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


In this [paper by Gangolli](http://www.numdam.org/article/AIHPB_1967__3_2_121_0.pdf), 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](https://files.eric.ed.gov/fulltext/EJ1093381.pdf), 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.



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> > **My question is, if $k(a,b) = \frac{\gcd^\*(a,b)}{a+b}$ is positive definite $\ge 0$.**
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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](https://www.researchgate.net/publication/326212690_On_the_positive_semi-definite_property_of_similarity_matrices), 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 $X\_a \cap X\_b = X\_{\gcd^\*(a,b)}$.



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> > Hence it remains (?) to find a measure $\mu$ on $G$ such that for all $X\_a$ we have:
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> > $$\mu(X\_a) = a$$
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> > Then we would have that $k(a,b)$ is positive definite!
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Of course, one does not need to follow this route, to prove the positive-definiteness of $k$, this is just an idea.


Thanks for your help!


Related:
[The abc-conjecture as an inequality for inner-products?](https://mathoverflow.net/questions/352054/the-abc-conjecture-as-an-inequality-for-inner-products)



The abc-conjecture over the positive rationals and Levy-Schoenberg kernels?
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! ;)


