Is there a non vanishing real analytic vector field $X$ on $S^3$ such that $X$ has an attractor periodic orbit(An asymptotically stable periodic orbit) ? What about the smooth case?

$\begingroup$ Did you try: Use that S^3 is a Lie group. So you have plenty of nonvanishing vector fields. Find one with a periodic orbit O, then homotope the field nearby to point inwards toward the orbit O. $\endgroup$– Artur JacksonCommented Apr 5, 2017 at 5:31
1 Answer
$S^3$ has the structure of a Lie group. Consider a left invariant vector field $X$. (Check that you can get closed orbits!) In a tubular neighborhood of a small amount of time around a closed orbit $o$, the vector field looks like a constant field on $\mathbb{R}^3\simeq D^2 \times I$. Consider the functionr $f = \beta\cdot d(\cdot,o)^2$, where $\beta$ is a smooth compactly supported bump function. The choice of unit vector $\nu$ pointing towards the central orbit $o$ from within this neighborhood is clear. Now set $X' = X + \epsilon\cdot f\cdot \nu$ for some choice of $\epsilon$.
For $X'$ we have the old orbit $o$, unperturbed. Moreover, nearby orbits will veer towards $o$. Thus now $o$ is a stable attractor. One checks that with $\epsilon$ small enough, $X'$ is still nonvanishing.

$\begingroup$ No problem. I believe this can be tidied up a bit! $\endgroup$ Commented Apr 5, 2017 at 12:09

$\begingroup$ So it is an smooth example yes? what about analytic case? $\endgroup$ Commented Feb 6, 2021 at 14:03