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This may be a silly question. I have learned to prove that circles are preserved at the infinitesimal scale, however, does this ALONE imply that circles are mapped as circles for the stereographic projection? Why/Why not.

swang
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1 Answers1

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Consider the map $z \mapsto z^2$, which is conformal on the punctured plane $\mathbb{C} \setminus \{0\}$. A circle in this domain is $C : |z-2| = 1$. Suppose $z \in C$, so $z = 2+e^{i \theta}$ for $\theta \in [0, 2\pi)$. We have

$$z^2 = (2+e^{i \theta})^2= 4 + 4e^{i \theta} + e^{2i \theta},$$

which is clearly not a circle. So the answer is no, conformality alone does not imply that circles are preserved. Conformality means that the angle between contours are preserved. So for example, if you have two orthogonal circles, they will be mapped to two orthogonal contours under a conformal map, albeit not necessarily to two new circles.

However, circles are preserved under stereographic projection, or more generally circles and lines are mapped to circles under the steoreographic projection, which can be shown by simply applying the formula for the steoreographic projection to a generalized circle.

Thusle Gadelankz
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  • OK, another silly question from me. Could you elaborate on how one "simply applies" the formula for the stereographic projection to a generalized circle? I have seen proofs with the reverse mapping function, is this the same one? Thanks in advance. – swang Feb 13 '21 at 10:17
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    Yes, it is usually done with the inverse mapping. The computations are easier. Consider a generalized circle $A(x^2+y^2) + Bx + Cy+ D$ in $\mathbb{R}^2$ ($A=0$ for a line and $A \neq 0$ for a circle) and apply the inverse map. You will see that the image on the Riemann-sphere has to be the intersection between a plane and the sphere, which is of course a circle. Furthermore, you can see that lines will be mapped to circles that pass through "the north pole" or $\infty$ if you will. – Thusle Gadelankz Feb 13 '21 at 10:23
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    I like your example. There is an interesting phenomenon here. Your example is a conformal bijection of the punctured plane. Now, any conformal bijection of the full plane $\mathbb C$ does preserve circles; indeed, such a map is a linear polynomial. Any conformal bijection of the Riemann sphere ALSO preserves circles; indeed these maps are Möbius. There is a pattern here. – Giuseppe Negro Feb 13 '21 at 10:48
  • I see the pattern you're pointing out, and I agree it is interesting. However I am not sure my example fits the pattern perfectly as my example is actually not a bijection of the punctured plane since it is not injective. However, it would be interesting if anyone could provide such an example. – Thusle Gadelankz Feb 13 '21 at 11:05
  • In the end I abandoned the train of thoughts of my previous comment. I don’t think it leads anywhere. The fact that both the stereographic projection and the Möbius maps preserve circles is probably just a consequence of the highly rigid structure of the Riemann sphere, rather than a general phenomenon of conformal geometry. Indeed, only the GREAT circles of the sphere are meaningful conformal objects, because they are geodesics. All the other circles do not have an abstract analogue. – Giuseppe Negro Feb 13 '21 at 11:15