Peng Zhou

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====== 2024-07-23 ======

  * Goal: understand Yetter-Drinfeld module, Drinfeld double and then Andy Manion's comment paper


===== What is YD-module =====
Let $H$ be a Hopf algebra, and $Rep(H)$ is the cat of finite dim rep. Let $(Y, \varphi)$ be such central element, then we have the following rep
$$ \varphi: H \otimes Y \to Y \otimes H  $$
As written this is a rep H morphism. We may restrict to $ \{1\} \otimes Y$, and get a linear map
$$ \tau: Y \to Y \otimes H. $$
This has the potential to be a (right) co-action of $H$ on $Y$. 

To check that this is indeed a co-action, we need to do splitting twice. either we split $Y$ twice, or the second split is on $H$. 

Consider the following example: $H = \C G$ where $G$ is a finite group. Let $(Y, \varphi)$ be given as a $G$-graded $G$-rep, namely $Y = V = \oplus_g V_g$, and for $v_g \in V_g$, $h \in G$,  we have $h \cdot v_g  \in V_{h g h^{-1}}$. Note that any $G$ rep can be given a trivial $G$-gradation. Now, given another $G$-rep $W$, we can do 
$$ \varphi: W \otimes V \to V \otimes W, \quad w \otimes v_g \mapsto v_g \otimes g \cdot w. $$
Now, we need to check that this commute with $G$-action. OK, it does. 

Now, we apply this to $W=H$, where $G$ acts on $\C G$ by left multiplication. Then we have 
$$ \tau: V \to V \otimes \C G, \quad v_g \to v_g \otimes g. $$
Now it is clear that this indeed is a co-action. 

Now, we should check that the action is compatible with the co-action. We faces the funny condition: 
$$ \tau(h v_g) = h v_g \otimes h g h^{-1}. $$
OK, this indeed works. I don't know the general rule for co-product, but at least here, for the co-commutative coproduct, this works. 

Let's try taking $H = O(G)$. Let $(Y, \varphi)$ be given as a $G$-graded $G$-rep. Then $O(G)$-action on $Y$ is pointwise multiplication. $O(G)$ has an interesting co-product, hence $Vec_G$ has an interesting monoidal structure, basically the grading multiply when they take product. Then, it is naturally true that, $W_1 \otimes W_2$ is not isomorphic to $W_2 \otimes W_1$: the grading won't match. Now comes $(Y, \varphi)$, say $Y = V = \oplus_g V_g$. Say $W = \C \delta_h$, a rank 1 skyscraper at position $h \in G$, then 
$$ (V \otimes W)_g = V_{gh^{-1}}, \quad (W \otimes V)_g = V_{h^{-1} g} $$
To have an isomorphism, we need to get for each $g$, an isomorphism $V_{gh^{-1}} \to V_{h^{-1} g}$, well, this is given by the action of $h^{-1}$ on $V$. 

Now, we consider $1 \in H$, this is represented as the constant function $1 \in O(G)$. We have $1 = \sum_h \delta_h$ (constant function is the sum of delta functions). So, we just need to consider what is $\delta_h \otimes v_g \mapsto ?$. Now this will go to $ (h v_g) \otimes \delta_h \in V_{hgh^{-1}} \otimes H_h$. Put these all together, we have
$$ \tau(v_g) = \sum_h (h v_g) \otimes \delta_h. $$
We first check that this is a co-action, we can do the comodule-split again, to get
$$ (\tau \otimes 1) \tau(v_g) = = \sum_{h_1, h_2} h_2 (h_1 v_g) \otimes \delta_{h_2} \otimes \delta_{h_1}. $$
Or, we can do the co-mult on $H$ to get
$$ (1 \otimes \Delta) \tau(v_g) = \sum_{h_1, h_2} (h_1 h_2 v_g) \otimes (\delta_{h_1} \otimes \delta_{h_2}).$$
So they match. Finally, we check the action-coaction compatibility. Somehow, for Hopf algebra, there is also action-coaction compatibility, that involves $S$. Given a $H$ element $h$ and $V$ element $v$, we can 
$$ \tau(h \cdot v) = (hv)_Y \otimes (hv)_H $$
and we can do split then combine, somehow we need to split $h$ to three parts, and $S$ the third part, we want to
$$ h_2 v_1 \otimes h_3 v_2 S^{-1}(h_1). $$
Now, let's test it out. Take $h = \delta_h$, $v=v_g$, then $\delta_h v_g = v_g$ if $h=g$ else $0$. Suppose $h=g$,then 
$$ \tau(\delta_g v_g) = \sum_h h v_g \otimes \delta_h $$
and 
$$ h_2 v_1 \otimes h_1 v_2 S(h_3) = \delta_{h g h^{-1}} (h v_g) \otimes \delta_{h_3} \delta_{h} \delta_{h_1^{-1}} $$
where $h_2 = h g h^{-1}$, $h_3=h$, $h_1^{-1} = h$. Then, $h_1 h_2 h_3 = g$. good, it works. Notice that it matters whether we are working with (left-right) YD module, or left-left YD module. 

Now we have seen that this works in the two examples, how do we prove it in general? The funny things is the anti-pode. It is sort like inverse