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

Let $\psi \colon M \to N$ be $C^\infty$. A smooth vector field X along $\psi$ (that is, $X \in C^\infty(M,T(N))$ and $\pi \circ X = \psi$) has local $C^\infty$ extensions in $N$ if given $m \in M$ there exists a neighborhood $U$ of $m$ and a neighborhood $V$ of $\psi(m)$ such that $\psi(U) \subset V$ and there is a $C^\infty$ vector field $Y$ on $V$ such that $Y \circ \psi|U = X|U$.

Does a $C^\infty$ vector field $X$ along an immersion $\psi \colon M \to N$ always have local $C^\infty$ extensions in $N$?


locked by S. Carnahan Jun 12 '11 at 8:47

closed as no longer relevant by Qiaochu Yuan, Jorge Vitório Pereira, S. Carnahan Jun 12 '11 at 8:47

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Why do you bother asking this (for the second time) if you don't read the answers? – Steven Landsburg May 10 '11 at 4:35
@Steven Landsburg it's not exactly the same question. The answer is yes, because for every point $m\in M$ there are local charts $\Phi$ and $\Psi$ of $M$ and $N$ respectively, around $m$ and $\psi (m)$ respectively, such that $\psi (\mathrm{Domain~of~}\Phi)\subset\mathrm{Domain~of~}\Psi$ and such that for all $(x_1,\dots,x_d)\in\mathbb{R}^d$, where $d$ is the dimension of $M$, $\Psi\psi\Phi^{-1}(x_1,\dots,x_d)=(x_1,\dots,x_d,0,\dots,0)$. It's a corollary to the local inversion theorem. In this chart you can extend your vector field along $\psi$ by letting it be constant on the "horizontal". – Olivier Bégassat May 10 '11 at 4:51
This looks like the third time a question like this has been asked. The answer is no for an immersion, because the map $\psi$ does not need to be one-to-one. In particular, if there are distinct points $p \ne q \in M$ such that $\psi(p) = \psi(q)$ but $X(p) \ne X(q)$, then there is no way to define a vector field $Y$ on any neighborhood of $\psi(p) = \psi(q)$. – Deane Yang May 12 '11 at 2:18
Dear Ralph, if you want to delete a question, please use the "delete" button instead of converting your question into a sequence of periods. – S. Carnahan May 25 '11 at 3:38
IMO the question should be closed as the question-asker appears to be a non-participant. – Ryan Budney Jun 11 '11 at 19:18

I've been thinking about extending the vector field using a bump function. I'm not entirely sure as to what is meant by the "horizontal".

There's no need for a bump function, since the desired extension is local (in the neighborhood of a point) anyway. – Deane Yang May 12 '11 at 14:02

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