1
$\begingroup$

On page 4 of the book "Differential Topology" (written by Amiya Mukherjee) the following is written:

[...] observe that the charts $(U,\phi)$ and $(U,\alpha\circ \phi )$, where $\alpha:\mathbb{R}^n\to \mathbb{R}^n$ is a diffeomorphism, are always compatible. In particular, taking $\alpha$ to be the translation which sends $\phi(p)$ to $0$, we can always suppose that every point $p\in M$ admits a coordinate chart $(U,\phi)$ such that $\phi(p)=0$. We may also suppose that $\phi(U)$ is a convex set, or the whole of $\mathbb{R}^n$.

In that book the word "diffeomorphism" means "$C^\infty$-diffeomorphism" and two charts $(U,\phi)$, $(V,\psi)$ are said to be compatible if $\psi \circ \phi ^{-1}:\phi(U\cap V)\to\psi(U\cap V)$ is a $C^\infty$-diffeomorphism.

My question is about the end of the above quote: "We may also suppose that $\phi(U)$ is a convex set, or the whole of $\mathbb{R}^n$".


Question: Given a chart $(U,\phi)$ how can I prove that exists a chart $(V,\psi)$ such that $(U,\phi)$ and $(V,\psi)$ are $C^\infty$-compatible and $\psi(V)=\mathbb{R}^n$?


I tried to use the questions below to answer my question but I couldn't.

$\endgroup$
1
$\begingroup$

The thing is that we need only to find a subset of $U$ such that the image under chart map $\phi$ is an open ball $B_r(0)$ in $\mathbb{R}^n$. After this, we can blow up the ball to $\mathbb{R}^n$ by a diffeomorphism. Suppose we have a chart $(U,\phi)$ with a point $p \in U$ having $\phi(p)=0 \in \phi(U)\subset \mathbb{R}^n$.

  • Let $V=\phi^{-1}(B_r(0))$ for some $r>0$ and $\psi :=\phi|_V$. Then $(V,\psi)$ is $C^{\infty}$-compatible with $(U,\phi)$ since it is just restriction of the larger chart.

  • Choose your favorite diffeomorphism $\alpha : B_r(0) \to \mathbb{R}^n$, we have new chart $(V,\alpha \circ \psi)$ with $(\alpha \circ \psi)(V) = \mathbb{R}^n$ and $C^{\infty}$-compatible with $(V,\psi)$ as you can verify it by yourself.

  • Therefore $(V,\alpha \circ \psi)$ $C^{\infty}$-compatible with $(U,\phi)$ with $\psi(V)= \mathbb{R}^n$.

$\endgroup$
2
  • $\begingroup$ One question: is there a $C^\infty$-diffeomorphism $\alpha:B_r(0)\to\mathbb{R}^n$? $\endgroup$
    – rfloc
    Jun 10 '20 at 2:48
  • $\begingroup$ @rfloc The accepted answer here math.stackexchange.com/questions/1025308/… provide a diffeomorphism from unit ball to $\mathbb{R}^n$. You can tweak it easily by scaling to any non-unit radius ball. $\endgroup$
    – Brown Bear
    Jun 10 '20 at 2:51

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.