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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
edited 2 days ago
Alfred Yerger
9,2761743
asked Aug 3 at 22:29
Better2BLucky
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up vote
7
down vote
favorite
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
edited 2 days ago
Alfred Yerger
9,2761743
asked Aug 3 at 22:29
Better2BLucky
535
add a comment |Â
up vote
7
down vote
favorite
up vote
7
down vote
favorite
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
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
edited 2 days ago
Alfred Yerger
9,2761743
asked Aug 3 at 22:29
Better2BLucky
535
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 |Â
add a comment |Â
1 Answer
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up vote
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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 $x = 3$
answered Aug 4 at 3:04
Alfred Yerger
9,2761743
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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 $x = 3$
answered Aug 4 at 3:04
Alfred Yerger
9,2761743
add a comment |Â
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 $x = 3$
answered Aug 4 at 3:04
Alfred Yerger
9,2761743
add a comment |Â
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 $x = 3$
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 $x = 3$
answered Aug 4 at 3:04
Alfred Yerger
9,2761743
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 |Â
add a comment |Â
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