# Twists of commutative Hopf algebras

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 .)

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

Could someone enlighten me?

• Since $H$ is commutative, isn't $\Delta^J$ equal to $\Delta$? – Julian Rosen Dec 15 '16 at 16:19
• You're perfectly right. – Yuji Tachikawa Dec 16 '16 at 0:34
• I don't understand your claim about twists being given by $H^2(G, k^{\times})$. I understand this group as classifying twisted group algebras in the sense of the algebras whose representations are projective representations of $G$, but twisted group algebras are not Hopf algebras so "twist" seems to be used in two different senses here. – Qiaochu Yuan Jan 18 '17 at 5:57

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