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Thread: Projection

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by manjohn12 View Post
    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?
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  3. #3
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    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.
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