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Cartan's Theorem B can be stated as follows:

Let $X$ be a space let $\mathcal{F}$ be a sheaf on $X$. Consider the following three conditions:

  1. $X$ is "simple";
  2. $\mathcal{F}$ is "nice";
  3. the sheaf cohomology $\mathrm{H}^p(X,\mathcal{F})$ is zero for all $p\geqslant1$.

Then the following hold:

  • (1) + (2) $\implies$ (3)
  • (3) + (2) $\implies$ (1) (where by "(3) + (2)" I mean "(3) holds for all $\mathcal{F}$ such that (2)")

where we can take ("simple", "nice") to be either (Stein manifold, coherent) or (affine scheme, quasi-coherent).

(Note that Theorem B is really just the fact that (1) + (2) $\implies$ (3); the other result is a sort of converse showing that this theorem is sharp, but I think it deserves to be a part of the theorem).

What I'd like to know is if there is a third option for ("simple", nice") that recovers some sort of $\mathbb{R}$-smooth statement. My guess would be that "simple" = contractible subset of $\mathbb{R}^n$. My only justification for this is that the notion of "simple" in the complex-analytic and complex-algebraic cases above correspond exactly to what one finds in the definition of a good cover, and in the smooth case one normally defines good covers to be those that are contractible with contractible intersections (if I'm not mistaken).

Edit: there has been some discussion in the comments about what sort of answer I'm looking for, and I agree that this question is rather vague, but maybe here's another way of phrasing it:

Cartan's Theorem B tells me that the analogy between analytic and algebraic geometry given by "Stein ↔︎ affine" and "coherent ↔︎ quasi-coherent" is, in the eyes of cohomology, a fair one. If I want to extend this analogy/dictionary to include smooth geometry, which words should I use?

context "simple" "nice"
complex-algebraic affine quasi-coherent
complex-analytic Stein coherent
real-analytic coherent analytic subvarieties of $\mathbb{R}^n$ coherent
smooth ? contractible ? ?

Edit 2: This MO question/answer pointed me to Cartan's paper Variétés analytiques réelles et variétés analytiques complexes, of which Theorem 3 gives a result for the real-analytic setting; I've added this to the table above.

Failing this, are there any other pairs that I can substitute in? Maybe something to recover a statement about rigid analytic geometry?

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    $\begingroup$ My memory may serve me incorrectly, but if I recall correctly, the real-valued $C^\infty$ functions on a (paracompact) $C^\infty$ manifold $X$ form a “fine” sheaf $\mathscr{F}$ (“fine” means any section on a closed set can be extended to a section on all of $X$), so it is acyclic: $H^p(X,\mathscr{F}) = 0$ for $p>0$. For the constant sheaf $\mathbb{R}$, on the other hand, obviously, $H^p(X,\mathbb{R}) = 0$ would hold only with extra assumptions, contractibility being a sufficient one. Which of these two situations are you more after? $\endgroup$
    – Gro-Tsen
    Commented May 30 at 15:31
  • $\begingroup$ hmm, good question. I think I would like the sheaves to be coherent if that is a well-behaved notion in the smooth setting. The fact that fine sheaves are acyclic on arbitrary (paracompact) manifolds makes me think they're not quite the right analogue, because I think the Stein <-> affine <-> contractible analogy is pretty strong (I'll update the question with more about this). $\endgroup$
    – Tim
    Commented May 30 at 19:27
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    $\begingroup$ I agree with Gro-Tsen's and Donu's comments. One way to think about this is that in homotopical/derived approaches to $C^\infty$ geometry, every manifold is affine. $\endgroup$
    – Dan Petersen
    Commented Jun 3 at 21:28
  • $\begingroup$ @DanPetersen Do you need paracompactness for affineness? For example, is the long line affine? $\endgroup$
    – Z. M
    Commented Jun 5 at 14:26

1 Answer 1

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It seems like Gro-Tsen's comment pretty much answers your main question, doesn't it? If "simple" = $C^\infty$ manifold, and "nice" = sheaf of modules over $C^\infty_X$. Then such a sheaf would be fine, so it would have no higher cohomology. If "simple" = contractible, then take "nice"= locally constant, and you get no higher cohomology. If there is something else you wanted, you should specify it.

I'm not sure that $(1) + (3) \Rightarrow (2)$ is reasonable. Injective sheaves would have no higher cohomology, but I'm guessing that you wouldn't consider them "nice".

Added To answer a question in the comments: If $X$ is a connected CW complex then I believe that Whitehead's theorem should imply that $X$ is contractible if and only if the higher cohomology of every local system is trivial. (For precise details, see David Speyer's answer to Converses to Cartan's Theorem B)

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    $\begingroup$ I think Dan's comment sort of touches at what I'm trying to ask (which I admit it pretty vague and not well defined), since if every manifold is affine (in some sense) then the analogy "simple = affine = arbitrary manifold" seems fair. But is the analogy "quasi-coherent = module over sheaf of smooth functions" a good one? I'll try to update the question with something a bit more along these lines. $\endgroup$
    – Tim
    Commented Jun 3 at 22:55
  • $\begingroup$ In the two answers that you/Gro-Tsen give, is the sharp version of the theorem true? i.e. if every locally constant sheaf is acyclic on a space then the space must be contractible? I guess I'd also like to know if this is really of roughly the same level of generalisation as Stein/affine case, i.e. I could give a weaker version of Cartan's Theorem B as "locally free sheaves are acyclic on Stein spaces", but the important thing is that it's actually true for coherent sheaves. If locally constant/contractible is "the right analogy", then is this the best we can say? $\endgroup$
    – Tim
    Commented Jun 3 at 23:03
  • $\begingroup$ Your comment about the Whitehead theorem making this strict is mirrored in math.stackexchange.com/questions/717018/… , which has a reference to a proof for simplicial sets, so I'll take it! (contractible, locally constant) is a nice (and surprising, to me) answer $\endgroup$
    – Tim
    Commented Jun 4 at 11:56
  • $\begingroup$ @Tim To some extent, quasi-coherent sheaves are modules, but you should look at liquid modules. If I understand correctly, this is what Scholze explained in his last lecture on analytic stacks. $\endgroup$
    – Z. M
    Commented Jun 5 at 14:36

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