It's incorrect, though.

The oriented real projective plane is a sphere, not a torus.

The projective points at infinity (one point for every 1-D angle (R mod 2pi)) form the equator of the sphere.

The T in T² is for "two-sided" , not Torus.

The torus explanation that Tangram gives doesn't make sense. In a pinhole projection, the horizontal and vertical infinites do not "wrap around" to meet. There is no meaningful "horizontal" and "vertical", the system is rotationally symmetric, which forms a hemisphere of curve it to make it compact. (Half sphere because you can only see one half of the space outside a pinhole camera)

https://en.m.wikipedia.org/wiki/Oriented_projective_geometry

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ChickenSando 7 months ago | parent | next [-]

You're right. I've messed up. The notation here is a bit misleading and I didn't dig as deep here as I should have. I'll fix the website soon.

	▲	erwincoumans 7 months ago \| parent [-]
		Thanks for sharing such an insightful article of complex material. Some follow-up showing how this helps optimization (gradient descent, Newton solvers?) would be great. >> For convenience of notation, we’ll drop the explicit parametrization of the curves and denote this vector shift as [...] FYI for clarity I wish the explicit parametrization was kept, even though it is more verbose.

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kevinventullo 7 months ago | parent | prev [-]

Indeed, I should have been a more critical reader!