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--- blog:2024-07-23 [2024/07/23 18:01] – created pzhou
+++ blog:2024-07-23 [2024/07/23 22:44] (current) – [What is YD-module] pzhou
@@ Line 21: / Line 21: @@
 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
+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