Sectional curvature of leaves of foliation
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Given a $k$- dimensional foliation $F$ of a riemannian $n$-manifold $M$, with the property that the leaves of the foliation have constant sectional curvature $s$, for some $s$, is it true that $M$ will also have the same constant sectional curvature?
Is the same true if sectional curvature is replaced by the Gaussian curvature?
If it's a well known result, any hint at proving this or a possible reference or a counter example otherwise, will be most welcome.
Thanks.
dg.differential-geometry mg.metric-geometry riemannian-geometry
add a comment |
up vote
7
down vote
favorite
Given a $k$- dimensional foliation $F$ of a riemannian $n$-manifold $M$, with the property that the leaves of the foliation have constant sectional curvature $s$, for some $s$, is it true that $M$ will also have the same constant sectional curvature?
Is the same true if sectional curvature is replaced by the Gaussian curvature?
If it's a well known result, any hint at proving this or a possible reference or a counter example otherwise, will be most welcome.
Thanks.
dg.differential-geometry mg.metric-geometry riemannian-geometry
3
If the leaves have dimension zero, it won't work.
– Ben McKay
Nov 26 at 9:41
add a comment |
up vote
7
down vote
favorite
up vote
7
down vote
favorite
Given a $k$- dimensional foliation $F$ of a riemannian $n$-manifold $M$, with the property that the leaves of the foliation have constant sectional curvature $s$, for some $s$, is it true that $M$ will also have the same constant sectional curvature?
Is the same true if sectional curvature is replaced by the Gaussian curvature?
If it's a well known result, any hint at proving this or a possible reference or a counter example otherwise, will be most welcome.
Thanks.
dg.differential-geometry mg.metric-geometry riemannian-geometry
Given a $k$- dimensional foliation $F$ of a riemannian $n$-manifold $M$, with the property that the leaves of the foliation have constant sectional curvature $s$, for some $s$, is it true that $M$ will also have the same constant sectional curvature?
Is the same true if sectional curvature is replaced by the Gaussian curvature?
If it's a well known result, any hint at proving this or a possible reference or a counter example otherwise, will be most welcome.
Thanks.
dg.differential-geometry mg.metric-geometry riemannian-geometry
dg.differential-geometry mg.metric-geometry riemannian-geometry
asked Nov 26 at 7:58
diptocal47
493
493
3
If the leaves have dimension zero, it won't work.
– Ben McKay
Nov 26 at 9:41
add a comment |
3
If the leaves have dimension zero, it won't work.
– Ben McKay
Nov 26 at 9:41
3
3
If the leaves have dimension zero, it won't work.
– Ben McKay
Nov 26 at 9:41
If the leaves have dimension zero, it won't work.
– Ben McKay
Nov 26 at 9:41
add a comment |
3 Answers
3
active
oldest
votes
up vote
9
down vote
The easiest counterexample is the Riemannian product $$M=Ftimes N$$
for $F$ a manifold of constant non-zero sectional curvature and $N$ an arbitrary Riemannian manifold of dimension $n-k$. ($M$ is foliated by the copies of $F$.)
Any plane that is the product of a line in $F$ with a line in $N$ will be locally isometric to ${mathbb R}^2$ and thus have sectional curvature equal to zero.
So the sectional curvature is not constant, even if the sectional curvature of $N$ were the same as that of $F$. (Which of course needn‘t be the case anyway.)
Thanks. Honestly wouldn't have thought of this.
– diptocal47
Nov 26 at 17:13
I‘m really impressed to get so many upvotes for a completely trivial example :-)
– ThiKu
2 days ago
add a comment |
up vote
8
down vote
The Hopf fibration on $S^3$ with the standard round metric gives a counterexample (one dimensional hence flat leaves).
In dimension 2 (where sectional=Gauss), take for example any non flat metric on the torus foliated by circles.
1
Thanks. the first example is a nice one.
– diptocal47
Nov 26 at 17:13
If you liked it, then feel free to accept the answer :)
– Raziel
Nov 26 at 18:44
Of course. By the way, is there any result pointing to the fact when my question has a positive answer? Just curious!
– diptocal47
Nov 27 at 9:20
If the leaves are totally geodesic, the sectional curvatures will be the same. If $sne 0$, then there are values of $n$ and $k$ for which this is also a necessary condition.
– Deane Yang
2 days ago
@daene, the Hopf fibration has totally geodesic leaves, but the leaves are flat and the total space is not.
– Raziel
2 days ago
|
show 1 more comment
up vote
2
down vote
One more counterexample: the hyperbolic $n$-space is foliated by horospheres with common center. Horospheres have zero curvature, the hyperbolic space curvature $-1$.
Yes, thought of this later on. Thanks for your input.
– diptocal47
yesterday
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
down vote
The easiest counterexample is the Riemannian product $$M=Ftimes N$$
for $F$ a manifold of constant non-zero sectional curvature and $N$ an arbitrary Riemannian manifold of dimension $n-k$. ($M$ is foliated by the copies of $F$.)
Any plane that is the product of a line in $F$ with a line in $N$ will be locally isometric to ${mathbb R}^2$ and thus have sectional curvature equal to zero.
So the sectional curvature is not constant, even if the sectional curvature of $N$ were the same as that of $F$. (Which of course needn‘t be the case anyway.)
Thanks. Honestly wouldn't have thought of this.
– diptocal47
Nov 26 at 17:13
I‘m really impressed to get so many upvotes for a completely trivial example :-)
– ThiKu
2 days ago
add a comment |
up vote
9
down vote
The easiest counterexample is the Riemannian product $$M=Ftimes N$$
for $F$ a manifold of constant non-zero sectional curvature and $N$ an arbitrary Riemannian manifold of dimension $n-k$. ($M$ is foliated by the copies of $F$.)
Any plane that is the product of a line in $F$ with a line in $N$ will be locally isometric to ${mathbb R}^2$ and thus have sectional curvature equal to zero.
So the sectional curvature is not constant, even if the sectional curvature of $N$ were the same as that of $F$. (Which of course needn‘t be the case anyway.)
Thanks. Honestly wouldn't have thought of this.
– diptocal47
Nov 26 at 17:13
I‘m really impressed to get so many upvotes for a completely trivial example :-)
– ThiKu
2 days ago
add a comment |
up vote
9
down vote
up vote
9
down vote
The easiest counterexample is the Riemannian product $$M=Ftimes N$$
for $F$ a manifold of constant non-zero sectional curvature and $N$ an arbitrary Riemannian manifold of dimension $n-k$. ($M$ is foliated by the copies of $F$.)
Any plane that is the product of a line in $F$ with a line in $N$ will be locally isometric to ${mathbb R}^2$ and thus have sectional curvature equal to zero.
So the sectional curvature is not constant, even if the sectional curvature of $N$ were the same as that of $F$. (Which of course needn‘t be the case anyway.)
The easiest counterexample is the Riemannian product $$M=Ftimes N$$
for $F$ a manifold of constant non-zero sectional curvature and $N$ an arbitrary Riemannian manifold of dimension $n-k$. ($M$ is foliated by the copies of $F$.)
Any plane that is the product of a line in $F$ with a line in $N$ will be locally isometric to ${mathbb R}^2$ and thus have sectional curvature equal to zero.
So the sectional curvature is not constant, even if the sectional curvature of $N$ were the same as that of $F$. (Which of course needn‘t be the case anyway.)
edited Nov 26 at 9:38
answered Nov 26 at 8:59
ThiKu
5,78211933
5,78211933
Thanks. Honestly wouldn't have thought of this.
– diptocal47
Nov 26 at 17:13
I‘m really impressed to get so many upvotes for a completely trivial example :-)
– ThiKu
2 days ago
add a comment |
Thanks. Honestly wouldn't have thought of this.
– diptocal47
Nov 26 at 17:13
I‘m really impressed to get so many upvotes for a completely trivial example :-)
– ThiKu
2 days ago
Thanks. Honestly wouldn't have thought of this.
– diptocal47
Nov 26 at 17:13
Thanks. Honestly wouldn't have thought of this.
– diptocal47
Nov 26 at 17:13
I‘m really impressed to get so many upvotes for a completely trivial example :-)
– ThiKu
2 days ago
I‘m really impressed to get so many upvotes for a completely trivial example :-)
– ThiKu
2 days ago
add a comment |
up vote
8
down vote
The Hopf fibration on $S^3$ with the standard round metric gives a counterexample (one dimensional hence flat leaves).
In dimension 2 (where sectional=Gauss), take for example any non flat metric on the torus foliated by circles.
1
Thanks. the first example is a nice one.
– diptocal47
Nov 26 at 17:13
If you liked it, then feel free to accept the answer :)
– Raziel
Nov 26 at 18:44
Of course. By the way, is there any result pointing to the fact when my question has a positive answer? Just curious!
– diptocal47
Nov 27 at 9:20
If the leaves are totally geodesic, the sectional curvatures will be the same. If $sne 0$, then there are values of $n$ and $k$ for which this is also a necessary condition.
– Deane Yang
2 days ago
@daene, the Hopf fibration has totally geodesic leaves, but the leaves are flat and the total space is not.
– Raziel
2 days ago
|
show 1 more comment
up vote
8
down vote
The Hopf fibration on $S^3$ with the standard round metric gives a counterexample (one dimensional hence flat leaves).
In dimension 2 (where sectional=Gauss), take for example any non flat metric on the torus foliated by circles.
1
Thanks. the first example is a nice one.
– diptocal47
Nov 26 at 17:13
If you liked it, then feel free to accept the answer :)
– Raziel
Nov 26 at 18:44
Of course. By the way, is there any result pointing to the fact when my question has a positive answer? Just curious!
– diptocal47
Nov 27 at 9:20
If the leaves are totally geodesic, the sectional curvatures will be the same. If $sne 0$, then there are values of $n$ and $k$ for which this is also a necessary condition.
– Deane Yang
2 days ago
@daene, the Hopf fibration has totally geodesic leaves, but the leaves are flat and the total space is not.
– Raziel
2 days ago
|
show 1 more comment
up vote
8
down vote
up vote
8
down vote
The Hopf fibration on $S^3$ with the standard round metric gives a counterexample (one dimensional hence flat leaves).
In dimension 2 (where sectional=Gauss), take for example any non flat metric on the torus foliated by circles.
The Hopf fibration on $S^3$ with the standard round metric gives a counterexample (one dimensional hence flat leaves).
In dimension 2 (where sectional=Gauss), take for example any non flat metric on the torus foliated by circles.
answered Nov 26 at 8:07
Raziel
1,84411324
1,84411324
1
Thanks. the first example is a nice one.
– diptocal47
Nov 26 at 17:13
If you liked it, then feel free to accept the answer :)
– Raziel
Nov 26 at 18:44
Of course. By the way, is there any result pointing to the fact when my question has a positive answer? Just curious!
– diptocal47
Nov 27 at 9:20
If the leaves are totally geodesic, the sectional curvatures will be the same. If $sne 0$, then there are values of $n$ and $k$ for which this is also a necessary condition.
– Deane Yang
2 days ago
@daene, the Hopf fibration has totally geodesic leaves, but the leaves are flat and the total space is not.
– Raziel
2 days ago
|
show 1 more comment
1
Thanks. the first example is a nice one.
– diptocal47
Nov 26 at 17:13
If you liked it, then feel free to accept the answer :)
– Raziel
Nov 26 at 18:44
Of course. By the way, is there any result pointing to the fact when my question has a positive answer? Just curious!
– diptocal47
Nov 27 at 9:20
If the leaves are totally geodesic, the sectional curvatures will be the same. If $sne 0$, then there are values of $n$ and $k$ for which this is also a necessary condition.
– Deane Yang
2 days ago
@daene, the Hopf fibration has totally geodesic leaves, but the leaves are flat and the total space is not.
– Raziel
2 days ago
1
1
Thanks. the first example is a nice one.
– diptocal47
Nov 26 at 17:13
Thanks. the first example is a nice one.
– diptocal47
Nov 26 at 17:13
If you liked it, then feel free to accept the answer :)
– Raziel
Nov 26 at 18:44
If you liked it, then feel free to accept the answer :)
– Raziel
Nov 26 at 18:44
Of course. By the way, is there any result pointing to the fact when my question has a positive answer? Just curious!
– diptocal47
Nov 27 at 9:20
Of course. By the way, is there any result pointing to the fact when my question has a positive answer? Just curious!
– diptocal47
Nov 27 at 9:20
If the leaves are totally geodesic, the sectional curvatures will be the same. If $sne 0$, then there are values of $n$ and $k$ for which this is also a necessary condition.
– Deane Yang
2 days ago
If the leaves are totally geodesic, the sectional curvatures will be the same. If $sne 0$, then there are values of $n$ and $k$ for which this is also a necessary condition.
– Deane Yang
2 days ago
@daene, the Hopf fibration has totally geodesic leaves, but the leaves are flat and the total space is not.
– Raziel
2 days ago
@daene, the Hopf fibration has totally geodesic leaves, but the leaves are flat and the total space is not.
– Raziel
2 days ago
|
show 1 more comment
up vote
2
down vote
One more counterexample: the hyperbolic $n$-space is foliated by horospheres with common center. Horospheres have zero curvature, the hyperbolic space curvature $-1$.
Yes, thought of this later on. Thanks for your input.
– diptocal47
yesterday
add a comment |
up vote
2
down vote
One more counterexample: the hyperbolic $n$-space is foliated by horospheres with common center. Horospheres have zero curvature, the hyperbolic space curvature $-1$.
Yes, thought of this later on. Thanks for your input.
– diptocal47
yesterday
add a comment |
up vote
2
down vote
up vote
2
down vote
One more counterexample: the hyperbolic $n$-space is foliated by horospheres with common center. Horospheres have zero curvature, the hyperbolic space curvature $-1$.
One more counterexample: the hyperbolic $n$-space is foliated by horospheres with common center. Horospheres have zero curvature, the hyperbolic space curvature $-1$.
answered Nov 27 at 14:24
Ivan Izmestiev
4,0181238
4,0181238
Yes, thought of this later on. Thanks for your input.
– diptocal47
yesterday
add a comment |
Yes, thought of this later on. Thanks for your input.
– diptocal47
yesterday
Yes, thought of this later on. Thanks for your input.
– diptocal47
yesterday
Yes, thought of this later on. Thanks for your input.
– diptocal47
yesterday
add a comment |
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3
If the leaves have dimension zero, it won't work.
– Ben McKay
Nov 26 at 9:41