311 lines
11 KiB
Plaintext
311 lines
11 KiB
Plaintext
---
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title: "Stereography, Algebraic, and Hyperspheres"
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description: |
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TODO
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format:
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html:
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html-math-method: katex
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jupyter: python3
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date: "2026-09-11"
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categories:
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- algebra
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---
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```{python}
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#| echo: false
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import sympy
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from IPython.display import Markdown
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from tabulate import tabulate
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x, x1, z = sympy.symbols("x x_1 z")
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```
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Topological Spheres
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-------------------
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In topology, hyperspheres are some of the primary spaces of interest.
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Spheres have a natural geometric definition: the locus of points which all have
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the same distance to the origin.
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Topologically, however, they're better-described inductively.
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First, notice that the equation for the circle depends on a single parameter *t*
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which ranges over the entire number line.
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There is also a point on the circle "at infinity", which "closes" the circle.
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Topologically, the resulting space is called the [https://mathworld.wolfram.com/One-PointCompactification.html](one-point compactification).
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In other words, the circle is the one-point compactification of the (open) line.
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For spheres, the same thing holds true, but the notion of "one-point" starts to become relevant.
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We map a 2-dimensional plane to the sphere, so there are two variables.
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But if one or both of these variables has a value of "infinity", then they are all said to describe the same point.
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Concretely, this gives the topological relation
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$$
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\mathbb{E}^{n} \cup \{ \infty \} \cong S^n
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$$
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### Algebraic Dual
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As a locus of points, the *n*-sphere exists within *n+1* dimensional space.
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But since the sphere is *n*-dimensional, a point on it is described by *n* coordinates,
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just like a point in *n*-dimensional space.
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We need a way to augment an *n*-dimensional vector with an extra dimension
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Fortunately, we have some direction from [the first post in this series](../1/),
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in which we derived an explicit map for the 2-, and 3-spheres.
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Namely, the result for 2-spheres was derived by the assertion
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$$
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o = {1 + {\bm v} \over 1 - {\bm v}} = a + {\bm u},
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\quad {\bm v} = is + jt,
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\quad i^2 = j^2 = -1
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\quad ij = -ji
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$$
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*i* and *j* are quaternions, which form a division algebra.
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This makes this expression legitimate, but not easy to generalize to higher dimensions[^1].
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[^1]: The limited number of division algebras is typically proved using algebraic topology
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through arguments that depend on spheres and quotient spaces thereof.
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Fortunately, the argument can be adjusted a little.
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[Geometric algebra](https://en.wikipedia.org/wiki/Geometric_algebra) gives some tools to generalize
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this argument to higher dimensions.
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In such an algebra, we have unit vectors $e_0 ... e_{n-1}$ and the following properties:
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- Scalars and vectors can be added and multiplied together,
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and all possibilities comprise the algebra
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- The product of a unit vector with itself is a scalar, generally chosen among -1, 0, or 1
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- Scalars commute, but the product of two different unit vectors anticommutes
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- e.g., $e_0 e_1 = - e_1 e_0$
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- Consequently, the square of the product is the negative of the product of the squares
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- e.g., $e_0 e_1 e_0 e_1 = - e_0 e_1 e_1 e_0 = - e_0^2 e_1^2$
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The square of a general vector ***v*** with components $x_k e_k$ is a scalar[^2].
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$$
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\begin{align*}
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{\bm v} &= e_0 x_0 + e_1 x_1 + ... e_{n-1} x_{n-1} = \sum_k e_k x_k
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\\
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{\bm v}^2 &= (\sum_k^{n-1} e_k x_k) (\sum_l^{n-1} e_l x_l)
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= \sum_k^{n-1} \sum_l^{n-1} e_k e_l x_k x_l
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\\
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&= \underset{\diagdown}{\sum_k^{n-1} e_k^2 x_k^2}
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+ \underset{◥}{ \sum_k^{n-1} \sum_{l > k} e_k e_l x_k x_l }
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+ \underset{◣}{ \sum_k^{n-1} \sum_{l < k} e_k e_l x_k x_l }
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\\
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&= \diagdown + ◥
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+ \sum_k^{n-1} \sum_{l > k} e_l e_k x_k x_l
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= \diagdown + ◥
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- \sum_k^{n-1} \sum_{l > k} e_k e_l x_k x_l
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\\
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&= \diagdown + ◥ - ◥
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= \diagdown
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\end{align*}
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$$
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For simplicity, assume that $e_k^2 = -1$ so that ${\bm v}^2 = - ||{\bm v}||$, its Euclidean norm,
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the sum of squares of the extent in each basis.
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Consider the expression
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$$
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\begin{align*}
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{1 + {\bm v} \over 1 - {\bm v}}
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&= \left( {1 + {\bm v} \over 1 - {\bm v}} \right)
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\left( {1 + {\bm v} \over 1 + {\bm v}} \right)
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= {(1 + {\bm v})^2 \over (1 - {\bm v})(1 + {\bm v})}
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\\
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&= {1 + 2{\bm v} + {\bm v}^2 \over 1 - {\bm v}^2}
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\\
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&= {1 - ||{\bm v}|| \over 1 + ||{\bm v}||} + {2{\bm v} \over 1 + ||{\bm v}||}
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= a + {\bm u}
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\end{align*}
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$$
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Similarly to quaternions, this is the sum of a vector and a scalar.
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If the scalar component is considered the extent in a new dimension,
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then the norm of the resulting vector is
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$$
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a^2 + ||{\bm u}|| = a^2 - {\bm u}^2
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$$
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According to this definition, we started with the ratio of two expressions with the same norm.
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This means that our resulting expression should have a norm of 1.
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$$
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||1 + {\bm v}|| = 1^2 - {\bm v}^2
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= 1^2 - ({\bm -v})^2 = ||1 - {\bm v}||
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$$
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Consequently,
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$$
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\begin{align*}
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a^2 - {\bm u}^2 &= 1
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\\
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\implies
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\stackrel{\text{Numerator of } a}{(1 + {\bm v}^2)^2}
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- \stackrel{\text{Numerator of } \bm u}{(2{\bm v})^2}
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&= \stackrel{\text{Common denominator}}{1 - {\bm v}^2}
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\end{align*}
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$$
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The final expression is always valid, no matter how many dimensions ***v*** has.
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Not only that, since *a* and ***u*** contain no vectors in the denominator, there are no concerns with the validity of division.
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The only reason we started from an expression which did contain vectors was to justify the requirement for anticommutativity.
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This is the denominator, so the dubious step of dividing by a vector has been backed up with pure algebra.
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Despite multiplication between two vectors being defined, dividing one vector by another is not.
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Inductivity
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-----------
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The previous topological description of spheres lacks a couple of things:
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- It does not make reference to lower-dimensional spheres
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- "Points at infinity", while intuitive, are logically suspect
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Fortunately, topology has an alternate description.
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The the 1-dimensional sphere is a little bit special.
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On a number line, there are two points equidistant to the origin,
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and these comprise the 0-sphere $S^0$.
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This can be turned into a 1-sphere $S^1$ (the circle) through a topological operation called
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[suspension](https://en.wikipedia.org/wiki/Suspension_%28topology%29), which connects
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all points in the space to two new, auxiliary points.
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Subsequently, we can take the circle and repeat the operation to build the 2-sphere $S^2$.
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In general,
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$$
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\text{Susp}(S^{n-1}) = S^n
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$$
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### Algebraic Dual, Part 2
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First, let's look at the first interesting case.
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We first definied the circle, or 1-dimensional sphere as
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$$
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{1 + it \over 1 - it}
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$$
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If the first coordinate remains fixed, then in most cases,
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the space looks two discrete points, or to wit, a 0-dimensional sphere.
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The remaining two points are in some sense "new" to the space.
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Examining the 2-dimensional sphere in the same way, at an intersecting plane,
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the space looks like a 1-dimensional sphere except at two points.
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This matches the inductive topological description of spheres one-for-one.
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Using the results of the previous section, we have a way to generalize *i* to any dimension.
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Coincidentally, this generalization is *also* inductive --
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if ***u*** is already a vector with norm 1, then the scalar component *a* must be 0.
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Further, this means that ${\bm u}^2 = -1$.
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This should sound familiar -- it matches the "unit quaternions"
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Intuitively, this means we can also describe a sphere by the equation:
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$$
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{1 + {\bm u}t \over 1 - {\bm u}t}
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= {1 - t^2 \over 1 + t^2} + {2t \over 1 + t^2}{\bm u}
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= a + b{\bm u}
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$$
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### Degree Maps
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Since ${\bm u}^2 = -1$, there's an interesting trick we can pull again.
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We wrapped the circle around itself twice in [the previous article](../2/) by
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simply squaring the same expression from the last article.
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That argument is only contingent upon one thing: the split between
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real and nonreal components, and the squaring of the unit nonreal to -1.
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When going from *i* to ***u***, the only thing that needs changing is replacing
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"real" with "scalar" and "nonreal" with "vector".
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$$
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_n o^m
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= ( a + b \cdot { {}_{n-1} {\bm u}} )^m
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= T_m(a) + b U_m(a) \cdot { {}_{n-1} {\bm u}}
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$$
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*T* and *U* here are the standard Chebyshev polynomials.
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This amounts to wrapping the sphere around itself any number of times as desired, *m*.
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*m* here is only really defined over positive integers here, since the Chebyshev polynomials
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are only defined over positive indices.
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The topological equivalent to this statement is
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$$
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\pi_n(S^n) = \Z
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$$
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This states that a map from the *n*-sphere to itself can be characterized by an integer,
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the *degree*.
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Maps sharing the same integer are considered to be *homotopic* to one another,
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and composing maps can be composed in the same way that the integers add.
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Telling us we can wrap an *n* sphere around itself, backwards and forwards any number of times.
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"Backwards" comes from antipodal map
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Degree of antipodal map is negative only if *n* is even
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But this just means we can look at a map where we negate only the vector components.
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Equator
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-------
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By describing spheres purely in terms of other spheres, we've taken care of the first problem.
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We still have another -- if we let a vector *v* range over the entirety of Euclidean space,
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we still have to include a point at infinity.
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By virtue of degree, we're still in the clear.
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The trick is actually the same from the previous post when integrating.
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For the circle, a degree 1 map wraps around once from $-\infty$ to $\infty$,
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and a degree 2 map wraps around once from -1 to 1.
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The -1 to 1 range in the degree 1 map describes only a semicircle, whose boundary is
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two points (the 0-sphere).
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One dimension up, to continue the analogy, we have a 1-sphere which bounds a hemisphere
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in the degree 1 map.
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The 1-sphere is just the equator of the sphere.
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The hemisphere replaces the range "from -1 to 1"; instead, we have a unit disc --
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geometrically, this consists of all vectors whose norm is less than 1.
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In fact, this still agrees with the 1 dimensional case.
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This analogy continues inductively to all *n*-dimensional spheres.
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Another way of seeing this is by looking at the equators of *n*-spheres.
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Using the inductive algebraic definition of the sphere, if the scalar component is 0, then
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the vector component is just the *n-1*-sphere.
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This can be thought of as the equator.
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The first coordinate expresses the condition that if **v** lies on the unit (hyper)sphere
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in the dimension below, then the coordinate is zero, and such points lie on an equator.
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:::
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USE THE ABOVE AS JUSTIFICATION FOR THIS SHIT.
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MAYBE START EARLIER, WHEN INDUCTION WAS BEING DISCUSSED?
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:::
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In the case of the 2-sphere, doubling the map by squaring quaternions produces a more interesting feature:
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all **v** lying on the unit (hyper)sphere in the dimension below get mapped to the same point, antipodal to 0.
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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.
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$$
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{ D^n / \partial D^n } = S^n
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$$
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