Suppose $ T \subset \mathbb{C} $. Show that the corresponding set $ S \subset \Sigma $ is

a. a circle if $ T $ is a circle.
b. a circle minus (0, 0, 1) if $ T $ is a line.

Here we are defining $ \Sigma $ to be the Riemann sphere, given by the set: $$ \Sigma = \left \{(\xi, \eta, \zeta) : \xi^{2} + \eta^{2} + (\zeta - \frac{1}{2})^{2} = \frac{1}{4} \right \} $$

To take a point from $ \mathbb{C} $ to $ \Sigma $ we can use the following:

$$ \xi = \frac{x}{x^{2} + y^{2} + 1}; \eta = \frac{y}{x^{2} + y^{2} + 1}; \zeta = \frac{x^{2} + y^{2}}{x^{2} + y^{2} + 1} $$

We define a circle on $ \Sigma $ to be the intersection of a plane of the form $ A\xi + B\eta + C\zeta = D $ with $ \Sigma $. We also know the converse of this problem is true, that the intersection above yeilds a set in $ \mathbb{C} $ with the following property:

$ (C - D)(x^{2} + y^{2}) + Ax + By = D $. As you can see, when C = D, then an equation for a line is yeilded, otherwise it is a circle.

I really am at a loss about how to solve this problem. The only thing I can think to do is to pick 3 points on a circle or radius $ r $ with center $ z_{0} $, use these points to find two vectors in $ \Sigma $, take their cross product to get a normal vector, use this normal vector to get a plane. Once I have the plane in form $ A\xi + B\eta + C\zeta = D $ then I could prove that the circle I had chosen corresponds exactly with $ (C - D)(x^{2} + y^{2}) + Ax + By = D $. Is there not an easier, less computation way to do this?

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    $\begingroup$ You really need to fix the LaTeX here by using $ signs - it is all but unreadable. $\endgroup$ – Ron Gordon Mar 19 '13 at 2:43
  • $\begingroup$ How do I fix the LaTeX using $ signs? This is my first time using this stack exchange, sorry. $\endgroup$ – Max Mar 19 '13 at 2:43
  • $\begingroup$ See meta.math.stackexchange.com/questions/5020/… -- That or just click edit on enough entries until you can figure it out $\endgroup$ – muzzlator Mar 19 '13 at 2:45
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    $\begingroup$ wrap the tex code between dollar signs. So, for instance, to get $\zeta = \xi$, write the text code: \zeta = \xi in between dollar signs. $\endgroup$ – Ittay Weiss Mar 19 '13 at 2:46
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    $\begingroup$ it's called stereographic projection. The proof with the fewest calculations is in Hilbert and Cohn-Vossen $\endgroup$ – Will Jagy Mar 19 '13 at 2:58

I know this is an old question, but here is a proof with essentially zero calculations.

[I'll distinguish the Plane and Riemann Sphere from other planes and spheres by capitalization.]

Let $p$ be the projection mapping from the Plane to the Riemann Sphere. Then $p$ is also an inversion about the north pole that maps the origin of the plane to the south pole. (This can be easily proven by similar triangles.) Note that any inversion maps any sphere not passing through the centre to a sphere. (To see why first prove that the inversion at $O$ that preserves $P$ such that $OP$ is tangent to a sphere also preserves the sphere, because any other point $Q$ on the sphere is mapped to the other intersection of $OQ$ with the sphere, since the plane through $OPQ$ intersects the sphere in a circle.) Now take any circle $C$ in the Plane. $p$ maps any sphere that contains $C$ and does not pass through the north pole to a sphere, and hence $p$ maps $C$ to the intersection of some sphere with the Sphere, which must be a circle!

Also note that $p$ maps any line (a generalized circle) in the Plane to a circle on the Sphere because the plane through the north pole and the line intersects the Sphere in a circle.

Exactly the same method shows the converse. Take any circle $C$ on the Sphere. If $C$ passes through the north pole, it lies on a plane through the north pole, which $p^{-1}$ preserves, and hence $p^{-1}$ maps $C$ to the intersection of that same plane with the Plane, which is a line. If $C$ does not pass through the north pole, it lies on some sphere that does not pass through the north pole, which $p^{-1}$ maps to a sphere, and hence $p^{-1}$ maps $C$ to the intersection of some sphere with the Plane, which is a circle.

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  • $\begingroup$ Also see math.stackexchange.com/a/1865181/21820 for a proof that $p$ preserves angles. $\endgroup$ – user21820 Jul 25 '16 at 12:34
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    $\begingroup$ Downvoter, if there is a mistake point it out and I will fix it. Otherwise, why did you downvote both answers for no reason? $\endgroup$ – user21820 Sep 14 '16 at 2:50

The type of map you describe is called a stereographic projection:


There are many resources out there which proves the thing you are looking for:

See http://www.geom.uiuc.edu/docs/doyle/mpls/handouts/node33.html for a proof that it maps circles to circles for a slightly different sphere, this will give you the basic idea

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  • $\begingroup$ I understand this proof, but it is going in the opposite direction that I am trying to solve. This proof shows the circles on the sphere are circles in the plane. I am trying to show that circles on the plane or circles on the sphere. $\endgroup$ – Max Mar 19 '13 at 11:05
  • $\begingroup$ Try to reverse engineer the process, the map is $1-1$ so it shouldnt be too hard undoing it. Otherwise just google, I remember finding a gazillion results when I googled last time. Hope that helps (otherwise I have actual books I can recommend) $\endgroup$ – muzzlator Mar 19 '13 at 11:14
  • $\begingroup$ I can't find any any proofs going in the opposite direction, and I can't figure out how to go backwards either. If you could provide some books that you know solve this, that would be helpful. $\endgroup$ – Max Mar 19 '13 at 11:28
  • $\begingroup$ Ponnusamy "Complex Variables with applications" devotes an entire subchapter to this matter. Thinking about it a bit more, reverse engineering it should be easy. Aha, here is a proof of the converse. As you can see, it's just a matter of playing with the coordinates you get in the other direction: people.maths.ox.ac.uk/earl/G2-lecture5.pdf $\endgroup$ – muzzlator Mar 19 '13 at 11:51
  • $\begingroup$ @Max: See my answer for both directions with no algebraic calculations at all. $\endgroup$ – user21820 Jul 28 '16 at 6:35

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