In the paper, page 28, Definition 4.2.1, the compatibility condition for a Yetter-Drinfeld module over $H$ is $$ h_{(1)} v_{(-1)} \otimes h_{(2)}.v_{(0)} = (h_{(1)}.v)_{(-1)}h_{(2)} \otimes (h_{(1)}.v)_{(0)}, v \in V, h \in H. $$ On the other hand, in the article, the compatibility condition for a Yetter-Drinfeld module over $H$ is $$ \delta(h.v) = h_{(1)} v_{(-1)} S(h_{(3)}) \otimes h_{(2)}.v_{(0)}, v \in V, h \in H. $$ Are the two conditions equivalent? Thank you very much.
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asked Jun 15, 2016 at 14:40
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Assume that the second equation holds. Then \begin{align*} \delta(h_{1}.v)=(h_{1}. v)_{-1}\otimes(h_{1}.v)_{0}=h_{1,1}v_{-1}Sh_{1,3}\otimes h_{1,2}. v_{0} \end{align*} and therefore $$ (h_{1}.v)_{-1}h_{2}\otimes(h_{1}.v)_{0}=h_{1,1}v_{-1}Sh_{1,3}h_{2}\otimes h_{1,2}.v_{0}=h_{1}v_{-1}\otimes h_{2}.v_{0}. $$ Conversely, assume that the first equation holds. Then \begin{align*} (m\otimes &id)(h_{11}v_{-1}\otimes Sh_2\otimes(h_{12}.v_0) )\\ &=(m\otimes id)\left( (h_{11}.v)_{-1}h_{12}\otimes Sh_2\otimes (h_{11}. v)_0 \right)\\ &=(h_1.v)_{-1}h_2Sh_3\otimes (h_1.v)_0\\ &=(h.v)_{-1}\otimes (h.v)_0. \end{align*}
answered Jun 15, 2016 at 19:48
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$\begingroup$ thank you very much. Why $ h_{1,1}v_{−1}Sh_{1,3}h_2 \otimes h_{1,2}.v_0=h_1 v_{−1} \otimes h_2.v_0 $? $\endgroup$ Commented Jun 16, 2016 at 1:01
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$\begingroup$ Because S is the antipode. $\endgroup$ Commented Jun 16, 2016 at 12:05
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3$\begingroup$ Use the coassociativity, we have $$ h_{11} v_{-1} S(h_{1,3}) h_2 \otimes h_{1,2}.v_0 \\ = h_1 v_{-1} S(h_3) h_4 \otimes h_2.v_0 \\ = h_1 v_{-1} \epsilon(h_3) \otimes h_2.v_0 \\ = h_1 v_{-1} \otimes \epsilon(h_3) h_2.v_0 \\ = h_1 v_{-1} \otimes h_2.v_0 $$ $\endgroup$ Commented Aug 6, 2016 at 7:46