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Let G be a Lie group and $H\subseteq G$ be a closed subgroup. It can be shown that $H$ has a unique differentiable structure such that the inclusion map $H\to G$ is an embedding of manifolds. The functor of points $h_G=\text{Hom}(-,G)$, $h_H=\text{Hom}(-,H)$ of the Lie groups G and H define group objects in the category of manifolds. The inclusion map $H\to G$ gives a natural transformation $h_H\to h_G$ which clearly makes $h_H$ into a subfunctor of $h_G$. Therefore we may take the quotient functor Q which takes each manifold M to the quotient $Q(M):=\text{Hom}(M,G)/\text{Hom}(M,H)$.

Is this functor $Q$ a sheaf on the Grothendieck pretopology where the coverings of an arbitary manifold M are given by the sets of arrows {$\iota_i:U_i \to M $ } such that each $\iota_i$ is a diffeomorphism of $U_i$ onto an open subset of M and $\bigcup \iota_i(U_i)=M$?

(An analogous question is often posed in the context of algebraic groups, however, I am interested in this differentiable setting.)

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    $\begingroup$ Surely you mean to ask if $G/H$ represents the sheafification of $Q$, yes? Since $p:G \rightarrow G/H$ is a principle fiber bundle for the left $H$-action (cf. end of Prop. 10, section 1.5, Ch. III of Bourbaki "Lie groups and Lie algebras"), $Q$ has sections locally over $G/H$ but generally $Q(G/H)$ is empty. Anyway, $Q$ is a sub-presheaf of the functor of points of $G/H$, and this inclusion is "locally surjective" in an evident sense (as $p$ is a principle $H$-bundle), so it computes the sheafification. (Historically the question is backwards: this case motivated the algebraic group case!) $\endgroup$
    – BCnrd
    Jun 28, 2010 at 20:56

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No; in fact any sheaf $F$ in this topology is a sheaf on the open subsets of $M$ in the standard (continuous) topology as open subsets of $M$ map diffeomorphically into $M$, and conversely any manifold mapping diffeomorphically into $M$ is isomorphic to an open subset of $M$ (its image). So if this were true the functor of global sections for sheaves on manifolds would be exact; but we know that this is not the case.

Added, (2/20/2011): I figured I would, as per BCnrd's comment, add an example where the quotient fails to be a sheaf, using connected Lie Groups. The construction is surprisingly elementary, and if I didn't make some silly mistake, I think quite nice!

Let $\rho: SO(3)\hookrightarrow U(n)$ be a faithful unitary representation of $SO(3)$; let $M=U(n)/SO(3)$. I first claim that the quotient map $U(n)\to M$ is a non-trivial principal $SO(3)$-bundle. Indeed $\pi_1(U(n))=\mathbb{Z}$, whereas $\pi_1(SO(3))=\mathbb{Z}/2\mathbb{Z}$, and thus $U(n)$ does not split topologically as a product of $M\times SO(3)$, as $\mathbb{Z}/2\mathbb{Z}$ is not a factor of $\mathbb{Z}$. So $U(n)\to M$ represents a non-trivial class in $H^1(M, SO(3))$, (which is a pointed set, corresponding to isomorphism classes of principal $SO(3)$-bundles over $M$, via the Cech construction).

But note that this class is the image of the canonical element $i$ of $$H^0(M, U(n)/SO(3))=\operatorname{Hom}(M, U(n)/SO(3))=\operatorname{Hom}(U(n)/SO(3), U(n)/SO(3))$$ (namely, the identity map), through the boundary map $H^0(M, U(n)/SO(3))\to H^1(M, SO(3))$. Thus $i$ maps to a non-trivial class, and so must not be in the image of the map $H^0(M, U(n))\to H^0(M, U(n)/SO(3))$.

Thus this last map is not surjective, and so the sheaf of $U(n)/SO(3)$ valued functions on $M$ is not the presheaf $\operatorname{Hom}(-, U(n))/\operatorname{Hom}(-, SO(3))$.

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  • $\begingroup$ Daniel, failure of exactness in general for sheaves of sets does not logically imply failure in a specific case. If $H$ is a direct (or semi-direct) factor of $G$ then it does work. To justify a counterexample one has to prove the non-exactness in a specific instance. A more interesting non-counterexample is if $H = K$ is a maximal compact subgroup of a Lie group $G$ with finite component group; see Theorem 3.1 in Chapter XV of Hochschild's "The structure of Lie groups". So some work is needed to prove that $G \rightarrow G/H$ has no section in some specific case. $\endgroup$
    – BCnrd
    Jun 28, 2010 at 22:05
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    $\begingroup$ Well some cases are quite easy, which is why I omitted an actual construction; e.g. $0\to \mathbb{Z}\to \mathbb{R}\to S^1\to 0$; here there is clearly no section. (Alternately, non-exactness follows by noting that the cohomology of $h_\mathbb{Z}$ is non-vanishing often whereas the cohomology of $h_\mathbb{R}$ does vanish, as it is fine.) My claim is of course not that these functors are never sheaves; just that they are often not sheaves. $\endgroup$
    – Daniel Litt
    Jun 28, 2010 at 22:24
  • $\begingroup$ Daniel, fair enough. The last sentence in your answer seemed to convey a stronger statement that you intended. (I had in mind trying to make examples with connected Lie groups, for which some more work is needed.) $\endgroup$
    – BCnrd
    Jun 28, 2010 at 22:43
  • $\begingroup$ @BCnrd: I've added an example with connected Lie Groups. $\endgroup$
    – Daniel Litt
    Feb 21, 2011 at 0:12

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