Why doesn't Mob(H) act transitively on the set P of pairs of distinct points of H

Question

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?

Answer 1

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 $\ell'$.

Answer 2

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.