# What does it mean to say a boundary is $C^k$?

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 ?

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Well, I can explain it to you but I can't guarantee it'll be a "nice" explanation. – Gigili May 28 '12 at 12:45
Roughly speaking, it means that locally the boundary behave like the graph of a $C^k$ function. – Davide Giraudo May 28 '12 at 12:50
@Gigili Please do so , i want to understand . Thanks – Theorem May 28 '12 at 12:52
How much differential geometry do you know? Do you know what a manifold is? – Najib Idrissi May 28 '12 at 13:15
@Ananda: To not beat around the bush, I think that people are pointing out that your demanding a 'nice' explanation rubs some people the wrong way. In general, this question feels as though you're ordering others around, which does not make me want to write an answer. We also have no idea what your background is. Really - how much differential geometry do you know? In what context is this question coming up? If you put more into your question, I'm sure more people will be willing to answer it. – mixedmath May 28 '12 at 14:49

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 $C^k$.

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