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We know that any 2D point $(x, y) $which represents as 3D homogeneous coordinates is of the form $(x, y, 1)$ which is the points of projective plane $P^2.$

If I use the same concepts for 3D points $(x, y, z)$ which represents as 4D homogeneous coordinates is of the form $(x, y, z, 1).$

My question is $(x, y, z, 1)$ could be the points of 3D projective space?

My second question is always homogenous coordinates are the points of projective space ?

N. B. -- $P^2$ is projective plane which contains the points of $\mathbb{R^2}$($\mathbb{R^2}$ points in $P^2$ can be represents as $(x, y, 1)$ ) and ideal points, is of the form $(x, y, 0)$ which is called points at infinity.

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  • $\begingroup$ Is $P^2$ supposed to be the real projective plane $RP^2$? $\endgroup$
    – lightxbulb
    Commented Nov 22, 2021 at 0:26
  • $\begingroup$ @lightxbulb explicitly $P^2=\mathbb{R^2}$+ ideal points. $\endgroup$
    – S. M.
    Commented Nov 22, 2021 at 3:14
  • $\begingroup$ Is the $+$ supposed to be union? Define "ideal points". Better yet - post the reference that you are using which employs this $P^2$ notation. $\endgroup$
    – lightxbulb
    Commented Nov 22, 2021 at 6:32
  • $\begingroup$ @lightxbulb ideal points is of form (x, y, 0) which is point at infinity. $\endgroup$
    – S. M.
    Commented Nov 22, 2021 at 7:01
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    $\begingroup$ The medium article explains your confusion. They define the term projective plane to be the plane $(x,y,1)$. It is true by their definition. This is usually not how the term is used as it refers to the real projective plane which is not just $(x,y,1)$. I could not find the definition of the medium article's author anywhere else (including the book that they cite). But by their definition $(x,y,z,1)$ would correspond to their definition of projective plane in 3D. I advise against using this terminology as I believe it is nonstandard and confusing (conflicts with the original term). $\endgroup$
    – lightxbulb
    Commented Nov 22, 2021 at 10:00

1 Answer 1

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So, a projective plane is a plane constructed such that parallel lines intersect at infinity. The plane defined by the equation (x, y, 1) is not projective. Just draw two parallel lines on that plane, and you will find that they never intersect, even under homogeneous transformations.

Now, homogeneous transformations can create projective planes. But such a plane would not have a fixed homogeneous coordinate of 1, because the homogeneous transform with a coordinate of 1 doesn't change the other coordinates. A projective plane in homogeneous space needs to be able to represent points at infinity, and that only happens with a homogeneous coordinate of zero.

So if a plane in homogeneous space does not pass through the 0 homogeneous coordinate, it cannot represent a projective plane.

Since the rest of your question is based on the assumption that (x, y, 1) is projective, it's kind of irrelevant.

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  • $\begingroup$ so what about (x, y, z, 1) in projective space? $\endgroup$
    – S. M.
    Commented Nov 21, 2021 at 20:33
  • $\begingroup$ your answer isn't according to my question. I am very much unsatisfied with your answer. $\endgroup$
    – S. M.
    Commented Nov 21, 2021 at 20:47
  • $\begingroup$ @B-spline: "so what about (x, y, z, 1) in projective space?" If (x, y, 1) isn't a projective plane, then why would you think that (x, y, z, 1) is a "projective space"? "I am very much unsatisfied with your answer." My answer is correct. It's not a projective plane, and the rest of your question is predicated on the assumption that it is. So the rest of your question is effectively moot. $\endgroup$
    – Nicol Bolas
    Commented Nov 22, 2021 at 0:00

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