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The answer is "yes". Consider $\mathbb S^1$ as the quotient $\mathbb R/\mathbb Z$. Your homeomorphism $f$ lifts to a homeomorphism $\phi : \mathbb R \to \mathbb R$ such that $\phi(x+1)=\phi(x)+1$. For …

To see why the second question cannot have a simple answer, it is sufficient to look at the local context near a fixed-point of a tangent-to-identity mapping, as Alexandre Eremenko suggests. By "a sim …

This question relates to this one and that one. Some background In the setting of discrete holomorphic dynamics (say, Julia sets) an irrational $\lambda$ is said to be well approximated by rational nu …

Quick answer to the first question: no, there is no reason why it should be analytic. Take e.g. the parametric vector field (written as a Lie derivative)$$X(x,y):=-x^3\partial_x+(y+\alpha x)\partial_y …

Consider a foliation of $\mathbb{R}^2$, say coming from the trajectories of a vector field $X$. Its orbit space (the quotient of $\mathbb{R}^2$ by the relation "lying on the same trajectory") is seldo …

The answer is 'no' for much the same reason that the OP indicates: the existence of a homoclinic or heteroclinic connection implies that neighboring trajectories are periodic. First, one needs to have …

While not pretending to answer the OP, the following is too long to fit in a comment while it might contain elements of interest to the poster. If $f$ is a convergent object (smooth or analytic), then …

First, this case is totally uninteresting regarding Hilbert XVI. Indeed, there are no limit cycles in such systems. The $\alpha / \omega$-limit of a trajectory is either a point or a non-isolated cycl …

The answer is "no". In fact, it is still "no" for germs of curves : generically, a germ of an analytic curve $\gamma : (\mathbb R,0)\rightarrow (\mathbb R^2,0)$ is not tangent to any polynomial vector …

I don't know offhand the answer of your first question, but I can answer the particular situation you describe afterwards : the holonomy is always trivial. First, notice that a compact leaf $L$ is e …

For your first question: $dy=y^2$ can be integrated by quadratures. The solutions are homographies. For your second question: no it is not true. By the uniformization theorem, the universal covering …

The period of the compact, non-singular orbit $\gamma$ is given by $$T(\gamma)=\oint_\gamma \tau$$ where $\tau$ is any differential $1$-form such that $\tau(F)=1$, e.g. $\tau:=\frac{\mathrm{d}y_2}{\mu …

Yes. Let $\mathcal{F}$ be the foliation (of arbitrary dimension in $\{1,\ldots,n-1\}$), which I assume regular and transversely continuous. It is sufficient to consider the case of a ball $V$ of rad …

If I understand correctly the setting (say $f$ is a germ at $0\in\mathbb C$ of a biholomorphic function), the answer is no in general, but yes generically. The diffeomorphismn $f$ can always be writte …

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 no …