---
title: "Knot or Unknot ?"
author: "Glenn Davis"
date: "`r Sys.Date()`"

header-includes:
#  - \usepackage{amsmath}
#  - \usepackage{amssymb}
#  - \usepackage{amsthm}
  - \usepackage{tikz}
#  - \usetikzlibrary{matrix,arrows,decorations.pathmorphing}     
#  - \usepackage{pgfplots}
   
#  rmarkdown::html_vignette:

output: 
  pdf_document:  
    toc: true
    toc_depth: 2
    number_sections: false
    
bibliography: bibliography.bib


# csl: iso690-numeric-brackets-cs.csl
csl: personal.csl
# csl: institute-of-mathematical-statistics.csl
# csl: transactions-on-mathematical-software.csl
vignette: >
  %\VignetteIndexEntry{Knot or Unknot ?}
  %\VignetteEngine{knitr::rmarkdown}
---


```{r setup, include=FALSE}
knitr::opts_chunk$set(echo = TRUE)
# old_opt = options( width=144 )
```

<br><br>


# A Natural Question

Let $A_n$ be the space of $n$ _or fewer_
disjoint arcs in the circle.
In the special case $n{=}0$, we define $A_0$ to be
the 2 improper arcs:  the empty arc and the full circle.
Thus $A_0$ has only 2 points, just like $\mathbb{S}^0 = \{-1,1\}$.
We have the inclusions:

\begin{center}
\begin{tikzpicture}[scale=1.75]

\node (A2) at (0,0) {$A_0$};
\node     at (0.5,0) {$\subseteq$};
\node (B2) at (1,0) {$A_1$};
\node     at (1.5,0) {$\subseteq$};
\node (C2) at (2,0) {$A_2$};
\node     at (2.5,0) {$\subseteq$};
\node (D2) at (3,0) {$...$};
\node     at (3.5,0) {$\subseteq$};
\node (E2) at (4,0) {$A_n$};
\node     at (4.5,0) {$\subseteq$};
\node (F2) at (5,0) {$A_{n+1}$};
\node     at (5.5,0) {$\subseteq$};
\node (G2) at (6,0) {$...$};
\end{tikzpicture}
\end{center}

We make $A_n$ into a metric space as follows.
Let $V$ and $W$ be collections of disjoint arcs in $A_n$, so $V , W \subseteq \mathbb{S}^1$.
Let $\mathbf{1}_{V}$ and  $\mathbf{1}_{W}$ be the indicator functions
for $V$ and $W$.
Define the metric
$d(V,W) := \int_{\mathbb{S}^1} \lvert \mathbf{1}_{V} - \mathbf{1}_{W} \rvert d\theta$.
So $A_n$ can be regarded as a subset of $L^1(\mathbb{S}^1)$ and it inherits a metric from that.
The metric can be viewed in another way; the symmetric difference of $V$ and $W$ is
also a collection of arcs, and $d(V,W)$ is the total length of the arcs in this
symmetric difference.
The corresponding functions in the package are `arcsdistance()` and `arcssymmdiff()`.

In the [**User Guide**](../doc/polarzonoid-guide.html) vignette
it is shown that there are homeomorphisms
\begin{equation}
A_n  ~~ \longleftrightarrow  ~~  \partial Z_n ~~ \longleftrightarrow ~~ \mathbb{S}^{2n}
\end{equation}
where $Z_n$ is the polar zonoid in $\mathbb{R}^{2n+1}$.
These inclusions and homeomorphisms induce embeddings
$\mathbb{S}^{2n} \hookrightarrow \mathbb{S}^{2n+2}$
in this large commutative diagram.

\begin{center}
\begin{tikzpicture}[scale=1.75]

\node (A2) at (0,2) {$A_0$};
\node     at (0.5,2) {$\subseteq$};
\node (B2) at (1,2) {$A_1$};
\node     at (1.5,2) {$\subseteq$};
\node (C2) at (2,2) {$A_2$};
\node     at (2.5,2) {$\subseteq$};
\node (D2) at (3,2) {$...$};
\node     at (3.5,2) {$\subseteq$};
\node (E2) at (4,2) {$A_n$};
\node     at (4.5,2) {$\subseteq$};
\node (F2) at (5,2) {$A_{n+1}$};
\node     at (5.5,2) {$\subseteq$};
\node (G2) at (6,2) {$...$};


\node (A1) at (0,1) {$\partial Z_0$};
\node     at (0.5,1) {$\hookrightarrow$};
\node (B1) at (1,1) {$\partial Z_1$};
\node     at (1.5,1) {$\hookrightarrow$};
\node (C1) at (2,1) {$\partial Z_2$};
\node     at (2.5,1) {$\hookrightarrow$};
\node (D1) at (3,1) {$...$};
\node     at (3.5,1) {$\hookrightarrow$};
\node (E1) at (4,1) {$\partial Z_n$};
\node     at (4.5,1) {$\hookrightarrow$};
\node (F1) at (5,1) {$\partial Z_{n+1}$};
\node     at (5.5,1) {$\hookrightarrow$};
\node (G1) at (6,1) {$...$};


\node (A0) at (0,0) {$\mathbb{S}^0$};
\node     at (0.5,0) {$\hookrightarrow$};
\node (B0) at (1,0) {$\mathbb{S}^2$};
\node     at (1.5,0) {$\hookrightarrow$};
\node (C0) at (2,0) {$\mathbb{S}^4$};
\node     at (2.5,0) {$\hookrightarrow$};
\node (D0) at (3,0) {$...$};
\node     at (3.5,0) {$\hookrightarrow$};
\node (E0) at (4,0) {$\mathbb{S}^{2n}$};
\node     at (4.5,0) {$\hookrightarrow$};
\node (F0) at (5,0) {$\mathbb{S}^{2n+2}$};
\node     at (5.5,0) {$\hookrightarrow$};
\node (G0) at (6,0) {$...$};

\draw [<->] (A0) to (A1);
\draw [<->] (A1) to (A2);
\draw [<->] (B0) to (B1);
\draw [<->] (B1) to (B2);
\draw [<->] (C0) to (C1);
\draw [<->] (C1) to (C2);
\draw [<->] (E0) to (E1);
\draw [<->] (E1) to (E2);
\draw [<->] (F0) to (F1);
\draw [<->] (F1) to (F2);

\end{tikzpicture}
\end{center}

The embeddings in the middle row are induced from the
inclusions on the top row.
The middle row is almost surely a Whitney stratification,
see @WhitneyStratification, but I have not checked this.
The embeddings in the bottom row are induce from those
in the middle row.
We will soon see that the embedding
\begin{equation}\label{embedding}
\mathbb{S}^{2n} \hookrightarrow \mathbb{S}^{2n+2}
\end{equation}
is *not* the standard one, which is formed by appending two zeros.
The codimension is 2,
which is the only codimension where knotting of spheres occurs
(at least in the PL category, see @Zeeman1963).
So it is natural to ask:
\begin{center}
Q: Is the embedding $\mathbb{S}^{2n} \hookrightarrow \mathbb{S}^{2n+2}$ unknotted ?
\end{center}
i.e. is it isotopic to the standard embedding
$\mathbb{S}^{2n} \subseteq \mathbb{S}^{2n+2}$ ?



\medskip

# Unknot

In this section we show that the embedding (\ref{embedding}) is *not* knotted.
For the purposes of this vignette, we use the standard order
for the basis of the trigonometric polynomials:
\begin{equation}
1, \cos(\theta),\sin(\theta), \cos(2\theta),\sin(2\theta),  ... ,\cos(n\theta),\sin(n\theta)  ~~~~~~~~ \theta \in [0,2\pi]
\end{equation}
and put $1$ at the beginning instead of the end,
as in the **polarzonoid** package.

This embedding is the composition:
\begin{equation}\label{comp1}
\mathbb{S}^{2n}  ~~  \to  ~~  \partial Z_n ~~ \hookrightarrow  ~~ \partial Z_{n+1} ~~ \to ~~ \mathbb{S}^{2n+2}
\end{equation}
Focus first on the middle embedding, which is the composition:
\begin{equation}\label{comp2}
\partial Z_n ~~ \to ~~  A_n  ~~ \subseteq  ~~  A_{n+1} ~~ \to  ~~ \partial Z_{n+1}
\end{equation}
If a point $p \in \partial Z_n$ maps to a set of disjoint arcs $a \in A_n$,
then this composition (\ref{comp2}) is:
\begin{equation}
p ~~ \mapsto ~~ \left( ~ p, \int_{a} \cos( (n+1)\theta ) d\theta ,  \int_{a} \sin( (n+1)\theta ) d\theta ~ \right)
\end{equation}\label{comp3}
And since $a$ is a function of $p$, (\ref{comp3}) can be written:
\begin{equation}\label{compv}
p ~~ \mapsto ~~ \left( ~ p, ~ v(p) ~ \right) ~~~~ \text{for a function} ~ v : \partial Z_n \to \mathbb{R}^2
\end{equation}
Returning now to (\ref{comp1}), it is convenient to translate 
$\partial Z_n$ and $\partial Z_{n+1}$ so their centers are at 0.
The center of $\partial Z_n$ is $(\pi, 0, ... , 0)$ so only the first coordinate
is changed, and (\ref{compv}) is still valid.
Denoting the centered sets by adding a prime $'$, the composition is now
\begin{equation}\label{comp4}
\mathbb{S}^{2n}  ~~  \to  ~~  \partial Z_n' ~~ \hookrightarrow  ~~ \partial Z_{n+1}' ~~ \to ~~ \mathbb{S}^{2n+2}
\end{equation}
After these translations, the first and last maps are simple multiplication and
division by positive real functions, and (\ref{comp4}) is:
\begin{equation}
u ~~ \mapsto ~~ \left( ~ \alpha(u) u, v( \alpha(u) u ) ~ \right)
~/~  \left(  \alpha^2(u) + \lvert v( \alpha(u) u ) \rvert ^2 ~ \right) ^{1/2} ~~~~~  \text{where} ~~ |u|=1  ~~ \text{and}  ~~ \alpha(u) > 0
\end{equation}
This can be simplified to
\begin{equation}\label{form}
u ~~ \mapsto ~~ \left( ~ \beta(u) u, ~ w(u) ~ \right) ~~~~~  \text{where} ~~ \beta(u) > 0 ~~ \text{and} ~~ w(u) \in \mathbb{R}^2 ~~ \text{and} ~~ \lvert w(u) \rvert < 1
\end{equation}
We are done if we can show:

**Theorem:**
Any embedding $f : \mathbb{S}^{2n} \hookrightarrow \mathbb{S}^{2n+2}$ that has the form
\begin{equation}
f(u) ~ = ~ \left( ~ \beta(u) u, ~ w(u) ~ \right) ~~~~~  \text{where} ~~ \beta(u) > 0 ~~ \text{and} ~~ w(u) \in \mathbb{R}^2 ~~ \text{and} ~~ \lvert w(u) \rvert < 1
\end{equation}
is isotopic to the standard embedding.

**Proof:**
First note that $\beta(u)$ and $w(u)$ are not independent of each other.
In fact we have:
\begin{equation}
1 ~=~ {\lvert f(u) \rvert}^2 ~=~ \beta(u)^2 \cdot 1^2 ~+~ {\lvert w(u) \rvert} ^2
\end{equation}
and so $\beta(u) = { \left( 1 - {\lvert w(u) \rvert} ^2 \right) }^{1/2}$
and $f$ depends only on $w : \mathbb{S}^{2n+2} \to \mathbb{R}^2$.
Now $\mathbb{R}^2$ is contractible so $w$ is homotopic to the 0 map.
In fact, if we let the homotopy parameter be $t \in [0,1]$,
we can define $w_t(u) := (1-t) w(u)$.
Note that $w_1$ is the 0 map,
and we always have $\lvert w_t(u) \rvert < 1$.
So each intermediate $w_t$ defines an intermediate $f_t$, which is an embedding.
And the final $f_1$ is the standard embedding: $f_1(u) = (u,0,0)$.
$\square$



\medskip

# Some Sample Calculations

```{r, echo=TRUE}
library( polarzonoid )
```

In this section, we verify that the induced embedding
$\mathbb{S}^6 ~ \hookrightarrow ~ \mathbb{S}^8$ has the correct form (\ref{form}),
for a few test cases.
Since the software functions put the constant term last (instead of first)
the indexes for $\mathbb{S}^6$ are 1,2,3,4,5,6,9 (omitting 7 and 8).

```{r, echo=TRUE}
idx = c(1:6,9)

#  make a random unit vector in S^6
set.seed(0)
u = rnorm(7) ;  u = u / sqrt( sum(u^2) )
# embed into S^8
up = spherefromarcs( arcsfromsphere(u), n=4 )
beta = up[idx] / u ; beta
range( diff(beta) )
```
So all the relevant coordinates are scaled by the same number, up to numerical truncation.

Now repeat this test many times.
```{r, echo=TRUE}
count = 50
umat = array( rnorm(count*7), dim=c(count,7) )
umat = umat / sqrt( rowSums(umat^2) )
upmat = t( apply( umat, 1, function(u) { spherefromarcs( arcsfromsphere(u), n=4 ) } ) )
betamat = upmat[ ,idx] / umat
delta = apply( betamat, 1, diff )
range( delta )
```
Once again, the relevant coordinates are scaled by the same number, up to numerical truncation.



\medskip

# References

<div id="refs"></div>





\medskip

# Session Information

This document was prepared
`r format(Sys.Date(), "%a %b %d, %Y")`
with the following configuration:

```
```{r, echo=FALSE, results='asis'}
# options(old_opt)
sessionInfo()
```
```