Timeline for How to show that Yetter-Drinfeld condition is equialent to $\Psi$ is a braiding
Current License: CC BY-SA 3.0
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Jul 11, 2016 at 19:54 | comment | added | Zahlendreher | Given the braiding as defined in your question, these conditions give that $\triangleleft$ is an action and $w\mapsto w_{(0)}\otimes w_{(1)}$ is a coaction. This alone does not apply the YD-condition. | |
| Jul 7, 2016 at 3:16 | comment | added | Jianrong Li | can the conditions $$\Psi_{V,W\otimes Z}=(id_W\otimes \Psi_{V,Z})(\Psi_{V,W}\otimes id_Z),$$ $$\Psi_{V\otimes W, Z}=(\Psi_{V,Z}\otimes id_W)(id_V\otimes \Psi_{W,Z})$$ imply the YD-condition? | |
| Jun 26, 2016 at 12:51 | comment | added | Zahlendreher | Think about from where to where the functors $\otimes$ and $\otimes^{op}$ go. In order to apply naturality, we need to have morphisms in the correct category, which here is $H$-modules. | |
| Jun 26, 2016 at 3:37 | comment | added | Jianrong Li | thank you very much. Why we need the condition that $\Psi_{V, V}$ is an $H$-module to apply the naturality? | |
| Jun 25, 2016 at 22:01 | comment | added | Zahlendreher | The braiding is a natural isomorphism $\Psi: \otimes \to \otimes^{op}$. It being natural means that we have the following relation for any morphisms $f:V\to V'$, $g:W\to W'$ of $H$-modules: $\Psi_{V',W'}(f\otimes g)=(g\otimes f)\Psi_{V,W}$. Now we apply this to the morphisms $f\otimes g=Id_V\otimes \Psi_{V,V}$. This can be done because $\Psi_{V,V}$ is a morphism of $H$-modules. | |
| Jun 25, 2016 at 14:07 | comment | added | Jianrong Li | thank you very much. How to apply the naturality of the braiding to $\Psi_{V, V}$ to prove that $\Psi_{V, V \otimes V} (1 \otimes \Psi_{V, V}) = (\Psi_{V, V} \otimes 1) \Psi_{V, V \otimes V}$? | |
| Jun 22, 2016 at 3:54 | vote | accept | Jianrong Li | ||
| Jun 21, 2016 at 14:40 | history | answered | Zahlendreher | CC BY-SA 3.0 |