 Dilaton equation from contracted Bianchi identity

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.