I have the generators of $sl(2,\mathbb{R})$ algebra
$$J_0=\begin{pmatrix}0&1\\ -1&0\end{pmatrix},\quad J_1=\begin{pmatrix}1&0\\ 0&-1\end{pmatrix},\quad J_2=\begin{pmatrix}0&1\\ 1&0\end{pmatrix}$$
They satisfy the commutation relations
$$[J_0,J_1]=-2J_2,\quad [J_0,J_2]=2J_1,\quad [J_1,J_2]=2J_0$$
Now, I need to find the matrix $M(a,b,c)$ such that I can linearize it and get the $J_i$ generators such that it is valid
$$J_i=\frac{\partial M}{\partial k}\Bigg|_{(a,b,c)=(0,0,0)}$$
where $k=a,b,c$ respectively for $i=0,1,2$. Plus matrix M has to be element of $SL(2,\mathbb{R})$, so $det(M)=1$ for all $a, b, c$.
I've tried manually, and I can find M, but usually I cannot get that $det(M)=1$.
Is there some kind of procedure step by step to do so? Some kind of system of equations or?
Any help will be appreciated.
If you have a Lie algebra basis $e_1, ..., e_n$, then the function $f=exp(e_1t_1)\cdot ...\cdot exp(e_nt_n)$ will satisfy $\frac{df}{dt_i}(0,...,0)=e_i$.
Or in general you can first construct one-variable functions $F_i(t)$ such that $F_i(0)=e$ and $F_i'(0)=e_i$. Then the product $F_1(t_1) \times ... \times F_n(t_n)$ will satisfy the desired relations.
In your case $$F_0(a)=\left( \begin{array}{cc} 1-a^2 & a \\ -a & 1 \\ \end{array} \right), F_1(b)=\left( \begin{array}{cc} e^{b} & 0 \\ 0 & e^{-b} \\ \end{array} \right), F_2(c)=\left( \begin{array}{cc} 1+c^2 & c \\ c & 1 \\ \end{array} \right)$$