微分几何/微分流形(2)
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Before stating the implicit function theorem, recall the following:
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
is invertible and \(F^{-1} : W\rightarrow V\) its inverse is smooth
- $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
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
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
in particular we have \(g(x_0)=y_0\)
- 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
Important examples of smooth manifolds are the following
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
Question: Are smooth submanifolds of \(\mathbb{R}^n\) actually smooth manifolds ?
Answer: Yes! use the IFT
Lemma
m-dimensional smooth submanifolds M of \(\mathbb{R}^n\) can be locally written as graphs of smooth maps
still need to find atlas on M
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
such that
Get local coordinates via:
cover M with open sets U as above
get charts on M:
\[ (\varphi = pr_{\mathbb{R}^m} \circ F|_{U\cap M}, U\cap M) \]
Yet another way to find examples of smooth manifolds is by taking the product of two manifolds:
Lemma
Let M with atlas \(\mathcal{A}\) and N with atlas \(\mathcal{B}\) be smooth manifolds. Then the **Cartesian product ** of M and N
together with the product atlas
is a smooth manifold.
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
- \(\mathbb{R}^m \cross \mathbb{R}^n \cong \mathbb{R}^{m+n}\)
Maps between manifolds are called smooth if they are smooth in local coordinates:
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,
is smooth. If f is invertible and its inverse \(f^{-1} : N \rightarrow M\) is smooth , f is called a diffeomorphism
- 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