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






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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.






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    1 Answer
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    1 Answer
    1






    active

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    active

    oldest

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    active

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    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






























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