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--- blog:2024-07-23 [2024/07/23 18:36] – pzhou
+++ blog:2024-07-23 [2024/07/23 22:44] (current) – [What is YD-module] pzhou
@@ Line 32: / Line 32: @@
 $$ \tau(v_g) = \sum_h (h v_g) \otimes \delta_h. $$
 We first check that this is a co-action, we can do the comodule-split again, to get
-$$ \sum_{h_1, h_2} h_2 (h_1 v_g) \otimes \delta_{h_2} \otimes \delta_{h_1}. $$
+$$ (\tau \otimes 1) \tau(v_g) = = \sum_{h_1, h_2} h_2 (h_1 v_g) \otimes \delta_{h_2} \otimes \delta_{h_1}. $$
 Or, we can do the co-mult on $H$ to get
-$$ \tau(v_g) = \sum_{h_1, h_2} (h_1 h_2 v_g) \otimes (\delta_{h_1} \otimes \delta_{h_2}).$$
+$$ (1 \otimes \Delta) \tau(v_g) = \sum_{h_1, h_2} (h_1 h_2 v_g) \otimes (\delta_{h_1} \otimes \delta_{h_2}).$$
+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