Closed 3D curve with linearly independent gradient
A curve $C$ in 3D can be described by the intersection of two surfaces, that is, $C={p in mathbb{R}^3: f_1(p)=0, f_2(p)=0}$. $f_1(p)=0$ and $f_2(p)=0$ are the implicit representations of two surface.
Now I want to ask, is it possible to find $f_1$ and $f_2$ such that the intersection is a simple closed curve, and $nabla f_1$ and $nabla f_2$ are linearly independent in $mathbb{R}^3$?
geometry differential-geometry algebraic-geometry surfaces curves
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A curve $C$ in 3D can be described by the intersection of two surfaces, that is, $C={p in mathbb{R}^3: f_1(p)=0, f_2(p)=0}$. $f_1(p)=0$ and $f_2(p)=0$ are the implicit representations of two surface.
Now I want to ask, is it possible to find $f_1$ and $f_2$ such that the intersection is a simple closed curve, and $nabla f_1$ and $nabla f_2$ are linearly independent in $mathbb{R}^3$?
geometry differential-geometry algebraic-geometry surfaces curves
add a comment |
A curve $C$ in 3D can be described by the intersection of two surfaces, that is, $C={p in mathbb{R}^3: f_1(p)=0, f_2(p)=0}$. $f_1(p)=0$ and $f_2(p)=0$ are the implicit representations of two surface.
Now I want to ask, is it possible to find $f_1$ and $f_2$ such that the intersection is a simple closed curve, and $nabla f_1$ and $nabla f_2$ are linearly independent in $mathbb{R}^3$?
geometry differential-geometry algebraic-geometry surfaces curves
A curve $C$ in 3D can be described by the intersection of two surfaces, that is, $C={p in mathbb{R}^3: f_1(p)=0, f_2(p)=0}$. $f_1(p)=0$ and $f_2(p)=0$ are the implicit representations of two surface.
Now I want to ask, is it possible to find $f_1$ and $f_2$ such that the intersection is a simple closed curve, and $nabla f_1$ and $nabla f_2$ are linearly independent in $mathbb{R}^3$?
geometry differential-geometry algebraic-geometry surfaces curves
geometry differential-geometry algebraic-geometry surfaces curves
asked Nov 30 '18 at 11:23
winstonwinston
517218
517218
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