Hyperbolic Rotation

Question

I'm studying the transformations of the upper half plane model of Hyperbolic Geometry. I got stuck studying the elliptic isometries (which are the hyperbolic rotations). Can anyone help me to find the explicit formula of an elliptic isometrie?

Answer 1

The upper half plane model is conformal: hyperbolic angles correspond to Euclidean angles (between tangents at the point of intersection). These angles are preserved (or reversed) under isometries, so you are looking at Möbius and anti-Möbius transformations. (So think about the plane in terms of complex numbers.) But elliptic transformations won't reverse, so you can restrict yourself to Möbius transformations. Furthermore, the set of ideal points is fixed under isometries, so you are dealing with Möbius transformations which fix the real axis, i.e. real Möbius transformations.

Of these real Möbius transformations, some are hyperbolic (translations), and some are elliptic (rotations). In between these two there are the parabolic transformations or limit rotations. One way to distinguish them is by their fixed points: a translation will fix the ideal points at the ends of the line of translation. (Remember that translation isn't parallel transport in hyperbolic geometry, so it fixes only a single line, not a family of parallel lines). So the corresponding Möbius transformation has two real fixed points. A rotation has a single hyperbolic fixed point, i.e. a point inside the upper half plane. But as the whole hyperbolic plane has a mirror image in the lower half plane, algebraically speaking, you'd get a pair of complex conjugates for the fixed points of the Möbius transformation. The limit case between these two is where the fixed points coincide in a single ideal point on the real axis.

A general real Möbius transformation has the form

$$z\mapsto \frac{az+b}{cz+d}$$

so for a fixed point you'd solve

\begin{align*} \frac{az+b}{cz+d}&=z\\ az+b&=(cz+d)z\\ az+b&=cz^2+dz\\ cz^2+(d-a)z-b&=0\\ z_{1,2}&=\frac{(a-d)\pm\sqrt{(d-a)^2+4bc}}{2c} \end{align*}

Assuming $a,b,c,d\in\mathbb R$ you get a rotation iff $$(d-a)^2+4bc<0$$ unless I made some mistake. If you want to have the identity as a special case of rotation (by a multiple of $2\pi$) you should handle that case explicitly.

Ideally you'd not use the fraction notation I used above, but homogeneous coordinates. That way you can also express the point at infinity, which is an ideal point for the upper half plane model as well. Using that the situation of $c=0$ make more sense. For $a=d$ you'd get parabolic transformations with the point at infinity fixed, while for $a\neq d$ and $c=0$ you get hyperbolic transformations with one of the fixed points at infinity, i.e. transformations which preserve the set of Euclidean lines in the model.

Edit: In a comment you indicate that you were looking for something more explicit, and mention angles. So perhaps you were looking for the transformation with angle $\varphi$ and center $x+iy$. That could be written as

$$z\mapsto f(z):=\frac {(x\sin\frac\varphi2-y\cos\frac\varphi2)z-(x^2+y^2)\sin\frac\varphi2} {(\sin\frac\varphi2)z-(x\sin\frac\varphi2+y\cos\frac\varphi2)}$$

which can probably found somewhere in the literature, but which I found from the matrix product

$$-\frac12\cdot\begin{pmatrix}x-iy&ix-y\\1&i\end{pmatrix}\cdot \begin{pmatrix}\exp(i\frac\varphi2)&0\\0&\exp(-i\frac\varphi2)\end{pmatrix} \cdot\begin{pmatrix}i&-ix+y\\-1&x-iy\end{pmatrix}$$

The first matrix, if interpreted as a Möbius transformation (i.e. either in the projective complex line $\mathbb{CP}^1$ using homogeneous coordinates, or written as a fraction in the usual way), maps the origin to the center of rotation, and the point at infinity to its complex conjugate. So if you have a rotation around the origin, this matrix takes the result to your desired position. The last matrix is the adjoint of the first, which is geometrically equivalent to the inverse. So taken together they move the desired center to the origin, let the middle matrix operate on that, then take the result pack to the desired center. The middle matrix is essentially a rotation by $\varphi$, but with the angle split between numerator and denominator as that makes it easier to obtain real entries in the final product. The second column of the first matrix was chosen with $i$ in its second coordinate to make the determinant of the matrix real, which also helped obtain real end results. The scalar factor at the beginning is for cosmetic reasons only, to cancel some common factors.

Let's verify what I just wrote. First of all, note that it's a real Möbius transformation, so it's an orientation-preserving isometry. Next, check that $x+iy$ is indeed a fixed point:

$$f(x+iy)=\frac {(x\sin\frac\varphi2-y\cos\frac\varphi2)(x+iy)-(x^2+y^2)\sin\frac\varphi2} {(\sin\frac\varphi2)(x+iy)-(x\sin\frac\varphi2+y\cos\frac\varphi2)} \\=\frac {(x\sin\frac\varphi2-y\cos\frac\varphi2)-(x-iy)\sin\frac\varphi2} {(\sin\frac\varphi2)(x+iy)-(x\sin\frac\varphi2+y\cos\frac\varphi2)}(x+iy) \\=\frac {x\sin\frac\varphi2-y\cos\frac\varphi2-x\sin\frac\varphi2+iy\sin\frac\varphi2} {x\sin\frac\varphi2+iy\sin\frac\varphi2-x\sin\frac\varphi2-y\cos\frac\varphi2}(x+iy) =x+iy$$

Now for the claimed rotation. If you consider an infinitesimal change in input position $z$, then the derivative of $f$ gives the corresponding change in output position. So the change in rotation can be observed from the derivative at the center of rotation:

$$\frac{\mathrm d\,f}{\mathrm dz}(x+iy)=\left(\cos\tfrac\varphi2+i\sin\tfrac\varphi2\right)^2=\left(\exp\left(i\tfrac\varphi2\right)\right)^2=\exp\left(i\varphi\right)$$

I skipped the computation of the derivative which I left to my computer algebra software. A multiplication by $\exp(i\varphi)$ does describe a rotation by $\varphi$, so the immediate neighbourhood of the fixed point gets rotated as intended.

Comparing the formula for $f$ with the considerations at the beginning of this post, you can find

\begin{align*} a &= x\sin\tfrac\varphi2-y\cos\tfrac\varphi2 \\ b &= -(x^2+y^2)\sin\tfrac\varphi2 \\ c &= \sin\tfrac\varphi2 \\ d &= -(x\sin\tfrac\varphi2+y\cos\tfrac\varphi2) \\ (d-a)^2+4bc &= -4y^2\sin^2\tfrac\varphi2 \end{align*}

So the discriminiant is negative as expected, unless $\sin\tfrac\varphi2=0$ which means $\varphi\in2\pi\mathbb Z$ in which case the “rotation” is in fact the identity.