Differentiable manifold testProving a set of $2times 3$ matrices is a manifold?Definition of a Manifold from...
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Differentiable manifold test
Proving a set of $2times 3$ matrices is a manifold?Definition of a Manifold from Guillemin PollackWhy the set $g^{-1}({0}) $ is not a differentiable manifold?Show that this is indeed a differentiable manifold with boundary.Compute the volume element in a differentiable manifold.Why is this set not a manifold?Implicit multivarible differentiable problem - $F(x,f(x))=1$Is this true for a continuously differentiable functionQuestion on the implicit function theorem and dimension.Invertibility of a function (Surjective/everywhere frechet)
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I was wondering how to test that the equations $x=2z^{4}+6$ and $y=z^{2}+1$ define a $C^{infty}$ differentiable manifold in $mathbb{R}^{3}$ with dimension $1$. Edit: I'm very familiar with the implicit function theorem, and I know that it has to be done using that, but I don't know how to relate it with a differentiable manifold.
multivariable-calculus
asked Mar 10 at 13:58
Jack TalionJack Talion
885
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add a comment |
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I was wondering how to test that the equations $x=2z^{4}+6$ and $y=z^{2}+1$ define a $C^{infty}$ differentiable manifold in $mathbb{R}^{3}$ with dimension $1$. Edit: I'm very familiar with the implicit function theorem, and I know that it has to be done using that, but I don't know how to relate it with a differentiable manifold.
multivariable-calculus
asked Mar 10 at 13:58
Jack TalionJack Talion
885
New contributor
Jack Talion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Your equations define a smooth curve on $mathbb R^3$ without intersections. Hence it is a smooth manifold.
$endgroup$
– Dog_69
Mar 10 at 14:30
add a comment |
$begingroup$
I was wondering how to test that the equations $x=2z^{4}+6$ and $y=z^{2}+1$ define a $C^{infty}$ differentiable manifold in $mathbb{R}^{3}$ with dimension $1$. Edit: I'm very familiar with the implicit function theorem, and I know that it has to be done using that, but I don't know how to relate it with a differentiable manifold.
multivariable-calculus
asked Mar 10 at 13:58
Jack TalionJack Talion
885
New contributor
Jack Talion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I was wondering how to test that the equations $x=2z^{4}+6$ and $y=z^{2}+1$ define a $C^{infty}$ differentiable manifold in $mathbb{R}^{3}$ with dimension $1$. Edit: I'm very familiar with the implicit function theorem, and I know that it has to be done using that, but I don't know how to relate it with a differentiable manifold.
multivariable-calculus
multivariable-calculus
asked Mar 10 at 13:58
Jack TalionJack Talion
885
New contributor
Jack Talion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 10 at 13:58
Jack TalionJack Talion
885
New contributor
Jack Talion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 10 at 14:13
Jack Talion
asked Mar 10 at 13:58
Jack TalionJack Talion
885
New contributor
Jack Talion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 10 at 13:58
Jack TalionJack Talion
885
asked Mar 10 at 13:58
Jack TalionJack Talion
885
885
New contributor
Jack Talion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Jack Talion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Jack Talion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
Your equations define a smooth curve on $mathbb R^3$ without intersections. Hence it is a smooth manifold.
$endgroup$
– Dog_69
Mar 10 at 14:30
add a comment |
1
$begingroup$
Your equations define a smooth curve on $mathbb R^3$ without intersections. Hence it is a smooth manifold.
$endgroup$
– Dog_69
Mar 10 at 14:30
1
1
$begingroup$
Your equations define a smooth curve on $mathbb R^3$ without intersections. Hence it is a smooth manifold.
$endgroup$
– Dog_69
Mar 10 at 14:30
$begingroup$
Your equations define a smooth curve on $mathbb R^3$ without intersections. Hence it is a smooth manifold.
$endgroup$
– Dog_69
Mar 10 at 14:30
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
The condition (in this case): let be $g:{Bbb R}^3longrightarrow{Bbb R}^2$. $g(x,y,z) = 0$ defines a submanifold if $Dg$ has maximal rank (2 in this case). See http://www.owlnet.rice.edu/~fjones/chap6.pdf.
Alternatively: (as suggested by Dog_69): the parametrization
$$x = 2z^4 + 6,$$
$$y = z^2 + 1,$$
$$z = z,$$
has rank 1.
answered Mar 10 at 14:22
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.6k42971
$endgroup$
add a comment |
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$begingroup$
The condition (in this case): let be $g:{Bbb R}^3longrightarrow{Bbb R}^2$. $g(x,y,z) = 0$ defines a submanifold if $Dg$ has maximal rank (2 in this case). See http://www.owlnet.rice.edu/~fjones/chap6.pdf.
Alternatively: (as suggested by Dog_69): the parametrization
$$x = 2z^4 + 6,$$
$$y = z^2 + 1,$$
$$z = z,$$
has rank 1.
answered Mar 10 at 14:22
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.6k42971
$endgroup$
add a comment |
$begingroup$
The condition (in this case): let be $g:{Bbb R}^3longrightarrow{Bbb R}^2$. $g(x,y,z) = 0$ defines a submanifold if $Dg$ has maximal rank (2 in this case). See http://www.owlnet.rice.edu/~fjones/chap6.pdf.
Alternatively: (as suggested by Dog_69): the parametrization
$$x = 2z^4 + 6,$$
$$y = z^2 + 1,$$
$$z = z,$$
has rank 1.
answered Mar 10 at 14:22
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.6k42971
$endgroup$
add a comment |
$begingroup$
The condition (in this case): let be $g:{Bbb R}^3longrightarrow{Bbb R}^2$. $g(x,y,z) = 0$ defines a submanifold if $Dg$ has maximal rank (2 in this case). See http://www.owlnet.rice.edu/~fjones/chap6.pdf.
Alternatively: (as suggested by Dog_69): the parametrization
$$x = 2z^4 + 6,$$
$$y = z^2 + 1,$$
$$z = z,$$
has rank 1.
answered Mar 10 at 14:22
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.6k42971
$endgroup$
The condition (in this case): let be $g:{Bbb R}^3longrightarrow{Bbb R}^2$. $g(x,y,z) = 0$ defines a submanifold if $Dg$ has maximal rank (2 in this case). See http://www.owlnet.rice.edu/~fjones/chap6.pdf.
Alternatively: (as suggested by Dog_69): the parametrization
$$x = 2z^4 + 6,$$
$$y = z^2 + 1,$$
$$z = z,$$
has rank 1.
answered Mar 10 at 14:22
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.6k42971
edited Mar 10 at 14:33
answered Mar 10 at 14:22
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.6k42971
answered Mar 10 at 14:22
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.6k42971
answered Mar 10 at 14:22
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.6k42971
34.6k42971
add a comment |
add a comment |
Jack Talion is a new contributor. Be nice, and check out our Code of Conduct.
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Jack Talion is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Your equations define a smooth curve on $mathbb R^3$ without intersections. Hence it is a smooth manifold.
$endgroup$
– Dog_69
Mar 10 at 14:30