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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 $1$.

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    $\begingroup$ If all have modulus less than 1, I know this from a paper of Kodaira about complex surfaces. $\endgroup$
    – Ben McKay
    Oct 21 '20 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
    Oct 21 '20 at 15:41
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    $\begingroup$ I think there can be resonances even in the case all eigenvalues are less than one in modulus. They can prevent linearization. $\endgroup$ Oct 23 '20 at 13:08
  • $\begingroup$ Arnaud, that's very interesting, if you can find an explicit example/reference, that would be great $\endgroup$
    – aglearner
    Oct 23 '20 at 13:10
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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.

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  • $\begingroup$ 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
    Oct 23 '20 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
    Oct 23 '20 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
    Oct 23 '20 at 15:04
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    $\begingroup$ 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
    Oct 23 '20 at 15:14

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