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To understand the relationship between $\mathbb{R}^3$ and the projective plane $P^2$, I asked this question.

From that answer, I understood that ${P}^2$ is basically $\mathbb{R}^2$ and points in infinity, and a line in $\mathbb{R}^3$ which passes through the origin gives a point in $P^2$, while a line in $P^2$ gives a plane in $\mathbb{R}^3$ which passes through the origin.

Also, a line in $P^2$ is a collection of points, and each of those points represent lines in $\mathbb{R}^3$ through the origin. If you consider all the lines together, they form a plane in $\mathbb{R}^3$ through the origin. For example, each point on the line at infinity in $P^2$ corresponds to a line through the origin in the $xy$-plane, and if we consider all of these lines together, they give us the $xy$-plane.

Moreover, from that answer I understand that a line in $\mathbb{R}^3$ does not give a line in $P^2$, and a line in $P^2$ does not give a line in $\mathbb{R}^3$.

But I don't understand this statement "a line in $\mathbb{R}^3$ is infinite in both directions, while a line in $P^2$ is a circle".

I'm a computer science student. To understand computer graphics, these basic concepts are very necessary. Please use simple terms which I can understand easily and not too many technical terms.

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    $\begingroup$ It might be "better" to say "a line in the projective plane is a projective line", i.e., an ordinary line with one point at infinity (which is a circle...). The diagram here may be helpful? $\endgroup$ Aug 2 '21 at 21:05
  • $\begingroup$ @Andrew D. Hwang But there is many technical term... Being a computer science student it is very difficult to understand... Could you explain very easily by which I could understand by on my basic knowledge mentioned above $\endgroup$
    – Saslok
    Aug 2 '21 at 21:12
  • $\begingroup$ It all boils down to $P^n$ being a compact space and $\mathbb R^n$ not being one. $\endgroup$ Aug 2 '21 at 21:30
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    $\begingroup$ It's topologically a circle, not geometrically. It's a one-point compactification of a line. You add an infinity to each end of the line, and since these two infinities are actually the same infinity, the two ends are practically glued so that the line forms a loop — a circle. $\endgroup$
    – md2perpe
    Aug 2 '21 at 22:08
  • $\begingroup$ @md2perpe could you make it as answer with easy explanation for naive beginners $\endgroup$
    – Saslok
    Aug 2 '21 at 22:13
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A line $\ell$ in $P^2$ corresponds to a plane $\mathcal{P}$ in $\mathbb{R}^3$ passing through the origin. Consider the following diagram representing such a plane $\mathcal{P}$.

enter image description here

The red dot is the origin in $\mathbb{R}^3$, and the grey lines are lines in $\mathcal{P}$ which pass through the origin; these correspond to points on the line $\ell$ in $P^2$. Now, the semicircle intersects each line in one point, except for the horizontal line, where there are two points of intersection indicated by the green dots. The line $\ell$ consists of points representing the lines in $\mathcal{P}$, as does the semi-circle, except for the two endpoints. To remedy this, we can glue the two points together (making only one point), so now each line in $\mathcal{P}$ corresponds to a unique point on the green curve. The result is a circle, and its points can be identified with the points of $\ell$. So we see that, topologically, $\ell$ is a circle.

Note that the line where the semi-circle begins and ends is not special. We could have drawn a semicircle beginning and ending on any line and the same argument would hold. In particular, the horizontal line has no intrinsic meaning (the plane can be rotated about the origin). However, if $\ell$ is not the line at infinity, there is a way to make a consistent choice of line to begin and end the semi-circle on. Namely, the line which is the intersection of $\mathcal{P}$ and the $xy$-plane; this corresponds in $P^2$ to the unique point of intersection between $\ell$ and the line at infinity.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Aloizio Macedo
    Aug 5 '21 at 11:01
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First make yourself clear, that you can "embed" a projective space $P^n$ into an euclidean space $E^{n+1}$, which is one dimension bigger.

In computer vision this happens on the fly when you are using homogeneus coordinates. So a point in $P$ corresponds to a line in $E$. And the example in your question is also an example of this.

Don't mix $P$ and $E$ and do not consider, that one may be a subspace of the other. In $P$ lines, planes, etc cannot be parallel, while in $E$ they can. Looking from a theoretical point of view at this, it makes things "easier". If you want to proof a theorem, like the theorem from Pappos-Pascal, you do not have to consider all the special cases, when lines are parallel.

But for us it is hard to imagine, since we are used to euclidean space. And it is hard to visualise it. For visualisation we use the embedding into an euclidean space. This kind of visualisation makes a circle out of the line at infinity.

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    $\begingroup$ I know you put embed in quotation marks, but just for the record, $P^2$ doesn't embed in $\mathbb{R}^3$. $\endgroup$ Aug 3 '21 at 0:36

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