Comment on "Twisted bialgebroids versus bialgebroids from Drinfeld twist"

TL;DR

This paper revises a previous proof about the isomorphism of twisted bialgebroids, removing restrictive assumptions and providing a more general and accurate demonstration applicable to broader cases.

Contribution

It offers a corrected, more general proof of the isomorphism of twisted bialgebroids without assuming quasitriangularity or special formulas for coaction and prebraiding.

Findings

Main result remains valid under general conditions.

Provides a proof without assuming quasitriangularity.

Corrects previous assumptions about coaction and prebraiding.

Abstract

A class of left bialgebroids whose underlying algebra $A\sharp H$ is a smash product of a bialgebra $H$ with a braided commutative Yetter--Drinfeld $H$-algebra $A$ has recently been studied in relation to models of field theories on noncommutative spaces. In [A. Borowiec, A. Pachol, ``Twisted bialgebroids versus bialgebroids from a Drinfeld twist'', J. Phys. A50 (2017) 055205] a proof has been presented that the bialgebroid $A_F\sharp H^F$ where $H^F$ and $A_F$ are the twists of $H$ and $A$ by a Drinfeld 2-cocycle $F = \sum F^1\otimes F^2$ is isomorphic to the twist of the bialgebroid $A\sharp H$ by the bialgebroid 2-cocycle $\sum 1\sharp F^1\otimes 1\sharp F^2$ induced by $F$. They assume $H$ is quasitriangular, which is reasonable for many physical applications. However the proof and the entire paper take for granted that the coaction and the prebraiding are both given by special formulas involving the R-matrix. There are counterexamples of Yetter--Drinfeld modules over quasitriangular Hopf algebras which are not of this special form. Nevertheless, the main result essentially survives. We present a proof with general coaction and the correct prebraiding, and even without the assumption of quasitriangularity.