Topological shape of the color spectrum

Question

Consider the HSV model of describing colors. Any color $c$ can be described as an element of $I^3$, where $\pi_1(c)$ describes hue, $\pi_2(c)$ describes saturation, and $\pi_3(c)$ describes brightness. Without thought, one might think the topological shape of the spectrum of colors is just $D^3$. However,

• A hue of zero describes the same color as a hue of one.
  1. $\ (0, s, b) \thicksim (1, s, b)$ for each $s, b \in I$.
• If a color has no saturation, it is uniquely described only by its brightness.
  2. $\ (h, 0, b) \thicksim (h', 0, b)$ for each $h, h', b \in I$.
• A brightness of zero or one is black or white respectively.
  3. $\ (h, s, 0) \thicksim (h', s', 0)$ for each $h, h', s, s' \in I$.
  4. $\ (h, s, 1) \thicksim (h', s', 1)$ for each $h, h', s, s' \in I$.

I believe there to be no further identifications, for if a color has some saturation, and is not black or white, it should be unique. So, what is the topological structure of $D^3$ after these identifications?

I think the answer is that the identifications $2, 3,$ and $4$ have no effect, and the outcome is a solid torus. But this doesn't make sense because many such "color spheres" exist.

Answer 1

It's not true that the identifications after the first have no effect. For one, the result is that the "loop" of the solid torus - the curve $f(t) = (t,s,b)$ - is homotopic to the identity map $f(t) = (t,s,0)$.

Instead, let's look at the identifications in the reverse order. We start with a cube, which has six faces: the $h=0$ face, the $h=1$ face, the $s=0$ face, the $s=1$ face, the $b=0$ face, and the $b=1$ face.

Identification 4 contracts the $b=1$ face to a point, giving us a pyramid.

Identification 3 contracts the $b=0$ face to a point, giving an awkward thing we can't draw as a polyhedron anymore, but which still has four $2$-sided faces meeting at the $b=0$ and $b=1$ points. (We could call this a hosohedron.)

Identification 2 contracts the $s=0$ face to a segment, so that now we have only the $h=0$, $h=1$, and $s=1$ faces left. They each meet along an edge (the $h=0$ and $h=1$ faces meet along the $s=0$ edge) and all three come together at the $b=0$ and $b=1$ points.

Finally, identification 1 identifies the two $h=0$ and $h=1$ faces. The result is a disk with only the $s=1$ face on the outside.