# Is $SL(2,\Bbb R)$ generated by $SO(2)$ and a single upper triangular element?

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 $$\operatorname{SL}(2,\mathbb{Z}) \subseteq \Gamma$$.

• This sure reminds me of the Iwasawa decomposition. Feb 20 at 21:37
• 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. Feb 20 at 21:42
• Yes, they generate. Feb 20 at 22:09
• 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. Feb 20 at 22:14
• @MoisheKohan If you can figure out the details, please post. I really need shut-eye! Feb 20 at 22:16

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 $$\lim_{n\to\infty} d(i, i+2n)=\infty.$$

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.