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Minimal surface with a “flat curve”


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$begingroup$


Let $Sigma^2 subseteq mathbb{R}^3$ be a complete minimal surface and let's assume that there exists a smooth regular curve $gamma colon I to Sigma$ such that $K(gamma(t)) = 0$ for all $t in I$, where $K$ is the Gauss curvature of $Sigma$.



Can I conclude that $Sigma$ is flat, i.e. a plane?










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Yes. Do you know how to interpret the Gauss map of a minimal surface as a holomorphic mapping to $Bbb CP^1$?
    $endgroup$
    – Ted Shifrin
    Mar 20 at 19:07










  • $begingroup$
    Right, but then how do you conclude? Moreover, what about the higher dimensional version of my question? I.e. if $Sigma^n subseteq mathbb{R}^{n+1}$ is a minimal hypersurfaces such that its second fundamental form vanishes along a certain $Gamma^{n-1} subseteq Sigma^n$, can I still conclude that $Sigma$ is flat? In this case I can't use complex analysis argument, right?
    $endgroup$
    – Math_tourist
    Mar 20 at 19:33






  • 1




    $begingroup$
    I don't know about the higher-dimensional situation. But with regard to the first question: This is the identity principle for holomorphic functions — on a connected set, if a holomorphic function is zero on a set of points with a limit point, then it is identically zero.
    $endgroup$
    – Ted Shifrin
    Mar 20 at 20:57










  • $begingroup$
    Ah right!!! Thanks a lot!!
    $endgroup$
    – Math_tourist
    Mar 20 at 21:03
















3












$begingroup$


Let $Sigma^2 subseteq mathbb{R}^3$ be a complete minimal surface and let's assume that there exists a smooth regular curve $gamma colon I to Sigma$ such that $K(gamma(t)) = 0$ for all $t in I$, where $K$ is the Gauss curvature of $Sigma$.



Can I conclude that $Sigma$ is flat, i.e. a plane?










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Yes. Do you know how to interpret the Gauss map of a minimal surface as a holomorphic mapping to $Bbb CP^1$?
    $endgroup$
    – Ted Shifrin
    Mar 20 at 19:07










  • $begingroup$
    Right, but then how do you conclude? Moreover, what about the higher dimensional version of my question? I.e. if $Sigma^n subseteq mathbb{R}^{n+1}$ is a minimal hypersurfaces such that its second fundamental form vanishes along a certain $Gamma^{n-1} subseteq Sigma^n$, can I still conclude that $Sigma$ is flat? In this case I can't use complex analysis argument, right?
    $endgroup$
    – Math_tourist
    Mar 20 at 19:33






  • 1




    $begingroup$
    I don't know about the higher-dimensional situation. But with regard to the first question: This is the identity principle for holomorphic functions — on a connected set, if a holomorphic function is zero on a set of points with a limit point, then it is identically zero.
    $endgroup$
    – Ted Shifrin
    Mar 20 at 20:57










  • $begingroup$
    Ah right!!! Thanks a lot!!
    $endgroup$
    – Math_tourist
    Mar 20 at 21:03














3












3








3


1



$begingroup$


Let $Sigma^2 subseteq mathbb{R}^3$ be a complete minimal surface and let's assume that there exists a smooth regular curve $gamma colon I to Sigma$ such that $K(gamma(t)) = 0$ for all $t in I$, where $K$ is the Gauss curvature of $Sigma$.



Can I conclude that $Sigma$ is flat, i.e. a plane?










share|cite|improve this question









$endgroup$




Let $Sigma^2 subseteq mathbb{R}^3$ be a complete minimal surface and let's assume that there exists a smooth regular curve $gamma colon I to Sigma$ such that $K(gamma(t)) = 0$ for all $t in I$, where $K$ is the Gauss curvature of $Sigma$.



Can I conclude that $Sigma$ is flat, i.e. a plane?







differential-geometry elliptic-equations minimal-surfaces






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 20 at 18:25









Math_touristMath_tourist

764




764








  • 1




    $begingroup$
    Yes. Do you know how to interpret the Gauss map of a minimal surface as a holomorphic mapping to $Bbb CP^1$?
    $endgroup$
    – Ted Shifrin
    Mar 20 at 19:07










  • $begingroup$
    Right, but then how do you conclude? Moreover, what about the higher dimensional version of my question? I.e. if $Sigma^n subseteq mathbb{R}^{n+1}$ is a minimal hypersurfaces such that its second fundamental form vanishes along a certain $Gamma^{n-1} subseteq Sigma^n$, can I still conclude that $Sigma$ is flat? In this case I can't use complex analysis argument, right?
    $endgroup$
    – Math_tourist
    Mar 20 at 19:33






  • 1




    $begingroup$
    I don't know about the higher-dimensional situation. But with regard to the first question: This is the identity principle for holomorphic functions — on a connected set, if a holomorphic function is zero on a set of points with a limit point, then it is identically zero.
    $endgroup$
    – Ted Shifrin
    Mar 20 at 20:57










  • $begingroup$
    Ah right!!! Thanks a lot!!
    $endgroup$
    – Math_tourist
    Mar 20 at 21:03














  • 1




    $begingroup$
    Yes. Do you know how to interpret the Gauss map of a minimal surface as a holomorphic mapping to $Bbb CP^1$?
    $endgroup$
    – Ted Shifrin
    Mar 20 at 19:07










  • $begingroup$
    Right, but then how do you conclude? Moreover, what about the higher dimensional version of my question? I.e. if $Sigma^n subseteq mathbb{R}^{n+1}$ is a minimal hypersurfaces such that its second fundamental form vanishes along a certain $Gamma^{n-1} subseteq Sigma^n$, can I still conclude that $Sigma$ is flat? In this case I can't use complex analysis argument, right?
    $endgroup$
    – Math_tourist
    Mar 20 at 19:33






  • 1




    $begingroup$
    I don't know about the higher-dimensional situation. But with regard to the first question: This is the identity principle for holomorphic functions — on a connected set, if a holomorphic function is zero on a set of points with a limit point, then it is identically zero.
    $endgroup$
    – Ted Shifrin
    Mar 20 at 20:57










  • $begingroup$
    Ah right!!! Thanks a lot!!
    $endgroup$
    – Math_tourist
    Mar 20 at 21:03








1




1




$begingroup$
Yes. Do you know how to interpret the Gauss map of a minimal surface as a holomorphic mapping to $Bbb CP^1$?
$endgroup$
– Ted Shifrin
Mar 20 at 19:07




$begingroup$
Yes. Do you know how to interpret the Gauss map of a minimal surface as a holomorphic mapping to $Bbb CP^1$?
$endgroup$
– Ted Shifrin
Mar 20 at 19:07












$begingroup$
Right, but then how do you conclude? Moreover, what about the higher dimensional version of my question? I.e. if $Sigma^n subseteq mathbb{R}^{n+1}$ is a minimal hypersurfaces such that its second fundamental form vanishes along a certain $Gamma^{n-1} subseteq Sigma^n$, can I still conclude that $Sigma$ is flat? In this case I can't use complex analysis argument, right?
$endgroup$
– Math_tourist
Mar 20 at 19:33




$begingroup$
Right, but then how do you conclude? Moreover, what about the higher dimensional version of my question? I.e. if $Sigma^n subseteq mathbb{R}^{n+1}$ is a minimal hypersurfaces such that its second fundamental form vanishes along a certain $Gamma^{n-1} subseteq Sigma^n$, can I still conclude that $Sigma$ is flat? In this case I can't use complex analysis argument, right?
$endgroup$
– Math_tourist
Mar 20 at 19:33




1




1




$begingroup$
I don't know about the higher-dimensional situation. But with regard to the first question: This is the identity principle for holomorphic functions — on a connected set, if a holomorphic function is zero on a set of points with a limit point, then it is identically zero.
$endgroup$
– Ted Shifrin
Mar 20 at 20:57




$begingroup$
I don't know about the higher-dimensional situation. But with regard to the first question: This is the identity principle for holomorphic functions — on a connected set, if a holomorphic function is zero on a set of points with a limit point, then it is identically zero.
$endgroup$
– Ted Shifrin
Mar 20 at 20:57












$begingroup$
Ah right!!! Thanks a lot!!
$endgroup$
– Math_tourist
Mar 20 at 21:03




$begingroup$
Ah right!!! Thanks a lot!!
$endgroup$
– Math_tourist
Mar 20 at 21:03










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