Let $X$ be an analytic vector field on a compact oriantable surface $S$ with volume form $\omega$. We denote the set of its singularities by $Z(X)$.

A local question

Is there an analytic vector field $Y$ on a neighborhood of $Z(X)$ such that $Z(Y)\subset Z(X)$ and $Z(Y)$ is a finite set. Moreover $\omega(Y(p),X(p))=0$ for all $p\in S$. That is: $X \parallel Y$ out of singularities of $X$?

A global question

Can we find an analytic vector field $Y$ as above, globally on whole $S$?

Motivation: I think the second question is implicitly used (and is needed) in the book "Finiteness theorem for Limit cycles" by YU.S. Ilyashenko.

In Fact my main motivation is the following question:

Main motivating question:

Is it easy to pass from statement $A$ to $B$ as follows?:

A: Every analytic vector field $X$ on $S^{2}$ has a finite number of limit cycles provided the singular set of $X$ is a finite set.

B: Every arbitrary analytic vector field on $S^{2}$ has a finit number of limit cycles.

Edit and update: There is a new version of "Finitness theorem of limit cycle whose abstract indicates to a new version of the proof of finitness theorem for analytic vector fields.



1 Answer 1


Disclaimer: The discussion below is about vector fields on $\mathbb RP(2)$, a non-orientable surface, and they do not answer the OP's questions.

The answer to the local question is no.

Consider on $\mathbb RP(2)$ the vector field defined on an affine chart $(x,y)\in \mathbb R^2$ by $v=x \partial_x + y \partial y$. The induced flow acts on $\mathbb R^2$ by homotheties, an therefore fixes the line at infinity $ell$. Hence $v$ extends to a vector field on $\mathbb RP(2)$ and contains $\ell$ in its zero set. Writing down explicitly how the vector field looks like at a neighborhood of point at the line at infinity, one sees that it vanishes in first order at it. The natural thing to do is to find a function vanishing at the line at infinity at first order and divide $v$ by it. But this cannot be done. A neighborhood of $\ell$ is isomorphic to the Moëbius band, with $\ell$ being the central circle. A function $f$ vanishing exactly on $\ell$ and only up to first order (i.e. $\ell$ is not contained in the critical set of $f$) would allow us to distinguish, through its sign, up and down the central circle of the Moëbius band.

A version of the local question is true.

The zeros of an analytic vector field, if not isolated, must contain an analytic curve. Locally this curve is defined by the vanishing of an analytic function. Dividing by the equation of the curve one gets another vector field with isolated zeros.

Anyway, I don't think that these issues cause any kind of problem.

In general a singular analytic foliation on a compact surface $S$ is not defined by a global analytic vector field. Usually one covers $S$ with open subsets where the foliation is defined by vector fields with isolated zeros, and ask that on non-empty intersections the vector fields differ by the multiplication with a nowhere vanishing analytic function.

  • $\begingroup$ thank you very much for your answer in non orientable case. is there a negative answer in orientable case,too?(As I asked in my question) $\endgroup$ Mar 23, 2014 at 4:17
  • $\begingroup$ Thanks again for your interesting construction on $\mathbb{R}P^{2}$. In this revised version I explain my main motivating question $\endgroup$ Mar 23, 2014 at 22:06
  • $\begingroup$ Could you please more explain on your statement that"I do not think that these issues cause any kind of problem" In page 306 of this paper, it is written that "we divide the component by a common factor" but this is about poincare compactification of polynomial mvec. field not arbitrary analytic vec. field :ams.org/journals/bull/2002-39-03/S0273-0979-02-00946-1/… $\endgroup$ Aug 10, 2014 at 7:44
  • $\begingroup$ Locally you can always divide by a common factor. If you do for a covering of your manifold then over the intersections the vector fields differ by a nowhere vanishing analytic function. At the end you get a vector field with coefficients in an analytic line bundle. That's what I meant. This works for any surface. Perhaps in the particular case of a sphere you always get a trivial line bundle at the end, but I am not sure. $\endgroup$ Aug 11, 2014 at 5:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.