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---
title: "Stereography, Algebraic, and Hyperspheres"
description: |
TODO
format:
html:
html-math-method: katex
jupyter: python3
date: "2026-09-11"
categories:
- algebra
---
```{python}
#| echo: false
import sympy
from IPython.display import Markdown
from tabulate import tabulate
x, x1, z = sympy.symbols("x x_1 z")
```
The Algebra Part
----------------
Though I alluded to the ability of the stereoscopic circle to generate the Chebyshev polynomials,
there is an important caveat which differs their use from typical spheres.
To review, the stereoscopic definition of the circle is:
$$
\begin{align*}
o_1(t) &= {1 + it \over 1 - it}
\\
&= c_1 + i s_1
= {1 - t^2 \over 1 + t^2} + i{2t \over 1 + t^2}
\end{align*}
$$
The second line decomposes the first into real and nonreal terms, which are each rational functions.
We can further define terms for the numerator and denominator:
$$
\begin{gather*}
x_1 = 1 - t^2
\qquad
y_1 = 2t
\qquad
d = 1 + t^2 = 2 - x_1
\\
o_1 = {z_1 \over d} = {x_1 \over d} + i{y_1 \over d}
\end{gather*}
$$
We have a recurrence relation for *o*, but it does not obey same relations as *z*:
$$
\begin{align*}
o_{n+2} &= 2c_1 o_{n+1} - o_n
\\
z_{n+2} &\stackrel{✗}{=} 2x_1 z_{n+1} - z_n
\end{align*}
$$
Fortunately, the correction is simple.
The denominator term $d^{n+2}$ can be multiplied through the top equation to produce:
$$
z_{n+2} = 2x_1 z_{n+1} - z_n d^2
$$
The only term that changes is the term lagging two terms behind, so the generating function *Z* is:
$$
\begin{align*}
O(x; o_1)
&= {1 + x(o_1 - 2 c_1) \over 1 - 2 c_1 x + x^2}
\\[10pt]
Z(x; z_1)
&= {1 + x(z_1 - 2 x_1) \over 1 - 2 x_1 x + \textcolor{red}{d^2} x^2}
\end{align*}
$$
We can express *d* in terms of $x_1$, so the terms of the series, like the one for *F*, have
- A real component which is a polynomial in $x_1$ (cf. $c_1$)
- An imaginary component which is the product of $y_1$ (cf. $s_1$) and a polynomial in $x_1$
$$
Z(x; z_1) = X(x; x_1) + i y_1 Y(x; x_1)
$$
Surprisingly, the polynomials in *Y* still factor cleanly,
like the [Chebyshev *U* polynomials](../../chebyshev/1/#tbl-chebyshevu).
```{python}
#| code-fold: true
#| label: tbl-newupolynomials
#| tbl-cap: "Table of numerator polynomials"
#| classes: plain
# cosine series
X = ( 1 - x1*x ) / ( 1 - 2*x1*x + (2 - x1)**2*x**2 )
# sine series
Y = x / ( 1 - 2*x1*x + (2 - x1)**2*x**2 )
def factor_sequence(polys, offset=0, symbol_name="p"):
ret = []
symbols = []
for i, poly in enumerate(polys):
new_poly = poly.copy()
old_factor = 1
for old, symbol in zip(ret, symbols):
q, r = sympy.div(new_poly, old)
if r == 0:
new_poly = q
old_factor *= symbol
if new_poly != 1:
ret.append(new_poly)
symbols.append(sympy.symbols(f"{symbol_name}_{i + offset}"))
yield poly, old_factor*new_poly.factor()
Markdown(tabulate(
[
[ n+1, "$" + sympy.latex(poly) + "$", sympy.Poly(unfactored, z).as_list() ]
for n, (unfactored, poly) in enumerate(
factor_sequence(
sorted(
[
i.subs(x,1).subs(x1, z).expand().factor()
for i in Y.series(x, n=11).args
][:-1],
key=lambda x: sympy.degree(x, z)
),
1
)
)
],
headers=[ "*n*", "$[x^n]Y(x; z) = p_n(z)$", "Coefficients (descending powers)" ],
numalign="left",
stralign="left",
))
```
Unfortunately, the sequence formed by the coefficients of the polynomials
does not appear in the OEIS.
Their factorizations appear to have the following traits:
- Like the Chebyshev *U* polynomials, they have "cyclotomic factoring" --
for the new term of index *n*, the factors can be separated into old factors
at indices of factors of *n* and new factors.
- If the index is even, then there is only one new monic, irreducible factor.
- If the index is odd, then there are two new irreducible factors
- If the index is prime or a prime power, the new factors are a monic and a non-monic
whose leading coefficient is that prime.
- Otherwise, the new factors are both monic.
The characterization of the leading terms of the new factor corresponds to
[OEIS A014963](https://oeis.org/A014963), which is related to cyclotomic polynomials.
### Similar Sequences
In fact, for a polynomial $q(z)$, it seems to be the case that the terms of
$$
Y(x; z) = {x \over 1 - 2 z x + q(z) x^2}
$$
tend to factor similarly.
Naturally, the *U* polynomials are the choice where *q = 1* and the new polynomials
are the choice when $q = (2 - z)^2$.
One can also write down a series for $z^n - 1$, which factor as the cyclotomic polynomials,
and also end up being generated by an order-2 recurrence.
$$
\begin{align*}
N(x; z) &= \sum_n (z^n - 1)x^n = {x(z - 1) \over 1 - (z + 1)x + zx^2}
\\
&= \sum_n \left ( x^n \prod_{d | n} \Phi_d(z) \right )
\end{align*}
$$
Actually, this shouldn't be terribly surprising.
For example, a simple result from generating functions tells us
that a series for the integers is:
$$
\begin{align*}
F(z) &= {1 \over 1 - z}
= \sum_n z^n
&& \text{All coefficients equal 1}
\\
F'(z) &= {1 \over ( 1 - z )^2 }
= \sum_n n z^{n - 1}
&& \text{Integers}
\\
z F'(z) &= {z \over 1 - 2z + z^2}
= \sum_n n z^n
&& \text{Integers matching powers}
\end{align*}
$$
The denominator being a quadratic polynomial means that the series terms *n*,
the integers, obey an order-2 recurrence, and factor in a similar way.
Obviously, the integers factor into primes by the fundamental theorem of arithmetic.
There's still an important distinction to be made about the polynomials, though.
Factoring a composite like 6 into 2 and 3 leaves an empty product behind, but
for polynomials, nonprime indices end up accumulate an "extra" factor.
Additionally (or rather, probably because of this), *all* factors of the index correspond
to a factor in the factorization, rather than pairing off as in integers.
For example, factoring 12 once gives either 3 and 4 or 2 and 6,
but the polynomial at index 12 includes polynomials at indices of all factors: 2, 3, 4, 6, and 12.
### Higher-order Recurrences
The integers also obey an order-3 recurrence:
$$
\begin{align*}
F(x) &= {x \over 1 - 2x + x^2} = {x(1 - x) \over (1 - 2x + x^2)(1 - x)}
\\
&= {x - x^2 \over 1 - 3x + 3x^2 - x^3}
\\[10pt]
&\equiv a_{n+3} = 3a_{n+2} - 3a_{n+1} + a_n
\end{align*}
$$
Another sequence that obeys similar factoring rules rules to the integers is
$$
G(x; z) = {x - x^2 \over 1 - z x + z x^2 - x^3}
$$
```{python}
#| code-fold: true
#| tbl-cap: "Table of G polynomials"
#| classes: plain
G = (x - x**2) / ( 1 - z*x + z*x**2 - x**3 )
Markdown(tabulate(
[
[ n+1, "$" + sympy.latex(poly) + "$", sympy.Poly(unfactored, z).as_list() ]
for n, (unfactored, poly) in enumerate(
factor_sequence(
sorted(
[
i.subs(x,1).subs(x1, z).expand().factor()
for i in G.series(x, n=11).args
][:-1],
key=lambda x: sympy.degree(x, z)
),
1,
"o"
)
)
],
headers=[ "*n*", "$[x^n]G(x; z) = o_n(z)$", "Coefficients (descending powers)" ],
numalign="left",
stralign="left",
))
```
There are a couple of things to note here.
Some of the factor polynomials here are the
[minimal polynomials of cosine](/posts/math/chebyshev/1/#tbl-cosinepolynomials)
The row where *n* = 5 is somewhat interesting.
Namely, $x^2 - x - 1$ has $\varphi$ (the golden ratio) as a root.
The other polynomial, $x^2 - 3x + 1$, has $\varphi^2$ as a root.
This seems to indicate that the other polynomial has roots which are
an algebraic expression of the other's, but I haven't bothered attempting a proof of this.
Unfortunately, peppering polynomials into the denominator and hoping that the same factorization
occurs isn't as easy as in the order-2 recurrence.
In fact, higher-order recurrences become more and more restrictive as $x^\bullet$ terms are added to
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.

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---
title: "Stereography, Algebraic, and Hyperspheres"
description: |
TODO
format:
html:
html-math-method: katex
jupyter: python3
date: "2026-09-11"
categories:
- algebra
---
```{python}
#| echo: false
import sympy
from IPython.display import Markdown
from tabulate import tabulate
x, x1, z = sympy.symbols("x x_1 z")
```
Topological Spheres
-------------------
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.
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.
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.
Concretely, this gives the topological relation
$$
\mathbb{E}^{n} \cup \{ \infty \} \cong S^n
$$
### 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.
[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:
- Scalars and vectors can be added and multiplied together,
and all possibilities comprise the algebra
- The product of a unit vector with itself is a scalar, generally chosen among -1, 0, or 1
- Scalars commute, but the product of two different unit vectors anticommutes
- 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., $e_0 e_1 e_0 e_1 = - e_0 e_1 e_1 e_0 = - e_0^2 e_1^2$
The square of a general vector ***v*** with components $x_k e_k$ is a scalar[^2].
$$
\begin{align*}
{\bm v} &= e_0 x_0 + e_1 x_1 + ... e_{n-1} x_{n-1} = \sum_k e_k x_k
\\
{\bm v}^2 &= (\sum_k^{n-1} e_k x_k) (\sum_l^{n-1} e_l x_l)
= \sum_k^{n-1} \sum_l^{n-1} e_k e_l x_k x_l
\\
&= \underset{\diagdown}{\sum_k^{n-1} e_k^2 x_k^2}
+ \underset{◥}{ \sum_k^{n-1} \sum_{l > k} e_k e_l x_k x_l }
+ \underset{◣}{ \sum_k^{n-1} \sum_{l < k} e_k e_l x_k x_l }
\\
&= \diagdown + ◥
+ \sum_k^{n-1} \sum_{l > k} e_l e_k x_k x_l
= \diagdown + ◥
- \sum_k^{n-1} \sum_{l > k} e_k e_l x_k x_l
\\
&= \diagdown + ◥ - ◥
= \diagdown
\end{align*}
$$
For simplicity, assume that $e_k^2 = -1$ so that ${\bm v}^2 = - ||{\bm v}||$, its Euclidean norm,
the sum of squares of the extent in each basis.
Consider the expression
$$
\begin{align*}
{1 + {\bm v} \over 1 - {\bm v}}
&= \left( {1 + {\bm v} \over 1 - {\bm v}} \right)
\left( {1 + {\bm v} \over 1 + {\bm v}} \right)
= {(1 + {\bm v})^2 \over (1 - {\bm v})(1 + {\bm v})}
\\
&= {1 + 2{\bm v} + {\bm v}^2 \over 1 - {\bm v}^2}
\\
&= {1 - ||{\bm v}|| \over 1 + ||{\bm v}||} + {2{\bm v} \over 1 + ||{\bm v}||}
= a + {\bm u}
\end{align*}
$$
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
$$
a^2 + ||{\bm u}|| = a^2 - {\bm u}^2
$$
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.
$$
||1 + {\bm v}|| = 1^2 - {\bm v}^2
= 1^2 - ({\bm -v})^2 = ||1 - {\bm v}||
$$
Consequently,
$$
\begin{align*}
a^2 - {\bm u}^2 &= 1
\\
\implies
\stackrel{\text{Numerator of } a}{(1 + {\bm v}^2)^2}
- \stackrel{\text{Numerator of } \bm u}{(2{\bm v})^2}
&= \stackrel{\text{Common denominator}}{1 - {\bm v}^2}
\end{align*}
$$
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 the denominator, so the dubious step of dividing by a vector has been backed up with pure algebra.
Despite multiplication between two vectors being defined, dividing one vector by another is not.
Inductivity
-----------
The previous topological description of spheres lacks a couple of things:
- It does not make reference to lower-dimensional spheres
- "Points at infinity", while intuitive, are logically suspect
Fortunately, topology has an alternate description.
The the 1-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
[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 $S^2$.
In general,
$$
\text{Susp}(S^{n-1}) = S^n
$$
### Algebraic Dual, Part 2
First, let's look at the first interesting case.
We first definied the circle, or 1-dimensional sphere as
$$
{1 + it \over 1 - it}
$$
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.
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.
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:
$$
{1 + {\bm u}t \over 1 - {\bm u}t}
= {1 - t^2 \over 1 + t^2} + {2t \over 1 + t^2}{\bm u}
= a + b{\bm u}
$$
### Degree Maps
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".
$$
_n o^m
= ( a + b \cdot { {}_{n-1} {\bm u}} )^m
= T_m(a) + b U_m(a) \cdot { {}_{n-1} {\bm u}}
$$
*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.
The topological equivalent to this statement is
$$
\pi_n(S^n) = \Z
$$
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.
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.
Equator
-------
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.
By virtue of degree, we're still in the clear.
The trick is actually the same from the previous post when integrating.
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).
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.
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.
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.
:::
USE THE ABOVE AS JUSTIFICATION FOR THIS SHIT.
MAYBE START EARLIER, WHEN INDUCTION WAS BEING DISCUSSED?
:::
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.
$$
{ D^n / \partial D^n } = S^n
$$