# Possibilities of an action of $S^1$ on a disk.

I'm dealing with actions of the circle over differentiable manifolds. In the book I'm reading, they use the fact that an action of $S^1$ over a disk has to be equivalent (there has to exist an equivariant diffeomorphism) to a rotation. Can someone give me a hint to prove this? Or maybe a reference where I can consult this result? It seems to be a "well known fact" but I haven't been able to find a place where they prove it.

-
Pick any point. Its orbit under $S^1$ is a loop (winding around an integer number of times in a fixed direction) or a point. The different orbits can't cross each other.... – Hurkyl Jul 4 '12 at 23:45
I think this might settle it for the case of the disk of dimension 2 but I'm failing to see this argument applied to the higher dimensional cases. I feel the orbits have too much space, so ones could go one way and others other way. I feel something more has to be said to conclude it is a rotation. – Sak Jul 5 '12 at 5:09
Ah, I hadn't realized you were interested in other dimensions. My intuition is that if the action sends a point around a loop, then the point drags a tubular neighborhood along with it. And the rotation within that tube drags a larger tubular neighborhood with it, and so forth. But I'm much less confident in turning that intuition into a proof; before trying to make it work I would first try to stumble about with homotopy groups. – Hurkyl Jul 5 '12 at 11:40
What is a disk of dimension higher than 2? – timur Aug 18 '12 at 2:13
Dumb question: Is it obvious that every $S^1$ action on a disc as a fixed point? The fact that the disc is a manifold with boundary is causing me difficulty. – Jason DeVito Aug 18 '12 at 2:51

Your question is a little imprecise but I'll take it to be the assertion that a smooth action of $S^1$ on $D^n$, i.e. a smooth homomorphism $S^1 \to Diff(D^n)$ is conjugate via a diffeomorphism of $D^n$ to a homomorphism $S^1 \to SO_n$.
First off, the generator of the motion is a vector field on $D^n$ which is tangent to $\partial D^n = S^{n-1}$. Since its tangent on the boundary, you can perturb the vector field near the boundary to be outward-pointing. Poincare-Hopf kicks in and tells you this vector field needs to have a zero on the interior, so your original vector field has a zero on the interior.
So in the orbit decomposition of $D^n$ you have a non-empty fixed point set. So all we need to do is show the fixed point set is isotopic to a linear subspace -- once you have that you know that the disc $D^n$ is just an $S^1$-equivariant tubular neighbourhood of that fixed point set, and so its characterized by its behaviour near the fixed point set, which is linear.
Your references are for only the $2$-dimensional case. – Ryan Budney Aug 29 '12 at 18:27