Using stereographic projection, you create a distorted image of the surface of a sphere, stretched out to cover an infinite plane. Each point on the sphere is mapped to a point in the equatorial plane by a projection ray starting at a pole of the sphere. Draw a circle on the sphere, e.g. by intersecting a plane with it. This circle is projected down to a circle in the equatorial plane.

Is there an intuitive way to see that circle are always mapped to circles? Here is a proof for stereographic projection being conformal, preserving angles – a closely related property. I stopped trying to follow it at this wise comment by the author:

By the way, sometimes it’s more confusing to try to follow someone else’s geometric proof than it is to construct the proof for yourself.

I did not search for more proofs for circles staying circles, but came up with my own – despite the risk.  An allegedly original proof might be either fallacious or trivial and well-known[*] (or, in the rare case of true originality, it may be plagiarized in the future by somebody else).

But even Richard Feynman struggled with a geometrical proof of Isaac Newton. In Physics on the Fringe, Margaret Wertheim writes:

Although it uses only rudimentary mathematical tools, it is a masterpiece of intricacy. So arcane is Newton’s proof that Feynman could not understand it. That is because in the age of calculus, physicists no longer learn much Euclidean geometry, which, like stonemasonry, has become something of a dying art. Feynman was rather surprised he couldn’t follow a piece of scientific reasoning three centuries old, and he seems to have taken that as a personal challenge. Because he couldn’t understand Newton’s proof, he decided to do a version for himself. The task nearly defeated him, and the result of his work, when it was finally published, occupies close to a hundred typewritten pages.

The sketch below should contain the whole proof (for the humble stereographic projection, not for Newton’s proof).

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My intention is to avoid thinking about anything three-dimensional and to reduce the problem to the usual spotting of suitable triangles in a plane. I got rid of the third dimension by spotting the elliptical cone made up of the projecting rays connecting the pole with the ‘source’ circle’ in the equatorial plane. The ‘target figure’ is also an intersection of the elliptical cone with another plane. It has to be an ellipse or a circle.

If target and source are both circles, there has to be some sort of symmetry. How many ways would be there to fit a circular disc into a given elliptical cone? If you already know one circle, the other one would have to be tilted by the same angle with respect to the symmetry axis of the cone. So, I only need to prove that the inclination angle of the target circle is the same as for the source circle.

In the sketch, I tried to outline the steps (in the right part, all the triangles inscribed in the ‘globe’), using numbers. The corresponding circles are perpendicular to the drawing plane, thus appear as lines. Starting point: I draw the symmetry axis of the cone, spot the two inclination angles α and β, and the opening angle of the cone, 2 times δ.

The constraint that these are connected by spherical projection is used in step two, when I spot the angle γ ‘outside of the cone’. Otherwise, it is just similar triangles, right angles and the sum of angles in a triangle:

1. Look at the right-angled triangle built from the plane of the circle on the sphere, and its inclination angle α. Spot the exterior angle 90 + α.
2. Keep the arc cut by the upper projection ray, move around the corner near the South Pole. Spot γ again, subtended by the same arc.
3. Complete the big triangle (that has the other instance of γ): As the hypotenuse is the diameter, the inscribed angle is a right angle, and the top angle is 90 – γ.
4. Look at the right-angled triangle built from the plane of the circle in the equatorial plane. Spot two similar, right-angled triangles (two sides of each triangle are perpendicular to each other, respectively). Thus β is rediscovered near the North Pole.
5. The top angle can be ‘partitioned’ in different ways; it can also be expressed as β + δ.
6. The two expressions for γ have to equal, δ and 90° cancel, and α = β

Step 7 is about the cone and why the target figure has to be circle if the source one is – making more precise what I have assumed intuitively: On the left, the cone is shown from the perspective of the ‘eye’ looking at the sphere (from the right).

The surfaces perpendicular to the cone’s axis are ellipses (parallel to the cone’s base area), these appear as lines from that perspective. So the images of the tilted circles are ellipses in the drawing. The inclination angle is the same (as just proved), thus both figures appear as ellipses with the same ratio of their principal axes. A circle can be detected by this exact aspect ratio. A true ellipse would appear as an ellipse, circle or line – in any case as something with a different aspect ratio.

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I think the difficulty with understanding geometrical proofs is that the presentation is ‘non-linear’, you cannot read a figure step-by-step like algebraic manipulations. Mathematical language is already ‘code’, and code is easier to read and follow than the ‘real thing’.

As a physics student you are enchanted when all the Euclidean geometry is finally encoded in linear algebra. No more looking at triangles, just writing down the vector equations and cranking the handle. Newton’s intricate proof is turned into a concise exercise of solving a simple differential equation (and spotting several clever substitutions of variables) – geometry replaced by calculus.

Analogue feedback loops in the real world become a software program whose source code can be followed line by line. It is easier to break down a thermodynamic system into little parts, model each as a (software) object, solve differential equations numerically, let objects interact by executing the code – and be surprised by the outcome. Been there, done that. The mystery of the effect of the feedback remains though: You ‘understand’ each little piece, yet you still struggle to keep the whole system in your head at once. If you run lots of simulations, you build up a table of rules of thumbs how the system ‘actually’ behaves – much like doing measurements in the real world.

I wonder how all that relates to the way in which software today can find novel mathematical proofs.

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Related posts – it’s interesting to discover something like a series in retrospect:

On Physics on the Fringe: Physics Paradoxers and Outsiders.

On struggling with geometrical proofs, incl. more quotes about Feynman from Wertheim’s book: Mastering Geometry is a Lost Art

I’ve some experience in getting a geometrical proof wrong first, in public: Revisiting the Enigma of the Intersecting Lines and That Pesky Triangle

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[*] Update 2021-11-10. It was the well-known option. Only now I had the idea to google for ‘stereographic projection’ combined with ‘elliptical cone’. And of course, somebody else came up with a very similar proof before me: http://www.mechanicaldust.com/papers/StereographicProjection_04.pdf.