description/proof of that for open subset of \(d_1\)-dimensional Euclidean \(C^\infty\) manifold, \(C^\infty\) map into \(d_2\)-dimensional Euclidean \(C^\infty\) manifold divided by never-zero \(C^\infty\) map into 1-dimensional Euclidean \(C^\infty\) manifold is \(C^\infty\)
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
Target Context
- The reader will have a description and a proof of 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 \(C^\infty\).
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:
\(\mathbb{R}^{d_1}\): \(= \text{ the Euclidean } C^\infty \text{ manifold }\)
\(U\): \(\in \{\text{ the open subsets of } \mathbb{R}^{d_1} \}\)
\(\mathbb{R}^{d_2}\): \(= \text{ the Euclidean } C^\infty \text{ manifold }\)
\(\mathbb{R}\): \(= \text{ the Euclidean } C^\infty \text{ manifold }\)
\(f_1\): \(: U \to \mathbb{R}^{d_2}\), \(\in \{\text{ the } C^\infty \text{ maps }\}\)
\(f_2\): \(: U \to \mathbb{R}\), \(\in \{\text{ the } C^\infty \text{ maps }\}\)
//
Statements:
\(\forall p \in U (f_2 (p) \neq 0)\)
\(\implies\)
\(f_1 / f_2: U \to \mathbb{R}^{d_2} \in \{\text{ the } C^\infty \text{ maps }\}\)
//
2: Natural Language Description
For the Euclidean \(C^\infty\) manifold, \(\mathbb{R}^{d_1}\), any open subset, \(U \subseteq \mathbb{R}^{d_1}\), the Euclidean \(C^\infty\) manifold, \(\mathbb{R}^{d_2}\), the Euclidean \(C^\infty\) manifold, \(\mathbb{R}\), any \(C^\infty\) map, \(f_1: U \to \mathbb{R}^{d_2}\), and any \(C^\infty\) map, \(f_2: U \to \mathbb{R}\), if \(f_2\) is never zero on \(U\), \(f_1 / f_2: U \to \mathbb{R}^{d_2}\) is a \(C^\infty\) map.
3: Proof
Whole Strategy: Step 1: prove the proposition that \(f_1 / f_2\) is \(C^1\) and the derivative is a \(C^\infty\) map divided by a never-zero \(C^\infty\) map; Step 2: prove by induction the proposition that \(f_1 / f_2\) is \(C^n\) and the \(n\)-th derivative is a \(C^\infty\) map divided by a never-zero \(C^\infty\) map; Step 3: conclude that proposition of this article.
As is described in Note for the definition of map from open subset of Euclidean \(C^\infty\) manifold into subset of Euclidean \(C^\infty\) manifold \(C^k\) at point, where \(k\) excludes \(0\) and includes \(\infty\), \(U_p\) cited in the definition can be taken to be \(U\). So, we do so hereafter.
Note that the superscripts like \(f^k\) hereafter except \((f_2)^2\) denote the \(k\)-th components, not powers.
Step 1:
Let us prove the proposition that \(f_1 / f_2\) is \(C^1\) and the derivative is a \(C^\infty\) map divided by a never-zero \(C^\infty\) map.
On \(U\), \(\partial_j (f_1 / f_2)^k = \partial_j ({f_1}^k / f_2) = ((\partial_j {f_1}^k) f_2 - {f_1}^k \partial_j f_2) / (f_2)^2\), which is continuous because \(\partial_j {f_1}^k\), \(f_2\), \({f_1}^k\), \(\partial_j f_2\), and \((f_2)^2\) are continuous and \((f_2)^2\) is never zero. \((\partial_j {f_1}^k) f_2 - {f_1}^k \partial_j f_2\) is \(C^\infty\), because \(\partial_j {f_1}^k\), \(f_2\), \({f_1}^k\), and \(\partial_j f_2\) are \(C^\infty\). \((f_2)^2\) is never-zero \(C^\infty\).
So, the proposition has been proved.
Step 2:
Let us prove by induction the proposition that \(f_1 / f_2\) is \(C^n\) and the \(n\)-th derivative is a \(C^\infty\) map divided by a never-zero \(C^\infty\) map.
For \(n = 1\), the proposition of Step 1 has proved it.
Let us suppose the proposition through \(n = n' - 1\) and think of \(n = n'\).
Let the \(n' - 1\)-th derivative be \(g_1 / g_2\). \(g_1 / g_2\) satisfies the condition of the proposition of Step 1, so, it is \(C^1\) and the derivative is a \(C^\infty\) map divided by a never-zero \(C^\infty\) map. That means that \(f_1 / f_2\) is \(C^{n'}\) and the \(n'\)-th derivative is a \(C^\infty\) map divided by a never-zero \(C^\infty\) map.
So, the proposition has been proved.
Step 3:
The proposition of Step 2 immediately implies the proposition of this article.
