# The abc-conjecture over the positive rationals and Levy-Schoenberg kernels?

I am continuing the "abc-adventure" and have a specific question, which needs some explanation:

In this paper by Gangolli, the term "Levy-Schoenberg" kernel is defined (Definition 2.3).

Consider the group of $$G = (\mathbb{Q}_{>0},\times)$$ of positive rationals. Then, in this paper by Boudreaux & Beslin, the $$\gcd$$ is extended to $$G$$, which I will call for short $$\gcd^*$$

Having a multiplicative function $$f:\mathbb{N} \rightarrow \mathbb{N}$$, one might extend it to $$G$$ via:

$$f^*\left(\frac{a}{b}\right) = \frac{f\left(\dfrac{a}{\gcd(a,b)}\right)}{f\left(\dfrac{b}{\gcd(a,b)}\right)}$$

Using:

$$\gcd^*(a,b)=1 \iff a,b \in \mathbb{N} \text{ and } \gcd(a,b)=1$$

then $$f^*$$ is multiplicative on $$G$$.

I will look at $$f=\operatorname{rad}$$, hence $$\operatorname{rad}^*$$ is the extension to $$G$$.

Using these "extensions" one might formulate the abc-conjecture over $$G$$.

It is not difficult to show, that it is equivalent to the abc-conjecture of the natural numbers.

My question is, if $$k(a,b) = \frac{\gcd^*(a,b)}{a+b}$$ is positive definite $$\ge 0$$.

Let $$d(a,b) = \sqrt{1-2k(a,b)}$$ and

$$f(a,b) = \frac{1}{2}\big(d(a,1)^2+d(b,1)^2-d(a,b)^2\big)$$

If $$k(a,b)$$ is positive definite over $$G$$, then $$d$$ is an Euclidean metric, and by the characterization of Schoenberg, $$f$$ is posivite definite.

Furthermore:

$$f(a,b) = f(b,a)$$

$$f(a,1) = 0 \quad \forall a \in G$$

$$r(a,b) := f(a,a)+f(b,b)-2f(a,b) = d(a,b)^2$$ is invariant under the action of $$G$$:

$$r(qa,qb) = r(a,b) \quad \forall q,a,b \in G$$

This makes $$f$$ by the definition (2.3) of the paper at the begining of the question to a "Levy-Schoenberg" kernel.

Of course replacing $$k(a,b)$$ with $$k(a,b) := \frac{1}{\operatorname{rad}^*\left(\frac{ab(a+b)}{\gcd^*(a,b)^3}\right)}$$ and using the abstract invariance property: $$k(qa,qb) = k(a,b) \quad \forall q,a,b \in G$$ we can construct more of these Levy-Schoenberg kernels, if the "rad"-function above is a positive definite kernel... which seems difficult to prove.

Why the question, if $$k(a,b)$$ is positive definite:

If we go back to the natural numbers, and define:

$$X_a :=$$ set of divisors of $$a$$. Then $$\mu(X) = \sum_{x \in X} \phi(x)$$ for every finite subset $$X \subset \mathbb{N}$$, and hence $$\mu(X_a) = a$$ and $$X_a \cap X_b = X_{\gcd(a,b)}$$, where $$\phi$$ is the Euler totient function.

Using this, one can prove that over the natural numbers, with the help of this paper by Nader, Bretto, Mourad and Abbas, that:

$$\frac{\gcd(a,b)}{a+b} = \frac{\mu(X_a \cap X_b)}{\mu(X_a)+\mu(X_b)}$$

is positive definite.

My idea was to do the same in the case $$G$$:

Let for $$a \in G$$ be defined $$X_a := \{ d | \gcd^*(a,d) = d \}$$ be the set of divisors of $$a$$, which does not need to be finite.

Then $$X_a \cap X_b = X_{\gcd^*(a,b)}$$.

Hence it remains (?) to find a measure $$\mu$$ on $$G$$ such that for all $$X_a$$ we have:

$$\mu(X_a) = a$$

Then we would have that $$k(a,b)$$ is positive definite!

Of course, one does not need to follow this route, to prove the positive-definiteness of $$k$$, this is just an idea.

I think I found an answer to the question above:

Let $$k(a,b)$$ be a (positive definite $$\ge 0$$, symmetric) kernel on $$\mathbb{N}\times \mathbb{N}$$ such that if $$k^*(a,b)$$ is a function on $$G \times G$$ then we have:

$$k^*(a,b) = k(a',b')$$ where $$a'=\frac{a}{\gcd^*(a,b)}, b'=\frac{b}{\gcd^*(a,b)}$$,

then $$k^*$$ is a kernel on $$G\times G$$.

Proof: Since $$k$$ is positive definite on $$\mathbb{N}\times \mathbb{N}$$, it follows that for $$a_i',b_i' \in \mathbb{N}$$ (which are pairwise coprime), the matrix:

$$k(a_i',b_i')=k^*(a_i,b_i)$$ is positive definte ($$i=1,\cdots,n$$ for some $$a_i,b_i \in G$$, $$n$$ a natural number).

Hence $$k^*$$ is positive definite.

Since $$k^*(a,b) = \frac{\gcd(a,b)}{a+b}$$ satisfies the assumption $$k^*(a,b) = k(a',b')$$, it follows that $$k^*$$ is a positive definite kernel.

Thanks for your patience, with my never-ending questions! ;)