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Minimal surface with a “flat curve”
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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
$endgroup$
add a comment |
$begingroup$
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?
differential-geometry elliptic-equations minimal-surfaces
$endgroup$
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
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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
add a comment |
$begingroup$
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?
differential-geometry elliptic-equations minimal-surfaces
$endgroup$
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
differential-geometry elliptic-equations minimal-surfaces
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
add a comment |
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
add a comment |
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$begingroup$
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