Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Take any natural number $n$, and factor it as $n=2^{e_1} 3^{e_2} 5^{e_3} ... p^{e_i}$, where $i$ is the $i$-th prime. Now map $n$ to the point $n \mapsto (e_1,e_2,\ldots,e_i,0,\ldots)$, where $i$ is the last prime in the factorization of $n$. For example, $$n=123456789 \mapsto (0,2,\ldots,1,\ldots,1,\ldots)$$ because $123456789=2^0 3^2 \cdots 3607^1 \cdots 3803^1$. So every number in $\mathbb{N}$ is mapped to a point in an arbitrarily high dimensional space. This mapping has the property that addition of the vectors corresponds to multiplication of the numbers.

My (extremely vague!) question is: Does this viewpoint helps gain any insights into the structure/properties of natural numbers? Do line/planes/curves in this space mark out numerically interesting regions? Perhaps allowing real or complex numbers? The numbers in $\mathbb{N}$ fill this infinite-dimensional space very sparsely.

This is (very!) far from my research expertise, so any comments/references/links would be appreciated.

share|improve this question
    
@KennyTM: Yes, of course, any extensions are welcome.! I am especially interested in what natural curves in this space imply for real numbers... –  Joseph O'Rourke Aug 20 '10 at 20:15
5  
This map looks like a Godel encoding. I think you will want to define this as a map into an infinite dimensional space rather than "arbitrarily high dimensional" (by simply putting zeroes after all nonzero powers of primes in the vector). I have heard of using some topology (Zariski/positive-information topology?) on this space but never geometry... interesting question! –  Tom Stephens Aug 20 '10 at 20:16
1  
Just a naïve comment in the midst of the knowledgeable responses, but I thought it worth mentioning that your map is a monoid isomorphism from the positive integers with multiplication to the countably infinite direct sum of the nonnegative integers with addition. This is just another way of saying that there are infinitely many primes and that each natural number has a unique prime factorization. –  Jonas Meyer Oct 31 '10 at 18:50
2  
This is such an interesting question. It suddenly occurs to me that we can define the angle between natural numbers; for example, to calculate the angle between $12$ and $8$, we first calculate the "inner product", like $\langle 12, 8\rangle = \langle 2^2 3^1,2^3 3^0 \rangle = (2 \cdot 3)+(1 \cdot 0) = 6$. Then we compute the "norm" of each number: $\|12\|^2 = \|2^2 3^1\| = 2^2+1^2 = 5$, so $\|12\| = \sqrt{5}$. Similarly $\|8\|=3.$ Hence $\cos \mathrm{angle}(12,8) = \frac{\langle 12,8\rangle}{\|12\| \cdot \|8\|} = \frac{6}{\sqrt{5}\cdot 3}.$ –  goblin Feb 21 at 18:36
    
So $\mathrm{angle}(12,8) \cong 0.073 \cdot 2\pi$. –  goblin Feb 21 at 18:37

4 Answers 4

up vote 11 down vote accepted

In algebraic number theory such rings of valuation vectors or adèles or repartitions play a key role in globally structuring local information for the global study of number fields. In particular, the topological properties of such adèle rings imply key results in algebraic number theory such as the structure of unit groups, finiteness of the class number, various results at the foundation of the Riemann-Roch theorem, etc. For example, see Tate's below review of a paper by Iwasawa (esp. the penultimate paragraph).

MR0053970 (14,849a) 10.0X
Iwasawa, Kenkichi. On the rings of valuation vectors. Ann. of Math. (2) 57, (1953). 331--356.

Let $K$ be a finite algebraic number field, or an algebraic function field of one variable over a finite constant field. The ring $R$ of valuation vectors over $K$ has the following properties: (1) $R$ is a semi-simple commutative ring with unit element 1; (2) $R$ is locally compact, but neither compact nor discrete; (3) $R$ has a subfield $K$ containing 1, such that $K$ is discrete in $R$ and the residue class space $R/K$ is compact.

In the first half of his paper the author proves conversely that any topological ring having these properties is the ring of valuation vectors over a field $K$ of the type described above. The main tool used is the notion of the norm $N(\sigma,G)$ of an automorphism $\sigma$ of a locally compact group $G$, which is the factor by which the Haar measure in $G$ is stretched by $\sigma$. It is multiplicative in $\sigma$; $N(\sigma,G)=N(\sigma,H)N(\sigma,G/H)$ if $\sigma H\subset H$; and $N(\sigma,G)=1$ if $G$ is compact or discrete. If $F$ is a nondiscrete locally compact field, then the function $N(\alpha,F)$ for $\alpha\in F^\ast$ gives the "normed" valuation of $F$. The intersection of all closed maximal ideals $M$ of $R$ is 0, and $N(x,R)=\prod_MN(x,R/M)$ for regular $x\in R$. In particular, if $\xi\in K^\ast$, then from (3), $N(\xi,R)=1$, so that the valuations $N(\xi,R/M)$ of $K$ satisfy the ``product formula.'' From this it follows by more or less familiar methods that $K$ is an arithmetic field and $R$ its valuation vector ring. The next section contains a proof that valuation vector rings do enjoy properties (1), (2), and (3).

Finally, the author shows that the two central theorems of algebraic number theory can be proved directly from the topological properties of $R$. The first of these is the self-duality of $R$ with respect to $K$, on which the analog of the Riemann-Roch theorem for number fields rests. If $\chi$ is any non-trivial additive character of $R$ vanishing on $K$, then the map $x\rightarrow\chi(xy)$ is an isomorphism of $R$ onto its character group, and $\chi(x\xi)=1$ for all $\xi\in K$ if and only if $x\in K$. The essential step in the proof is to show that the kernel of the map $x\rightarrow\chi(xy)$ is a compact ideal of $R$, and is therefore 0. The second theorem states the compactness of the group of idèle classes of norm 1, which is equivalent to the finiteness of class number and unit theorem. The idèle group $J$ is the multiplicative group of regular elements of $R$. It is a locally compact group in the topology for which the convergence of $a_i$ in $J$ means the convergence of $a_i$ and $a_i{}^{-1}$ in $R$. The map $a\rightarrow N(a,R)$ is a homomorphism of $J$ with kernel $J_1$, and the compactness of $J_1/K^\ast$ is proved by a Minkowski-type argument using Haar measure.

Throughout the paper, the case of function fields over non-finite constant fields is treated in parallel, using linear compactness instead of compactness.
Reviewed by J. T. Tate

share|improve this answer
1  
What a terrifically detailed response to my ridiculously imprecise question! Adèle rings Indeed! –  Joseph O'Rourke Aug 20 '10 at 21:34

This is a really broad question, so I'll just point you to two closely related notions. (About which, I should admit, I know very little.)

In the function field setting, the objects you're mapping into go by the name of divisors. Divisors are an important tool in algebraic geometry and the theory of Riemann surfaces, e.g. this is how people talk about Riemann-Roch. Back in the number field setting one can still define divisors and even ask for an arithmetic analogue of Riemann-Roch, which is described in the relevant chapter of Neukirch (I don't have it on hand at the moment). This type of stuff is the domain of Arakelov theory.

Alternately, instead of just taking the exponents of the prime factorization you could embed the integers into their adele ring; this is, for example, the modern way to do class field theory.

share|improve this answer
    
A wonderfully rich response, well beyond what my vague question deserves. I will learn much from exploring your links. Thanks so much! –  Joseph O'Rourke Aug 20 '10 at 20:42

A friend of mine and me were discussing this possibility some years ago. This vectorial space with infinite dimension is what we called the "encoded space". There are some interesting facts, for instance, prime numbers in $\mathbb{N}$ are in the sphere of radius 1 in the enconded space. Also, you can encode far beyond natural numbers. In fact, you can encode the whole set of $\mathbb{Q}$, rational numbers if you write negative numbers inside the coordinates of the vector. But the thing is that you can even encode any root of a rational number if you add the rational numbers inside the vector. For instance:

$$ 12 = 2^2·3 \rightarrow (2, 1, 0, 0,...) = (2, 1 \rangle $$ $$ \frac{12}{7} = 2^2·3·7^{-1} \rightarrow (2, 1, 0, -1 \rangle $$ $$ \sqrt[7]{\frac{12}{5}} = (2^2·3·5^{-1})^{\frac{1}{7}} =2^{\frac{2}{7}}·3^{\frac{1}{7}}·5^{-\frac{1}{7}} \rightarrow (\frac{2}{7}, \frac{1}{7}, -\frac{1}{7} \rangle $$

Also, some transcendental numbers can be encoded if you add to the vector some irrational algebraic numbers:

$$ 2^{\sqrt{2}} \rightarrow (\sqrt{2}\rangle $$

There is no problem if you try to encode these kind of numbers, BUT the really hard problem arrives when you try to encode the usual summation between real numbers. For instance, you can encode $\frac{1}{2}$ and also $\frac{\sqrt{5}}{2}$, but you cannot encode this:

$$ \frac{1}{2} + \frac{ \sqrt{5}}{2} $$

I've been trying to find a way to encode this but dealing with summation codification is difficult.


I think that the most interesting thing is that it's easy to prove that the exponent-vectors and the encoded space behave as a vector space. It can easily proved (we did some years ago). The most important implication of this is that you can take all the theorems from linear algebra and apply them to the natural numbers or the rational numbers. SPOILER: we try to get the equivalent theorems from algebra to $\mathbb{N}$ and $\mathbb{Q}$ and it was a little bit disappointing (pretty obvious facts).

share|improve this answer

This doesn't answer your question, but I just wanted to offer an algebraic/categorical viewpoint on what's going on here. In particular, I want to answer the question:

Where does this map into infinite-dimensional space come from?

First, some notation: given a set $X$:

  • write $\mathbb{N}^X$ for the set of all functions $\mathbb{N} \leftarrow X$
  • $\mathbb{N}^X_\mathrm{fin}$ for the set of all functions $\mathbb{N} \leftarrow X$ that are finitely supported.

That is, $\mathbb{N}^X_\mathrm{fin}$ is the set of all functions $f : \mathbb{N} \leftarrow X$ such that $\{x \in X \mid 0_\mathbb{N}=f(x)\}$ is cofinite. Now thinking of $\mathbb{N}$ as an additively denoted commutative monoid, it is clear that $\mathbb{N}^X$ is also an additively denoted commutative monoid with respect to pointwise addition. We can think of $\mathbb{N}^X$ as consisting of all $\mathbb{N}$-valued multisets in $X$.

Since $\mathbb{N}^X_\mathrm{fin}$ is a submonoid of $\mathbb{N}^X$, therefore, it can be viewed as a commutative monoid in its own right; its elements are precisely those $\mathbb{N}$-valued multisets in $X$ that are finitely supported. Now it is well known the commutative monoid $\mathbb{N}^X_\mathrm{fin}$ satisfies the universal property of "free (additively denoted) commutative monoid on $X$." In fact, not only is $X$ a basis of $\mathbb{N}^X_\mathrm{fin},$ but in fact, it is the only basis. Hence:

Proposition. Every commutative monoid $C$ has at most one basis.

(The concept "idempotent commutative monoid" also has this remarkable property.)

Of course, this holds even if we change from additive to multiplicative notation; so, writing $\mathbb{N}^\times$ for the set $\mathbb{N} \setminus \{0\}$ viewed a commutative monoid with respect to multiplication, we may conclude that $\mathbb{N}^\times$ has at most one basis. Now let $Q$ denote an arbitrary subset of $\mathbb{N}$. There is function

$$\pi_Q : \mathbb{N}^\times \leftarrow \mathbb{N}^Q_\mathrm{fin}$$

given by taking the product; for example, if $Q=\{2,4,93\},$ then $$\pi_Q(\{2,2,4\}) = 2 \times 2 \times 4.$$

Now clearly, $\pi_Q$ is always a homomorphism; for example, $$\pi_Q(\{2\} + \{2,4\}) = \pi_Q(\{2,2,4\}) = 2 \times 2 \times 4 = 2 \times (2 \times 4) = \pi_Q(\{2\}) \times \pi_Q(\{2,4\}).$$

Furthermore:

Lemma. For all $Q \subseteq \mathbb{N}^\times$, $Q$ is a basis for $\mathbb{N}^\times$ iff $\pi_Q$ is an isomorphism.

(But why? I'm a little confused about this point. Hmmmm.)

Anyway, we conclude that there is at most one $Q \subseteq \mathbb{N}^\times$ for which $\pi_Q$ is an isomorphism. Now write $\mathbb{P}$ for the subset of $\mathbb{N}^\times$ consisting of all prime numbers. We have:

Theorem. (Fundamental theorem of arithmetic for $\mathbb{N}$). The function $\pi_\mathbb{P}$ is an isomorphism. Hence, $\mathbb{N}^\times$ is free, and its unique basis is given by the set of prime numbers.

The map described in your question, which lands us in infinite-dimensional space, is $\pi_\mathbb{P}^{-1}$.

In summary:

  1. An element of $\mathbb{N}^\times$ can always be made into a finitely-supported $\mathbb{N}$-valued multiset of prime numbers by applying $\pi_\mathbb{P}^{-1}.$
  2. A finitely-supported $\mathbb{N}$-valued multiset of prime numbers can always be made into an element of $\mathbb{N}^\times$ by applying $\pi_\mathbb{P}.$
  3. These operations are homomorphisms, and inverses.
share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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