Examples and non-examples of Riemannian foliations - MathOverflow

Examples and non-examples of Riemannian foliations

2012-02-06T17:45:47Z

Recall a tranverse metric on a (regular) foliated manifold $(M,F)$ is a positive symmetric $C^\infty (M)$-bilinear form $g$ such that

1) $Ker(g_x)=T_x F$

2) It is invariant with respect to lie derivtives along vector fields tangent to the foliation.

I know that not every foliation $(M,F)$ admits such a tranverse metric, however, I would like to know some simple examples of when this fails. I do know that if the foliation arises as the fibers of a sumbersion, then it always admits a transverse metric, however I would also like to know some examples of foliations not of this form which DO admit a tranverse metric. Thank you!

Answer by Robert Bryant for Examples and non-examples of Riemannian foliations

2012-02-07T01:21:05Z

For your second kind of example, consider the foliation of the unit $3$-sphere that is the integral curves of the vector field $$ X = p\left(x^1\frac{\partial\ }{\partial x^0 } -x^0\frac{\partial\ }{\partial x^1 }\right) + q\left(x^2\frac{\partial\ }{\partial x^3 } -x^3\frac{\partial\ }{\partial x^2 }\right), $$ where $p$ and $q$ are relatively prime integers. This has a transverse metric, but it is not the fibers of any submersion from the $3$-sphere to a $2$-manifold.

As for things that don't admit transverse metrics at all, you want a foliation such that the holonomy of the leaves (actually, it's enough to have one such leaf) is not compact. A good example of this is the foliation of the unit circle bundle of a compact surface of negative curvature by the tangential lifts of geodesics of the metric.

Answer by R W for Examples and non-examples of Riemannian foliations

2012-02-07T02:24:51Z

Another (actually quite close) example of a foliation without a transverse metric: the stable foliation of the geodesic flow on the unit tangent bundle of a negatively curved manifold.

Answer by Pablo Lessa for Examples and non-examples of Riemannian foliations

2012-02-07T17:28:58Z

By the corollary to theorem 4 in Kaimanovich's two page paper "Brownian motion on foliations: Entropy, Invariant measures, mixing" any foliation by leaves of sub-exponential volume growth must have a transverse invariant measure. An example of this is the foliation of the unit tangent bundle of $d$-dimensional hyperbolic space by the normal vectors to each horocycle (i.e. the strong stable foliation of the geodesic flow).

There is a $3$ dimensional manifold foliated by $2$-dimensional leaves without any transverse invariant measures which I think is due to Hirsch (I like it because because one can visualize it in Euclidean $3$ space easily).

To construct it take a solid torus $T = D \times S$ ($D$ a disk and $S$ the unit circle in the complex plane) and a solenoid map $f: T \to T$ which we take to be $s \mapsto s^2$ in the $S$ coordinate. Consider the foliation of $T \setminus f(T)$ by sets of the form $D \times {s}$ (each of these is a disk minus two smaller disks). Then identify the outer and inner boundaries of $T \setminus f(T)$ using $f$.

You get a compact foliated manifold (without boundary). Depending on whether the the $S$ coordinate of a point is eventually periodic under $s \mapsto s^2$ or not, the leaf the point belongs to can look like a $2$-sphere minus a Cantor set or a Torus minus a Cantor set. All leaves are dense.

The pasting you did makes is so that each time you want to go through a boundary torus you either square or take a square root of the $S$ coordinate.

In particular the holonomy along any curve which crosses the outer torus exactly once from the inside towards the outside is $s \mapsto s^2$ in the $S$ coordinate. Using this fact you can show there is no transverse invariant measure.