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Superperiods

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
$\mathfrak{p} \in \mathbb{R}$ is called a superperiod if it can explicitly be shown to be an area of a polygon, not necessarily regular or simple, in the Euclidean space $\mathbb{R}^2$ with length of all sides in $\mathbb{N \cup \{0\}}$ and all interior angles being rational in degrees, i.e., in $\pi \mathbb{Z}$ in radians.

The collection of superperiods are denoted by $\mathfrak{W}$. The operation $+$ acts over them as union of areas by proper adjunction to leave the side lengths integer (which is always possible). The identity thus is the unique $0$-dimensional object. Such operation have, quite trivially, the associativity property.

For example, $\mathbb{Z} \subset \mathfrak{W}$ trivially follows by considering the areas of rectangles in $\mathbb{R}^2$, with angles all $\pi/4$. A more general result is that $\mathbb{Z} + \frac{1}{2} \subset \mathfrak{W}$, as $1/2$ is a superperiod which can be seen by considering a $+$ing two parallelograms with sides all $1$ and the smaller of the angles being $\pi/6$.

It's a matter of fact that parallelogramic tessellation provides are sufficient to represent any (abelian) algebraic integer which follows directly from Kronecker-Weber theorem, thus giving $\mathbb{A^{\!*}} \subset \mathfrak{W}$. Generalized Gau$\beta$ sums can be readily used to develop polygons (parallelogramic tessellations, mostly, and note how every even degree regular polynomial can be tessellated like this). The main breakdown here is to get the subtraction of areas correctly, which I believe can be shown to be always possible.

Now $\mathbb{Q} \subset \mathfrak{W}$ doesn't seem always true (if true, we are working with nothing but $\bar{\mathbb{Q}}$). For example, is $\frac{1}{3}$ in $\mathfrak{W}$? Only SF known this. Now, I conjecture that it isn't. The main back-ups behind this is that $\sin^{-1}\left(\frac{1}{3}\right)$ is likely to have no closed form, let alone being $\pi$-linearly dependent. Perhaps someone would like to have a go at this one?

The post doesn't really particularly asks very much, but rather elaborates an idea which may or may not be developed by someone else. Nevertheless, feel free to have a go! What happens when restriction on $\mathbb{R}^2$ is lifted? Is multiplication of superperiods always a superperiod? Can this be more rigorously stated like periods by multiple integrals on domains in $\mathbb{R}$?

As a remark, a fair warning: this isn't completely a geometry work. It might take a lot of transcendental number theory as well as complicated field and ring theory to get a good idea on what $\mathfrak{W}$ is. So please be advised before jumping on, folks!
 
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mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
Is $1/3$ in $\mathfrak W$?
Well, No.

$\text{arcsin}(1/3)$ is irrational over $\Bbb Q(\pi)$. In particular, $\text{arcsin}(z)$ is $\Bbb Q$-algebraic for only finitely many cases, i.e., when

$$x \in \pm\{0, 1, 1/2\}$$

This is known as Niven's theorem.