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

This paper discusses general stuff. Very readable, except the notation needs some getting used to. The setup is as following : $X$ is some stable $\infty$-category. No need to assume compactly generated. $U$ is some subcategory which is both reflexive and coreflexive, which means $j_*$ admits both left-adjoint $j^*$ and right-adjoint $j^!$ (I don't know which term corresponds to which condition)

We define $Z^L$ and $Z^R$ as subcategory of $X$. $Z^L = Z^\vee$, is something $\Hom(Z^L, U) = 0$. Similarly, $Z^R = Z^\wedge$ satisfies $\Hom(U, Z^R)=0$.

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 $Z^R$. 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.

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 $1$.