6
$\begingroup$

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.

Thanks for your help!

Related: The abc-conjecture as an inequality for inner-products?

$\endgroup$
5
$\begingroup$

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! ;)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy