• Comonad
• todo list

$\gdef\ccal{\mathcal{C}}$ $\gdef\dcal{\mathcal{D}}$ $\gdef\lra{\leftrightarrow}$ $\gdef\xto{\xrightarrow}$ $\gdef\Om{\Omega}$

What a comonad? Given two categories $$ L: \ccal \lra \dcal: R$$ We can form the comonad $\Om = LR \in End(\dcal)$, which have $$ \epsilon: \Om \to 1_\dcal, \quad L \eta R: \Om \to \Om \circ \Om $$

What is a comodule over $\Om$? We need an object $X$ in $\dcal$, with the structure map $$ a: X \to \Om X $$ that satisfies conditions like $$ X \xto{a} \Om X \xto{\epsilon} X == id_X $$ and one more condition.... $$ X \xto{a} \Om X \rightrightarrows \Om \Om X $$ where the second step is like, either stretch out $\Om$, or pull out from $X$.

That's a lot of conditions, you see. It is not easy to be a comodule.

Consider sheaves on two points map to one point, we have functors $L = p_!, R = p^!, \Om = LR$ $$ \Om(V) = V \oplus V. $$ What's the structure maps on $\Om$?

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 $(T_1, T_2) = (0, 1)$.

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 $L (V,W) \mapsto LRL(V,W)$ $$ V \oplus W \to (V_1 \oplus W_1) \oplus (V_2 \oplus W_2). $$ so there is only one reasonable thing to do.