I have two questions regarding the inversion across the unit circle in the hyperbolic plane.

Recall that the hyperbolic plane is a metric space consisting of the open half-plane

$$\mathbb{H}^2 = {z \in \mathbb{C}; \Im(z)>0}$$

and that for a curve $\gamma$ parametrized by a differentiable vector-valued function $t \mapsto (x(t),y(t)), a \leq t \leq b$ the hyperbolic length is given by,

$$l_{hyp}(\gamma) = \int_{a}^{b} \frac{\sqrt{x'(t)^2 + y'(t)^2}}{y(t)} dt$$

Lastly, the hyperbolic distance between two points $P$ and $Q$ in the hyperbolic plane, $d_{hyp}(P,Q)$ is the infimum of the hyperbolic lengths of all piecewise differentiable curves $\gamma$ going from $P$ to $Q$.

Question 1 [SOLVED]

What is the relationship between the standard inversion in rectangular and polar coordinates?

The standard inversion is an isometry of $(\mathbb{H}^2,d_{hyp})$ that is defined by

$$\varphi(x,y) = (\frac{x}{x^2 + y^2},\frac{y}{x^2 + y^2})$$

I am having trouble understanding the motivation for this isometry and why it is introduced so I thought that it may have a more clean representation in polar form. The following is my attempt to define the standard inversion in polar form:

Let $x=r \cos(\theta)$ and $y= r \sin(\theta)$, then we have that $$\varphi(r, \theta) = \left(\frac{r \cos(\theta)}{(r \cos(\theta))^2 + (r \sin(\theta))^2}, \frac{r \sin(\theta)}{(r \cos(\theta))^2 + (r \sin(\theta))^2}\right)$$ By $\sin^2(\theta) + \cos^2(\theta) = 1$ we then have $$\varphi(r, \theta) = \left(\frac{r \cos(\theta)}{r^2}, \frac{r \sin(\theta)}{r^2}\right)$$ $$\varphi(r, \theta) = \left(\frac{\cos(\theta)}{r}, \frac{\sin(\theta)}{r}\right)$$

Yet, when I look up the standard inversion in polar form, it is defined as

$$\varphi(r,\theta) = \left(\frac{1}{r},\theta\right)$$

Where have I gone wrong?

Question 2 In the proof that the standard inversion is an isometry of $(\mathbb{H}^2,d_{hyp})$ I have two questions. Here is the proof given in the book I am reading through, "Low-Dimensional Geometry: From Euclidean Surfaces to Hyperbolic Knots" by Francis Bonahon:

The image $\varphi(\gamma)$ under the inversion $\varphi$ is parametrized by $t \mapsto (x_{1}(t),y_{1}(t)), a \leq t \leq b$, with $x_{1}(t)=\frac{x(t)}{x(t)^2 + y(t)^2}$ and $y_{1}(t)= \frac{y(t)}{x(t)^2 + y(t)^2}$. Then, $$x_{1}'(t)=\frac{(y(t)^2 - x(t)^2)x'(t) - 2x(t)y(t)y'(t)}{(x(t)^2 + y(t)^2)^2}$$ $$y_{1}'(t)=\frac{(x(t)^2 - y(t)^2)y'(t) - 2x(t)y(t)x'(t)}{(x(t)^2 + y(t)^2)^2}$$ so that after simplifications, $$x_{1}'(t)^2 + y_{1}'(t)^2 = \frac{x'(t)^2 + y'(t)^2}{(x(t)^2 + y(t)^2)^2}$$ It follows that, $$l_{hyp}(\varphi(\gamma)) = \int_{a}^{b} \frac{\sqrt{x_{1}'(t)^2 + y_{1}'(t)^2}}{y_{1}(t)} dt$$ $$l_{hyp}(\varphi(\gamma)) = \int_{a}^{b} \frac{\sqrt{x'(t)^2 + y'(t)^2}}{y(t)} dt$$ $$l_{hyp}(\varphi(\gamma)) = l_{hyp}(\gamma)$$ Therefore, the standard inversion is an isometry of $(\mathbb{H}^2,d_{hyp})$.

My problem with the proof is that I do not understand how $x_{1}'(t)$ and $y_{1}'(t)$ are arrived at? I understand the motivation, the completion of the proof, and how showing that if the hyperbolic length of the curve under the inversion is the same as the hyperbolic length of the curve, then the inversion is an isometry. Additionally, I could not complete the "so that after simplifications" part and I think that is just an algebra problem where I didn't see something clever.

Conclusion In short, I am looking for an answer that points out my error in converting the standard inversion in rectangular coordinates to polar coordinates and provides the correct way for Question 1. For Question 2, I am looking for an explanation of how $x_{1}'(t)$ and $y_{1}'(t)$ are arrived at and how the step in the proof after "so that after simplifications" is accomplished (full algebra not necessary).

Thank you for any responses.


1 Answer 1


for question 1, $$r=\sqrt{x^2+y^2}\to\sqrt{\frac{x^2}{(x^2+y^2)^2}+\frac{y^2}{(x^2+y^2)^2}}=\sqrt{\frac{1}{x^2+y^2}}=\frac{1}{r}$$ and theta is unchanged (what your wrote mixes the coordinate systems, you wrote $x'(r,\theta),y'(r,\theta)$ instead of $r'(r,\theta), \theta'(r,\theta)$ where primes denote the inversion)

for question 2, we have (' denoting differentiation wrt $t$ and with $u=x/(x^2+y^2), v=y/(x^2+y^2)$) $$u'=\frac{x}{x^2+y^2}'=\frac{(x^2+y^2)x'-x(2xx'+2yy')}{(x^2+y^2)^2}=\frac{(y^2-x^2)x'+2yy'}{(x^2+y^2)^2}$$ similarly $$v'=\frac{y}{x^2+y^2}'=\frac{(x^2-y^2)y'+2xx'}{(x^2+y^2)^2}$$ so that the metric changes as you described above (i guess the answer to your question is they used the quotient rule and simplified).

  • $\begingroup$ Thank you for the clarification, any ideas for the second question? $\endgroup$ Commented Dec 28, 2011 at 23:21

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