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enter image description here

I found this:

The tesseract is a four dimensional cube. It has 16 edge points $v=(a,b,c,d)$, with $a,b,c,d$ either equal to $+1$ or $-1$. Two points are connected, if their distance is $2$. Given a projection $P(x,y,z,w)=(x,y,z)$ from four dimensional space to three dimensional space, we can visualize the cube as an object in familiar space. The effect of a linear transformation like a rotation $$ R(t)=\pmatrix{1&0&0&0\\0&1&0&0&\\0&0&\cos(t)&\sin(t)\\0&0&-\sin(t)&\cos(t)} $$ in $4D$ space can be visualized in $3D$ by viewing the points $v(t) = P R(t) v$ in $\mathbb R^3$.

So how does $P$ actually look?

draks ...
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1 Answers1

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The particular $P$ used to make your animation loop looks like projection away from a point directly above the the $4$-cube. Intuitively, put a light source at $\mathbf{c} = (0, 0, 0, h)$ and map each point $\mathbf{x} = (x, y, z, w)$ with $w \neq h$ to its "shadow", the intersection of $\mathbf{R}^3 \times \{0\}$ with the ray from $\mathbf{c}$ through $\mathbf{x}$: $$ P(x, y, z, w) = \frac{h}{h - w}(x, y, z). $$

If we assume the $4$-cube has edges of length $2$ and is centered at the origin (i.e., it's the $1$-skeleton of the $4$-fold Cartesian product $[-1, 1]^{4}$), then $h \approx 3$: When the $4$-cube is aligned with the coordinate axes (so the projection looks like a small cube nested in a large cube, with corresponding vertices joined), the "top face" looks about twice as large as the "bottom face".

Andrew D. Hwang
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  • I think I begin to see: We have a scaling factor $\frac h{h-w}$ and a $z$ component that changes with $t$ accoriding to $R(t)$. But then I would expect two coordinates of each vertex to just move in and out by the scaling. In the gif it looks like they move on orbits... – draks ... Mar 23 '14 at 22:20
  • If you look down the axis of the cube along which the orbiting motion occurs (i.e., along the axis most nearly pointing toward the viewer), those two coordinates do just move in and out radially. :) – Andrew D. Hwang Mar 23 '14 at 23:23