I need a explanation on what does it mean to say a boundary is $C^k$. Can anyone help me please.
And also need some explanation on how to straighten boundary ?
What does it mean to say a boundary is $C^k$?
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0@mixedmath: sorry for posing question in such a way. I will edit the question. I know very less of differential geometry . But i am trying to learn this text : http://www.thehcmr.org/issue1_2/poincare_lemma.pdf . – 2012-05-28
1 Answers
I don't know how much following will help you. But have it:
Let $M$ be any topological space such that each point $p\in M$ has a neighborhood $U\subset M$, homeomorphic to some open set of $[0,\infty)\times \mathbb R^{n-1}$ or open set of $\mathbb R^n$ via homeomorphism $\phi$. We say that $(U,\phi)$ is a chart around $p$. If every point has a such neighborhood, we say $M$ as topological manifold with boundary. Now as in wikipedia,(for terminology click on the link)
A smooth(C^k) manifold with boundary is a topological manifold with boundary equipped with an equivalence class of atlases whose transition maps are all smooth ($C^k$ ).
Difference with smooth manifold(C^k) and smooth manifold(C^k) with boundary is that for defining smooth manifold with boundary, we are allowing chart from open set of $[0,\infty)\times \mathbb R^{n-1}$ as well as open set of $\mathbb R^n$.
Here main point is that: If $p\in M$ has a neighborhood $U$ which is diffeomorphic ($\phi$) to some open set of $[0,\infty)\times \mathbb R^{n-1}$ such at $\phi(p)=0$, then there doesn't exists any coordinate neighborhood of $p$ which can be diffeomorphic to open set contained in interior of $[0,\infty)\times \mathbb R^{n-1}$. Hence following definition makes sense:
Now define the boundary of smooth manifold with boundary as: $\partial M:=\{p\in M: \text{ there is a coordinate chart }(U,\phi) \text{such that } \phi(p)=0 \}$
We can prove that $\partial M$ is topological manifold of dimensional $n-1$. If this manifold has differential structure of $C^k$, we say that boundary is of $C^k$.