# Is a "global period" similar to a "local period"?

Let $$v\in\mathcal{C}^1(\mathbb{R}^n,\mathbb{R}^n)$$ $$(n\geq 2)$$ a vector field, such that the set $$E=\{v=0\}$$ is a manifold of dimension $$n-2$$. Assume that for every $$x\in\mathbb{R}^n-E$$, the trajectory $$(x_t)_{t\geq 0}$$ such that $$x_0=x$$ and $$\dot{x}_t=v(x_t)$$ for $$t\geq 0$$, is periodic, with period $$T(x)$$, such that $$T\in\mathcal{C}^0(\mathbb{R}^n-E,\mathbb{R}_+)$$. Assume that for every $$x\in E$$ there exist an antisymmetric matrix $$A(x)$$ of rank 2 such that $$A\in\mathcal{C}^0\big(E, \mathcal{A}_n(\mathbb{R})\big)$$ (for the natural topologies of these spaces) and that for a $$x'\in\mathbb{R}^n-E$$ neighbor of $$x$$

$$v(x')=A(x)x'+o(x'-x).$$

Question: Does $$T$$ have a continuous extension defined over $$\mathbb{R}^n$$?

• What do you mean? Typically (for instance, if $A$ has some eigenvalue with modulus >1) the period must go to infinity as the point approaches $E$, do you want the function $1/T$ rather than $T$? Still, for the linear vector.field in the plane $x'=-y$ and $y'=x$ this does not hold. Jun 15, 2021 at 1:29
• For $\omega>0$, the planar vector field $x'=-\omega y$ and $y'=\omega x$ has for solution $(x,y)(t)=\mathcal{R}_{\omega t}(x_0,y_0)$ where $\mathcal{R}_{\theta}$ is the rotation matrix with angle $\theta$. So if $(x_0,y_0)\neq (0,0)$, then $T(x_0,y_0)=2\pi/\omega$, so $T$ has a continuous extension such that $T(0,0)=2\pi/\omega$. This does not depend on the modulus of the eigenvalues of $A_{(0,0)}$ (which is $\omega$). Jun 15, 2021 at 13:40