Consider the subgroup $\Gamma < \operatorname{SL}(2,\mathbb{R})$ generated by the element $$ \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} $$ and all elements of $\operatorname{SO}(2)$. Is $\Gamma = \operatorname{SL}(2,\mathbb{R})$?

Some calculations show that $$ \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \in \Gamma $$ and hence $\operatorname{SL}(2,\mathbb{Z}) \subseteq \Gamma$.

  • $\begingroup$ This sure reminds me of the Iwasawa decomposition. $\endgroup$ Feb 20 at 21:37
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    $\begingroup$ Too late an hour for me to think straight, but I would check out how $\Gamma$ acts on the upper half-plane. $SO(2)$ is the stabilizer of $i$, so all you need is $\Gamma$ to act transitively on the upper half plane. $\endgroup$ Feb 20 at 21:42
  • $\begingroup$ Yes, they generate. $\endgroup$ Feb 20 at 22:09
  • $\begingroup$ If I correctly reconstructed enough of (my memory of) that action, then the orbit of $z\in H$ under $SO(2)$ is a circle with its center at a point of the form $it, t>1$, that also intersects the unit circle orthogonally (in other words, stable under inversion w.r.t. the unit circle). The upper half plane is surely a union of such circles, and it sure feels like you can go from any large enough circle to any other using the translation by one map (= your extra generator). Sorry, it's past midnight here. $\endgroup$ Feb 20 at 22:14
  • $\begingroup$ @MoisheKohan If you can figure out the details, please post. I really need shut-eye! $\endgroup$ Feb 20 at 22:16

1 Answer 1


Theorem. $\Gamma=SL(2, {\mathbb R})$.

Proof. I will be using complex numbers to describe points of the upper half-plane ${\mathbb H}^2=\{z\in {\mathbb C}: Im(z)>0\}$. Let $d$ denote the hyperbolic distance on ${\mathbb H}^2$. The group $G=SL(2, {\mathbb R})$ acts isometrically and transitively on ${\mathbb H}^2$ via linear-fractional transformations, so that the stabilizer of the point $i\in {\mathbb H}^2$ is the subgroup $K=SO(2)< SL(2, {\mathbb R})$. Thus, in order to show that the subgroup $\Gamma$ equals $G$ it suffices to prove that $\Gamma$ acts transitively on ${\mathbb H}^2$. A (continuous) curve $c: [0,\infty)\to {\mathbb H}^2$ is called proper if $$ \lim_{t\to\infty} d(c(0), c(t))=\infty. $$

Lemma. Suppose that there exists a proper curve $c: [0,\infty)\to {\mathbb H}^2$ contained in the $\Gamma$-orbit of $i$ in ${\mathbb H}^2$. Then the $\Gamma$-orbit of $i$ equals ${\mathbb H}^2$, i.e. $\Gamma$ acts transitively on ${\mathbb H}^2$.

Proof. Let $D=d(i, c(0))$. By the intermediate value theorem, continuity and properness of $c$, for every $r\in [D,\infty)$ there exists $t$ such that $d(i, c(t))=r$. Take $z\in {\mathbb H}^2$ such that $d(z,i)=r\ge D$ and find $t$ as above. Then, since $K$ acts transitively on each hyperbolic circle centered at $i$, there exists $k\in K$ suh that $z=k(c(t))$. Since $c(t)\in \Gamma i$, it follows that $z\in \Gamma i$ as well. Thus, the complement in ${\mathbb H}^2$ to the open hyperbolic disk $B(i, D)$ centered at $i$ and of radius $D$ is contained in $\Gamma i$. It remains to prove that $B(i, D)$ is contained in $\Gamma i$. There exists $n\in {\mathbb N}$ such that $$ \gamma^n= \left[\begin{array}{cc} 1&n\\ 0&1 \end{array}\right] $$ satisfies $d(i, \gamma^n(i))\ge 2D$. Therefore, the image under $\gamma^n$ of the disk $B(i, D)$ is contained in $$ {\mathbb H}^2\setminus B(i,D)\subset \Gamma i. $$ It follows that $B(i,D)\subset \Gamma i$ as well. qed

It is left to find a curve $c$ as in the lemma. For $n\in {\mathbb N}$ consider the hyperbolic circles $S(i+2n,D)$ centered at $i+2n$ and of the radius $D=d(i, i+1)$ (where, of course, $i+1=\gamma(i)$). Since $C=S(i, D)$ is $K(i+1)\subset \Gamma i$ and $S(i+2n,D)=\gamma^{2n}(C)\subset \Gamma i$, it follows that $$ U=\bigcup_{n\in {\mathbb N}} S(i+2n,D)\subset \Gamma i. $$ The above sequence of circles has the property that any two consecutive circles $S(i+2n,D), S(i+2n+2,D)$ have nonempty intersection (in two points, one of which is $i+2n+1$) and $$ \lim_{n\to\infty} d(i, i+2n)=\infty. $$

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Thus, there is a proper curve $c: [0,\infty)\to U$ such that $c(0)=i+1$. Hence, we found the required curve $c$. By applying the Lemma we conclude that $\Gamma i={\mathbb H}^2$. qed

Remark. There is nothing special about the element $\gamma$ in this proof, essentially the same proof goes through for every nonelliptic element of $SL(2, {\mathbb R})$, i.e. an element generating an unbounded cyclic subgroup. I am not sure what happens if one uses an elliptic element which is not in $SO(2)$. Most likely, one still generates the entire $SL(2, {\mathbb R})$, but the proof would have to be different.


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