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I am aware of a few analytical calculations showing that the stereographic projection sends circles on the sphere to circles on the equatorial plane. There are related questions here.

What about a geometrical proof? I have found one somewhere that goes as follows.

Let $O$ be the center of the sphere and $S$ the pole doing the projection. Choose a circle $C$ on the sphere lying on a plane $P$, and denote its center $A$. The rays joining $S$ to the circle $C$ cross the equatorial plane $P'$ at some points, which we want to prove form another circle. Let $M$ and $N$ be the two points on the circle $C$ such that they belong to the $SOA$ plane. Their stereographic projection on the equatorial plane are denoted by $M'$ and $N'$.

enter image description here

The angles noted in the attached figure can easily be proven equal, leading to the two angles $SMN$ and $SN'M'$ being equal. This means that the planes $P$ and $P'$ are symmetrical to each other by reflection across the axis $SA$ (shown as an arrow). Now, the proof follows saying that the "cone" from $S$ to $C$ has the same symmetry, so that its intersections with the two planes $P$ and $P'$ are "equivalent". As a consequence, the two intersections are circles, and this proves the theorem.

But wait. To me, this is not really a "cone". But fine, it does not need to be. However, it does not seem symmetrical across $SA$: taking the symmetric point of $N$ or $M$ will clearly not land on the same "cone". So that would mean that the two shapes made by intersecting with $P$ and $P'$ should not be equivalent; consequently, stereographic projection would not send circles to circles.

Where is my error?

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    $\begingroup$ there is a proof by picture in Hilbert and Cohn-Vossen, Geometry and the Imagination. I will see if I can find the correct pictures on line. Alright, mostly pages 248-251 $\endgroup$ – Will Jagy Dec 30 '16 at 0:58
  • $\begingroup$ can't have everything; they do not show page 250, but do 248, 249, 251 books.google.com/… $\endgroup$ – Will Jagy Dec 30 '16 at 1:05
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    $\begingroup$ Are you aware that, by sterographic projection, the image of the center is not the center of the image ? Said otherwise, the centers of the circle on the sphere and the center of the corresponding circle on the equatorial plane are (generally) not aligned with $S$. $\endgroup$ – Jean Marie Dec 30 '16 at 20:41
  • $\begingroup$ @JeanMarie, I did not know, but what does it change for this question? $\endgroup$ – fffred Dec 31 '16 at 17:26
  • $\begingroup$ It was just a remark, without direct incidence on the question. $\endgroup$ – Jean Marie Dec 31 '16 at 18:33
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The cone is perhaps an oblique one, but it is a cone we are looking at: based on the circle $C$ from plane $P$, with summit $S$.
(Note however, that being oblique only means, from other perspective, that it's a straight cone based on another conic section.)

Now the main point is that - as seen in current picture - the plane $P'$ meets this cone in the same angle as $P$ does. So the conic section of $P'$ will be similar to that of $P$, so will be a circle.

The wording is not too lucky: I guess the symmetry is about the cone itself.

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  • $\begingroup$ I can accept that this definition of a cone, but I still don't see that symmetry in this case. The angles might be the same for the two planes, but they are on opposite sides of the cone, so that a symmetry is required. You say that it is a straight cone: what does this mean? Does it imply its symmetry across an axis? $\endgroup$ – fffred Dec 30 '16 at 10:46
  • $\begingroup$ @fffred: The sphere is symmetric "front to back" (i.e., in the plane of your diagram), so the cone has elliptical cross-sections in planes orthogonal to its axis $SA$, and one axis of these ellipses is perpendicular to the plane of the diagram. The relevant symmetry is, as you say, reflection in $SA$ (i.e., reflection in the plane perpendicular to the plane of the diagram and containing $SA$); the cone is preserved because its cross sections are ellipses, and the reflection plane bisects each along an axis. $\endgroup$ – Andrew D. Hwang Mar 6 '17 at 12:03

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