Example of a Sheaf (on the site of smooth manifolds) with Nontrivial Cohomology on $\mathbb{R}^n$? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T06:52:48Z http://mathoverflow.net/feeds/question/94258 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/94258/example-of-a-sheaf-on-the-site-of-smooth-manifolds-with-nontrivial-cohomology-o Example of a Sheaf (on the site of smooth manifolds) with Nontrivial Cohomology on $\mathbb{R}^n$? Jesse Wolfson 2012-04-16T23:48:50Z 2012-04-18T14:11:50Z <p>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? </p> <p>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. </p> <p>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 $\mathbb{R}^n$. </p> <p>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.</p> http://mathoverflow.net/questions/94258/example-of-a-sheaf-on-the-site-of-smooth-manifolds-with-nontrivial-cohomology-o/94286#94286 Answer by John Hubbard for Example of a Sheaf (on the site of smooth manifolds) with Nontrivial Cohomology on $\mathbb{R}^n$? John Hubbard 2012-04-17T11:46:17Z 2012-04-17T11:46:17Z <p>A long time ago (1972) I proved that the first cohomology of the sheaf of nash functions on an interval is infinite-dimensional.</p> <p>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 $U$.</p> <p>This is proved in the paper</p> <p><a href="http://www.math.cornell.edu/~hubbard/CohomologyNashSheaves.pdf" rel="nofollow">http://www.math.cornell.edu/~hubbard/CohomologyNashSheaves.pdf</a>.</p> <p>John Hubbard </p> http://mathoverflow.net/questions/94258/example-of-a-sheaf-on-the-site-of-smooth-manifolds-with-nontrivial-cohomology-o/94351#94351 Answer by John Hubbard for Example of a Sheaf (on the site of smooth manifolds) with Nontrivial Cohomology on $\mathbb{R}^n$? John Hubbard 2012-04-18T03:13:42Z 2012-04-18T03:13:42Z <p>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 C^2 ] which your are welcome to think of as $R^4$. </p> <p>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.</p> <p>I could spell t out if you are interested.</p> <p>John Hubbard </p>