This is a continuation of my earlier, rather vague question.

I am interested in studying the action of the Möbius group $PGL(2,\mathbb{C})$, on the circlines in the extended complex plane $\mathbb{C} \cup \infty$.

A circline is, by definition any circle or line in $\mathbb{C}$. Circlines are parametrised by $\mathbb{C}^2 \times \mathbb{R}^+$. For $(v,w,\lambda)$ in there, the corresponding circline is given by

$ \{ z \in \mathbb{C} \cup \infty : |\frac{z-\alpha}{z-\beta} |= \lambda \}$

Note: this paramterisation is far from injective and I don't know if it's regular in a way, like a fibre bundle or something.

The action of the Möbius group isn't particularly hard to write down. The map $w = \frac{az+b}{cz+d}$ sends $\{ z : |\frac{z-\alpha}{z-\beta} |= \lambda \}$ to

$\{ w :|\frac{\frac{dw-b}{a-cw} - \alpha}{\frac{dw-b}{a-cw} - \beta}| = \lambda \} $

which can be simplified as in Priestley's book, to the required form.

Now, there are three ways I know of looking at the extended complex plane:

  • $\mathbb{C} \cup \infty$ with some funky laws of arithmetic
  • $S^2 \subseteq \mathbb{R}^3$
  • $\mathbb{C}P^1$

The books I'm learning from treat Möbius transformations and circlines almost exclusively in the first way. This question stems from me going out of my way to avoid doing so.

Now $\mathbb{C} \cup \infty \cong S^2$ by stereographic projection:

$\psi : S^2 \to \mathbb{C} \cup \infty$

$\psi(x,y,z) = \frac{x+iy}{1-z}$

I am under the impression (but haven't proved) that this map turns circles to circlines. This allows me to think of circlines as something more natural. In an answer to my earlier question, it was shown that the space of circles on $S^2$ is parametrised (in fact, double covered) by $S^2 \times (-1,1)$ as intersections with planes.

The actual space of circles on $S^2$ is given by $M = S^2 \times (-1,1) / \sim$ where $(x,y) \sim (-x,-y)$.

My question is, how does the Möbius group act on this manifold. Is it possible to write it down an actual map: $PGL(2,\mathbb{C}) \times M \to M$? (Note: the above snippet from Priestley's book shows that this is certainly possible when working with the other parametrisation)

The approach I tried was to find the inverse image of a circle under the projection map $\pi : \mathbb{C}^2 \setminus 0 \to S^2$, seeing what $GL(2,\mathbb{C})$ does to it and then quotienting it back to circle.

The actual map $\pi : \mathbb{C}^2 \setminus 0 \to \mathbb{R}^3$ is given by

$\pi(z,w) = (2\frac{\mathrm{Re} (z\overline{w})}{|z|^2 + |w|^2}, 2\frac{\mathrm{Im} (z\overline{w})}{|z|^2 + |w|^2}, \frac{|z|^2 - |w|^2}{|z|^2 + |w|^2})$

so unfortunately I chickened out of brute force calculating anything.

The above map is the Hopf-fibration, precomposed with the normalisation map $\mathbb{C}^2 \setminus 0 \to S^3$. So it really is just the composition of two fibre bundle projections, and a fibre bundle itself, with fibres $\mathbb{C} \setminus 0$. Thus the inverse image of a circle would be expected to be something like a fat torus or a twisted variant thereof.

This is about as far as my mathematical ability allowed me to proceed, but I'm hoping someone can help me write down the action $PGL(2,\mathbb{C}) \times M \to M$. Thanks for reading, and apologies about the length of this question.

  • $\begingroup$ You're parametrizing circlines as Apollonian circles. It would be nice to include a link to your earlier question about that $S^2\times(-1,1)$ parametrization, and repeat the core idea of that parametrization. In any case, your question seems remotely related to this question of mine where I, too, asked about how a Möbius transformation acts on the set of all circlines, although I had parametrized my circles differently. $\endgroup$ – MvG Dec 7 '14 at 13:03
  • $\begingroup$ The video Möbius Transformations Revealed (also mentioned in this question might be a good thing to keep in mind as well. $\endgroup$ – MvG Dec 7 '14 at 13:52
  • $\begingroup$ @MvG, thanks for the links and th video. fyi, my earlier question is linked in the opening sentence $\endgroup$ – user40167 Dec 7 '14 at 16:06

This is a complete rewrite of this answer, following an approach which is very different from my first attempt.

Homogeneous parametrization

Suppose you have a point $x=(x_1,x_2,x_3)$ on a circle on the sphere. That means it belongs to the intersection of the sphere and a plane, or in formula, $\lVert x\rVert=1$ and $\langle x,n\rangle=d$. This $n$ is the normal vector of the plane, and $d$ encodes the distance from the origin. If you choose $n\in S^2$ then you can choose $d\in(-1,1)$, so this is your parametrization of circles on the sphere.

Now I'll homogenize things, and move to $\mathbb{RP}^3$. This will allow me to express the stereographic projection in a nice way. For homogeneous coordinates, we represent the unit sphere by the diagonal matrix $S=\operatorname{diag}(1,1,1,-1)$. A point with homogeneous coordinates $x$ lies on that sphere if it satisfies $x^TSx=0$. We also combine the parameters of the plane, writing $p=(n_1,n_2,n_3,-d)$ so a point $x$ lies on the plane $p$ if $\langle x,p\rangle=0$.

Stereographic projection of a point

We'll also need a stereographic projection. For that we'll embed the plane at $z=-1$ and use the point with homogeneous coordinates $N=(0,0,1,1)^T$ as the center of projection. At first we'll map from the plane to the sphere. So suppose now you have a point $(x,y)^T$ in the plane. In $\mathbb{RP}^3$ that point would have coordinates $P=(x,y,-1,1)^T$. To project that onto the sphere, you compute $Q=P^TSP\cdot N - 2N^TSP\cdot P$. This is a a point on the line $PN$ since it is a linear combination of these two. It's distinct from $N$ since it has a non-zero coefficient in front of $P$. And it's on the sphere as you can verify by computing $Q^TSQ$. So the image of $P$ on the sphere is $Q$. In coordinates:

$$P=\begin{pmatrix}x\\y\\-1\\1\end{pmatrix}\mapsto \begin{pmatrix}4x\\4y\\x^2 + y^2 - 4\\x^2 + y^2 + 4\end{pmatrix}=Q$$

Stereographic projection of a circle

Next I'll project a circle with center $(x,y)$ and radius $r$ in the plane up to the sphere. One way to do this is by projecting three points, say $(x+r,y)$, $(x-r,y)$ and $(x,y+r)$. Projecting these onto the sphere gives three points in space, and three (non-collinear) points in space uniquely define a plane. In this case, the plane in question can be parametrized using

$$p=(4x, 4y, x^2 + y^2 - r^2 - 4, -x^2 - y^2 + r^2 - 4)^T \tag1$$

You can check this by ensuring $\langle Q,p\rangle=0$ for $Q$ being the image of each of the three points I just mentioned. In fact I found that vector by taking these three homogeneous coordinate vectors and computing all $3\times3$ minors, in suitable order and with alternating signs. Any multiple of the vector above will describe the same plane, since those coordinates are homogeneous. The representative above has the special property of $p_3+p_4=-8$. So if you start with any representative $p$, scaling that by $\frac{-8}{p_3+p_4}$ will lead to a representative of the form above.

This allows us to determine the description of the circle in the $z=-1$ plane from the parameters of the plane intersecting the sphere:

$$x=\frac{-2p_1}{p_3+p_4}\qquad y=\frac{-2p_2}{p_3+p_4} \qquad x^2+y^2-r^2=\frac{-4(p_3-p_4)}{p_3+p_4}$$

Applying the Möbius transformation

Now I want to use this question of mine and this answer I received. One beautiful way of representing circles in the complex plane is using hermitian matrices of the form


or multiples thereof. That approach is completely different from the Apollonian circles you describe in your question, but for this application it's very useful since you can get one such multiple directly from the parameters of your plane:


To this you can apply your Möbius transformation


using conjugation, as detailed in the answer I mentioned:


Here $M^*$ denotes the conjugate transpose of $M$.

Back to the sphere

Now we want to project that circle back to the sphere. We could normalize the matrix by dividing it by its lower right entry. Or we could take the matrix as it is, labeling its entries as follows:


Then you have

$$x'=\frac{u}{t} \qquad y'=\frac{v}{t} \qquad x'^2+y'^2-r'^2=\frac{s}{t}$$

Plug these back into equation $(1)$:

$$p'\sim\left(4\frac ut, 4\frac vt, \frac st - 4, -\frac st - 4\right)^T$$

Multiply that by $t$ to avoid the divisions:


If you really want to, you can also normalize this in such a way that $p_1'^2+p_2'^2+p_3'^2=1$, i.e. the normal vector lies on the sphere. But I'd prefer not to do any divisions along the whole way.

Combining everything

If you wanted to, you could combine everything to a single big operation. The step from $p$ to $C$ is linear in $p$. The step from $C$ to $C'$ is quadratic in $M$, but still linear in $C$. The step from $C'$ to $p'$ is linear in $C'$ again. So the whole transformation from $p$ to $p'$ can be written as a $\mathbb R^{4\times 4}$ matrix, with coefficients which are quadratic in the real and imaginary parts of $a$ through $d$, the entries of your Möbius transformation:

$${\tiny\begin{pmatrix} 8b_rc_r+8b_ic_i+8a_rd_r+8a_id_i& 8b_ic_r-8b_rc_i-8a_id_r+8a_rd_i& 16a_rc_r+16a_ic_i-4b_rd_r-4b_id_i& -16a_rc_r-16a_ic_i-4b_rd_r-4b_id_i\\ 8b_ic_r-8b_rc_i+8a_id_r-8a_rd_i& -8b_rc_r-8b_ic_i+8a_rd_r+8a_id_i& 16a_ic_r-16a_rc_i-4b_id_r+4b_rd_i& -16a_ic_r+16a_rc_i-4b_id_r+4b_rd_i\\ 4a_rb_r+4a_ib_i-16c_rd_r-16c_id_i& -4a_ib_r+4a_rb_i+16c_id_r-16c_rd_i& 4\lvert a\rvert^2-\lvert b\rvert^2-16\lvert c\rvert^2+4\lvert d\rvert^2& -4\lvert a\rvert^2-\lvert b\rvert^2+16\lvert c\rvert^2+4\lvert d\rvert^2\\ -4a_rb_r-4a_ib_i-16c_rd_r-16c_id_i& 4a_ib_r-4a_rb_i+16c_id_r-16c_rd_i& -4\lvert a\rvert^2+\lvert b\rvert^2-16\lvert c\rvert^2+4\lvert d\rvert^2& 4\lvert a\rvert^2+\lvert b\rvert^2+16\lvert c\rvert^2+4\lvert d\rvert^2 \end{pmatrix}}$$

But perhaps this is more readable here if you don't write it as a matrix:

\begin{align*} p_1' &= (8b_rc_r+8b_ic_i+8a_rd_r+8a_id_i)p_1 \\ &+ (8b_ic_r-8b_rc_i-8a_id_r+8a_rd_i)p_2 \\ &+ (16a_rc_r+16a_ic_i-4b_rd_r-4b_id_i)p_3 \\ &+ (-16a_rc_r-16a_ic_i-4b_rd_r-4b_id_i)p_4 \\ p_2' &= (8b_ic_r-8b_rc_i+8a_id_r-8a_rd_i)p_1 \\ &+ (-8b_rc_r-8b_ic_i+8a_rd_r+8a_id_i)p_2 \\ &+ (16a_ic_r-16a_rc_i-4b_id_r+4b_rd_i)p_3 \\ &+ (-16a_ic_r+16a_rc_i-4b_id_r+4b_rd_i)p_4 \\ p_3' &= (4a_rb_r+4a_ib_i-16c_rd_r-16c_id_i)p_1 \\ &+ (-4a_ib_r+4a_rb_i+16c_id_r-16c_rd_i)p_2 \\ &+ (4\lvert a\rvert^2-\lvert b\rvert^2-16\lvert c\rvert^2+4\lvert d\rvert^2)p_3 \\ &+ (-4\lvert a\rvert^2-\lvert b\rvert^2+16\lvert c\rvert^2+4\lvert d\rvert^2)p_4 \\ p_4' &= (-4a_rb_r-4a_ib_i-16c_rd_r-16c_id_i)p_1 \\ &+ (4a_ib_r-4a_rb_i+16c_id_r-16c_rd_i)p_2 \\ &+ (-4\lvert a\rvert^2+\lvert b\rvert^2-16\lvert c\rvert^2+4\lvert d\rvert^2)p_3 \\ &+ (4\lvert a\rvert^2+\lvert b\rvert^2+16\lvert c\rvert^2+4\lvert d\rvert^2)p_4 \end{align*}

where $a=a_r+ia_i, b=b_r+ib_i, c=c_r+ic_i, d=d_r+id_i$ are the entries of your Möbius transformation.

  • $\begingroup$ Thanks for your answer. Unfortunately you lose me at step 1. Why are we working with $\mathbb{R}P^3$ at all? What is the purpose rest of the computation? $\endgroup$ – user40167 Dec 8 '14 at 16:11
  • $\begingroup$ The purpose of the whole computation is to find the image of a point on the sphere under the Möbius transformation. So we have stereographic projection from sphere to plane, Möbius transformation there, then stereographic projection back to the sphere. To model these stereographic projections, $\mathbb{RP}^3$ is my environment of choice. Steps 5 through 7 should correspond to your $\pi$, but I haven't verified that. $\endgroup$ – MvG Dec 8 '14 at 16:48
  • $\begingroup$ @krey: I completely rewrote my answer. I guess it should be much more useful now. Please let me know whether it is understandable the way I wrote it. $\endgroup$ – MvG Dec 8 '14 at 21:08

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