Analytic conjugacy of vanishing holonomy groups implies analytic conjugacy of foliations i am reading and trying to do some exercises and problems of the book Lectures on Analytic Differential Equations- Y. Ilyashenko, S. Yakovenko.
I can not solve the problem 11.6 that says  
Consider nondicritical foliations having at most three hyperbolic singularities on the exceptional divisor after blow-up.
Prove that for such singularities analytic conjugacy of the vanishing holonomy groups implies the conjugacy of the foliations.
In such case the vanishing holonomy group is always integrable?     
Since the singularities are hyperbolic (the eigenvalues of the linearization matix are linearly independent over $\mathbb R$) then by the Poincaré-Dulac theorem you can think that the foliation near every singularity is equivalent to its linear part.
I can't see if for such foliations the vanishing holonomy group is always integrable, because if it is, i think i can solve the problem.
Thanks   
 A: No, in general the holonomy group is not solvable. Yet, this is how I'd tackle the exercise. (By the way, as it is posed the exercise cannot be solved: you need to assume that each eigenratio of both foliations agree, since holonomy conjugacy only provides equality up to $\mathbb Z$ and said eigenratios are local analytic invariants. I give a counterexample below.)
Since the singularities are hyperbolic and the foliation is not dicritical, the saturation of a transverse disc $\Sigma$ by the foliation is a neighbourhood of the exceptional divisor, except for the (at most) three transverse separatrices. Therefore if the holonomy groups (computed on $\Sigma$) are conjugate then the conjugacy extends to a fibered conjugacy between foliations by the classical path-lifting technique of Mattei-Moussu. This extension can also be performed up to the missing separatrices (at this stage one needs the eigenratio of the singularities to agree for both foliations). By blowing down, one obtains a conjugacy between the foliations as expected.

Let me give a counterexample to the exercise in its given form. Let $\mathcal F_{a,b,c}$ be the foliation defined by the differential $1$-form $$\mathrm d(\log(x^ay^b(x+y)^c))$$ with $a, b, c\notin \mathbb R$. Clearly the foliation satisfies the hypothesis of the exercise, and its vanishing holonomy group is the linear group $<h\exp(2i\pi a),h\exp(2i\pi b),h\exp(2i\pi c)>$. Therefore shifting $(a,b,c)$ by any element of $\mathbb Z^3$ gives identical groups. Since the eigenratios of the blown-up foliation are given by $(a,b,c)$ this proves that $\mathcal F_{a,b,c}$ and $\mathcal F_{a+k_a,b+k_b,c+k_c}$ cannot be conjugate as soon as one $k_*$ is nonzero.
