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Consider the ideal quadrilateral $Q_x$ comprising geodesics $S_1$ between $0$ and $\infty$, $S_2$ between $x$ and $\infty$, $T_1$ between $1$ and $x$, and $T_2$ between $1$ and $0$.

We have hyperbolic isometries $A$ and $B$ that send $S_1$ to $T_1$ and $S_2$ to $T_2$ respectively.

Call the domain bounded by the quadrilateral $D$. Then prove that both $A$ and $B$ move $D$ disjoint from itself, that is $A.D\cap D = \emptyset$ and $B.D\cap D=\emptyset$.

I have managed to find out the isometries $A$ and $B$. But I don't know how to proceed with showing that they move $D$ disjoint from itself.

Any help is kindly appreciated. Thank you.

Edit:

$D$ is the domain bounded by the ideal quadrilateral $Q_x$. The sides are not included in $D$. The isometries $A$ and $B$ have to be hyperbolic and such that the commutator $\left[A,B\right]$ is a parabolic isometry fixing $\infty$.

In my attempt, I have found possible isometries $A$ and $B$.

$A=\frac{1}{\sqrt{x-1}}\left(\begin{matrix} x & 1 \\ 1 & 1 \end{matrix}\right)$

$B=\left(\begin{matrix} 0 & 1 \\ -1 & x+1 \end{matrix}\right)$

$\left[A,B\right]= A^{-1}B^{-1}AB=\left(\begin{matrix} -1 & 2x+2 \\ 0 & -1 \end{matrix}\right)$, which is parabolic and fixes $\infty$.

I cannot proceed further.

WhySee
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  • Your question needs some improvement. Is $Q_x$ the same as $D$? Are the sides included in $D$? The isometries $A$ and $B$ are not uniquely specified by the question, and no quantifiers on $A$ and $B$ are specified, leading one to ask: Do you mean for all choices of $A$ and $B$, or only for some choices of $A$ and $B$? – Lee Mosher Mar 30 '18 at 14:46
  • @LeeMosher I get you. I shall edit my question with the necessary clarifications. Please do check back. Thank you. – WhySee Mar 30 '18 at 18:34

1 Answers1

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Let's start with the isometry $A$, which takes the line $S_1$ to the line $T_1$.

Let's examine how $A$ behaves with respect to various different types of orientations.

I observe that your $A$ preserves orientation on the hyperbolic plane because the determinant is positive. I also observe that $A$ takes $0 \to 1$ and $\infty \to x$, and therefore $A$ takes the "upward" orientation on $S_1$ to the "rightward" orientation on $T_1$. Putting these together, and using the right-hand-rule, it follows that $A$ takes the rightward transverse orientation on $S_1$ to the downward transverse orientation on $T_1$, and therefore $A$ takes the rightward half plane of $S_1$ to the downward half plane of $T_1$. Since $D$ is contained in the rightward half plane of $S_1$ it follows that $A \cdot D$ is contained in the downward half plane of $T_1$. But also, $D$ is contained in the upward half plane of $T_1$ and thereore $A \cdot D \cap D = \emptyset$.

A similar orientation analysis of $B$ shows that $B$ takes the leftward half plane of $S_2$ to the downward half plane of $T_2$, and by the same argument we get $B \cdot D \cap D = \emptyset$.

Lee Mosher
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  • Brilliant! Very enlightening indeed! Thanks a lot! But I was wondering if we could take a point in the intersection and then arrive at a contradiction using the explicit form of the isometry? – WhySee Mar 30 '18 at 19:26
  • There is a slight typo in the last line. It should be $B.D \cap D = \emptyset$. By mistake you have written $B$ there. :) – WhySee Mar 30 '18 at 19:47
  • The form of the isometry is not particularly relevant. For example, you could use two parabolic isometries $F$ and $G$, where $F$ fixes the point $0$ and takes $S_1$ to $T_2$, whereas $G$ fixes the point $x$ and takes $S_2$ to $T_1$. You would get similar conclusions using the same orientation argument: $F . D \cap D = \emptyset$, $G . D \cap D = \emptyset$. – Lee Mosher Mar 30 '18 at 21:11
  • Yeah. It makes sense. Thanks a lot. :) – WhySee Mar 30 '18 at 21:23