I was thinking about the nature of continuous actions of the circle group $C^1$ on the plane $ℝ^2$ with $n$ points removed. Obviously, there are plenty of circular actions when $n=1$ that leave no point fixed. When $n=0$, I can prove that there is at least one fixed point by the following argument: Consider $X$, the orbit of a point $x∈ℝ^2$ together with its interior. This is a compact set, so the function assigning to a point the diameter of its orbit space attains a mimimum value at $y∈X$. Then there is a contradiction unless $y$ is fixed by the circular action.
It seems to me that when $n≥2$, it's not even possible to have a nontrivial circular action. Can this be proven?
I think I've figured out a solution for $n≥2$ (and using the property that the intersection of nested nonempty compact sets is nonempty, it gives an alternate proof of the case when $n=0$): First, we will prove the existence of fixed points. Let $x,y$ be nonfixed points. If we draw a line from $x$ to $y$ through nonfixed points, then this gives rise to a free homotopy of the orbit spaces that doesn't pass through at least one of those orbit spaces' interiors (here we mean to include punctures in the interior). By the nontrivial fundamental group of the singly punctured plane, this implies the interiors of the nontrivial orbit spaces form a totally ordered set under inclusion (Lemma 1). As well, again due to homotopy, we may conclude that the puncture points contained inside the interior of any nontrivial orbit space are always the same (Lemma 2).
Now, consider our action as being on the $(n+1)$-punctured Riemann sphere. We may assume that the so-called "interior" (choose one orientation of them) of the nontrivial orbit spaces excludes $≥2$ of these punctures. Then, taking the union of all those interiors forms a simply connected subset (use Lemma 1), and which is seen to be all of the Riemann sphere except for the $≥2$ punctures (use Lemma 2) and the fixed points. Thus, we have proven the existence of (uncountably many) fixed points.
Let $z$ be a puncture or fixed point contained in the interior of the included punctures. If we reorient the sphere so that $z$ becomes the point at infinity and reinstitute the plane metric, then this says that the nonfixed orbit spaces have a diameter at least that of the distance $ε$ between the closest two (formerly) excluded punctures. But now consider the set of fixed points: It must be closed and open (it is the pullback of the continuous diameter function of either $[0,ε/2)$ or $[0,ε/2]$), hence the whole action is trivial.$~\square$
I would be interested if anyone could make this elementary proof more elegant, or find another.