Prove $Phi$ is a diffeomorphism












0












$begingroup$


Let $d in mathbb N$.



Define $Phi:]0, infty[ times { x in mathbb R^{d-1}: |x|<1}to {y in mathbb R^d:y_{1}>0}, (r,x) mapsto r(sqrt{1-|x|^2},x)$



Show that $Phi$ is a diffeomorphism.



Ideas:



$1.$ Bijectivity:



$1.1$ Injectivity:
Let $x,y in mathbb R^d$ and $r,s in ]0, infty[$ whereby $x neq y$ or $r neq s$



if $x neq y$ is clear that $Phi(r,x)neq Phi(r,y)Rightarrow$ injectivity



So let $s neq r$ and $Phi(s,x)=s(sqrt{1-|x|^2},x) neqPhi(r,x)=r(sqrt{1-|x|^2},x) Rightarrow$ injectivity



$1.2$ Surjectivity: Let $z in {y in mathbb R^d:y_{1}>0}$ it is clear that there is an $r in ]0, infty[$ so that $r sqrt{1-|x|^2}=z_{1}Rightarrow$ surjectivity



$Rightarrow$ Bijectivity



On the issue of Differentiability of $Phi$ as well as Differentiability $Phi^{-1}$ I am lost, as this is the first time doing this...



Any ideas, corrections, tips?






real-analysis derivatives diffeomorphism







share|cite|improve this question









$endgroup$












  • $begingroup$
    Your injectivity proof does not work. You also have to consider $rneq s$ and $xneq y$ while $Phi(r,x)= Phi(s,y)$. For the differentability you can use the fact that a function is differentiable iff each coordinate is differentiable. An easy way to check differentiabilty of the inverse function is to use the inverse function theorem. Otherwise you have to find the inverse and check by hand that it is differentiable.
    $endgroup$
    – Severin Schraven
    Jan 4 at 17:28










  • $begingroup$
    So $Phi$ is not injective?
    $endgroup$
    – SABOY
    Jan 4 at 18:30










  • $begingroup$
    It is, but your proof is not complete.
    $endgroup$
    – Severin Schraven
    Jan 4 at 23:15










  • $begingroup$
    Also your proof of surjectivity is not clear at all. If you change $r$, you also change the second coordinate.
    $endgroup$
    – Severin Schraven
    Jan 4 at 23:25










  • $begingroup$
    On the issue of injectivity where $ r neq s$ and $x neq y$ do you mean we generate a contradiction after assuming $Phi(r,x) = Phi(s,y)$? Because assuming $r(sqrt{1-|x|^2},x)=s(sqrt{1-|y|^2},y)$ that would mean $r(sqrt{1-|x|^2})=s(sqrt{1-|y|^2})$ and that $rx_{i}=sy_{i}, forall i in {1,...,n}$. I have no idea how to generate a contradiction from this
    $endgroup$
    – SABOY
    Jan 4 at 23:59
















0












$begingroup$


Let $d in mathbb N$.



Define $Phi:]0, infty[ times { x in mathbb R^{d-1}: |x|<1}to {y in mathbb R^d:y_{1}>0}, (r,x) mapsto r(sqrt{1-|x|^2},x)$



Show that $Phi$ is a diffeomorphism.



Ideas:



$1.$ Bijectivity:



$1.1$ Injectivity:
Let $x,y in mathbb R^d$ and $r,s in ]0, infty[$ whereby $x neq y$ or $r neq s$



if $x neq y$ is clear that $Phi(r,x)neq Phi(r,y)Rightarrow$ injectivity



So let $s neq r$ and $Phi(s,x)=s(sqrt{1-|x|^2},x) neqPhi(r,x)=r(sqrt{1-|x|^2},x) Rightarrow$ injectivity



$1.2$ Surjectivity: Let $z in {y in mathbb R^d:y_{1}>0}$ it is clear that there is an $r in ]0, infty[$ so that $r sqrt{1-|x|^2}=z_{1}Rightarrow$ surjectivity



$Rightarrow$ Bijectivity



On the issue of Differentiability of $Phi$ as well as Differentiability $Phi^{-1}$ I am lost, as this is the first time doing this...



Any ideas, corrections, tips?






real-analysis derivatives diffeomorphism







share|cite|improve this question









$endgroup$












  • $begingroup$
    Your injectivity proof does not work. You also have to consider $rneq s$ and $xneq y$ while $Phi(r,x)= Phi(s,y)$. For the differentability you can use the fact that a function is differentiable iff each coordinate is differentiable. An easy way to check differentiabilty of the inverse function is to use the inverse function theorem. Otherwise you have to find the inverse and check by hand that it is differentiable.
    $endgroup$
    – Severin Schraven
    Jan 4 at 17:28










  • $begingroup$
    So $Phi$ is not injective?
    $endgroup$
    – SABOY
    Jan 4 at 18:30










  • $begingroup$
    It is, but your proof is not complete.
    $endgroup$
    – Severin Schraven
    Jan 4 at 23:15










  • $begingroup$
    Also your proof of surjectivity is not clear at all. If you change $r$, you also change the second coordinate.
    $endgroup$
    – Severin Schraven
    Jan 4 at 23:25










  • $begingroup$
    On the issue of injectivity where $ r neq s$ and $x neq y$ do you mean we generate a contradiction after assuming $Phi(r,x) = Phi(s,y)$? Because assuming $r(sqrt{1-|x|^2},x)=s(sqrt{1-|y|^2},y)$ that would mean $r(sqrt{1-|x|^2})=s(sqrt{1-|y|^2})$ and that $rx_{i}=sy_{i}, forall i in {1,...,n}$. I have no idea how to generate a contradiction from this
    $endgroup$
    – SABOY
    Jan 4 at 23:59














0












0








0





$begingroup$


Let $d in mathbb N$.



Define $Phi:]0, infty[ times { x in mathbb R^{d-1}: |x|<1}to {y in mathbb R^d:y_{1}>0}, (r,x) mapsto r(sqrt{1-|x|^2},x)$



Show that $Phi$ is a diffeomorphism.



Ideas:



$1.$ Bijectivity:



$1.1$ Injectivity:
Let $x,y in mathbb R^d$ and $r,s in ]0, infty[$ whereby $x neq y$ or $r neq s$



if $x neq y$ is clear that $Phi(r,x)neq Phi(r,y)Rightarrow$ injectivity



So let $s neq r$ and $Phi(s,x)=s(sqrt{1-|x|^2},x) neqPhi(r,x)=r(sqrt{1-|x|^2},x) Rightarrow$ injectivity



$1.2$ Surjectivity: Let $z in {y in mathbb R^d:y_{1}>0}$ it is clear that there is an $r in ]0, infty[$ so that $r sqrt{1-|x|^2}=z_{1}Rightarrow$ surjectivity



$Rightarrow$ Bijectivity



On the issue of Differentiability of $Phi$ as well as Differentiability $Phi^{-1}$ I am lost, as this is the first time doing this...



Any ideas, corrections, tips?






real-analysis derivatives diffeomorphism







share|cite|improve this question









$endgroup$




Let $d in mathbb N$.



Define $Phi:]0, infty[ times { x in mathbb R^{d-1}: |x|<1}to {y in mathbb R^d:y_{1}>0}, (r,x) mapsto r(sqrt{1-|x|^2},x)$



Show that $Phi$ is a diffeomorphism.



Ideas:



$1.$ Bijectivity:



$1.1$ Injectivity:
Let $x,y in mathbb R^d$ and $r,s in ]0, infty[$ whereby $x neq y$ or $r neq s$



if $x neq y$ is clear that $Phi(r,x)neq Phi(r,y)Rightarrow$ injectivity



So let $s neq r$ and $Phi(s,x)=s(sqrt{1-|x|^2},x) neqPhi(r,x)=r(sqrt{1-|x|^2},x) Rightarrow$ injectivity



$1.2$ Surjectivity: Let $z in {y in mathbb R^d:y_{1}>0}$ it is clear that there is an $r in ]0, infty[$ so that $r sqrt{1-|x|^2}=z_{1}Rightarrow$ surjectivity



$Rightarrow$ Bijectivity



On the issue of Differentiability of $Phi$ as well as Differentiability $Phi^{-1}$ I am lost, as this is the first time doing this...



Any ideas, corrections, tips?






real-analysis derivatives diffeomorphism




real-analysis derivatives diffeomorphism






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 4 at 11:57









SABOYSABOY

656311




656311












  • $begingroup$
    Your injectivity proof does not work. You also have to consider $rneq s$ and $xneq y$ while $Phi(r,x)= Phi(s,y)$. For the differentability you can use the fact that a function is differentiable iff each coordinate is differentiable. An easy way to check differentiabilty of the inverse function is to use the inverse function theorem. Otherwise you have to find the inverse and check by hand that it is differentiable.
    $endgroup$
    – Severin Schraven
    Jan 4 at 17:28










  • $begingroup$
    So $Phi$ is not injective?
    $endgroup$
    – SABOY
    Jan 4 at 18:30










  • $begingroup$
    It is, but your proof is not complete.
    $endgroup$
    – Severin Schraven
    Jan 4 at 23:15










  • $begingroup$
    Also your proof of surjectivity is not clear at all. If you change $r$, you also change the second coordinate.
    $endgroup$
    – Severin Schraven
    Jan 4 at 23:25










  • $begingroup$
    On the issue of injectivity where $ r neq s$ and $x neq y$ do you mean we generate a contradiction after assuming $Phi(r,x) = Phi(s,y)$? Because assuming $r(sqrt{1-|x|^2},x)=s(sqrt{1-|y|^2},y)$ that would mean $r(sqrt{1-|x|^2})=s(sqrt{1-|y|^2})$ and that $rx_{i}=sy_{i}, forall i in {1,...,n}$. I have no idea how to generate a contradiction from this
    $endgroup$
    – SABOY
    Jan 4 at 23:59


















  • $begingroup$
    Your injectivity proof does not work. You also have to consider $rneq s$ and $xneq y$ while $Phi(r,x)= Phi(s,y)$. For the differentability you can use the fact that a function is differentiable iff each coordinate is differentiable. An easy way to check differentiabilty of the inverse function is to use the inverse function theorem. Otherwise you have to find the inverse and check by hand that it is differentiable.
    $endgroup$
    – Severin Schraven
    Jan 4 at 17:28










  • $begingroup$
    So $Phi$ is not injective?
    $endgroup$
    – SABOY
    Jan 4 at 18:30










  • $begingroup$
    It is, but your proof is not complete.
    $endgroup$
    – Severin Schraven
    Jan 4 at 23:15










  • $begingroup$
    Also your proof of surjectivity is not clear at all. If you change $r$, you also change the second coordinate.
    $endgroup$
    – Severin Schraven
    Jan 4 at 23:25










  • $begingroup$
    On the issue of injectivity where $ r neq s$ and $x neq y$ do you mean we generate a contradiction after assuming $Phi(r,x) = Phi(s,y)$? Because assuming $r(sqrt{1-|x|^2},x)=s(sqrt{1-|y|^2},y)$ that would mean $r(sqrt{1-|x|^2})=s(sqrt{1-|y|^2})$ and that $rx_{i}=sy_{i}, forall i in {1,...,n}$. I have no idea how to generate a contradiction from this
    $endgroup$
    – SABOY
    Jan 4 at 23:59
















$begingroup$
Your injectivity proof does not work. You also have to consider $rneq s$ and $xneq y$ while $Phi(r,x)= Phi(s,y)$. For the differentability you can use the fact that a function is differentiable iff each coordinate is differentiable. An easy way to check differentiabilty of the inverse function is to use the inverse function theorem. Otherwise you have to find the inverse and check by hand that it is differentiable.
$endgroup$
– Severin Schraven
Jan 4 at 17:28




$begingroup$
Your injectivity proof does not work. You also have to consider $rneq s$ and $xneq y$ while $Phi(r,x)= Phi(s,y)$. For the differentability you can use the fact that a function is differentiable iff each coordinate is differentiable. An easy way to check differentiabilty of the inverse function is to use the inverse function theorem. Otherwise you have to find the inverse and check by hand that it is differentiable.
$endgroup$
– Severin Schraven
Jan 4 at 17:28












$begingroup$
So $Phi$ is not injective?
$endgroup$
– SABOY
Jan 4 at 18:30




$begingroup$
So $Phi$ is not injective?
$endgroup$
– SABOY
Jan 4 at 18:30












$begingroup$
It is, but your proof is not complete.
$endgroup$
– Severin Schraven
Jan 4 at 23:15




$begingroup$
It is, but your proof is not complete.
$endgroup$
– Severin Schraven
Jan 4 at 23:15












$begingroup$
Also your proof of surjectivity is not clear at all. If you change $r$, you also change the second coordinate.
$endgroup$
– Severin Schraven
Jan 4 at 23:25




$begingroup$
Also your proof of surjectivity is not clear at all. If you change $r$, you also change the second coordinate.
$endgroup$
– Severin Schraven
Jan 4 at 23:25












$begingroup$
On the issue of injectivity where $ r neq s$ and $x neq y$ do you mean we generate a contradiction after assuming $Phi(r,x) = Phi(s,y)$? Because assuming $r(sqrt{1-|x|^2},x)=s(sqrt{1-|y|^2},y)$ that would mean $r(sqrt{1-|x|^2})=s(sqrt{1-|y|^2})$ and that $rx_{i}=sy_{i}, forall i in {1,...,n}$. I have no idea how to generate a contradiction from this
$endgroup$
– SABOY
Jan 4 at 23:59




$begingroup$
On the issue of injectivity where $ r neq s$ and $x neq y$ do you mean we generate a contradiction after assuming $Phi(r,x) = Phi(s,y)$? Because assuming $r(sqrt{1-|x|^2},x)=s(sqrt{1-|y|^2},y)$ that would mean $r(sqrt{1-|x|^2})=s(sqrt{1-|y|^2})$ and that $rx_{i}=sy_{i}, forall i in {1,...,n}$. I have no idea how to generate a contradiction from this
$endgroup$
– SABOY
Jan 4 at 23:59










1 Answer
1






active

oldest

votes


















1












$begingroup$

Hint: The easiest way is to find the inverse function. For this we compute
$$ vert Phi(r,x) vert = r $$
as
$$ vert Phi(r,x) vert = vert r vert sqrt{(1-vert x vert^2)+ x_1^2 + dots + x_{d_1}^2} = vert r vert sqrt{1- vert x vert^2 + vert x vert^2} = vert r vert = r. $$
Hence, the inverse function is (for $y=(y_1, dots, y_d)$)
$$ Psi (y_1, dots, y_d) = left(vert y vert, frac{1}{vert y vert} y_2, dots, frac{1}{vert y vert} y_d right).$$






share|cite|improve this answer









$endgroup$













    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',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    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%2f3061571%2fprove-phi-is-a-diffeomorphism%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Hint: The easiest way is to find the inverse function. For this we compute
    $$ vert Phi(r,x) vert = r $$
    as
    $$ vert Phi(r,x) vert = vert r vert sqrt{(1-vert x vert^2)+ x_1^2 + dots + x_{d_1}^2} = vert r vert sqrt{1- vert x vert^2 + vert x vert^2} = vert r vert = r. $$
    Hence, the inverse function is (for $y=(y_1, dots, y_d)$)
    $$ Psi (y_1, dots, y_d) = left(vert y vert, frac{1}{vert y vert} y_2, dots, frac{1}{vert y vert} y_d right).$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Hint: The easiest way is to find the inverse function. For this we compute
      $$ vert Phi(r,x) vert = r $$
      as
      $$ vert Phi(r,x) vert = vert r vert sqrt{(1-vert x vert^2)+ x_1^2 + dots + x_{d_1}^2} = vert r vert sqrt{1- vert x vert^2 + vert x vert^2} = vert r vert = r. $$
      Hence, the inverse function is (for $y=(y_1, dots, y_d)$)
      $$ Psi (y_1, dots, y_d) = left(vert y vert, frac{1}{vert y vert} y_2, dots, frac{1}{vert y vert} y_d right).$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Hint: The easiest way is to find the inverse function. For this we compute
        $$ vert Phi(r,x) vert = r $$
        as
        $$ vert Phi(r,x) vert = vert r vert sqrt{(1-vert x vert^2)+ x_1^2 + dots + x_{d_1}^2} = vert r vert sqrt{1- vert x vert^2 + vert x vert^2} = vert r vert = r. $$
        Hence, the inverse function is (for $y=(y_1, dots, y_d)$)
        $$ Psi (y_1, dots, y_d) = left(vert y vert, frac{1}{vert y vert} y_2, dots, frac{1}{vert y vert} y_d right).$$






        share|cite|improve this answer









        $endgroup$



        Hint: The easiest way is to find the inverse function. For this we compute
        $$ vert Phi(r,x) vert = r $$
        as
        $$ vert Phi(r,x) vert = vert r vert sqrt{(1-vert x vert^2)+ x_1^2 + dots + x_{d_1}^2} = vert r vert sqrt{1- vert x vert^2 + vert x vert^2} = vert r vert = r. $$
        Hence, the inverse function is (for $y=(y_1, dots, y_d)$)
        $$ Psi (y_1, dots, y_d) = left(vert y vert, frac{1}{vert y vert} y_2, dots, frac{1}{vert y vert} y_d right).$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 5 at 0:59









        Severin SchravenSeverin Schraven

        6,4251935




        6,4251935






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061571%2fprove-phi-is-a-diffeomorphism%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Bressuire

            Image
            .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
            Read more

            Cabo Verde

            Image
              Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
            Read more

            Gyllenstierna

            Image
            .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
            Read more