Proof of the Isomorphism between: $SL(2,\mathbb R) \times SL(2, \mathbb R) \cong SO^+(2,2)$ 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?
 A: 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.
$$\Psi:SL_2(\mathbb{R})\times SL_2(\mathbb{R})\to SO^+(2,2).$$
