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Does such a sheaf of abelian groups exist? If not, is there a reference or a proof? Does such a sheaf of non-abelian groups exist?

I realized recently that while I've taken it for granted that there wasn't such a sheaf on $\mathbb{R}^n$, I only have proofs that specific sheaves are acyclic.

EDIT: The site of smooth manifolds is the category of smooth manifolds, endowed with the Grothendieck topology generated by defining surjective submersions to be "covers". In more down to earth language, I'm asking whether there is a sheaf, defined naturally and intrinsically for all smooth manifolds, which has nontrivial cohomology on $\mathbb{R}^n$.

The motivation for asking this is that I'm trying to understand the role of good covers in differential topology, i.e. can we always calculate cohomology of any sheaf just by picking a good cover and calculating the Cech cohomology on the cover, or do we in general need to pass to the limit over all covers? If all sheaves on the smooth site have vanishing cohomology on $\mathbb{R}^n$, then good covers always work. I'm trying to understand whether this is the case and why or why not.

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    $\begingroup$ Sure, take a circle in $S^1\subset \mathbb{R}^2$, and extend the constant sheaf on it by zero. Its first cohomology is the same as for the circle, which is nonzero. $\endgroup$
    – Donu Arapura
    Commented Apr 17, 2012 at 0:00
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    $\begingroup$ @David, for any non-zero group $G$, the constant sheaf with values in $G$ on $S^1 \subset \RR^2$ has first cohomology equal to $G$. $\endgroup$
    – Dustin Cartwright
    Commented Apr 17, 2012 at 7:11
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    $\begingroup$ David, it's as Dustin says. My point really was that a sheaf on $\mathbb{R"^n$ is at least as complicated as its support. $\endgroup$
    – Donu Arapura
    Commented Apr 17, 2012 at 9:45
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    $\begingroup$ Sorry, I should have clarified: I'm looking for sheaves on the site of smooth manifolds which have nontrivial cohomology on $\mathbb{R}^n$. Donu, your point about support suggests that there be also be a counterexample of this type, but I haven't put my hands on it yet. $\endgroup$
    – Jesse Wolfson
    Commented Apr 17, 2012 at 12:57
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    $\begingroup$ I like the question, but I am not sure I really understand it. What is exactly the "site of smooth manifold". I mean what topology of you put on it ? $\endgroup$
    – Joël
    Commented Apr 18, 2012 at 2:05

2 Answers 2

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A long time ago (1972) I proved that the first cohomology of the sheaf of nash functions on an interval is infinite-dimensional.

As a specific example, consider the cover of $(-2,2)$ by $U_1=(-2,1)$ and $U_2=(-1,2)$ and the 1-cocycle $\alpha=\sqrt{1-x^2}$ on $U=U_1\cap U_2=(-1,1)$. Then $\alpha$ is not a coboundary: there are no Nash functions $\beta_1$ on $U_1$ and $\beta_2$ on $U_2$ such that [ \beta_1-\beta_2=\alpha ] on $U$.

This is proved in the paper

http://www.math.cornell.edu/~hubbard/CohomologyNashSheaves.pdf.

John Hubbard

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  • $\begingroup$ Hi John, thanks for the answer and the link. I'll take a look at it. $\endgroup$
    – Jesse Wolfson
    Commented Apr 17, 2012 at 13:00
  • $\begingroup$ Hi John, I read your article. I hadn't heard of Nash manifolds before, so this was a nice example to start with. As I clarified above, what I'd really like is an example of a sheaf on the site of smooth manifolds which has nontrivial cohomology on $\mathbb{R}^n$. Is it the case that every smooth manifold is Nash, and your example applies? My hunch would be no, but it at least seems every compact smooth manifold is diffeomorphic to a Nash manifold . . . $\endgroup$
    – Jesse Wolfson
    Commented Apr 17, 2012 at 13:48
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    $\begingroup$ Nash proved ("Real algebraic manifolds", 1952) that every smooth compact manifold is diffeomorphic to a component (in the classical topology) of a real-algebraic set. Tognoli ("Su una congettura di Nash", 1973) proved that every smooth compact manifold $M$ is diffeomorphic to a real-algebraic set $V$. One can improve this result, I think it is in the book of Akbulut and King "Topology of real algebraic sets," as follows: The variety defining $V$ can be taken to be nonsingular. In any case, every such $M$ admits a Nash structure. (I do not know if it is unique though.) $\endgroup$
    – Misha
    Commented Apr 18, 2012 at 0:18
  • $\begingroup$ See also en.wikipedia.org/wiki/Nash_functions#Nash_manifolds $\endgroup$
    – Misha
    Commented Apr 18, 2012 at 0:18
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    $\begingroup$ Shiota proved that all Nash manifolds are tame, so yes, some smooth open manifolds are not Nash. $\endgroup$
    – Misha
    Commented Apr 18, 2012 at 3:50
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Another example, actually a lot more important, is to take a Hartogs figure, such at $$ X:=D\times \{0\}\cup \partial D \times D \subset \mathbb C^2 $$ which your are welcome to think of as $\mathbb R^4$.

Take a neighborhood $U$ of $X$. Then the first cohomoloy of a small neighborhood $U$ with values in the sheaf of holomorphic functions is not zero.

I could spell t out if you are interested.

John Hubbard

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