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

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
7
down vote

favorite
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

























    up vote
    7
    down vote

    favorite
    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























      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













      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












      share|cite|improve this question




      share|cite|improve this question








      edited 2 days ago









      Alfred Yerger

      9,2761743




      9,2761743









      asked Aug 3 at 22:29









      Better2BLucky

      535




      535




















          1 Answer
          1






          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























            Your Answer




            StackExchange.ifUsing("editor", function ()
            return StackExchange.using("mathjaxEditing", function ()
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            );
            );
            , "mathjax-editing");

            StackExchange.ready(function()
            var channelOptions =
            tags: "".split(" "),
            id: "69"
            ;
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function()
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled)
            StackExchange.using("snippets", function()
            createEditor();
            );

            else
            createEditor();

            );

            function createEditor()
            StackExchange.prepareEditor(
            heartbeatType: 'answer',
            convertImagesToLinks: true,
            noModals: false,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            );



            );








             

            draft saved


            draft discarded


















            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2871551%2fis-normalizing-automorphism-in-mathbbrn-by-jacobian-still-an-automorphism%23new-answer', 'question_page');

            );

            Post as a guest

















            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 $x = 3$






            share|cite|improve this answer



























              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

























                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















                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















                share|cite|improve this answer



                share|cite|improve this answer








                edited Aug 4 at 3:11


























                answered Aug 4 at 3:04









                Alfred Yerger

                9,2761743




                9,2761743






















                     

                    draft saved


                    draft discarded


























                     


                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function ()
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2871551%2fis-normalizing-automorphism-in-mathbbrn-by-jacobian-still-an-automorphism%23new-answer', 'question_page');

                    );

                    Post as a guest
































































                    Comments

                    Post a Comment

                    Popular posts from this blog

                    What is the equation of a 3D cone with generalised tilt?

                    Image
                    Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite What is the equation of a 3D cone with generalized tilt? I've noticed that in most equations given to represent a cone, there is no parameter which defines the tilt of the cone in 3D space and that most of them have their point at the origin $(0,0,0)$ - I was wondering if anyone could give me a more generalized cone equation for a cone in any position in 3D space 3d share | cite | improve this question edited Jul 25 '16 at 0:05 ervx 9,425 3 13 36 asked Jul 24 '16 at 15:38 Charlene 11 2 2 Welcome to Math.SE! Could you please explain a few things to help people give an answer useful to you: 1. What do you mean by "generalized tilt"? 2. Are you asking about a right circular cone? Does the angle at the vertex matter? 3. What form of answer (implicit, parametric...?) are you looking for? 4. If this is homework, what tools do you have available, an...
                    Read more

                    Relationship between determinant of matrix and determinant of adjoint?

                    Image
                    Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/…...
                    Read more

                    Color the edges and diagonals of a regular polygon

                    Image
                    Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne...
                    Read more