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.


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    $\begingroup$ If the leaves have dimension zero, it won't work. $\endgroup$ – Ben McKay Nov 26 '18 at 9:41

The easiest counterexample is the Riemannian product $$M=F\times N$$ 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.)

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  • $\begingroup$ Thanks. Honestly wouldn't have thought of this. $\endgroup$ – diptocal47 Nov 26 '18 at 17:13
  • $\begingroup$ I‘m really impressed to get so many upvotes for a completely trivial example :-) $\endgroup$ – ThiKu Nov 28 '18 at 8:50

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.

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    $\begingroup$ Thanks. the first example is a nice one. $\endgroup$ – diptocal47 Nov 26 '18 at 17:13
  • $\begingroup$ If you liked it, then feel free to accept the answer :) $\endgroup$ – Raziel Nov 26 '18 at 18:44
  • $\begingroup$ Of course. By the way, is there any result pointing to the fact when my question has a positive answer? Just curious! $\endgroup$ – diptocal47 Nov 27 '18 at 9:20
  • $\begingroup$ If the leaves are totally geodesic, the sectional curvatures will be the same. If $s\ne 0$, then there are values of $n$ and $k$ for which this is also a necessary condition. $\endgroup$ – Deane Yang Nov 28 '18 at 2:30
  • $\begingroup$ @daene, the Hopf fibration has totally geodesic leaves, but the leaves are flat and the total space is not. $\endgroup$ – Raziel Nov 28 '18 at 7:14

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

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  • $\begingroup$ Yes, thought of this later on. Thanks for your input. $\endgroup$ – diptocal47 Nov 28 '18 at 17:23

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