Transitivity of mobius groups

Question

I am trying to understand the transitivity of m\"obius groups. I know that mob$(\mathbb{\bar{C}})$ acts transitively on the set of triples of distinct points. Does that mean that it also acts transitively on pairs of distinct points? I know that mob$(\mathbb{H})$ is not transitive on pairs of points of $\mathbb{H}$. I don't understand why this is for mob($\mathbb{H}$)? Also, I can't figure out if mob$(\mathbb{C})$ acts transitively on pairs of points. I understand that mob$(\mathbb{C})$ fixes infinity but that's it. Thanks.

1) Does mob$(\mathbb{\bar{C}})$ act transitively on pairs of distinct points?

2) Why doesn't mob$(\mathbb{H})$ act transitively on pairs of distinct points?

3) Does mob$(\mathbb{C})$ act transitively on pairs of distinct points?

Answer 1

Let's try out this question with $\text{mob}(\mathbb{C})$: does it act transitively on pairs?

I'm going to simplify the formulas in your comment for elements of $\text{mob}(\mathbb{C})$: they have the form $az+b$ or $a\bar z + b$ where $a \ne 0$, $a,b \in \mathbb{C}$ (the denominator can always be divided into the other two coefficients).

So, given two pairs of points $w_1 \ne w_2$ and $z_1 \ne z_2 \in \mathbb{C}$, we are asking whether there exists $a,b \in \mathbb{C}$ such that $a \ne 0$ and such that \begin{align*} a w_1 + b &= z_1 \\ a w_2 + b &= z_2 \end{align*} Notice that I'm using only one of the two types of elements of $\text{mob}(\mathbb{C})$, namely the one of the form $aw+b$, and I'm ignoring the one of the form $a\bar w + b$; as it turns out, the question can be completely resolved by the first form.

Thinking of this as two linear equations in the unknowns $a,b$, it's not hard to solve. Subtract the two equations to get $$a(w_1-w_2)=z_1-z_2 $$ Since $w_1 \ne w_2$ it follows that $w_1-w_2 \ne 0$ and so we can divide: $$a = \frac{z_1-z_2}{w_1-w_2} $$ Notice that this value of $a$ is not equal to zero, as is required.

Now plug this formula for $a$ into $$b = z_1 - a w_1 $$ we get $$b = z_1 - \frac{z_1-z_2}{w_1-w_2} w_1 $$ And since we have successfully solved these equations for the unknowns $a$ and $b$ we conclude that yes, $\text{mob}(\mathbb{C})$ does act transitively on pairs.

Each of the other questions can be resolved in the same way, assuming you know the general form of elements in each case. The answers may differ: some do act transitively on pairs, some don't. But all can be resolved by algebra like this one.

Answer 2

(1) If $\mathrm{Mob}(\widehat{\mathbb{C}})$ acts transitively on ordered triples of distinct points, then it follows that it acts transitively on ordered pairs of distinct points. Indeed, let $(a,b)$ and $(x,y)$ be any two pairs of distinct points (so $a\ne b$ and $x\ne y$), then simply pick some $c\ne a,b$ and some $z\ne x,y$ and by hypothesis there is a $g$ such that $g(a,b,c)=(x,y,z)$, implying $g(a,b)=(x,y)$.

Acting transitively on ordered $k$-tuples of points is called $k$-transitivity. This argument generalizes to show that $k$-transitive implies $r$-transitive for all $r<k$, so each level of transitivity is stronger than the last. In fact, $\mathrm{Mob}(\widehat{\mathbb{C}})$ acts sharply transitively on $3$-tuples of distinct points, meaning it acts $3$-transitively but not $4$-transitively. This can be deduced from the fact that it acts regularly on the space of $3$-tuples of points from Riemann sphere: the unique Mobius transformation which sends $(a,b,c)$ to $(0,1,\infty)$ is the cross-ratio $(z-a)(b-c)~/~(z-c)(b-a)$.

(First, you need $z-a$ in the numerator to send $a\mapsto 0$, then you need $z-c$ in the denominator to send $c\mapsto\infty$, and finally need to normalize by $(b-a)/(b-c)$ so that $b\mapsto1$.) I encourage you to try your hand at explaining why the cross ratio is the unique transformation that does this, why this implies the action is regular, and why this regular action on triples implies the original action on points is not $4$-transitive. Heuristically, specifying a $3$-tuple is three (complex variable) degrees of freedom, and specifying an element of $\mathrm{PGL}(2,\mathbb{C})$ involves three complex numbers, but this is not a proof by itself.

(3) Yes, $\mathrm{Mob}(\mathbb{C})$ acts transitively on pairs of distinct points. This can be explained geometrically: given any two pairs $(a,b)$ and $(x,y)$, we can first apply multiplication by a real scalar $r>0$ to ensure the first pair of points $(ra,rb)$ is the same distance apart as the second pair $(x,y)$; then, we can multiply by $e^{i\theta}$ for the angle $\theta$ so as to make the line segment from $re^{i\theta}a$ to $re^{i\theta}b$ is the same direction as from $x$ to $y$; and thirdly, we can add a complex number $u$ that shifts the first pair to the second pair, so $re^{i\theta}a+u=x$ and $re^{i\theta}b+u=y$.

(2) Now suppose you can't multiply by anything but positive real numbers (otherwise it would rotate the upper half-plane to a different half-plane) and can't add by anything except real numbers. Trying to get from $(a,b)$ to $(x,y)$, you could multiply by $r$ so that $ra$ and $x$ are at the same height, and then add $s$ so that $ra+s=x$, but then there's nothing you can do to ensure $rb+s=y$.

(Only now reading the comments do I see you include conjugation as a Mobius transformation; you will have to add more to the arguments above in that case.)