5
$\begingroup$

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?

$\endgroup$
4
$\begingroup$

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.