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Let $H$ be a bialgebra and $A$ a $H$-module algebra. The cross product $A \rtimes H$ is defined as follows. As a vector space $A \rtimes H = A \otimes H$. The multiplication on $A \rtimes H$ is defined as follows: \begin{align} (a \otimes h)(b \otimes g) = \sum a(h_{(1)}.b) \otimes h_{(2)}g, \end{align} where $a, b \in A$, $h, g \in H$, $\Delta(h) = \sum h_{(1)} \otimes h_{(2)}$.

Is there a comultiplication $\Delta$ on $A \rtimes H$ such that the cross product $A \rtimes H$ is a bialgebra? If $\Delta$ does not always exist, under what conditions there is a comultiplication $\Delta$ on $A \rtimes H$ such that $A \rtimes H$ is a bialgebra? Thank you very much.

Edit: we add one more condition: suppose that $A$ is a bialgebra. In this case, is there a comultiplication $\Delta$ on $A \rtimes H$ such that the cross product $A \rtimes H$ is a bialgebra? If $\Delta$ does not always exist, under what conditions there is a comultiplication $\Delta$ on $A \rtimes H$ such that $A \rtimes H$ is a bialgebra? Thank you very much.

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    $\begingroup$ Take $H$ trivial. Then you're asking whether $A$ always has a bialgebra structure, and there's no reason that should be true. $\endgroup$
    – Qiaochu Yuan
    Commented May 15, 2016 at 17:24
  • $\begingroup$ @Qiaochu Yuan, thank you very much. I edited the post. $\endgroup$
    – Jianrong Li
    Commented May 16, 2016 at 1:41
  • $\begingroup$ And you don't want the assumptions of (double) bosonization? $\endgroup$
    – AHusain
    Commented May 16, 2016 at 1:48
  • $\begingroup$ @AHusain, thank you very much. Maybe we want the assumptions of double bosonization. I am not very familiar with double bosonization. $\endgroup$
    – Jianrong Li
    Commented May 16, 2016 at 1:54
  • $\begingroup$ @JianrongLi I think you've said you had Majid's book before, so do you want an explanation as an answer? $\endgroup$
    – AHusain
    Commented May 16, 2016 at 2:02

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In the case that $A$ is a braided Hopf algebra in $_H^H YD$ you can form $A \# H$ with the multiplication you already wrote down. The coproduct uses the fact that this kind of $A$ also has a left $H$ coaction as well given by $\delta r^{(2)} = r^{(2)}_{(-1)} \otimes r^{(2)}_{(0)}$

$$ \Delta ( r \# h) = ( r^{(1)} \# r^{(2)}_{(-1)} h_{(1)} ) \otimes ( r^{(2)}_{(0)} \# h_{(2)}) $$

If you want finite dimensional $H$, then you can say in terms of modules for $D(H)$.

Chapters 6 and 7 of Foundations of Quantum Group Theory discusses extension problems such as these. See there for compatibility conditions too. You might also look in http://www.sciencedirect.com/science/article/pii/S0021869384710118

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  • $\begingroup$ in the paper, Theorem 2.4, the coproduct is defined as $$\Delta_{B \rtimes H} = \Psi_{B, H} \circ (\Delta_B \otimes \Delta_H).$$ This is different from your formula $$ \Delta ( r \# h) = ( r^{(1)} \# r^{(2)}_{(-1)} h_{(1)} ) \otimes ( r^{(2)}_{(0)} \# h_{(2)}). $$ How could we know which coproduct should we use? $\endgroup$
    – Jianrong Li
    Commented May 17, 2016 at 2:01
  • $\begingroup$ now I understand. The two formulas are the same. $\endgroup$
    – Jianrong Li
    Commented Jun 1, 2016 at 6:56

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