Mixing
Second fundamental form of a hypersurface with a defining function
$begingroup$
The matrix of the second fundamental form of a parametrized hypersurface is given by
$$h_{ij}=-langle N,r_{ij}rangle,$$
where $N$ is the unit normal vector and $r$ is the parametrization of the hypersruface (say $M$ of $mathbb{R}^N$). If the same hypersurface has a defining function $rho:mathbb{R}^Ntomathbb{R}$, with $left| nabla rhoright|=1$ along $M$, then the unit normal is given by
$$N = nablarho.$$
The shape operator is given by
$$S_p = -dN_p = -Hessian_p(rho).$$
Is the matix of the second fundamental form of $M$ at $p$ is the same as the hessian of $rho$? How does the quadratic form of the second fundamentatl form look like?
differential-geometry
asked Jan 9 at 17:14
user888888888888888888888user888888888888888888888
157116
$endgroup$
add a comment |
$begingroup$
The matrix of the second fundamental form of a parametrized hypersurface is given by
$$h_{ij}=-langle N,r_{ij}rangle,$$
where $N$ is the unit normal vector and $r$ is the parametrization of the hypersruface (say $M$ of $mathbb{R}^N$). If the same hypersurface has a defining function $rho:mathbb{R}^Ntomathbb{R}$, with $left| nabla rhoright|=1$ along $M$, then the unit normal is given by
$$N = nablarho.$$
The shape operator is given by
$$S_p = -dN_p = -Hessian_p(rho).$$
Is the matix of the second fundamental form of $M$ at $p$ is the same as the hessian of $rho$? How does the quadratic form of the second fundamentatl form look like?
differential-geometry
asked Jan 9 at 17:14
user888888888888888888888user888888888888888888888
157116
$endgroup$
$begingroup$
No. Note that the Hessian is an $Ntimes N$ matrix, and the second fundamental form should be an $(N-1)times (N-1)$ matrix (and you have to tell me with respect to what basis for the tangent space you're computing that matrix). Nevertheless, this answer may be of interest.
$endgroup$
– Ted Shifrin
Jan 9 at 22:02
add a comment |
$begingroup$
The matrix of the second fundamental form of a parametrized hypersurface is given by
$$h_{ij}=-langle N,r_{ij}rangle,$$
where $N$ is the unit normal vector and $r$ is the parametrization of the hypersruface (say $M$ of $mathbb{R}^N$). If the same hypersurface has a defining function $rho:mathbb{R}^Ntomathbb{R}$, with $left| nabla rhoright|=1$ along $M$, then the unit normal is given by
$$N = nablarho.$$
The shape operator is given by
$$S_p = -dN_p = -Hessian_p(rho).$$
Is the matix of the second fundamental form of $M$ at $p$ is the same as the hessian of $rho$? How does the quadratic form of the second fundamentatl form look like?
differential-geometry
asked Jan 9 at 17:14
user888888888888888888888user888888888888888888888
157116
$endgroup$
The matrix of the second fundamental form of a parametrized hypersurface is given by
$$h_{ij}=-langle N,r_{ij}rangle,$$
where $N$ is the unit normal vector and $r$ is the parametrization of the hypersruface (say $M$ of $mathbb{R}^N$). If the same hypersurface has a defining function $rho:mathbb{R}^Ntomathbb{R}$, with $left| nabla rhoright|=1$ along $M$, then the unit normal is given by
$$N = nablarho.$$
The shape operator is given by
$$S_p = -dN_p = -Hessian_p(rho).$$
Is the matix of the second fundamental form of $M$ at $p$ is the same as the hessian of $rho$? How does the quadratic form of the second fundamentatl form look like?
differential-geometry
differential-geometry
asked Jan 9 at 17:14
user888888888888888888888user888888888888888888888
157116
asked Jan 9 at 17:14
user888888888888888888888user888888888888888888888
157116
asked Jan 9 at 17:14
user888888888888888888888user888888888888888888888
157116
asked Jan 9 at 17:14
user888888888888888888888user888888888888888888888
157116
asked Jan 9 at 17:14
user888888888888888888888user888888888888888888888
157116
157116
$begingroup$
No. Note that the Hessian is an $Ntimes N$ matrix, and the second fundamental form should be an $(N-1)times (N-1)$ matrix (and you have to tell me with respect to what basis for the tangent space you're computing that matrix). Nevertheless, this answer may be of interest.
$endgroup$
– Ted Shifrin
Jan 9 at 22:02
add a comment |
$begingroup$
No. Note that the Hessian is an $Ntimes N$ matrix, and the second fundamental form should be an $(N-1)times (N-1)$ matrix (and you have to tell me with respect to what basis for the tangent space you're computing that matrix). Nevertheless, this answer may be of interest.
$endgroup$
– Ted Shifrin
Jan 9 at 22:02
$begingroup$
No. Note that the Hessian is an $Ntimes N$ matrix, and the second fundamental form should be an $(N-1)times (N-1)$ matrix (and you have to tell me with respect to what basis for the tangent space you're computing that matrix). Nevertheless, this answer may be of interest.
$endgroup$
– Ted Shifrin
Jan 9 at 22:02
$begingroup$
No. Note that the Hessian is an $Ntimes N$ matrix, and the second fundamental form should be an $(N-1)times (N-1)$ matrix (and you have to tell me with respect to what basis for the tangent space you're computing that matrix). Nevertheless, this answer may be of interest.
$endgroup$
– Ted Shifrin
Jan 9 at 22:02
add a comment |
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$begingroup$
No. Note that the Hessian is an $Ntimes N$ matrix, and the second fundamental form should be an $(N-1)times (N-1)$ matrix (and you have to tell me with respect to what basis for the tangent space you're computing that matrix). Nevertheless, this answer may be of interest.
$endgroup$
– Ted Shifrin
Jan 9 at 22:02