I've seen this result posed in several places as an exercise, and I've also seen the proof for $SL_n(\mathbb{R})$, for example in this answer. But I'm not quite able to follow the details.
Can someone work out the details of the $n=2$ case in the most elementary way possible?
Thanks in advance for any help.
Let $A \in Sl_2(\mathbb{R})$ with $A =\begin{bmatrix} a & b \\ c & d \end{bmatrix}$.Observe that there is a natural map $g:\mathbb{R}^4 \to Sl_2(\mathbb{R})$ where we take;
$$(x_1,x_2,x_3,x_4) \mapsto \begin{bmatrix} x_1 & x_2 \\ x_3 & x_4 \end{bmatrix}$$
$$\\$$
This map is an isomorphism. Now consider $f = det: Mat_2(\mathbb{R}) \to \mathbb{R}$. We can identify any matrix with its respective coordinate $(x_1,x_2,x_3,x_4)$ i.e $f: \mathbb{R}^4 \to \mathbb{R}$. Now we have ;
$$\\$$
$$f(x_1,x_2,x_3,x_4) = x_1x_4 - x_2x_3$$
$$\\$$
which is smooth. Observe also that $f^{-1}(1) = Sl_2(\mathbb{R})$ and so by the level-set theorem, if we can show $1$ is a regular value of $f$ then we are done. This amounts to showing that $Df$ has rank $1$. Remember that if $g: \mathbb{R}^k \to \mathbb{R}$ then $Dg(X) = \langle g_{x_1}(X), g_{x_2}(X),...,g_{x_m}(X) \rangle$. Therefore if we can show that one of the directional derivatives $f_{x_i} \not = 0$ then the differential is non-zero.
$$\\$$
For that we take;
$$\frac{d}{dt} det(A+At)|_{t=0} = \frac{d}{dt} det \left(\begin{matrix} (1+t)(a) & (1+t)b \\ (1+t)c & (1+t)d \end{matrix} \right)|_{t=0} = \frac{d}{dt} (1+t)^2 det(A)|_{t=0} = 2$$
Therefore $f$ has max rank and by the theorem $f^{-1}(1) = Sl_2(\mathbb{R})$ is a smooth manifold of dimension $n^2-1$ where $n =2$.