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I've got trouble to understand the concept of a "smooth structure" associated to a submanifold.

Let $\mathbb H^k:=\mathbb R^{k-1}\times[0,\infty)$. Say $M\subseteq\mathbb R^d$ is a $k$-dimensional embedded submanifold

  1. without boundary if $M$ is locally homeomorphic to $\mathbb R^k$;
  2. with boundary if $M$ is locally homeomorphic to $\mathbb H^k$.

If I didn't make a mistake, (1.) should imply (2.): If $x\in M$, then (since $\mathbb R^d$ is locally compact) there is a homeomorphism $\varphi$ from a compact neighborhood $\Omega$ of $x$ onto an open subset $U$ of $\mathbb R^k$. Now $\varphi-\inf_\Omega\varphi_k$ is a homeomorphism from $\Omega$ onto $U-\inf_\omega\varphi_k\subseteq\mathbb H^k$.

Now $(\Omega,\phi)$ is called a $k$-dimensional chart of $M$ if $\Omega$ is an open subset of $M$ (equipped with the subspace topology) and $\phi$ is a homeomorphism from $\Omega$ onto an open subset of $\mathbb R^k$ or $\mathbb H^k$. In the first case, it is called an interior chart and if in the second case it additionally holds $\phi(\Omega)\cap\partial\mathbb H^k=\emptyset$, then it is called a boudary chart.

If $(\Omega_i,\phi_i)$ is a $k$-dimensional chart of $M$, then $(\Omega_1,\phi_1)$ and $(\Omega_2,\phi_2)$ are called $C^\alpha$-compatible, if $\phi_2\circ\phi_1^{-1}:\phi_1(\Omega_1\cap\Omega_2)\to\phi_2(\Omega_1\cap\Omega_2)$ is a $C^\alpha$-diffeomorphism.$^1$ Now an atlas $\mathcal A$ for $M$ is a collection of charts whose domain cover $M$ and $\mathcal A$ is called $C^\alpha$-atlas if any two of its charts are $C^\alpha$-compatible.

Now my question is: If I got such an atlas $\mathcal A$, is it somehow possible to show that the charts itself are $C^\alpha$-diffeomorphism?

What I also want to know: If $x\in M$, then a function $f$ from $M$ into a Banach space is called $C^\alpha$-differentiable at $x$, if there is a chart $(\Omega,\phi)$ of $M$ with $x\in\Omega$ and $f\circ\phi^{-1}$ is $C^\alpha$-differentiable at $\phi(x)$. How strongly does this notion depend on the particular chart? I guess we can show that it also follows that $f\circ\psi^{-1}$ is $C^\alpha$-differentiable at $\psi(x)$ for any other chart $\psi$ which is $C^\alpha$-compatible to $\phi$; but can we show more?

EDIT: It seems like I've found the claim in this book, but I cannot really follow the argumentation given there:

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$^1$ Is this notion well-defined even when one of the $(\Omega_i,\phi_i)$ is a boundary chart? Note that I say that a function on an arbitrary subset of $\mathbb R^d$ is differentiable, if it is the restriction of a differentiable map on an open subset of $\mathbb R^d$.

0xbadf00d
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  • Your "If I didn't make a mistake" paragraph does seem to contain some mistakes. First, the implication goes the other way: if $x \in M$ has a chart that satisfies (1) then $x$ has a (possibly different) chart that satisfies (2.); but the converse is false. Also, referring to $\Omega$ as a "compact" neighborhood is confusing; generally speaking compactness should not be required of the domain of a chart. – Lee Mosher Jul 08 '20 at 14:46
  • Also, when you write "...is it somehow possible to show that the charts itself are $C^\alpha$-diffeomorphism", you have included what it means for two charts to be $C^\alpha$-compatible but you have not included what it means for one chart to be a $C^\alpha$-diffeomorphism. – Lee Mosher Jul 08 '20 at 14:48
  • @LeeMosher Thank you for your comment. I did intend to write that "(1.) implies (2.)";. So, that was only a typo. And I didn't require for a general chart that its domain is a compact neighborhood of some point. The idea is that by (1.) there is a open neighborhood $\Omega$ of $x$ and a homeomorphism of $\Omega$ onto an open subset of $\mathbb R^k$. And by local compactness of $\mathbb R^d$, we may assume that $\Omega$ is a compact neighborhood. This is crucial, since otherwise $\inf_\Omega\varphi_k$ might not be finite. – 0xbadf00d Jul 08 '20 at 15:19

1 Answers1

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I think the answer in ''no'' since the topological manifold has no intrinsic differentiable structure. Thus you can not define ''differentiability'' both ways. Hence charts can not be diffeomorphisms themselves.

astro
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  • You got me wrong: As I wrote, I assume that we are given an atlas $\mathcal A$ and the question is whether we can show that the charts **belonging to $\mathcal A$ are diffeomorphisms. By definition of $\mathcal A$, we only know that the transition between two such charts are diffeomorphisms; not if the charts itself are diffeomorphisms. – 0xbadf00d Jul 08 '20 at 15:21
  • Then I guess you didn't understand my answer. – astro Jul 08 '20 at 15:59
  • Observe that the preimage of the $\phi$'s may be differentiable but the $\phi$'s themselves can not, according to what I explained in the answer. This can only have an affirmative answer if the manifolds considered are open subsets of $\mathbb{R}^k$ for some $k$, but for general abstract manifolds there is no way to fulfill the definition of differentiabilty straight from $M$. – astro Jul 08 '20 at 16:11
  • Please take a look at Proposition 6.10 here: https://books.google.com/books?id=xQsTJJGsgs4C&pg=PA63&lpg=PA63&dq=%22an+open+subset+of+a+Euclidean+space+can+serve+as+a+coordinate+map%22&source=bl&ots=yPvJMPDvO8&sig=ACfU3U30Q2rvOxks__7BNn8V_aEOgYNu9A&hl=de&sa=X&ved=2ahUKEwi977vxjb7qAhXCURUIHR6jDngQ6AEwAHoECAgQAQ#v=onepage&q=%22an%20open%20subset%20of%20a%20Euclidean%20space%20can%20serve%20as%20a%20coordinate%20map%22&f=false. Isn't it precisely claiming that the charts are diffeomorphisms? – 0xbadf00d Jul 08 '20 at 16:48
  • As you know, everything in mathematics can be made true depending on how we define things and what are the things that we define. In this case, since pages 48 (definition of a manifold) and page 53 (definition of a differentiable structure) are not available, I couldn't know what the author is talking about. But I am certain that you can not say a chart is a diffeomorphism under the standard definition. – astro Jul 08 '20 at 20:34
  • I wouldn't want to be speculative. But, if for the author ''smooth'' is equivalent to be an element of the atlas, then every chart will be a diffeomorphism trivially. But again, this is far from regular usage of the term. I will advise to go over ''Riemannian Geometry'' by Manfredo Do Carmo and see the subtleties of the definition. Cheers. – astro Jul 08 '20 at 20:52
  • I've edited the question and added the relevant definitions of the book. Please take a look. – 0xbadf00d Jul 09 '20 at 04:24
  • I'm not familiar with general manifolds, but meanwhile I think the main difficulty arises by the fact that, generally, the manifold is a Hausdorff space and so not known to be included in an Euclidean space. I guess that's why one needs to consider $(\varphi\circ\gamma)'(0)$ instead of $\gamma'(0)$ for a chart $\varphi$ and a curve $\gamma$ in the manifold. However, the situation in my question is more elementary, since $M$ is a submanifold of $\mathbb R^d$. – 0xbadf00d Jul 09 '20 at 05:02
  • I will take care of reading the definitions above in a carefull manner asa I have some time, but if your context is that all manifolds are taken as subsets of some $\mathbb{R}^d$ then the charts can be defined as diffeomorphisms since they are defined over a set which has more structure so differentiability can be defined straight from $M$ with the structure it inherits from $\mathbb{R}^d$. – astro Jul 09 '20 at 13:17
  • Yes, that's a fact I wasn't aware of before. I think this question describes the answer I'm searching for in a better way then. – 0xbadf00d Jul 09 '20 at 13:19
  • I think the questions are quite not the same. :) – astro Jul 09 '20 at 13:25
  • @0xbadf00d As I conjectured earlier, according to his definition of smoothness, charts are trivially smooth since all charts $\phi$ give rise to the identity in $\mathbb{R}^n$ under the composition $\phi \circ \phi^{-1}$. So in this particular context charts are always smooth, which I confess is somewhat comfortable to live with. Regards! – astro Jul 09 '20 at 13:31
  • I get the point that $\phi\circ\phi^{-1}$ is the identity, which is clearly smooth, but how does it follow that $\phi$ itself is smooth? – 0xbadf00d Jul 09 '20 at 13:49
  • It trivially fulfills definition 6.1 – astro Jul 09 '20 at 14:04
  • Sorry, I missed the fact that he defines the notion of being smooth in a way which conflicts the usual use of this terminology when $M$ is embedded in a Banach space. – 0xbadf00d Jul 09 '20 at 14:50