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Ali Taghavi
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Suppose we have a compact three-manifold $M$, a codimension-one foliation $F$ of $M$, and a one-form on $\alpha$, chosen so that $F$ is tangent to $\alpha = 0$. We deduce that $\alpha \wedge d\alpha = 0$. Also, there is some one-form $\beta$ so that $d\alpha = \alpha \wedge \beta$.

The Godbillon-Vey invariant of $F$ is defined as the cohomology class of $\beta \wedge d\beta$. This class is independent of the choices made (at least up to sign).

Then we have the following question which is raised by Etienne Ghys:

Conjecture: The Godbillon-Vey invariant is a topological invariant.

See the paper É. Ghys, L’invariant de Godbillon–Vey, Astérisque (1989) 177–178, Exp. No. 706, Séminaire Bourbaki, Vol. 1988/89, 155–181

That is, two topologically equivalent foliations have the same Godbillon-Vey class.

What are some updates on this conjecture?

Suppose we have a compact three-manifold $M$, a codimension-one foliation $F$ of $M$, and a one-form on $\alpha$, chosen so that $F$ is tangent to $\alpha = 0$. We deduce that $\alpha \wedge d\alpha = 0$. Also, there is some one-form $\beta$ so that $d\alpha = \alpha \wedge \beta$.

The Godbillon-Vey invariant of $F$ is defined as the cohomology class of $\beta \wedge d\beta$. This class is independent of the choices made (at least up to sign).

Then we have the following:

Conjecture: The Godbillon-Vey invariant is a topological invariant.

That is, two topologically equivalent foliations have the same Godbillon-Vey class.

What are some updates on this conjecture?

Suppose we have a compact three-manifold $M$, a codimension-one foliation $F$ of $M$, and a one-form on $\alpha$, chosen so that $F$ is tangent to $\alpha = 0$. We deduce that $\alpha \wedge d\alpha = 0$. Also, there is some one-form $\beta$ so that $d\alpha = \alpha \wedge \beta$.

The Godbillon-Vey invariant of $F$ is defined as the cohomology class of $\beta \wedge d\beta$. This class is independent of the choices made (at least up to sign).

Then we have the following question which is raised by Etienne Ghys:

Conjecture: The Godbillon-Vey invariant is a topological invariant.

See the paper É. Ghys, L’invariant de Godbillon–Vey, Astérisque (1989) 177–178, Exp. No. 706, Séminaire Bourbaki, Vol. 1988/89, 155–181

That is, two topologically equivalent foliations have the same Godbillon-Vey class.

What are some updates on this conjecture?

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Ali Taghavi
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Suppose we have a compact three-manifold $M$, a codimension-one foliation $F$ of $M$, and a one-form on $\alpha$, chosen so that $F$ is tangent to $\alpha = 0$. We deduce that $\alpha \wedge d\alpha = 0$. Also, there is some one-form $\beta$ so that $d\alpha = \alpha \wedge \beta$.

The Godbillon-Vey invariant of $F$ is defined as the cohomology class of $\beta \wedge d\beta$. This class is independent of the choices made (at least up to sign).

Then we have the following:

Conjecture: The Godbillon-Vey invariant is a topological invariant.

That is, two topologically equivalent foliations have the same Godbillon-Vey class.

What are some updates on this conjecture?

Suppose we have a three-manifold $M$, a codimension-one foliation $F$ of $M$, and a one-form on $\alpha$, chosen so that $F$ is tangent to $\alpha = 0$. We deduce that $\alpha \wedge d\alpha = 0$. Also, there is some one-form $\beta$ so that $d\alpha = \alpha \wedge \beta$.

The Godbillon-Vey invariant of $F$ is defined as the cohomology class of $\beta \wedge d\beta$. This class is independent of the choices made (at least up to sign).

Then we have the following:

Conjecture: The Godbillon-Vey invariant is a topological invariant.

That is, two topologically equivalent foliations have the same Godbillon-Vey class.

What are some updates on this conjecture?

Suppose we have a compact three-manifold $M$, a codimension-one foliation $F$ of $M$, and a one-form on $\alpha$, chosen so that $F$ is tangent to $\alpha = 0$. We deduce that $\alpha \wedge d\alpha = 0$. Also, there is some one-form $\beta$ so that $d\alpha = \alpha \wedge \beta$.

The Godbillon-Vey invariant of $F$ is defined as the cohomology class of $\beta \wedge d\beta$. This class is independent of the choices made (at least up to sign).

Then we have the following:

Conjecture: The Godbillon-Vey invariant is a topological invariant.

That is, two topologically equivalent foliations have the same Godbillon-Vey class.

What are some updates on this conjecture?

typesetting, formatting
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Sam Nead
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Suppose we have a 3 manifoldthree-manifold $M$ with, a codimension-1one foliation tangent to $\alpha=0$ for$F$ of $M$, and a 1one-form on $\alpha$, chosen so that $F$ is tangent to $\alpha = 0$. ThenWe deduce that $\alpha \wedge d\alpha =0$$\alpha \wedge d\alpha = 0$. Also, and $\exists$ a 1there is some one-form $\beta $ with$\beta$ so that $d\alpha =\alpha \wedge \beta$$d\alpha = \alpha \wedge \beta$. 

The Godbillon-Vey invariantGodbillon-Vey invariant of the foliation$F$ is defined as the cohomology class of $\beta \wedge d\beta$. This class is independent of the choices made, at (at least up to sign).

The conjecture about this invariant isThen we have the following:

Conjecture: The Godbillon-Vey invariant is a topological invariant. 

That is, two topologically equivalent foliations have the same Godbillon-Vey class.

What are some updates on this conjectureconjecture?

Suppose we have a 3 manifold $M$ with a codimension-1 foliation tangent to $\alpha=0$ for a 1-form $\alpha$. Then $\alpha \wedge d\alpha =0$, and $\exists$ a 1-form $\beta $ with $d\alpha =\alpha \wedge \beta$. The Godbillon-Vey invariant of the foliation is defined as the cohomology class of $\beta \wedge d\beta$. This class is independent of the choices made, at least up to sign.

The conjecture about this invariant is the following:

Conjecture: The Godbillon-Vey invariant is a topological invariant. That is, two topologically equivalent foliations have the same Godbillon-Vey class.

What are some updates on this conjecture?

Suppose we have a three-manifold $M$, a codimension-one foliation $F$ of $M$, and a one-form on $\alpha$, chosen so that $F$ is tangent to $\alpha = 0$. We deduce that $\alpha \wedge d\alpha = 0$. Also, there is some one-form $\beta$ so that $d\alpha = \alpha \wedge \beta$. 

The Godbillon-Vey invariant of $F$ is defined as the cohomology class of $\beta \wedge d\beta$. This class is independent of the choices made (at least up to sign).

Then we have the following:

Conjecture: The Godbillon-Vey invariant is a topological invariant. 

That is, two topologically equivalent foliations have the same Godbillon-Vey class.

What are some updates on this conjecture?

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Daniel Asimov
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Daniel Asimov
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Ali Taghavi
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Ali Taghavi
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