# Geometry of the space of circles in the Euclidean plane

We know that Mobius transformations, $z\to\frac{az+b}{cz+d}$, permutes circles and lines in the Euclidean plane, $(\mathbb{R}^2, dx^2 + dy^2 )$.

It may even be possible to write an explicit formula for the general Mobius action on any given circle:

$$|z-z_0|=r \mapsto \left|\frac{az+b}{cz+d} -z_0 \right|=r$$

Such a space can be generated by translations $z \mapsto z + z_1$, rotations $z \mapsto \omega z$, dilations $z \mapsto rz$ and inversions $\displaystyle z = \frac{1}{\overline{z}}$. The action on all circles is clear except for the last case:

$$|z-z_0|=r \mapsto \left| z - \frac{\tfrac{1}{2}z_0}{|z_0|^2 - r^2} \right|= \frac{\tfrac{1}{2}r}{|z_0|^2 - r^2}$$

Is there a more succinct way to write this transformation as a Lie group action?

Here, the Mobius group $G = PSL(2, \mathbb{C}) \simeq SO(1,3)^+$ should act on on the space of circles (and lines) in the plane $\mathbb{R}^2$.

This question also leads me to wonder what the space of circles in the Euclidean plane should be. Naively, the circles are a copy of $\mathbb{C} \times \mathbb{R}^+$ which looks like it could possibly be Hyperbolic space $\mathbb{H}^3$, in which case there would even be a natural metric.

However, the lines all have infinite radius. In which case, we still have three parameters, now a copy of $\mathbb{C} \times S^1$ identifying each line with the closest point to the origin and its direction. The $S^1$ behaving something like the circle at infinity (YouTube).

So now my space of circles is $\big(\mathbb{C}\times \mathbb{R}^+ \big) \cup \big(\mathbb{C}\times S^1 \big)$. How does that happen?

In fact this space should be $\big(\mathbb{C}\times \mathbb{R}^+ \big) \cup \big(S^1 \times \mathbb{R}^+ \big)$.

• are you sure about this "$\mathbb{C}\times S^1$"? – Hao Chen Dec 11 '14 at 16:29
• The space of lines by itself is actually a copy of the Möbius strip. In particular, it is 2-dimensional. – Dylan Thurston Dec 11 '14 at 16:36
• @HaoChen not at all, maybe $\mathbb{C} \times \mathbb{R}P^1$? I don't specify oriented or non-oriented circles. My main question is just if the inversion operation extends to an action on the "moduli space" of all circles on the extended plane $\hat{\mathbb{C}}$ – john mangual Dec 11 '14 at 16:40
• @johnmangual. Two references are Introduction to Möbius Differential Geometry by Hertrich-Jeromin and Lie Sphere Geometry by Cecil. Personally, I prefer oriented circles and allow Mobius transformations to reverse the orientation, then the answer of Bryant is good enough. – Hao Chen Dec 11 '14 at 16:43
• This question on MSE also discusses the space of all circles and lines, how to parametrize them and how Möbius transformations act on them. In my answer there I used input from this question I myself had asked here on MO before. Those other posts might contribute various facets to this picture. – MvG Dec 11 '14 at 20:18

Things will simplify if you just consider the circles on the Riemann sphere $S^2 = \mathbb{C}\cup\{\infty\}$, for your space is simply the space of circles on the sphere (with the lines in $\mathbb{C}$ just becoming the circles through $\infty$). Thus, your extended space of circles becomes $M^3 = \mathrm{PSL}(2,\mathbb{C})/H$ where $H\simeq\mathrm{PSL}(2,\mathbb{R})$ is the subgroup of linear fractional transformations that preserve a given circle, say $\mathbb{R}\cup\{\infty\}$. (See correction below. $\mathrm{PSL}(2,\mathbb{R})$ is actually the subgroup that preserves the oriented circle $\mathbb{R}\cup\{\infty\}$, when I actually need to divide by the $2$-component subgroup that preserves this circle but not necessarily its orientation.)

Because $\mathrm{PSL}(2,\mathbb{R})\simeq\mathrm{SO}(2,1)^+$, it follows easily that this homogeneous space $M^3$ is naturally a Lorentzian manifold, but not naturally a Riemannian manifold.

If you want to think of $M$ as $\mathrm{SO}(3,1)^+/\mathrm{SO}(2,1)^+$, then you can see it as the space of oriented, Lorentzian $3$-planes in $\mathbb{R}^{3,1}$, i.e., it is identifiable with the quadratic hypersurface $\mathbb{S}^{2,1}\subset \mathbb{R}^{3,1}$ consisting of those vectors $v$ that satisfy $v\cdot v = 1$. The invariant Lorentzian metric on it is the one that it inherits as a submanifold of $\mathbb{R}^{3,1}$.

Correction: As Dylan points out, I should have divided by the subgroup $H\subset \mathrm{PSL}(2,\mathbb{C})$ generated by $\mathrm{PSL}(2,\mathbb{R})$ and the linear fractional transformation $f(z) = -z$, so that $H$ is the subgroup that preserves $\mathbb{R}\cup\{\infty\}$, but not necessarily its orientation. This means that the correct answer is $M = \mathrm{SO}(3,1)^+/H$, and this is actually realized not as $\mathbb{S}^{2,1}$ as described above, but as its quotient by the involution $v\mapsto-v$, which is actually $\mathbb{RP}^{2,1}\subset \mathbb{RP}^3$. Of course, it's still a connected Lorentzian $3$-manifold. (Whether some would call it "de Sitter space", I don't know.)

• We also call it "de Sitter space". – Hao Chen Dec 11 '14 at 16:23
• You're actually describing the double cover of the correct answer, which is also the space of oriented circles. – Dylan Thurston Dec 11 '14 at 16:38
• @DylanThurston: Oh, yes, you are right. I always think oriented circles, but, of course, the subgroup preserving a circle need not preserve the orientation, so it's actually the quotient by the orientation-reversing involution. I'll fix the answer above. – Robert Bryant Dec 11 '14 at 17:03

Robert's answer is complete as far as I'm concerned and should be accepted. However some people may be wondering about the Lorentzian geometry of the space of circles and I ran out of space making a comment.

Well it's more interesting if we take oriented circles, because then the manifold has a causal structure and we can discuss whether one circle is in the future or past of another circle. The way I would represent the orientation is to choose an interior for each circle, which I'll interpret as an open disc in the 2-sphere.

Then we can say that circle A is in the future of circle B, if the interior of A is contained in the interior of B. Two circles are null-like (light-like) separated if the interior of one is contained in the interior of the other and the circles are tangent.

• There are more: two circles are tangent from outside if their Lorentzian inner product is $-1$; two circles intersect orthogonally if the inner product is $0$. – Hao Chen Dec 11 '14 at 17:55
• Good point, if they intersect, then the inner product should be the cosine of the angle between them. When the interiors don't intersect then the inner product can be expressed in terms of hyperbolic functions, though it gets a bit trickier to state. – James Griffin Dec 12 '14 at 12:05
• There's a "separation function" due to Boyd, describing how close two spheres are or how deep they intersect. The inner product is exactly the negation of separation. – Hao Chen Dec 12 '14 at 12:17
• That sounds interesting, could you tell me which Boyd, or provide a reference? I assume the "separation function" is related in the notion of distance, where you view the two circles as boundaries of 2-planes in hyperbolic 3-space and then take the shortest geodesic from one to the other (this is what I had in mind). – James Griffin Dec 12 '14 at 16:57
• Looking through your publications I was able to answer my question, though if you have any comments I'd like to hear them: David W. Boyd, possible reference, "The osculatory packing of a three dimensional sphere". – James Griffin Dec 12 '14 at 17:16

About the second part of the question: the space of circles and lines in the plane, or equivalently the space of circles on the Riemann sphere, is homeomorphic to the complement of a ball in the real projective $3$-space.

(This had been given to me as an exercise about ten years ago by Alberto Verjovsky). Consider the unit sphere in $\mathbb{R}^3$, and to each of its circle, associate the tip of the cone which is tangent to the sphere along this circle. If this cone is in fact a cylinder, then it only means its tip is at infinity, in $\mathbb{R}\mathrm{P}^3$. The above claim follows.

• That's a nice proof, you can also see it immediately from Robert's answer above, the projective space of $\mathbb{R}^{3,1}$ is decomposed via its inner product into three pieces, depending on whether vectors have strictly positive, strictly negative or zero norm. The negative vectors form an open ball (and a model for hyperbolic space), the "0" vectors are its boundary, so the positive vectors are $\mathbb{R}P^3$ take a closed ball. – James Griffin Dec 12 '14 at 17:05

Disclaimer: my original posting made wrong statements about geometry, but Ī opted to preserve and fix it because a different approach ultimately resulted in similar conclusions than of other answers.

Addressing the second part (Euclidean) of the question.

There is very obvious parametrization of (Euclidean) circles: each has the centre (a point on the plane) and radius (a number), so as a smooth manifold their space is E2 × (0, +∞) (if we exclude all degenerate cases). Let’s use respective real coordinates $(x,y,r)$. To address the question of its geometry one first should look on symmetries acting on it.

Which transformations of Euclidean plane preserve all circles? Obviously, similarity transformations (although not only). The next interesting question is a nature of their action, and the following observation will help us:

Any similarity transformation preserves quadratic form (or symmetric 0,2-tensor) fields $\frac{1}{r^2}(dx^2 + dy^2)$ and $\frac{1}{r^2}dr^2$ on the space of circles.

How could we guess that they should combine to a pseudo-Riemannian structure? There can be at least two arguments. One, “global”, is to consider inversion about a circle (that preserves geometry of circles) and look what happened to quadratic forms. Another, “local”, is to consider paths in the space of circles. Let’s consider such smooth path $φ(t)$ that $\dot φ(0) ≠ 0$, and look whether will $φ(t)$ for small $t$ intersect $φ(0)$. If at $t=0$: $\dot r > 0$ and $(\dot r)^2 > (\dot x)^2 + (\dot y)^2$ ($r$, $x$, and $y$ refer to respective coordinates of $φ(t)$), then for some $ε>0$ circles $φ(t)$ for $0<t<ε$ will lie outside $φ(0)$, without intersection. Likewise, if at $t=0$: $\dot r < 0$ and $(\dot r)^2 > (\dot x)^2 + (\dot y)^2$, then for some $ε>0$ circles $φ(t)$ for $0<t<ε$ will lie inside $φ(0)$. It suggests that the correct quadratic form must be conformally equivalent to $dr^2 - dx^2 - dy^2$, that (together with preservation requirements) yields $\frac{1}{r^2}(dr^2 - dx^2 - dy^2)$, the difference of original quadratic forms.

What about paths where $(\dot r)^2 < (\dot x)^2 + (\dot y)^2$? They are different wrt geometry of circles. In such paths “neighbouring” circles intersect each other.

This posting doesn’t contain a proof that any circles-preserving (conformal or anticonformal) transformation of Euclidean plane will preserve said pseudo-Riemannian form (that other answers imply). It only derives from “elementary” geometric considerations that $\frac{1}{r^2}(dr^2 - dx^2 - dy^2)$ might be the correct structure on the manifold of Euclidean circles. It also suggests some twistor-like semantics for all this thing, namely, that null geodesics in the circles’ space are families of tangent circles on the plane. Ī hope this “elementary” approach can be helpful for somebody.