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--- blog:2023-04-17 [2023/04/18 06:03] – 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 12: / Line 14: @@
 Consider the original example. Consider  $\C[x]/(x) = O_0$ . Does  $U$  map to it? This module $\C[x][1/x] $is a colimit of$ \C[x] $. To define a map out of a colimit, is to define a map from every member to it, in a consistent way. Say, if$ 1 \in \C[x,1/x] $maps to$ 1 \in \C[x]/x $, then we need to ask where does$ 1/x $maps to? No place, so we are dead, so$ U$ does not map to it.
-Does this map to  $U$ ? I don't think so, this would give $Hom(\C[x]/(x), U) = U \xto{x} U = 0 $. So, either way, this guys is in$ Z^L $and$ Z^R $. How about$ \C[[x] ] =\lim \C[x]/(x^n)$? It is complete. It must be in  $Z^R$ . Consider the $Hom(U, Z^R)$,
+Does this map to  $U$ ? I don't think so, this would give $Hom(\C[x]/(x), U) = U \xto{x} U = 0 $. So, either way, this guys is in$ Z^L $and$ Z^R $. How about$ \C[[x] ] =\lim \C[x]/(x^n)$? It is complete. It must be in  $Z^R$ . We should have
+ $j_!j^! \to id \to i_\wedge i^\wedge$
+where  $i^\wedge -| i_\wedge$ , so  $i_\wedge$  only admit left-adjoint, so we say  $Z^\wedge= Z^R$  is reflexive, meaning 'pullback to it' is a left-adjoint.
+Similarly, we get
+ $i_\vee i^\vee \to id \to j_* j^*$
+consist of things set theoretically supported on  $Z$ .
+I don't know the difference.
+===== A-torision and A-completion =====
+{{ :blog:dwyer-greenlees.pdf |Dwyer-Greenlees: A-torsion and A-completion. }}
+Let  $R$  be a ring (maybe not commutative). Let  $A$  be a chain complex of  $R$ -module. Define
+  * A-nil objects, as the right-orthogonal of  $A$ . Call this subcategory as  $C$ . This is like
+ $L: R-mod \to C = R-mod / \langle A \rangle$  then use the right-adjoint  $\iota: C \to R$  to embed it back.
+  * The left-orthogonal of  $C$ , is called A-torsion objects, they can be built by cobble up  $A$  and its suspension. We use  $\langle A \rangle$  to denote it. For example $\C[x,x^{-1}] / \C[x]$, the singular terms. It is a torsion module, meaning each element in it is torsion. (we don't say that it is killed by everyone).
+  * The right-orthogonal of  $C$ . There are the  $A$ -completion objects. For example,  $C = \Z[1/p]$ ,  $A = \Z/p$ . What do I mean by  $A$ -completion? $$\Z_p^\wedge = \lim (\cdots \to \Z/(p^n) \to \Z/(p^{n-1}) \to  \cdots \to \Z/p).$$ We know that $Hom(C, \Z/(p^n)) = 0  $. Right slot plays nicely with limit, hence$ Hom(C, \Z_p^\wedge) = 0$
+  * Q: is it true that $Hom(\Z_p^\wedge, C) $may not be zero? Yes, say$ Hom(\Z_p^\wedge, \Z_p^\wedge[1/p])$
+  * Q: is it true that  $Hom(C, \langle A \rangle)$  may not be zero? well, you are mapping to the colimit, even though each term you get zero, but the colimit is something large, and we have the tautological map $$ \C[x,1/x] \to \C[x,1/x]/\C[x] $$
+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  $A \oplus A[-1]$ .
+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  $1$ .