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We know that the category of the (quasi-)coherent sheaves on a smooth projective variety $X$ determine the variety (aka. Gabriel–Rosenberg reconstruction theorem) and the derived category of coherent sheaves on a smooth projective variety $X$ with $\omega_X$ ample or anti-ample determine the variety (aka. Bondal–Orlov reconstruction theorem).

Do we have any known results about whether the category of algebraic vector bundles (or locally free sheaves) $\mathbf{Vect}_a(X)$ can determine a variety? $$\mathbf{Vect}_a(X)\cong \mathbf{Vect}_a(Y)\implies X\cong Y$$ Does it work when the equivalence is between categories, $k$-linear categories or a $k$-linear tensor categories?

Edit: refer to Martin Brandenburg in the comment, the affine case is algebraic and one can show it by corresponding results in algebra.

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    $\begingroup$ It will make a lot of difference whether you regard $\text{Vect}_a(X)$ as just a category, or a $\mathbb{C}$-linear category, or a $\mathbb{C}$-linear tensor category, so you should specify that. $\endgroup$
    – Neil Strickland
    Commented Nov 13, 2022 at 19:41
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    $\begingroup$ Of course you want to consider schemes with the resolution property, which means there are enough locally free sheaves. Projective schemes have this property, though. But the main problem here is that exact sequences in $\mathbf{Vect}(X)$ can be "weird", by which I mean that they don't have to be exact in $\mathbf{Qcoh}(X)$. For example, $2 : \mathbb{Z} \to \mathbb{Z}$ is an epimorphism in the category of all locally free $\mathbb{Z}$-modules, in fact all torsionfree $\mathbb{Z}$-modules, but of course not in the category of all $\mathbb{Z}$-modules. This already nukes many "obvious proofs". $\endgroup$
    – Martin Brandenburg
    Commented Nov 14, 2022 at 4:45
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    $\begingroup$ The affine case is easy since a commutative ring R is not just (this is well-known) isomorphic to the center of the category of R-modules, but - with the same proof - also of the category of finitely generated projective R-modules. (ncatlab.org/nlab/show/center) $\endgroup$
    – Martin Brandenburg
    Commented Nov 14, 2022 at 13:01
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    $\begingroup$ Some unfinished thoughts: Let $X$ be a scheme with $\mathbf{Vect}(X) \simeq \mathbf{Vect}(\mathbb{P}^n)$ as $k$-linear tensor categories. There is an invertible object $\mathcal{L} \in \mathbf{Vect}(X)$ and morphisms $s_0,\dotsc,s_n : \mathcal{O}_X \to \mathcal{L}$ which correspond to $\mathcal{O}(1)$ and $t_0,\dotsc,t_n$ under the equivalence. Clearly $(s_0,\dotsc,s_n) : \mathcal{O}_X^n \to \mathcal{L}$ is an epi in $\mathbf{Vect}(X)$, but in order to construct $f : X \to \mathbb{P}^n$ with the universal property of $\mathbb{P}^n$, we need that it is an epi in $\mathbf{Qcoh}(X)$. Unclear! $\endgroup$
    – Martin Brandenburg
    Commented Nov 14, 2022 at 17:14
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    $\begingroup$ @DanielLoughran A category is more than just its objects $\endgroup$
    – Martin Brandenburg
    Commented Nov 15, 2022 at 1:14

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$\newcommand\Vect{\mathit{Vect}}\newcommand\Hom{\mathit{Hom}}$At least the birational tyie of a smooth projective variety can be recovered from the monoidal category of vector bundles on it. (the previous version of this answer claimed that the isomorphism class can be recovered but now I don't think that the original argument works)

For any line bundle $L$ on $X$ there is a dominant rational map from $X$ to $R(X,L):=\operatorname{Proj}\bigoplus H^0(X,L^n) $ which is an isomorphism if $L$ is very ample. In the monoidal category $\Vect(X)$ the invertible objects are precisely the line bundles, and the structure sheaf is the unit object, so given the category $\Vect(X)$ we can recover the collection of schemes $R(X,L)$ for all line bundles $L$ (because $H^0(X,L^n)=\Hom_{\Vect(X)}(O_X,L^{\otimes n})$), though we are not being told which of these arise from ample line bundles.

Let's discard all $R(X,L)$ that are not of finite type. Among the remaining ones all the $R(X,L)$s that have maximal dimension are birational to $X$, so the birational type of $X$ can be recovered from $(Vect(X),\otimes)$.

It is unclear to me right now how to recover $X$ itself: the issue is that an equivalence $(Vect(X),\otimes)\simeq (Vect(Y),\otimes)$ might a priori carry an ample line bundle on $X$ to a non-ample (though necessarily big) line bundle on another variety $Y$.

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  • $\begingroup$ Unfortunately I lack the knowledge about algebraic geometry to understand this properly. But it seems that you overcome the problem that the property of being ample cannot be formulated within the category $\mathbf{Vect}(X)$ (again, the epimorphism issue!) by finding other conditions on the associated graded ring? $\endgroup$
    – Martin Brandenburg
    Commented Nov 15, 2022 at 1:18
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    $\begingroup$ @MartinBrandenburg Maybe it is helpful to say that for your example of a variety $X$ with $Vect(X)=Vect(\mathbb{P}^n)$ what I'm suggesting comes down to simply turning your attempted argument around. Let's choose a very ample line bundle $L$ on $X$, it is carried to some $O(k)$ under this equivalence. The graded rings corresponding to $L$ and to $O(k)$ are isomorphic so we necessarily have $k>0$, and this proves that $X\simeq \mathbb{P}^n$ because $O(k)$ is very ample. $\endgroup$
    – SashaP
    Commented Nov 15, 2022 at 1:35
  • $\begingroup$ Why is X isomorphic to P^n here? Sorry for the stupid question. $\endgroup$
    – Martin Brandenburg
    Commented Nov 15, 2022 at 2:10
  • $\begingroup$ @MartinBrandenburg If $L$ is an ample line bundle on a projective variety $X$ then $X$ is isomorphic to $Proj \oplus H^0(X,L^n)$, see e.g. stacks.math.columbia.edu/tag/01Q1 Hence if we are able to prove that an equivalence $Vect(X)\simeq Vect(Y)$ carries an ample line bundle $L$ on $X$ to an ample line bundle $M$ on $Y$ then $X\simeq Proj \oplus H^0(X,L^n)\simeq Proj \oplus H^0(Y,M^n)\simeq Y$. This is the situation in this example because an ample line bundle on $X$ is necessarily carried to an ample line bundle on $\mathbb{P}^n$. $\endgroup$
    – SashaP
    Commented Nov 15, 2022 at 10:50
  • $\begingroup$ Though I realized that my general argument is not quite correct: I was too hasty to claim that we can figure out which of the briationallly equivalent varieties $R(X,L)$ is $X$ itself, I edited the answer accordingly, so the original question is still open. $\endgroup$
    – SashaP
    Commented Nov 15, 2022 at 10:51

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