Isometric embedding of a genus g surface












5












$begingroup$


Can a genus $g$ surface with constant negative curvature and $g>1$ be isometrically embedded in $mathbb{R}^4?$






dg.differential-geometry riemannian-geometry riemann-surfaces metric-embeddings







share|cite|improve this question











$endgroup$












  • $begingroup$
    Is there any example or counter example?
    $endgroup$
    – GAUTAM NEELAKANTAN MEMANA
    Mar 20 at 8:53






  • 4




    $begingroup$
    Relevant: math.stackexchange.com/questions/1528046/…
    $endgroup$
    – Aknazar Kazhymurat
    Mar 20 at 8:57










  • $begingroup$
    So, is this an open problem?
    $endgroup$
    – GAUTAM NEELAKANTAN MEMANA
    Mar 20 at 8:59






  • 3




    $begingroup$
    You might be interested in this question and its answers.
    $endgroup$
    – Michael Albanese
    Mar 20 at 12:00
















5












$begingroup$


Can a genus $g$ surface with constant negative curvature and $g>1$ be isometrically embedded in $mathbb{R}^4?$






dg.differential-geometry riemannian-geometry riemann-surfaces metric-embeddings







share|cite|improve this question











$endgroup$












  • $begingroup$
    Is there any example or counter example?
    $endgroup$
    – GAUTAM NEELAKANTAN MEMANA
    Mar 20 at 8:53






  • 4




    $begingroup$
    Relevant: math.stackexchange.com/questions/1528046/…
    $endgroup$
    – Aknazar Kazhymurat
    Mar 20 at 8:57










  • $begingroup$
    So, is this an open problem?
    $endgroup$
    – GAUTAM NEELAKANTAN MEMANA
    Mar 20 at 8:59






  • 3




    $begingroup$
    You might be interested in this question and its answers.
    $endgroup$
    – Michael Albanese
    Mar 20 at 12:00














5












5








5


1



$begingroup$


Can a genus $g$ surface with constant negative curvature and $g>1$ be isometrically embedded in $mathbb{R}^4?$






dg.differential-geometry riemannian-geometry riemann-surfaces metric-embeddings







share|cite|improve this question











$endgroup$




Can a genus $g$ surface with constant negative curvature and $g>1$ be isometrically embedded in $mathbb{R}^4?$






dg.differential-geometry riemannian-geometry riemann-surfaces metric-embeddings




dg.differential-geometry riemannian-geometry riemann-surfaces metric-embeddings






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 20 at 8:57









Sean Lawton

3,97622248




3,97622248










asked Mar 20 at 8:49









GAUTAM NEELAKANTAN MEMANAGAUTAM NEELAKANTAN MEMANA

264




264












  • $begingroup$
    Is there any example or counter example?
    $endgroup$
    – GAUTAM NEELAKANTAN MEMANA
    Mar 20 at 8:53






  • 4




    $begingroup$
    Relevant: math.stackexchange.com/questions/1528046/…
    $endgroup$
    – Aknazar Kazhymurat
    Mar 20 at 8:57










  • $begingroup$
    So, is this an open problem?
    $endgroup$
    – GAUTAM NEELAKANTAN MEMANA
    Mar 20 at 8:59






  • 3




    $begingroup$
    You might be interested in this question and its answers.
    $endgroup$
    – Michael Albanese
    Mar 20 at 12:00


















  • $begingroup$
    Is there any example or counter example?
    $endgroup$
    – GAUTAM NEELAKANTAN MEMANA
    Mar 20 at 8:53






  • 4




    $begingroup$
    Relevant: math.stackexchange.com/questions/1528046/…
    $endgroup$
    – Aknazar Kazhymurat
    Mar 20 at 8:57










  • $begingroup$
    So, is this an open problem?
    $endgroup$
    – GAUTAM NEELAKANTAN MEMANA
    Mar 20 at 8:59






  • 3




    $begingroup$
    You might be interested in this question and its answers.
    $endgroup$
    – Michael Albanese
    Mar 20 at 12:00
















$begingroup$
Is there any example or counter example?
$endgroup$
– GAUTAM NEELAKANTAN MEMANA
Mar 20 at 8:53




$begingroup$
Is there any example or counter example?
$endgroup$
– GAUTAM NEELAKANTAN MEMANA
Mar 20 at 8:53




4




4




$begingroup$
Relevant: math.stackexchange.com/questions/1528046/…
$endgroup$
– Aknazar Kazhymurat
Mar 20 at 8:57




$begingroup$
Relevant: math.stackexchange.com/questions/1528046/…
$endgroup$
– Aknazar Kazhymurat
Mar 20 at 8:57












$begingroup$
So, is this an open problem?
$endgroup$
– GAUTAM NEELAKANTAN MEMANA
Mar 20 at 8:59




$begingroup$
So, is this an open problem?
$endgroup$
– GAUTAM NEELAKANTAN MEMANA
Mar 20 at 8:59




3




3




$begingroup$
You might be interested in this question and its answers.
$endgroup$
– Michael Albanese
Mar 20 at 12:00




$begingroup$
You might be interested in this question and its answers.
$endgroup$
– Michael Albanese
Mar 20 at 12:00










2 Answers
2






active

oldest

votes


















11












$begingroup$

The Nash-Kuiper Theorem implies the answer is yes if you only require the embedding to be of class $C^1$. However, I believe the actual visualization problem for $ggeq 2$ (that is, producing the embedding in this case) is an open problem (unlike the visualization of the $C^1$-embedding of the flat torus, which has a $C^infty$-isometric embedding into $mathbb{R}^4$).



Since the smallest known $C^infty$-embedding for the hyperbolic plane is $mathbb{R}^6$ I would guess the answer is no for a genus 2 hyperbolic surface (but as far as I know it is open). Note it is a theorem of Hilbert that the hyperbolic plane cannot be $C^r$-embedded into $mathbb{R}^3$ for $rgeq 2$. Later Efimov generalized this to closed hyperbolic surfaces.



I believe these facts and references may be found in:



Isometric Embedding of Riemannian Manifolds in Euclidean Spaces by Qing Han, and Jia-Xing Hong.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    There are visualizations of the Nash--Kuiper embeddings here: hevea-project.fr
    $endgroup$
    – John Pardon
    Mar 20 at 14:30










  • $begingroup$
    @JohnPardon Thanks for the link. However, I already knew about the flat tori and Nash spheres from Hevea; I referenced one in my answer. It is the hyperbolic ones for closed surfaces of $ggeq 2$ that are not presently visualized (as far as I know).
    $endgroup$
    – Sean Lawton
    Mar 20 at 14:33








  • 1




    $begingroup$
    That's a confusing place to put the link: the smooth flat embedding of the torus in R^4 and the C^1 flat embedding of the torus in R^3 aren't related at all.
    $endgroup$
    – John Pardon
    Mar 20 at 14:38






  • 1




    $begingroup$
    There isn't really any difference between the $g=1$ and $ggeq 2$ cases for the Nash--Kuiper construction of C^1 isometric embeddings; their method works for an arbitrary metric without regard to its curvature. I'm fairly certain that if one can render a flat torus in R^3, then one can render any surface with any metric in R^3, though the people at Hévéa might know better. (+1 btw)
    $endgroup$
    – John Pardon
    Mar 20 at 14:45








  • 1




    $begingroup$
    @JohnPardon Sure, I know the method works theoretically in general, but actually doing it is computationally complicated (which is why it hasn't been done).
    $endgroup$
    – Sean Lawton
    Mar 20 at 14:48



















5












$begingroup$

I don't know the answer to your question, and I wouldn't be surprised if it were open. You may be interested to know that any compact Riemannian two-manifold $(V, g)$ admits a $C^{infty}$ isometric embedding $V to mathbb{R}^5$. See Gromov's Partial Differential Relations, pages 298 - 303.






share|cite|improve this answer











$endgroup$














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    2 Answers
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    2 Answers
    2






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    active

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    11












    $begingroup$

    The Nash-Kuiper Theorem implies the answer is yes if you only require the embedding to be of class $C^1$. However, I believe the actual visualization problem for $ggeq 2$ (that is, producing the embedding in this case) is an open problem (unlike the visualization of the $C^1$-embedding of the flat torus, which has a $C^infty$-isometric embedding into $mathbb{R}^4$).



    Since the smallest known $C^infty$-embedding for the hyperbolic plane is $mathbb{R}^6$ I would guess the answer is no for a genus 2 hyperbolic surface (but as far as I know it is open). Note it is a theorem of Hilbert that the hyperbolic plane cannot be $C^r$-embedded into $mathbb{R}^3$ for $rgeq 2$. Later Efimov generalized this to closed hyperbolic surfaces.



    I believe these facts and references may be found in:



    Isometric Embedding of Riemannian Manifolds in Euclidean Spaces by Qing Han, and Jia-Xing Hong.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      There are visualizations of the Nash--Kuiper embeddings here: hevea-project.fr
      $endgroup$
      – John Pardon
      Mar 20 at 14:30










    • $begingroup$
      @JohnPardon Thanks for the link. However, I already knew about the flat tori and Nash spheres from Hevea; I referenced one in my answer. It is the hyperbolic ones for closed surfaces of $ggeq 2$ that are not presently visualized (as far as I know).
      $endgroup$
      – Sean Lawton
      Mar 20 at 14:33








    • 1




      $begingroup$
      That's a confusing place to put the link: the smooth flat embedding of the torus in R^4 and the C^1 flat embedding of the torus in R^3 aren't related at all.
      $endgroup$
      – John Pardon
      Mar 20 at 14:38






    • 1




      $begingroup$
      There isn't really any difference between the $g=1$ and $ggeq 2$ cases for the Nash--Kuiper construction of C^1 isometric embeddings; their method works for an arbitrary metric without regard to its curvature. I'm fairly certain that if one can render a flat torus in R^3, then one can render any surface with any metric in R^3, though the people at Hévéa might know better. (+1 btw)
      $endgroup$
      – John Pardon
      Mar 20 at 14:45








    • 1




      $begingroup$
      @JohnPardon Sure, I know the method works theoretically in general, but actually doing it is computationally complicated (which is why it hasn't been done).
      $endgroup$
      – Sean Lawton
      Mar 20 at 14:48
















    11












    $begingroup$

    The Nash-Kuiper Theorem implies the answer is yes if you only require the embedding to be of class $C^1$. However, I believe the actual visualization problem for $ggeq 2$ (that is, producing the embedding in this case) is an open problem (unlike the visualization of the $C^1$-embedding of the flat torus, which has a $C^infty$-isometric embedding into $mathbb{R}^4$).



    Since the smallest known $C^infty$-embedding for the hyperbolic plane is $mathbb{R}^6$ I would guess the answer is no for a genus 2 hyperbolic surface (but as far as I know it is open). Note it is a theorem of Hilbert that the hyperbolic plane cannot be $C^r$-embedded into $mathbb{R}^3$ for $rgeq 2$. Later Efimov generalized this to closed hyperbolic surfaces.



    I believe these facts and references may be found in:



    Isometric Embedding of Riemannian Manifolds in Euclidean Spaces by Qing Han, and Jia-Xing Hong.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      There are visualizations of the Nash--Kuiper embeddings here: hevea-project.fr
      $endgroup$
      – John Pardon
      Mar 20 at 14:30










    • $begingroup$
      @JohnPardon Thanks for the link. However, I already knew about the flat tori and Nash spheres from Hevea; I referenced one in my answer. It is the hyperbolic ones for closed surfaces of $ggeq 2$ that are not presently visualized (as far as I know).
      $endgroup$
      – Sean Lawton
      Mar 20 at 14:33








    • 1




      $begingroup$
      That's a confusing place to put the link: the smooth flat embedding of the torus in R^4 and the C^1 flat embedding of the torus in R^3 aren't related at all.
      $endgroup$
      – John Pardon
      Mar 20 at 14:38






    • 1




      $begingroup$
      There isn't really any difference between the $g=1$ and $ggeq 2$ cases for the Nash--Kuiper construction of C^1 isometric embeddings; their method works for an arbitrary metric without regard to its curvature. I'm fairly certain that if one can render a flat torus in R^3, then one can render any surface with any metric in R^3, though the people at Hévéa might know better. (+1 btw)
      $endgroup$
      – John Pardon
      Mar 20 at 14:45








    • 1




      $begingroup$
      @JohnPardon Sure, I know the method works theoretically in general, but actually doing it is computationally complicated (which is why it hasn't been done).
      $endgroup$
      – Sean Lawton
      Mar 20 at 14:48














    11












    11








    11





    $begingroup$

    The Nash-Kuiper Theorem implies the answer is yes if you only require the embedding to be of class $C^1$. However, I believe the actual visualization problem for $ggeq 2$ (that is, producing the embedding in this case) is an open problem (unlike the visualization of the $C^1$-embedding of the flat torus, which has a $C^infty$-isometric embedding into $mathbb{R}^4$).



    Since the smallest known $C^infty$-embedding for the hyperbolic plane is $mathbb{R}^6$ I would guess the answer is no for a genus 2 hyperbolic surface (but as far as I know it is open). Note it is a theorem of Hilbert that the hyperbolic plane cannot be $C^r$-embedded into $mathbb{R}^3$ for $rgeq 2$. Later Efimov generalized this to closed hyperbolic surfaces.



    I believe these facts and references may be found in:



    Isometric Embedding of Riemannian Manifolds in Euclidean Spaces by Qing Han, and Jia-Xing Hong.






    share|cite|improve this answer











    $endgroup$



    The Nash-Kuiper Theorem implies the answer is yes if you only require the embedding to be of class $C^1$. However, I believe the actual visualization problem for $ggeq 2$ (that is, producing the embedding in this case) is an open problem (unlike the visualization of the $C^1$-embedding of the flat torus, which has a $C^infty$-isometric embedding into $mathbb{R}^4$).



    Since the smallest known $C^infty$-embedding for the hyperbolic plane is $mathbb{R}^6$ I would guess the answer is no for a genus 2 hyperbolic surface (but as far as I know it is open). Note it is a theorem of Hilbert that the hyperbolic plane cannot be $C^r$-embedded into $mathbb{R}^3$ for $rgeq 2$. Later Efimov generalized this to closed hyperbolic surfaces.



    I believe these facts and references may be found in:



    Isometric Embedding of Riemannian Manifolds in Euclidean Spaces by Qing Han, and Jia-Xing Hong.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Mar 20 at 14:42

























    answered Mar 20 at 9:08









    Sean LawtonSean Lawton

    3,97622248




    3,97622248








    • 1




      $begingroup$
      There are visualizations of the Nash--Kuiper embeddings here: hevea-project.fr
      $endgroup$
      – John Pardon
      Mar 20 at 14:30










    • $begingroup$
      @JohnPardon Thanks for the link. However, I already knew about the flat tori and Nash spheres from Hevea; I referenced one in my answer. It is the hyperbolic ones for closed surfaces of $ggeq 2$ that are not presently visualized (as far as I know).
      $endgroup$
      – Sean Lawton
      Mar 20 at 14:33








    • 1




      $begingroup$
      That's a confusing place to put the link: the smooth flat embedding of the torus in R^4 and the C^1 flat embedding of the torus in R^3 aren't related at all.
      $endgroup$
      – John Pardon
      Mar 20 at 14:38






    • 1




      $begingroup$
      There isn't really any difference between the $g=1$ and $ggeq 2$ cases for the Nash--Kuiper construction of C^1 isometric embeddings; their method works for an arbitrary metric without regard to its curvature. I'm fairly certain that if one can render a flat torus in R^3, then one can render any surface with any metric in R^3, though the people at Hévéa might know better. (+1 btw)
      $endgroup$
      – John Pardon
      Mar 20 at 14:45








    • 1




      $begingroup$
      @JohnPardon Sure, I know the method works theoretically in general, but actually doing it is computationally complicated (which is why it hasn't been done).
      $endgroup$
      – Sean Lawton
      Mar 20 at 14:48














    • 1




      $begingroup$
      There are visualizations of the Nash--Kuiper embeddings here: hevea-project.fr
      $endgroup$
      – John Pardon
      Mar 20 at 14:30










    • $begingroup$
      @JohnPardon Thanks for the link. However, I already knew about the flat tori and Nash spheres from Hevea; I referenced one in my answer. It is the hyperbolic ones for closed surfaces of $ggeq 2$ that are not presently visualized (as far as I know).
      $endgroup$
      – Sean Lawton
      Mar 20 at 14:33








    • 1




      $begingroup$
      That's a confusing place to put the link: the smooth flat embedding of the torus in R^4 and the C^1 flat embedding of the torus in R^3 aren't related at all.
      $endgroup$
      – John Pardon
      Mar 20 at 14:38






    • 1




      $begingroup$
      There isn't really any difference between the $g=1$ and $ggeq 2$ cases for the Nash--Kuiper construction of C^1 isometric embeddings; their method works for an arbitrary metric without regard to its curvature. I'm fairly certain that if one can render a flat torus in R^3, then one can render any surface with any metric in R^3, though the people at Hévéa might know better. (+1 btw)
      $endgroup$
      – John Pardon
      Mar 20 at 14:45








    • 1




      $begingroup$
      @JohnPardon Sure, I know the method works theoretically in general, but actually doing it is computationally complicated (which is why it hasn't been done).
      $endgroup$
      – Sean Lawton
      Mar 20 at 14:48








    1




    1




    $begingroup$
    There are visualizations of the Nash--Kuiper embeddings here: hevea-project.fr
    $endgroup$
    – John Pardon
    Mar 20 at 14:30




    $begingroup$
    There are visualizations of the Nash--Kuiper embeddings here: hevea-project.fr
    $endgroup$
    – John Pardon
    Mar 20 at 14:30












    $begingroup$
    @JohnPardon Thanks for the link. However, I already knew about the flat tori and Nash spheres from Hevea; I referenced one in my answer. It is the hyperbolic ones for closed surfaces of $ggeq 2$ that are not presently visualized (as far as I know).
    $endgroup$
    – Sean Lawton
    Mar 20 at 14:33






    $begingroup$
    @JohnPardon Thanks for the link. However, I already knew about the flat tori and Nash spheres from Hevea; I referenced one in my answer. It is the hyperbolic ones for closed surfaces of $ggeq 2$ that are not presently visualized (as far as I know).
    $endgroup$
    – Sean Lawton
    Mar 20 at 14:33






    1




    1




    $begingroup$
    That's a confusing place to put the link: the smooth flat embedding of the torus in R^4 and the C^1 flat embedding of the torus in R^3 aren't related at all.
    $endgroup$
    – John Pardon
    Mar 20 at 14:38




    $begingroup$
    That's a confusing place to put the link: the smooth flat embedding of the torus in R^4 and the C^1 flat embedding of the torus in R^3 aren't related at all.
    $endgroup$
    – John Pardon
    Mar 20 at 14:38




    1




    1




    $begingroup$
    There isn't really any difference between the $g=1$ and $ggeq 2$ cases for the Nash--Kuiper construction of C^1 isometric embeddings; their method works for an arbitrary metric without regard to its curvature. I'm fairly certain that if one can render a flat torus in R^3, then one can render any surface with any metric in R^3, though the people at Hévéa might know better. (+1 btw)
    $endgroup$
    – John Pardon
    Mar 20 at 14:45






    $begingroup$
    There isn't really any difference between the $g=1$ and $ggeq 2$ cases for the Nash--Kuiper construction of C^1 isometric embeddings; their method works for an arbitrary metric without regard to its curvature. I'm fairly certain that if one can render a flat torus in R^3, then one can render any surface with any metric in R^3, though the people at Hévéa might know better. (+1 btw)
    $endgroup$
    – John Pardon
    Mar 20 at 14:45






    1




    1




    $begingroup$
    @JohnPardon Sure, I know the method works theoretically in general, but actually doing it is computationally complicated (which is why it hasn't been done).
    $endgroup$
    – Sean Lawton
    Mar 20 at 14:48




    $begingroup$
    @JohnPardon Sure, I know the method works theoretically in general, but actually doing it is computationally complicated (which is why it hasn't been done).
    $endgroup$
    – Sean Lawton
    Mar 20 at 14:48











    5












    $begingroup$

    I don't know the answer to your question, and I wouldn't be surprised if it were open. You may be interested to know that any compact Riemannian two-manifold $(V, g)$ admits a $C^{infty}$ isometric embedding $V to mathbb{R}^5$. See Gromov's Partial Differential Relations, pages 298 - 303.






    share|cite|improve this answer











    $endgroup$


















      5












      $begingroup$

      I don't know the answer to your question, and I wouldn't be surprised if it were open. You may be interested to know that any compact Riemannian two-manifold $(V, g)$ admits a $C^{infty}$ isometric embedding $V to mathbb{R}^5$. See Gromov's Partial Differential Relations, pages 298 - 303.






      share|cite|improve this answer











      $endgroup$
















        5












        5








        5





        $begingroup$

        I don't know the answer to your question, and I wouldn't be surprised if it were open. You may be interested to know that any compact Riemannian two-manifold $(V, g)$ admits a $C^{infty}$ isometric embedding $V to mathbb{R}^5$. See Gromov's Partial Differential Relations, pages 298 - 303.






        share|cite|improve this answer











        $endgroup$



        I don't know the answer to your question, and I wouldn't be surprised if it were open. You may be interested to know that any compact Riemannian two-manifold $(V, g)$ admits a $C^{infty}$ isometric embedding $V to mathbb{R}^5$. See Gromov's Partial Differential Relations, pages 298 - 303.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Mar 20 at 11:50

























        answered Mar 20 at 11:38









        Michael AlbaneseMichael Albanese

        7,88155393




        7,88155393






























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            Biblatex bibliography style without URLs when DOI exists (in Overleaf with Zotero bibliography)

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            2 2 I have a Zotero bibliography in Overleaf. I use biblatex . usepackage[backend=biber,citestyle=authoryear]{biblatex} addbibresource{zoteroALH.bib} Bibliography entries look like: @book{de_saint-gervais_uniformization_2016, title = {Uniformization of {Riemann} {Surfaces}}, isbn = {978-3-03719-145-3}, url = {https://www.ems-ph.org/books/book.php?proj_nr=198&srch=series%7Chem}, urldate = {2019-02-03}, author = {de Saint-Gervais, Paul Henri}, year = {2016}, doi = {10.4171/145} } In the bibliography down the paper the reference shows like this: There is the DOI, great, but also the URL which is unnecessary when the DOI is there. How can I remove the URL when the DOI is present? biblatex bibliographies overleaf zote
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            ComboBox Display Member on multiple fields

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            2 I'm trying to set the .DisplayMember property of a ComboBox in C#, but I want to bind it to multiple columns in the .DataSouce . My SQL looks like this: SELECT PersNbr, PersFirstName, PersMiddleName, PersLastName FROM Pers WHERE PersNbr = :persNbr; I'm saving this query in a DataTable so each column selected has it's own column in the Datatable . I want to make the .DisplayMember a combination of PersFirstName + PersMiddleName + PersLastName so their full name appears like this: comboBox.DisplayMemeber = "PersFirstName" + "PersMiddleName" + "PersLastName" I know I can just to this on the query: SELECT PersNbr, (PersFirstName || PersMiddleName || PersLastName) PersName and then just do this: comboBox.DisplayMember = "PersName"; but I
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            Is it possible to collect Nectar points via Trainline?

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            2 Since I am not eligible for any Railcard, the only available discount/bonus scheme is Nectar Card for me. GWR, SWR, LNER and Virgin Trains allow to collect Nectar points and I have already linked a card to my GWR account. A friend suggested me to use Trainline App, because it can search for bus tickets either. I had a look at the application, but could not see any option for Nectar. Is there a way to collect Nectar points when I buy tickets via Trainline? uk trains loyalty-programs trainline share | improve this question asked Dec 2 '18 at 15:50 ahmedus 3,224 5 19 49
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