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Where H is the hyperbolic plane.

I am stuck because we know mob(H) acts transitively on the set of hyperbolic lines and we we know two points make up a hyperbolic line so Why doesn't Mob(H) act transitively on the set P of pairs of distinct points of H?enter image description here

MathIsHard
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2 Answers2

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To say that $\operatorname{Möb}(\mathbb{H})$ acts transitively on lines means that if $\ell$ and $\ell'$ are lines, there is some $f\in\operatorname{Möb}(\mathbb{H})$ such that $f(\ell)=\ell'$. To say that $\operatorname{Möb}(\mathbb{H})$ acts transitively on pairs of distinct points means that if $a,b,a',$ and $b'$ are points with $a\neq b$ and $a'\neq b'$, there is some $f\in\operatorname{Möb}(\mathbb{H})$ such that $f(a)=a'$ and $f(b)=b'$.

Let's try to implement the argument you have in mind: given that $\operatorname{Möb}(\mathbb{H})$ acts transitively on lines, we want to prove it acts transitively on pairs of distinct points. So let $a,b,a',$ and $b'$ be points such that $a\neq b$ and $a'\neq b'$. Let $\ell$ be the line through $a$ and $b$ and let $\ell'$ be the line through $a'$ and $b'$. We then know there is some $f\in \operatorname{Möb}(\mathbb{H})$ such that $f(\ell)=\ell'$. But we can't conclude from this that $f(a)=a'$ and $f(b)=b'$! All we know is that $f(a)$ and $f(b)$ are two points on the line $\ell'$, but they don't have to be $a'$ and $b'$.

To put it another way, two points determine a line, but it's not true that two points "make up" a line: a line has more than two points! So just because $f$ maps $\ell$ to $\ell'$ does not mean $f$ maps two chosen points on $\ell$ to two chosen points on $\ell'$.

Eric Wofsey
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  • Thank you so much. That makes sense. – MathIsHard Mar 15 '17 at 00:52
  • What about the reverse argument? Given that we have a function that maps a,b on l to a',b' on l', why can't we assume that the function maps points on l bijectively to l'? Since every hyperbolic line is part of a circle in $\mathbb{\bar{C}}$ and we know we can find a function that maps three specific points to three other points. Assuming the third point may or may not be in the upper half plane. But regardless, we are taking the two points to the other two points by the function. – MathIsHard Mar 15 '17 at 00:58
  • You can achieve $\operatorname{M\ddot{o}b}(\mathbb{H})$ by using \operatorname{M\ddot{o}b}(\mathbb{H}). – Michael Albanese Mar 15 '17 at 00:59
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Besides the answer of @EricWofsey, you can see this clearly using metric considerations.

$H$ is a metric space and Möbius transformations of $H$ preserve the metric. To word this differently, every Möbius transformation of $H$ is an isometry of $H$. To put this in symbols, if $d(\cdot,\cdot)$ denotes the metric on $H$ (where $d(a,b)$ is the length of the shorted path with endpoints $a,b$), if $f$ is a Möbius transformation, and if $a,b$ are distinct points in $\mathbb{H}$, then $d(f(a),f(b)) = d(a,b)$.

So if $c,d \in H$, and if $d(a,b) \ne d(c,d)$, it follows that no Möbius transformation takes the pair $(a,b)$ to the pair $(c,d)$.

Here's an analogy. You might make exactly the same argument in the Euclidean plane: "We know rigid motions of the Euclidean plane act transitively on the set of lines in the Euclidean plane, and we know two points determine a line. So why don't rigid motions act transitively on the set of pairs of points of the Euclidean plane?" The metric answer is the same: because rigid motions preserve distance.

Lee Mosher
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