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The group $\mathrm{PSL}_2(\mathbf{Q})$ of fractional linear transformations $x \mapsto (ax+b)/(cx+d)$ such that $a,b,c,d \in \mathbf{Q}$ and $ad-bc = 1$ acts on $\mathbf{Q} \cup \lbrace \infty\rbrace$. The weakest version of my question is the following:

Is there a short proof that no element of $\mathrm{PSL}_2(\mathbf{Q})$ acts transitively on $\mathbf{Q} \cup \lbrace \infty\rbrace$?

It seems obvious to me that no element of $\mathrm{PSL}_2(\mathbf{Q})$ can have finitely many orbits on $\mathbf{Q} \cup \lbrace\infty\rbrace$, but I have not been able to prove this.

Is this stronger claim true, and if so, does it have a short proof?

Some remarks and motivation: (1) The subgroup of $\mathrm{PSL}_2(\mathbf{Q})$ generated by the maps $x \mapsto x+1$ and $x \mapsto x/(x+1)$ acts transitively on $\mathbf{Q} \cup \lbrace \infty\rbrace$; these elements are used to generate the Calkin–Wilf tree of rational numbers.

(2) By conjugating by elements of $\mathrm{PGL}_2(\mathbf{Q})$ it suffices to look at maps of the form $g : x \mapsto 1/(b-x)$. If $|b| > 2$ then $g$ is conjugate in $\mathrm{PGL}_2(\mathbf{R})$ to a map of the form $x \mapsto \lambda x$. It follows that the iterates of $g$, starting at the initial value $x_0$ are of the form

$$ \frac{1}{b-\frac{1}{b-\ldots\frac{1}{b-x_0}}} $$

and converge to the continued fraction $1/(b-1/(b- \ldots ))$. The iterates of $g^{-1}$ behave similarly. So in this case each orbit of $\langle g \rangle \le \mathrm{PSL}_2(\mathbf{Q})$ has precisely two limit points in $\mathbf{R}$, namely the two roots of $x^2-bx+1$. Clearly this implies that $\langle g \rangle$ has infinitely many orbits on $\mathbf{Q} \cup \lbrace \infty\rbrace$. However if $|b| < 2$ then I have not been able to make this approach work.

(3) If $\mathbf{Q}$ is replaced with the finite field $\mathbf{F}_p$ then is not too hard to show that no element of $\mathrm{PSL}_2(\mathbf{F}_p)$ acts transitively on $\mathbf{F}_p \cup \lbrace \infty \rbrace$: see Lemma 8.2 in this paper with John Britnell. This leads to a proof of the result for $\mathbf{Q}$ by reduction mod $p$.

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If your matrix is scalar it acts trivially on the projective line, so assume it's not scalar; hence it's conjugate in $\text{GL}_2(\mathbf{Q})$ to a companion matrix $A=\pmatrix{0 & -1 \cr 1 & z}$.

If $z$ is integer with $|z|\ge 3$ then the dynamics of $A$ on the real projective line $\mathbf{P}^1(\mathbf{R})$ has a repelling point $a_-$ and a attractive point $a_+$; in particular, if $K$ is an infinite compact subset of $\mathbf{P}^1(\mathbf{R})\smallsetminus (a_-,a_+)$, then its intersection with each $A$-orbit $\{A^nx:n\in\mathbf{Z}\}$ is finite and hence $K$ intersects infinitely many $A$-orbits. Since $K$ can be chosen to be contained in $\mathbf{P}^1(\mathbf{Q})$, it follows that $A$ admits infinitely many orbits on $\mathbf{P}^1(\mathbf{Q})$. If $|z|=2$ a similar argument works (with $a_+=a_-$). If $|z|<2$ then $A$ has finite order.

Assume now $z$ is not an integer. Let $p$ be a prime divisor of the denominator of $z$. Let $|z|_p$ be the norm of $z$ in $\mathbf{Q}_p$. An easy calculation shows that the function $x\mapsto -1/(x+z)$ maps $\mathbf{Z}_p$ into itself and is $(1/|z|_p^2)$-contracting on $\mathbf{Z}_p$, so it admits a unique fixed point which I denote by $a_+$. Thus $A$ admits a north-south dynamics on $\mathbf{P}^1({\mathbf{Q}_p})$, with repelling and attracting points $a_-$ and $a_+$. So the same argument as in the real case, using density of $\mathbf{P}^1({\mathbf{Q}})$ in $\mathbf{P}^1({\mathbf{Q}_p})$ shows that $A$ has infinitely many orbits. I assumed (as you do) that $A$ has determinant 1 but the argument carries over with minor modification for arbitrary $A$.

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  • $\begingroup$ @Yves Thank you for your short and elegant proof. (I checked the easy calculation you left as an exercise.) It's interesting that all the proofs so far depend on some form of reduction mod $p$. I wonder if there is a more elementary approach. $\endgroup$
    – Mark Wildon
    Jul 31, 2012 at 15:51
  • $\begingroup$ Well, it's a bit intrinsically part of the picture (although other approches make this less explicit). For instance, the argument shows that given any element $w$ in $P^1(\mathbf{Q})$ (not fixed by $A$), there is a neighbourhood $V$ of $w$ in $P^1(\mathbf{K})$ in which elements are pairwise in distinct $\langle A\rangle$-orbits. Here $\mathbf{K}$ is the completion of $\mathbf{Q}$ ($\mathbf{R}$ or $\mathbf{Q}_p$) on which $A$ has a north-south dynamics. $\endgroup$
    – YCor
    Jul 31, 2012 at 20:10
  • $\begingroup$ @YvesCornulier: You say in your answer that $K$ intersects finitely many $A$-orbits. Surely you meant to say that $K$ intersects each $A$-orbit finitely many times. $\endgroup$
    – KConrad
    May 24, 2014 at 18:54
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If the eigenvalues of $g$ are irrational, then they generate a quadratic extension field $K$ of $\mathbb Q$. Then the desired statement is that in the group $K^\star/\mathbb Q^\star$ the subgroup generated by a single element does not have finite index. But $K^\star/\mathbb Q^\star$ is not finitely generated, since there are infinitely many primes that split in $K$.

Edit: In more detail, you then have a basis for $K$ such that the map $K\to K$ given by multiplication by an eigenvalue corresponds to the matrix $g$. The cases with rational eigenvalues are easy.

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  • $\begingroup$ @Tom: I can't see why the result follows from your observation that $K^\times/\mathbb{Q}^\times$ has no cyclic subgroups of finite index. I agree there is a basis for $K^2$ such that the map on $K \cup \lbrace\infty\rbrace$ becomes $x \mapsto \lambda^2 x$, where $\lambda$ is an eigenvalue of the original matrix. The copy of $\mathbb{Q}^2$ in $K^2$ is conjugated to $\lbrace (−x+\lambda y,x−\lambda^{-1}y)/(\lambda−\lambda^{-1}):x,y \in \mathbb{Q} \rbrace$. So we are no longer interested in orbits on $\mathbb{Q}^2$, but orbits on $\lbrace(-x+\lambda y)/(x-\lambda^{-1}y):x,y\in\mathbb{Q}\rbrace$. $\endgroup$
    – Mark Wildon
    Jul 31, 2012 at 15:01
  • $\begingroup$ I am thinking of $K$ as a rational vector space of dimension two. The points in the projective line associated with this vector space are equivalence classes of non-zero vectors. These vectors are the elements of $K^\times$. The equivalence classes are elements of $K^\times/\mathbb Q^\times$. The action of $g$ is multiplication by a certain element of the group $K^\times/\mathbb Q^\times$, namely the element represented by $\lambda$. $\endgroup$
    – Tom Goodwillie
    Jul 31, 2012 at 16:55
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The stronger claim is indeed true. Assume $g \in PSL_2(\mathbb{Q})$ has infinite order, equivalently $|trace(g)| \ge 2$. If necessary replace $g$ by its square to make the trace positive. The action of $g$ on the circle $\mathbb{R} \cup {\infty}$ comes in one of two flavors, depending on whether $trace(g) = 2$ or $>2$. When $trace(g)=2$ then $g$ has a single fixed point in $\mathbb{R} \cup {\infty}$ and acts properly discontinuously on the complementary line (topologically conjugate to a translation acting on the real number line). When $trace(g)>2$ then $g$ has two fixed points and acts properly discontinuously on each of the two complementary lines. In either case, since $\mathbb{Q} \cup {\infty}$ is dense in each such line, $g$ has infinitely many orbits.

Edit: Yves' comment shows that there is a hole in this approach due to infinite order elliptic elements in $SL(2,\mathbb{Q})$. Not sure whether it can be patched it up over $\mathbb{R}$, but Yves' answer, which for the case $trace(g)<2$ uses north-south dynamics over $\mathbb{Q}_p$, does the trick very nicely.

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    $\begingroup$ I'm afraid the rational matrix $\pmatrix{4/5 & -3/5\cr 3/5 & 4/5 \cr}$ has trace less than 2 but has infinite order. $\endgroup$
    – YCor
    Jul 31, 2012 at 6:26
  • $\begingroup$ Oops, I succumbed to $SL_2(\mathbb{Z})$ thinking there... $\endgroup$
    – Lee Mosher
    Jul 31, 2012 at 13:24
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Consider the matrix $A=\begin{pmatrix}a&b \cr c&d\end{pmatrix}$ in ${\rm SL}(2,\mathbb Z)$. Then you're asking if setting $\begin{pmatrix}p_n\cr q_n\end{pmatrix}=A^n\begin{pmatrix}p\cr q\end{pmatrix}$ all of ${\mathbf P}^2(\mathbb Q)$. To see that it's not, notice that there are a handful of cases:

$A$ is elliptic if it has a pair of eigenvalues on the unit circle. These are necessarily roots of unity. In this case, $A$ has finite order, and so it's obvious that there are infinitely many orbits.

$A$ is parabolic if it has a repeated root of $\pm 1$. In this case, $A^n\begin{pmatrix}p\cr q\end{pmatrix}$ converges in ${\mathbf P}^2(\mathbb Q)$ to a fixed direction (the generalized eigenvector) unless $\begin{pmatrix}p\cr q\end{pmatrix}$ was in the direction of the eigenvector

Similarly if $A$ is hyperbolic, there are two possible limit directions.

This rules out the transitivity you were looking for.

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    $\begingroup$ We can't reduce to the case that the element lies in $\text{PSL}_2(\mathbb{Z})$ (it suffices to exhibit an integer matrix with coprime entries whose determinant is not $\pm 1$). $\endgroup$
    – Qiaochu Yuan
    Jul 31, 2012 at 0:47
  • $\begingroup$ Hmmm... I guess so. Not sure this changes the rest of the argument though? $\endgroup$
    – Anthony Quas
    Jul 31, 2012 at 1:23
  • $\begingroup$ @Anthony The tricky case is the elliptic one. Of course I agree with your argument for $SL(2,\mathbb{Z})$. $\endgroup$
    – Mark Wildon
    Jul 31, 2012 at 18:57

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