# How can I show that circles in the complex plane correspond to circles on the Riemann sphere? How about lines?

Suppose $T \subset \mathbb{C}$. Show that the corresponding set $S \subset \Sigma$ is

a. a circle if $T$ is a circle.
b. a circle minus (0, 0, 1) if $T$ is a line.

Here we are defining $\Sigma$ to be the Riemann sphere, given by the set: $$\Sigma = \left \{(\xi, \eta, \zeta) : \xi^{2} + \eta^{2} + (\zeta - \frac{1}{2})^{2} = \frac{1}{4} \right \}$$

To take a point from $\mathbb{C}$ to $\Sigma$ we can use the following:

$$\xi = \frac{x}{x^{2} + y^{2} + 1}; \eta = \frac{y}{x^{2} + y^{2} + 1}; \zeta = \frac{x^{2} + y^{2}}{x^{2} + y^{2} + 1}$$

We define a circle on $\Sigma$ to be the intersection of a plane of the form $A\xi + B\eta + C\zeta = D$ with $\Sigma$. We also know the converse of this problem is true, that the intersection above yeilds a set in $\mathbb{C}$ with the following property:

$(C - D)(x^{2} + y^{2}) + Ax + By = D$. As you can see, when C = D, then an equation for a line is yeilded, otherwise it is a circle.

I really am at a loss about how to solve this problem. The only thing I can think to do is to pick 3 points on a circle or radius $r$ with center $z_{0}$, use these points to find two vectors in $\Sigma$, take their cross product to get a normal vector, use this normal vector to get a plane. Once I have the plane in form $A\xi + B\eta + C\zeta = D$ then I could prove that the circle I had chosen corresponds exactly with $(C - D)(x^{2} + y^{2}) + Ax + By = D$. Is there not an easier, less computation way to do this?

• You really need to fix the LaTeX here by using $signs - it is all but unreadable. Commented Mar 19, 2013 at 2:43 • How do I fix the LaTeX using$ signs? This is my first time using this stack exchange, sorry.
– Max
Commented Mar 19, 2013 at 2:43
• See meta.math.stackexchange.com/questions/5020/… -- That or just click edit on enough entries until you can figure it out Commented Mar 19, 2013 at 2:45
• wrap the tex code between dollar signs. So, for instance, to get $\zeta = \xi$, write the text code: \zeta = \xi in between dollar signs. Commented Mar 19, 2013 at 2:46
• it's called stereographic projection. The proof with the fewest calculations is in Hilbert and Cohn-Vossen Commented Mar 19, 2013 at 2:58

I know this is an old question, but here is a proof with essentially zero calculations.

[I'll distinguish the Plane and Riemann Sphere from other planes and spheres by capitalization.]

Let $p$ be the projection mapping from the Plane to the Riemann Sphere. Then $p$ is also an inversion about the north pole that maps the origin of the plane to the south pole. (This can be easily proven by similar triangles.) Note that any inversion maps any sphere not passing through the centre to a sphere. (To see why first prove that the inversion at $O$ that preserves $P$ such that $OP$ is tangent to a sphere also preserves the sphere, because any other point $Q$ on the sphere is mapped to the other intersection of $OQ$ with the sphere, since the plane through $OPQ$ intersects the sphere in a circle.) Now take any circle $C$ in the Plane. $p$ maps any sphere that contains $C$ and does not pass through the north pole to a sphere, and hence $p$ maps $C$ to the intersection of some sphere with the Sphere, which must be a circle!

Also note that $p$ maps any line (a generalized circle) in the Plane to a circle on the Sphere because the plane through the north pole and the line intersects the Sphere in a circle.

Exactly the same method shows the converse. Take any circle $C$ on the Sphere. If $C$ passes through the north pole, it lies on a plane through the north pole, which $p^{-1}$ preserves, and hence $p^{-1}$ maps $C$ to the intersection of that same plane with the Plane, which is a line. If $C$ does not pass through the north pole, it lies on some sphere that does not pass through the north pole, which $p^{-1}$ maps to a sphere, and hence $p^{-1}$ maps $C$ to the intersection of some sphere with the Plane, which is a circle.

• Also see math.stackexchange.com/a/1865181/21820 for a proof that $p$ preserves angles. Commented Jul 25, 2016 at 12:34
• Downvoter, if there is a mistake point it out and I will fix it. Otherwise, why did you downvote both answers for no reason? Commented Sep 14, 2016 at 2:50

The type of map you describe is called a stereographic projection:

http://en.wikipedia.org/wiki/Stereographic_projection

There are many resources out there which proves the thing you are looking for:

See http://www.geom.uiuc.edu/docs/doyle/mpls/handouts/node33.html for a proof that it maps circles to circles for a slightly different sphere, this will give you the basic idea

• I understand this proof, but it is going in the opposite direction that I am trying to solve. This proof shows the circles on the sphere are circles in the plane. I am trying to show that circles on the plane or circles on the sphere.
– Max
Commented Mar 19, 2013 at 11:05
• Try to reverse engineer the process, the map is $1-1$ so it shouldnt be too hard undoing it. Otherwise just google, I remember finding a gazillion results when I googled last time. Hope that helps (otherwise I have actual books I can recommend) Commented Mar 19, 2013 at 11:14
• I can't find any any proofs going in the opposite direction, and I can't figure out how to go backwards either. If you could provide some books that you know solve this, that would be helpful.
– Max
Commented Mar 19, 2013 at 11:28
• Ponnusamy "Complex Variables with applications" devotes an entire subchapter to this matter. Thinking about it a bit more, reverse engineering it should be easy. Aha, here is a proof of the converse. As you can see, it's just a matter of playing with the coordinates you get in the other direction: people.maths.ox.ac.uk/earl/G2-lecture5.pdf Commented Mar 19, 2013 at 11:51
• @Max: See my answer for both directions with no algebraic calculations at all. Commented Jul 28, 2016 at 6:35

In this answer I proved that non-degenerate circles in $$\mathbb{R}^2$$ are mapped by the inverse of the stereographic projection to non-degenerate circles on the unit sphere that do not pass through the northern pole. I will now prove that lines in $$\mathbb{R}^2$$ are mapped by the inverse of the stereographic projection to non-degenerate circles on the unit sphere that do pass through the northern pole.

(I) Preliminaries

I will use the following notations (most of which are copied from the above-mentioned answer of mine).

Assumption

Let $$a$$, $$b$$, and $$c$$ be real numbers such that $$a^2 + b^2 \neq 0$$.

Definitions

• $$\Sigma$$ is defined to be the unit sphere in the Euclidean space, i.e. $$\Sigma := \big\{(x,y,z)\in\mathbb{R}^3 : x^2+y^2+z^2 = 1\big\}.$$

• $$N$$ is defined to be the northern pole of the unit sphere, i.e. $$N := (0,0,1)$$.

• $$S$$ is defined to be the stereographic projection, i.e. the function $$S:\Sigma\setminus\{N\}\rightarrow\mathbb{R}^2$$ satisfying $$S\big((x,y,z)\big) = \Big(\frac{x}{1-z},\frac{y}{1-z}\Big)$$ for every $$(x,y,z) \in \Sigma\setminus\{N\}$$.

• $$R$$ is defined to be the function $$R:\mathbb{R}^2\rightarrow\mathbb{R}^3$$ satisfying $$R\big((x,y)\big) = \Big(\frac{2x}{x^2+y^2+1}, \frac{2y}{x^2+y^2+1}, \frac{x^2+y^2-1}{x^2+y^2+1}\Big)$$ for every $$(x,y) \in \mathbb{R}^2$$.

• $$\operatorname{Id}_\Sigma$$ and $$\operatorname{Id}_{\mathbb{R}^2}$$ are defined to be the identity functions on $$\Sigma$$ and on $$\mathbb{R}^2$$, respectively. (We will only really need $$\operatorname{Id}_\Sigma$$.)

• We define $$L$$ to be the line $$\big\{(x,y) \in \mathbb{R}^2 : ax + by = c\big\}$$.

Fact

As I did in the above mentioned answer, I take the following fact for granted, without proving it:

• $$S$$ and $$R$$ are inverses, i.e. (a) $$R\circ S = \operatorname{Id}_\Sigma$$, and (b) $$\operatorname{Img} R \subseteq \Sigma\setminus\{N\}$$ and $$S\circ R = \operatorname{Id}_{\mathbb{R}^2}$$. Note that (b) implies, in particular, that $$R$$ is injective. (The proof will only use the facts that $$R\circ S = \operatorname{Id}_\Sigma$$, that $$\operatorname{Img} R \subseteq \Sigma\setminus\{N\}$$, and that $$R$$ is injective.)

(II) A formal statement of the problem

There is a plane $$\Pi \subseteq \mathbb{R}^3$$ which passes through $$N$$ and whose distance from the origin is strictly less than $$1$$, such that $$R[L] = (\Sigma\cap\Pi)\setminus\{N\}$$.

(III) Proof

From Analytic Geometry, a plane $$\Pi \subseteq \mathbb{R}^3$$ is any set satisfying $$\Pi = \big\{(x,y,z) \in \mathbb{R}^3 : Ax + By + Cz + D = 0\big\}$$ for some $$A, B, C, D \in \mathbb{R}$$ such that $$A^2 + B^2 + C^2 \neq 0$$.

Define \begin{align*} A &:= \frac{a}{2},\\ B &:= \frac{b}{2}, \end{align*} and define $$C$$ and $$D$$ to be the unique real numbers satisfying \begin{align*} C + D &= 0\\ C - D &= c. \end{align*}\tag{*} (Linear Algebra tells us that this system of equations has a unique solution $$\begin{bmatrix}C\\D\end{bmatrix} \in \mathbb{R}_{2\times1}$$, because the matrix of coefficients $$\begin{bmatrix}1 & 1\\1 & -1\end{bmatrix}$$ is invertible.)

Define $$\Pi := \big\{(x,y,z) \in \mathbb{R}^3 : Ax + By + Cz + D = 0\big\}$$.

Note:

• $$A^2 + B^2 + C^2 \neq 0$$, since otherwise we would have $$a = b = 0$$, in contradiction to the assumption that $$a^2 + b^2 \neq 0$$. Hence, by the first paragraph of this proof, $$\Pi$$ is a plane in $$\mathbb{R}^3$$.

• $$N \in \Pi$$, as can be readily verified from the definition of $$\Pi$$, using $$(*)$$.

• I claim that in order to show that $$\Pi$$'s distance from the origin is strictly less than $$1$$, it suffices to show that $$R[L] = (\Sigma\cap\Pi)\setminus\{N\}$$. Indeed, if $$R[L] = (\Sigma\cap\Pi)\setminus\{N\}$$, then in particular $$R[L] \subseteq \Sigma\cap\Pi$$, hence the fact that $$R$$ is injective implies that the set $$\Sigma\cap\Pi$$ includes at least two points, and therefore $$\Pi$$ passes through $$\Sigma$$'s interior, which implies, by $$\Sigma$$'s definition, that $$\Pi$$'s distance from the origin is strictly less than $$1$$.

It remains to show that $$R[L] = (\Sigma\cap\Pi)\setminus\{N\}$$. We will do so by showing that $$R[L] \subseteq (\Sigma\cap\Pi)\setminus\{N\}$$ and that $$(\Sigma\cap\Pi)\setminus\{N\} \subseteq R[L]$$.

Firstly we show that $$R[L] \subseteq (\Sigma\cap\Pi)\setminus\{N\}$$. By the fact that $$\operatorname{Img} R \subseteq \Sigma\setminus\{N\}$$ it suffices to show that $$R[L] \subseteq \Pi$$. Let $$(x,y) \in L$$. Then, by the definitions of $$R$$ and of $$\Pi$$ it suffices to show that $$A\frac{2x}{x^2+y^2+1} + B\frac{2y}{x^2+y^2+1} + C\frac{x^2+y^2-1}{x^2+y^2+1} + D = 0,$$ which, by algebraic manipulations, can be shown to be equivalent to $$2Ax + 2By + (C+D)(x^2+y^2) = C-D,$$ which, by the definitions of $$A$$, $$B$$, $$C$$, and $$D$$, reduces to $$ax + by = c.$$ Since, by selection, $$(x,y) \in L$$, the last equation holds by the definition of $$L$$.

Secondly we show that $$(\Sigma\cap\Pi)\setminus\{N\} \subseteq R[L]$$. Let $$p \in (\Sigma\cap\Pi)\setminus\{N\}$$. Then, in particular, $$p \in \mathbb{R}^3$$, and we can denote its coordinates by $$x$$, $$y$$, and $$z$$, respectively, so that $$p = (x,y,z)$$. Note that since, by selection, $$p \in (\Sigma\cap\Pi)\setminus\{N\}$$, we have $$p \in \Sigma\setminus\{N\}$$. It follows from the definition of $$S$$ that $$p \in \operatorname{dom} S$$, and we can therefore define $$q := S(p)$$. By the fact that $$R\circ S = \operatorname{Id}_\Sigma$$, it follows that $$p = R\big(S(p)\big) = R(q)$$. Therefore, it suffices to show that $$q \in L$$, i.e., by the definition of $$q$$, that $$S\big((x,y,z)\big) \in L$$, i.e., by the definitions of $$S$$ and of $$L$$, that $$a\frac{x}{1-z} + b\frac{y}{1-z} = c,$$ which, by algebraic manipulations can be shown to be equivalent to $$ax + by = c(1-z).\tag{**}$$ (Since, as we saw above, $$p \in \Sigma\setminus\{N\}$$, it follows that $$z \neq 1$$.)

By algebraic manipulations and by using the definitions of $$A$$, $$B$$, $$C$$, and $$D$$, $$(**)$$ can be shown to be equivalent to $$2(Ax + By + Cz + D) = (C + D)(1 + z).\tag{***}$$ Since, by $$p$$'s selection, $$(x,y,z) \in \Pi$$, it follows from the definition of $$\Pi$$ that $$(*\!*\!*)$$ reduces to $$0 = (C+D)(1+z)$$. The last equation is valid by $$(*)$$. Q.E.D.

Related posts

• In this answer I proved that non-degenerate circles in $$\mathbb{R}^2$$ are mapped by the inverse of the stereographic projection to non-degenerate circles on the unit sphere that do not pass through the northern pole.

• In this answer I clarified a step in a proof from this paper. That paper shows that non-degenerate circles on the unit sphere are mapped by the stereographic projection to either lines or to non-degenerate circles in $$\mathbb{R}^2$$, depending on whether the original circle passes or no through the northern pole, respectively.