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Minimal surface with a “flat curve”


Curvature of a non-compact complete surfaceFor a closed plane curve, showing some inequalities.Proof check for critical point definition with mean curvatureShow $gamma(I)$ is a regular parametrized curve in $S$Is a zero mean curvature submanifold, with a flat open subset, flat everywhere?Minimal surface having a Jordan curve as boundaryWhen is a minimal surface a graph and not area-minimizing?Constant normal vector along surface curve implies straight lineMinimal surface between two parallel circle of same radius : why is it a surface of revolution?Compactness for Stable Minimal Surfaces













3












$begingroup$


Let $Sigma^2 subseteq mathbbR^3$ 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 $Sigma^n subseteq mathbbR^n+1$ 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 $Sigma^2 subseteq mathbbR^3$ 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 $Sigma^n subseteq mathbbR^n+1$ 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 $Sigma^2 subseteq mathbbR^3$ 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 $Sigma^2 subseteq mathbbR^3$ 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 $Sigma^n subseteq mathbbR^n+1$ 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 $Sigma^n subseteq mathbbR^n+1$ 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 $Sigma^n subseteq mathbbR^n+1$ 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 $Sigma^n subseteq mathbbR^n+1$ 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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