For the stereographic coordinate $Z=\tan(\theta/2)e^{i\varphi}$, the round metric is
$$
ds^2=\frac{4R^2,dZ,d\bar Z}{(1+Z\bar Z)^2}.
$$
Substitution into the sigma-model action gives
$$
S=\frac{R^2}{\pi\alpha'}\int d^2\sigma\,
\frac{\partial_\alpha Z\partial^\alpha\bar Z}{(1+Z\bar Z)^2}
=\frac1{\lambda^2}\int d^2\sigma\,
\frac{\partial_\alpha Z\partial^\alpha\bar Z}{(1+Z\bar Z)^2},
$$
so
$$
\boxed{\lambda^2=\frac{\pi\alpha'}{R^2}}.
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 306 / 4 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-306/4/b/solution