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Let $x \in \mathbb{Q}^+$. Then $x$ can be expressed uniquely as a product over primes:

$$x = \prod_{p \text{ prime}} p^{\nu_p(x)}$$

where $\nu_p(x)$ is the p-adic valuation of $x$. Let $\nu(x) = \langle \nu_2(x), \nu_3(x), \ldots \rangle \in \mathbb{Z}^{\oplus \mathbb{N}}$. Then $\nu(x) + \nu(y) = \nu(xy)$ and $a \nu(x) = \nu(x^a)$. Thus the positive rationals can be viewed as a module where addition and multiplication of two module elements correspond to multiplication and exponentiation of their corresponding rationals. Suppose we define the following "inner product":

$$\nu(x) \cdot \nu(y) = \sum_{p \text{ prime}} \nu_p(x) \nu_p(y)$$

This gives rise to a notion of orthogonality, angles, lengths, and volumes between positive rationals. My question is this: Do this "inner product" and the aforementioned notions have any significance, from the perspective of number theory? Have they been studied before?

For example: if $x,y \in \mathbb{Z}^+$ then $\nu(x) \cdot \nu(y) =0$ iff $x$ and $y$ are coprime. Furthermore, if $x \in \mathbb{Z}^+$ then $\lVert \nu(x) \rVert_1 = \Omega(x)$ and $\lVert \nu(x) \rVert_0 = \omega(x)$.

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    $\begingroup$ Of course the insight that the fundamental theorem of arithmetic implies that the positive rationals are isomorphic to the free abelian group $\bigoplus_p \mathbb{Z}$ generated by the set of primes is useful and important, but I don't see any especial significance to these observations of the OP beyond that. They might sometimes lend themselves to notational convenience, of course. $\endgroup$
    – Todd Trimble ♦
    Mar 18, 2018 at 18:53
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    $\begingroup$ $\lVert v \rVert_1$ and $\lVert v \rVert_0$ aren't inner product based notions, are they? For evaluating the significance of the inner product itself, and not the other structure around, we'd want to see significance for $\lVert v \rVert_2$, surely. $\endgroup$
    – Sridhar Ramesh
    Mar 18, 2018 at 23:26
  • $\begingroup$ That $x, y \in \mathbb{Z}^{+}$ are coprime iff the vectors they correspond to are orthogonal according to this inner product is pleasing, but the same would be true for any inner product of the form $\nu(x) \cdot \nu(y) = \displaystyle \sum_{p \text{ prime}} w_p \nu_{p}(x) \nu_p(y)$ for some fixed choice of positive weights $w_{p}$ for each prime. To justify the constant weighting inner product as significant vs. the alternatives, it would be good to see this notion of orthogonality as in some sense significant even when dealing with rationals with a mix of positive and negative valuations. $\endgroup$
    – Sridhar Ramesh
    Mar 18, 2018 at 23:36
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    $\begingroup$ @SridharRamesh Right. According to this inner product, $\nu(pq) \cdot \nu(pq^{-1}) = 0$ for any primes $p,q$. $\endgroup$
    – user76284
    Mar 18, 2018 at 23:49

2 Answers 2

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Here is a high-brow way to interpret your construction.

Recall that the ideles $\mathbb{A}^\times$ of $\mathbb{Q}$ are defined to be the restricted direct product of $\mathbb{R}^\times$ and all $\mathbb{Q}_p^\times$ with respect to the subgroups $\mathbb{Z}_p^\times$.

There is a natural "diagonal" embedding $\mathbb{Q}^\times \subset \mathbb{A}^\times$. Moreover, we can combine all $p$-adic valuations together to get a homomorphism $$\mathbb{A}^\times \to \bigoplus_p \mathbb{Z}.$$

Your "inner product" is then just given by the composition $$\mathbb{Q}^\times \times \mathbb{Q}^\times \subset \mathbb{A}^\times \times \mathbb{A}^\times \to \bigoplus_p \mathbb{Z} \times \bigoplus_p \mathbb{Z} \to \bigoplus_p \mathbb{Z} \to \mathbb{Z},$$ where the penultimate map is component wise multiplication and the last map is the obvious sum.

How does this help? Well it illustrates how your construction can be put into the bigger picture of the more natural ideles, which are certainly a very important part of number theory.

But I don't think your construction has "any significance, from the perspective of number theory". Number theorists tend to work with the more natural diagonal embedding $\mathbb{Q}^\times \subset \mathbb{A}^\times$ into the ideles, as the ideles have more structure (e.g. form a topological group)

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    $\begingroup$ Shouldn't the penultimate map here be component-wise multiplication? $\endgroup$
    – Sridhar Ramesh
    Mar 18, 2018 at 23:16
  • $\begingroup$ Do you know of any identities which might involve this inner product, like ones that involve $\lVert \nu(x) \rVert_1 = \Omega(x)$ and $\lVert \nu(x) \rVert_0 = \omega(x)$? $\endgroup$
    – user76284
    Mar 28, 2018 at 5:50
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You can find the same idea from Graham, Knuth, and Patashnik's book Concrete Mathematics on page 115 and some related results.

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