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I believe that showing that every element of $\text{Mob}^+(\mathbb{H})$ is the product of an even number of inversions is quite straightforward (barring a few lemmas here and there). I do not; however, know how I should go about showing that every element of $\text{Mob}^+(\mathbb{H})$ is the product of two inversions?

For your reference:

An element $f$ of $\text{Mob}^-(\mathbb{H})$ having the form: $f(z) = \frac{a \overline{z} + b}{c \overline{z} + d}$ (here where $a, b, c, d \in \mathbb{R}$ and $a d - b c = -1$), is said to be an inversion iff $a + d = 0$.

DrPhD
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  • Maybe you could be interested by this question of mine. – Jean Marie Oct 05 '21 at 21:46
  • Hint: First prove that every orientation preserving Euclidean isometry is a composition of two reflections. – Moishe Kohan Oct 05 '21 at 21:54
  • This is probably a silly question, but why is a map of this type called an inversion? In what way is it an inversion? – user387394 Oct 06 '21 at 13:47
  • @Greg Markowsky It is an old name coined around 200 years ago. This name is justified by the fact that the image of a figure close to the circle of inversion looks like the mirror image of the initial figure. – Jean Marie Oct 06 '21 at 14:05
  • I can understand that, but what I mean is, why does the condition a+d=0 mean that it has a circle of inversion? – user387394 Oct 06 '21 at 14:34
  • @MoisheKohan supposing that I have shown this, would it not have to also be the case that the product of two reflections is also a reflection (which is not true in general)? – DrPhD Oct 07 '21 at 07:09
  • No, product of two reflections is never a reflection. – Moishe Kohan Oct 07 '21 at 16:18
  • @GregMarkowsky If you work it out, you'll find that $a+d=0$ is true iff $f \in \mathrm{Mob}^-(\mathbb H)$ is its own inverse. – Magma Oct 08 '21 at 12:39

1 Answers1

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You should have as a lemma, or can easily prove, that an element of $\mathrm{Mob}^-(\mathbb H)$ with a fixed point in the upper half-plane is an inversion, and that an element of $\mathrm{Mob}^+(\mathbb H)$ with two fixed points in the upper half-plane is the identity.

Given any element $f \in \mathrm{Mob}^+(\mathbb H)$, can you find inversions $g, h \in \mathrm{Mob}^-(\mathbb H)$ such that $g \circ f$ has a fixed point and $h \circ g \circ f$ has two?

Magma
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  • I think that I would choose g to be the unique inversion mapping f(p) to p. Then I would choose h to be an inversion such that p lies on the line/circle of inversion. Would this work? – DrPhD Oct 08 '21 at 14:52
  • $h$ needs to satisfy more than just fixing $p$, and $g$ is not necessarily unique if $f(p) = p$, but otherwise yes. – Magma Oct 08 '21 at 21:52
  • Yes, of course. Although, I don't think that the uniqueness/non-uniqueness of g really matters. However, I am unsure as to how I should go about showing that the final composition of all three functions will fix a second point. Will I need to provide a specific construction of g and h w.r.t. f? – DrPhD Oct 09 '21 at 13:37
  • The construction itself is rather easy. To construct $g$, simply pick any point $p$ with $f(p) \neq p$ and let $g$ be the unique inversion mapping $f(p)$ to $p$. To construct $h$, pick any point $q$ with $g(f(q)) \neq q$ and let $h$ be the unique inversion mapping $g(f(q))$ to $q$. Conveniently, $h$ automatically also fixes $p$, but you need to prove that. – Magma Oct 09 '21 at 20:37
  • I understand the argument; however, I do not see how h automatically also fixes p. – DrPhD Oct 11 '21 at 11:58
  • I believe that perhaps the best approach would be to proceed by a proof by cases considering whether g, h are inversions in a line/circle. – DrPhD Oct 11 '21 at 17:08
  • Since hyperbolic Mobius transformations preserve hyperbolic distance, $p$ lies on the hyperbolic perpendicular bisector of $q$ and $g(f(q))$, which is the hyperbolic line of inversion of $h$. – Magma Oct 11 '21 at 22:01