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I know that not necessarily you can extend a smooth vector field defined over a subset of a manifold to ALL of the maniffold, but, can you extend it at least to an open set? (Of course I'm talking about smooth extensions)

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What is a smooth vector field over a subset which is not already open? –  Qiaochu Yuan Jun 29 '11 at 0:35
    
Why did you vote down? Thank you for your insight –  Chu Jun 29 '11 at 22:27

3 Answers 3

up vote 11 down vote accepted

You're probably already aware of the result that not every smooth vector field defined on an open subset admits a smooth extension to a larger open set. For example, the vector field $V(x) = \frac{1}{x(x-1)}\ \mathbf{e_1}$ on $(0,1)\subset\mathbb{R}$ admits no smooth extension to a larger open set.

On the other hand, one usually defines a smooth vector field on a non-open subset $A$ as one that admits a smooth extension in a neighborhood of every point of $A$. (More precisely, a smooth vector field $V$ on a set $A$ is one that for every $p \in A$, there is an open set $U$ containing $p$ and a smooth extension $\widetilde{V}$ defined on $U$ whose restriction to $U \cap A$ agrees with $V$.)

However, it is a fact that every smooth vector field $V$ defined on an embedded submanifold $S \subset M$ that is closed as a subset of $M$ admits a smooth extension to all of $M$. I don't remember how the proof goes offhand, but it probably involves a partition of unity argument.

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It involves a partition of unity argument, as well as the tubular neighborhood theorem. –  Aaron Mazel-Gee Jun 29 '11 at 2:13

As Qiaochu has hinted at, the very definition of smoothness of a vector field $V$ on a (presumably smooth) manifold $M$ on an arbitrary subset $S$ is that there is already some open neighborhood $U \supset S$ on which there is a smooth vector field $V'$ restricting to $V$ on $S.$ So the answer is yes, by definition.

edit: oops! I got scooped. :)

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Yes, I agree with Jesse Madnick, you must use unity partitions and define such that field. Look in the Lee book "Introductions to smooth manifolds".

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This does not answer the question. Comments can be left on a question or answer when you have sufficient reputation. –  robjohn Feb 12 '13 at 20:07

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