MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.