# Boolean ring of unitary divisors / Structure of unitary divisors?

I hope this question is appropriate for MO:

Let $$n$$ be a natural number, $$U_n := \{ d | d \text{ divides } n, \gcd(d,n/d)=1\}$$ be the set of unitary divisors.

We can make $$U_n$$ to a boolean ring:

$$a \oplus b := \frac{ab}{\gcd(a,b)^2} = \frac{\operatorname{lcm}(a,b)}{\gcd(a,b)}$$ and $$a \otimes b := \gcd(a,b)$$

Let $$\Pi(n) := \{ p | p \text{ is prime}, p| n\}$$ be the set of prime divisors of $$n$$. We can define a topology on this set where the open sets are

$$\{ \Pi(d) | d \text{ divides } n \}$$

then $$\Pi(\operatorname{rad}(ab)) = \Pi(a) \cup \Pi(b)$$ and $$\Pi(\gcd(a,b)) = \Pi(a) \cap \Pi(b)$$

where $$\operatorname{rad}(x) = \prod_{p|x}p$$ is the radical of $$x$$.

To each open set $$U$$ we define a number

$$\operatorname{rad}(U):= \prod_{p \in U}p$$

The open sets build also a boolean ring with:

$$U \oplus V := U \Delta V$$ where $$\Delta$$ denotes the symmetric difference, and $$U \otimes V := U \cap V$$

Then $$\operatorname{rad}$$ is a isomorphism of boolean rings:

$$\operatorname{rad}(U \oplus V) = \operatorname{rad}(U) \oplus \operatorname{rad}(V)$$ $$\operatorname{rad}(U \otimes V) = \operatorname{rad}(U) \otimes \operatorname{rad}(V)$$ Also $$\operatorname{rad}(\emptyset) = 1$$, where $$1$$ is the zero in $$U_{\operatorname{rad}(n)}$$ and $$\operatorname{rad}(\Pi(n)) = \operatorname{rad}(n)$$, where $$\operatorname{rad}(n)$$ is the one in $$U_{\operatorname{rad}(n)}$$.

Furthermore, since $$k(a,b) = \frac{\gcd(a,b)^2}{ab} = \frac{1}{a\oplus b}$$ is a positive definite function on the natural numbers and a simililarity, we can embedd this boolean ring $$U_n$$ isometrically in Euclidean space $$\mathbb{R}^{2^{\omega(n)}}$$ (on the sphere of radius one with center $$0$$) where $$\omega(n)$$ counts the distinct prime divisors of $$n$$ and we can define a distance between two unitary divisors:

$$d(a,b) = \sqrt{k(a,a)+k(b,b)-2k(a,b)} = \sqrt{2(1-\frac{1}{a\oplus b})}$$

Also for all $$a,b,c \in U_n$$ we have:

$$k(c\oplus a , c \oplus b ) = k(a,b)$$

My (soft) question is this:

Is this of any use for anything, maybe in number theory? :) Thanks for your help.

This is too long for a comment, so I am writing an answer after two years. :-) Finally something useful.

Idea: Integer composition <-> Subsets of a finite set <-> Bolean ring of unitary divisors

Integer compositions: https://en.wikipedia.org/wiki/Composition_(combinatorics)

The integer compositions could be useful in algorithmic composition/music for manipulation of durations in a bar:

(For instance: One could measure with nearest neighbors how far / near two bars are from the durations (/integer compositions) perspective using the kernel above.)

Algorithmic composition:

To every bar associate to the durations of the bar the composition of an integer.

For example:

Durations of bar:                1/4,1/4,1/8,1/8,1/4

composition of the integer n=8:   2,   2,  1 , 1 , 2 ( 2+2+1+1+2  = 8)

subset of {1,2,..,n-1} :         {2,4,5,6}

unitary divisors of P_{n-1} = 2*3*...*p_{n-1}:     3*7*11*13 = 3003

In case one needs to compute a direct embedding (without requiring to compute the expensive Cholesky decomposition) of these feature vectors, here is an example how to do it:

Let $$e_d$$ be the $$d$$-th standard-basis vector in the Hilbert space $$H=l_2(\mathbb{N})$$. Let $$h(n) = J_2(n)$$ be the second Jordan totient function. Define:

$$\phi(n) = \frac{1}{n} \sum_{d|n}\sqrt{h(d)} e_d$$.

Then we have:

$$\left < \phi(a),\phi(b) \right > = \frac{\gcd(a,b)^2}{ab}=:k(a,b)$$

The vectors $$\phi(a_i)$$ are linearly independent for each finite set $$a_1,\cdots,a_n$$ of natural numbers, since

$$\det(G_n) = \prod_{i=1}^n \frac{h(a_i)}{a_i^2}$$ is not zero, where $$G_n$$ denotes the Gram matrix.

Here is some sagemath code which does the translation between unitary divisors of the primorial numbers and integer compositions:

Sage Cell Server with code

• A piece of music generated with the method above and p-adic numbers: youtube.com/watch?v=YE4gObYHU60 Commented Mar 23, 2022 at 21:02