Transforming the point-split normal ordering changes the subtraction as well as the two derivatives. Expanding $w(z\pm\delta/2)$ to third order gives
$$
\widetilde T(w)=\left(\frac{dw}{dz}\right)^{-2}
\left[T(z)-\frac1{12}\{w,z\}\right],
$$
where the <Schwarzian derivative> is
$$
\boxed{R(w;z)=\frac1{12}\{w,z\}
=\frac1{12}\left(\frac{w'''(z)}{w'(z)}
-\frac32\frac{w''(z)^2}{w'(z)^2}\right)}.
$$
The inhomogeneous Schwarzian term means that $T$ is not a primary operator. It reflects the central charge $c=1$ of one free scalar.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 306 / 2 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-306/2/d/solution