From eb245ce11e81bd9fa6ffc8af0bc886f9c3b238eb Mon Sep 17 00:00:00 2001 From: queue-miscreant Date: Mon, 21 Sep 2026 12:57:17 -0500 Subject: [PATCH] rambling that I need to rewrite --- posts/math/stereo/3/index.qmd | 400 ++++++++++++++++++++++++++++++ posts/math/stereo/3/inductive.qmd | 310 +++++++++++++++++++++++ 2 files changed, 710 insertions(+) create mode 100644 posts/math/stereo/3/index.qmd create mode 100644 posts/math/stereo/3/inductive.qmd diff --git a/posts/math/stereo/3/index.qmd b/posts/math/stereo/3/index.qmd new file mode 100644 index 0000000..cdf50eb --- /dev/null +++ b/posts/math/stereo/3/index.qmd @@ -0,0 +1,400 @@ +--- +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. diff --git a/posts/math/stereo/3/inductive.qmd b/posts/math/stereo/3/inductive.qmd new file mode 100644 index 0000000..02d9e44 --- /dev/null +++ b/posts/math/stereo/3/inductive.qmd @@ -0,0 +1,310 @@ +--- +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. + +$$ +{ D^n / \partial D^n } = S^n +$$ +