Three point function in CFT
Reason for deriving the 3-point function
Searching for a derivation of the three-point function constraints in CFT online I have realised that there is no derivation of the 3-point function. Most authors derive the 2-point function and leave the other one as an exercise to the reader.
In trying to derive the 3-point function I realised that there are some subtleties I don't fully understand.
Derivation:
We define the 3-point correlation function as$$ \langle\mathcal{O}_1(x_1)\mathcal{O}_2(x_2)\mathcal{O}_3(x_3)\rangle=f(x_1,x_2,x_3) $$ Translational symmetry
We impose the condition that $$ \left(\frac{\partial}{\partial x_1}+\frac{\partial}{\partial x_2}+\frac{\partial}{\partial x_3}\right)f(x_1,x_2,x_3)=0 $$ From the derivtion of the 2-point function I understand that this gives $$ f=f(X_{12}^\mu X_{13}^\mu X_{23}^\mu) $$ where $X_{ij}=(x_i-x_j)^\mu$, but can we show how this is the case from the translational symmetry condition imposed above?
Lorentz group symmetry
Here if we assume the above we have $$ f=f(X_{12}^2 X_{13}^2 X_{23}^2) $$ Dilatation symmetry
This is where I'm not quite sure how to proceed. We should satisfy the following (correct me if I'm wrong) $$ \left(x_1^\mu \frac{\partial}{\partial x_1^\mu}+x_2^\mu \frac{\partial}{\partial x_2^\mu}+x_3^\mu \frac{\partial}{\partial x_3^\mu}+\Delta_1+\Delta_2+\Delta_3\right)f(X_{12}^2 X_{13}^2 X_{23}^2)=0 $$ Not sure how I can solve this equation.
Special Conformal symmetry
Finally we must satisfy $$ (-2x_{1\mu}\Delta_1-2x_{2\mu}\Delta_2-2x_{3\mu}\Delta_3+k_{1\mu}+k_{2\mu}+k_{3\mu})f(X_{12}^2 X_{13}^2 X_{23}^2)=0 $$ with $$k_{i\mu}=x^2\frac{\partial}{\partial x_{i\mu}}-2x_\mu x^\nu\frac{\partial}{\partial x_{i\nu}}$$
Which again, I'm not sure how to calculate.
Disclaimer
I think it would be useful to have this derivation available with all the steps carefully shown so I'm hoping that with several hints I can complete this. 818