Infinite dimensional lattice for integers and the Riemann hypothesis?

Question

It is known that for each finite set of primes $p$ we have: $\log(p)$ are linear independent over the rational numbers.

We have $\log(ab) = \log(a)+\log(b)$ and $\log(n) = \sum_{p |n}v_p(n) \log(p)$.

We define $\psi(n) := \sum_{p |n}v_p(n) e_p$ where $e_p$ is the $p$-th standard basis vector of the Hilbert space of sequences.

for $q = a/b \in \mathbb{Q}_{>0}$ we define:

$$\psi(a/b) := \psi(a/\gcd(a,b))-\psi(b/\gcd(a,b))$$.

We have for $n \in \mathbb{N}$:

$$|\psi(n)|^2 = \sum_{p|n} v_p(n)^2 \text{ is a natural number}$$

Furthermore the function

$$K(a,b) = \left < \psi(a),\psi(b)\right > = \sum_{p|\gcd(a,b)} v_p(a)v_p(b)$$

is a positive definite kernel on the natural numbers.

Let us define the $\infty$-dimensional lattice $\Gamma$ as:

$\Gamma:= {\psi(q)| q \in \mathbb{Q}_{>0}, |\psi(q)|^2 \equiv 0 \mod(2)}$$

and let us also define:

$$\eta(n) := (-1)^{|\psi(n)|^2}= \prod_{p|n}(-1)^{v_p(n)^2}$$

For $\psi(a),\psi(b) \in \Gamma$ we have:

$$|\psi(ab)|^2 = |\psi(a)+\psi(b)|^2 = |\psi(a)|^2+|\psi(b)|^2+2K(a,b) \equiv 0+0+0 \equiv 0 \mod(2)$$

Therefore $\psi(a)+\psi(b) \in \Gamma$ and $\psi(1) \in \Gamma$.

The lattice shares some properties with the Leech-lattice and the E8 lattice:

1. It is unimodular in the sense that: For a finite set of primes, the gram matrix $(K(p,q))_{p,q \text{ is prime}}$ has $\det=1$.

2. It is even: $\forall x \in \Gamma: |x|^2 \equiv 0 \mod(2)$

3. For all nonzero vectors in the lattice the squared norm is at least $2$.

My questions are these:

1) Probably an easy question: 

$$\eta(mn) = \eta(m) \eta(n) \forall m,n\in \mathbb{N}$$

2) Does it satisfy a Riemann-Hypothesis type of random variable equality: 

$$\forall \epsilon > 0 : \lim_{N\rightarrow \infty} \frac{1}{N^{1/2+\epsilon}}\sum_{n=1}^N \eta(n) = 0$$

3) Is there a relationship to the Riemann Hypothesis and this lattice?

Here is some Sagemath code for experiments.

And here is some picture concerning the 2. question:

Edit: Let $\lambda(n)=(-1)^{\Omega(n)}$ be the Liouville funciton.

Then because $v_p(n)^2\equiv v_p(n) \mod(2)$, we have indeed:

$$\eta(n)= \lambda(n)$$

This answers the first question since it is known that $\lambda$ is multiplicative.

This also answers the second and third question.

Modified question: Is this lattice known in literature?

Thans for your help.

Answer 1

A maybe useful point of view, is to use the Ehrhart polynomial of a polytope, namely the simplex, to count the numbers as points in a simplex:

Let $Q_N:=$ Polytope of $(\{\psi(p)|1\le p \le N, p \text{ prime }\})$

Then the polytope generated by the primes is the simplex, and it is known ( "Computing the continuous discretely" Matthias Beck, Sinai Robins, Second edtion, p.31 ff) that the Ehrhart polynomial is given by:

$$L(Q_N,t) = \operatorname{binomial}(d+t,d)$$

where $d= \pi(N)$ is the dimension of the simplex, and it is equal to the number of primes $\le N$, which is denoted by $\pi(N)$. Let $\hat{\pi}(n):=\{p \text{ prime}: p|n \}$ and let $p_n:=n$-th prime.

Since $L(Q_N,t)$ counts the points $\psi(n)=(x_1,\cdots,x_d)$ in the dilated polytope $t Q_N$ and those points have cooridnates $\ge 0$ with $x_1+\cdots+x_d \le t$, we conclude that, by observing that the sum of the coordinates correspond to $\Omega(n) := \sum_{p|n} v_p(n)$:

$$L(Q_N,t) = |\{\psi(n) : \Omega(n) \le t , \hat{\pi}(n) \subset \hat{\pi}(N!), 1 \le n \le p_{\pi(N)}^t\}|$$

The condition $\hat{\pi}(n) \subset \hat{\pi}(N!)$ is for making sure we look at those numbers which have only prime divisors $\le N$.

We also observe that:

$$L(Q_N,t)-L(Q_N,t-1) = |\{ n : 1 \le n \le p_d^t, \Omega(n)=t, \hat{\pi}(n) \subset \hat{\pi}(N!) \}| = |A_{N,t}|$$

where I have defined:

$$A_{N,t} = \{ n : 1 \le n \le p_d^t, \Omega(n)=t, \hat{\pi}(n) \subset \hat{\pi}(N!) \}$$

We can now define and try to evalute the following sum:

$$F(N,t):= \sum_{k=0}^t \sum_{n \in A_{N,k}} \lambda(n)$$ $$=\sum_{k=0}^t (-1)^k ( L(Q_N,k)-L(Q_N,k-1) )$$ $$=\sum_{k=0}^t (-1)^k ( \operatorname{binomial}(d+k,d)-\operatorname{binomial}(d+k-1,d) )$$

which according to Wolfram Alpha is equal to:

$$=((-1)^t \operatorname{binomial}(d+t+1,d) F_{2,1}(1,d+t+2;t+2;-1)+2^{-d-1})-2^{-d-1}(2^{d+1}(-1)^t \operatorname{binomial}(d+t,d)F_{2,1}(1,d+t+1;t+1;-1))$$ $$=((-1)^t \operatorname{binomial}(\pi(N)+t+1,\pi(N)) F_{2,1}(1,\pi(N)+t+2;t+2;-1)+2^{-\pi(N)-1})-2^{-\pi(N)-1}(2^{\pi(N)+1}(-1)^t \operatorname{binomial}(\pi(N)+t,\pi(N))F_{2,1}(1,\pi(N)+t+1;t+1;-1))$$

where $F_{2,1}(a,b;c;z)$ is the hypergeometric function.

I must admit that the formula is not really "nice" but I think it should be useful, because using the RHS of the last equality with $d$ could maybe be extended to other values for $d,t$ other than the natural numbers.

I can also not see, how to relate the sum $F(N,t)$ to the Liouville sum:

$$\sum_{n=1}^N \lambda(n)$$

The naive idea is that in the limit $N\rightarrow \infty, t \rightarrow \infty$, the two sets:

$$\mathbb{N}:= \lim_{N\rightarrow \infty}\{1,\cdots,N\} \text{ and } \lim_{N,t\rightarrow \infty} A_{N,t} $$

should be intuitively equal.

There is also an equivalent formulation of the Riemann Hypothesis with the points on a different polytope, but while it uses the Ehrhardt polynomials, which in this case are very complicated at $t=1$ I can not see how to derive an analytic formula in this case.

Here you can find some sanity check in SageMaths.

def Omega(n):
    return sum(valuation(n,p) for p in prime_divisors(n))

def lhsK(N,t,k):
    d = prime_pi(N)
    return sum((-1)**Omega(n) for n in range(1,nth_prime(d)**t+1) if set(prime_divisors(n)).issubset(set(primes(N+1))) and Omega(n)==k)

def lhsList(N,k):
    d = prime_pi(N)
    return [n for n in range(1,nth_prime(d)**k+1) if set(prime_divisors(n)).issubset(set(primes(N+1))) and Omega(n)==k]


def rhs0(N,t):
    d = prime_pi(N)
    return (t-d)*binomial(d-t-1,d)/(d+1)+(t-d-1)*(binomial(d-t,d)+d+1)/(d+1)

def rhsK(N,k):
    d = prime_pi(N)
    return ((-1)**k*(binomial(d+k,d)-binomial(d+k-1,d)))

#https://doc.sagemath.org/html/en/reference/functions/sage/functions/hypergeometric.html

def bino(x,y):
    return gamma(x+1)/(gamma(y+1)*gamma(x-y+1))


def rhsH(N,t):
    d = prime_pi(N)
    return ((-1)**t*bino(d+t+1,d)*hypergeometric([1,d+t+2],[t+2],-1)+2**(-d-1))-2**(-d-1)*(2**(d+1)*(-1)**t*bino(d+t,d)*hypergeometric([1,d+t+1],[t+1],-1))


def lhs(N,t):
    d = prime_pi(N)
    return sum((-1)**k*(bino(d+k,d)-bino(d+k-1,d)) for k in range(int(t)+1)) 
    #return sum((-1)**Omega(n) for n in range(1,nth_prime(d)**t+1) if set(prime_divisors(n)).issubset(set(primes(N+1))))

def rhs(N,t):
    d = prime_pi(N)
    return sum(rhsK(N,k) for k in range(t+1))

    
for N in range(4,240):
    for t in range(1,14):
        print(N,t,rhsH(N,t).n(),lhs(N,t).n())

Edit: Let $B_{N,t} = \{n : \Omega(n) \le t , \hat{\pi}(n) \subset \hat{\pi}(N!), 1 \le n \le p_{\pi(N)}^t\}$, so that

$$\operatorname{binomial}(d+t,d) = L(Q_N,t) = |B_{N,t}|$$

The arguments above show that:

$$\sum_{n \in B_{N,t}} \lambda(n) = F(N,t) = $$

$$=((-1)^t \operatorname{binomial}(d+t+1,d) F_{2,1}(1,d+t+2;t+2;-1)+2^{-d-1})-2^{-d-1}(2^{d+1}(-1)^t \operatorname{binomial}(d+t,d)F_{2,1}(1,d+t+1;t+1;-1))$$

hence dividing by the number of points under which the sum runs, we get:

$$\frac{1}{|B_{N,t}|} \sum_{n \in B_{N,t}} \lambda(n) = \frac{1}{|B_{N,t}|} F(N,t) = $$ $$=(-1)^t \frac{d}{t+1}F_{2,1}(1,d+t+2;t+2;-1)+(-1)^t F_{2,1}(1,d+t+2;t+2;-1)+\frac{1}{2^{d+1} \operatorname{binomial}(d+t,d)}+(-1)^{t+1}F_{2,1}(1,d+t+1;t+1;-1)$$

where $d:=\pi(N)$.

Notice the similarity to the prime number theorem:

We have in both cases sets $C_N$ such that:

$$\lim_{N\rightarrow \infty} \frac{1}{|C_N|} \sum_{n \in C_N} \lambda(n) = 0$$

The case $C_N := \{1,\cdots,N\}$ corresponds by Landau equivalently to the prime number theorem.

We observe that:

$$\lim_{N \rightarrow \infty} B_{N,t} = \{n \in \mathbb{N} | \Omega(n) \le t \} =: B_{\infty,t}$$

and that

$$\lim_{t \rightarrow \infty}(\lim_{N\rightarrow \infty} B_{N,t} ) = \lim_{t \rightarrow \infty} B_{\infty,t} = \lim_{t \rightarrow \infty} \{n \in \mathbb{N} | \Omega(n) \le t \} = \mathbb{N}$$

Similarily: $$\lim_{N \rightarrow \infty}(\lim_{t\rightarrow \infty} B_{N,t} ) = \lim_{N \rightarrow \infty} B_{N,\infty} = \lim_{N \rightarrow \infty} \{n \in \mathbb{N} | \hat{\pi}(n) \subset \hat{\pi}(N!)\} = \mathbb{N}$$

Hence we have:

$$\lim_{t \rightarrow \infty, N \rightarrow \infty} B_{N,t} = \mathbb{N}$$

(Here we have used that there are infinitely many primes.)

Now let us put $C_N := B_{N,\pi(N)}$.

Then one empirical observation, which I was not able to prove directly, is:

$$\lim_{N \rightarrow \infty} \frac{1}{|C_N|} \sum_{n \in C_N} \lambda(n) =^? 0$$

which reminds a little bit on the prime number theorem where $C_N:=\{1,\cdots,N\}$.