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In algebraic and analytic geometry, vector bundles are usually interpreted as locally free sheaves of modules (over the structure sheaves). They are in particular examples of quasi-coherent sheaves. If the bundle is of finite rank, then the sheaf is actually coherent and this is good for certain cohomology groups to have finite rank for example.

I think the equivalence of vector bundles and locally free sheaves holds as well for the categories of topological spaces and smooth manifolds, and locally free sheaves are in particular quasi-coherent. The question is, when are they coherent? maybe it is best to ask first about the structure sheaf itself. In algebraic geometry the structure sheaf is coherent for neotherian schemes and in analytic geometry the structure sheaf of a complex manifold is coherent. What about smooth manifolds for example?

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You may like to look at §6.3 this paper. It seems that there are problems with both Hartshorne's and Grothendieck's definition of coherent sheaf. – Zhen Lin Sep 18 '12 at 17:23
up vote 10 down vote accepted

Since coherence is a local property and is preserved by finite direct sums, the general question reduces to the structure sheaf. But this is coherent only in trivial cases.

If $M$ is a smooth manifold whose connected components have positive dimension, then $C^{\infty}_M$ is not coherent.

For a proof, see Prop. 7.3.8 in the course notes by Andrew Lewis on sheaf theory. By a projection argument, it suffices to deal with $M=\mathbb{R}$. If $f$ is a smooth function, which is $>0$ on $\mathbb{R}_{>0}$ and $=0$ on $\mathbb{R}_{\leq 0}$, then the kernel of the multiplication map $f : C^{\infty}_{\mathbb{R}} \to C^{\infty}_{\mathbb{R}}$ is not of finite type: Loot at the stalk at $0$. The kernel $I$ is given by those germs of smooth functions vanishing on $\mathbb{R}_{\geq 0}$. If $\mathfrak{m}$ is the maximal ideal of functions vanishing at $0$, then we have $I = \mathfrak{m} I$ by Taylor expansion. Since $I \neq 0$, Nakayama's Lemma shows that $I$ is not finitely generated.

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If you're using the [sheaf of continuous sections] and [etale space] functors they do not give an equivalence of these categories. The problem is that locally constant sheaves are covering spaces, and vector bundles aren't covering spaces. They're not even (topologically) etale in general over the base space.

Calling a sheaf coherent over a space that isn't a scheme is silly, since [the definition of coherence] implicitly uses the structure sheaf of the space. A manifold doesn't really have a structure sheaf(it does have a natural sheaf of smooth functions, but I don't know that it makes much sense to talk about coherent modules over such a sheaf of rings, as I don't think there is a good, canonical, functorial way to take a module of the global section and 'tilde'(to use Hartshorne notation) it into a sheaf.)

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Hartshorne's definition really only works for locally Noetherian schemes. If $X$ is an arbitrary ringed space, a sheaf $\mathscr{F}$ on $X$ is coherent if it is finite type and if for every open $U$ and every morphism $\alpha:\mathscr{O}_U^n\rightarrow\mathscr{F}\vert_U$, $\ker(\alpha)$ is finite type. When $X$ is a locally Noetherian scheme, this coincides with quasi-coherent plus finite type, and with Hartshorne's definition. – Keenan Kidwell Sep 18 '12 at 16:44
The definition Keenan gives is the original definition of coherent sheaf, as defined for complex manifolds (which are obviously not locally noetherian in general). The sheaf of smooth functions on a manifold is a very good notion of structure sheaf – in fact, if the manifold is compact, then the underlying topological space can be recovered as MaxSpec of the global sections. The equivalence of categories between vector bundles (of finite rank) and locally free sheaves (of finite rank) is, of course, not induced by the éspace étalé functor – a more sophisticated approach is required. – Zhen Lin Sep 18 '12 at 16:52
Zhen, can you give me a reference to hunt down? – John Stalfos Sep 18 '12 at 18:15

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