# Compatibility conditions for Yetter-Drinfeld modules

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.

## 1 Answer

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*}

• thank you very much. Why $h_{1,1}v_{−1}Sh_{1,3}h_2 \otimes h_{1,2}.v_0=h_1 v_{−1} \otimes h_2.v_0$? – Jianrong Li Jun 16 '16 at 1:01
• Because S is the antipode. – Leandro Vendramin Jun 16 '16 at 12:05
• Use the coassociativity, we have $$h_{11} v_{-1} S(h_{1,3}) h_2 \otimes h_{1,2}.v_0 \\ = h_1 v_{-1} S(h_3) h_4 \otimes h_2.v_0 \\ = h_1 v_{-1} \epsilon(h_3) \otimes h_2.v_0 \\ = h_1 v_{-1} \otimes \epsilon(h_3) h_2.v_0 \\ = h_1 v_{-1} \otimes h_2.v_0$$ – Jianrong Li Aug 6 '16 at 7:46