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--- blog:2023-04-17 [2023/04/18 18:55] – [A-torision and A-completion] 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 36: / Line 38: @@
 Now that we know these are different stuff. Torsion is left-orthogonal, but can receive map; completion is right-orthogonal, but can map out from.
-The next thing we say is that, both categories are equivalent to $E-mod$, where $E = End(A)$. Say $A=\C[x]/(x)$, then $E$ is the exterior algebra, of $A \oplus A[-1]$. (are there any subtleties?)
+The next thing we say is that, both categories are equivalent to $E-mod$, where $E = End(A)$. Say $A=\C[x]/(x)$, then $E$ is the exterior algebra, of $A \oplus A[-1]$.
+There is a mysterious duality functor. We define $A^\vee = Hom_R(A, R)$. We know $A$ is a left R-mod. This has used up all things we can put on $A$. But we can act on $R$ from the right. Given a function $f \in Hom(A, R)$, we can act on it by an element $r \in R$. By $(rf)(a)= r(f(a))$, but this is equal to $f(ra)$. However, this $(rf)$ is no longer a left R-morphism, since for any $s \in R$, we need $s(rf)(a) = (rf)(sa)$, which is saying $sr=rs$. Not always true!
+The better way is to act on the right, using $(fr) (a) := (f(a))r$. This ways, when we check $s(fr)(a) = (fr)(sa)$, we just need to unpack both side as $s f(a) r = f(sa) r $.
+About $A$ as an $End(A)$-mod, is it a left or right $E$-mod? By default, when given two endomorphism $f_1$, $f_2$, we compose $f_1 f_2$ means we apply $f_2$ first, so that is $(f_1 f_2)(a) = f_1(f_2(a))$. Unless the definition of composition is defined differently. And, so $Hom_R(A, M)$ is a right $E$-mod, because given any $\phi \in Hom_R(A,M)$, we can define $(\phi f)(a) = \phi(f(a))$, this commute with $R$-action, indeed, $r(\phi f)(a) = r \phi(f(a)) = \phi( r (f(a)) = \phi(f(ra)) $. So, in general, $Hom(A,B)$ is a $(End(B),End(A))$ bimodule.
+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 $1$.