3
$\begingroup$

Let $v$ be a holomorphic vector field defined in a neighbourhood of $0$ on $\mathbb C^n$ with an isolated zero at $0$. Let $\sum_{i,j}{a_{ij}}z_i\frac{\partial}{\partial z_j}$ be the linear term of $v$ and suppose that the matrix $a_{ij}$ is invertible and all its eigenvalues have modulus different from $1$. Is it true that for some holomorphic coordinates $w_i$ in a neighbourhood of $0$ we have $v=\sum_{i,j}{a_{ij}}w_i\frac{\partial}{\partial w_j}$?

If yes, where could I find such a statement? If not, what would be a counterexample? I am happy to assume that the eigenvalues of $A$ all have modulus less than $1$.

Improve this question
$\endgroup$
4
  • 3
    $\begingroup$ If all have modulus less than 1, I know this from a paper of Kodaira about complex surfaces. $\endgroup$
    – Ben McKay
    Commented Oct 21, 2020 at 15:16
  • $\begingroup$ That's nice! What is the exact reference? You can leave it as an answer (and not a comment) if you wish $\endgroup$
    – aglearner
    Commented Oct 21, 2020 at 15:41
  • 1
    $\begingroup$ I think there can be resonances even in the case all eigenvalues are less than one in modulus. They can prevent linearization. $\endgroup$
    – Arnaud Chéritat
    Commented Oct 23, 2020 at 13:08
  • $\begingroup$ Arnaud, that's very interesting, if you can find an explicit example/reference, that would be great $\endgroup$
    – aglearner
    Commented Oct 23, 2020 at 13:10

1 Answer 1

Reset to default
4
$\begingroup$

A relevant result, but not a complete answer. Vladimir Arnol'd, Geometrical Methods in the Theory of Ordinary Differential Equations, p. 181:

A zero of a vector field is in the Poincare domain if the origin is not in the convex hull of the eigenvalues of the linearization.

A resonance of a zero of a vector field is an equation $\lambda_i=\sum_{j\ne i} m_j \lambda_j$ among eigenvalues of the linearization of the vector field about the zero, where $\{m_j\}$ are integers, $m_j\ge 0$, and $\sum_j m_j\ge 2$.

Theorem of Poincare: If the eigenvalues of the linearization of a formal analytic vector field are nonresonant, then it can be reduced to a linear vector field by a formal analytic change of variables. If the eigenvalues of the linear part of a holomorphic vector field at a singular point belong to the Poincare domain and are nonresonant, then the field is biholomorphically equivalent to its linear part in the neighborhood of the singular point.

Improve this answer
$\endgroup$
4
  • $\begingroup$ I am curious, in the case when all the eigenvalues of $A$ are smaller than $1$ in modulus, can it happen (say when there are resonants) that the mapping is not holomorphically conjugate to a linear one in a small neighbourhood of $0$? $\endgroup$
    – aglearner
    Commented Oct 23, 2020 at 12:58
  • $\begingroup$ Sorry, I gave the result for maps, not for vector fields. Here is the correct version for vector fields. $\endgroup$
    – Ben McKay
    Commented Oct 23, 2020 at 14:59
  • $\begingroup$ If you rescale a resonant vector field, you get resonances with eigenvalues as large or as small as you like. $\endgroup$
    – Ben McKay
    Commented Oct 23, 2020 at 15:04
  • 1
    $\begingroup$ The dynamics of a linear vector field with resonances are going to repeat for infinitely many values of complex time, but when you throw in some resonances, you break that, so you can't get back to the linear case, since the linearization is an invariant. $\endgroup$
    – Ben McKay
    Commented Oct 23, 2020 at 15:14

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .