Are the Drinfeld doubles of twist equivalent Hopf algebras twist equivalent?

Question

Let $H_1$ and $H_2$ be finite dimensional Hopf algebras that are twist equivalent, i.e. $H_2$ is obtained from $H_1$ using a Drinfeld twist. My question is: are the Drinfeld doubles $D(H_1)$ and $D(H_2)$ twist equivalent as quasitriangular Hopf algebras? If so, how to explicitly construct the Drinfeld twist relating the two?

My guess is that $D(H_1)$ and $D(H_2)$ are twist equivalent, based on the following reasoning: it is known that two Hopf algebras $H_1,H_2$ are twist equivalent iff their category of representations $Rep(H_1)$ and $Rep(H_2)$ are equivalent as fusion categories. But $Rep(D(H))$ is the Drinfeld center of $Rep(H)$, which is a categorical construction, so $Rep[D(H_1)]$ and $Rep[D(H_2)]$ should also be equivalent (as braided fusion categories?), implying that $D(H_1)$ and $D(H_2)$ are twist equivalent. Is this correct?

Answer 1

That's a qualified yes.

Your argument is essentially correct, but you need to specify some things. Are the Hopf algebras finite dimensional? Is the base field algebraically closed of characteristic zero? If you're ok with these assumptions, then the following is true.

The assignment $H\mapsto(\,\mathrm{Rep}(H)\,,\,\mathrm{Forg}\,)$ gives a bijection between (finite dimensional) Hopf algebras up to isomorphism, and fusion categories equipped with a fiber functor (up to monoidal equivalence respecting the functor). Here by a fiber functor, I mean a linear, exact, monoidal functor whose target is the category of vector spaces.

Note: A large portion of the literature uses the term fiber functor to refer to a symmetric functor between symmetric categories, but we are not assuming any such symmetric structure here.

A given fusion category can admit multiple fiber functors, but all of these functors must be naturally isomorphic as mere functors. Thus the only difference between fiber functors lies in the monoidal structure. If $(F,J_i):\mathcal C\to\mathrm{Vec}$ are two fiber functors, corresponding to Hopf algebras $H_i$, then the comparison $J:=J_1J_2^{-1}$ is the twist that realizes the desired twist equivalence, i.e. $H_1\cong H_2^J$. The twist can be recorded explicitly by realizing $\mathcal C$ as $\mathrm{Rep}(H_2)$, and computing $J_{H_{2},H_2}(1\otimes1)$ as an element of $H_2\otimes H_2$.

The forgetful functor $\mathcal Z(\mathcal C)\to\mathcal C$ is monoidal, and thus we can compose it with our two fiber functors to produce two separate fiber functors for the Drinfel'd center. These fiber functors will realize $\mathcal Z(\mathcal C)$ as $\mathrm{Rep}\big(D(H_i)\big)$. This is Tannaka-Krein reconstruction, i.e. the inverse bijection to the one described at the beginning. In order to write down the desired twist, you would need to compute $J'_{D(H_{2}),D(H_2)}(1\otimes1)$ as an element of $D(H_2)\otimes D(H_2)$, where $J'$ is the comparison of the two fiber functors for $\mathcal Z(\mathcal C)$ that we built by composing with the forgetful functor.

This argument establishes that $D(H_1)\cong D(H_2)^{J'}$ as Hopf algebras. To see that they are isomorphic as quasitriangular Hopf algebras, just notice that the quasitriangular structure on $D(H_i)$ is (as an $R$-matrix) built out of the braiding on $\mathcal Z(\mathcal C)$ and the given fiber functor. Thus the twist $J'$ is precisely the element that intertwines the two quasitriangular structures.

Answer 2

Actually, Proposition 3.2 of Etingof & Gelaki 99' gives an explicit construction of the Drinfeld twist $J$ relating $D(H_1)$ and $D(H_2)$ and is quite simple: let $H_2=H_1^J$. Since $H_1$ is a Hopf subalgebra of $D(H_1)$, the Drinfeld twist $J \in H_1 \otimes H_1$ can be regarded as an element $J \in D(H_1) \otimes D(H_1)$, and it is straightforward to see that $J$ is a Drinfeld twist of $D(H_1)$. Proposition 3.2 then shows that $D(H_1)^J\cong D(H^J_1)=D(H_2)$ as quasitriangular Hopf algebras.