0
$\begingroup$

So I've read Loring Tu's Introduction to Manifolds and now am looking at Guillemin and Pollack's Differential Topology. Tu uses "intrinsic" definitions of smooth manifolds and the like i.e. a certain kind of topological space with a maximal atlas and such. G&P consider manifolds as subsets of $\mathbb{R}^N$ for some $N$ s.t. at each point the manifold is locally diffeomorphic to $\mathbb{R}^k$ for some fixed $k$ (i.e. there exists an open (in the manifold $M$) neighborhood $U$ of a point $x$ that is diffeomorphic to an open set $V$ in $\mathbb{R}^k$; diffeomorphic is a a smooth bijection with a smooth inverse, see below for smooth on non-open subsets of Euclidean space). Whatever, there is some theorem that every manifold can be embedded into Euclidean space (which I haven't seen a proof of but that's besides the point) and anyway the only manifolds I care about I constructed as subsets of some large ambient Euclidean space anyway, and clearly a G&P manifold is a Tu manifold.

I want to understand the correspondence between the definitions for manifolds defined intrinsically and as subsets of Euclidean space. In this question, I'm considering the definition of a smooth map. Let's first go with Tu's definition. Let $M$, $N$ be manifolds and $f: M \to N$ a continuous function. Let $p \in M$. Say $f$ is smooth at $p$ if there are charts $(U, \phi), (V,\psi)$ s.t. $p \in U, f(p) \in V$ and $\psi \circ f \circ \phi^{-1}: \phi(f^{-1}(V) \cap U) \to \mathbb{R}^n$ is smooth. $f$ is smooth if $f$ is smooth at $p$ for every $p \in M$.

Now G&P. Let $X \subseteq \mathbb{R}^N$ for some $N$. Call $f: X \to \mathbb{R}^m$ smooth if for all $x \in X$ there exists an open neighborhood $U\subseteq \mathbb{R}^n$ of $x$ and an extension of $f|_{U \cap X}$ to $U$ that is smooth. Then for $Y \subseteq \mathbb{R}^m$, we have $f: X \to Y$ is smooth if it is smooth as a function $f: X \to \mathbb{R}^m$.

It is easy to verify G&P smooth implies Tu smooth for embedded submanifolds of Euclidean space. How does one show the converse, i.e. supposing a function between two embedded submanifolds of Euclidean space is Tu smooth, why is it G&P smooth. This seems to involve constructing extensions of smooth functions.

$\endgroup$
4
  • $\begingroup$ What is the definition of "locally diffeomorphic to $\mathbb R^k$ for some $k$" in G&P? By the way, it cannot be a point having this property. $\endgroup$
    – Paul Frost
    Jan 17 '20 at 0:31
  • $\begingroup$ I made some edits to clarify; the wording was bad. Note though that the section you are referring to is primarily for context and not necessarily integral to the proper question. $\endgroup$ Jan 17 '20 at 0:42
  • $\begingroup$ The easiest way to get local smooth extensions is to apply the immersion theorem to get local coordinates $(x_1,\dots,x_N)$ on $\Bbb R^N$ so that the piece of $X$ looks like $x_{k+1}=\dots=x_N=0$. Given a function $f(x_1,\dots,x_k)$, the obvious extension is $F(x_1,\dots,x_N) = f(x_1,\dots,x_k)$. (Geometrically, we're taking a tubular neighborhood and extending $f$ by letting it be constant, say, on normal cross-sections.) $\endgroup$ Jan 17 '20 at 5:00
  • $\begingroup$ The defnition in G&P is essential. See my answer. $\endgroup$
    – Paul Frost
    Jan 17 '20 at 12:02
0
$\begingroup$

Your question is essentially covered by my answer to Equivalent definition of a tangent space? However, my answer did not focus on the concept of a smooth map, so let me do it properly here.

First note that property "Tu-smooth at $x \in X$" does not depend on the choice of charts around $x$ and $f(x)$. This is a basic theorem.

G&P define a manifold as a subset $X \subset \mathbb R^N$ if for each $x \in X$ there exist an open neighborhood $U$ in $X$ and a diffeomorphism $\psi : U \to V$ onto an open $V \subset \mathbb R^k$. Here "diffeomorphism" means that $\psi$ is a bijection such that both $\psi, \psi^{-1}$ are "G&M-smooth" in the sense defined in your question.

The first thing we have to do is to endow a G&M-manifold with a differentiable structure in the sense of Tu. As an atlas take the family of all local diffeomorphisms $\psi : U \to V$ as above. Now have a look at the above link. You will see that for each $z \in U$ there exist an open neigborhood $W$ of $z$ in $\mathbb R^N$, $W \cap X \subset U$, and a smooth extension $F : W \to V$ of $\psi \mid_{W \cap X}$.

Now the transition map between charts $\psi : U \to V$ and $\psi' : U' \to V'$ is given by $$\psi' \circ \psi : \psi^{-1}(U \cap U') \to \psi'(U \cap U') .$$ But this map is smooth: We know that $\overline{\psi^{-1}} : U \to \mathbb R^N, \overline{\psi^{-1}}(\zeta) = \psi^{-1}(\zeta)$, is smooth, and that around each $z = \psi(\zeta)$ with $\zeta \in \psi^{-1}(U \cap U')$ we have a smooth extension $F' : W' \to U'$ of $\psi' \mid_{W' \cap X}$ for some open neigborhood $W'$ of $\psi(\zeta)$ in $\mathbb R^N$. We may assume that $W' \subset U \cap U'$. This shows that $\zeta$ has an open neighborhood conatined in $\psi^{-1}(U \cap U')$ on which $\psi' \circ \psi$ agrees with $F \circ \overline{\psi^{-1}}$ which is smooth.

Thus our atlas is smooth and generates a differentiable structure on $X$.

It is then easy to see that the inclusion $j : X \to \mathbb R^N$ is Tu-smooth. In fact, let $\psi : U \to V$ be a chart in our atlas. On $\mathbb R^N$ take the identity as a chart. Then $id \circ j \circ \psi^{-1} = \overline{\psi^{-1}}$ which is smooth.

Now let us verify that Tu-smooth maps are G&P-smooth. Let $f : X \to Y$ be Tu-smooth at $x \in X$. Then also $\bar f = j \circ f : X \to \mathbb R^N$ is Tu-smooth. This means that for any chart $\psi : U \to V$ around $x$ tha map $\bar f \circ \psi^{-1} : V \to \mathbb R^N$ is smooth. There is a smooth extension $F : W \to V$ of $\bar f \mid_{W\cap X}$, where $W$ is some open neigborhood of $x \in \mathbb R^N$. Thus also $ \bar f \circ \psi^{-1} \circ F : W \to \mathbb R^N$ is smooth. Clearly it is an extension of $\bar f \mid_{W\cap X}$.

$\endgroup$
1
  • $\begingroup$ I appreciate the answer, but I was still left confused about some things (in particular the third to last sentence asserts the existence of a smooth extension, but I don't see why one necessarily exists, at least in the context given). I have spent a good chunk of time demonstrating the correspondence between Tu's definitions and G&P's definitions to my own satisfaction; I'm going to attach my write up as a new answer. $\endgroup$ Jan 19 '20 at 23:20
0
$\begingroup$

I was still confused after looking at Paul's answer and wanted to just work through everything from first principles to make sure I understood. I've written up a somewhat lengthy but very explicit demonstration of the correspondence between Tu's and G&P's definitions of smooth manifolds, smooth maps, tangent spaces, and differentials. I will leave a link to the pdf of the write up here for anyone who might want it:

https://drive.google.com/file/d/1WvfwaKvKLp13Swioqy0ElLb7zLxisi_D/view?usp=sharing

$\endgroup$

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.