Mixing
Do these two definitions of hypersurfaces coincide?
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According to Wikipedia and some other sources, a hypersurface is a manifold of dimension $n-1$ embedded in an ambient space of dimension $n$ which is simple to understand.
Now here https://homepages.warwick.ac.uk/staff/C.M.Elliott/DziEll13a.pdf the Definition 2.1 has a different definition of a hypersurface:
Let $k in mathbb N cup{infty}$. $Gamma subset mathbb R^{n+1}$ is called a hypersurface if for each point $x_0 in Gamma$ there exists an open set $U subset mathbb R^{n+1}$ containing $x_0$ and a function $phi in C^k(Gamma)$ with $nabla phi neq 0$ on $Gamma cap U$ and such that $$U cap Gamma={x in U ,vert, phi(x)=0 }$$
Does this have a synonymous meaning as the definition from say Wikipedia? If so, how? I don't have any intuition what the latter definition should mean.
manifolds surfaces
asked Jan 2 at 20:34
TeslaTesla
855426
$endgroup$
add a comment |
$begingroup$
According to Wikipedia and some other sources, a hypersurface is a manifold of dimension $n-1$ embedded in an ambient space of dimension $n$ which is simple to understand.
Now here https://homepages.warwick.ac.uk/staff/C.M.Elliott/DziEll13a.pdf the Definition 2.1 has a different definition of a hypersurface:
Let $k in mathbb N cup{infty}$. $Gamma subset mathbb R^{n+1}$ is called a hypersurface if for each point $x_0 in Gamma$ there exists an open set $U subset mathbb R^{n+1}$ containing $x_0$ and a function $phi in C^k(Gamma)$ with $nabla phi neq 0$ on $Gamma cap U$ and such that $$U cap Gamma={x in U ,vert, phi(x)=0 }$$
Does this have a synonymous meaning as the definition from say Wikipedia? If so, how? I don't have any intuition what the latter definition should mean.
manifolds surfaces
asked Jan 2 at 20:34
TeslaTesla
855426
$endgroup$
2
$begingroup$
The latter definition means "locally cut out by a single equation"; the two should be equivalent by the implicit function theorem.
$endgroup$
– Qiaochu Yuan
Jan 2 at 20:48
$begingroup$
Qiaochu's comment is definitely true, though it is opaque to me what the authors mean by $Bbb R^infty$ here; if it is not supposed to mean the Hilbert space $ell^2(Bbb N)$ or some Banach space you do not get an implicit function theorem.
$endgroup$
– Mike Miller
Jan 2 at 20:51
$begingroup$
thanks. what is the benefit of defining it in such way? And what does "locally cut out be a single equation" mean?
$endgroup$
– Tesla
Jan 2 at 21:37
1
$begingroup$
This is how you find them in practice. And it means exactly what your formula does: there's a single function to R, with 0 as a regular value, so that your submanifold is the zero set.
$endgroup$
– Mike Miller
Jan 3 at 3:08
add a comment |
$begingroup$
According to Wikipedia and some other sources, a hypersurface is a manifold of dimension $n-1$ embedded in an ambient space of dimension $n$ which is simple to understand.
Now here https://homepages.warwick.ac.uk/staff/C.M.Elliott/DziEll13a.pdf the Definition 2.1 has a different definition of a hypersurface:
Let $k in mathbb N cup{infty}$. $Gamma subset mathbb R^{n+1}$ is called a hypersurface if for each point $x_0 in Gamma$ there exists an open set $U subset mathbb R^{n+1}$ containing $x_0$ and a function $phi in C^k(Gamma)$ with $nabla phi neq 0$ on $Gamma cap U$ and such that $$U cap Gamma={x in U ,vert, phi(x)=0 }$$
Does this have a synonymous meaning as the definition from say Wikipedia? If so, how? I don't have any intuition what the latter definition should mean.
manifolds surfaces
asked Jan 2 at 20:34
TeslaTesla
855426
$endgroup$
According to Wikipedia and some other sources, a hypersurface is a manifold of dimension $n-1$ embedded in an ambient space of dimension $n$ which is simple to understand.
Now here https://homepages.warwick.ac.uk/staff/C.M.Elliott/DziEll13a.pdf the Definition 2.1 has a different definition of a hypersurface:
Let $k in mathbb N cup{infty}$. $Gamma subset mathbb R^{n+1}$ is called a hypersurface if for each point $x_0 in Gamma$ there exists an open set $U subset mathbb R^{n+1}$ containing $x_0$ and a function $phi in C^k(Gamma)$ with $nabla phi neq 0$ on $Gamma cap U$ and such that $$U cap Gamma={x in U ,vert, phi(x)=0 }$$
Does this have a synonymous meaning as the definition from say Wikipedia? If so, how? I don't have any intuition what the latter definition should mean.
manifolds surfaces
manifolds surfaces
asked Jan 2 at 20:34
TeslaTesla
855426
asked Jan 2 at 20:34
TeslaTesla
855426
asked Jan 2 at 20:34
TeslaTesla
855426
asked Jan 2 at 20:34
TeslaTesla
855426
asked Jan 2 at 20:34
TeslaTesla
855426
855426
2
$begingroup$
The latter definition means "locally cut out by a single equation"; the two should be equivalent by the implicit function theorem.
$endgroup$
– Qiaochu Yuan
Jan 2 at 20:48
$begingroup$
Qiaochu's comment is definitely true, though it is opaque to me what the authors mean by $Bbb R^infty$ here; if it is not supposed to mean the Hilbert space $ell^2(Bbb N)$ or some Banach space you do not get an implicit function theorem.
$endgroup$
– Mike Miller
Jan 2 at 20:51
$begingroup$
thanks. what is the benefit of defining it in such way? And what does "locally cut out be a single equation" mean?
$endgroup$
– Tesla
Jan 2 at 21:37
1
$begingroup$
This is how you find them in practice. And it means exactly what your formula does: there's a single function to R, with 0 as a regular value, so that your submanifold is the zero set.
$endgroup$
– Mike Miller
Jan 3 at 3:08
add a comment |
2
$begingroup$
The latter definition means "locally cut out by a single equation"; the two should be equivalent by the implicit function theorem.
$endgroup$
– Qiaochu Yuan
Jan 2 at 20:48
$begingroup$
Qiaochu's comment is definitely true, though it is opaque to me what the authors mean by $Bbb R^infty$ here; if it is not supposed to mean the Hilbert space $ell^2(Bbb N)$ or some Banach space you do not get an implicit function theorem.
$endgroup$
– Mike Miller
Jan 2 at 20:51
$begingroup$
thanks. what is the benefit of defining it in such way? And what does "locally cut out be a single equation" mean?
$endgroup$
– Tesla
Jan 2 at 21:37
1
$begingroup$
This is how you find them in practice. And it means exactly what your formula does: there's a single function to R, with 0 as a regular value, so that your submanifold is the zero set.
$endgroup$
– Mike Miller
Jan 3 at 3:08
2
2
$begingroup$
The latter definition means "locally cut out by a single equation"; the two should be equivalent by the implicit function theorem.
$endgroup$
– Qiaochu Yuan
Jan 2 at 20:48
$begingroup$
The latter definition means "locally cut out by a single equation"; the two should be equivalent by the implicit function theorem.
$endgroup$
– Qiaochu Yuan
Jan 2 at 20:48
$begingroup$
Qiaochu's comment is definitely true, though it is opaque to me what the authors mean by $Bbb R^infty$ here; if it is not supposed to mean the Hilbert space $ell^2(Bbb N)$ or some Banach space you do not get an implicit function theorem.
$endgroup$
– Mike Miller
Jan 2 at 20:51
$begingroup$
Qiaochu's comment is definitely true, though it is opaque to me what the authors mean by $Bbb R^infty$ here; if it is not supposed to mean the Hilbert space $ell^2(Bbb N)$ or some Banach space you do not get an implicit function theorem.
$endgroup$
– Mike Miller
Jan 2 at 20:51
$begingroup$
thanks. what is the benefit of defining it in such way? And what does "locally cut out be a single equation" mean?
$endgroup$
– Tesla
Jan 2 at 21:37
$begingroup$
thanks. what is the benefit of defining it in such way? And what does "locally cut out be a single equation" mean?
$endgroup$
– Tesla
Jan 2 at 21:37
1
1
$begingroup$
This is how you find them in practice. And it means exactly what your formula does: there's a single function to R, with 0 as a regular value, so that your submanifold is the zero set.
$endgroup$
– Mike Miller
Jan 3 at 3:08
$begingroup$
This is how you find them in practice. And it means exactly what your formula does: there's a single function to R, with 0 as a regular value, so that your submanifold is the zero set.
$endgroup$
– Mike Miller
Jan 3 at 3:08
add a comment |
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$begingroup$
The latter definition means "locally cut out by a single equation"; the two should be equivalent by the implicit function theorem.
$endgroup$
– Qiaochu Yuan
Jan 2 at 20:48
$begingroup$
Qiaochu's comment is definitely true, though it is opaque to me what the authors mean by $Bbb R^infty$ here; if it is not supposed to mean the Hilbert space $ell^2(Bbb N)$ or some Banach space you do not get an implicit function theorem.
$endgroup$
– Mike Miller
Jan 2 at 20:51
$begingroup$
thanks. what is the benefit of defining it in such way? And what does "locally cut out be a single equation" mean?
$endgroup$
– Tesla
Jan 2 at 21:37
1
$begingroup$
This is how you find them in practice. And it means exactly what your formula does: there's a single function to R, with 0 as a regular value, so that your submanifold is the zero set.
$endgroup$
– Mike Miller
Jan 3 at 3:08