I recently got to know about the existence of the so-called Godbillon–Vey invariant, and I am interested in its relationship with foliation theory in 3-manifolds. I briefly recall here the definition:

Suppose that $\mathcal{F}$ is a codimension-$1$ foliation of a $3$-manifold $M$, and suppose that $\mathcal{F}$ is defined as the kernel of a $1$-form $\alpha$. We have that $d\alpha\wedge\alpha=0$ and this implies that there exists a $1$-form $\theta$ such that $d\alpha=\theta\wedge\alpha$ and now one can define the $3$-form $\theta\wedge d\theta$. This form is closed and its cohomology class does not depend on the choice of $\theta$, and it is called the Godbillon–Vey invariant of $\mathcal{F}$.

I got the idea that this invariant should tell us something about the transverse topology of the foliation, and I have some questions about it.

  • Does the vanishing of the Godbillon–Vey invariant imply something about the leaf space of $\mathcal{F}$? More precisely, can we say whether this space is Hausdorff or not?

  • Is there some known technique to calculate this invariant?

  • Is there some generalization of this invariant for foliation that are only $C^{\infty,0}$? That is to say, the leaves are smooth but the tangent plane bundle to the foliation is only continuous.

Any reference is appreciated.


2 Answers 2


You should check out Thurston, William, Noncobordant foliations of $S^3$. Bull. Amer. Math. Soc. 78 (1972), 511–514.

He constructs foliations with arbitrary real-valued GV invariants. In this paper, he described the GV invariant as "measuring the helical wobble" of a foliation.


There's a discussion of the history of the Godbillon-Vey invariant in Hurder's Dynamic's & the Godbillon-Vey Class. He recalls:

Reinhart and Wood ... gave a formula expressing the the Godbillon-Vey class of a foliation of a Riemannian 3-manifold in terms of the curvatures of the leaves and normal bundle, which can be interpreted as [Thurston's] helical wobble.

He doesn't expressly detail it, but there is a reference. The early history of the subject suggested that when the invariant did not vanish there were leaves of exponential growth. This was conjectured by Moussu & Pelletier and Sullivan and was proved for increasingly broader classes of foliations. Finally Duminy proved in 1986 in a unpublished manuscript that:

Theorem: If $F$ is a codimension one, $C^2$-foliation of a compact manifold $M$ with non-trivial Godbillon measure $g_F$ , then $F$ must have a hyperbolic resilient leaf, and hence there is an open subset of $M$ consisting of leaves with exponential growth.

Here, Duminy 'factored' the Godbillon-Vey invariant into the Godbillon measure measure constructed from the leaf cohomology class $[\eta] \in H^1(M, F)$. The other 'factor' is the Vey class, $[d\eta] \in H2(M/F)$. The theorem was generalised to higher codimension:

THEOREM: Let $F$ be a $C^1$ foliation of codimension $q ≥ 1$ such that almost every leaf has subexponential growth rate. Then the Godbillon measure $g_F = 0$. If $F$ is $C^2$ then the Godbillon-Vey class $GV (F)=0$.

(See Kotschick's paper, Godbillon-Vey invariants for families of foliations for the generalisation of the Godbillon-Vey invariant to higher codimension).

The results above are for $C^2$ foliations with smooth leafs but they do mention an unproven conjecture (as of '00) which Hurder terms 'very surprising if true' is that the pullback of the invariant by a continuous map should be invariant. But they note it was proven under additional regularity: by Raby under $C^1$-diffeomorphism whilst Hurder and Katok showed independently that if the 'conjugancy map' and its inverse are absolutely continuous, then the conclusion is also true.

Finally Connes has shown that the Godbillon-Vey class can be calculated from the flow of weights for a natural von Neumann algebra associated to the foliation. In fact, if the class does not vanish then this algebra has factor of type III.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.