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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!

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You can take a foliation of a 2-torus by lines of an irrational slope. –  Ian Agol Feb 7 '12 at 3:48

3 Answers 3

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.

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The holonomy group of any leaf is a discrete group, so what do you mean by compact? Did you mean finite? By the way, do you have a reference for this non-existence claim? Thanks! –  David Carchedi Feb 7 '12 at 11:59
    
@David: Yes, the holonomy of each leaf is discrete in this case. If the leaf isn't closed, it's trivial and, if the leaf is closed, it's generated by a single element. However, that single element will not preserve any metric if it has a real eigenvalue of modulus greater than $1$, which is the case for closed geodesics on a surface of negative curvature. This is is thoroughly discussed under the topic of Anosov flows, which you should Google for a full explanation. The Wikipedia page on this is quite explicit. –  Robert Bryant Feb 7 '12 at 12:58
    
@Robert: I am a little confused. The holonomy group of any leaf of any (regular) foliated manifold is discrete, by definition, since it is a quotient of the fundamental group. Are you speaking of something else? Are you claiming that if I have a foliated manifold $\left(M,F\right)$ and a leaf which is not compact (and I am working here without corners or boundaries) then its holonomy group has to have a single generator? If so, why is this true and where can I read this? –  David Carchedi Feb 7 '12 at 14:14
    
Also, are you claiming that if there is such a non-compact leaf such that the generator of the holonomy group induces an endomorphism of the tangent space with a real eigenvalue with modulus greater than 1, then there cannot be a tranverse metric on the foliated manifold? Any references, or clarifications would be most helpful. Thank you! –  David Carchedi Feb 7 '12 at 14:15
    
@David: You have misread what I wrote or aren't taking into account the fact that I'm talking about the particular example that I gave, in which the leaves are $1$-dimensional and thus are either circles or copies of $\mathbb{R}$. In the former case, the holonomy is generated by a single element (corresponding to the generator of $\pi_1$), in the later case, the holonomy is trivial, in this particular example. The claim about a generator with real eigenvalue greater than $1$ obstructing a transverse metric seems obvious to me, so I'm not sure what you are not getting. –  Robert Bryant Feb 7 '12 at 15:55

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.

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Could you please give a reference for this claim? and also, why is that it does not have a transverse metric? –  Camilo Angulo Feb 7 '12 at 15:51

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.

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