Sectional curvature of leaves of foliation












7














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







share|cite|improve this question


















  • 3




    If the leaves have dimension zero, it won't work.
    – Ben McKay
    Nov 26 at 9:41
















7














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







share|cite|improve this question


















  • 3




    If the leaves have dimension zero, it won't work.
    – Ben McKay
    Nov 26 at 9:41














7












7








7


2





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







share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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














  • 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










3 Answers
3






active

oldest

votes


















9














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.)






share|cite|improve this answer























  • 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
    Nov 28 at 8:50



















8














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.






share|cite|improve this answer

















  • 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
    Nov 28 at 2:30












  • @daene, the Hopf fibration has totally geodesic leaves, but the leaves are flat and the total space is not.
    – Raziel
    Nov 28 at 7:14



















3














One more counterexample: the hyperbolic $n$-space is foliated by horospheres with common center. Horospheres have zero curvature, the hyperbolic space curvature $-1$.






share|cite|improve this answer





















  • Yes, thought of this later on. Thanks for your input.
    – diptocal47
    Nov 28 at 17:23











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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









9














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.)






share|cite|improve this answer























  • 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
    Nov 28 at 8:50
















9














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.)






share|cite|improve this answer























  • 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
    Nov 28 at 8:50














9












9








9






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.)






share|cite|improve this answer














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.)







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 26 at 9:38

























answered Nov 26 at 8:59









ThiKu

5,89711933




5,89711933












  • 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
    Nov 28 at 8:50


















  • 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
    Nov 28 at 8:50
















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
Nov 28 at 8:50




I‘m really impressed to get so many upvotes for a completely trivial example :-)
– ThiKu
Nov 28 at 8:50











8














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.






share|cite|improve this answer

















  • 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
    Nov 28 at 2:30












  • @daene, the Hopf fibration has totally geodesic leaves, but the leaves are flat and the total space is not.
    – Raziel
    Nov 28 at 7:14
















8














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.






share|cite|improve this answer

















  • 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
    Nov 28 at 2:30












  • @daene, the Hopf fibration has totally geodesic leaves, but the leaves are flat and the total space is not.
    – Raziel
    Nov 28 at 7:14














8












8








8






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.






share|cite|improve this answer












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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 26 at 8:07









Raziel

1,85411324




1,85411324








  • 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
    Nov 28 at 2:30












  • @daene, the Hopf fibration has totally geodesic leaves, but the leaves are flat and the total space is not.
    – Raziel
    Nov 28 at 7:14














  • 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
    Nov 28 at 2:30












  • @daene, the Hopf fibration has totally geodesic leaves, but the leaves are flat and the total space is not.
    – Raziel
    Nov 28 at 7:14








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
Nov 28 at 2:30






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
Nov 28 at 2:30














@daene, the Hopf fibration has totally geodesic leaves, but the leaves are flat and the total space is not.
– Raziel
Nov 28 at 7:14




@daene, the Hopf fibration has totally geodesic leaves, but the leaves are flat and the total space is not.
– Raziel
Nov 28 at 7:14











3














One more counterexample: the hyperbolic $n$-space is foliated by horospheres with common center. Horospheres have zero curvature, the hyperbolic space curvature $-1$.






share|cite|improve this answer





















  • Yes, thought of this later on. Thanks for your input.
    – diptocal47
    Nov 28 at 17:23
















3














One more counterexample: the hyperbolic $n$-space is foliated by horospheres with common center. Horospheres have zero curvature, the hyperbolic space curvature $-1$.






share|cite|improve this answer





















  • Yes, thought of this later on. Thanks for your input.
    – diptocal47
    Nov 28 at 17:23














3












3








3






One more counterexample: the hyperbolic $n$-space is foliated by horospheres with common center. Horospheres have zero curvature, the hyperbolic space curvature $-1$.






share|cite|improve this answer












One more counterexample: the hyperbolic $n$-space is foliated by horospheres with common center. Horospheres have zero curvature, the hyperbolic space curvature $-1$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 27 at 14:24









Ivan Izmestiev

4,0731238




4,0731238












  • Yes, thought of this later on. Thanks for your input.
    – diptocal47
    Nov 28 at 17:23


















  • Yes, thought of this later on. Thanks for your input.
    – diptocal47
    Nov 28 at 17:23
















Yes, thought of this later on. Thanks for your input.
– diptocal47
Nov 28 at 17:23




Yes, thought of this later on. Thanks for your input.
– diptocal47
Nov 28 at 17:23


















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