Derivation of the Riemannian metric tensor
$begingroup$
Firstly, I would like to say that I didn't do a course in Riemannian geometry, so my doubt is probably something that is not clear for me from the basis of the Riemannian geometry, which I'm self-studying.
I'm reading Interior estimates for hypersurfaces moving by mean curvature and I'm trying understand the statement:
"Now note that $nabla u = omega - langle nu , omega rangle$ and $nabla (|textbf{x}|^2 - u^2) = 2(textbf{x} - langle textbf{x} , nu rangle nu - u nabla u)$",
where $M$ is an hypersurface of dimension $n$ which evolves under Mean Curvature Flow by the immersion $F: M times [0,T) longrightarrow mathbb{R}^{n+1}$, $textbf{x} = F(p,t)$, $u := langle textbf{x} , omega rangle$, $nu$ is the normal unit vector and $omega$ is a fixed vector on $mathbb{R}^{n+1}$ such that $langle omega, nu rangle > 0$.
First, I think the authors wanted to write $nabla u = omega - langle nu, omega rangle nu$, i.e., $nabla u$ it's the projection of $omega$ on tangent space of $M$.
I tried compute $nabla u$ and I found $nabla u = langle nabla textbf{x}, omega rangle$, which is the length of the projection of $omega$ on tangent space of $M$. I did this computation by the fact that $u$ is a function defined on $M$, by the compatibility of the tensor metric, but I found some problems because my computation of $nabla u$ doesn't match with the computation of $nabla u$ did by the authors and the other problem is that the compatibility of the tensor metric it's relationed with the derivation of a function along to a vector field by relation
$$nabla_X langle Y,Z rangle = X langle Y,Z rangle = langle nabla_X Y,Z rangle + langle Y, nabla_X Z rangle,$$
but I don't have explicitly along what vector field $u$ is being differentiated. I think this vector field defined on $M$ is arbitrary and this explain why the direction was omitted in the differentiation of $u$. I had these same problems when I tried compute $nabla |textbf{x}|^2$ by an analagous argument.
I would like to know how compute $nabla u$ and $nabla |textbf{x}|^2$. Thanks in advance!
differential-geometry riemannian-geometry
asked Dec 18 '18 at 19:46
GeorgeGeorge
850615
$endgroup$
add a comment |
$begingroup$
Firstly, I would like to say that I didn't do a course in Riemannian geometry, so my doubt is probably something that is not clear for me from the basis of the Riemannian geometry, which I'm self-studying.
I'm reading Interior estimates for hypersurfaces moving by mean curvature and I'm trying understand the statement:
"Now note that $nabla u = omega - langle nu , omega rangle$ and $nabla (|textbf{x}|^2 - u^2) = 2(textbf{x} - langle textbf{x} , nu rangle nu - u nabla u)$",
where $M$ is an hypersurface of dimension $n$ which evolves under Mean Curvature Flow by the immersion $F: M times [0,T) longrightarrow mathbb{R}^{n+1}$, $textbf{x} = F(p,t)$, $u := langle textbf{x} , omega rangle$, $nu$ is the normal unit vector and $omega$ is a fixed vector on $mathbb{R}^{n+1}$ such that $langle omega, nu rangle > 0$.
First, I think the authors wanted to write $nabla u = omega - langle nu, omega rangle nu$, i.e., $nabla u$ it's the projection of $omega$ on tangent space of $M$.
I tried compute $nabla u$ and I found $nabla u = langle nabla textbf{x}, omega rangle$, which is the length of the projection of $omega$ on tangent space of $M$. I did this computation by the fact that $u$ is a function defined on $M$, by the compatibility of the tensor metric, but I found some problems because my computation of $nabla u$ doesn't match with the computation of $nabla u$ did by the authors and the other problem is that the compatibility of the tensor metric it's relationed with the derivation of a function along to a vector field by relation
$$nabla_X langle Y,Z rangle = X langle Y,Z rangle = langle nabla_X Y,Z rangle + langle Y, nabla_X Z rangle,$$
but I don't have explicitly along what vector field $u$ is being differentiated. I think this vector field defined on $M$ is arbitrary and this explain why the direction was omitted in the differentiation of $u$. I had these same problems when I tried compute $nabla |textbf{x}|^2$ by an analagous argument.
I would like to know how compute $nabla u$ and $nabla |textbf{x}|^2$. Thanks in advance!
differential-geometry riemannian-geometry
asked Dec 18 '18 at 19:46
GeorgeGeorge
850615
$endgroup$
add a comment |
$begingroup$
Firstly, I would like to say that I didn't do a course in Riemannian geometry, so my doubt is probably something that is not clear for me from the basis of the Riemannian geometry, which I'm self-studying.
I'm reading Interior estimates for hypersurfaces moving by mean curvature and I'm trying understand the statement:
"Now note that $nabla u = omega - langle nu , omega rangle$ and $nabla (|textbf{x}|^2 - u^2) = 2(textbf{x} - langle textbf{x} , nu rangle nu - u nabla u)$",
where $M$ is an hypersurface of dimension $n$ which evolves under Mean Curvature Flow by the immersion $F: M times [0,T) longrightarrow mathbb{R}^{n+1}$, $textbf{x} = F(p,t)$, $u := langle textbf{x} , omega rangle$, $nu$ is the normal unit vector and $omega$ is a fixed vector on $mathbb{R}^{n+1}$ such that $langle omega, nu rangle > 0$.
First, I think the authors wanted to write $nabla u = omega - langle nu, omega rangle nu$, i.e., $nabla u$ it's the projection of $omega$ on tangent space of $M$.
I tried compute $nabla u$ and I found $nabla u = langle nabla textbf{x}, omega rangle$, which is the length of the projection of $omega$ on tangent space of $M$. I did this computation by the fact that $u$ is a function defined on $M$, by the compatibility of the tensor metric, but I found some problems because my computation of $nabla u$ doesn't match with the computation of $nabla u$ did by the authors and the other problem is that the compatibility of the tensor metric it's relationed with the derivation of a function along to a vector field by relation
$$nabla_X langle Y,Z rangle = X langle Y,Z rangle = langle nabla_X Y,Z rangle + langle Y, nabla_X Z rangle,$$
but I don't have explicitly along what vector field $u$ is being differentiated. I think this vector field defined on $M$ is arbitrary and this explain why the direction was omitted in the differentiation of $u$. I had these same problems when I tried compute $nabla |textbf{x}|^2$ by an analagous argument.
I would like to know how compute $nabla u$ and $nabla |textbf{x}|^2$. Thanks in advance!
differential-geometry riemannian-geometry
asked Dec 18 '18 at 19:46
GeorgeGeorge
850615
$endgroup$
Firstly, I would like to say that I didn't do a course in Riemannian geometry, so my doubt is probably something that is not clear for me from the basis of the Riemannian geometry, which I'm self-studying.
I'm reading Interior estimates for hypersurfaces moving by mean curvature and I'm trying understand the statement:
"Now note that $nabla u = omega - langle nu , omega rangle$ and $nabla (|textbf{x}|^2 - u^2) = 2(textbf{x} - langle textbf{x} , nu rangle nu - u nabla u)$",
where $M$ is an hypersurface of dimension $n$ which evolves under Mean Curvature Flow by the immersion $F: M times [0,T) longrightarrow mathbb{R}^{n+1}$, $textbf{x} = F(p,t)$, $u := langle textbf{x} , omega rangle$, $nu$ is the normal unit vector and $omega$ is a fixed vector on $mathbb{R}^{n+1}$ such that $langle omega, nu rangle > 0$.
First, I think the authors wanted to write $nabla u = omega - langle nu, omega rangle nu$, i.e., $nabla u$ it's the projection of $omega$ on tangent space of $M$.
I tried compute $nabla u$ and I found $nabla u = langle nabla textbf{x}, omega rangle$, which is the length of the projection of $omega$ on tangent space of $M$. I did this computation by the fact that $u$ is a function defined on $M$, by the compatibility of the tensor metric, but I found some problems because my computation of $nabla u$ doesn't match with the computation of $nabla u$ did by the authors and the other problem is that the compatibility of the tensor metric it's relationed with the derivation of a function along to a vector field by relation
$$nabla_X langle Y,Z rangle = X langle Y,Z rangle = langle nabla_X Y,Z rangle + langle Y, nabla_X Z rangle,$$
but I don't have explicitly along what vector field $u$ is being differentiated. I think this vector field defined on $M$ is arbitrary and this explain why the direction was omitted in the differentiation of $u$. I had these same problems when I tried compute $nabla |textbf{x}|^2$ by an analagous argument.
I would like to know how compute $nabla u$ and $nabla |textbf{x}|^2$. Thanks in advance!
differential-geometry riemannian-geometry
differential-geometry riemannian-geometry
asked Dec 18 '18 at 19:46
GeorgeGeorge
850615
asked Dec 18 '18 at 19:46
GeorgeGeorge
850615
asked Dec 18 '18 at 19:46
GeorgeGeorge
850615
asked Dec 18 '18 at 19:46
GeorgeGeorge
850615
asked Dec 18 '18 at 19:46
GeorgeGeorge
850615
850615
add a comment |
add a comment |
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