Let $M$ be a compact manifold of dimension $m \geq 2$. Show that there exists an injective immersion of $\mathbb{R}$ in $M$, whose image is not the trajectory of any flow.
I know how to do it for $M=\mathbb R^2$ (of course this is not compact!). I just take $\gamma:\mathbb R\rightarrow \mathbb{R}^2$ defined by $\gamma(t)=(e^{-t},\sin t)$. From an easy computation this is clearly an injective immersion. Suppose by contradiction, $\gamma$ is the flow line of the global flow on $\mathbb{R}^2$ then let $\Phi: \mathbb{R} \times \mathbb{R}^2\rightarrow \mathbb{R}^2$ and let $\Phi(t,x)=\varphi_t(x)$. Then we know that $\dot\gamma(t)=(d\varphi)_{\gamma(t)}(\dot\gamma(0)).$ If we take $t_n=n\pi$ we get that $\lim_{n \rightarrow \infty}\dot\gamma(t_n)=(d\varphi)_{(0,0)}(\dot\gamma(0))$ and on the other hand $\lim_{n \rightarrow \infty}\dot\gamma(t_n)=\lim(e^{-n\pi},\cos n\pi)$ which does not have a limit. Contradiction. It is clear that we can generalize for $\mathbb{R}^n$.
But how do we do it for $M$ compact? Thank you.
Let $\gamma : \mathbb R \to \mathbb R^m$ be your "clear generalization". Fix a chart $\xi : \mathbb R^m \to M$ and consider the injective immersion $\tau = \xi \circ \gamma : \mathbb R \to M$. If $\tau$ was a flow line of a vector field $X$ we would have $\dot \tau(t) = D\xi(\dot \gamma(t)) = X(\xi(\gamma(t)))$. Applying $D\xi^{-1}$ to this we would get $$ \dot \gamma(t) = D \xi^{-1}(X(\xi(\gamma(t)))) = (\xi^* X)(\gamma(t));$$ i.e. $\gamma$ is a flow line of the pullback vector field $\xi^* X$ on $\mathbb R^m$, a contradiction. (Note that this pullback is well defined only because $\xi$ is a local diffeomorphism.)