Let $V=\sum_{i=1}^{n}a_i(z_1,\ldots z_n)\frac{\partial}{\partial z_i}$ be a holomorphic vector field defined on a neighborhood $U\subset \mathbb{C}^n$ of the origin, such that the common zero point of $a_i$ are the origin, i.e. $V$ has isolated singularity. We know that if $V$ is a radial vector field $\sum_{i=1}^{n}z_i\frac{\partial}{\partial z_i}$, then there are infinite separatrix through the origin. My question is: is this the only case that admit infinite separatrix? If there are other types of vector fields which have infinite separatrix, what is the condition?
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No, a lot more vector fields have infinitely many separatrices. They are called «dicritical». For instance, you can change locally the analytic coordinates $(z_1,\ldots,z_n)$, the vector field will not be radial anymore but still it will admit infinitely many separatrices (all oft its trajectories, actually). Aside this trivial case, you can determine if the vector field is dicritical by performing a (succesive chain of) blow-up of the singularity. This is an algebraic process and the dicriticality condition amounts to the vanishing of some term in the power series expansion in the new coordinates. This is classical stuff for $n=2$ (Seidenberg algorithm) and tending-to-classical for $n=3$ (Cano, Panazzolo-MacQuillan). For $n>3$ not much is known I think.
answered Nov 7, 2015 at 1:32
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$\begingroup$ +1 for the interesting answer. What about some degenerate case for example $\begin{cases}\dot x=y^{3}\\ \dot y=-x^{3}+\frac{1}{2}x^{2}y^{2} \end{cases}$. I think this is a real center which does not admit a real analytic first integral. Does this system have infinte number of seperatrices? $\endgroup$ Commented Nov 14, 2015 at 10:30
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$\begingroup$ Well, I don't know, try to blow the singularity up by setting $x=ty$ then $y=sx$. Dicritic components may (or may not) appear in complex coordinates. $\endgroup$ Commented Nov 14, 2015 at 14:01
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$\begingroup$ @ Teyssier: Thanks for your answer. There is another definition of ``dicritical", which says that if we blow-up the singularity, the exceptional divisor is not invariant by the induced foliation. Are these two definitions equivalent? $\endgroup$ Commented Nov 17, 2015 at 12:14
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$\begingroup$ In dimension 2 they are. As for greater dimension, I don't know. Try looking for the names I mentioned. $\endgroup$ Commented Nov 17, 2015 at 13:18