Let X be a vector field on $\mathbb {T}^2$, we say that $\varphi: \mathbb {R} \to \mathbb {T}^2$ is a periodic orbit of $X $, if $\varphi $ is a periodic function and $\varphi'(t) = X (\varphi(t)), \forall t \in \mathbb {R} $.

Let $H: T\mathbb{T}^2 \to \mathbb{T}^2 \times\mathbb{R}^2$ be a parallelization of $\mathbb{T}^2$, i.e; $H$ is a smooth function such that $H(p,v) = (p, A(p) v)$, where $A(p)$ is a isomophism between $T_p\mathbb{T}^2$ and $\mathbb{R}^2$ definided in the following way:

Consider $\mathbb{T}^2 = \mathbb{R}^2 / \mathbb{Z}^2$ (flat torus), then for each point $[(x,y)]\in\mathbb{T}^2$ there are two loops $[(x+t,y)]$ and $[(x,y+t)]$, we define $A([(x,y)])$ as the unique linear transformation between $T_{[(x,y)]} \mathbb {T}^2$ and $\mathbb {R}^2$ satisfying

$$A([(x,y)]) \left.\frac{d}{dt} [(x+t,y)]\right|_{t=0} = (1,0) \quad\text{and,}\quad A([(x,y)]) \left.\frac{d}{dt} [(x,y+t)]\right|_{t=0} = (0,1). $$

I'm having trouble to solve the following exercise (This exercise can be found on page 6 of the book "The dynamics of vector fields in dimension 3-Etienne Ghys").

Exercise: Suppose $X$ is a non-singular vector field on $\mathbb{T}^2$, and consider $X^∗$ as the map $$ X^*: \pi_1\left(\mathbb{T}^2, x_0\right) \to \pi_1 \left(\mathbb{R}^2 \setminus \{0\}, A(x_0)X(x_0) \right)$$ $$\left[\alpha(t)\right] \mapsto \left[A(\alpha(t)) X(\alpha(t))\right] $$ If $X^*$ is a non-trivial homomorphism then $X$ has a periodic orbit.

I was trying to show that if $X$ doesn't admit a periodic orbit, then for every loop $\varphi$ we have $A(\varphi(t)) X(\varphi(t))$ homotopic to a constant curve, but I wasn't able to conclude such thing.

Can anyone help me or can give me some hint?


I noticed that if there exists $\alpha_1,\alpha_2:\mathbb{S}^{1}\to \mathbb{T}^2$, such that $[\alpha_1]$ and $[\alpha_2]$ are generators of $\pi_1(\mathbb{T}^2,x_0)$ and $\{ \alpha_i'(t),X(\alpha_i(t))\}$ is a basis of $T_{\alpha_i(t)} \mathbb{T}^2$, then $A(\alpha_i(t)) X(\alpha_i(t))$ is homotopic to $A(\alpha_i(t)) \alpha'_i (t) $ (the homotopy $$\tilde{H}(s,t) = A(\alpha_i(s))\left( (1-t) X(\alpha_i(s)) + t \alpha_i'(s) \right) $$ do the job).

And is relatively easy to demonstrate that $A(\alpha_i(t)) \alpha'_i (t)$ is homotopic to the constant map. (I realized that this homotopy doesn't work because it doesn't fix the ends of the interval)

Does anyone know if $X$ is a non-singular vector field on $\mathbb{T}^2$ without periodic orbits then there are $\alpha_1$ and $\alpha_2$ as described above?

I think it's important to inform that in the chapter that this exercise is contained there is the following theorem

Theorem: Every non-singular $\mathcal{C}^2$ vector field on a compact surface that has no periodic orbits is topologically equivalent to a linear flow on $\mathbb{T}^2$ with irrational slope

I think that the key to solving the exercise is in the above theorem but I wasn't able to figure out how to do this.

  • $\begingroup$ My impression is that your own reply in mathoverflow.net/questions/303616/… suffices. No need for more regular conjugacies. $\endgroup$
    – John B
    Commented Jun 27, 2018 at 20:03
  • $\begingroup$ @JohnB Sorry, I'm a bit confused. Are you saying that the fact of $X$ is equivalent to the irrational flow on $\mathbb{T}^2$ imply this result? $\endgroup$ Commented Jun 27, 2018 at 20:49
  • $\begingroup$ Sure, the irrational flow in linear in the flat metric. $\endgroup$
    – John B
    Commented Jun 27, 2018 at 20:53
  • $\begingroup$ @JohnB ok, if $X$ is a non-vanishing vector field and has no periodic orbits then $X$ equivalent to the irrational flow then $X$ has no periodic orbits. But how this assertion implies that the map $X^*$ is the trivial homomorphism? What I'm trying to say is that isn't obvious to me why this result (about the equivalence of vector fields) solves the exercise. Could you explain me, please? $\endgroup$ Commented Jun 27, 2018 at 21:08


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