Bellaard.com

# Living on the n-sphere

By: Gijs Bellaard

## Definitions

We define the n-sphere $$S^n$$ as all the points in $$\bbR^{n+1}$$ with unit euclidean length: $S^n := \{ x \in \bbR^{n+1} \mid ||x|| = 1 \}$ A great circle $$C$$ is defined as the intersection of the n-sphere with a plane $$H \subseteq \bbR^{n+1}$$: $C(H) = S^n \cap H$ By writing $$H$$ as the span of two vectors $$a, b \in S^n$$ that are orthogonal $$\Ang{a,b} = 0$$, where $$\Ang{\cdot, \cdot}$$ is the standard dot product, allows us to parametrize $$C$$: $C(a, b) = \{ a \cos t + b \sin t \mid t \in [0, 2\pi ]\}$ One can confirm for himself, as an exercise, that this parametrization is indeed correct. The angle $$\alpha$$ between two vectors on the n-sphere is exactly the distance between them on the n-sphere. Using this the dot product can also be written as $\Ang{x,y} = ||x|| \cdot ||y|| \cdot \cos \alpha = 1 \cdot 1 \cdot \cos d(x,y) = \cos d(x,y)$ which reveals the metric on the n-sphere to be $d(x,y) = \arccos \Ang{x,y}$ again, one can confirm this to be an actual metric by himself. Now that we have a metric we can define a ball $$B$$ on the n-sphere to be all the points that are at most some distance $$r \leq \pi$$ away from a center $$c$$: $B(c, r) := \{ x \in S^n \mid d(x,c) \leq r \} = \{ x \in S^n \mid \Ang{x,c} \geq \cos r \}$ We can also define the "great donut" $$D$$ as all the points that are at most some distance $$r \leq \pi/2$$ away from a great circle $$C$$: $D(C, r) := \{ x \in S^n \mid d(x,C) \leq r\}$ where the distance to a set $$\Omega$$ is simply defined as: $d(x,\Omega) := \inf_{o \in \Omega} d(x,o)$ The distance/angle $$\alpha$$ between the great circle $$C(a,b)$$ and $$x$$ is the same as the angle between $$x$$ and its projection $$p$$ onto $$H(a,b)$$. This means that, together with the fact that the origin, $$x$$ and $$p$$ form a right angle triangle, that: $\cos \alpha = ||p|| = \sqrt{\Ang{x,a}^2 + \Ang{x,b^2}}$

this allows us to rewrite the distance to a great circle as: $d(x, C) = \arccos \sqrt{\Ang{x,a}^2 + \Ang{x,b}^2}$ All in all, this means a great donut is also: $D(a,b,r) := \{ x \in S^n \mid \Ang{x,a}^2 + \Ang{x,b}^2 \geq \cos^2 r\}$

## Intersectors

A ray $$r(t)$$ on the n-sphere which is basically just a parametrized great circle: $r(t) := o \cos t + d \sin t$ where we interpret $$o$$ as the start(origin) of the ray and $$d$$ as the direction it's going, with again $$o,d \in S^n$$ and $$\Ang{o,d}=0$$.

### Ball

We are interested in when a ray intersects a ball $$B(c,r)$$: $r(t) \in \partial B$ where $$\partial B$$ is the boundary of $$B$$. By definition of the ball this is equivalent to: $\Ang{r(t),c} = \cos r$ which can be expanded to: $\Ang{o,c} \cos t + \Ang{d,c} \sin t = \cos r$ this is an equation of the form $$A\cos t + B\sin t = C$$ which can be solved analytically. How this is done exactly will be explained later.

### Great Donut

We are interested in when a ray intersects a great donut $$D(a, b, r)$$: $r(t) \in \partial D$ By definition of the great donut this is equivalent to: $\Ang{r(t),a}^2 + \Ang{r(t),b}^2 = \cos^2 r$ We introduce $$\tilde{o} := ( \Ang{o, a}, \Ang{o, b} ) \in \bbR^2$$ and similarly $$\tilde{d}$$. The equation above can then be expanded to: $\Ang{\tilde{o},\tilde{o}}\cos^2 t + 2\Ang{\tilde{o},\tilde{d}}\cos t \sin t + \Ang{\tilde{d},\tilde{d}}\sin^2 t = \cos ^2 r$ This is an equation of the form $$A\cos^2 t + 2B\cos t\sin t + C \sin^2 t= D$$ which can be solved analytically. How this is done exactly will be explained later.

## Solving Trigonometric Equations

### The one of the Ball

We are trying to solve the following equation for $$t$$: $A\cos t + B\sin t = C$ The trick is to interpret $$(B,A)$$ as a point in $$\bbR^2$$ and write it in polar coordinates: $B = r \cos \theta, \quad A = r \sin \theta$ substituting this back in the equation gives: $r(\sin \theta \cos t + \cos \theta \sin t) = C$ applying the sum formula for sine: $r \sin(\theta + t) = C$ and so: $t = \arcsin(C/r) - \theta$ Of course this equation doesn't always allow for a solution, this happens when $$|C/r| > 1$$.

### The one of the Great Donut

We are trying to solve the following equation for $$t$$: $A\cos^2 t + 2B\cos t\sin t + C \sin^2 t= D$ using: $\cos^2t = \frac{1 + \cos 2t}{2}, \quad \sin^2 t = \frac{1 - \cos 2 t}{2}, \quad 2 \cos t \sin t = \sin 2 t$ we can transform this equation into one of the form: $A\cos u + B\sin u = C$ where $$u = 2t$$, which we can solve as seen above.

## Spherical Universe

One might have heard about the idea that the universe is not flat be spherical, which means the universe is like $$S^3$$. I was wondering how this would actually look so I made the following shader using the math described here. To control the camera use WASD+QE+Space+Shift+Mouse.