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This is an improved version of this question. Maybe it should be an edit -- I'm not sure what the MO convention is.

I'm curious, more or less, how much information one can get out of the derived category of coherent sheaves on a variety which are trivialized outside a finite set. Specifically, suppose $X$ is a smooth, proper variety of finite type over $\mathbb{C}$. Consider the category $\mathcal{C}=\text{Coh}(X)_{\text{codim }0}$ of coherent sheaves which are trivial (in the sense of being isomorphic to a power of the constant sheaf) outside a finite set of $\mathbb{C}$-points of $X$. This is a symmetric monoidal category with inner Homs. Suppose we do not know $X$, but know the category $\mathcal{C}$, together with derived data (the derived category of complexes whose cohomology is in $\mathcal{C}$, together with derived tensor products, Homs and most importantly global sections). Suppose in addition we are given the set of points of $X$ together with their formal neighborhoods, and know for every sheaf in $\mathcal{C}$ how it can be patched out of a globally trivial sheaf together with patching data at some finite set of formal neighborhoods.

How much information about $X$ does the category $\mathcal{C}$ contain? In particular can $X$ be recovered from $\mathcal{C}$?

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  • $\begingroup$ How do you define $Hom$'s in your category? Is it a full subcategory of $Coh(X)$? $\endgroup$
    – Sasha
    Commented Mar 24, 2013 at 15:42
  • $\begingroup$ Yes, of course (otherwise it'd be boring in dimension $>1$). $\endgroup$
    – Dmitry Vaintrob
    Commented Mar 24, 2013 at 17:46
  • $\begingroup$ My hunch is you are missing a lot in dimension $>1$. The reason is because the homs do not contain much global data: If $F$ is trivial outside $S$, and $G$ is trivial outside $T$, then on $X - S - T$, the $Hom(F,G)$ is just a vector space of dimension $dim (f) \times dim(G)$. Adding in $S$ and $T$ can lead to increases and decreases in the dimension, but nothing that illuminates the global geometry of the situation. And higher homs do not tell you much more than $H^k(X,\mathcal O_X)$, which of course is not a complete set of invariants. $\endgroup$
    – Will Sawin
    Commented Mar 25, 2013 at 3:51
  • $\begingroup$ Can't we reconstruct $X$ as a component of the moduli space of objects of $\mathcal{C}$, namely the component parameterizing ideal sheaves of a point. $\endgroup$
    – Jim Bryan
    Commented Mar 27, 2013 at 6:23

1 Answer 1

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Let $F$ be a sheaf which is trivial on $X - S$. This means that $j^*F = O_{X-S}^{\oplus n}$ for some $n$, where $j$ is the embedding of $X - S$ into $X$. Note that by Hartogs theorem one has $j_*O_{X-S} = O_X$, so by adjunction one has a morphism $F \to j_*j^*F = O_X^{\oplus n}$ which is an isomorphism on $X - S$. So, this means that such $F$ is a modification of the trivial bundle in the finite number of points. In other words, $Coh(X)_{codim 0}$ is the extenion closed abelian subcategory of $Coh(X)$ generated by $O_X$ and all structure sheaves of points.

EDIT. Let us see what the information we have in this category. First, we have the cohomology of $O_X$. Second, we have a bunch of structure sheaves of points, and this part of the category depends only on the dimension of the variety. Finally, we have multiplication maps $$ Ext^\bullet(O_X,O_S)\otimes Ext^\bullet(O_S,O_X) \to Ext^\bullet(O_X,O_X) $$ and $$ Ext^\bullet(O_X,O_S)\otimes Ext^\bullet(O_S,O_X) \to Ext^\bullet(O_S,O_S). $$ Since $S$ is $0$ dimensional the first factor lives in degree $0$ and the second in degree $n = \dim X$, so the only thing we have are the maps $$ Hom(O_X,O_S)\otimes Ext^n(O_S,O_X) \to Ext^n(O_X,O_X) $$ and $$ Hom(O_X,O_S)\otimes Ext^n(O_S,O_X) \to Ext^n(O_S,O_S). $$ Now let $S= x$ be a point. By Serre duality these maps correspond to the maps $$ Hom(O_X,O_x)\otimes Hom(O_X,\omega_X) \to Ext^n(O_X,\omega_X\otimes O_x) $$ and $$ Hom(O_X,O_x)\otimes Hom(O_x,O_x) \to Hom(O_X,O_x). $$ So, the second map contains no information, while the first gives the canonical morphism $$ X \to P(H^0(\omega_X)^*) = P(H^n(O_X)). $$

So, on one hand, if $H^n(O_X) = 0$ you cannot reconstruct anything. On the other hand, if the canonical class is very ample, by reconstrcting the anticanonical image you reconstruct all $X$. I am not quite sure what goes on in the intermediate cases.

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    $\begingroup$ Hm... I agree that it seems like something is lost in dimension $>1$, but not sure your argument is correct, or else I'm not following it. On a curve, if you take the extension closed abelian subcategory of $(D^b)Coh(X)$ generated by $O_X$ and all structure sheaves of points, you get all of $Coh$, which certainly lets you recover $X$. So despite the fact that Homs and Exts between objects in the subcategory <i>consisting</i> of $O_X$ and skyscraper sheaves depend only on the coherent cohomology, once you start taking extensions things can become more subtle. How does Hartog change this? $\endgroup$
    – Dmitry Vaintrob
    Commented Mar 26, 2013 at 2:54
  • $\begingroup$ Dmitry, you are completely right. I edited the answer, now you see that in some cases you can reconstruct a lot of geometry! $\endgroup$
    – Sasha
    Commented Mar 27, 2013 at 4:19
  • $\begingroup$ Thanks Sasha! Neat. I agree that when coherent cohomology is trivial, we get no information. However are you sure that there's no data if only $H^n(O)=0$? For example what if we compose your first map with the multiplication map $Hom(O_X,O_S)\otimes Hom(O_X, O_S)\to Hom(O_X, O_S)$ to get a map $Hom(O_X,O_S)\otimes Hom(O_X,O_S)\otimes Ext^n(O_S,O_X)\to Ext^n(O_X,O_X)$. Then if $H^n(O_X)=0$ then this product will of course be zero, but there will be a well-defined and possibly nonzero Massey product. Is it clear that this Massey product will be zero? $\endgroup$
    – Dmitry Vaintrob
    Commented Mar 29, 2013 at 6:49

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