In fact I already calculate the dual space with a formula, but I did'd understand some steps of the formula.
So, I want to calculate the dual space of The lie algebra of $SL(2,R)$. Knowing that $B=\left(\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} , \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \right)$, using the trace paring and that the matrix in the dual is the transpose of matrix in the lie algebra, we obtain that the dual basis is the B basis transpose.
The thing I did not understand was why the transpose are the dual.
This is a general fact from linear algebra. If $e_1, \ldots , e_n$ is a basis of a vector space $V$, then $V^{\ast}$ has the dual basis consisting of linear functions $f_1, \ldots, f_n$ defined by $f_i(e_j) = 0$ for all $i$ and $j$ such that $i\neq j$, and $f_i(e_i) = 1$ for all $i$. If a linear transformation $\phi$ is given by a $n × n$-matrix $A$, $\phi(x) = Ax$, then the matrix of $\phi^{\ast}$ in the dual basis is the transpose $A^T$ .
A representation $ρ : \mathfrak{g} \rightarrow \mathfrak{gl}(V )$ defines the dual (or contragredient) representation $\rho^{\ast} : \mathfrak{g}\rightarrow \mathfrak{gl}(V^{\ast})$ as follows: $a\in \mathfrak{g}$ sends a linear function $f(x)$ to $−f(\rho(a)x)$, or $a \mapsto −\rho(a)^T$ in the matrix form, with the transpose matrix. Hence for $\mathfrak{g}=\mathfrak{sl}_2$ and the natural representation $\rho$ of it (the identity map on trace zero matrices), the dual representation is just given by the negative of the transposed matrices.