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Consider a countable family of finite-rank vector bundles $V_k$ over a finite-dimensional smooth manifold $M$. The direct limit of such a family is still a topological vector bundle even though it may have infinite rank.

The first question is: Is there a natural structure of smooth manifold on the total space of $\varinjlim V_k$?

And the second one: Is it true, that the functor of smooth sections commutes with direct limit? I.e, is it true that $$\mathcal{C}^\infty(M,\varinjlim V_k) = \varinjlim \mathcal{C}^\infty(M,V_k)$$

Or turning things upside down -- is there a choice of smooth structure such that the above equation holds? If so, how does it look like?

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What category are you working in? If it's finite rank vector bundles under inclusion, then the sequence of vector bundles has to stabilize at some point to a fixed vector bundle of finite rank. So the equation holds trivially. –  Deane Yang Mar 12 '12 at 10:10
    
You are right, the setting is also not very clear to me. I've changed my question accordingly. –  Vít Tuček Mar 12 '12 at 12:08
    
I'm in way over my head here, but I am under the impression that the definition of a direct limit already implies a topological structure, where smooth sections are those that remain inside some finite dimensional subbundle. So the equation still holds more or less by definition. –  Deane Yang Mar 12 '12 at 13:20

1 Answer 1

up vote 7 down vote accepted

The answers are: Yes and No-but-yes-if-M-is-compact.

  1. Kriegl and Michor's A Convenient Setting for Global Analysis describes how to put a smooth structure on an arbitrary locally convex topological vector space, say $V$, by looking first at the smooth curves in $V$ (these can be unambiguously defined). This works for $V = \lim V_k$ where the $V_k$ are finite dimensional. As your family is countable, this is just $\sum_{\mathbb{N}} \mathbb{R}$. Smooth curves are continuous, and therefore an important property of this structure is that if $c \colon \mathbb{R} \to \sum_{\mathbb{N}} \mathbb{R}$ is smooth then $c([a,b])$ is contained in a finite dimensional subspace.

  2. And that's really the key to the second answer. If $M$ is compact, then its image lies in a finite dimensional subspace of $\lim V_k$, whence inside one of the $V_k$. If $M$ is not compact then it will be possible to find a smooth map $M \to \lim_k V_k$ which does not lie in any subspace and so the limit will not commute with the mapping space construction. Note that it is enough to show that such a map exists for $M = \mathbb{R}$, whereupon you take curves $\alpha_k \colon [0,1] \to V_k$ which map to $0$ on their endpoints, are infinitely slow there, and such that the image of $\alpha_k$ contains a basis for $V_k$. Then concatenate these paths together to get a smooth path $\alpha \colon \mathbb{R} \to \lim V_k$ with image not contained in any finite dimensional subspace.

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In the same spirit as Andrew's answer: if you want to understand the general context for commutation of direct limits with functors of the form Hom(X,-), you could try looking up "finitely presentable object". –  Tom Leinster Mar 12 '12 at 15:24
    
Thank you guys! –  Vít Tuček Mar 12 '12 at 16:23

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