Linearization of singular foliation in the plane - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T18:34:55Z http://mathoverflow.net/feeds/question/87050 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/87050/linearization-of-singular-foliation-in-the-plane Linearization of singular foliation in the plane Klaus Niederkrüger 2012-01-30T18:06:32Z 2012-01-30T18:06:32Z <p>Hello,</p> <p>I would like to obtain a smooth model around a singular point of a foliation (in my case the model should be the linearization of the foliation). It seems that the answer could be hidden in Grobmann-Hartman, and I apologize for not having read the proof.</p> <p>Here is my problem:</p> <p>I have a singular foliation on $\mathbb{R}^2$ given by the smooth $1$-form</p> <p>$$\beta = a(x,y) dx + b(x,y) dy$$</p> <p>and I assume that $(0,0)$ is a singular point.</p> <p>Additionally I suppose that the matrix $$\begin{pmatrix} \partial_x b &amp; \partial_y b \\ - \partial_x a &amp; -\partial_y a \end{pmatrix}$$ does not have purely imaginary eigenvalues. (In fact, I have $d\beta \ne 0$ so that in the worst case, I could have a 0 eigenvalue, but I assume for now that this is not the case).</p> <p>Since the kernel of $\beta$ is spanned by the vector field $$X = b \partial_x - a \partial_y,$$ I can use the Hartman-Grobmann theorem to say that my foliation is homeomorphic to its linearization, but I would like to have a diffeomorphism.</p> <p>I have seen that there exist counter-examples for vector fields to be smoothly conjugated to its linearization, but on the other hand, I'm not interested in the time parameter of the trajectories and I would only like to map the foliation onto the linearized foliation. Could it be possible that I still get a diffeomorphism in my situation? I would bet that there must be a reference, where this has already been worked out, but googling only confused me (it spoke of resonance conditions and other things).</p> <p>Thank you very much for any response. </p> <p>Best Klaus</p>