# Is the hypersurface of class $C^k$ a $C^k$-differentiable manifold?

In Folland's Introduction to Partial Differential Equations:

A subset $S$ of ${\mathbb R}^n$ is called a hypersurface of class $C^k$($1\leq k\leq\infty$) if for every $x_0\in S$ there is an open set $V\subset{\mathbb R}^n$ containing $x_0$ and a real-valued function $\phi\in C^k(V)$ such that $\nabla\phi$ is nonvanishing on $S\cap V$ and $$S\cap V=\{x\in V:\phi(x)=0\}.$$

In this case, by the implicit function theorem we can solve the equation $\phi(x)=0$ near $x_0$ for some coordinate $x_i$---for convenience, say $i=n$---to obtain $$x_n=\psi(x_1,\dots,x_{n-1})$$ for some $C^k$ function $\psi$. A neighborhood of $x_0$ in $S$ can then be mapped to a piece of the hyperplane $x_n=0$ by the $C^k$ transformation $$x\to(x',x_n-\psi(x'))\qquad (x'=(x_1,\dots,x_{n-1}))$$ This same neighborhood can also be represented in parametric form as the image of an open set in ${\mathbb R}^{n-1}$(with coordinate $x'$) under the map $$x'\to(x',\psi(x')).$$

Here is my question:

Is the above definition equivalent to say that $S$ is a $C^k$-differentiable manifold?

I learned from S.S. Chern 's Lectures on Differential Geometry the definition as following:

Suppose $M$ is an m-dimensional topological manifold. If a given set of coordinate charts ${\mathcal A} = \{(U,\phi_U),(V,\phi_V),(W,\phi_W),\cdots\}$ on $M$ satisfies the following conditions, then we call ${\mathcal A}$ a $C^r$-differentiable structure on $M$:
1). $\{U,V,W,\cdots\}$ is an open covering of $M$;
2). any two coordinate charts in ${\mathcal A}$ are $C^r$-compatible;
3). ${\mathcal A}$ is maximal, i.e., if a coordinate chart $(\tilde{U},\phi_{\tilde{U}})$ is $C^r$-compatible with all coordinate charts in ${\mathcal A}$, then $(\tilde{U},\phi_{\tilde{U}})\in{\mathcal A}$.
If a $C^r$-differentiable structure is given on $M$, then $M$ is called a $C^r$-differentiable manifold.

1) is not hard to find, 3) can be obtained once one has a covering by compatible charts. I am not able to get 2).

For some particular case, e.g., $S=S^1$ in ${\mathbb R}^{2}$, the answer is yes. However, for arbitrary $S$, I don't know how to find a covering of $S$ by compatible charts.

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@Ryan: Thanks for your response in MO. I move the question here. –  Jack Jun 11 '11 at 14:44

The version of the Implicit Function Theorem that I like best is a little simpler than the one you're using. It goes like this.

Let $U \subset \mathbb R^n$ be open and $f : U \to \mathbb R^m$ be a $C^k$-function on $U$.

Given $p \in U$ and $Df_p : \mathbb R^n \to \mathbb R^m$ be an onto linear function (equivalently the derivative matrix has rank $m$).

Then there exists a function $\phi : \mathbb R^n \to \mathbb R^{n-m}$ such that $f \oplus \phi : U \to \mathbb R^m \oplus \mathbb R^{n-m} \equiv \mathbb R^n$ is $C^k$ and its derivative at $p$ is an isomorphism (equiv. invertible matrix). So the implicit function theorem applies. Technically my definition is $(f \oplus \phi)(x) = (f(x), \phi(x))$. Usually $\phi$ is chosen to be a linear isomorphism between the kernel of $Df_p$ and $\mathbb R^{n-m}$.

So $f \oplus \phi_{|W}$ is a diffeomorphism onto its image, in particular $(f \oplus \phi)_{|W}^{-1}$ when restricted to $(f\oplus \phi)(W)$ intersected with $\mathbb R^m \times \{0\}^{n-m}$ is the parametrization of $f^{-1}(f(p))$ in the neighbourhood $W$ of $p$. i.e. the function you get from your version of the Implicit Function Theorem.

Anyhow, this formalism provides you with diffeomorphisms. And being a diffeomorphism is a transitive relation under composition. So if you get two charts this way, you can check compatibility by extending them from the `parametrization' ${(f \oplus \phi)_{|W}^{-1}}_{|\mathbb R^m \times \{0\}^{n-m}}$ to the diffeomorphism $(f \oplus \phi)_{|W}^{-1}$.

Also note that since the inverse function theorem preserves the order of differentiability, everything above is at least $C^k$.

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Thanks for your answer. I think your version of the Implicit Function Theorem is the same as this, right? –  Jack Jun 11 '11 at 16:29
I don't think so but perhaps I missed the point where they convert from $f$ to $f \oplus \phi$ i.e. replacing a function with another function whose domain and range have the same dimension. –  Ryan Budney Jun 11 '11 at 16:37
You can find this version in Spivak's "Calculus on Manifolds" also many introductory manifolds textbooks, like Conlon's. –  Ryan Budney Jun 11 '11 at 16:38
Actually, the version of the implicit theorem I posted here is also from Folland's Introduction to PDE. As I understand, what matters here is the dimension of ${\mathbb R}^m$ in your answer. When $m=1$, I will get the version I put. –  Jack Jun 11 '11 at 16:44