Buscher Isometric Background Transform

In the chart of the Definition write $X^{\mu }={\lambda }^{\mu }{x}^{\mu }$ with fixed positive lengths ${\lambda }^{\mu }$ and ${\lambda }^y=R$. If $G_{X,\mu \nu }^{(s)}$ and $B_{X,\mu \nu }$ are the components in the length-valued $X$ coordinates, then $G_{x,\mu \nu }^{(s)}={\lambda }^{\mu }{\lambda }^{\nu }{G}_{X,\mu \nu }^{(s)}$ and $B_{x,\mu \nu }={\lambda }^{\mu }{\lambda }^{\nu }{B}_{X,\mu \nu }$ have length-squared dimension; setting $\widehat{G}=G_x^{(s)}\mathrm{/}{\alpha }^{\mathrm{\prime }}$ and $\widehat{B}=B_x\mathrm{/}{\alpha }^{\mathrm{\prime }}$ turns [F1]'s matter action into $\frac{1}{4\pi }\int [\sqrt{h}{h}^{ab}{\widehat{G}}_{\mu \nu }{\mathrm{\partial }}_a{x}^{\mu }{\mathrm{\partial }}_b{x}^{\nu }+i{\epsilon }^{ab}{\widehat{B}}_{\mu \nu }{\mathrm{\partial }}_a{x}^{\mu }{\mathrm{\partial }}_b{x}^{\nu }]$. Let ${\xi }^a$, $a=1,2$, be dimensionless local worldsheet coordinates. In a local gauge for the free shift $y\mapsto y+ϵ$, introduce a real worldsheet one-form $A=A_{a}d{\xi }^a$, with dimensionless components and transformation $A\mapsto A-dϵ$, and use $D_{a}y={\mathrm{\partial }}_{a}y+A_a$. Let the dimensionless Lagrange multiplier $\tilde{y}$ have period $2\pi$. With ${\epsilon }^{12}=+1$, the Euclidean gauged parent action is

The multiplier term is equivalently $\frac{i}{2\pi }\int d^2\xi {\epsilon }^{ab}\tilde{y}{\mathrm{\partial }}_a{A}_b$; its normalization corresponds to the source's dimensionful multiplier $\tilde{\theta }={\alpha }^{\mathrm{\prime }}\tilde{y}$. Integrating over $\tilde{y}$ imposes $dA=0$, and on a contractible patch $A=d\theta$ can be removed by the shift gauge symmetry. This is only the local gauged parent action; no global holonomy or path-integral equivalence is asserted.

Fix the shift gauge $y=0$ in the parent action of step 1.1. Varying its algebraic Gaussian field $A_a$ gives, with ${\widehat{G}}_{yy}=E_{yy}>0$,

where ${\epsilon }_a{}^b:=h_{ac}{\epsilon }^{cb}$. Substituting into the parent action and collecting the coefficients of the symmetric metric contraction and antisymmetric two-form contraction gives the dual matrix blocks $E_{yy}^{\mathrm{\prime }}=1\mathrm{/}{E}_{yy}$, $E_{yi}^{\mathrm{\prime }}=E_{yi}\mathrm{/}{E}_{yy}$, $E_{iy}^{\mathrm{\prime }}=-E_{iy}\mathrm{/}{E}_{yy}$, and $E_{ij}^{\mathrm{\prime }}=E_{ij}-E_{iy}{E}_{yj}\mathrm{/}{E}_{yy}$, with $E=\widehat{G}+\widehat{B}$. This is a local Gaussian elimination; its functional-determinant contribution is treated separately by [F2].

Writing $E^{\mathrm{\prime }}={\widehat{G}}^{\mathrm{\prime }}+{\widehat{B}}^{\mathrm{\prime }}$ with ${\widehat{G}}^{\mathrm{\prime }}$ symmetric and ${\widehat{B}}^{\mathrm{\prime }}$ antisymmetric gives ${\widehat{G}}_{yi}^{\mathrm{\prime }}=(E_{yi}^{\mathrm{\prime }}+E_{iy}^{\mathrm{\prime }})\mathrm{/}2={\widehat{B}}_{yi}\mathrm{/}{\widehat{G}}_{yy}$ and ${\widehat{B}}_{yi}^{\mathrm{\prime }}=(E_{yi}^{\mathrm{\prime }}-E_{iy}^{\mathrm{\prime }})\mathrm{/}2={\widehat{G}}_{yi}\mathrm{/}{\widehat{G}}_{yy}$. Taking the symmetric and antisymmetric parts of $E_{ij}^{\mathrm{\prime }}$ gives exactly the displayed $ij$ rules, because $E_{iy}={\widehat{G}}_{yi}-{\widehat{B}}_{yi}$ and $E_{yj}={\widehat{G}}_{yj}+{\widehat{B}}_{yj}$.

By the adopted regularized measure input [F2], the Gaussian determinant changes the curvature coupling by $-\frac{1}{8\pi }\int \sqrt{h}{R}^{(2)}\log {\widehat{G}}_{yy}$. Combining it with [F1]'s coefficient $\frac{1}{4\pi }\int \sqrt{h}\mathrm{\Phi }{R}^{(2)}$ is equivalent to replacing $\mathrm{\Phi }$ by $\mathrm{\Phi }-\frac{1}{2}\log {\widehat{G}}_{yy}$, the dilaton rule in the Definition. This is the specified local measure correction; no global equivalence is inferred.