The abc conjecture and positive definite kernel on the natural numbers with inductive Cholesky decomposition.

This is a transcription of the following MathOverflow question, found here. One formulation of the abc-conjecture is: $$\forall a,b \in \mathbb{N}: \frac{a+b}{\gcd(a,b)} < \operatorname{rad}\left ( \frac{ab(a+b)}{\gcd(a,b)^3}\right )^2$$ Let us define: $$K(a,b) := \frac{2(a+b)}{\gcd(a,b)\operatorname{rad}\left ( \frac{ab(a+b)}{\gcd(a,b)^3}\right )^2}$$ I have checked in Sagemath up to $n=100$ that the matrix $$(K(i,j))_{1 \le i,j \le n} \text{ is positive definite }$$ and up to $1\le a,b \le 100$ that : $$K(a,b)\le 1$$ In the Encyclopedia of Distances the word "similarity" is described as: A similarity $s:X\times X \rightarrow \mathbb{R}$ is defined in the Encyclopedia of Distances as:

1. $s(x,y) \ge 0 \forall x,y \in X$
2. $s(x,y) = s(y,x) \forall x,y \in X$
3. $s(x,y) \le s(x,x) \forall x,x \in X$
4. $s(x,y) = s(x,x) \iff x=y$

So one possible formulation of the empirical observation above might be: $$K \text{ is a pos. def. kernel and a sim. function with } K(a,a)=1$$ From this formulation the abc conjecture in the variant above follows, because: $$K(a,b) \le K(a,a)=1 \rightarrow \frac{(a+b)}{\gcd(a,b)\operatorname{rad}\left ( \frac{ab(a+b)}{\gcd(a,b)^3}\right )^2} \le \frac{1}{2} < 1$$ Since in the definition of the similarity function, some of the properties are easy to prove for $K$ such as 1. and 2. and from 3. follows abc, I am asking if it is possible to prove $\require{enclose}\enclose{horizontalstrike}{\text{ property 4. and / or}}$ that $K$ is positive definite function? Thanks for your help.
Edit: I think that $4.$ can be proved by considering that if $K(x,y)=1$ and assuming there exists a prime $p$ which divides $x$ but not $y$, leads to a contradiction.
Second edit: The abc conjecture (in the version above) would follow, if one could prove that the function $$K(a,b) := \frac{2(a+b)}{\gcd(a,b)\operatorname{rad}\left ( \frac{ab(a+b)}{\gcd(a,b)^3}\right )^2}$$ is positive definite over the natural numbers.
Proof: If $K$ is a positive definite function over the natural numbers, then by the Moore-Aronszajn theorem , there exists a Hilbert space $H$ and a feature mapping $\phi: \mathbb{N} \rightarrow H$ such that: $$K(a,b) = \left < \phi(a),\phi(b) \right >_{H}$$ But by Cauchy-Schwarz inequality we get: $$K(a,b) = |K(a,b)| = |\left < \phi(a), \phi(b) \right >_{H}|$$ $$\le |\phi(a)||\phi(b)| = \sqrt{K(a,a)K(b,b)} = \sqrt{1 \cdot 1} = 1$$ But then we get by division of $2$ and application of the last inequality: $$\frac{(a+b)}{\gcd(a,b)\operatorname{rad}\left ( \frac{ab(a+b)}{\gcd(a,b)^3}\right )^2} \le \frac{1}{2} < 1$$ hence it would follow that: $$\forall a,b \in \mathbb{N}: \frac{a+b}{\gcd(a,b)} < \operatorname{rad}\left ( \frac{ab(a+b)}{\gcd(a,b)^3}\right )^2$$ Modified question: Is there a way to speed up the computations to try and test if the Gram matrix for $n=1,\cdots,10000$ is positive definite? Here is my code so far, but it is rather slow:

# inductive cholesky decomposition
def rad(n):
    return prod(prime_divisors(n))


def kk(a,b):
    return (2*(a+b)*a**1*b**1/gcd(a,b))*1/rad(a*b*(a+b)/gcd(a,b)**3)**2

def notPositiveDefinite(a,b):
    return rad(a*b/gcd(a,b))

#kk = notPositiveDefinite

def grammat(kk,n):
    return matrix([[kk(a,b) for a in range(1,n+1)] for b in range(1,n+1)],ring=QQ)


PHI = dict([])

def phi(n,useN=False):
    if n in PHI.keys():
        return PHI[n]
    
    if n==1:
        if not useN:
            x = vector([sqrt(kk(1,1))])
        else:
            x = vector([sqrt(kk(1,1)+0.0)])
        PHI[n] = x
        return x
    else:
        w = CC(n-1)**(-1)*vv(n-1)
        lw = list(w)
        if not useN:
            lw.append(sqrt(kk(n,n)-w.norm()**2))
        else:
            lw.append(sqrt(kk(n,n)-w.norm().n()**2).n())
        x = vector(lw)
        PHI[n]= x
        return x

def vv(n):
    return vector([kk(i,n+1) for i in range(1,n+1)])
    

CCDict = dict([])    

def CC(n):
    if n in CCDict.keys():
        return matrix(CCDict[n])
    mm = []
    for i in range(1,n+1):
        vi = list(phi(i))
        if len(vi) < n:
            vi.extend((n-i)*[0])
        mm.append(vi)
    M= matrix(mm)
    CCDict[n] = mm
    return M
    
    

for n in range(1,200):
    print(n,all(x in RR for x in phi(n,useN=True)))
    #print(grammat(kk,n).rational_form())
```

Comment on code:
It is based on inductive Cholesky decomposition as described below and keeps track in a dictionary if the $\phi(n)$ or $C_n$ have already been computed. It uses float approximation for greater speed and in the last step we check if each component of the vector $\phi(n)$ is a real number. If it is for all $1 \le k \le n$ then we conclude that the Gram matrix of $n$ is positive definite. If it is not, then at least the Gram matrix for $n$ is not positive definite:
Inductive Cholesky decomposition to find an embedding $\phi(n)$ for each number $n$:
Equation 1): $$\phi(n+1) = \left(\begin{array}{r} {C_{n}^{-1} v_n} \\ \sqrt{k(n+1,n+1)-{\left| {C_{n}^{-1} v_n} \right|}^{2} } \end{array}\right)$$ Equation 2): $$v_n = \left(\begin{array}{r} k(1,n+1) \\ k(2,n+1) \\ \cdots\\ k(n,n+1)\\ \end{array}\right)$$ Equation 3): $$C_n = \left(\begin{array}{r} \phi(1)^T \\ \phi(2)^T \\ \cdots\\ \phi(n)^T\\ \end{array}\right)$$ Equation 4): $$\phi(1) := \sqrt{k(1,1)} e_1$$ If we assume that the function $k$ is positive definite, then this inductive method, should give us the vectors $\phi(a),\phi(b)$ such that under the normal inner product we have: $$k(a,b) = \left < \phi(a), \phi(b) \right >$$ . Edit (11.03.2024): This conjecture about the positive semidefiniteness of $K$ is wrong as the determinant of $G_{192}$ is < $0$

Application of Ehrhart polynomials for counting of numbers and lattices for natural numbers.

This is a transcription of the MO question written here: It is known that $\log(p)$ is independent as a vector over the rational numbers where $p$ is a prime. 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 function. 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? Thanks for your help. ------------------------------------------------------------------------------------------------- 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 edition, 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 coordinates $\ge 0$ 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 evaluate 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 SageMath. 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 $$\lim_{t \rightarrow \infty}(\lim_{N\rightarrow \infty} B_{N,t} ) = \lim_{t \rightarrow \infty} B_{\infty,t} = \lim_{t \rightarrow \infty} \{n \in \mathbb{N} | \Omega(n) \le t \} = \mathbb{N}$$ Similarly: $$\lim_{N \rightarrow \infty}(\lim_{t\rightarrow \infty} B_{N,t} ) = \lim_{N \rightarrow \infty} B_{N,\infty} = \lim_{N \rightarrow \infty} \{n \in \mathbb{N} | \hat{\pi}(n) \subset \hat{\pi}(N!)\} = \mathbb{N}$$ 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 $C_N:=\{1,\cdots,N\}$.