# How to show that Yetter-Drinfeld condition is equialent to $\Psi$ is a braiding

Let $H$ be a Hopf algebra and $V$ a right $H$-module and right $H$-comodule. The module $V$ is a Yetter-Drinfeld module over $H$ if and only if \begin{align} ( v \triangleleft h_{(2)} )_{(0)} \otimes h_{(1)} ( v \triangleleft h_{(2)} )_{(1)} = ( v_{(0)} \triangleleft h_{(1)} ) \otimes v_{(1)} h_{(2)}. \quad (1) \end{align}

Let $\Psi: U \otimes W \to W \otimes U$ be a braiding given by \begin{align} \Psi(u \otimes w) = w_{(0)} \otimes ( u \triangleleft w_{(1)} ). \end{align} Then \begin{align} \Psi_{12} \Psi_{23} \Psi_{12} = \Psi_{23} \Psi_{12} \Psi_{23}. \end{align} The algebra $H$ is in the right-right Yetter-Drinfeld category $YD^H_H$. Suppose that $V \in YD^H_H$. Let $\Psi: V \otimes H \to H \otimes V$ be a braiding. Let $u \in H, v \in V, h \in H$. Then \begin{align} & \Psi_{12} \Psi_{23} \Psi_{12} (u \otimes v \otimes h) \\ & = h_{(0)} \otimes (v_{(0)} \triangleleft h_{(1)}) \otimes ( u \triangleleft v_{(1)} h_{(1)} ) \quad (2) \\ & \Psi_{23} \Psi_{12} \Psi_{23} (u \otimes v \otimes h) \\ & = h_{(0)} \otimes (v \triangleleft h_{(2)})_{(0)} \otimes ( u \triangleleft h_{(1)} ( v \triangleleft h_{(2)} )_{(1)} ). \quad (3) \end{align} How to show that $\Psi$ is a braiding is equavalent to the compatibility condition (1)? My problem is: how to remove $u$ in (2) and (3)? Thank you very much.

The YD-condition is not equivalent to $\Psi$ being a braiding. The condition (1) is related to $\Psi$ being a morphism of $H$-modules (see the more detailed answer to your other question Reference request: compatibility conditions of four versions of Yetter-Drinfeld modules).

The Quantum Yang-Baxter equation is a consequence of the action and the coaction conditions (and the YD-condition), which are of course part of the YD-structure. These give the two hexagon axioms of a braiding (see standard references on braided monoidal categories for these conditions). $$\Psi_{V,W\otimes Z}=(id_W\otimes \Psi_{V,Z})(\Psi_{V,W}\otimes id_Z),$$ $$\Psi_{V\otimes W, Z}=(\Psi_{V,Z}\otimes id_W)(id_V\otimes \Psi_{W,Z}).$$ However, the two hexagon axioms are stronger than the quantum Yang-Baxter equation $$\Psi_{12}\Psi_{23}\Psi_{12}=\Psi_{23}\Psi_{12}\Psi_{23}.$$ This is proved by observing that $\Psi_{12}\Psi_{23}=\Psi_{V,V\otimes V}.$ This allows us to rewrite the right right hand side as $\Psi_{V,V\otimes V}(id_V\otimes \Psi_{V,V})$, but by naturality of the braiding applied to $\Psi_{V,V}$ (which is a morphism of $H$-modules by the YD-condition (1) and hence naturality applies) this equals $( \Psi_{V,V}\otimes id_V)\Psi_{V,V\otimes V}$ which can be translated into the left hand side by the same reasoning.

• thank you very much. How to apply the naturality of the braiding to $\Psi_{V, V}$ to prove that $\Psi_{V, V \otimes V} (1 \otimes \Psi_{V, V}) = (\Psi_{V, V} \otimes 1) \Psi_{V, V \otimes V}$? – Jianrong Li Jun 25 '16 at 14:07
• 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. – Zahlendreher Jun 25 '16 at 22:01
• thank you very much. Why we need the condition that $\Psi_{V, V}$ is an $H$-module to apply the naturality? – Jianrong Li Jun 26 '16 at 3:37
• 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. – Zahlendreher Jun 26 '16 at 12:51
• 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? – Jianrong Li Jul 7 '16 at 3:16

We can also prove that Yetter-Drinfeld condition implies that $\Psi$ is a braiding as follows.

Let $u, v, w \in V$. Then

\begin{align} & \Psi_{12} \Psi_{23} \Psi_{12} (u \otimes v \otimes w) \\ & = \Psi_{12} \Psi_{23}( (u_{(-1)}.v) \otimes u_{(0)} \otimes w ) \\ & = \Psi_{12} ( (u_{(-1)}.v) \otimes (u_{(0)})_{(-1)}.w \otimes (u_{(0)})_{(0)} ) \\ & = ( u_{(-1)}.v )_{(-1)}.( ( (u_{(0)})_{(-1)} ).w ) \otimes (u_{(-1)}.v)_{(0)} \otimes ( u_{(0)} )_{(0)} \\ & = ( (u_{(-1)})_{(1)}.v )_{(-1)}.( (u_{(-1)})_{(2)}.w ) \otimes ((u_{(-1)})_{(1)}.v)_{(0)} \otimes u_{(0)} \\ & = ( ((u_{(-1)})_{(1)}.v )_{(-1)} (u_{(-1)})_{(2)} ).w \otimes ((u_{(-1)})_{(1)}.v)_{(0)} \otimes u_{(0)}, \end{align} \begin{align} & \Psi_{23} \Psi_{12} \Psi_{23} (u \otimes v \otimes w) \\ & = \Psi_{23} \Psi_{12} (u \otimes v_{(-1)}.w \otimes v_{(0)}) \\ & = \Psi_{23} ((u_{(-1)}v_{(-1)}).w \otimes u_{(0)} \otimes v_{(0)}) \\ & = (u_{(-1)} v_{(-1)}).w \otimes (u_{(0)})_{(-1)}.v_{(0)} \otimes ( u_{(0)} )_{(0)} \\ & = ( (u_{(-1)})_{(1)} v_{(-1)}).w \otimes (u_{(-1)})_{(2)}.v_{(0)} \otimes u_{(0)}. \end{align}

Therefore the Yetter-Drinfeld condition \begin{align} & (h_{(1)}.v )_{(-1)} h_{(2)} \otimes (h_{(1)}.v)_{(0)}\\ & = h_{(1)} v_{(-1)} \otimes h_{(2)}.v_{(0)}. \end{align} implies that (every $u_{(-1)}$ is some $h \in H$) \begin{align} & ((u_{(-1)})_{(1)}.v )_{(-1)} (u_{(-1)})_{(2)} \otimes ((u_{(-1)})_{(1)}.v)_{(0)} \otimes u_{(0)} \\ & = (u_{(-1)})_{(1)} v_{(-1)} \otimes (u_{(-1)})_{(2)}.v_{(0)} \otimes u_{(0)}. \end{align} Therefore \begin{align} & \Psi_{12} \Psi_{23} \Psi_{12} (u \otimes v \otimes w) = \Psi_{23} \Psi_{12} \Psi_{23} (u \otimes v \otimes w). \end{align}

• According to the previous answer, braiding requirements are strictly stronger than the identity you prove here (qYBe), no? – მამუკა ჯიბლაძე Mar 4 '17 at 8:10
• @მამუკაჯიბლაძე, yes, Yetter-Drinfeld condition implies that $\Psi$ is a braiding. Maybe $\Psi$ is a braiding does not imply Yetter-Drinfeld condition. – Jianrong Li Mar 4 '17 at 8:32
• I meant different thing - that $\Psi_{12}\Psi_{23}\Psi_{12}=\Psi_{23}\Psi_{12}\Psi_{23}$ is weaker than braiding condition – მამუკა ჯიბლაძე Mar 4 '17 at 9:42