# Conformal Casimir as Differential Operator

my question regards equation (165) of [1], namely, how to write the conformal Casimir as a differential operator in the "usual" $$z,\bar{z}$$ coordinates. If one inspects the definition of the Casimir operator as $$$$C = -\frac{1}{2}L^{ab}L_{ab},$$$$ where $$L_{ab}$$ are the embedded $$SO(d+1,1)$$ generators defined in (38) of [1], it is possible to write the action of the Casimir in the "12" Channel of a 4pt function of identical scalars ($$\langle \phi(x_{1})\phi(x_{2})\phi(x_{3})\phi(x_{4}) \rangle$$), where $$$$C\phi(x_{1})\phi(x_{2}) = -\frac{1}{2}\left(\mathcal{L}^{ab}_{1}+\mathcal{L}^{ab}_{2}\right)\left(\mathcal{L}_{ab,1}+\mathcal{L}_{ab,2}\right)\phi(x_{1})\phi(x_{2}),$$$$ where subscript $$i$$ means the operator acting on coordinates $$x_{i}$$ (162) of [1]. Given the representation of the scalar field in respect to the conformal group generators (scattered throughout [1]) \begin{align} &[P_{\mu},\phi(x)] = \partial_{\mu}\phi(x),\\ &[D,\phi(x)] = \left( x\cdot\partial + \Delta \right)\phi(x),\\ &[M_{\mu\nu},\phi(x)] = -\left(x_{\mu}\partial_{\nu} - x_{\nu}\partial_{\mu} \right)\phi(x) ,\\ &[K_{\mu},\phi(x)] = \left(2 x_{\mu}(x\cdot\partial) - x^{2}\partial_{\mu} + 2\Delta x_{\mu} \right)\phi(x), \end{align} one can easily read $$[L_{ab},\phi(x)]$$ from the above (I assumed the SO(d+1,1) metric as $$\text{diag}(-1,1,\cdots,1)$$, with the "-1" sign in the -1 component). By acting as $$$$L^{ab}L_{ab}\phi(x_{1})\phi(x_{2}) = [L^{ab},[L_{ab},\phi(x_{1})\phi(x_{2})]],$$$$ I can indeed generate differential operators of first and second order depending on $$x_{1},x_{2},\partial_{1},\partial_{2}$$. My problem now lies in how to explicitly arrive at (165), and I believe it has to do with the choice of conformal frame. In section 5.1 of [1], the authors define the conformal frame as \begin{align} &x_{1} = (0,0,\cdots,0)\\ &x_{2} = (x,y,0,\cdots,0)\\ &x_{3} = (1,0,\cdots,0)\\ &x_{4}\to\infty. \end{align} Does it mean that I need to take $$x_{1}^{\mu} = 0^{\mu}$$ and $$\frac{\partial}{\partial x_{1}^{\mu}} = 0$$ in the Casimir operator in order to reproduce (168)? Which would yield only derivatives in respect to the $$x_{2}$$ point, which can be transformed into ($$z,\bar{z}$$) via a complex plane identification? Then I should act in (158) to recover $$\mathcal{D}$$ (168)?

If someone has a better understanding on how to reproduce the desired equation, I would much appreciate the help.

[1] David Simmons-Duffin, TASI Lectures on the Conformal Bootstrap, https://arxiv.org/abs/1602.07982

1. Act on eq. (158) with the commutators and work you way through the prefactor $$1/(x_{12}^{2\Delta_\phi} x_{34}^{2\Delta_\phi})$$ to see how the Casimir operator acts on the function $$g_{\Delta, \ell}(u,v)$$. You should obtain a second-order differential equation in $$\partial/\partial x_1$$ and $$\partial/\partial x_2$$.
2. Use the chain rule to figure out how to write $$\partial/\partial x_1$$ and $$\partial/\partial x_2$$ in terms of $$\partial/\partial z$$ and $$\partial/\partial \bar{z}$$. This means computing $$\partial z/\partial x_1$$ from the definition of $$z$$ (and so on for $$\bar{z}$$ and $$x_2$$) in eq. (187) of [1].
3. Write you result in terms of $$z$$ and $$\bar{z}$$. If you don't do mistakes this part should be very simple.
• Point 1 can be agreed upon, the problem is point 2. If I am not mistaken, $z,\bar{z}$ coordinates are only defined in the conformal frame, in which the point $x_{1}$ is fixed to the origin. Which means there is no coordinate transformation from $x_{1}$ to $z,\bar{z}$, in fact $z,\bar{z}$ are the only freedom you have ($x_{2}$ coordinates). Having said that, how do we treat derivatives $\frac{\partial}{\partial x^{\mu}_{1}}$ in this frame? Jan 26, 2022 at 18:39
• $z$ and $\bar{z}$ are defined in terms of $x_1$, $x_2$, $x_3$ and $x_4$ in eq. (187). I'm editing my answer to include this point.