2
$\begingroup$

Let $M$ be a differentiable manifold, $U\subseteq M$ an open neighborhood of $p\in M$ and $f:U\to\mathbb{R}$ differentiable. Then there exists $F:M\to\mathbb{R}$ differentiable such that $F=f$ in a neighborhood of $p$.

I found this problem in a book, and it's solved in the case $M=\mathbb{R}^n$, but it says the general case is easy by using charts, but I don't understand it yet.

This is what I have tried. Suppose $\dim M=n$ and take a chart $(V,\phi)$ with $p\in V$. Then $$f\circ\phi^{-1}:\phi (U\cap V)\to \mathbb{R}$$

is differentiable, where $\phi(U\cap V)$ is a neighborhood of $\phi(p)$ in $\mathbb{R}^n$.

In $\mathbb{R}^n$, we can now find $F:\mathbb{R}^n\to\mathbb{R}$ such that $F=f\circ\phi^{-1}$ in some neighborhood of $\phi (p)$, say $V_0\subseteq\phi(U\cap V)$.

From here I can't see how to obtain a differentiable function $M\to\mathbb{R}$. First I thougt something like $F\circ\phi$, but it's not defined in the whole manifold $M$.

Any hint? Thank you.

$\endgroup$
1
$\begingroup$

Hint multiply f by a cut off function g such that the support of g is contained in the domain of a chart which contains $p$ and whose value at $p$ is 1.

$\endgroup$
  • $\begingroup$ these are often called "bump" functions. $\endgroup$ – James S. Cook Feb 29 '16 at 3:04
  • $\begingroup$ We can take the cutt off function to be constant equal to 1 in a neighborhood of $p$ so that the resulting function $F$ is equal to $f$ in this neighborhood. $\endgroup$ – Eduardo Longa Feb 29 '16 at 3:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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