# When is a chart of a submanifold not only a homeomorphism, but a diffeomorphism?

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:

$$^1$$ 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 $$\mathbb R^d$$.

• 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 8 '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 8 '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 8 '20 at 15:19

• 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 8 '20 at 15:21
• 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 8 '20 at 16:11