Under a finite conformal coordinate change $z'=f(z)$ and $\bar z'=\bar f(\bar z)$, a <primary operator> of <conformal weight> $(h,\widetilde h)$ obeys
$$
\mathcal O'(z',\bar z')
=\left(\frac{df}{dz}\right)^{-h}
\left(\frac{d\bar f}{d\bar z}\right)^{-\widetilde h}
\mathcal O(z,\bar z).
$$
Equivalently, its <operator product expansion> with the <holomorphic stress-energy tensor> has singular part
$$
T(z)\mathcal O(w,\bar w)
\sim\frac{h\,\mathcal O(w,\bar w)}{(z-w)^2}
+\frac{\partial\mathcal O(w,\bar w)}{z-w}.
$$
The antiholomorphic stress tensor gives the analogous formula with $\widetilde h$.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 306 / 3 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-306/3/a/solution