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--- blog:2023-04-17 [2023/04/18 19:33] – pzhou
+++ blog:2023-04-17 [2023/06/25 15:53] (current) – external edit 127.0.0.1
@@ Line 3: / Line 3: @@
 ===== Two adjoints =====
 Say $X = \C$ and $U = \C^*$, $j: U \to \C$. Do you know what is $j^*$ and $j^!$ of $O_X$?
+Actually, I don't really care about the $j^!$ (since it is a right-adjoint).
 https://arxiv.org/pdf/1607.02064.pdf
@@ Line 45: / Line 47: @@
 So, what should I say, $Hom(A, -)$, because $A$ is compact, preserves both colimit and limit. So, there are two adjoints $Mod-End(A)$ back to $R-mod$. This is like $i^*$, for a closed guy. There is the $- \otimes_E A$ (since $R$-action commute with $E$-action on $A$, so we have a left $R$-mod). There is the $Hom_{mod-E}(A^\vee, -)$, where $A^\vee$ is a right $R$-mod, so dualize, we get a left $R$-mod.
+So, what is a $mod-E$? suppose $E$ is this cdga, $A \oplus A[-1]$. How to compute $Hom_E(A,A)$? We may compute free resolution of $A$, by
+$$ \cdots E[-2] \to E[-1] \to  E \to A \to 0 $$
+Then, when we hom this free resolution to $A$, we get back $A \oplus A \oplus A \cdots $. Somehow, this infinty should be a projective limit, we should map out of a colimit, and if we truncate the resolution, it should map to the longer resolution. So, function on a fatter point (meaning longer resolution), maps to function on a smaller point. That's why we end up with the completion. $\C[[x] ]$.
+Now, What is $A^\vee = Hom(A, R)$? It is somehow $A$ shifted up by $1$.