Reference: riemannian metric with operator norm arbitrarily small compact manifolds












3












$begingroup$


I read this posts:



uniform equivalence of norms induced by riemannian-metrics



and



supremum of operator norm of the differential



So, it is natural to ask the following. Let $f: M to M$ to be a $C^{1}$ local diffeomorphism map in a compact finite dimensional manifold $M$, fixed $beta >0$. Does there exist a Riemannian metric on $TM$ such that $|df_{x}| leq beta$ ?



I suspect that the answer is negative, I tried to solve this for only one point of manifold, but I stuck. Does someone know any reference related with this topic?



Thanks






differential-geometry riemannian-geometry







share|cite|improve this question









$endgroup$












  • $begingroup$
    Hint: Consider the map $f(x)=2x$ on $M={mathbb R}$ and $beta=1$.
    $endgroup$
    – Moishe Cohen
    Dec 25 '18 at 14:35










  • $begingroup$
    $mathbb{R}$ is not a compact space
    $endgroup$
    – Pedro do Norte
    Dec 25 '18 at 22:55










  • $begingroup$
    It does not matter, take the 1-point compactification of ${mathbb R}$, which is the circle.
    $endgroup$
    – Moishe Cohen
    Dec 26 '18 at 1:25










  • $begingroup$
    I did not realize it. Could explain how does we translate $f$ from $mathbb{R}$ to the circle? Which inner product does there exist in $mathbb{R}$? Thanks
    $endgroup$
    – Pedro do Norte
    Dec 26 '18 at 1:40
















3












$begingroup$


I read this posts:



uniform equivalence of norms induced by riemannian-metrics



and



supremum of operator norm of the differential



So, it is natural to ask the following. Let $f: M to M$ to be a $C^{1}$ local diffeomorphism map in a compact finite dimensional manifold $M$, fixed $beta >0$. Does there exist a Riemannian metric on $TM$ such that $|df_{x}| leq beta$ ?



I suspect that the answer is negative, I tried to solve this for only one point of manifold, but I stuck. Does someone know any reference related with this topic?



Thanks






differential-geometry riemannian-geometry







share|cite|improve this question









$endgroup$












  • $begingroup$
    Hint: Consider the map $f(x)=2x$ on $M={mathbb R}$ and $beta=1$.
    $endgroup$
    – Moishe Cohen
    Dec 25 '18 at 14:35










  • $begingroup$
    $mathbb{R}$ is not a compact space
    $endgroup$
    – Pedro do Norte
    Dec 25 '18 at 22:55










  • $begingroup$
    It does not matter, take the 1-point compactification of ${mathbb R}$, which is the circle.
    $endgroup$
    – Moishe Cohen
    Dec 26 '18 at 1:25










  • $begingroup$
    I did not realize it. Could explain how does we translate $f$ from $mathbb{R}$ to the circle? Which inner product does there exist in $mathbb{R}$? Thanks
    $endgroup$
    – Pedro do Norte
    Dec 26 '18 at 1:40














3












3








3





$begingroup$


I read this posts:



uniform equivalence of norms induced by riemannian-metrics



and



supremum of operator norm of the differential



So, it is natural to ask the following. Let $f: M to M$ to be a $C^{1}$ local diffeomorphism map in a compact finite dimensional manifold $M$, fixed $beta >0$. Does there exist a Riemannian metric on $TM$ such that $|df_{x}| leq beta$ ?



I suspect that the answer is negative, I tried to solve this for only one point of manifold, but I stuck. Does someone know any reference related with this topic?



Thanks






differential-geometry riemannian-geometry







share|cite|improve this question









$endgroup$




I read this posts:



uniform equivalence of norms induced by riemannian-metrics



and



supremum of operator norm of the differential



So, it is natural to ask the following. Let $f: M to M$ to be a $C^{1}$ local diffeomorphism map in a compact finite dimensional manifold $M$, fixed $beta >0$. Does there exist a Riemannian metric on $TM$ such that $|df_{x}| leq beta$ ?



I suspect that the answer is negative, I tried to solve this for only one point of manifold, but I stuck. Does someone know any reference related with this topic?



Thanks






differential-geometry riemannian-geometry




differential-geometry riemannian-geometry






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 25 '18 at 12:18









Pedro do NortePedro do Norte

857




857












  • $begingroup$
    Hint: Consider the map $f(x)=2x$ on $M={mathbb R}$ and $beta=1$.
    $endgroup$
    – Moishe Cohen
    Dec 25 '18 at 14:35










  • $begingroup$
    $mathbb{R}$ is not a compact space
    $endgroup$
    – Pedro do Norte
    Dec 25 '18 at 22:55










  • $begingroup$
    It does not matter, take the 1-point compactification of ${mathbb R}$, which is the circle.
    $endgroup$
    – Moishe Cohen
    Dec 26 '18 at 1:25










  • $begingroup$
    I did not realize it. Could explain how does we translate $f$ from $mathbb{R}$ to the circle? Which inner product does there exist in $mathbb{R}$? Thanks
    $endgroup$
    – Pedro do Norte
    Dec 26 '18 at 1:40


















  • $begingroup$
    Hint: Consider the map $f(x)=2x$ on $M={mathbb R}$ and $beta=1$.
    $endgroup$
    – Moishe Cohen
    Dec 25 '18 at 14:35










  • $begingroup$
    $mathbb{R}$ is not a compact space
    $endgroup$
    – Pedro do Norte
    Dec 25 '18 at 22:55










  • $begingroup$
    It does not matter, take the 1-point compactification of ${mathbb R}$, which is the circle.
    $endgroup$
    – Moishe Cohen
    Dec 26 '18 at 1:25










  • $begingroup$
    I did not realize it. Could explain how does we translate $f$ from $mathbb{R}$ to the circle? Which inner product does there exist in $mathbb{R}$? Thanks
    $endgroup$
    – Pedro do Norte
    Dec 26 '18 at 1:40
















$begingroup$
Hint: Consider the map $f(x)=2x$ on $M={mathbb R}$ and $beta=1$.
$endgroup$
– Moishe Cohen
Dec 25 '18 at 14:35




$begingroup$
Hint: Consider the map $f(x)=2x$ on $M={mathbb R}$ and $beta=1$.
$endgroup$
– Moishe Cohen
Dec 25 '18 at 14:35












$begingroup$
$mathbb{R}$ is not a compact space
$endgroup$
– Pedro do Norte
Dec 25 '18 at 22:55




$begingroup$
$mathbb{R}$ is not a compact space
$endgroup$
– Pedro do Norte
Dec 25 '18 at 22:55












$begingroup$
It does not matter, take the 1-point compactification of ${mathbb R}$, which is the circle.
$endgroup$
– Moishe Cohen
Dec 26 '18 at 1:25




$begingroup$
It does not matter, take the 1-point compactification of ${mathbb R}$, which is the circle.
$endgroup$
– Moishe Cohen
Dec 26 '18 at 1:25












$begingroup$
I did not realize it. Could explain how does we translate $f$ from $mathbb{R}$ to the circle? Which inner product does there exist in $mathbb{R}$? Thanks
$endgroup$
– Pedro do Norte
Dec 26 '18 at 1:40




$begingroup$
I did not realize it. Could explain how does we translate $f$ from $mathbb{R}$ to the circle? Which inner product does there exist in $mathbb{R}$? Thanks
$endgroup$
– Pedro do Norte
Dec 26 '18 at 1:40










1 Answer
1






active

oldest

votes


















2












$begingroup$

I will consider $S^1$ as the one-point compactification of the real line, $S^1={mathbb R}cup{infty}$ where the smooth atlas is given by the identity chart on ${mathbb R}$ and the chart given by the map $xmapsto x^{-1}$ on ${mathbb R}cup{infty}- {0}$. The map $f: S^1to S^1$ given by $f(x)=2x, xne infty$, $f(infty)=infty$ is a diffeomorphism.



If you do not like this, just take the standard unit circle in the complex plane and let $f$ be the restriction to $S^1$ of a suitable linear-fractional transformation preserving the unit disk.



Now, suppose that $g$ is any Riemannian metric on $S^1$; take $beta=1$. The derivative $df_0$ of $f$ at $0$ is again the multiplication by $2$. Hence, $||df_0||=2$ where the norm is taken with respect to the metric $g$. But $1<2$.



Hence, the answer to your question is indeed negative.



Here is a more general argument. Let $M$ be any smooth compact connected manifold and let $f: Mto M$ be any degree 1 smooth map, for instance, the identity map. Take $0<beta<1$. Then for any Riemannian metric $g$ on $M$ there exists a point $xin M$ such that $||df_x||ge 1>beta$. The reason is that otherwise the volume of $f(M)$ is strictly less than the volume of $M$, which contradicts the degree assumption.






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%2f3052063%2freference-riemannian-metric-with-operator-norm-arbitrarily-small-compact-manifo%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









    2












    $begingroup$

    I will consider $S^1$ as the one-point compactification of the real line, $S^1={mathbb R}cup{infty}$ where the smooth atlas is given by the identity chart on ${mathbb R}$ and the chart given by the map $xmapsto x^{-1}$ on ${mathbb R}cup{infty}- {0}$. The map $f: S^1to S^1$ given by $f(x)=2x, xne infty$, $f(infty)=infty$ is a diffeomorphism.



    If you do not like this, just take the standard unit circle in the complex plane and let $f$ be the restriction to $S^1$ of a suitable linear-fractional transformation preserving the unit disk.



    Now, suppose that $g$ is any Riemannian metric on $S^1$; take $beta=1$. The derivative $df_0$ of $f$ at $0$ is again the multiplication by $2$. Hence, $||df_0||=2$ where the norm is taken with respect to the metric $g$. But $1<2$.



    Hence, the answer to your question is indeed negative.



    Here is a more general argument. Let $M$ be any smooth compact connected manifold and let $f: Mto M$ be any degree 1 smooth map, for instance, the identity map. Take $0<beta<1$. Then for any Riemannian metric $g$ on $M$ there exists a point $xin M$ such that $||df_x||ge 1>beta$. The reason is that otherwise the volume of $f(M)$ is strictly less than the volume of $M$, which contradicts the degree assumption.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      I will consider $S^1$ as the one-point compactification of the real line, $S^1={mathbb R}cup{infty}$ where the smooth atlas is given by the identity chart on ${mathbb R}$ and the chart given by the map $xmapsto x^{-1}$ on ${mathbb R}cup{infty}- {0}$. The map $f: S^1to S^1$ given by $f(x)=2x, xne infty$, $f(infty)=infty$ is a diffeomorphism.



      If you do not like this, just take the standard unit circle in the complex plane and let $f$ be the restriction to $S^1$ of a suitable linear-fractional transformation preserving the unit disk.



      Now, suppose that $g$ is any Riemannian metric on $S^1$; take $beta=1$. The derivative $df_0$ of $f$ at $0$ is again the multiplication by $2$. Hence, $||df_0||=2$ where the norm is taken with respect to the metric $g$. But $1<2$.



      Hence, the answer to your question is indeed negative.



      Here is a more general argument. Let $M$ be any smooth compact connected manifold and let $f: Mto M$ be any degree 1 smooth map, for instance, the identity map. Take $0<beta<1$. Then for any Riemannian metric $g$ on $M$ there exists a point $xin M$ such that $||df_x||ge 1>beta$. The reason is that otherwise the volume of $f(M)$ is strictly less than the volume of $M$, which contradicts the degree assumption.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        I will consider $S^1$ as the one-point compactification of the real line, $S^1={mathbb R}cup{infty}$ where the smooth atlas is given by the identity chart on ${mathbb R}$ and the chart given by the map $xmapsto x^{-1}$ on ${mathbb R}cup{infty}- {0}$. The map $f: S^1to S^1$ given by $f(x)=2x, xne infty$, $f(infty)=infty$ is a diffeomorphism.



        If you do not like this, just take the standard unit circle in the complex plane and let $f$ be the restriction to $S^1$ of a suitable linear-fractional transformation preserving the unit disk.



        Now, suppose that $g$ is any Riemannian metric on $S^1$; take $beta=1$. The derivative $df_0$ of $f$ at $0$ is again the multiplication by $2$. Hence, $||df_0||=2$ where the norm is taken with respect to the metric $g$. But $1<2$.



        Hence, the answer to your question is indeed negative.



        Here is a more general argument. Let $M$ be any smooth compact connected manifold and let $f: Mto M$ be any degree 1 smooth map, for instance, the identity map. Take $0<beta<1$. Then for any Riemannian metric $g$ on $M$ there exists a point $xin M$ such that $||df_x||ge 1>beta$. The reason is that otherwise the volume of $f(M)$ is strictly less than the volume of $M$, which contradicts the degree assumption.






        share|cite|improve this answer









        $endgroup$



        I will consider $S^1$ as the one-point compactification of the real line, $S^1={mathbb R}cup{infty}$ where the smooth atlas is given by the identity chart on ${mathbb R}$ and the chart given by the map $xmapsto x^{-1}$ on ${mathbb R}cup{infty}- {0}$. The map $f: S^1to S^1$ given by $f(x)=2x, xne infty$, $f(infty)=infty$ is a diffeomorphism.



        If you do not like this, just take the standard unit circle in the complex plane and let $f$ be the restriction to $S^1$ of a suitable linear-fractional transformation preserving the unit disk.



        Now, suppose that $g$ is any Riemannian metric on $S^1$; take $beta=1$. The derivative $df_0$ of $f$ at $0$ is again the multiplication by $2$. Hence, $||df_0||=2$ where the norm is taken with respect to the metric $g$. But $1<2$.



        Hence, the answer to your question is indeed negative.



        Here is a more general argument. Let $M$ be any smooth compact connected manifold and let $f: Mto M$ be any degree 1 smooth map, for instance, the identity map. Take $0<beta<1$. Then for any Riemannian metric $g$ on $M$ there exists a point $xin M$ such that $||df_x||ge 1>beta$. The reason is that otherwise the volume of $f(M)$ is strictly less than the volume of $M$, which contradicts the degree assumption.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 26 '18 at 2:04









        Moishe CohenMoishe Cohen

        47.9k344110




        47.9k344110






























            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%2f3052063%2freference-riemannian-metric-with-operator-norm-arbitrarily-small-compact-manifo%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

            To store a contact into the json file from server.js file using a class in NodeJS

            Image
            0 I have created three files -contactPresentation.js, contactService.js, contacts.json. I will run the contactService.js file. And it will navigate to contactPresentation.js file that will only choose an option to either add or display a file. If add(), then add the data into add of service file.And the data will be added to the contacts.json file . And if display(), then display the display() of service file. And the display will take the data from contacts.json file. However, i know my code is flawed. Please correct it . I am also getting an error - this.contacts.push(new ContactService(userName, contactNumber, emailId)); ^ TypeError: Cannot read property 'push' of undefined at ContactService.add (e:NodeJSProjectcontactManager2contactService.js:57:23) at new C
            Read more

            Redirect URL with Chrome Remote Debugging Android Devices

            Image
            1 How can i redirect an URL with Chrome Remote Debugging Android Devices? like Requestly Chrome extension on desktop. google-chrome debugging url redirect testing share | improve this question edited Nov 24 '18 at 8:16 sachinjain024 14.7k 23 73 129 asked Nov 23 '18 at 16:09 Momen Momen 91 10
            Read more

            Dieringhausen

            Image
            Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
            Read more