Navigating though the surface of a hypersphere in a computer game

Solution 1:

The $M$ you asked in your final question is in general non-unique, given the data you specified. The way to see this is as follows:

Given a point $(x,y,z,w)$ and another $(x',y',z',w')$, add to it the third point $(0,0,0,0)$ in four dimensional space. Assuming that the two original points do not coincide and that they are not antipodes, this three points are not collinear and so defines a plane in four dimensional space. So you can take a rotation in that plane that carries $(x,y,z,w)$ to $(x',y',z',w')$ while fixing the directions perpendicular to that plane fixed. Call this transformation $M_1$.

However, since your ambient space is four dimensional, the directions perpendicular to the given plane also is two dimensional (4-2 = 2). So you can equally take an arbitrary rotation in that plane which fixes all directions perpendicular to it. Call such a rotation $O$. Then you can check that $M_2 = OM_1$ also sends $(x,y,z,w)$ to $(x',y',z',w')$.

To be explicit. Assume your initial coordinate is $(1,0,0,0)$ and the final coordinate is $(0,1,0,0)$. Then we have

$$ M_1 = \begin{pmatrix} 0 & -1 & 0 & 0\\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{pmatrix}$$

If you take $O$, parametrized by $\theta$, to be

$$ O(\theta) = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0\\ 0 & 0 & \cos\theta & \sin\theta \\ 0 & 0 & -\sin\theta & \cos\theta\end{pmatrix}$$

then you can check that $O(\theta)M_1$ will send $(1,0,0,0)$ to $(0,1,0,0)$ for any $\theta$, but the matrices $O(\theta)M_1$ are all different for $\theta$ in the range $0 \leq \theta < 2\pi$.

So what does this mean physically? Moving "forward" in Euclidean space is not as simple an issue as you think. The analogue of the different $O(\theta)$ in 3 dimension Euclidean space corresponds to rotating the whole space along the axis of travel! In other words, imagine you have a spaceship in Euclidean space and it spins (with the axis of spinning in the same direction of its travel) while it moves forward. This is what $O(\theta)$ captures, the spinning.

Without getting too much into the gory details, I will note that this ambiguity lies in the heart of differential/Riemannian geometry, and is closely connected with the notion of parallel transport.

In any case, using a bit of differential geometry, you can see that the matrix $M_1$ defined above is the "correct" notion of translation is you do not want "spinning".

Okay, enough about that. How to actually implement this? Let's say that the viewer is sitting at coordinates $(x,y,z,w)$. And let's say the viewer is facing in the direction of $(\delta x, \delta y, \delta z, \delta w)$ with $\delta x^2 + \delta y^2 + \delta z^2 + \delta w^2 = 1$. Notice that since the direction the viewer is facing is tangential to the hypersphere, it must be perpendicular to the coordinates, that is

$$ x\cdot \delta x + y\cdot \delta y + z \cdot \delta z + w \cdot \delta w = 0$$

Given these two vectors, you can use some linear algebra to complete them to an orthonormal basis of four dimensional space (which can be obtained by solving some linear equations based on orthogonality to the two given vectors), call the two additional vectors $(a,b,c,d)$ and $(a',b',c',d')$. Then your translation matrix should be given by

$$ M(\phi) = \begin{pmatrix} x & \delta x & a & a'\\ y & \delta y & b & b' \\ z & \delta z & c & c' \\ w & \delta w & d & d'\end{pmatrix} \begin{pmatrix} \cos(\phi) & \sin(\phi) & 0 & 0\\ -\sin(\phi) & \cos(\phi) & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{pmatrix} \begin{pmatrix} x & y & z & w\\ \delta x & \delta y & \delta z & \delta w \\ a & b & c & d \\ a' & b' & c' & d'\end{pmatrix} $$

A quick explanation: by fixing the orthonormal basis, we can use it to construct an orthogonal transformation to a coordinate system in which the viewer sits at $(1,0,0,0)$ and is facing in the $(0,1,0,0)$ direction. Then all we need to do is to conjugate back the transportation matrix for that situation. The $\phi$ parameter measures the (angular) distance you travelled on the hypersphere.