5
$\begingroup$

I'm trying to prove the first part of Proposition 5.47 of Lee's Smooth Manifolds, which is left to the reader. It says

Suppose $M^m$ is a smooth manifold, and $f\colon M\to\mathbb{R}$ smooth. For each regular value $b$ of $f$, the sublevel set $f^{-1}(-\infty,b]$ is a regular domain, that is, a properly embedded codimension $0$ submanifold with boundary.

First, $f^{-1}(\infty,b)$ is open, hence an embedded submanifold of codimension $0$. Also, $f^{-1}(-\infty,b]$ is closed in $M$, so if $f^{-1}(-\infty,b]$ is a embedded submanifold, it is in fact a properly embedded submanifold of codimension $0$.

I want to show $S:=f^{-1}(-\infty,b]$ satisfies the local $m$-slice condition. If $p\in f^{-1}(-\infty,b)$, then since this set is open, we can find a chart $(U,\varphi)$ around $p$ in $S$. But then $\varphi(S\cap U)=\varphi(U)$, so $(U,\varphi)$ is an $m$-slice chart around $p$.

I suspect $f^{-1}(b)$ is the boundary of $S$. Since $f^{-1}(b)$ is a regular level set, it is a properly embedded submanifold of dimension $m-1$ in $M$. I could then find an $m-1$ slice chart $(U,\varphi)$ in $M$ for $f^{-1}(b)$, so that $$ \varphi(f^{-1}(b)\cap U)=\{(x^1,\dots,x^m)\in\varphi(U):x^m=0\} $$

I want to try to modify it somehow to a chart such that $$ \varphi(U\cap S)=\{(x^1,\dots,x^m)\in\varphi(U):x^m\geq 0\} $$ to show it is an $m$-dimensional half slice. Is there maybe a way to restrict to a precompact open set, so that the coordinate functions achieve a mimnimum, and then just shift the coordinate map so the last coordinate is always nonnegative?

$\endgroup$
2
$\begingroup$

By submersion theorem, after suitable coordinate transformation, your map should look like: $$ f:(x^1,\dots,x^m)\mapsto x^m $$ in a neighborhood for each regular point in $f^{-1}(b)$. And the claimed conclusion is obvious: you only need to worry about two types of points, the interior points and the boundary points. For an interior point $x\in M$,i.e. $f(x)<b$, everything is fine: $f^{-1}((f(x)-\varepsilon,f(x)+\varepsilon))$ is the desired neighborhood for $x$, which (possibly after shrinking) is homeomorphic to $\mathbb R^m$. For a boundary point, on the other hand, you need to show the existence of a neighborhood homeomorphic to $\mathbb H^m=\{x^m\leq0\}$, which is also automatic under the aforementioned canonical form.

$\endgroup$
  • $\begingroup$ I think I see now, thanks. $\endgroup$ – Clara Aug 15 '14 at 23:51

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.