Nothing
Knot or Unknot ?"
In polarzonoid: Compute Maps and Properties of Polar Zonoids
knitr::opts_chunk$set(echo = TRUE)
# old_opt = options( width=144 )
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 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
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).
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.
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
\medskip
Session Information
This document was prepared
r format(Sys.Date(), "%a %b %d, %Y")
with the following configuration:
```r
# options(old_opt)
sessionInfo()
```
Try the polarzonoid package in your browser
Any scripts or data that you put into this service are public.
polarzonoid documentation built on June 13, 2025, 9:08 a.m.
R Package Documentation
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We want your feedback!
Note that we can't provide technical support on individual packages. You should contact the package authors for that.
knitr::opts_chunk$set(echo = TRUE) # old_opt = options( width=144 )
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 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
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).
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.
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
\medskip
Session Information
This document was prepared
r format(Sys.Date(), "%a %b %d, %Y")
with the following configuration:
```r # options(old_opt) sessionInfo()
```
Try the polarzonoid package in your browser
Any scripts or data that you put into this service are public.
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