Set $Z=\lambda\delta Z$. Expansion about the north pole gives
$$
S=\int d^2\sigma\left[
\partial_\alpha\delta Z\partial^\alpha\delta\bar Z
-2\lambda^2\delta Z\delta\bar Z
\partial_\alpha\delta Z\partial^\alpha\delta\bar Z+O(\lambda^4)\right].
$$
Thus $\langle\delta Z(k)\delta\bar Z(-k)\rangle=1/k^2$ up to the Fourier convention, and the leading four-point vertex is $-2\lambda^2$ times the scalar product of the two differentiated momenta, summed over assignments of differentiated legs.

Contracting the two undifferentiated fields in this vertex gives a one-loop tadpole proportional to
$$
I(\Lambda,\mu)=\int_{\mu<|k|<\Lambda}\frac{d^2k}{(2\pi)^2}\frac1{k^2}
=\frac1{2\pi}\log\frac\Lambda\mu.
$$
It produces the divergent metric correction $-2\lambda^2I\,\partial\delta Z\partial\delta\bar Z$, up to the allowed overall-normalization ambiguity. Geometrically this is the one-loop term $\beta^G_{\mu\nu}\propto\alpha'R_{\mu\nu}$. Since the round sphere has positive Ricci curvature, the isolated $S^2$ sigma model is not conformal. Therefore $S^2\times N$ is not a bosonic-string background unless contributions from $N$, a dilaton, an antisymmetric tensor, or further corrections cancel the sphere beta function.

Solution

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

 Solution

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