# Can you extend vector fields on a manifold?

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)

• 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 – Sak Jun 29 '11 at 22:27

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.
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.