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--- blog:2023-03-10 [2023/03/11 05:12] – pzhou
+++ blog:2023-03-10 [2023/06/25 15:53] (current) – external edit 127.0.0.1
@@ Line 31: / Line 31: @@
 出了趟门，得瑟了。我对$\Om(V)$说。$V$说，不是我的错，$\Om$对谁都这样。
+$\Om$ automatically comes with structure maps
+$$ \Om(V) \to V, \quad LR V \to V, \quad id_{RV}: RV \to RV $$
+now, why we have adjoint? I think it must be the summation $ V \oplus V \to V$,  there is nothing better than summation. It generalizes to many points.
+and we have another one
+$$ \Om(V) \to \Om \Om V, \quad V\oplus V \to V\oplus V \oplus V\oplus V. $$
+How does this work? We know that $LR(V) \to L (RL) R(V)$, so if we have $(W_1, W_2) \in \ccal$, what does $W \to RL(W)$ do? $RL(W) = (W_1 \oplus W_2,W_1 \oplus W_2)$, so it must be the.
+Let's do a bit marking. $\Om(V) = V e_1 \oplus V e_2$, so we keep track of which vector space is sitting at which point. So, we have
+$$ R(V) \to RLR(V), \quad (V_1, V_2) \to (V_1 \oplus V_2, V_1 \oplus V_2) $$
+so when they get pushed down, we have
+$$ V_1 \oplus V_2 \to V_{11} \oplus V_{12} \oplus V_{21} \oplus V_{22} $$
+and this map is
+$$ V_1 \to V_{11}, \quad V_2 \to V_{22} $$
+the subscript is the vector space's travel passport, where she keeps with her, and collect a stamp whenever she vist upstairs. (each time, we get $p^!$, she would split up, the universe bifurcate).
+Let's try to build some $\Om$ comodule, basically, we need to decide how to map
+$$ V \to \Om V = V_1 \oplus V_2 $$
+suppose we have some linear transformation $v \mapsto T_1 v \oplus T_2 v$.
+Assum $V$ is rank $1$. Then $T_i$ are scalar, and we have $T_1 + T_2 = 1$.
+Then, we check the $V \to \Om V \to \Om \Om V$, if we use $\Om$ to split, then we have
+$$ v \mapsto (T_1 v, T_2 v) \mapsto (T_1 v, 0; 0, T_2 v) $$
+if we use co-action of $V$, then we get
+$$ v \mapsto (T_1 v, T_2 v) \mapsto (T_1 T_1 v, T_1 T_2 v; T_2 T_1 v, T_2^2 v) $$
+hence, we need to have $T_i^2 = T_i, T_1 T_2=0$. The only chance here is that $(T_1, T_2)= (1,0)$ or $(T_1, T_2) = (0, 1)$.
+That's only for rank-1 comodule. How about
+You know what is the easiest way to get comodule? Just take the image of $L$. Suppose we have $(V, W)$ upstairs, and we get $V \oplus W$ downstairs. Then, we need to know how does $L (V,W) \mapsto LRL(V,W)$
+$$ V \oplus W \to (V_1 \oplus W_1) \oplus (V_2 \oplus W_2), (v,w) \mapsto ( (v,0), (0, w)) $$
+so there is only one reasonable thing to do.
+Summary, what is the comonad structure? It is equip $\Om V$ with the trail of paths (here two paths) as it moves up and down.
+What is the comodule structure? It is dividing $v$ into $v=v_1 + v_2$. Some sort of idempotent projection.
+==== Another example ====
+Consider the skeleton of a cross, and consider the localization to the vertical line and horizontal line.
+Danny says, these two localization functors preserves limit (left exact), hence satisfies Barr-Beck.
+when you kill a linking disk (cocore to some non-compact component of a skeleton), everybody will do negative Reeb flow to snap on the skeleton again. How do I know if that motion preserves limit?
+The word 'preserve limit' is too abstract. (why do I want to preserve limit? and what kind of limit do I care? finite limit? infinite limit? homotopy limit?)
+So, don't worry about the words if you don't know what it means. Let's just see, if you can understand the comodule.
+So, we understand $\dcal$, which is just $Vect \times Vect$. And the comodules $\Omega$ is doing what? It is
+$$ \Omega(V_1 ,  V_2) = (V_1 + V_2[1], V_2 + V_1) $$
+so it is clear, what is the co-unit map. And what is $\Om (V_1 ,  V_2) \to \Om \Om (V_1 ,  V_2)$? First, what is $