Is normalizing automorphism in $mathbbR^n$ by Jacobian still an automorphism?

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Let $n in mathbbN$.
Suppose $f: mathbbR^n to mathbbR^n$ is a $C^infty$ diffeomorphism. Let $J_f: mathbbR^n to textM (n, mathbbR)$ be the Jacobian of $f$. Is the map $F: mathbbR^n to mathbbR^n$ defined by $F(x) = [J_f(x)]^-1cdot f(x)$ a $C^infty$ diffeomorphism?

calculus real-analysis differential-topology

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edited 2 days ago

Alfred Yerger

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up vote
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Let $n in mathbbN$.
Suppose $f: mathbbR^n to mathbbR^n$ is a $C^infty$ diffeomorphism. Let $J_f: mathbbR^n to textM (n, mathbbR)$ be the Jacobian of $f$. Is the map $F: mathbbR^n to mathbbR^n$ defined by $F(x) = [J_f(x)]^-1cdot f(x)$ a $C^infty$ diffeomorphism?

calculus real-analysis differential-topology

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edited 2 days ago

Alfred Yerger

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asked Aug 3 at 22:29

Better2BLucky

535

add a comment | 

up vote
7
down vote

favorite

2

up vote
7
down vote

favorite

2

2

Let $n in mathbbN$.
Suppose $f: mathbbR^n to mathbbR^n$ is a $C^infty$ diffeomorphism. Let $J_f: mathbbR^n to textM (n, mathbbR)$ be the Jacobian of $f$. Is the map $F: mathbbR^n to mathbbR^n$ defined by $F(x) = [J_f(x)]^-1cdot f(x)$ a $C^infty$ diffeomorphism?

calculus real-analysis differential-topology

share|cite|improve this question

edited 2 days ago

Alfred Yerger

9,2761743

asked Aug 3 at 22:29

Better2BLucky

535

Let $n in mathbbN$.
Suppose $f: mathbbR^n to mathbbR^n$ is a $C^infty$ diffeomorphism. Let $J_f: mathbbR^n to textM (n, mathbbR)$ be the Jacobian of $f$. Is the map $F: mathbbR^n to mathbbR^n$ defined by $F(x) = [J_f(x)]^-1cdot f(x)$ a $C^infty$ diffeomorphism?

calculus real-analysis differential-topology

share|cite|improve this question

edited 2 days ago

Alfred Yerger

9,2761743

asked Aug 3 at 22:29

Better2BLucky

535

share|cite|improve this question

share|cite|improve this question

edited 2 days ago

Alfred Yerger

9,2761743

edited 2 days ago

Alfred Yerger

9,2761743

edited 2 days ago

Alfred Yerger

9,2761743

9,2761743

asked Aug 3 at 22:29

Better2BLucky

535

asked Aug 3 at 22:29

Better2BLucky

535

asked Aug 3 at 22:29

Better2BLucky

535

535

add a comment | 

add a comment | 

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I don't think this is even true for $mathbbR$.

Consider the function $x^3 + e^x$. This function is clearly smooth and bijective, and the inverse is smooth by repeated application of the inverse function theorem, as the derivative has no zeros, and the inverse function theorem can be iterated to arbitrarily many derivatives.

However, for us here, the Jacobian is just the derivative, and your map $F$ is just $f/f'$. But for us, this is $fracx^3 + e^x3x^2 + e^x$ which is not injective. In particular, it attains the value $1$ at $0$, but also at $x = 3$

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edited Aug 4 at 3:11

answered Aug 4 at 3:04

Alfred Yerger

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

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active

oldest

votes

up vote
7
down vote



accepted

I don't think this is even true for $mathbbR$.

Consider the function $x^3 + e^x$. This function is clearly smooth and bijective, and the inverse is smooth by repeated application of the inverse function theorem, as the derivative has no zeros, and the inverse function theorem can be iterated to arbitrarily many derivatives.

However, for us here, the Jacobian is just the derivative, and your map $F$ is just $f/f'$. But for us, this is $fracx^3 + e^x3x^2 + e^x$ which is not injective. In particular, it attains the value $1$ at $0$, but also at $x = 3$

share|cite|improve this answer

edited Aug 4 at 3:11

answered Aug 4 at 3:04

Alfred Yerger

9,2761743

add a comment | 

up vote
7
down vote



accepted

I don't think this is even true for $mathbbR$.

Consider the function $x^3 + e^x$. This function is clearly smooth and bijective, and the inverse is smooth by repeated application of the inverse function theorem, as the derivative has no zeros, and the inverse function theorem can be iterated to arbitrarily many derivatives.

However, for us here, the Jacobian is just the derivative, and your map $F$ is just $f/f'$. But for us, this is $fracx^3 + e^x3x^2 + e^x$ which is not injective. In particular, it attains the value $1$ at $0$, but also at $x = 3$

share|cite|improve this answer

edited Aug 4 at 3:11

answered Aug 4 at 3:04

Alfred Yerger

9,2761743

add a comment | 

up vote
7
down vote



accepted

up vote
7
down vote



accepted

I don't think this is even true for $mathbbR$.

Consider the function $x^3 + e^x$. This function is clearly smooth and bijective, and the inverse is smooth by repeated application of the inverse function theorem, as the derivative has no zeros, and the inverse function theorem can be iterated to arbitrarily many derivatives.

However, for us here, the Jacobian is just the derivative, and your map $F$ is just $f/f'$. But for us, this is $fracx^3 + e^x3x^2 + e^x$ which is not injective. In particular, it attains the value $1$ at $0$, but also at $x = 3$

share|cite|improve this answer

edited Aug 4 at 3:11

answered Aug 4 at 3:04

Alfred Yerger

9,2761743

I don't think this is even true for $mathbbR$.

Consider the function $x^3 + e^x$. This function is clearly smooth and bijective, and the inverse is smooth by repeated application of the inverse function theorem, as the derivative has no zeros, and the inverse function theorem can be iterated to arbitrarily many derivatives.

However, for us here, the Jacobian is just the derivative, and your map $F$ is just $f/f'$. But for us, this is $fracx^3 + e^x3x^2 + e^x$ which is not injective. In particular, it attains the value $1$ at $0$, but also at $x = 3$

share|cite|improve this answer

edited Aug 4 at 3:11

answered Aug 4 at 3:04

Alfred Yerger

9,2761743

share|cite|improve this answer

share|cite|improve this answer

edited Aug 4 at 3:11

edited Aug 4 at 3:11

edited Aug 4 at 3:11

answered Aug 4 at 3:04

Alfred Yerger

9,2761743

answered Aug 4 at 3:04

Alfred Yerger

9,2761743

answered Aug 4 at 3:04

Alfred Yerger

9,2761743

9,2761743

add a comment | 

add a comment | 

 

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