Leading metric-dilaton Weyl condition (${\overline{\beta }}_{ab}^g={\alpha }^{\prime }(R_{ab}+2{\nabla }_a{\nabla }_b\phi )$)

For the conventional Euclidean dilaton coupling $+(4\pi {)}^{-1}\int \sqrt{h}R\phi$ and no antisymmetric background, the metric worldsheet Weyl anomaly vanishes at first order in ${\alpha }^{\prime }$ when $R_{ab}+2{\nabla }_a{\nabla }_b\phi =0$. The one-loop Ricci term follows from the background field expansion of a string sigma model. Under a Weyl transformation, the dilaton coupling varies as $-(2\pi {)}^{-1}\int \sqrt{h}\omega {\square }_h\phi$; the embedding equations replace this by the target Hessian contribution. Full Weyl invariance also requires the dilaton curvature coefficient, including the central-charge deficit, to vanish. See the Lorentzian dilaton coupling sign convention when translating an action with a specified Lorentzian sign.

The conventional coefficient agrees with the background equations in Tong's string theory notes, Section 7.2.3.

Diverging $R_{ab}+2{\nabla }_a{\nabla }_b\phi =0$ and using the contracted Bianchi identity and Ricci identity gives ${\nabla }_b(R+4{\square }_g\phi -4|\nabla \phi {|}^2)=0$. The trace equation $R+2{\square }_g\phi =0$ then gives ${\square }_g\phi -2|\nabla \phi {|}^2=C$, with $C$ constant on each connected component. Zero does not follow from the tensor equation alone. The additional dilaton anomaly coefficient fixes $C=(D-26)/(3{\alpha }^{\prime })$ at leading order for the bosonic string without an antisymmetric background. In critical dimension it is zero; noncritical backgrounds have a central-charge-deficit term.

In flat target space, $\phi =q_a{X}^a$ has zero Hessian and therefore solves the leading metric-dilaton Weyl condition for every constant covector $q$. But $\square \phi -2|\nabla \phi {|}^2=-2q^2$, which need not vanish. This demonstrates the integration constant left by the contracted Bianchi identity. Full quantum Weyl invariance fixes the allowed $q^2$ through the central charge, as in a linear dilaton conformal field theory; the metric equation alone does not.

For the dilaton equation from contracted Bianchi identity, put $F=e^{-2\phi }$. The chain rule gives ${\square }_{g}F=F(-2{\square }_g\phi +4|\nabla \phi {|}^2)=-2CF$. Thus the exponential obeys a linear Klein-Gordon equation on the given background, although the original dilaton equation is nonlinear and the string-frame metric remains coupled to the dilaton. With the opposite named-field convention $\phi =-\Phi$, this variable is $F=e^{2\Phi }$.