Consider an action $S$ depending on fields $\phi_{i}$, where the index $i$ labels both the field type, the component and the spacetime point. Add a term quadratically proportional to the field equations $S_{i}\equiv \delta S/\delta \phi _{i}$ and define the modified action

\begin{equation}

\phantom{(1)}\qquad\qquad\qquad S^{\prime }(\phi _{i})=S(\phi _{i})+S_{i}F_{ij}S_{j}, \qquad\qquad\qquad (1)

\end{equation}

where $F_{ij}$ is symmetric and can contain derivatives acting to its left and to its right. Summation over repeated indices (including the integrationover spacetime points) is understood. Then there exists a field redefinition

\begin{equation}

\phantom{(1)}\qquad\qquad\qquad\phi _{i}^{\prime }=\phi _{i}+\Delta _{ij}S_{j}, \qquad\qquad\qquad (2)

\end{equation}

with $\Delta _{ij}$ symmetric, such that, perturbatively in $F$ and to all orders in powers of $F$,

\begin{equation}

\phantom{(1)}\qquad\qquad\qquad S^{\prime }(\phi _{i})=S(\phi _{i}^{\prime }). \qquad\qquad\qquad (3)

\end{equation}

**Proof**

The condition (3) can be written as

\begin{eqnarray}

&S(\phi _{i})+S_{i}F_{ij}S_{j}=S(\phi _{i}+\Delta _{ij}S_{j})=

\\ &\qquad\qquad\qquad=S(\phi_{i})+\sum_{n=1}^{\infty }\frac{1}{n!}S_{k_{1}\cdots

k_{n}}\prod_{l=1}^{n}(\Delta _{k_{l}m_{l}}S_{m_{l}}),

\end{eqnarray}

after a Taylor expansion, where $S_{k_{1}\cdots k_{n}}\equiv \delta^{n}S/(\delta \phi _{k_{1}}\cdots \delta \phi _{k_{n}})$. This equality is verified if

\begin{equation}

\phantom{(1)}\qquad\Delta _{ij}=F_{ij}-\Delta _{ik_{1}}\left[ \sum_{n=2}^{\infty }\frac{1}{n!} S_{k_{1}k_{2}k_{3}\cdots k_{n}}\prod_{l=3}^{n}(\Delta

_{k_{l}m_{l}}S_{m_{l}})\right] \Delta _{k_{2}j}, \qquad (4)

\end{equation}

where the product is meant to be equal to unity when $n=2$. Equation (4) can be solved recursively for $\Delta $ in powers of $F$. $\Box$

The first terms of the solution are

\begin{equation}

\phantom{(1)}\qquad\qquad\qquad \Delta _{ij}=F_{ij}-\frac{1}{2}F_{ik_{1}}S_{k_{1}k_{2}}F_{k_{2}j}+\cdots

\qquad\qquad\qquad (5)

\end{equation}

The theorem is very general. It works both for local and non-local theories. If $S(\phi _{i})$ and $F_{ij}$ are perturbatively local, namely they can be perturbatively expanded so that every order of the expansion is local, the field redefinition (2) and the action $S^{\prime }(\phi _{i})$ are perturbatively local. If both $S(\phi _{i})$ and $F_{ij}$ are local, in general (2) and $S^{\prime }(\phi _{i})$ are only perturbatively local. Actually, the resummation of the expansion can produce a non-local field redefinition. Finally, if $S(\phi _{i})$ and $F_{ij}$ are local or perturbatively local at the classical level, then (2) and $S^{\prime }(\phi _{i})$ are perturbatively local at the classical level.

This theorem was proved in

D. Anselmi, *Renormalization and causality violations in* *classical gravity coupled with quantum matter*,

06A1 Renorm

JHEP 0701 (2007) 062 | DOI: 10.1088/1126-6708/2007/01/062

and arXiv:hep-th/0605205.

A known situation where the theorem applies is the three-dimensional $U(1)$ gauge theory. The field equations of the Chern-Simons action

\[

S(A)=\frac{1}{2\bar{\alpha}}\int \varepsilon ^{\mu \nu \rho }F_{\mu

\nu }A_{\rho }

\]

are $F^{\mu \nu }=0$, so there exists a field redefinition $A_{\mu }^{\prime

}(A,\alpha /\bar{\alpha})$ such that

\begin{equation}

S^{\prime }(A)=S(A^{\prime }), \label{abi}

\end{equation}

where $S^{\prime }$ is the sum of the Chern-Simons action plus the square of the field strength,

\[

S^{\prime }(A)=\frac{1}{\bar{\alpha}}\int \varepsilon ^{\mu \nu

\rho }F_{\mu \nu }A_{\rho }-\frac{1}{4\alpha }\int F_{\mu \nu }F^{\mu \nu }.

\]

In pure gravity, the theorem just proved ensures that there exists a field redefinition that maps a class of higher-derivative theories into the Einstein theory. For example, there exists a field redefinition $g\rightarrow g^{\prime }(g,a,b)$ such that

\begin{equation}

S_{\text{HD}}(g)=S_{\text{E}}(g^{\prime }),

\end{equation}

where

\begin{eqnarray}

S_{\text{HD}}(g) &=&\frac{1}{2\kappa ^{2}}\int \sqrt{-g}\left( R(g)+aR_{\mu

\nu }R^{\mu \nu }(g)+bR^{2}(g)\right) , \\

S_{\text{E}}(g) &=&\frac{1}{2\kappa ^{2}}\int \sqrt{-g}R(g)

\end{eqnarray}

Indeed, the terms $R_{\mu \nu }R^{\mu \nu }$ and $R^{2}$ are quadratically proportional to the field equations of the action $S_{\text{E}}(g)$. The lowest-order contributions to the map are, from (2) and (5),

\begin{equation}

g_{\mu \nu }^{\prime }(g,a,b)=g_{\mu \nu }-aR_{\mu \nu }+\frac{1}{2}%

(a+2b)g_{\mu \nu }R+\mathcal{O}(a^{2},b^{2},ab).

\end{equation}