# Projection

• Jan 18th 2009, 01:25 PM
manjohn12
Projection
Proposition: Let $S$ be a circle on $\displaystyle \Sigma$ and let $T$ be its projection on $\mathbb{C}$. Then (a) if $S$ contains $(0,0,1)$, $T$ is a line; (b) if $S$ does not contain $(0,0,1)$, $T$ is a circle.

If $S$ contains $(0,0,1)$ 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 $(0,0,1)$ is a circle completely contained on the sphere. Thus its projection is a circle.
• Jan 18th 2009, 01:37 PM
ThePerfectHacker
Quote:

Originally Posted by manjohn12
Proposition: Let $S$ be a circle on $\displaystyle \Sigma$ and let $T$ be its projection on $\mathbb{C}$. Then (a) if $S$ contains $(0,0,1)$, $T$ is a line; (b) if $S$ does not contain $(0,0,1)$, $T$ is a circle.

If $S$ contains $(0,0,1)$ 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 $(0,0,1)$ 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 $\Sigma$ then it passes through the point at infinity. Thus, when you map it onto $\mathbb{C}$ (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 $\Sigma$ 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?
• Jan 18th 2009, 01:42 PM
manjohn12
Yes. E.g. the circle is homeomorphic to $[0,1]$. I was thinking that since $(0,0,1)$ 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 $\mathbb{C}$.

Or maybe think of lines as infinitely long circles.