If a subset $N$ of a manifold $M$ can be written as $f^{-1}(\{0\})$ being $f:M \longrightarrow \mathbb{R}$ a differentiable function, can I conclude that $N$ is a submanifold of $M$?
Asked
Active
Viewed 243 times
2 Answers
0
Not necessarily. Consider for example $M=\mathbb R^2$ and $f(x,y)=xy$. Or, for a more dramatic example, $$ f(x,y) = \begin{cases}0 & x\le 0 \\ ye^{-1/x} & x > 0 \end{cases}$$
hmakholm left over Monica
- 286,031
0
Oh no, you can't: every closed subset of $\mathbb R^n$ is the zero set of a $\mathcal C^\infty$ map $f:\mathbb R^n\to \mathbb R$.
Unbelievable? Maybe.
True? Definitely: this a theorem of Whitney, proved in Madsen-Tornehave page 224.
Georges Elencwajg
- 150,790
-
IMO it is somewhat intuitive, once you think of the distance function $$d(x, A) = \inf_{y \in A} d(x,y)$$ You just need to plug it into something to smooth out the kink on the boundary; e.g. squaring the distance if you just want it to be once differentiable. Or am I totally wrong and this doesn't work? – Nov 19 '14 at 19:32
-
Dear @Hurkyl, I don't know if your idea works but Whitney's proof is rather more complicated. – Georges Elencwajg Nov 19 '14 at 19:47
-
I bet the complication comes from considering an arbitrary manifold rather than just $\mathbf{R}^n$. – Nov 19 '14 at 22:17