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I am not a rigid analytic geometrer, so I apologise if the question is trivial, but I can't find an answer anywhere myself. I'm trying to understand in what ways (rigid) analytic geometry compares to the two geometries I know much better: complex-analytic, and algebraic. For me, one of the (many) big distinctions between these two is how often the resolution property holds. In brief, resolving a coherent sheaf by locally free sheaves is generally impossible the complex-analytic case and generally possible in the complex-algebraic. But I don't know what happens in the rigid analytic case.

Question. Let $X$ be a rigid analytic space and $\mathscr{F}$ a coherent sheaf on $X$. Is it true that $\mathscr{F}$ can be globally resolved by a complex of locally free sheaves on $X$? (feel free to add "mild" conditions on $X$, e.g. analogous to paracompact or Noetherian).

(Here I'm really using words that I don't know how to use: I say "rigid analytic space" because that's what I've heard of, but if instead I should be saying "Berkovich space" or "adic space" then I'd still be interested in an answer. The main reason I'd like to hear "rigid analytic" is just because I've had quite a few people tell me that this is something I should care about for other reasons.)

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  • $\begingroup$ Could you explain a bit how complex analytic spaces usually fail to have resolution property? For example, what kinds of coherent sheaves do not receive a surjection from a locally free sheaf? $\endgroup$
    – Z. M
    Commented Jun 27 at 14:23
  • $\begingroup$ There's a famous example of Voisin of a coherent sheaf on a 3-torus (I think). It is true that any projective manifold has the resolution property, so you really need to look at non-algebraic things to see this though. $\endgroup$
    – Tim
    Commented Jun 27 at 15:17
  • $\begingroup$ Do you have any conceptual reason for the example that you mentioned? I want to see what obstruction leads to this. $\endgroup$
    – Z. M
    Commented Jun 27 at 18:33
  • $\begingroup$ That's a good question, but I'm not sure I have a very good answer. In some sense, I suppose it's the lack of partitions of unity that make the holomorphic case more difficult than e.g. the smooth or algebraic case. I think mathoverflow.net/questions/188271/… is asking exactly the same question, but this has yet to receive any answers... $\endgroup$
    – Tim
    Commented Jun 27 at 19:22
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    $\begingroup$ I would think that not having ample line bundles (on nonprojective compact complex manifolds) means you don't have any reasonable notion of twisted holomorphic functions, and hence would expect many fewer morphisms between coherent sheaves in general (for example since such morphisms give sections of determinant bundles). $\endgroup$
    – David Ben-Zvi
    Commented Jun 29 at 3:21

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