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The dimensions of a set of three axes can be arranged in two ways; left or right handed. Cartesian co-ordinates are by convention always oriented to comply with the right-hand rule. It would seem this rule can be thought of as a cyclic transformation of order 3 which takes us from one axis to the next.

What are the analogues for left- and right- handedness in higher dimensions? Particularly in infinite dimensions?

We know that three is geometrically special arising out of the parallelisability of three-sphere, and the only higher dimensional space in which this happens again is 7-dimensions so is there only an analogue in 7-dimensions or can the concept be extended to other spaces?

In particular... and this is just a bit of background perhaps not material to the question. I'm interested in a space I'm constructing to study number theory in which every axis represents a prime number and every point along that axis represents an increment in the power of that prime number, so along the first axis we have 2, 4, 8, ... and on the 2nd axis we have 3, 9, 27, ... By this means every co-ordinate in the infinite-dimensional space represents a unique natural number given by the product of its co-ordinates. If the points along each axis are then spaced according to their square root, Pythagorus theorem guarantees that all points in the infinite-dimensional space are well-ordered by their distance from the origin, which is equal to the square root of the log of natural number they represent.

What I want to bring some understanding to, is how the process of counting in this space is described by some translation from any $x$ to $x+1$, and whether there might be some sense to be made of the rotation between axes that takes place with each increment.

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$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Basis}{\mathbf{e}}$"Handedness" has a natural generalization to arbitrary finite-dimensional vector spaces, or to finite-dimensional smooth manifolds, see also chirality and orientability. Martin Gardner's The New Ambidextrous Universe may be worth a look.

Let $V$ be a finite-dimensional real vector space, and $B = (\Basis_{j})_{j=1}^{n}$ an ordered basis. Given an ordered basis $B' = (\Basis_{j}')_{j=1}^{n}$, there exists a unique linear isomorphism $T:V \to V$ satisfying $T(\Basis_{j}) = \Basis_{j}'$ for each $j$. The basis $B'$ defines the same orientation as $B$ if $\det T > 0$, and defines the opposite orientation as $B$ if $\det T < 0$.

Note carefully that orientation (or handedness, or chirality) is not an intrinsic property of a single ordered basis; instead, it's a relationship between two ordered bases. Often we fix an ordered basis $B$ and declare it to be "positively-oriented", which really means that an arbitrary $B'$ is positively oriented if and only if $B'$ has the same orientation as $B$. (According to Gardner, this issue confused Kant for some time.)

This concept generalizes to finite-dimensional vector spaces over an ordered field. If your space is infinite-dimensional, you may have trouble defining the determinant of an isomorphism. If your field of scalars is not ordered, there's no way to compare $\det T$ with $0$.

In your situation, you're using the bijection from the positive integers to the set of finite sequences of non-negative integers given by the uniqueness of prime factorization: Letting $p_{j}$ denote the $j$th prime number, and letting $(k_{j})_{j=1}^{\infty}$ denote a sequence of non-negative integers with at most finitely many non-zero, you have $$ N \sim (k_{j})_{j=1}^{\infty} \quad\text{iff}\quad N = \prod_{j=1}^{\infty} p_{j}^{k_{j}}. \tag{1} $$ You can view each such sequence as an element of $\Reals^{\infty}$, the vector space of finite sequences of reals.

Your idea of considering the mapping $N \mapsto N + 1$ and the induced action on sequences of exponents in (1) doesn't appear to induce an isomorphism of the ambient vector space $\Reals^{\infty}$, however, so it's not clear whether the preceding definition of orientation will be useful to you.

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  • $\begingroup$ Thanks Andrew. Do I understand your statement "doesn't appear to induce an isomorphism..." correctly? Are you are saying a cyclic translation from one axis to the next must be an isomorphism and there isn't one? I think this is because some numbers are prime and sit on the axes while others are compound and sit off the axes. If I understand correctly then from $p_j$ to $p_{j+1}$ does induce an isomorphism. Also, why can each sequence be seen as an element of $R^\infty$? $\endgroup$ – user334732 Aug 7 '16 at 17:28
  • $\begingroup$ By "doesn't appear to induce an isomorphism", I meant that the mapping $N \mapsto N+1$ induces a (bijective) mapping on sequences of exponents, but this is not the restriction of a linear operator. Note, incidentally, that mapping each prime to the next also does not induce an isomorphism, because $2$ (or, its coordinate axis) isn't in the image. Finally, each sequence of exponents may be viewed as real sequence, hence an element of $\Reals^{\infty}$. $\endgroup$ – Andrew D. Hwang Aug 7 '16 at 21:03
  • $\begingroup$ Interesting about mapping one prime to the next not being an isomorphism because as I imagine rotating the axes putting the 5 axis where 3 was, 3 where 2 was, we would need to put 2 where the "infiniteth" prime was. But if we extend the axes to negative exponents does this not suggest we could we project the axes onto an analogue of the projective real line which extends from $\frac{1}{p_{\infty}}$ to $p_{\infty}$ and then have an isomorphism? $\endgroup$ – user334732 Aug 8 '16 at 6:17
  • $\begingroup$ In the "current picture", the set of positive integers is identified with the set of sequences of non-negative integers with only finitely many non-zero terms. In that sense, there's not exactly a "$2$-axis", just a "first coordinate axis", which corresponds to the prime $2$ via (1). (Separately, and to re-emphasize, even if you construct an automorphism of $\Reals^{\infty}$, you may (or may not) have trouble defining a determinant.) $\endgroup$ – Andrew D. Hwang Aug 8 '16 at 10:49
  • $\begingroup$ Could we equally use $\mathbb{N}^{\infty}$ or $\mathbb{Q}^{\infty}$ since every natural number is the product of a list of $n$th primes where $n\in\mathbb{N}$? Is there a reason why you've chosen $\mathbb{R}^{\infty}$? I'm guessing $\mathbb{R}^\infty$ is easier to work with? I understand det T>0 to mean increasing through the primes and det T<0 to be moving from one prime to the next smallest. With respect to the determinant, it looks like we are looking at an infinite set of sums to infinity which may well not converge and may require switching to p-adics; do I understand correctly? $\endgroup$ – user334732 Aug 8 '16 at 15:29

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