cube/zenzicubi.co
cube/zenzicubi.co
1
0
Fork 0
Code Issues Pull Requests Actions Packages Projects Releases Wiki Activity

improved rambling, need to add images

Browse Source
This commit is contained in:
queue-miscreant 2026-09-29 06:00:23 -05:00
parent eb245ce11e
commit 3f0ef867d2
3 changed files with 378 additions and 281 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

4
posts/math/stereo/1/index.qmd
View File

@ -485,7 +485,7 @@ $$
While there is some cancellation in the denominator, both products are rather messy.
To get more cancellation, we can add some nice algebraic properties between *i* and *j*.
If we let i and j be *anticommutative* (meaning that $ij = -ji$) then $stij$ cancels with $stji$.
If we let *i* and *j8 be *anticommutative* (meaning that $ij = -ji$) then $stij$ cancels with $stji$.
So that the denominator is totally real (and therefore can be guaranteed to divide),
we can also assert that $i^2$ and $j^2$ are both real.
Then the expression becomes
@ -533,7 +533,7 @@ If you know a little group theory, you might know there are only two nonabelian
In the latter group, *j* and *k* are both imaginary, but square to 1 (*i* still squares to -1)[^4].
[^4]: I'm being a bit careless with the meanings of "1" and "-1" here.
Properly, these are the group identity and another group element which commutes with all others.
Properly, these are the group identity and another group element of order 2 which commutes with all others.
Changing the sign of one (or both) of the imaginary squares in the expression $h / h^{*}$ above
switches the multiplicative structure from quaternions to the dihedral group.

540
posts/math/stereo/3/inductive.qmd
View File

@ -1,5 +1,5 @@
---
title: "Stereography, Algebraic, and Hyperspheres"
title: "Stereographic Hyperspheres"
description: |
TODO
format:
@ -9,74 +9,116 @@ jupyter: python3
date: "2026-09-11"
categories:
- algebra
- topology
- geometric algebra
---
```{python}
#| echo: false
import sympy
from IPython.display import Markdown
from tabulate import tabulate
In [the first post of this series](../1/), we explored the application of the quaternions
to rotation in three dimensions.
Rotation is one problem, but the quaternions and the characterization of the sphere given
correspond to another: how do we parameterize higher-dimensional spheres?
x, x1, z = sympy.symbols("x x_1 z")
```
It's relatively easy to describe spheres implicitly using coordinates.
The natural definition is the locus of points which all have the same distance to the origin
in Euclidean space.
In other words, a point $(x_0, x_1, x_2, ... x_n)$ is on a unit hypersphere
in *n*+1-dimensional space if
$$
x_0^2 + x_1^2 + x_2^2 + ... + x_n^2 = 1
$$
Topological Spheres
-------------------
Guidance from Lower Dimensions
------------------------------
In topology, hyperspheres are some of the primary spaces of interest.
Spheres have a natural geometric definition: the locus of points which all have
the same distance to the origin.
Topologically, however, they're better-described inductively.
Because points on the sphere are constrained by an equation, there is one fewer degree of freedom
than a general point in the space they occupy.
Hence, a sphere in *n+1*-dimensional space is itself *n*-dimensional, and is termed an *n*-sphere.
First, notice that the equation for the circle depends on a single parameter *t*
which ranges over the entire number line.
There is also a point on the circle "at infinity", which "closes" the circle.
Topologically, the resulting space is called the [https://mathworld.wolfram.com/One-PointCompactification.html](one-point compactification).
In other words, the circle is the one-point compactification of the (open) line.
As a basic example, the complex unit circle is a 1-sphere in the 2-dimensional complex plane:
For spheres, the same thing holds true, but the notion of "one-point" starts to become relevant.
We map a 2-dimensional plane to the sphere, so there are two variables.
But if one or both of these variables has a value of "infinity", then they are all said to describe the same point.
$$
o(t) = {1 + it \over 1 - it}
= {1 - t^2 \over 1 + t^2} + i{2t \over 1 + t^2}
$$
Concretely, this gives the topological relation
The explicit map for the 2-sphere is similar; we have two parameters and
have two "nonreal"s *i* and *j*, which turned out to be quaternions.
$$
o_2(s, t) = {1 + is + jt \over 1 - is - jt}
= {1 - s^2 - t^2 \over 1 + s^2 + t^2} + i{2s \over 1 + s^2} + j{2t \over 1 + t^2}
$$
These are valid constructions because division works for both complex numbers and quaternions.
### Topological Insights
In above equation for a circle, we assign values to *t* from a number line,
a 1-dimensional Euclidean space.
More precisely, the line is the imaginary axis $it$ in the numerator.
We also include an extra point "at infinity".
This same point is approached regardless of whether *t* is negative or positive,
and "closes" the circle.
$$
\begin{align*}
o(\infty) &\approx {1 + i\infty \over 1 - i\infty}
\approx {-\infty \over \infty}
\approx -1
\\
o(-\infty) &\approx {1 - i\infty \over 1 + i\infty}
\approx {\infty \over -\infty}
\approx -1
\end{align*}
$$
For (2-)spheres, a similar statement holds true.
Rather than a line, we range over the imaginary plane $is + jt$, a 2-dimensional Euclidean space.
If one or both of the parameters *s* or *t* has a value of "infinity",
then they seem to describe the same point.
$$
\begin{align*}
o_2(s, \infty) &\approx {1 + is + j\infty \over 1 - is - j\infty}
\approx {\infty \over -\infty}
\approx -1
\\
o_2(\infty, t) &\approx {1 + i\infty + jt \over 1 - i\infty - jt}
\approx {\infty \over -\infty}
\approx -1
\end{align*}
$$
In both expressions, the point at infinity contains no "nonreals" like *i* or *j*.
In another sense, the real space is the extra dimension into which the sphere extends as a surface.
Topologically, this description of the resulting space is called the
[one-point compactification](https://mathworld.wolfram.com/One-PointCompactification.html).
In other words, the circle is the one-point compactification of the line,
and in general, an *n* sphere is the one-point compactification of Euclidean *n*-space.
$$
\mathbb{E}^{n} \cup \{ \infty \} \cong S^n
$$
![]()
### Algebraic Dual
As a locus of points, the *n*-sphere exists within *n+1* dimensional space.
But since the sphere is *n*-dimensional, a point on it is described by *n* coordinates,
just like a point in *n*-dimensional space.
We need a way to augment an *n*-dimensional vector with an extra dimension
Fortunately, we have some direction from [the first post in this series](../1/),
in which we derived an explicit map for the 2-, and 3-spheres.
Namely, the result for 2-spheres was derived by the assertion
$$
o = {1 + {\bm v} \over 1 - {\bm v}} = a + {\bm u},
\quad {\bm v} = is + jt,
\quad i^2 = j^2 = -1
\quad ij = -ji
$$
*i* and *j* are quaternions, which form a division algebra.
This makes this expression legitimate, but not easy to generalize to higher dimensions[^1].
[^1]: The limited number of division algebras is typically proved using algebraic topology
through arguments that depend on spheres and quotient spaces thereof.
Fortunately, the argument can be adjusted a little.
### Invariance of Dimension
The topological definition seems to imply that our construction shouldn't care about
how many dimensions are in the space.
In fact, when constructing the 2-sphere, all we cared about was that *i* and *j* anti-commute
to get cancellation.
[Geometric algebra](https://en.wikipedia.org/wiki/Geometric_algebra) gives some tools to generalize
this argument to higher dimensions.
In such an algebra, we have unit vectors $e_0 ... e_{n-1}$ and the following properties:
In an *n*-dimensional algebra, we have unit vectors $e_0, e_1, ..., e_{n-1}$
and the following properties:
- Scalars and vectors can be added and multiplied together,
and all possibilities comprise the algebra
@ -85,8 +127,10 @@ In such an algebra, we have unit vectors $e_0 ... e_{n-1}$ and the following pro
- e.g., $e_0 e_1 = - e_1 e_0$
- Consequently, the square of the product is the negative of the product of the squares
- e.g., $e_0 e_1 e_0 e_1 = - e_0 e_1 e_1 e_0 = - e_0^2 e_1^2$
- Division by anything other than scalars is undefined
The square of a general vector ***v*** with components $x_k e_k$ is a scalar[^2].
A consequence is that the square of a general vector ***v*** with components $x_k e_k$ is a scalar.
This can be seen by arranging the components of the product after distributing as a square:
$$
\begin{align*}
@ -98,76 +142,94 @@ $$
&= \underset{\diagdown}{\sum_k^{n-1} e_k^2 x_k^2}
+ \underset{◥}{ \sum_k^{n-1} \sum_{l > k} e_k e_l x_k x_l }
+ \underset{◣}{ \sum_k^{n-1} \sum_{l < k} e_k e_l x_k x_l }
\\
&= \diagdown + ◥
+ \sum_k^{n-1} \sum_{l > k} e_l e_k x_k x_l
= \diagdown + ◥
- \sum_k^{n-1} \sum_{l > k} e_k e_l x_k x_l
\\
&= \diagdown + ◥ - ◥
= \diagdown
\end{align*}
$$
For simplicity, assume that $e_k^2 = -1$ so that ${\bm v}^2 = - ||{\bm v}||$, its Euclidean norm,
the sum of squares of the extent in each basis.
Consider the expression
Due to anticommutativity, we can cancel the upper and lower triangles, leaving only the diagonal.
$$
\begin{align*}
{1 + {\bm v} \over 1 - {\bm v}}
&= \left( {1 + {\bm v} \over 1 - {\bm v}} \right)
\left( {1 + {\bm v} \over 1 + {\bm v}} \right)
= {(1 + {\bm v})^2 \over (1 - {\bm v})(1 + {\bm v})}
◣ &= \sum_k^{n-1} \sum_{l < k} e_k e_l x_k x_l
= \sum_k^{n-1} \sum_{k < l} e_l e_k x_l x_k
\\
&= {1 + 2{\bm v} + {\bm v}^2 \over 1 - {\bm v}^2}
\\
&= {1 - ||{\bm v}|| \over 1 + ||{\bm v}||} + {2{\bm v} \over 1 + ||{\bm v}||}
= a + {\bm u}
&= - \sum_k^{n-1} \sum_{l > k} e_k e_l x_k x_l
= - ◥
\\[10pt]
&\implies \diagdown + ◥ + ◣ = \diagdown + ◥ - ◥ = \diagdown
\end{align*}
$$
Similarly to quaternions, this is the sum of a vector and a scalar.
If the scalar component is considered the extent in a new dimension,
then the norm of the resulting vector is
To align with the prior examples *i* and *j*, we'll assume that $e_k^2 = -1$ for all *k*.
This means that ${\bm v}^2 = - ||{\bm v}||$, the sum of squares of the extent
in each basis (or Euclidean norm).
### Being Hyperrational
Finally, we can consider an expression analogous to the one from which we derived
the 1- and 2-spheres.
Suppose that a vector and a scalar are added together, as $a + {\bm v}$.
If this point is on a sphere and the scalar component is considered the extent in a new dimension,
then the norm of the entire quantity should be
$$
a^2 + ||{\bm u}|| = a^2 - {\bm u}^2
||a + {\bm v}|| = a^2 + ||{\bm v}|| = a^2 - {\bm v}^2 = 1
$$
According to this definition, we started with the ratio of two expressions with the same norm.
This means that our resulting expression should have a norm of 1.
As a vector ***u*** ranges over *n*-dimensional space, the expression...
$$
||1 + {\bm v}|| = 1^2 - {\bm v}^2
= 1^2 - ({\bm -v})^2 = ||1 - {\bm v}||
o_n({\bm u}) = {1 + {\bm u} \over 1 - {\bm u}}
$$
Consequently,
...seems to be a ratio between two distinct quantities with the same norm,
since $1^2 - {\bm u}^2 = 1^2 - (-{\bm u})^2$.
This expression is actually ill-defined since there is a vector in the denominator,
but we can use a conjugation trick to clear it:
$$
\begin{align*}
a^2 - {\bm u}^2 &= 1
{1 + {\bm u} \over 1 - {\bm u}}
&= \left( {1 + {\bm u} \over 1 - {\bm u}} \right)
\left( {1 + {\bm u} \over 1 + {\bm u}} \right)
= {(1 + {\bm u})^2 \over (1 - {\bm u})(1 + {\bm u})}
\\
&= {1 + 2{\bm u} + {\bm u}^2 \over 1 - {\bm u}^2}
\\
&= {1 - ||{\bm u}|| \over 1 + ||{\bm u}||} + {2{\bm u} \over 1 + ||{\bm u}||}
= a + {\bm v}
\end{align*}
$$
The quantity in the denominator of both components is always a scalar and greater than zero,
so there are no concerns about the validity of division.
We can also show that the norm of this expression is 1, as desired:
$$
\begin{align*}
a^2 - {\bm v}^2 &= 1
\\
\implies
\stackrel{\text{Numerator of } a}{(1 + {\bm v}^2)^2}
- \stackrel{\text{Numerator of } \bm u}{(2{\bm v})^2}
&= \stackrel{\text{Common denominator}}{1 - {\bm v}^2}
\stackrel{\text{Numerator of } a}{(1 + {\bm u}^2)^2}
- \stackrel{\text{Numerator of } \bm v}{(2{\bm u})^2}
&= \stackrel{\text{Common denominator}}{1 - {\bm u}^2}
\end{align*}
$$
The final expression is always valid, no matter how many dimensions ***v*** has.
Not only that, since *a* and ***u*** contain no vectors in the denominator, there are no concerns with the validity of division.
The only reason we started from an expression which did contain vectors was to justify the requirement for anticommutativity.
This is true no matter how many dimensions ***v*** has[^1], justifying our earlier abuse of notation.
This is the denominator, so the dubious step of dividing by a vector has been backed up with pure algebra.
[^1]: Technically, this should only hold for finitely many dimensions.
The $\infty$-sphere, composed of vectors with only finitely many nonzero components,
is probably also valid under this construction, but I haven't proven this.
Despite multiplication between two vectors being defined, dividing one vector by another is not.
:::{}
TODO: Chebyshev relation on this form of the sphere is also valid!
:::
Inductivity
-----------
Inducing an Alternative
-----------------------
The previous topological description of spheres lacks a couple of things:
@ -176,10 +238,10 @@ The previous topological description of spheres lacks a couple of things:
Fortunately, topology has an alternate description.
The the 1-dimensional sphere is a little bit special.
The 0-dimensional sphere is a little bit special.
On a number line, there are two points equidistant to the origin,
and these comprise the 0-sphere $S^0$.
This can be turned into a 1-sphere $S^1$ (the circle) through a topological operation called
This can (topologically) be turned into a 1-sphere $S^1$ (the circle) by an operation called
[suspension](https://en.wikipedia.org/wiki/Suspension_%28topology%29), which connects
all points in the space to two new, auxiliary points.
Subsequently, we can take the circle and repeat the operation to build the 2-sphere $S^2$.
@ -190,121 +252,271 @@ $$
\text{Susp}(S^{n-1}) = S^n
$$
![]()
### Algebraic Dual, Part 2
First, let's look at the first interesting case.
Let's look at the first interesting case.
We first definied the circle, or 1-dimensional sphere as
$$
{1 + it \over 1 - it}
o(t) = {1 + it \over 1 - it}
= {1 - t^2 \over 1 + t^2} + i{2t \over 1 + t^2}
$$
If the first coordinate remains fixed, then in most cases,
the space looks two discrete points, or to wit, a 0-dimensional sphere.
The remaining two points are in some sense "new" to the space.
If the real part remains fixed, then in most cases,
the space looks two discrete points; to wit, a 0-dimensional sphere.
The remaining two points 1 and -1 are in some sense "new" to the space.
Examining the 2-dimensional sphere in the same way, at an intersecting plane,
the space looks like a 1-dimensional sphere except at two points.
![]()
This matches the inductive topological description of spheres one-for-one.
Similarly, when we intersect the 2-sphere with a plane along a line of latitude,
the space looks like a 1-dimensional sphere except at two points, also 1 and -1.
Using the results of the previous section, we have a way to generalize *i* to any dimension.
Coincidentally, this generalization is *also* inductive --
if ***u*** is already a vector with norm 1, then the scalar component *a* must be 0.
Further, this means that ${\bm u}^2 = -1$.
This should sound familiar -- it matches the "unit quaternions"
![]()
Intuitively, this means we can also describe a sphere by the equation:
If we create a 2D vector with components in the real and imaginary parts of *o*,
we can immediately create an expression for the 2-sphere:
$$
{1 + {\bm u}t \over 1 - {\bm u}t}
= {1 - t^2 \over 1 + t^2} + {2t \over 1 + t^2}{\bm u}
= a + b{\bm u}
\begin{align*}
{\bm w}_1(t) &= {1 - t^2 \over 1 + t^2} e_0 + {2t \over 1 + t^2} e_1
\\[10pt]
\varsigma_2(s,t) &= {1 + {\bm w}_1(t)s \over 1 - {\bm w}_1(t)s}
= {1 - s^2 \over 1 + s^2} + {2s \over 1 + s^2} {\bm w}_1(t)
\end{align*}
$$
Note that if *s* is exchanged with *-s*, then the scalar part remains the same,
but the vector part, which corresponds to latitudinal circles, is negated.
In effect, this means that if *s* is allowed to range over negative numbers,
we will produce two duplicate circles.
### Degree Maps
This process can be continued indefinitely -- at each stage,
$\varsigma_k$[^2] describes a *k*-dimensional unit sphere.
It be converted to a pure vector ${\bm w}_k$ by multiplying the scalar component
with a new unit vector $e_k$.
In this form, ${\bm w}_k^2 = -1$ for any *k*-dimensional algebra[^3].
This provides an inductive construction parallel to the topological one.
Since ${\bm u}^2 = -1$, there's an interesting trick we can pull again.
We wrapped the circle around itself twice in [the previous article](../2/) by
simply squaring the same expression from the last article.
That argument is only contingent upon one thing: the split between
real and nonreal components, and the squaring of the unit nonreal to -1.
When going from *i* to ***u***, the only thing that needs changing is replacing
"real" with "scalar" and "nonreal" with "vector".
[^2]: For "σφαίρα", sphere. I'm using ς in hope that it'll be less prone to confusion with "o".
[^3]: This should sound familiar from the first post -- it matches the "unit quaternions".
$$
_n o^m
= ( a + b \cdot { {}_{n-1} {\bm u}} )^m
= T_m(a) + b U_m(a) \cdot { {}_{n-1} {\bm u}}
\begin{align*}
{\bm w}_k(x_0, x_1, ..., x_{k-1})
&= \text{Scalar}(\varsigma_k(x_0, x_1, ..., x_{k-1}))e_k
\\
&+ \text{Vector}(\varsigma_k(x_0, x_1, ..., x_{k-1}))
\\
\varsigma_{k+1}(x_0, x_1, ..., x_{k-1}, x_k)
&= {1 + {\bm w}_k(x_0, x_1, ..., x_{k-1})x_k \over 1 - {\bm w}_k(x_0, x_1, ..., k_{k-1})x_k}
\end{align*}
$$
*T* and *U* here are the standard Chebyshev polynomials.
This amounts to wrapping the sphere around itself any number of times as desired, *m*.
*m* here is only really defined over positive integers here, since the Chebyshev polynomials
are only defined over positive indices.
When the new parameter $x_k$ is 0 or $\infty$, the vector part collapses,
and we get either 1 or -1, the "new points" of the suspension.
$$
\begin{align*}
\varsigma_{k+1}(..., 0)
&= {1 + {\bm w}_k(...)\cdot 0 \over 1 - {\bm w}_k(...) \cdot 0}
- {1 \over 1} = 1
\\
\varsigma_{k+1}(..., \infty)
&\approx {1 + {\bm w}_k(...)\cdot \infty \over 1 - {\bm w}_k(...) \cdot \infty}
\approx {\infty \over -\infty} \approx -1
\end{align*}
$$
The phenomenon of duplicate latitudinal spheres is a recurring one in dimensions greater than 1.
This can be seen from the 0-sphere being unique among spheres --
as two discrete, disconnected points, negative numbers are needed for the expected duplication.
More directly, this means that the parameter attached to the 1D case ($x_0$) ranges over
positive and negative numbers, but all others range over only positve numbers.
:::{TODO}
We could also construct $\bm w$ from the results in the previous section.
:::
Multiple Wrappings
------------------
One feature of the complex rational circle mentioned in [the previous article](../2/)
was that its powers correspond to going around multiple times.
Conveniently, a similar fact holds for *n*-spheres in general.
Starting with the scalar/vector form of the sphere, we can square the sphere and apply
the fact that the difference of squares of each part is constant:
$$
\begin{align*}
o_n &= a + {\bm v}
\\
o_n^2 &= (a + {\bm v})^2 = a^2 + {\bm v}^2 + 2a{\bm v}
\\
&= a^2 + {\bm v}^2 + \textcolor{red}{(a^2 - {\bm v}^2 - 1 = 0)} + 2a{\bm v}
\\
&= 2a^2 - 1 + 2a{\bm v} = 2a(a + {\bm v}) - 1 = 2a o_n - 1
\end{align*}
$$
This gives the familiar recurrence relation...
$$
o_n^{m+2} = 2a o_n^{m+1} - o_n^m
$$
...and thus a sphere can be wrapped around itself any number of times, as given by:
$$
o_n^m = T_m(a) + U_{m-1}(a){\bm u}
$$
*T* and *U* are the standard Chebyshev polynomials.
### Negative Indices and Beyond
The topological equivalent to this statement is
$$
\pi_n(S^n) = \Z
H_n(S^n) = \Z
$$
This states that a map from the *n*-sphere to itself can be characterized by an integer,
the *degree*.
Maps sharing the same integer are considered to be *homotopic* to one another,
and composing maps can be composed in the same way that the integers add.
More directly, a map from the *n*-sphere to itself can be characterized by an integer,
the *degree*, and these compose as integers add.
Since this is an integer, there's the notion maps in an opposite direction
which correspond to negative degrees.
This seems to align with the behavior of the exponent in $o_n^m$.
However, we've only defined *m* over positive integers;
after all, $o_n$ contains a vector, so we can't really divide by it.
Telling us we can wrap an *n* sphere around itself, backwards and forwards any number of times.
"Backwards" comes from antipodal map
Degree of antipodal map is negative only if *n* is even
But this just means we can look at a map where we negate only the vector components.
Fortunately, it's pretty easy to make sense of this.
Since we have a recurrence relation for the powers of the sphere, we
can extend it backwards to define it over negative indices[^4].
$$
\begin{align*}
o_n^1 &= 2a o_n^{0} - o_n^{-1}
\\
a + {\bm v} &= 2a - o_n^{-1}
\\
o_n^{-1} &= a - {\bm v}
\end{align*}
$$
[^4]: The same argument holds for the Chebyshev polynomials.
In general, $T_{-n}(x) = T_n(x)$ and $U_{-1} = 0$, $U_{-n}(x) = -U_{n-2}(x)$ for
the standard indexing of *U*.
If anything, this is another argument that this indexing of *U* isn't very well-suited,
since if $U_0 \stackrel{\Delta}{=} 0$, it follows that $U_{-n}(x) = -U_{n}(x)$.
This actually aligns with what we'd expect according to adding powers, since:
$$
o_n^1 \cdot o_n^{-1} = (a + {\bm v})(a - {\bm v}) = a^2 - {\bm v}^2 = 1 = o_n^0
$$
Equator
-------
### Degrees and Induction
By describing spheres purely in terms of other spheres, we've taken care of the first problem.
We still have another -- if we let a vector *v* range over the entirety of Euclidean space,
we still have to include a point at infinity.
In the inductive case, since ${\bm w}^2 = -1$, we can pull a similar trick.
Replacing *i* with ***w***, the only thing that needs changing from the previous article is
replacing "real" with "scalar" and "nonreal" with "vector".
By virtue of degree, we're still in the clear.
The trick is actually the same from the previous post when integrating.
$$
\varsigma_n^m
= ( c + s \cdot { {\bm w}_{n-1}} )^m
= T_m(c) + s U_m(c) \cdot {\bm w}_{n-1}
$$
For the circle, a degree 1 map wraps around once from $-\infty$ to $\infty$,
and a degree 2 map wraps around once from -1 to 1.
The -1 to 1 range in the degree 1 map describes only a semicircle, whose boundary is
two points (the 0-sphere).
Similarly,
One dimension up, to continue the analogy, we have a 1-sphere which bounds a hemisphere
in the degree 1 map.
The 1-sphere is just the equator of the sphere.
The hemisphere replaces the range "from -1 to 1"; instead, we have a unit disc --
geometrically, this consists of all vectors whose norm is less than 1.
In fact, this still agrees with the 1 dimensional case.
$$
\begin{align*}
\varsigma_n^{-1} &= ( c - s \cdot { {\bm w}_{n-1}} )
\\
\varsigma_n^{1} \cdot \varsigma_n^{-1}
&= ( c + s \cdot {{\bm w}_{n-1}} )( c - s \cdot {{\bm w}_{n-1}} )
\\
&= c^2 - s^2 {\bm w}_{n-1}^2 = c^2 + s^2
\\
&= 1 = \varsigma_n^0
\end{align*}
$$
This analogy continues inductively to all *n*-dimensional spheres.
Another way of seeing this is by looking at the equators of *n*-spheres.
Using the inductive algebraic definition of the sphere, if the scalar component is 0, then
the vector component is just the *n-1*-sphere.
This can be thought of as the equator.
Degree 2 and the Equator
------------------------
The first coordinate expresses the condition that if **v** lies on the unit (hyper)sphere
in the dimension below, then the coordinate is zero, and such points lie on an equator.
The prior discussion about degree gives us the tools to address "points at infinity".
As a reminder, if we let a vector ***u*** range over the entirety of Euclidean space,
the sphere is only closed by allowing such an extra point.
:::
USE THE ABOVE AS JUSTIFICATION FOR THIS SHIT.
MAYBE START EARLIER, WHEN INDUCTION WAS BEING DISCUSSED?
:::
We can get rid of this point for the circle by considering a degree 2 map rather than a degree 1 map.
The former wraps around once $-\infty$ to $\infty$, while the latter wraps around once from -1 to 1.
Conveniently, -1 and 1 are both the points at which the real part of the degree 1 map becomes 0.
Points in this range lie on the semicircle bounded by these two points (which form a 0-sphere).
In the case of the 2-sphere, doubling the map by squaring quaternions produces a more interesting feature:
all **v** lying on the unit (hyper)sphere in the dimension below get mapped to the same point, antipodal to 0.
In other words, the sphere in the next dimension is obtained by considering all points on the boundary of a ball to be the same.
![]()
For higher-dimensional spheres we just need to replace "real part" with "scalar part".
For a sphere $o_n({\bm u}_n)$, this is precisely when ${\bm u}_n$ has a norm of 1.
$$
o_n({\bm u}_n)
= {1 - ||{\bm u_n}|| \over 1 + ||{\bm u_n}||}
+ {2{\bm u_n} \over 1 + ||{\bm u_n}||}
= {1 - 1 \over 1 + 1} + {2 \over 1 + 1}{\bm u_n}
= {\bm u}_n
$$
Let *n* = 2 so we can plot it.
Then we're actually talking about place in which the unit sphere in 3D space looks like the unit circle.
Specifically, this unit circle can be considered an equator bounding the hemisphere around the scalar 1.
![]()
In general ${\bm u}_n$ has a norm of 1 exactly when it's a point on the equatorial *n-1*-sphere.
Conveniently, under the degree 2 map, this equatorial sphere collapses to a single point.
$$
o_n({\bm u}_n)^2
= {\bm u}_n^2 = -1
$$
This point was formerly the image of the point at infinity,
eliminating its necessity in the description of the *n*-sphere.
### Closing the Sphere
Of course, this comes with another topological analogue.
Another description of the *n*-sphere is by taking the boundary of an *n*-dimensional disc
and collapsing its boundary to a single point.
$$
{ D^n / \partial D^n } = S^n
$$
This exactly aligns with the behavior of the equator when going from the degree 1 to the degree 2 map.
If ${\bm u}_n$ has a norm of less than or equal to 1, then it lies within a unit disc.
This unit disc gets sent by $o_n$ to the aforementioned "hemisphere around the scalar 1",
and when fed to $o_n^2$, it produces the *n*-sphere.
Closing
-------
There are a couple of things that I still want to explore here.
One is the composition of one-point spheres and inductive spheres.
This structure should describe an (unbounded) lattice, where every "one-point" parametrization
has no lesser element, but has a greater element as an element in an inductive parametrization.
Parametrizations should be considered the same up to a symmetric transformation of coordinates.
There's also a lot of interesting topological arguments to nail down.
One of these is the degree of the antipodal map.
Another is constructing explicit homotopies purely from algebra.

115
posts/math/stereo/3/index.qmd → posts/math/stereo/4/index.qmd
View File

@ -283,118 +283,3 @@ In fact, higher-order recurrences become more and more restrictive as $x^\bullet
the numerator and denominator.
If there is a rule to determine what relation must be obeyed between the coefficients
for factorization to occur, it is not obvious, especially as the order grows.
Higher-dimensional Spheres
--------------------------
We derived an explicit map for the 2-sphere (or rather, the 3-sphere, since the description ended up matching the quaternions)
in [the first post in this series](../1/).
It's quite easy to generalize the argument to higher dimensions.
In *n* dimensions, assume that we have unit vectors $e_0 ... e_{n-1}$.
Placing these vectors within a [geometric algebra](https://en.wikipedia.org/wiki/Geometric_algebra)
gives some promising properties:
- Vectors can be multiplied like ordinary numbers, and even added to ordinary numbers
- The product of a vector with itself can be chosen among -1, 0, or 1
- The product of two vectors anticommutes (e.g., $e_0 e_1 = - e_1 e_0$)
- Consequently, the square of a product is the negative of the product of the squares (e.g., $e_0 e_1 e_0 e_1 = - e_0 e_1 e_1 e_0 = - e_0^2 e_1^2$)
The square of a general vector with components $x_k e_k$ is a scalar, its norm.
$$
\begin{align*}
{\bm v} &= e_0 x_0 + e_1 x_1 + ... e_{n-1} x_{n-1} = \sum_k e_k x_k
\\
{\bm v}^2 &= (\sum_k^{n-1} e_k x_k) (\sum_l^{n-1} e_l x_l)
= \sum_k^{n-1} \sum_l^{n-1} e_k e_l x_k x_l
\\
&= \underset{\diagdown}{\sum_k^{n-1} e_k^2 x_k^2}
+ \underset{◥}{ \sum_k^{n-1} \sum_{l > k} e_k e_l x_k x_l }
+ \underset{◣}{ \sum_k^{n-1} \sum_{l < k} e_k e_l x_k x_l }
\\
&= \diagdown + ◥
+ \sum_k^{n-1} \sum_{l > k} e_l e_k x_k x_l
= \diagdown + ◥
- \sum_k^{n-1} \sum_{l > k} e_k e_l x_k x_l
\\
&= \diagdown + ◥ - ◥
= \diagdown
\end{align*}
$$
For simplicity, assume that $e_k^2 = -1$ so that ${\bm v}^2 = - ||{\bm v}||$,
the sum of squares of the extent in each basis.
Despite multiplication between two vectors being defined, dividing one vector by another is not.
Ignoring this, consider the expression
$$
\begin{align*}
{1 + {\bm v} \over 1 - {\bm v}}
&= \left( {1 + {\bm v} \over 1 - {\bm v}} \right)
\left( {1 + {\bm v} \over 1 + {\bm v}} \right)
= {(1 + {\bm v})^2 \over (1 - {\bm v})(1 + {\bm v})}
\\
&= {1 + 2{\bm v} + {\bm v}^2 \over 1 - {\bm v}^2}
\\
&= {1 - ||{\bm v}|| \over 1 + ||{\bm v}||} + {2{\bm v} \over 1 + ||{\bm v}||}
\end{align*}
$$
Similarly to quaternions, this is the sum of a vector and a scalar.
If the scalar component is considered the extent in a new dimension,
then the norm of the resulting vector is
$$
a^2 + ||u|| = a^2 - u^2 = (a + u)(a - u)
$$
This is actually an inductive hypothesis.
This forces us to choose two things: the norm we use is Euclidean, and each new unit vector squares to -1.
For the sphere, focusing just on the numerator
$$
(1 - ||v||)^2 - u^2 = (1 - ||v||)^2 - (2v)^2 = 1 - 2||v|| + ||v||^2 - 4v^2 = 1 - 2||v|| + ||v||^2 + 4||v|| = 1 + 2||v|| + ||v||^2 = (1 + ||v||)^2
$$
This is the denominator, so the dubious step of dividing by a vector has been backed up with pure algebra.
### Degree Maps
If $\bm u$ is a vector with norm 1.
The first coordinate expresses the condition that if **v** lies on the unit (hyper)sphere
in the dimension below, then the coordinate is zero, and such points lie on an equator.
In the case of the 2-sphere, doubling the map by squaring quaternions produces a more interesting feature:
all **v** lying on the unit (hyper)sphere in the dimension below get mapped to the same point, antipodal to 0.
In other words, the sphere in the next dimension is obtained by considering all points on the boundary of a ball to be the same.
$$
{ D^n / \partial D^n } = S^n
$$
This applies generally.
This can be made more topological by doubling the sphere, but we don't know how to do that generally,
A classical homotopy result informs
$$
\pi_n(S^n) = \Z
$$
Telling us we can wrap an *n* sphere around itself, backwards and forwards any number of times.
It's annoying to do this explicitly without a generic way to describe higher dimensional spheres.
The relation seems to be:
$$
_n o_m = \left( T_m(o_{1,0}), o_{1,[1:n]}U(o_{1,0}) \right)
$$
Just like in the circle. This is extraordinarily convenient.