Constraining two-point functions of boundary operators on the disk 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!
 A: On the disk boundary an operator can be dragged around the circle, so the ordering is only defined up to cyclic permutations. For two operators there is only one such ordering, so the two correlators are equal. (For three operators there are two cyclic orderings, etc.)
For two operators you can see this on the half-plane by taking the $PSL(2,R)$ transformation
$$ \tau \mapsto \frac{\tau}{1-2\tau} $$
which, on the boundary, takes 1 to -1 and 0 to itself.
Edit: Indeed, as commented, this only works for primary operators. Let me suggest an idea for descendants. I am not very sure about it because I did not check the details carefully.
Suppose that $O_1$ is primary and $O_2'$ is a direct descendent of a primary $O_2$. Then $\langle O_1(0) O_2'(1) \rangle$ can be written as a contour integral of $\langle O_1(0) O_2(1) T(z) \rangle$ with a tight contour around $O_2$. I am ignoring a possible factor of $(z-1)$ to some power, and am closing the contour using the doubling trick. Applying the $PSL(2,R)$ transformation will take this to $\langle O_1(0) O_2(-1) T(z') \rangle$. This is non-trivial: even though $T$ is not a tensor operator when $c \ne 0$, the Schwarzian derivative of this transformation vanishes, so in this case $T$ does transform like a primary operator.
The new contour in $z'$ will surround $-1$ with an opposite orientation. This gives $\langle O_1(0) O_2'(-1) \rangle$, up to a possible sign (an additional sign may come from the factor I mentioned). This means the two correlators are at least related, if not equal.
If the argument is correct then I think the extension to any two operators is easy.
