I want to do a proof that $SL(2,\mathbb R)\times SL(2, \mathbb R) \cong SO^+(2,2)$. My idea was to use the same Argument as in this Question. So I wanted to begin with the Basis of the Lie algebra $\mathfrak{sl(2,\mathbb R)} \times \mathfrak{sl(2, \mathbb R)}$, but i am not sure how to do it. I know that the basis of $\mathfrak{sl(2, \mathbb R)}$ is $$ \mathfrak{sl(2,\mathbb R)} = span \left\{ \left(\begin{matrix} 0 & 1\\ 0 & 0 \end{matrix}\right), \left(\begin{matrix} 0 & 0\\ 1 & 0 \end{matrix}\right), \left(\begin{matrix} 1 & 0\\ 0 & -1 \end{matrix}\right) \right\}. $$ But i dont know what to do with this result. I read, that $$ \left(\begin{matrix} 1 & 0\\ 0 & 1 \end{matrix}\right), \left(\begin{matrix} 1 & 0\\ 0 & -1 \end{matrix}\right), \left(\begin{matrix} 0 & 1\\ -1 & 0 \end{matrix}\right), \left(\begin{matrix} 0 & 1\\ 1 & 0 \end{matrix}\right) (*) $$ Is a basis of $\mathfrak{sl(2,\mathbb R)} \times \mathfrak{sl(2, \mathbb R)}$ wich gives for the following bilinear form: $$ \langle x, y \rangle := tr(x\cdot wy^Tw^{-1}) \qquad \qquad w := \left(\begin{matrix} 0 & -1\\ 1 & 0 \end{matrix} \right) $$ the desired signature $2,-2,2,-2$. On the other hand, this question, $\dim_{\mathbb{R}}(\mathfrak{so(2,2)}) = 6$. Therefore $\dim_{\mathbb R}(\mathfrak{sl(2, \mathbb R) \times sl(2, \mathbb R)})$ should also be $6$. Was my conclusion in this question wrong, and the statement, $\dim_{\mathbb{R}}(\mathfrak{so(2,2)}) = 6$ is wrong, or is $(*)$ not a basis of $\mathfrak{sl(\mathbb R, 2)\times sl(\mathbb R,2)}$?
What am i missing, and can someone help me, to write this proof in a more explicit way?
Let us consider the usual sporadic isogenies related to the one you ask. They are associated with quadratic forms of different signatures.
The case of the Lorentz group and the $2:1$ isogeny from its universal cover: $$SL_2(\mathbb{C})\twoheadrightarrow SO^{\uparrow}_+(1,3)$$
The case of the quaternion algebra, the double cover $SU(2)\to SO(3)$ and the total group of proper isometries of the Euclidean quaternion algebra (with its quaternionic norm as the associated quadratic form):
$$\Psi:SU(2)\times SU(2)/\{\pm 1\}\cong SO(\mathbb{H})$$ given by $\Psi(A,B)(X)=AXB^{\dagger}$, where all quaternions are written in their standard complex $2$-dimensional representation (i.e. general unitary matrices when nonzero).
In the case of signature $(2,2)$ we have a nice way of presenting this quadratic space which suits us very well. That is, just write a $2 \times 2$-matrix and compute its determinant; this indeed corresponds to a neutral quadratic form on $\mathbb{R}^4$.
Now, it suffices to write down the action of $SL_2(\mathbb{R})^2$ on $M_2(\mathbb{R})$ similarly to the one above, that is:
$\Psi(A,B)(X):=AXB^{-1}$ defines a left action on $M_2(\mathbb{R})$ via proper isometries with respect to the quadratic form $A\mapsto \det A$, which in turn furnishes a double cover, which is the one you were looking for.
Note that both double covers have isomorphic complexifications. Hope this helps.
$$\Psi:SL_2(\mathbb{R})\times SL_2(\mathbb{R})\to SO^+(2,2).$$