Discriminant of elliptic curve (Frey-Hellegouarch), j-invariant and positive definite kernels, similarities?

Question

Consider the Frey-Hellegouarch curve given $a,b$ natural numbers:

$$y^3= x(x-\frac{a}{\gcd(a,b)})(x+\frac{b}{\gcd(a,b)})$$

Then the discriminant is given by $\Delta = \Delta(a,b) = 16 \left(\frac{ab(a+b)}{\gcd(a,b)^3}\right)^2$.

A similarity $s:X\times X \rightarrow \mathbb{R}$ is defined in 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. $s(x,y) = s(x,x) \iff x=y$

A positive definite kernel $k$ is a positive definite function on some set $X$.

A "kernel-similarity" is a function $f$ which is a kernel and a similiarity.

One can prove that $\frac{1}{16 \left ( \frac{ab(a+b)}{\gcd(a,b)^3}\right )^2}$ is a kernel-similarity.

If $f$ is a "kernel-similarity" such that $f(a,a) = f(b,b) \forall a,b \in X$ then:

$$d(a,b) = \sqrt{f(a,a) + f(b,b) - 2f(a,b)}$$

is a metric space which can be embedded to a Hilbert space. (Or for finite $X$ can be embedded to Euclidean space)

It is conjectured that $\frac{1}{\operatorname{rad}(\frac{ab(a+b)}{\gcd(a,b)^3})}$ is a positive definite kernel over the natural numbers. (One can prove that it is a similarity over the natural numbers.). This similarity has to do with the "conductor" of the elliptic curve above. (Although I am not familiar with elliptic curves, but trying to read about them).

My 1). question is if there are other families of elliptic curves indexed by $a,b$ such that the reciprocal of the determinant over all the family of considered elliptic curves:

$$\frac{1}{\Delta(a,b)}=k(a,b)=s(a,b)$$

is a "kernel-similarity" over the natural numbers.

This would allow to apply the theory of Hilbert spaces over the family of elliptic curves.

Thanks for your help.

My 2. question is:

What properties does the above defined elliptic curve have?

( I am planning to read an introductory book on elliptic curves, but could not find much information about the Frey elliptic curve on the net).

Here is some SAGEMATH code to experiment with (https://sagecell.sagemath.org/):

def frey_curve(a,b):
    return EllipticCurve([0,(b-a)/gcd(a,b),0,-a*b/gcd(a,b)**2,0])

def rad(n):
    return prod(prime_divisors(n))

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

from itertools import product

X = list(product(range(1,20+1),range(1,20+1)))

for a,b in X:
    E = frey_curve(a,b)
    print(a,b)
    print(E.conductor())
    print("quotient disc.=")
    print(E.discriminant()/((a*b*(a+b)/gcd(a,b)**3)**2)==16)
    print(E.conductor()%(1/kk(a,b)))
    print(E.gens())
    print(E)
    try:
        print(E.database_attributes())
    except:
        pass

Thanks for your help!

Edit: The j-invariant is given by:

$$j(a,b) = \frac{16^2(a^2+ab+b^2)^3}{a^2b^2(a+b)^2}$$

It seems that:

$$\frac{1}{j(a,b)}$$

is positive definite.

Question: Is $$\frac{1}{j(a,b)}$$ a positive definite function over the natural numbers?

If the answer to this question is yes, then we could project all natural numbers $n$ on some, say two numbers $x,y$ and visualise the curve after this projection, or apply differential geometry of curves to analyze these projections: Here is the visualization for $x=2,y=3$ and $n=1,\cdots,200$:

Edit: After waiting some time, also asked at MSE: https://math.stackexchange.com/questions/3725860/j-invariant-of-elliptic-curve-frey-hellegouarch-discriminant-and-positive-def

Answer 1

The function $1/j(a,b)$ is positive definite over the natural numbers as can be seen by observing that

$$1/(a+b) = \int_0^1 x^{a+b-1}dx$$

is positive definite (known as a Hilbert matrix). And then we compute:

It is sufficient to show that $\frac{(a+b)^2}{(a^2+ab+b^2)^3}$ is a positive definite function over the natural numbers.

But we have:

$$\frac{(a+b)^2}{(a^2+ab+b^2)^3} = \frac{(a+b)^2}{((a+b)^2-ab)^3} = \frac{1}{(a+b)^6}\frac{(a+b)^2}{(1-\frac{ab}{(a+b)^2})^3}=$$

$$=\frac{1}{(a+b)^4}\frac{1}{(1-\frac{ab}{(a+b)^2})^3}$$

Hence we see that it is sufficient to show that:

$$\frac{1}{1-\frac{ab}{(a+b)^2}}$$

is a positive definite function over the natural numbers:

But by the geometric series, we have:

$$\frac{1}{1-\frac{ab}{(a+b)^2}} = \sum_{k=0}^{\infty} (\frac{ab}{(a+b)^2})^k$$

from which one sees that the function is positive definite over the natural numbers.

I have used the following known facts about kernels:

If $k,K$ are positive definite kernels on some set $X$ and $\lambda>0$, so are:

$$k(x,y)+K(x,y), k(x,y)\cdot K(x,y), \lambda \cdot k(x,y)$$

and so is:

$$\sum_{i=0}^{\infty} k(x,y)^i$$.