# Sectional curvature of leaves of foliation

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.

• If the leaves have dimension zero, it won't work. – 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.)

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

• Thanks. the first example is a nice one. – diptocal47 Nov 26 '18 at 17:13
• If you liked it, then feel free to accept the answer :) – Raziel Nov 26 '18 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 '18 at 9:20
• 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. – Deane Yang Nov 28 '18 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 '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 $$-1$$.

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