The classical elimination in parts b and c discarded the field-dependent <Gaussian functional integral> determinant. At one loop that determinant is a local curvature coupling and supplies the <Buscher rules> dilaton shift
$$
\boxed{\widetilde\Phi=\Phi-\frac12\log V}.
$$
The metric alone consequently need not obey $\widetilde R_{\mu\nu}=0$. The relevant leading <sigma-model beta function> in the dual background instead contains
$$
\beta^{\widetilde G}_{\mu\nu}
=\alpha'\left(\widetilde R_{\mu\nu}
+2\widetilde\nabla_\mu\widetilde\nabla_\nu\widetilde\Phi\right)+\cdots.
$$
The new dilaton term cancels the failure of the dual metric to be Ricci-flat, so the complete metric-dilaton background remains conformal and physically <T-duality>[T-dual] to the original one.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 306 / 4 / d / Solution

{c}

The Buscher rules invert the metric component along an Abelian isometry and transform the other metric and B-field components. The Gaussian functional determinant also shifts the dilaton; when the metric is block diagonal with $G_{zz}=V$, $\widetilde\Phi=\Phi-\tfrac12\log V$.


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 Incoming links: Buscher rules