A conceptual understanding of transmutations (and bosonizations) of (braided) Hopf algebras

Consider a coquasitriangular Hopf-algebra $(H,\mu,\eta,\Delta,\epsilon, S)$ over a field $\mathbb F$ with characteristic zero and the braided monoidal category $\mathcal C$ of $H$-right-comodules. We explicitly denote the coquasitriangular form of $H$ by $r$ and its convolution inverse by $r'$. A lengthy calculation yields the result that $H$ can be "transmutated" into an Hopf-algebra object in $\mathcal C$ by considering it as a comodule algebra over itself by the coadjoint coaction and redefining its multiplication and antipode. That is: $(H,\overline \mu,\eta, \Delta,\epsilon, \overline S)$ with \begin{align*} \overline \mu &: H \otimes_\mathcal C H \rightarrow H , h\otimes g \mapsto h_{(2)}g_{(2)} r(S(h_{(1)})h_{(3)} \otimes S(g_{(1)}))\\ \overline S &: H \rightarrow H , S(h_{(2)})r(S^2(h_{(3)})S(h_{(1)}) \otimes h_{(4)}) \end{align*} and comodule structure given by \begin{align*} \delta: H \rightarrow H \otimes_\mathbb F H, h \mapsto h_{(2)}\otimes S(h_{(1)})h_{(3)} \end{align*} is a Hopf-algebra object in $\mathcal C$.

What I would be interested in is a conceptual understanding of this method and some of its properties:

1. Why is the coaction set to be the coadjoint coaction ?
2. Is there a general technique involved in finding the fitting multiplication and antipode?
3. What properties of the original Hopf algebra are preserved? 3a) Is there a connection between the category of modules over $\overline H$ and the category of (Yetter-Drinfeld) modules over $H$?
4. Can this process be iterated?
5. Is this construction functorial?

1. First, it is maybe worth mentioning that the transmutated (or covariantized Hopf algebra) can be motivated by generalizing a result for (co)commutative Hopf algebras). Observe thatin the case where $H$ is commutative, there is no need to covariantize the product as $H$ already becomes a module algebra with respect to the coadjoint coaction.

This is why the coadjoint action is used. Coquasi-triangular conceptually means "almost commutative" for a Hopf algebra.

Another conceptual use of $\underline{H}$ is that it gives a way to view the Drinfeld double of Hopf algebras as a bosonozation. This, again, generalizes a result where the transmutation is not necessary for (co)commutative Hopf algebras and involves the (co)adjoing (co)action. See Majid's book [Foundations of Quantum Group Theory, Theorem 7.4.5] or the original papers.

1. Yes, it is much more natural to understand such a construction after learning about a variation of Tannaka--Krein reconstruction within braided monoidal categories. An exposition of this can be found in Majid's book [Foundations of Quantum Group Theory] in Chapter 9.4. It may be worth reading the earlier sections of Chapter 9 first.

The point of the transmutation is then to find a braided Hopf algebra $\underline{H}$ within the category of $H$-comodules such that modules over it within this category capture all $H$-comodules. This can be done and the answer is the structure you describe. $\underline{H}$ is $H$ itself as a coalgebra, but it will need this new, "transmutated" product.

1. $\underline{H}$ will be "braided commutative" in some sense, see Majid's book, 9.4.10.

2. This is a good question. I suppose not in an obvious way on $H$. What can be done is to study the Hopf algebra bosonization (Radford biproduct) $\underline{H}\rtimes H$ and then covariantize this, as it should again be dual quasi-triangular due to $\underline{H}$ is in some sense braided commutative. But this would have to be investigated more carefully.

3. You could look at the construction of $B(H_1,H)$ for a Hopf algebra inclusion $H_1\to H$ in Majid's book. The result is a braided Hopf algebra in $H_1$-comodules. This is not a full answer to functoriality of course, but it goes into that direction.

• I don't think that in general $\underline{H}-comod(H-comod)\cong H-comod$ is true in general. Take as an example $H=\Bbbk[x]$. Then the coadjoint coaction is trivial (due to commutativity and cocommutativity). Also by commutativity, $\underline{\Bbbk[y]}=\Bbbk[y]$, where I denote the generator by $y$ to avoid confusion. Now, the category $\underline{H}-comod(H-comod)$ is equivalent to $\Bbbk[x,y]-comod$, which is not equivalent to $\Bbbk[x]-comod$. – Zahlendreher Jul 29 '17 at 13:25
• What is true though is that there is a functor $H-comod\to \underline{H}-comod(H-comod)$. The coaction of $\underline{H}$ is the same as the original one of $H$. – Zahlendreher Jul 29 '17 at 13:26
• What is quite interesting though is that $\underline{H}-mod(H-comod)$ is equivalent to the category of Yetter-Drinfeld modules, the center of $H-comod$. – Zahlendreher Jul 29 '17 at 13:26
• This is really interesting. I modified my original question accordingly because this deserves to be added to the answer. thank you for sharing your insights. – SeHa Jul 29 '17 at 14:00