$\text{SL}(2,\mathbb{R})$ is homeomorphic to $S^1 \times \mathbb{R}^2$

Question

Using the left action of $\text{SO}(2)$ on $\text{SL}(2,\mathbb{R})$, show that $\text{SL}(2,\mathbb{R})$ is homeomorphic to $S^1 \times \mathbb{R}^2$

I tried defining the action by $\psi: \text{SO}(2) \times \text{SL}(2,\mathbb{R}) \to \text{SL}(2,\mathbb{R})$ by $(A,B) \mapsto AB$.

The kernel of this map are pairs $(A,A^{-1})$ such that $A \in \text{SO}(2)$.

Clearly this map is surjective, so I wanted to show that I can now find a canonical homeomorphism $\phi: \text{SO}(2) \times \text{SL}(2,\mathbb{R})/ \ker \psi \to S^1 \times \mathbb{R}^2$. For instance, maybe the map $(A,B)\ker\psi \mapsto (A,(\cos xb_{1,1} -\sin xb_{1,2}, \sin xb_{2,1} +\cos xb_{2,2}))$ where $A$ is rotation by $x$.

However, I'm not entirely sure this is well defined, and even if so, if it is injective and open.

Using the theory of Lie groups, how should I approach this problem?

@ThomasShelby I saw this indeed, but I don't think this is the way intended to approach this problem here. For instance, the surjectivity of the map follows from a theorem that's quite specific. I'm thinking there may be a way to derive this using covers, or some properties of simply connected groups. — Mariah, May 27 '19 at 18:05
Do you know about the Iwasawa decomposition (aka KAN decomposition)? — Moishe Kohan, May 27 '19 at 18:24
@MoisheKohan no, unfortunately. This is the content of the link Thomas attached. — Mariah, May 27 '19 at 18:25
OK, then do you know about the action of $SL(2,R)$ on the upper half plane? Or on the space of positive definite quadratic forms in two variables? You have to know something in order to solve this problem. — Moishe Kohan, May 27 '19 at 18:28
@MoisheKohan alright, I'm not familiar with these(..), but thank you for the references. — Mariah, May 27 '19 at 18:43
Possible duplicate of $M= { A \in Mat_{2 \times 2}{\mathbb{R}}| \det(A)=1 }$ is homeomorphic to $S^{1} \times \mathbb{R}^{2}$ General statement: $SL(n,\mathbb R)$ diffeomorphic to $SO(n) \times \mathbb R^{n(n+1)/2-1}$? — YuiTo Cheng, May 27 '19 at 23:35

Jean Marie · Answer 1 · 2019-05-27T20:20:37.000

Do you know QR decomposition ?

It splits any matrix $M$ under the form $M=QR$ where $Q$ is orthogonal and $R$ is upper triangular with positive entries on its diagonal (thus with $\det R >0$). This factorization is known to exist and to be unique if $M$ is invertible.

Now consider $2 \times 2$ matrices $M$ with determinant $1$ (this can be assumed if $M \in SL_2(\mathbb{R})$ ; see remark below). As $\det(R)=\det(M)/\det(Q)=\pm 1$ (in fact $+1$), this decomposition takes the form :

$$\begin{pmatrix}\cos \theta & -\sin \theta\\\sin \theta & \ \ \ \cos\theta\end{pmatrix}\begin{pmatrix}a&b\\0&1/a\end{pmatrix} \ \ \text{for certain real numbers} \ \theta \in [0,2\pi],a>0,b \tag{1}$$

One may object that we have taken a rotation matrix for $Q$ and not left the possibility of a symmetry matrix. But the latter case is impossible due to the fact that $\det M = \det Q \times \det R >0$.

(1) represents the looked-for homeomorphism.

Remark : there is another way to consider $M \in SL_2(\mathbb{R})$ as an equivalence class of invertible defined-up-to-a-multiple-constant matrix ; in this case, we can take as representative of this class a matrix that has determinant $1$, and we are back to the property above.

Iwasawa decomposition mentionned by @Moishe Kohan is linked to QR decomposition ; see for example this (high level) article in connection (your question) with Lie groups.

$\text{SL}(2,\mathbb{R})$ is homeomorphic to $S^1 \times \mathbb{R}^2$

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