# Projection

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

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

If \$\displaystyle S \$ contains \$\displaystyle (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 \$\displaystyle (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 \$\displaystyle \Sigma\$ then it passes through the point at infinity. Thus, when you map it onto \$\displaystyle \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 \$\displaystyle \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 \$\displaystyle [0,1] \$. I was thinking that since \$\displaystyle (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 \$\displaystyle \mathbb{C} \$.

Or maybe think of lines as infinitely long circles.