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Let $M \subseteq \mathbb{R}^{n}$ be an embedded (smooth) submanifold of dimension $m$, with $c \in M$. Show that there is an open set $U \subseteq \mathbb{R}^{n}$ containing $c$, and a smooth map $f : U \cap \mathbb{R}^{m} \to \mathbb{R}^{n-m}$, such that $U \cap M = \{ (x, f(x)) \in \mathbb{R}^{m} \times \mathbb{R}^{n-m} \equiv \mathbb{R}^{n} \,|\, x \in U \cap \mathbb{R}^{m} \}$.

We've been given two hints: we can assume that both $c = 0^{n}$ and $T_{c}M = \mathbb{R}^{m} \times \{ 0^{n-m} \}$ (euclidean isometry of $\mathbb{R}^{n}$), and that we should consider the inverse of a chart, postcomposed with projection to $\mathbb{R}^{m}$, and apply the inverse function theorem.

I'm not quite sure what the use the inverse function theorem on, since the inverse of a chart (say $\phi_{\alpha}$) postcomposed with projection to $\mathbb{R}^{m}$ (say $\pi$) is a map $$ \pi \,\circ\, \phi_{\alpha} : \mathbb{R}^{m} \to \mathbb{R}^{m}. $$ The inverse function theorem formulation that I have is the following.

Suppose $M$, $N$ are smooth $n$-manifolds, and that $f : M \to N$ is smooth. Suppose that $x \in M$ and that $d_{x}f$ is invertible. Then there are open sets $U_{1} \subseteq M$ and $U_{2} \subseteq N$ with $x \in U_{1}$ such that $f|U_{1}$ is a diffeomorphism to $U_{2}$.

The inverse function theorem requires a smooth map between two manifolds of the same dimension, but the above map isn't necessarily smooth (nor do I see how it would help get the required map or open set).

I'm sure that this is fairly simply but I am totally stuck on it.

Bilbottom
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1 Answers1

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You write

The inverse function theorem requires a smooth map between two manifolds of the same dimension, but the above map isn't necessarily smooth (nor do I see how it would help get the required map or open set).

Let's start with the second part. Suppose you knew that there was an open set $U' \subset R^m$ such that $h = \pi \circ \phi_a: U \to V' \subset \Bbb R^m$ was a diffeomorphism. Then $U = \phi_a(U') \times \Bbb R^{n-m}$ would be a pretty good candidate for the $U$ that you're looking for.

So what about the first part? $\pi$ is certainly smooth, and $\phi_a$ is required to be smooth in most definitions of a manifold that I know. So you have a smooth map, as far as I can tell. Or am I missing something here?

John Hughes
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  • You're certainly right about the first part --- the map isn't necessarily a diffeomorphism is what I meant, my apologies! I am still unsure where to go from here: $h$ is a map $\mathbb{R}^{m} \to \mathbb{R}^{m}$, not $\mathbb{R}^{m} \to \mathbb{R}^{n-m}$, so I don't see how this would give us the structure of $U \cap M = { (x, f(x)) ,|, x \in U \cap \mathbb{R}^{m}$, particularly since $\phi_{\alpha}(U')$ is a subset of $\mathbb{R}^{n}$ and so I am unclear as to why the construction $U = \phi_{\alpha}(U') \times \mathbb{R}^{n-m}$ would be the correct candidate. – Bilbottom Jan 03 '18 at 13:19
  • @Bilbottom Sorry to dredge this up, but at the risk of being seen as exceedingly dumb, are you absolutely sure you were told to "postcompose with projection to $R^m$" and not "postcompose with projection to $R^{n-m}$? Of course this question is six years old now and I could easily understand this comment being an annoyance, but I'm in the middle of researching this topic and just had to post it. – user167131 Mar 02 '24 at 16:07
  • @user167131 no need to apologise -- though, I don't think I ever solved this so I can't say for sure whether it should be $R$ or $R^{n-m}$ – Bilbottom Mar 03 '24 at 17:54