5

enter image description hereLet $\Gamma$ be a group generated by two matrices as follows:

$\Gamma:= \bigg\langle \begin{bmatrix}1&0\\3&1\end{bmatrix},\begin{bmatrix}13&12\\12&13\end{bmatrix} \bigg\rangle$

For any $\begin{bmatrix}a&b\\c&d\end{bmatrix}\in \Gamma$, and for any $x\in \mathbb{R}$, we define an action: $\begin{bmatrix}a&b\\c&d\end{bmatrix}(x):=\dfrac{ax+b}{cx+d}$.

Let $\Gamma(1)$ be the set of images of the action of $\Gamma$ on 1. Now we define a function $f$ on the Cartesian product $\Gamma(1)\times \Gamma(1)$ as follows:$f(x,y)= 0$ if $(x \cdot y)>0$, and $f(x,y)= \sqrt{-x.y}$ otherwise. I knew that: $f(\Gamma(1)\times \Gamma(1)) \subset [0,1]$.

My question: Is $f(\Gamma(1)\times \Gamma(1))$ dense in $[0,1]$?

I hope someone can help me or give me any hints for this question. Thank you so much for your help!

Student
  • 75
  • If x and y are both negative, then your function f maps into the complex? – RSpeciel Jul 16 '19 at 15:20
  • It maps to 0 as in my definition. But we can also define a function which maps to the complex. – Student Jul 16 '19 at 15:28
  • 1
    sorry I misunderstood your notation, fixing it now... – RSpeciel Jul 16 '19 at 15:29
  • 2
    Your choice of generators is quite peculiar. What is the motivation for the question? Did you check discreteness of the subgroup or if it projects to a lattice in $PSL(2,R)$? And why the geometric measure theory tag? – Moishe Kohan Jul 16 '19 at 17:52
  • 1
    Dear Kohan, thannk you for your comment. The motivation is to give an example of surface with boundary which can be cover by the set of bi-infinite geodesics. This is the link: https://math.stackexchange.com/questions/3275012/on-a-pair-of-pants-complete-geodesics-do-not-cover-the-whole-surface. The group I chose is the Fuchsian group associated to a pair of pants with 2 cusps and one geodesic boundary of length $2\log5$. – Student Jul 16 '19 at 20:52
  • I think it is related to cantor set arithmetic. This is a reference: https://faculty.math.illinois.edu/~reznick/jsabrjtt.pdf –  Jul 17 '19 at 11:54
  • Actually, we can define function $f$ in many ways. I think we should begin with simpler functions for instance: $f(x,y)=x.y$ or $f(x,y)=\frac{x+y}{2}$. – Student Jul 17 '19 at 13:52

0 Answers0