微分几何/微分流形（2）

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Before stating the implicit function theorem, recall the following:

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Inverse function theorem

Let \(U \subset \mathbb{R}^n\) be open , \(F :U \rightarrow \mathbb{R}^n\) be smooth, and assume that the Jacobi matrix $ dF|_p$ is invertible for some \(p \in \mathbb{R}^n\) .Then $\exists open\ nbhs. V \subset \mathbb^n $of \(p\) and $W \subset \mathbb^n $ of \(F(p)\),such that

\[ F : V \rightarrow W \]

is invertible and \(F^{-1} : W\rightarrow V\) its inverse is smooth

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• $F $ is not necessarily globally invertible
• there are version of the inverse function theorem for e.g. analytic maps , holomorphic functions, Frechet-differentiable maps between Banach spaces, and of course for smooth maps between smooth manifolds

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Inverse function theorem (smooth version)

Let \(f : \mathbb{R}^n \cross \mathbb{R}^m \rightarrow \mathbb{R}^m , (x,y)\mapsto f(x,y)\) , be smooth, \(f(p) = 0\) for \(p = (x_0,y_0) \in \mathbb{R}^n \cross \mathbb{R}^m\),and assume that that the Jacobi matrix of f with respect to y at p

\[ d_yf|_p = \left[ \matrix{ \frac{df_1}{dy^1}(p) &\cdots& \frac{df_1}{dy^m}(p)\\ \vdots &\ddots & \vdots\\ \frac{df_m}{dy^1}(p) &\cdots& \frac{df_m}{dy^m}(p) }\right] \]

is invertible. Then \(\exists\) open nbhs. \(U\subset \mathbb{R}^n\) of \(x_0\) and \(V \subset \mathbb{R}^m\) of $y_0, s.t. \exists $ a unique smooth map $g : U \rightarrow V $ fulfilling

\[ f(x,y)=0, x\in U , u \in V \Leftrightarrow y=g(x) \]

in particular we have \(g(x_0)=y_0\)

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• there are other versions of the IFT , e.g. if f is analytic g will be analytic, or a version of the IFT for Banach spaces
• the IFT follows from the inverse function theorem

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Important examples of smooth manifolds are the following

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Definition

An \(m < n\) -dimensional smooth submanifold of \(\mathbb{R}^n\) is a subset \(M \subset \mathbb{R}^n\) ,such that $\forall p 'in M \exists $ open nbh. \(U \subset \mathbb{R}^n\) of p and a smooth map $ f : U \rightarrow \mathbb^$ with Jacob matrix of maximal rank n-m for all points in U fulfilling

\[ M \cap U =\{x\in U| f(x)=0 \} \]

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Question: Are smooth submanifolds of \(\mathbb{R}^n\) actually smooth manifolds ?

Answer: Yes! use the IFT

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Lemma

m-dimensional smooth submanifolds M of \(\mathbb{R}^n\) can be locally written as graphs of smooth maps

\[ g : V \rightarrow \mathbb{R}^{n-m} , V\subset \mathbb{R}^m \ open \]

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still need to find atlas on M

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Corollary

For any point p in an m-dim. smooth submfd. $M\subset \mathbb^n \exists \((after possibly reordering coordinates on\)\mathbb^n$) open nbh. \(U \subset \mathbb{R}^n\) of p and a smooth invertible map with smooth inverse

\[ F : U \rightarrow \mathbb{R}^n \]

such that

\[ F|_{U\cap M} : (x^1 , … , x^n) \mapsto (x^1,…,x^m,0,…,0). \]

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Get local coordinates via:

• cover M with open sets U as above

• get charts on M:

  \[ (\varphi = pr_{\mathbb{R}^m} \circ F|_{U\cap M}, U\cap M) \]

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Yet another way to find examples of smooth manifolds is by taking the product of two manifolds:

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Lemma

Let M with atlas \(\mathcal{A}\) and N with atlas \(\mathcal{B}\) be smooth manifolds. Then the **Cartesian product ** of M and N

\[ M \cross N \]

together with the product atlas

\[ \mathcal{A} \cross \mathcal{B} = \{ (\varphi \cross \psi , U \cross V) | (\varphi , U)\in \mathcal{A} , (\psi, V) \in \mathcal{B}\} \]

is a smooth manifold.

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Proof: ……

Remark: The product atlas is in general not maximal.

Examples:

• \(S^1 \cross S^1 = T^2\) ,the 2-torus
• \(S^1 \cross \mathbb{R}\) , the cylinder
• \(\mathbb{R}^m \cross \mathbb{R}^n \cong \mathbb{R}^{m+n}\)

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Maps between manifolds are called smooth if they are smooth in local coordinates:

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Definition

Let M and N be smooth manifolds. A continuous map $f: M\rightarrow N \(is called **smooth** if for all charts\)(\varphi ,U)\(of *M* ,\)(\psi ,V)$ of N,

\[ \psi \circ f \circ \varphi^{-1} : \varphi(U \cap f^{-1}(V)\rightarrow \psi (V)) \]

is smooth. If f is invertible and its inverse \(f^{-1} : N \rightarrow M\) is smooth , f is called a diffeomorphism

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• a map \(f : U \rightarrow V\) between open sets U of \(\mathbb{R}^n\) and V of \(\mathbb{R}^m\) is smooth in the sense of real analysis if and only if it is smooth in the sense above where U and V are viewed as smooth manifolds equipped with the restriction of the canonical coordinates
• the set of diffeomorphisms on an $n \ge $1-dim. smooth manifold M , Diff(M), forms an infinite dimensional Lie group