[Math] Parallel Lines Intersecting in the Projective Plane

projective-geometryprojective-spacevisualization

My question is about visualizing projective space, in particular the real projective plane $\mathbb{P}^2(\mathbb{R})$. I know there are different ways to define this space, but in each we can say that "two parallel lines intersect." If you type projective space into Wikipedia, or look it up in a textbook, a lot of the time you will see an image of two train tracks. The tracks are parallel, but as they go off into the distance (towards infinity) they appear to be approaching one another.

I believe I understand why we can say that two lines which are parallel in $\mathbb{R}^3$ intersect in $\mathbb{P}^2(\mathbb{R})$ (at least with the equivalence class construnction) but I do not see how this is related to that visual image of the two tracks appearing to approach a common point.

Maybe it is naive, but my question is: why do parallel lines, when looked at as in the image of the train tracks, appear to approach a common point? What is the mathematical reason for this?

Best Answer

Suppose that your eye is at the origin, and the "canvas" on which you draw is in front of your eye, and is the $z = 1$ plane. Then the point $(2, 3, 5)$ in space will project, in your "drawing" or "seeing" of the world, to the point $(2/5, 3/5, 1)$ in the $z = 1$ plane. In general, any point $(x, y, z)$ will project to $(x/z, y/z, 1)$.

Now consider two parallel lines; the first one consisting of all points of the form $(1, t, 4)$ and the second consisting of points $(3, t, 2)$. The projections of these to the drawing plane will consist of points of the form $(1/4, t/4, 1)$ and $(3/2, t/2, 1)$, respectively, i.e., they'll still be parallel.

Now look at the lines $(-1, 0, t)$ and $(1, 0, t)$. These project to $(-1/t, 0, 1)$ and $(1/t, 0, 1)$, which "meet" at the point $(0, 0, 1)$ when $t$ goes to $\infty$.

Does that help at all?