What is the k-linear structure on the derived infinity category of quasi-coherent sheaves?

Question

Let $f : X \overset{f}{\rightarrow} Y \overset{g}{\rightarrow} \mathrm{Spec} (k)$ be morphisms of schemes (feel free to add any hypothesis necessary). Let $\mathrm{QCoh}(Y)$ denote the derived (stable) $\infty$-category of quasi-coherent sheaves, which carries a natural symmetric monoidal structure $\mathrm{QCoh}(Y)^\otimes \rightarrow \mathrm{Fin}_*$ whose tensor product is the derived tensor product of quasi-coherent sheaves. Then one can extend $f^* : \mathrm{QCoh}(Y) \rightarrow \mathrm{QCoh}(X)$ to a symmetric monoidal functor with a lax monoidal right adjoint $f_*$ and, of course, we have a similar situation for $g$.

Now, my question is twofold:

1. How do I define the $\mathrm{QCoh}(k)$-linear structure of $\mathrm{QCoh}(Y)$? I can think of two ways one could approach this: Firstly, one could exhibit $\mathrm{QCoh}(Y)^{\otimes, L} \rightarrow \mathrm{QCoh}(Y)^\otimes$ as tensored over itself (see 2.1.3 of DAGII) and them pull this map back along $g^* : \mathrm{QCoh}(k)^\otimes \rightarrow \mathrm{QCoh}(Y)^\otimes$ to obtain a map $M \rightarrow \mathrm{QCoh}(k)^\otimes$. Secondly, one could perhaps use a symmetric version $\mathbf{Mod}^\otimes$ of Lurie's $\infty$-operad $\mathbf{LM}^\otimes$, but here I fail to construct a corresponding coCartesian fibration $M' \rightarrow \mathbf{Mod}^\otimes$ of $\infty$-operads. Perhaps we do not even need an $\infty$-operad, and something like a coCartesian fibration $M'' \rightarrow \Delta^1 \times \mathrm{Fin}_*$ would do the trick?

2. Whatever approach chosen above should be robust enough to first realize $f^*$ as $\mathrm{QCoh}(k)$-linear and subsequently its adjoints $f_*$ and $f^!$ (where now additional hypotheses on $f$ are required, I'm sure), by applying some criterion for relative adjoints as developed in Higher Algebra, Section 7.3.2

I hope that someone can clear up my confusion. Thank you for your time!

Answer 1

For any symmetric monoidal $\infty$-category $\mathcal C$ and commutative algebra $A\in \mathcal C$ (you're supposed to think of $\mathcal C$ being $Cat_\infty$ itself, or maybe $Pr^L$; and $A= QCoh(Spec(k)) = Mod_k$), there is an equivalence: $CAlg(\mathcal C)_{A/}\simeq CAlg(Mod_A(\mathcal C))$, where the right hand side is to be interpreted as you expect if $\mathcal C$ has enough appropriate colimits, and can be made sense of operadically in general).

This in particular means that if you have a diagram $A\to B\to C$ of commutative algebras, then $B,C$ can be made into $A$-modules canonically (in fact into commutative $A$-algebras, but you can forget most of that structure), and the map $B\to C$ can be made $A$-linear canonically.

You can apply this to $QCoh(-)$ applied to $X\to Y\to Spec(k)$ and get a canonical $Mod_k$-linear structure on $f^*:QCoh(Y)\to QCoh(X)$. By naturality, this coincides with the approach you describe in (1).

Now, given a symmetric monoidal $\infty$-category $A$, and an $A$-linear functor $f: M\to N$, if $f$ admits a right adjoint $f^R$, then $f^R$ is canonically lax-$A$-linear - this follows from the theory of relative adjunctions in Higher Algebra, but concretely boils down to the existence of (coherent, natural, compatible etc.) maps $a\otimes f^R(n)\to f^R(a\otimes m)$, which are simply the usual projection maps, the mate of $f(a\otimes f^R(m))\cong a\otimes ff^R(m)\to a\otimes m$, where the first equivalence is the $A$-linearity of $f$, and the second map is the co-unit of the $f\dashv f^R$-adjunction. It then becomes a property for $f^R$ to be $A$-linear, namely the property that these projection maps be isomorphisms.

A cool thing to note is that this projection map is always an isomorphism when $a\in A$ is dualizable. So in your situation, $f_*$ is always $Perf(k)$-linear. If furthermore $f_*$ preserves colimits, then because $Mod_k$ is generated under colimits by $Perf(k)$, it follows that $f_*$ is $Mod_k$-linear; and this is essentially an "if and only if" (if $f_*$ is $Mod_k$-linear, one can prove that it preserves arbitrary coproducts and thus arbitrary colimits). This happens quite often, e.g. if $X,Y$ are qcqs, but presumably more generally too.

I'm not sure what $f^!$ is in an algebro-geometric context, but pesumably some similar game can be played there.