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How does one give a proof that is

  • short; and
  • strictly within the bounds of secondary-school geometry

that the stereographic projection

  • is conformal; and
  • maps circles to circles?
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Mapping circles to circles can be proven using pure algebra. I'm not sure about conformality though... – Zhen Lin Nov 29 '11 at 21:50
@ZhenLin : I can believe that, and I think I could do it, but I'd still like to see a geometric rather than algebraic proof. – Michael Hardy Nov 29 '11 at 22:54
How "correct" do you want the proof? If you wave your hands a little bit, mapping circles to circles plus the fact that latitudes and longitudes remain orthogonal by the projection should imply conformality. – Willie Wong Nov 30 '11 at 10:28
I don't want it to be so "correct" that checking details of how it fits in to a rigorous axiom system distracts from the point. But it should be amenable to being made "correct" if necessary. – Michael Hardy Nov 30 '11 at 18:11
I think this geometric proof that stereographic projection maps circles to circles should fit the bill. – t.b. Dec 13 '11 at 7:13
up vote 1 down vote accepted

I'm not sure what is in the post by TheoBromine, it says I need plugins to see all the media on the page. But a full and very careful proof, with five figures, is in Geometry and the Imagination by David Hilbert and Stephan Cohn-Vossen, translated by P. Nemenyi, see AMAZON AND GIGGLEBOOKS on pages 248-251. Also available reprinted by the AMS in 1998 or so.

P.S. Giggle let me see all of pages 248-251.

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The proof I linked to is essentially the same as the one in Hilbert--Cohn-Vossen. Here's a direct link to the relevant pages. – t.b. Dec 13 '11 at 7:47
Very good. Thanks for the link. – Will Jagy Dec 13 '11 at 20:34

It's plain that a circle on a sphere is projected into an oval cone, which when cut by a plane parallel to the circle gives a circle.

(edit: oops - it's not obvious that this code is a symmetrical oval, that the radii at the top and the bottom are the same. Drat.)

Its also plain that there is another way to cut the cone to get a circle - using a plane that is the same as the plane of the circle but reflected about the axis of the cone (angled the opposite way, if you like.

Demonstrating that this second plane is parallel to the plane tangent to the projection point is left as an exercise for the reader :) .

Proving conformality involves a similar construction, I think, but instead of a projected cone you are projecting a pair of planes that pass through the intersection point of then lines on the sphere.

As is obvious, I haven't quite gotten there myself.

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