I'm having difficulties proving lemma 3 of Milnor, Topology from the differentiable point of view. The lemma intends to give us a way to generate examples of mainfold with boundary. The lemma states the following:

Let $M$ be a smooth manifold without boundary and let $g:M \to \mathbb{R}$ have $0$ as regular value. The set of $x$ in $M$ with $g(x) \geq 0$ is a smooth manifold, with a boundary equal to $g^{-1}(0)$.

What I have so far:

I supposed $g$ is a smooth function because of the existence of regular value for $g$ and the general context of the book.

Because $g$ is smooth, $g^{-1}((0,+\infty))$ is open in $M$. Therefore, $g^{-1}((0,+\infty))$ is a smooth manifold of dimension $m$, as is $M$. By a preceding lemma, we know $g^{-1}(0)$ is a smooth manifold of dimension $m-1$. In that case, for every point $x \in g^{-1}(0)$ there exists an open neighborhood of $x$ diffeomorphic to an open set in $\mathbb{R}^{m-1}$.

To conclude:

To conclude the proof that $g^{-1}([0,+\infty))$ is a smooth manifold, I need to show that we can map with a diffeomorphism every small enought neighborhood in $g^{-1}([0,+\infty))$ to $V \cap H^{m}$, with the points of the boundary $g^{-1}(0)$ mapped to $\partial H^m$. Here, $H^m$ is the half-space $\{(x_1,...,x_m) \in \mathbb{R}^m\vert x_m \geq 0 \}$ and $V$ is an open set of $\mathbb{R}^m$.

I do not know how to proceed from here. Any help is greatly appreciated.


There are several ways of showing this; here's a sketch of one path:

Fix $p\in g^{-1}(\{0\})$. Since you have already showed that $g^{-1}(\{0\})$ is an embedded codimension $1$ submanifold, you can choose local coordinates $(x^1,\cdots,x^{n-1})$ with domain $U$, where $p\in U\subseteq g^{-1}(\{0\})$. Shrinking $U$ as necessary, we can arbitrarily extend the $x^i$ to smooth functions $M\to\mathbb{R}$. One can then use inverse function theorem to show that $(x^1,\cdots,x^{n-1},g)$ forms a coordinate chart on $M$ when restricted to a sufficiently small neighborhood of $p$, and restricting this map to $g^{-1}([0,\infty))$will give the desired boundary chart.

  • $\begingroup$ Thank you for your answer, I have a few questions. i) I am not familiar with the $(x^{1},...,x^{n-1})$ notation. Are the exponent indexes of the functions $x$? ii)How do you know you can appply the inverse function theorem on $(x^1,...,x^{n-1},g)$? $\endgroup$
    – Momo
    May 20 at 17:36
  • $\begingroup$ @Momo i) Yes , they're just indices.; each is a local smooth function $U\to\mathbb{R}$. ii) The point is that, taken together, the functions $(x^1,\cdots,x^{n-1},g)$ form a map $M\to\mathbb{R}^n$, and the differential of this map is invertible at $p$. This can be shown using the fact that $x^i$ are coordinate functions for $g^{-1}(\{0\})$ at $p$, and that $p$ is a regular point of $g$. $\endgroup$
    – Kajelad
    May 20 at 18:05

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.