$SO(1, d + 1)$ symmetry of the Exact RG equation

TL;DR

This paper demonstrates that the ERG evolution operator in quantum field theory possesses an $SO(1,d+1)$ symmetry regardless of the cutoff function form, extending previous results that required specific cutoff choices.

Contribution

It proves the $SO(1,d+1)$ symmetry of the ERG evolution operator for any cutoff function, generalizing prior work limited to special cutoff forms.

Findings

ERG evolution operator has $SO(1,d+1)$ symmetry for any cutoff

Generators of conformal transformations depend on cutoff function

Full Wilson action's ERG operator can be transformed to exhibit the symmetry

Abstract

There is a method for constructing from first principles, a holographic bulk dual action in Euclidean $AdS_{d+1}$ space for a $d$-dimensional Euclidean CFT on the boundary, starting from the Polchinski's Exact Renormalization Group (ERG) equation that describes the RG evolution of the interaction part of the boundary Wilson action. The bulk action in $AdS_{d+1}$ has an $SO(1,d+1)$ symmetry and is obtained from the evolution operator of the Polchinski's ERG equation by a map that involves a field redefinition and requires a $\textit{special}$ form of the UV cutoff function in the ERG equation. In this paper, we show that for $\textit{any form}$ of the cutoff function, the ERG evolution operator has an $SO(1,d+1)$ symmetry. The generators of the special conformal transformation depend on the cutoff function. For the special cutoff function that maps to $AdS$ space, the transformations have the standard form of $AdS$ isometry. We also show that the ERG evolution operator for the $\textit{full}$ Wilson action can be put in the same form as the Polchinski's ERG equation by a field redefinition and consequently also has an $SO(1,d+1)$ symmetry for any cutoff function.