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I have a function $f:\mathbb{R}^3\to \mathbb{R}P^2$ and I'd like to visualize it as an assignment of a color to each point in space. But I'm having trouble coming up with a good mapping from $\mathbb{R}P^2$ to colors. It's easy to come up with good assignments of color to each point on a cylinder or sphere using the HSV color space (pictured below) but I'm not sure the best way to do it for the projective plane.

In other words I'd like a function from $RP^2$ to $S^1 \times [0,1]$ (hue and saturation) which is

  • continuous;
  • injective to the extent possible;
  • surjective to the extent possible.

Is there a good such map?

enter image description here

Sammy Black
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user7530
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    If you accept a Mobius strip intsead of $S^1\times [0,1]$ it would be easy. If you just remove one point in ${\bf R}P^2$,, you get a Mobius strip. – Thomas Jan 09 '24 at 08:09
  • You could use one for the sphere and just apply the quotient to $\Bbb Rp^2$. This would mean each colour is assigned to exactly two points. You could do the same for a colour mapping of the upper hemisphere, for $\Bbb Rp^2$ is a quotient of that as well. It's also a square quotient, so you could colour a square.. – FShrike Jan 09 '24 at 09:15
  • Any such map is nullhomotopic. Intuitively, you would want to use a 4-dimensional color-space, if there is such a thing. https://math.stackexchange.com/questions/2158230/embedding-rp2-into-r4 – Julius J. Jan 09 '24 at 15:23
  • @FShrike I am not sure what you mean? Won't I get two colors per point (rather than two points per color)? How do I pick between the two? – user7530 Jan 09 '24 at 16:43
  • @user7530 You're quite right, my apologies, it is two colours per point. If you use the hemisphere or square quotient models then duplication only happens at the image of the boundary, so usually isn't a problem, and I suppose you could blend the two colours when the issue arises – FShrike Jan 09 '24 at 17:13
  • @Thomas I am OK with a multiple-cover of $S^1 \times [0,1]$ but the problem is the removal of a point: the color will not appear continuous at that point. – user7530 Jan 09 '24 at 17:49

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