# Diametrically opposite points go to diametrically opposite points under stereographic projection [closed]

I asked this question in MSE here: http://math.stackexchange.com/questions/184524/diametrically-opposite-points-go-to-diametrically-opposite-points-under-stereogr but I didn’t get an answer. I really need to solve this problem and believe me this is not a homework.

Suppose $P_1$ and $P_2$ are two diametrically opposite points of a circle $C$ in the complex plane and suppose $\sigma (z)$ denotes the image of $z\in\mathbb{C}$ on Riemann sphere due to the inverse of the stereographic projection. Then I need to prove that $\sigma (P_1)$ and $\sigma (P_2)$ are also diametrically opposite points of the circle $\sigma (C)$.

Intuitively this is obvious but I need to prove this fact algebraically (i.e. using coordinate geometry ).

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## closed as off-topic by Ricardo Andrade, Yemon Choi, Stefan Waldmann, Neil Strickland, Felipe VolochNov 24 '14 at 15:54

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What do you mean by "a circle"? If you mean circle centered at the origin, then the statement is obviously true since $\sigma$ conjugates rotations about origin in the complex plane to rotations about vertical axis in the 2-sphere. (E.g., since $\sigma$ is a Moebius transformation). If "a circle" means an arbitrary round circle, then this is of course false. Also, what is "coordinate geometry"? Do you mean "computing everything in Cartesian coordinates"? Then just take the formula for $\sigma$ and do the computation yourself, it will be rather ugly, of course. –  Misha Aug 23 '12 at 6:29
I meant circle centred at any arbitrary point, is it really false then ? Basically I want a proper proof without using geometric intution. –  Pritam Majumder Aug 23 '12 at 6:40
Yes, it is utterly false in the sense that it is false for every circle which is not centered at the origin. Proof of this is both proper and geometric, and reduces to the following: If $\tau$ is an orientation-preserving elliptic Moebius transformation of the round 2-sphere which fixes two non-antipodal points, then $\tau$ cannot be a rotation along any round circle. By the way, somebody greatly misinformed you that "proper" proofs cannot use geometry or be aided by intuition. You may want to read Hilbert and Cohn Vossen "Geometry and Imagination". –  Misha Aug 23 '12 at 7:05

Let $C$ be a round circle in $R^2$ centered at $p$. Let $g$ be a rotation fixing $p$; then, as a Moebius transformation of $R^2\cup \infty$, $g$ has two fixed points $p, \infty$. Let $g^\sigma$ be the congutate of $g$ by $\sigma$. Then $g^\sigma$ is again a Moebius transformation of $S^2$ which preserves the circle $C'=\sigma(C)$ and fixes the points $\sigma(\infty)=N$ and $p'=\sigma(p)$, where $N$ is the north pole of $S^2$. Now, suppose that $g$ has order $2$, i.e., sends each $q\in C$ to the diametrically opposite point. Then $g^\sigma$ also has order $2$.

Note that saying that $\sigma$ sends opposite points on $C$ to opposite points on $C'$, just means that $g^\sigma$ has the property that $g^\sigma(q)=\hat{q}$ (here $\hat{q}\in C'$ denotes the point diametrically opposite to $q$ on $C'$). Under this assumption, $g^\sigma|C'$ would act as a rigid order 2 rotation $R$ on $C'$. Extend $R$ to the entire $S^2$. Since $R$ agrees with $g^\sigma$ on $C'$, and both are orientation-preserving Moebius transformations, $g^\sigma=R$. In particular, $R$ fixes $N$ and $p'$. Thus, $p'=-N$. However, $\sigma(-N)=0$, thus, $p=0$. Therefore, $C$ has center at the origin.

This shows that if $C$ is a circle satisfying the property you want, then $C$ is centered at the origin. The converse is clear too: If $p=0$ then $p'=-N$ and, hence, $g^\sigma$ is the rigid rotation of order $2$ around the vertical axis. Clearly, $g^\sigma(C')=C'$. Thus, $g^\sigma$ sends each point on $C'$ to its opposite.

This proof is intuitive, geometric, rigorous and computation-free.

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If I'm interpreting this correctly, this is false. (Misha pointed this out in the comments, too.)

Stereographic projection sends $(x,y,z)$ on the unit sphere centered at the origin to $(\frac{x}{1-z},\frac{y}{1-z})$. Take the circle $C$ centered at $(1,0,0)$ passing through $(1/\sqrt2,\pm 1/\sqrt2,0)$ and $(1/\sqrt2,0,\pm 1/\sqrt2)$. The images of these $4$ points are $(1/\sqrt2,\pm1/\sqrt2)$ (the equator is fixed in space) and $(\sqrt2 \pm 1,0)$.

The latter pair of images have distance $2$, and they are a diameter, while the former have distance $\sqrt 2$ and are $90$ degrees apart. The image of $C$ is the circle of radius $1$ centered at $(\sqrt2,0).$

It should be intuitive that this fails by considering circles near those passing through $(0,0,1).$ The projections of these circles become very large, but the projections of diametrically opposed points away from the North Pole do not become large, so they can't be diametrically opposed on the large image.

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