Metaplectic and Bipartite representations of SL(2,R)

Question

I have a question about two different oscillator representations of the group $SL(2,\mathbb{R})$. By oscillator representation I mean one using creation and annihilation operators. I want to know if they are equivalent or not. Here are the two representations \begin{equation} B^0 = \frac{a^{\dagger}a - b^\dagger b}{2},\quad B^- = \frac{a^2-(b^{\dagger})^2}{2},\quad B^+ = \frac{(a^\dagger)^2-b^2}{2},\quad \end{equation} and \begin{equation} L^0 = \frac12 (a_1^\dagger a_1 + a_2^\dagger a_2 + 1),\quad L^+ = a_1^\dagger a_2^\dagger,\quad L^- = a_1 a_2, \end{equation} where every oscillator $a,b,a_1,a_2$ has the standard commutation relation \begin{equation} [a,a^\dagger] = 1,\quad [b,b^\dagger] = 1, \quad [a_i,a_j^\dagger] = \delta_{ij}, \end{equation} and the commutators of different oscillators are all zero. One can check that those two representations indeed form the $SL(2,\mathbb{R})$ algebra. \begin{equation} [X^-,X^+] = 2X^0, \quad [X^0,X^-] = -X^-, \quad [X^0,X^+] = X^+, \end{equation} where $X$ can be $B$ or $L$. The question then is, is there any map $(a,b) \leftrightarrow (a_1,a_2)$ that maps the $B$ operators into the $L$ ones. For example my first guess was \begin{align} a &= \frac{a_1+a_2}{\sqrt{2}},\\ b^\dagger &= \frac{a_1-a_2}{\sqrt{2}}, \end{align} which indeed maps $B^\pm$ into $L^\pm$. However the $[b,b^\dagger]$ commutator has the wrong sign and the $B^0$ operators does not map into $L^0$. If anyone has any idea it would be greatly appreciated. Thank you!