Here, $$b(n)$$ is not nearly as nice as the function $$1$$ that appears in the zeta function. In particular, one can say the following---in increasing order of sophistication.

1. The zeta function having a pole at $$1$$ is essentially related to $$\sum_n 1/n$$ diverges---this is relatively easy to prove---and implies that there are infinitely many primes. Understanding $$\sum_n b(n)/n$$ is much harder, and---as pointed out---is dependent on sieve theoretic bounds. Showing that $$\sum_n b(n)/n^a$$ diverges for some $$0 shows that there are infinitely many twin primes, and this is the original---hard---problem. In contrast, showing $$\sum_n 1/n^a$$ diverges for $$a<1$$ is "trivial".
2. One cannot apply partial summation to the Dirichlet series of $$F(s)$$ to get analytic continuation. Unlike the zeta function where $$\sum_{n\le x} 1 = [x]$$ is easily understood.
3. The coefficients for zeta---and other $$L$$-functions---arise from certain representations---this means that $$\zeta(s)$$ can be expressed as the Mellin transform of some automorphic form, and thereby can be continued everywhere. Moreover, the automorphy implies the functional equation for $$\zeta(s)$$. In this context, this is equivalent to Poisson summation and Poisson summation---and other Fourier analytic methods---appear to not help with the coefficients $$b(n)$$.
4. On a more structural level, it is impossible for $$F(s)$$ to be analytically continued with a standard functional equation of degree $$1$$---that would imply $$F(s)$$ belongs to the Selberg class, and all degree $$1$$ Selberg class have been classified.

The natural conjecture would be that $$F(s)$$ cannot be meromorphically continued to the whole complex plane---and I think this can probably be proven assuming certain conjectures. The overall answer is that proving certain fundamental properties of $$\zeta(s)$$ does not depend on any knowledge about primes, while even basic facts about $$F(s)$$ necessitate knowledge about twin primes---so that understanding $$F(s)$$ goes back to the original hard problem of understanding twin primes. A more exact count for twin primes was mentioned. In this context, perhaps it's useful to take a look at the following.

https://en.wikipedia.org/wiki/Twin_prime#First_Hardy%E2%80%93Littlewood_conjecture

Or the more general Hardy-Littlewood conjectures which do state an exact asymptotic for these things.

If you are really energetic, you can have a look at Soundararajan's Bulletin paper.

https://arxiv.org/abs/math/0605696

• Can you please elaborate? Perhaps give an expression for the twin prime zeta function and show why this is difficult? Commented Oct 13, 2018 at 11:34
• Thank you for this excellent answer. For info, I think (8) should be a sum instead of a product. Otherwise thank you very much! Commented Oct 15, 2018 at 7:22