When is a sheaf of smooth functions acylic? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T10:12:33Z http://mathoverflow.net/feeds/question/29512 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/29512/when-is-a-sheaf-of-smooth-functions-acylic When is a sheaf of smooth functions acylic? Konrad Waldorf 2010-06-25T13:23:57Z 2010-07-09T14:50:12Z <p>Let $G$ be a Lie group, and let $\underline{G}$ denote the sheaf of smooth $G$-valued maps, i.e. for a smooth manifold $M$ we have $G(M) = C^\infty(M,G)$. </p> <p>What are conditions on $G$ that imply that $\underline{G}$ is acyclic, i.e. the sheaf cohomology $H^n(M,\underline{G})=0$ for all smooth manifolds $M$ and all $n>0$? </p> <p>It is clear that soft, flabby or fine sheaves are acyclic. I am interested in concrete conditions on the group $G$, e.g. like smooth contractibility. </p> <p>EDIT: Daniel's answer below answers my question in the case that $G$ is abelian, using the classification of abelian Lie groups. So let us concentrate on the case that $G$ is non-abelian. The condition I am looking for is supposed to imply the vanishing of the set $H^1(M,\underline{G})$. This set can be defined for example via Cech cohomology. Its geometrical meaning is that it classifies principal $G$-bundles over $M$ up to isomorphism. </p> http://mathoverflow.net/questions/29512/when-is-a-sheaf-of-smooth-functions-acylic/29516#29516 Answer by Daniel Litt for When is a sheaf of smooth functions acylic? Daniel Litt 2010-06-25T14:10:06Z 2010-07-09T14:50:12Z <p>For Abelian $G$ (that is, the product of a torus with $\mathbb{R}^n$), an argument identical to macbeth's comment gives that $H^n(M, \underline{G})=0$ for all $M, n>0$ iff $G\simeq \mathbb{R}^n$).</p> <p>Explicitly, in the case $G\simeq \mathbb{R}^n$ the sheaf in question is fine; otherwise, if $G\simeq \mathbb{R}^n\times (S^1)^k$ then it fits into an exact sequence $0\to \mathbb{Z}^k\to \mathbb{R}^{n+k}(M)\to \underline{G}\to 0$, giving the claim.</p> <hr> <p>Added (7/7/2010): Having thought a bit about the non-Abelian case, I thought I'd add another non-vanishing theorem. </p> <p><strong>Theorem.</strong> Let $G$ be a Lie group admitting a faithful unitary representation, with $\pi_1(G)\neq 0, \mathbb{Z}$. Then there exists $M$ with $H^1(M, \underline{G})\neq 0$.</p> <p><strong>Proof.</strong> Let $\rho: G\to U(n)$ be the given faithful unitary representation, and let $M=U(n)/G$. Then $U(n)$ is a $G$-bundle over $M$, and it is non-trivial as $\pi_1(U(n))=\mathbb{Z}$ wheareas $\pi_1(G)$ cannot be a factor of $\mathbb{Z}$ by assumption. That is, $U(n)\not\simeq G\times M$ as $\pi_1(U(n))\not\simeq \pi_1(G)\times \pi_1(M)$. $\square$</p> <p>This holds for e.g. compact Lie groups with the appropriate fundamental group; it seems likely that this argument can be strengthened by e.g. considering higher homotopy groups or using other results on the existence of faithful representations.</p> <hr> <p>Added (7/9/2010): I don't know why I didn't mention it before, but replacing "unitary" with "complex" in the theorem above gives the same result for e.g. complex connected semisimple Lie groups, by an identical proof. In this case the manifold $M$ constructed in the proof cannot be guaranteed to be compact however.</p>