I'm trying to understand the constraints on the disk CFT correlation function $\langle O_1(y_1)O_2(y_2)\rangle$, where the $O_i$'s are boundary operators that are not necessarily primary. It's a well-known fact that the corresponding sphere correlator is determined up to an overall constant, but I seem to be getting two independent constants in the case of the disk.
Let me just quickly give the argument that I've come up with. For $y_{1}>y_2$, we can make the $PSL(2,R)$ transformation $y'=(y_1-y_2)y+y_2$, under which $(\infty,1,0)\mapsto(\infty,y_1,y_2)$. This gives \begin{align*} \langle O_1(y_1)O_2(y_2)\rangle=(y_1-y_2)^{-2(h_1+h_2)}\langle O_1(1)O_2(0)\rangle. \end{align*} For $y_2>y_1$, we instead transform $y'=(y_2-y_1)y+y_1$, giving \begin{align*} \langle O_1(y_1)O_2(y_2)\rangle=(y_2-y_1)^{-2(h_1+h_2)}\langle O_1(0)O_2(1)\rangle. \end{align*} Putting them together, \begin{align*} \langle O_1(y_1)O_2(y_2)\rangle=|y_1-y_2|^{-2(h_1+h_2)}(\langle O_1(1)O_2(0)\rangle\theta(y_1-y_2)+\langle O_1(0)O_2(1)\rangle\theta(y_2-y_1)). \end{align*}
Now, for primary operators it's straightforward to show that $\langle O_1(1)O_2(0)\rangle=\langle O_1(0)O_2(1)\rangle$, but I don't see why (or if) this is true for nonprimaries. Are there just two independent constants in this case?
Thanks for your help!