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By the Gabriel-Rosenberg-Brandenberg-Gabber Theorem, sufficiently nice schemes $X$ are determined up to isomorphism by the category $\text{QCoh}(X)$.

My question: for sufficiently nice $X$, can you determine $\text{Aut}(X)$ from $\text{QCoh}(X)$?

Kevin Casto's answer: Yes of course, because by GRBG you can determine $X$, then take its automorphism group.

Any automorphism $\varphi$ induces an autoequivalence of $\text{QCoh}(X)$, so we get a homomorphism $\text{Aut}(X)\to\text{Aut}(\text{QCoh}(X))$ which is probably injective, but not surjective.

My question 2: How do you characterise the image of $\text{Aut}(X)$ inside $\text{Aut}(\text{QCoh}(X))$? What examples do we get if we replace quasicoherent sheaves with other interesting categories?

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  • $\begingroup$ I mean, since by the theorem you mention you can determine $X$ up to isom, then you can determine $\mathrm{Aut}(X)$ from $X$-up-to-isom, so the answer is yes. But it sounds like what you're actually asking is if there's a more direct way to determine/characterize $\mathrm{Aut}(X)$, for instance as a subset of $\mathrm{Aut}(\mathrm{QCoh}(X))$ as you mention. $\endgroup$ Commented Aug 1 at 18:33

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If you remember the tensor product then in nice cases $X \mapsto \text{QCoh}(X)$ is fully faithful, even for some nice stacks; this is a generalization of Tannaka-Krein duality due to Lurie in Tannaka Duality for Geometric Stacks , see e.g. the discussion here. In particular, with the appropriate hypotheses $\text{Aut}(X)$ is automorphisms of $\text{QCoh}(X)$ as a tensor category. For schemes there are results due to Brandenburg in Tensorial schemes and Tensor categorical foundations of algebraic geometry.

Let me indicate how this works for affine schemes since this is the only case I understand. So the question is about recovering $\text{Aut}(R)$ from $\text{Mod}(R)$ where $R$ is a commutative ring. Here the "reconstruction theorem" is really easy: $R$ can be recovered from $\text{Mod}(R)$ as its center, namely the ring of endomorphisms of the identity functor $\text{id}_{\text{Mod}(R)}$. However, this is in some sense a coincidence; what the center actually computes is the zeroth Hochschild cohomology $HH^0(R)$, which happens to be isomorphic to $R$ in this discrete setting, but this is no longer true in the derived setting, eg.

So here is a better reconstruction theorem for affine schemes. To set the stage, if $f : R \to S$ is a morphism of commutative rings, it induces a symmetric monoidal cocontinuous functor $f_{\ast} : \text{Mod}(R) \to \text{Mod}(S)$ given by extension of scalars. This is of course the pullback of quasicoherent sheaves $\text{QCoh}(\text{Spec } R) \to \text{QCoh}(\text{Spec } S)$ along the corresponding morphism $\text{Spec } S \to \text{Spec } R$.

"Tannaka-Krein duality for affine schemes": If $R, S$ are two commutative rings, then the category of symmetric monoidal cocontinuous functors $\text{Mod}(R) \to \text{Mod}(S)$ is equivalent to the set of morphisms $f : R \to S$ (in particular, it is discrete as a category).

This version of the reconstruction theorem is trivial on objects, since $\text{Mod}(R)$ as a tensor category knows its monoidal unit $R$, whose endomorphisms are $R$, and this is still true on a derived level. IIRC there is a general pattern that goes something like: if $R$ is an $E_n$-algebra then it can be reconstructed as an $E_n$-algebra from $\text{Mod}(R)$ as an $E_{n-1}$-algebra.

Sketch. Let $F : \text{Mod}(R) \to \text{Mod}(S)$ be symmetric monoidal cocontinuous. By the Eilenberg-Watts theorem, $F(-) \cong M \otimes_R (-)$ where $M$ is an $(R, S)$-bimodule (uniquely determined by $F$, it is $F(R)$). Since $F$ is monoidal it must in particular preserve the monoidal unit, meaning that $M \cong S$ as an $S$-module. We have $\text{End}_S(S) \cong S$, so the $R$-module structure on $M$ comes from a morphism $f : R \to \text{End}_S(S) \cong S$. But this says exactly that $F$, as a functor, is isomorphic to $f_{\ast}$. Now there is some cleaning-up to do to check that $F$ is isomorphic to $f_{\ast}$ as a symmetric monoidal functor, etc. $\Box$

This is "$2$-affine algebraic geometry"; we are thinking of $\text{QCoh}$ itself as a categorified ring, where the multiplication is $\otimes$ and the addition is colimits. (In this discrete setting I think we actually only need to keep track of the monoidal unit but the symmetric monoidal structure would be relevant to recovering $E_{\infty}$ morphisms between $E_{\infty}$-ring spectra rather than $E_1$ morphisms, or something like that.)

So we recover $\text{Aut}(R)$ as automorphisms of $\text{Mod}(R)$ as a tensor category. Here is an example of what happens if we forget the tensor product: $\text{Mod}(R)$ has automorphisms as a cocomplete category ("$\text{QCoh}(\text{Spec } \mathbb{Z})$-module") which are not monoidal, given by tensoring with any nontrivial invertible module, so for example if $R = \mathcal{O}_K$ then $\text{Mod}(R)$ has non-monoidal automorphisms given by the nontrivial elements of the ideal class group. More generally, by Eilenberg-Watts the group of automorphisms is the group of invertible $(R, R)$-bimodules, which are some funny mix of automorphisms of $R$, invertible modules, and maybe weirder stuff.

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  • $\begingroup$ Amazing, thanks! As for other examples - do you happen to know if this result holds for noncommutative rings, so that we can say what Aut of say $\mathcal{D}_X\text{-Mod}$ or $U_q(\mathfrak{g})\text{-Mod}$ is? $\endgroup$
    – Pulcinella
    Commented Aug 2 at 9:46
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    $\begingroup$ @Pulcinella: Eilenberg-Watts is true for noncommutative rings so we get that the automorphisms of $\text{Mod}(R)$ for $R$ an arbitrary ring are given by invertible $(R, R)$-bimodules. I have no concrete sense of what these are for $R = U(\mathfrak{g})$, though. If $R$ is a $K$-algebra and you want $K$-linear automorphisms then these correspond to bimodules over $K$ (meaning the left and right actions of $K \subset R$ agree, or equivalently we consider $R \otimes_K R^{op}$-modules). For D-modules we could hope to prove a relative Eilenberg-Watts theorem. $\endgroup$ Commented Aug 2 at 17:56

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