It's probably a really naive question, but I didn't find any references. Given a category $V$ and we know that it is equivalent to the category $\mathbf{Vect}(X)$ of smooth vector bundles over a smooth manifold $X$, under what assumptions could we determine $X$? Or is this generally hopeless?
Differential Geometry – Reconstructing Base Manifold from Smooth Vector Bundles
ct.category-theorydg.differential-geometryvector-bundles
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I think that the most interesting part of your question is the part you put in parentheses!
(and these should satisfy some compatibility condition)
What are your compatibility conditions? That is everything here. If you specify the correct conditions, you may find that all your definitions collapse to just one.
I have an issue with Konrad's answer (which I doubt very much that he will be surprised to hear me express!). Whenever I heard words like "Grothendieck topology" or "sheaves" or encoding similar ideas then I feel that something's been lost. I don't like the idea that "smooth" is just "really nice continuity". "Smooth" sits alongside continuity and can be expressed in a different way which is extremely simple: takes smooth curves to smooth curves.
Of course, I would say that, as everyone by now presumably knows that I prefer Frolicher spaces to the other variants (like Chen spaces or diffeological spaces, see generalized spaces for links). It is interesting that Chen's third definition (by my count) was stronger than his eventual sheaf condition and was more along the lines of "a map is smooth if enough tests say that it is smooth".
But Frolicher spaces have a problem, which is that it is extremely difficult to prise them away from being a set-based theory. The compatibility condition is so strong that it forces an underlying set. I'd really like to figure out how to make this extension, and I know that Urs would as well. If I could just encourage you and Konrad over to the nLab to play around with these ideas to see how they could work ...
If you want to study The Smooth, The Whole Smooth, and Nothing But The Smooth, then you should do so and not flirt continually with continuity. The stronger compatibility condition means that more stuff is smooth than you first thought (witness my recent question on this) and that makes it interesting! The unexpected happens, so study it!
This isn't much of an answer so far, it's more of a commentary on your question which (as is usual for me) is too long for an actual comment. So let me end with an actual answer (which I freely confess that I stole from a rabbi):
That is such a great question, why on earth would you want an answer?
As Piotr remarks, these kind of questions quickly lead to studying reflexive sheaves. I would add that one also better get acquainted with Serre's condition $S_2$. For more on this see this and this MO answers.
Perhaps the first remark is that besides the singularities of the surface $X$ you also have to take into account the singularities of the sheaf you are considering.
You are asking about normal surfaces. Normal implies $S_2$ and pretty much everything that follows is OK for $S_2$ in arbitrary dimensions.
A coherent sheaf on an open set can always be extended as a coherent sheaf on the ambient space. Furthermore if $X$ is $S_2$ and $j:U\hookrightarrow X$ is an open set such that $\mathrm{codim}_X(X\setminus U)\geq 2$, then a coherent sheaf $\mathscr F$ on $X$ such that $\mathscr F|_U$ is locally free is reflexive if and only if $$\mathscr F \simeq j_*(\mathscr F|_U).$$ Notice that this means that if $X$ is $S_2$, then a locally free sheaf $\mathscr E$ on $U$ can be extended as a locally free sheaf to $X$ if and only if $j_*\mathscr E$ is locally free. (The point is, that since $X$ is $S_2$, $j_* \mathscr E$ is reflexive and if there is a locally free sheaf extending $\mathscr E$, then that is also a reflexive sheaf which agrees with $j_*\mathscr E$ on an open set with at least codimension $2$ complement, so they have to agree).
This also tells you how to produce locally free sheaves that cannot be extended as a locally free sheaf: Take any reflexive sheaf that is not locally free, then take the open set where it is locally free. By the above, this sheaf on that open set is a locally free sheaf that does not have an extension as a locally free sheaf on the ambient space. Piotr's example is probably the simplest such sheaf.
So now the question is: When is a reflexive sheaf locally free?
In some contexts one defines the singularity set of a coherent sheaf $\mathscr F$ as the locus where it is not locally free. Then there are various results that say that the singularity set of certain sheaves have to be at least such and such codimension. Here is a short list of those:
Sample statement: If $\mathscr F$ is bluh, then the singularity set of $\mathscr F$ has codimension at least boo.
Actual statement If $X$ is smooth, we can substitute the following into the above sentence:
- If bluh = coherent, then boo = $1$.
- If bluh = torsion-free, then boo = $2$.
- If bluh = reflexive, then boo = $3$.
You may recognize theorems that you know as simple consequences of these:
Thm 1 For any coherent sheaf there exists a non-empty open set on which it is (locally) free.
Remark You may also note that this does not require $X$ smooth, but that's essentially because the conclusion is invariant under first restricting to the open set where $X$ is smooth.
Thm 2 A torsion-free sheaf on a smooth curve is locally free.
Remark We know that this fails if the variety is either not smooth or has dimension at least $2$.
Thm 3 A reflexive sheaf on a smooth surface is locally free.
Remark This is the reason why the statement you mention at the start is true. If $X$ is a smooth surface, then a sheaf that's locally free on the complement of finitely many points pushes forward to a reflexive sheaf which by this theorem has to be locally free. Piotr's example or any Weil divisor that's not Cartier shows that this fails if the surface is not factorial. (Meaning that every local ring is a UFD. This is a weaker condition than being smooth.) It is also true that a reflexive sheaf of rank $1$ on a smooth variety is always locally free, so you need to go to higher ranks to get a counter-example.
To complete the picture here is an example of a reflexive sheaf on a smooth $3$-fold that is not locally free.
Example Consider the Euler sequence of $\mathbb P^3$: $$ 0\to \mathscr O_{\mathbb P^3}\to \bigoplus_{i=0}^3 \mathscr O_{\mathbb P^3} (1) \to T_{\mathbb P^3} \to 0 $$ and consider (one of) the induced maps $$ \alpha: \mathscr O_{\mathbb P^3} (1) \to T_{\mathbb P^3}. $$ This is clearly injective and let the cokernel of $\alpha$ be $\mathscr F$. The original (Euler) sequence shows that $\alpha$ cannot be a vector bundle embedding and hence $\mathscr F$ cannot be locally free. I leave it for you to prove that it is reflexive. (This is not absolutely trivial, but it is a good exercise).
So, to answer your question, the statement you cite is not true for singular varieties, at least not in the way it is stated. You can cook up some versions that work by adding additional assumptions. In any case, these are likely the notions that will help you do that.
Finally, here are some references:
- On reflexive sheaves I would recommend Hartshorne's series of papers Stable reflexive sheaves I-II-III.
- You can read about the singularity set and other interesting stuff in Chapter 2 of Okonek-Schneider-Spindler's Vector bundles on complex projective spaces
Best Answer
Yes, it is possible to recover the manifold through the following steps: