Understanding projective plane conceptually (page-342, Road to Reality by Roger Penrose) 


In the above picture, I am a bit confused how it turns out parallel lines seems to meet in the artist's potrait. Could someone explain in simple words why the roads which don't intersect in the ambient world do intersect in the painting?
So far, I get the idea that when drawing the painting , a line is drawn from the artist's eye to the object in 3-D space and the point on the canvas is the intersection of the line with it. However, the converging line thing is still tripping me up.
Edit: For future readers, I found this Deviant Art page by Nisio very helpful. You have to click the picture to zoom, but they go over the details of it. Also check out this video (also helpful).
 A: One place the mathematics is discussed: Points at infinity where last element in homogeneous vector is $0$?.
More geometrically, an artist's field of view may be modeled either by a sphere of rays or by a projective plane of lines through the eye. We may as well pick the unit sphere centered at the origin of Euclidean three-space.

An open hemisphere $H$ of the sphere model corresponds to a dense open set $P$ of the projective plane, and the boundary great circle $C$ of $H$ maps to the line at infinity with respect to $P$ by definition. The golden circle shown lies in the equatorial plane $Z = 0$, and the plane $P$ being visualized is $Z = -1$, which corresponds to the lower hemisphere.
Straight lines in $P$ map to the sphere by radial projection from the eye, so their images are great circles. Since great circles intersect on the sphere, the images of lines in $P$ also intersect, even if the lines are parallel as shown. By definition, parallel lines in $P$ do not intersect in $P$. The points of intersection on the sphere therefore line on the boundary great circle, a.k.a., the line at infinity with respect to $P$.
In Penrose's drawing, the artist's canvas might be the plane $Y = 1$ (not shown here). Projecting the great circles to that plane gives a pair of crossing lines, compare the railroad tracks in the linked question.

Added: The diagram below shows the same picture without the sphere, and with the projective lines shown as affine planes. To emphasize,

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*Each colored affine line through the origin (Eye) represents a projective point.

*Each affine plane through the origin represents a projective line.

*We're choosing the line at infinity to be the set of gold affine lines lying in the affine plane $Z = 0$.


The plane $Z = -1$, by contrast, does not pass through the origin.

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*Its points correspond to projective points: Each point $p$ of the plane $Z = -1$ determines a unique affine line $\ell$ through the origin and $p$, and $\ell$ intersects $Z = -1$ precisely at $p$ rather than being contained in the plane $Z = -1$.

*Its affine lines (the two parallel blue lines, for example, modeling the edges of Penrose's roadway) correspond to projective lines, represented by the shaded affine planes.

*The entire affine plane $Z = -1$ corresponds to the "finite" Euclidean part of the projective plane with respect to the projective line at infinity, here chosen to be the affine plane $Z = 0$.

*The two parallel affine lines in the affine plane $Z = -1$ intersect in the projective plane because the affine planes that represent them intersect along an affine line through the origin, i.e., at a point of the projective plane. The intersection of the two slanted affine planes lies in the affine plane $Z = 0$, a.k.a., the projective line at infinity. In that sense, the parallel affine/Euclidean lines intersect at infinity.

