diff --git a/posts/math/stereo/1/index.qmd b/posts/math/stereo/1/index.qmd index 7e00740..a626369 100644 --- a/posts/math/stereo/1/index.qmd +++ b/posts/math/stereo/1/index.qmd @@ -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. diff --git a/posts/math/stereo/3/inductive.qmd b/posts/math/stereo/3/inductive.qmd index 02d9e44..5953873 100644 --- a/posts/math/stereo/3/inductive.qmd +++ b/posts/math/stereo/3/inductive.qmd @@ -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. diff --git a/posts/math/stereo/3/index.qmd b/posts/math/stereo/4/index.qmd similarity index 67% rename from posts/math/stereo/3/index.qmd rename to posts/math/stereo/4/index.qmd index cdf50eb..632189b 100644 --- a/posts/math/stereo/3/index.qmd +++ b/posts/math/stereo/4/index.qmd @@ -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.