Thread: Projection

Proposition: Let be a circle on and let be its projection on . Then (a) if contains , is a line; (b) if does not contain , is a circle.


If contains then the circle in not entirely contained on the sphere. Is this an intuitive reason why the projection is a line? Whereas a circle that does not contain is a circle completely contained on the sphere. Thus its projection is a circle.

Originally Posted by manjohn12

Proposition: Let be a circle on and let be its projection on . Then (a) if contains , is a line; (b) if does not contain , is a circle.


If contains then the circle in not entirely contained on the sphere. Is this an intuitive reason why the projection is a line? Whereas a circle that does not contain is a circle completely contained on the sphere. Thus its projection is a circle.

I cannot think of a great intuitive explanation why lines and circles go to lines and circles, I think you need to do some algebraic calculations. However, this is how I remember it. If a circle passes through (0,0,1) on then it passes through the point at infinity. Thus, when you map it onto (i.e. project) it means the image becomes unbounded. Since circle are bounded it cannot be a circle. It has to be a line then. And if a circle is on and not through (0,0,1) it means it stays away from the point at infinity, therefore it stays bounded. Thus, such a circle gets projected back onto a circle.

For intuition you might want to think in 2d. Imagine a circle sitting on top of the real number line. Can you visualize this?

Yes. E.g. the circle is homeomorphic to . I was thinking that since is the point at infinity, a circle that contains it also contains points not on the sphere. The part that is contained on the sphere can be "stretched" out into a line on .

Or maybe think of lines as infinitely long circles.