I have a dumb question.


Given a Hopf algebra $H$, take an invertible element $J\in H\otimes H$ and define $\Delta^J=J^{-1} \Delta J$. This becomes a new coproduct when $J$ satisfies a certain condition. Such $J$ are called twists, and let's denote the new Hopf algbera obtained this way by $H^J$. Two twists are called gauge equivalent when there is an invertible element $u \in H$ such that $J\_2=\Delta(v) J\_1 (v^{-1}\otimes v^{-1})$. Then $H^{J\_1}$ and $H^{J\_2}$ are isomorphic.


Now, let's take $H=k^G$ (i.e. the function algebra on $G$ with point wise multiplication). I understand that the twists up to gauge equivalent are given by $H^2(G,k^\times)$.


What I don't understand is the following. Given $J\in H^2(G,k^\times)$, $(k^G)^J$ is still commutative. So it should still be a function algebra on a group, and the group in question should be $G$ itself, by considering the representations of the Hopf algebra. So I believe $(k^G)^J$ is isomorphic to $k^G$. (And there is a statement to this effect in Proposition 1.36.5 and Remark 1.36.6 in [this pdf](https://ocw.mit.edu/courses/mathematics/18-769-topics-in-lie-theory-tensor-categories-spring-2009/lecture-notes/MIT18_769S09_lec07.pdf) .)


But I can't construct an explicit isomorphism, mapping $\Delta^J$ to $\Delta$. 


Could someone enlighten me?



Twists of commutative Hopf algebras
This wa really dumb. As Julian says in the comment above, $\Delta^J$ is equal to $\Delta$, so there was nothing to be done. 


I posted the question around the midnight, just before going to bed, and I'm writing this on my way to work. This shows we shouldn't think too hard before going to sleep, we are prone to do dumb things when sleepy...


And there was in fact [another MO question](https://mathoverflow.net/questions/53251/twist-of-a-group-hopf-algebra) which was an exact duplicate of my dumb quesition! Argh!


