description/proof of that for Euclidean \(C^\infty\) manifold, open ball is diffeomorphic to whole space
Topics
About: \(C^\infty\) manifold
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
Starting Context
- The reader knows a definition of Euclidean \(C^\infty\) manifold.
- The reader knows a definition of diffeomorphism between arbitrary subsets of \(C^\infty\) manifolds with boundary.
- The reader admits the proposition that for any open subset of the \(d_1\)-dimensional Euclidean \(C^\infty\) manifold, any \(C^\infty\) map into the \(d_2\)-dimensional Euclidean \(C^\infty\) manifold divided by any never-zero \(C^\infty\) map into the 1-dimensional Euclidean \(C^\infty\) manifold is \(C^\infty\).
Target Context
- The reader will have a description and a proof of the proposition that for any Euclidean \(C^\infty\) manifold, any open ball is diffeomorphic to the whole space.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(\mathbb{R}^d\): \(= \text{ the Euclidean } C^\infty \text{ manifold }\)
\(B_{p, \epsilon}\): \(= \text{ the open ball around } p \text{ on } \mathbb{R}^d\)
\(f\): \(: B_{p, \epsilon} \to \mathbb{R}^d, p' \mapsto (p' - p) / \sqrt{\epsilon^2 - \Vert p' - p \Vert^2}\)
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Statements:
\(B_{p, \epsilon} \cong_f \mathbb{R}^d\), where \(\cong_f\) denotes being diffeomorphic by \(f\)
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2: Natural Language Description
For the Euclidean \(C^\infty\) manifold, \(\mathbb{R}^d\), any open ball around any \(p \in \mathbb{R}^d\), \(B_{p, \epsilon} \subseteq \mathbb{R}^d\), is diffeomorphic to \(\mathbb{R}^d\) by \(f: B_{p, \epsilon} \to \mathbb{R}^d, p' \mapsto (p' - p) / \sqrt{\epsilon^2 - \Vert p' - p \Vert^2}\).
3: Proof
Whole Strategy: Step 1: see that \(f\) is bijective; Step 2: get the inverse of \(f\); Step 3: see that \(f\) is \(C^\infty\); Step 4: see that \(f^{-1}\) is \(C^\infty\).
\(f\) is well-defined, because \(0 \lt \epsilon^2 - \Vert p' - p \Vert^2\) on \(B_{p, \epsilon}\).
Step 1:
Let us see that \(f\) is injective.
Let \(p', p'' \in B_{p, \epsilon}\) be any such that \(p' \neq p''\).
There are 2 cases: Case 1: \(\Vert p' - p \Vert \neq \Vert p'' - p \Vert\); Case 2: \(\Vert p' - p \Vert = \Vert p'' - p \Vert\).
Let us think of Case 1.
\(\Vert (p' - p) / \sqrt{\epsilon^2 - \Vert p' - p \Vert^2} \Vert^2 = \Vert (p' - p) \Vert^2 / (\epsilon^2 - \Vert p' - p \Vert^2) = 1 / (\epsilon^2 / \Vert p' - p \Vert^2 - 1)\). Likewise, \(\Vert (p'' - p) / \sqrt{\epsilon^2 - \Vert p'' - p \Vert^2} \Vert^2 = 1 / (\epsilon^2 / \Vert p'' - p \Vert^2 - 1)\). Obviously, \(1 / (\epsilon^2 / \Vert p' - p \Vert^2 - 1) \neq 1 / (\epsilon^2 / \Vert p'' - p \Vert^2 - 1)\), which means that \(\Vert f (p') \Vert^2 \neq \Vert f (p'') \Vert^2\), which implies that \(f (p') \neq f (p'')\).
Let us think of Case 2.
The directions of \(p' - p\) and \(p'' - p\) are different. So, the directions of \(f (p')\) and \(f (p'')\) are different. So, \(f (p') \neq f (p'')\).
So, \(f\) is injective.
Let us see that \(f\) is surjective.
Let \(p'' \in \mathbb{R}^d\) be any. Let us see that there is a \(p' \in B_{p, \epsilon}\) such that \(f (p') = p''\).
Let \(p' := (p'' / \sqrt{1 + \Vert p'' \Vert^2}) \epsilon + p\). \(f (p') = (p'' / \sqrt{1 + \Vert p'' \Vert^2}) \epsilon / \sqrt{\epsilon^2 - \Vert (p'' / \sqrt{1 + \Vert p'' \Vert^2}) \epsilon \Vert^2} = p'' / \sqrt{1 + \Vert p'' \Vert^2} \epsilon / \sqrt{\epsilon^2 - (\Vert p'' \Vert^2 / (1 + \Vert p'' \Vert^2)) \epsilon^2} = p'' / \sqrt{1 + \Vert p'' \Vert^2} \epsilon / \sqrt{(\epsilon^2 + \epsilon^2 \Vert p'' \Vert^2 - \epsilon^2 \Vert p'' \Vert^2) / (1 + \Vert p'' \Vert^2)} = p'' / \sqrt{1 + \Vert p'' \Vert^2} \epsilon / \sqrt{\epsilon^2 / (1 + \Vert p'' \Vert^2)} = p''\).
Step 2:
By Step 1, \(f^{-1}: p'' \mapsto (p'' / \sqrt{1 + \Vert p'' \Vert^2}) \epsilon + p\).
Step 3:
Let us see that \(f\) is \(C^\infty\).
\(p' - p\) is obviously \(C^\infty\).
\(\epsilon^2 - \Vert p' - p \Vert^2\) is obviously \(C^\infty\).
\(g: x \mapsto \sqrt{x}\) is \(C^\infty\) on \(\mathbb{R} / \{0\}\), because \(d g / d x = 1 / 2 x^{- 1 / 2}\), \(d^2 g / d x^2 = 1 / 2 (- 1 / 2) x^{- 3 / 2}\), ..., while \(x \neq 0\).
So, the composition, \(\sqrt{\epsilon^2 - \Vert p' - p \Vert^2}\), is \(C^\infty\).
So, \(f\) is \(C^\infty\), by the proposition that for any open subset of the \(d_1\)-dimensional Euclidean \(C^\infty\) manifold, any \(C^\infty\) map into the \(d_2\)-dimensional Euclidean \(C^\infty\) manifold divided by any never-zero \(C^\infty\) map into the 1-dimensional Euclidean \(C^\infty\) manifold is \(C^\infty\).
Step 4:
Let us see that \(f^{-1}\) is \(C^\infty\).
\(p'' \epsilon\) is obviously \(C^\infty\).
\(1 + \Vert p'' \Vert^2\) is obviously \(C^\infty\).
\(g: x \mapsto \sqrt{x}\) is \(C^\infty\) on \(\mathbb{R} / \{0\}\) as before.
So, the composition, \(\sqrt{1 + \Vert p'' \Vert^2}\), is \(C^\infty\).
\(p'' / \sqrt{1 + \Vert p'' \Vert^2} \epsilon\) is \(C^\infty\), by the proposition that for any open subset of the \(d_1\)-dimensional Euclidean \(C^\infty\) manifold, any \(C^\infty\) map into the \(d_2\)-dimensional Euclidean \(C^\infty\) manifold divided by any never-zero \(C^\infty\) map into the 1-dimensional Euclidean \(C^\infty\) manifold is \(C^\infty\).
\(f^{-1}\) is obviously \(C^\infty\).
