# A diagram for understanding action/coaction compatibility in a Yetter-Drinfeld module

For a Hopf algebra $H$ with antipode $S$, let $M$ be a left $H$-module with the action $h \otimes m \mapsto \rho(h,m)$, and also a left $H$-comodule with coaction $\delta \colon m \mapsto m^{(-1)} \otimes m^{(0)}$. For $M$ to be a Yetter-Drinfeld module, it must satisfy the compatibility condition $$\delta(\rho(h,m)) = h_{(1)}m^{(-1)} S(h_{(3)}) \otimes \rho(h_{(2)},m^{(0)})\,.$$ Is there a natural commutative diagram to draw that illustrates this compatibility condition? I've included my attempt below in a CW answer. Also, is there any more reason behind this condition besides "it's the condition we need to be true to get the nice braiding to work out in the Yetter-Drinfeld category?".

## migrated from math.stackexchange.comJan 29 '18 at 17:16

This question came from our site for people studying math at any level and professionals in related fields.

• As for the last question: The category of Yetter-Drinfeld modules of $H$ is equivalent to the category of modules over the Drinfeld double $D(H)$. But I guess this merely shifts your question to why the Drinfeld double is defined the way it is. – Mathematician 42 Feb 3 '18 at 6:06

$$\require{AMScd} \begin{CD} H \otimes M @>{\rho}>> M @>{\delta}>> H \otimes M \\ @V{\Delta^2 \otimes \delta}VV @. @AA{m^2 \otimes \rho}A \\ H^{\otimes 4} \otimes M @>>{\mathbb{1} \otimes T \otimes \mathbb{1}}> H^{\otimes 4} \otimes M @>>{\mathbb{1}\otimes\mathbb{1}\otimes S \otimes\mathbb{1}\otimes\mathbb{1}}> H^{\otimes 4} \otimes M \end{CD}$$
In this diagram, the Hopf algebra is associative and coassociative so the maps $$\Delta^2 \colon H \to H \otimes H \otimes H \quad\text{and}\quad m^2 \colon H \otimes H \otimes H \to H$$ are well defined. The symbol $H^{\otimes 4}$ is short for $H \otimes H \otimes H \otimes H$, and $T$ is an "outer twist" map where $T(a \otimes b \otimes c) = c \otimes b \otimes a$ . Or if you prefer, if $\tau$ is the usual twist map $\tau(a \otimes b) = b \otimes a$, then we can write $T$ as $(\mathbb{1}\otimes{\tau})(\tau\otimes\mathbb{1})(\mathbb{1}\otimes{\tau})$, which I think relates this more closely to the braiding we want in the category.
The equivalent way of writing the Yetter-Drinfeld compatibility condition between left action and left coaction described in this post can be easily translated into a commutative diagram. It looks like this $$(\nabla\otimes\rho)(1\otimes\tau\otimes 1)(1\otimes\delta)=(\nabla^\mathrm{op}\otimes 1)(1\otimes\delta)(1\otimes\rho)(\Delta^\mathrm{op}\otimes 1)\,.$$ Similary equations work for $${}_H\mathcal{YD}^H$$ instead of $${}_H^H\mathcal{YD}$$, see for example Kassel, Definition IX.5.1, page 220. These equations tell us how $$\rho$$ and $$\delta$$ "commute". We need precisely this condition to compute the Drinfeld center of $${}_H\mathrm{Mod}$$.