When a manifold is isometric to a product manifold?
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Let $M$ be a Riemannian smooth manifold of dimension p and $phi$ a smooth submersion from $M$ to some other smooth manifold $N$ of dimension q. Denote by $mathcal{F}$ the foliation of leaves $phi^{-1}(z)$ for $zin phi(M)$. When can we say that $M$ is isomorphic to a product manifold of the form $Ltimes S$ where $L$ is a leave of $mathcal{F}$ passing through a point $x_0$ and $S$ is a transversal leave passing through the same point $x_0$.
A classical condition in ([1], Theorem 4.4 section 4.4) says that if $M$ is complete simply connected and if $mathcal{F}$ is parallel then $M$ is isomorphic to such product manifold. But I don't know what condition $phi$ should satisfy to ensure that $mathcal{F}$ is parallel?
Are there more general conditions on $phi$ to ensure that $M$ can be seen as a product manifold?
[1] A. Bejancu and H. R. Farran. Foliations and Geometric Structures. Springer Science & Business Media, Jan. 2006. Google-Books-ID: NtqEFj4nYecC.
differential-geometry riemannian-geometry foliations
asked Dec 13 '18 at 20:02
Michael ArbelMichael Arbel
163
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add a comment |
$begingroup$
Let $M$ be a Riemannian smooth manifold of dimension p and $phi$ a smooth submersion from $M$ to some other smooth manifold $N$ of dimension q. Denote by $mathcal{F}$ the foliation of leaves $phi^{-1}(z)$ for $zin phi(M)$. When can we say that $M$ is isomorphic to a product manifold of the form $Ltimes S$ where $L$ is a leave of $mathcal{F}$ passing through a point $x_0$ and $S$ is a transversal leave passing through the same point $x_0$.
A classical condition in ([1], Theorem 4.4 section 4.4) says that if $M$ is complete simply connected and if $mathcal{F}$ is parallel then $M$ is isomorphic to such product manifold. But I don't know what condition $phi$ should satisfy to ensure that $mathcal{F}$ is parallel?
Are there more general conditions on $phi$ to ensure that $M$ can be seen as a product manifold?
[1] A. Bejancu and H. R. Farran. Foliations and Geometric Structures. Springer Science & Business Media, Jan. 2006. Google-Books-ID: NtqEFj4nYecC.
differential-geometry riemannian-geometry foliations
asked Dec 13 '18 at 20:02
Michael ArbelMichael Arbel
163
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2
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Something's wrong with your setup. If $phi$ is an embedding, then the "leaves" $phi^{-1}(z)$ are just single points, so the foliation is trivial.
$endgroup$
– Jack Lee
Dec 14 '18 at 0:24
$begingroup$
thank you Jack Lee for pointing this out, it was a mistake, $phi$ needs to be a submersion.
$endgroup$
– Michael Arbel
Dec 15 '18 at 10:44
add a comment |
$begingroup$
Let $M$ be a Riemannian smooth manifold of dimension p and $phi$ a smooth submersion from $M$ to some other smooth manifold $N$ of dimension q. Denote by $mathcal{F}$ the foliation of leaves $phi^{-1}(z)$ for $zin phi(M)$. When can we say that $M$ is isomorphic to a product manifold of the form $Ltimes S$ where $L$ is a leave of $mathcal{F}$ passing through a point $x_0$ and $S$ is a transversal leave passing through the same point $x_0$.
A classical condition in ([1], Theorem 4.4 section 4.4) says that if $M$ is complete simply connected and if $mathcal{F}$ is parallel then $M$ is isomorphic to such product manifold. But I don't know what condition $phi$ should satisfy to ensure that $mathcal{F}$ is parallel?
Are there more general conditions on $phi$ to ensure that $M$ can be seen as a product manifold?
[1] A. Bejancu and H. R. Farran. Foliations and Geometric Structures. Springer Science & Business Media, Jan. 2006. Google-Books-ID: NtqEFj4nYecC.
differential-geometry riemannian-geometry foliations
asked Dec 13 '18 at 20:02
Michael ArbelMichael Arbel
163
$endgroup$
Let $M$ be a Riemannian smooth manifold of dimension p and $phi$ a smooth submersion from $M$ to some other smooth manifold $N$ of dimension q. Denote by $mathcal{F}$ the foliation of leaves $phi^{-1}(z)$ for $zin phi(M)$. When can we say that $M$ is isomorphic to a product manifold of the form $Ltimes S$ where $L$ is a leave of $mathcal{F}$ passing through a point $x_0$ and $S$ is a transversal leave passing through the same point $x_0$.
A classical condition in ([1], Theorem 4.4 section 4.4) says that if $M$ is complete simply connected and if $mathcal{F}$ is parallel then $M$ is isomorphic to such product manifold. But I don't know what condition $phi$ should satisfy to ensure that $mathcal{F}$ is parallel?
Are there more general conditions on $phi$ to ensure that $M$ can be seen as a product manifold?
[1] A. Bejancu and H. R. Farran. Foliations and Geometric Structures. Springer Science & Business Media, Jan. 2006. Google-Books-ID: NtqEFj4nYecC.
differential-geometry riemannian-geometry foliations
differential-geometry riemannian-geometry foliations
asked Dec 13 '18 at 20:02
Michael ArbelMichael Arbel
163
asked Dec 13 '18 at 20:02
Michael ArbelMichael Arbel
163
edited Dec 15 '18 at 10:43
Michael Arbel
asked Dec 13 '18 at 20:02
Michael ArbelMichael Arbel
163
asked Dec 13 '18 at 20:02
Michael ArbelMichael Arbel
163
asked Dec 13 '18 at 20:02
Michael ArbelMichael Arbel
163
163
2
$begingroup$
Something's wrong with your setup. If $phi$ is an embedding, then the "leaves" $phi^{-1}(z)$ are just single points, so the foliation is trivial.
$endgroup$
– Jack Lee
Dec 14 '18 at 0:24
$begingroup$
thank you Jack Lee for pointing this out, it was a mistake, $phi$ needs to be a submersion.
$endgroup$
– Michael Arbel
Dec 15 '18 at 10:44
add a comment |
2
$begingroup$
Something's wrong with your setup. If $phi$ is an embedding, then the "leaves" $phi^{-1}(z)$ are just single points, so the foliation is trivial.
$endgroup$
– Jack Lee
Dec 14 '18 at 0:24
$begingroup$
thank you Jack Lee for pointing this out, it was a mistake, $phi$ needs to be a submersion.
$endgroup$
– Michael Arbel
Dec 15 '18 at 10:44
2
2
$begingroup$
Something's wrong with your setup. If $phi$ is an embedding, then the "leaves" $phi^{-1}(z)$ are just single points, so the foliation is trivial.
$endgroup$
– Jack Lee
Dec 14 '18 at 0:24
$begingroup$
Something's wrong with your setup. If $phi$ is an embedding, then the "leaves" $phi^{-1}(z)$ are just single points, so the foliation is trivial.
$endgroup$
– Jack Lee
Dec 14 '18 at 0:24
$begingroup$
thank you Jack Lee for pointing this out, it was a mistake, $phi$ needs to be a submersion.
$endgroup$
– Michael Arbel
Dec 15 '18 at 10:44
$begingroup$
thank you Jack Lee for pointing this out, it was a mistake, $phi$ needs to be a submersion.
$endgroup$
– Michael Arbel
Dec 15 '18 at 10:44
add a comment |
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$begingroup$
Something's wrong with your setup. If $phi$ is an embedding, then the "leaves" $phi^{-1}(z)$ are just single points, so the foliation is trivial.
$endgroup$
– Jack Lee
Dec 14 '18 at 0:24
$begingroup$
thank you Jack Lee for pointing this out, it was a mistake, $phi$ needs to be a submersion.
$endgroup$
– Michael Arbel
Dec 15 '18 at 10:44